![Most wells produce oil through tubing strings, mainly
because a tubing string provides good sealing performance
and allows the use of gas expansion to lift oil. The Ameri-
can Petroleum Institute (API) defines tubing size using
nominal diameter and weight (per foot). The nominal
diameter is based on the internal diameter of the tubing
body. The weight of tubing determines the tubing outer
diameter. Steel grades of tubing are designated H-40, J-55,
C-75, L-80, N-80, C-90, and P-105, where the digits repre-
sent the minimum yield strength in 1,000 psi. The min-
imum performance properties of tubing are given in
Chapter 9 and Appendix B.
The ‘‘wellhead’’ is defined as the surface equipment set
below the master valve. As we can see in Fig. 1.7, it
includes casing heads and a tubing head. The casing head
(lowermost) is threaded onto the surface casing. This can
also be a flanged or studded connection. A ‘‘casing head’’
is a mechanical assembly used for hanging a casing string
(Fig. 1.8). Depending on casing programs in well drilling,
several casing heads can be installed during well construc-
tion. The casing head has a bowl that supports the casing
hanger. This casing hanger is threaded onto the top of the
production casing (or uses friction grips to hold the cas-
ing). As in the case of the production tubing, the produc-
tion casing is landed in tension so that the casing hanger
actually supports the production casing (down to the
freeze point). In a similar manner, the intermediate cas-
ing(s) are supported by their respective casing hangers
(and bowls). All of these casing head arrangements are
supported by the surface casing, which is in compression
and cemented to the surface. A well completed with three
casing strings has two casing heads. The uppermost casing
head supports the production casing. The lowermost cas-
ing head sits on the surface casing (threaded to the top of
the surface casing).
Most flowing wells are produced through a string of
tubing run inside the production casing string. At the
surface, the tubing is supported by the tubing head (i.e.,
the tubing head is used for hanging tubing string on the
production casing head [Fig. 1.9]). The tubing head sup-
ports the tubing string at the surface (this tubing is landed
on the tubing head so that it is in tension all the way down
to the packer).
The equipment at the top of the producing wellhead is
called a ‘‘Christmas tree’’ (Fig. 1.10) and it is used to
control flow. The ‘‘Christmas tree’’ is installed above the
tubing head. An ‘‘adaptor’’ is a piece of equipment used to
join the two. The ‘‘Christmas tree’’ may have one flow
outlet (a tee) or two flow outlets (a cross). The master
valve is installed below the tee or cross. To replace a master
valve, the tubing must be plugged. A Christmas tree consists
of a main valve, wing valves, and a needle valve. These valves
are used for closing the well when needed. At the top of the
tee structure (on the top of the ‘‘Christmas tree’’), there is a
pressure gauge that indicates the pressure in the tubing.
Gas Cap
Oil
Figure 1.4 A sketch of a gas-cap drive reservoir.
Oil and Gas
Reservoir
Figure 1.5 A sketch of a dissolved-gas drive reservoir.
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![2.1 Introduction
Properties of crude oil and natural gas are fundamental for
designing and analyzing oil and gas production systems in
petroleum engineering. This chapter presents definitions of
these fluid properties and some means of obtaining these
property values other than experimental measurements.
Applications of the fluid properties appear in subsequent
chapters.
2.2 Properties of Oil
Oil properties include solution gas–oil ratio (GOR),
density, formation volume factor, viscosity, and compress-
ibility. The latter four properties are interrelated through
solution GOR.
2.2.1 Solution Gas–Oil Ratio
‘‘Solution GOR’’ is defined as the amount of gas (in
standard condition) that will dissolve in unit volume of
oil when both are taken down to the reservoir at the
prevailing pressure and temperature; that is,
Rs ¼
Vgas
Voil
, (2:1)
where
Rs ¼ solution GOR (in scf/stb)
Vgas ¼ gas volume in standard condition (scf)
Voil ¼ oil volume in stock tank condition (stb)
The ‘‘standard condition’’ is defined as 14.7 psia and
60 8F in most states in the United States. At a given reservoir
temperature, solution GOR remains constant at pressures
above bubble-point pressure. It drops as pressure decreases
in the pressure range below the bubble-point pressure.
Solution GOR is measured in PTV laboratories.
Empirical correlations are also available based on data
from PVT labs. One of the correlations is,
Rs ¼ gg
p
18
100:0125(
API)
100:00091t
!1:2048
(2:2)
where gg and 8API are defined in the latter sections, and
p and t are pressure and temperature in psia and 8F,
respectively.
Solution GOR factor is often used for volumetric oil
and gas calculations in reservoir engineering. It is also used
as a base parameter for estimating other fluid properties
such as density of oil.
2.2.2 Density of Oil
‘‘Density of oil’’ is defined as the mass of oil per unit
volume, or lbm=ft3
in U.S. Field unit. It is widely used in
hydraulics calculations (e.g., wellbore and pipeline per-
formance calculations [see Chapters 4 and 11]).
Because of gas content, density of oil is pressure depen-
dent. The density of oil at standard condition (stock tank
oil) is evaluated by API gravity. The relationship between
the density of stock tank oil and API gravity is given
through the following relations:
API ¼
141:5
go
À 131:5 (2:3)
and
go ¼
ro,st
rw
, (2:4)
where
8API ¼ API gravity of stock tank oil
go ¼ specific gravity of stock tank oil, 1 for freshwater
ro,st ¼ density of stock tank oil, lbm=ft3
rw ¼ density of freshwater, 62:4 lbm=ft3
The density of oil at elevated pressures and temperatures
can be estimated on empirical correlations developed by a
number of investigators. Ahmed (1989) gives a summary
of correlations. Engineers should select and validate the
correlations carefully with measurements before adopting
any correlations.
Standing (1981) proposed a correlation for estimating
the oil formation volume factor as a function of solution
GOR, specific gravity of stock tank oil, specific gravity of
solution gas, and temperature. By coupling the mathemat-
ical definition of the oil formation volume factor with
Standing’s correlation, Ahmed (1989) presented the fol-
lowing expression for the density of oil:
ro ¼
62:4go þ 0:0136Rsgg
0:972 þ 0:000147 Rs
ffiffiffiffiffi
gg
go
r
þ 1:25t
!1:175
, (2:5)
where
t ¼ temperature, 8F
gg ¼ specific gravity of gas, 1 for air.
2.2.3 Formation Volume Factor of Oil
‘‘Formation volume factor of oil’’ is defined as the volume
occupied in the reservoir at the prevailing pressure and
temperature by volume of oil in stock tank, plus its dis-
solved gas; that is,
Bo ¼
Vres
Vst
, (2:6)
where
Bo ¼ formation volume factor of oil (rb/stb)
Vres ¼ oil volume in reservoir condition (rb)
Vst ¼ oil volume in stock tank condition (stb)
Formation volume factor of oil is always greater than
unity because oil dissolves more gas in reservoir condition
than in stock tank condition. At a given reservoir tempera-
ture, oil formation volume factor remains nearly constant
at pressures above bubble-point pressure. It drops as pres-
sure decreases in the pressure range below the bubble-
point pressure.
Formation volume factor of oil is measured in PTV labs.
Numerous empirical correlations are available based on
data from PVT labs. One of the correlations is
Bo ¼ 0:9759 þ 0:00012 Rs
ffiffiffiffiffi
gg
go
r
þ 1:25t
!1:2
: (2:7)
Formation volume factorof oil is oftenused for oil volumet-
riccalculationsandwell-inflowcalculations.Itisalsousedas
a base parameter for estimating other fluid properties.
2.2.4 Viscosity of Oil
‘‘Viscosity’’ is an empirical parameter used for describing
the resistance to flow of fluid. The viscosity of oil is of
interest in well-inflow and hydraulics calculations in oil
production engineering. While the viscosity of oil can be
measured in PVT labs, it is often estimated using empirical
correlations developed by a number of investigators
including Beal (1946), Beggs and Robinson (1975), Stand-
ing (1981), Glaso (1985), Khan (1987), and Ahmed (1989).
A summary of these correlations is given by Ahmed
(1989). Engineers should select and validate a correlation
with measurements before it is used. Standing’s (1981)
correlation for dead oil is expressed as
mod ¼ 0:32 þ
1:8 Â 107
API4:53
360
t þ 200
A
, (2:8)
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![«3 ¼ 120(A0:9
À A1:6
) þ 15(B0:5
À B4:0
) (2:26)
Tpc0 ¼ Tpc À «3(corrected Tpc) (2:27)
Ppc0 ¼
PpcTpc0
Tpc þ B(1 À B)«3
(corrected ppc) (2:28)
Correlations with impurity corrections for mixture
pseudo-criticals are also available (Ahmed, 1989):
ppc ¼ 678 À 50(gg À 0:5) À 206:7yN2
þ 440yCO2
þ 606:7yHsS (2:29)
Tpc ¼ 326 þ 315:7(gg À 0:5) À 240yN2
À 83:3yCO2
þ 133:3yH2S: (2:30)
Applications of the pseudo-critical pressure and
temperature are normally found in petroleum engineer-
ing through pseudo-reduced pressure and temperature
defined as
ppr ¼
p
ppc
(2:31)
Tpr ¼
T
Tpc
: (2:32)
2.3.3 Viscosity of Gas
Dynamic viscosity (mg) in centipoises (cp) is usually used in
petroleum engineering. Kinematic viscosity (ng) is related
to the dynamic viscosity through density (rg),
ng ¼
mg
rg
: (2:33)
Kinematic viscosity is not typically used in natural gas
engineering.
Direct measurements of gas viscosity are preferred for a
new gas. If gas composition and viscosities of gas com-
ponents are known, the mixing rule can be used to deter-
mine the viscosity of the gas mixture:
mg ¼
P
(mgiyi
ffiffiffiffiffiffiffiffiffiffiffi
MWi
p
)
P
(yi
ffiffiffiffiffiffiffiffiffiffiffi
MWi
p
)
(2:34)
Viscosity of gas is very often estimated with charts or
correlations developed based on the charts. Gas viscosity
correlation of Carr et al. 1954 involves a two-step pro-
cedure: The gas viscosity at temperature and atmospheric
pressure is estimated first from gas-specific gravity and
inorganic compound content. The atmospheric value is
then adjusted to pressure conditions by means of a correc-
tion factor on the basis of reduced temperature and pres-
sure state of the gas. The atmospheric pressure viscosity
(m1) can be expressed as
m1 ¼ m1HC þ m1N2
þ m1CO2
þ m1H2S, (2:35)
where
m1HC ¼ 8:188 Â 10À3
À 6:15 Â 10À3
log (gg)
þ (1:709 Â 10À5
À 2:062 Â 10À6
gg)T, (2:36)
m1N2
¼ [9:59 Â 10À3
þ 8:48 Â 10À3
log (gg)]yN2
, (2:37)
m1CO2
¼ [6:24 Â 10À3
þ 9:08 Â 10À3
log (gg)]yCO2
, (2:38)
m1H2S ¼ [3:73 Â 10À3
þ 8:49 Â 10À3
log (gg)]yH2S, (2:39)
Dempsey (1965) developed the following relation:
mr ¼ ln
mg
m1
Tpr
¼ a0 þ a1ppr þ a2p2
pr þ a3p3
pr þ Tpr(a4 þ a5ppr
þ a6p2
pr þ a7p3
pr) þ T2
pr(a8 þ a9ppr þ a10p2
pr
þ a11p3
pr) þ T3
pr(a12 þ a13ppr þ a14p2
pr
þ a15p3
pr), (2:40)
where
a0 ¼ À2:46211820
a1 ¼ 2:97054714
a2 ¼ À0:28626405
a3 ¼ 0:00805420
a4 ¼ 2:80860949
a5 ¼ À3:49803305
a6 ¼ 0:36037302
a7 ¼ À0:01044324
a8 ¼ À0:79338568
a9 ¼ 1:39643306
a10 ¼ À0:14914493
a11 ¼ 0:00441016
a12 ¼ 0:08393872
a13 ¼ À0:18640885
a14 ¼ 0:02033679
a15 ¼ À0:00060958
Thus, once the value of mr is determined from the right-
hand side of this equation, gas viscosity at elevated pres-
sure can be readily calculated using the following relation:
mg ¼
m1
Tpr
emr (2:41)
Other correlations for gas viscosity include that of Dean
and Stiel (1958) and Lee et al. (1966).
Example Problem 2.3 A 0.65 specific–gravity natural gas
contains 10% nitrogen, 8% carbon dioxide, and 2%
hydrogen sulfide. Estimate viscosity of the gas at
10,000 psia and 1808F.
Solution Example Problem 2.3 is solved with the spread-
sheet Carr-Kobayashi-Burrows-GasViscosity.xls, which is
attached to this book. The result is shown in Table 2.3.
2.3.4 Gas Compressibility Factor
Gas compressibility factor is also called ‘‘deviation factor’’
or ‘‘z-factor.’’ Its value reflects how much the real gas
deviates from the ideal gas at a given pressure and tem-
perature. Definition of the compressibility factor is
expressed as
z ¼
Vactual
Videal gas
: (2:42)
Introducing the z-factor to the gas law for ideal gas results
in the gas law for real gas as
pV ¼ nzRT, (2:43)
where n is the number of moles of gas. When pressure p is
entered in psia, volume V in ft3
, and temperature in 8R, the
gas constant R is equal to10.73
psia À ft3
mole ÀR
.
Gas compressibility factor can be determined on the basis
of measurements in PVT laboratories. For a given amount
of gas, if temperature is kept constant and volume is mea-
sured at 14.7 psia and an elevated pressure p1, z-factor can
then be determined with the following formula:
z ¼
p1
14:7
V1
V0
, (2:44)
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![3.1 Introduction
Reservoir deliverability is defined as the oil or gas produc-
tion rate achievable from reservoir at a given bottom-hole
pressure. It is a major factor affecting well deliverability.
Reservoir deliverability determines types of completion
and artificial lift methods to be used. A thorough knowl-
edge of reservoir productivity is essential for production
engineers.
Reservoir deliverability depends on several factors in-
cluding the following:
. Reservoir pressure
. Pay zone thickness and permeability
. Reservoir boundary type and distance
. Wellbore radius
. Reservoir fluid properties
. Near-wellbore condition
. Reservoir relative permeabilities
Reservoir deliverability can be mathematically modeled on
the basis of flow regimes such as transient flow, steady
state flow, and pseudo–steady state flow. An analytical
relation between bottom-hole pressure and production
rate can be formulated for a given flow regime. The
relation is called ‘‘inflow performance relationship’’
(IPR). This chapter addresses the procedures used for
establishing IPR of different types of reservoirs and well
configurations.
3.2 Flow Regimes
When a vertical well is open to produce oil at production
rate q, it creates a pressure funnel of radius r around the
wellbore, as illustrated by the dotted line in Fig. 3.1a. In
this reservoir model, the h is the reservoir thickness, k is
the effective horizontal reservoir permeability to oil, mo is
viscosity of oil, Bo is oil formation volume factor, rw is
wellbore radius, pwf is the flowing bottom hole pressure,
and p is the pressure in the reservoir at the distance r from
the wellbore center line. The flow stream lines in the
cylindrical region form a horizontal radial flow pattern
as depicted in Fig. 3.1b.
3.2.1 Transient Flow
‘‘Transient flow’’ is defined as a flow regime where/when
the radius of pressure wave propagation from wellbore has
not reached any boundaries of the reservoir. During tran-
sient flow, the developing pressure funnel is small relative
to the reservoir size. Therefore, the reservoir acts like an
infinitively large reservoir from transient pressure analysis
point of view.
Assuming single-phase oil flow in the reservoir, several
analytical solutions have been developed for describing the
transient flow behavior. They are available from classic
textbooks such as that of Dake (1978). A constant-rate
solution expressed by Eq. (3.1) is frequently used in pro-
duction engineering:
pwf ¼ pi À
162:6qBomo
kh
 log t þ log
k
fmoctr2
w
À 3:23 þ 0:87S
, (3:1)
where
pwf ¼ flowing bottom-hole pressure, psia
pi ¼ initial reservoir pressure, psia
q ¼ oil production rate, stb/day
mo ¼ viscosity of oil, cp
k ¼ effective horizontal permeability to oil, md
h ¼ reservoir thickness, ft
t ¼ flow time, hour
f ¼ porosity, fraction
ct ¼ total compressibility, psiÀ1
rw ¼ wellbore radius to the sand face, ft
S ¼ skin factor
Log ¼ 10-based logarithm log10
Because oil production wells are normally operated at
constant bottom-hole pressure because of constant well-
head pressure imposed by constant choke size, a constant
bottom-hole pressure solution is more desirable for well-
inflow performance analysis. With an appropriate inner
boundary condition arrangement, Earlougher (1977)
developed a constant bottom-hole pressure solution,
which is similar to Eq. (3.1):
q ¼
kh( pi À pwf )
162:6Bomo log t þ log
k
fmoctr2
w
À 3:23 þ 0:87S
,
(3:2)
which is used for transient well performance analysis in
production engineering.
Equation (3.2) indicates that oil rate decreases with flow
time. This is because the radius of the pressure funnel, over
which the pressure drawdown (pi À pwf ) acts, increases
with time, that is, the overall pressure gradient in the
reservoir drops with time.
For gas wells, the transient solution is
qg ¼
kh[m( pi) À m(pwf )]
1; 638T log t þ log
k
fmoctr2
w
À 3:23 þ 0:87S
, (3:3)
where qg is production rate in Mscf/d, T is temperature in
8R, and m(p) is real gas pseudo-pressure defined as
m( p) ¼
ðp
pb
2p
mz
dp: (3:4)
rw
r
pwf
p
mo Bok
h
q
(a)
k mo Bo
p
pwf
r
(b)
rw
Figure 3.1 A sketch of a radial flow reservoir model: (a)
lateral view, (b) top view.
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![The real gas pseudo-pressure can be readily determined
with the spreadsheet program PseudoPressure.xls.
3.2.2 Steady-State Flow
‘‘Steady-state flow’’ is defined as a flow regime where the
pressure at any point in the reservoir remains constant
over time. This flow condition prevails when the pressure
funnel shown in Fig. 3.1 has propagated to a constant-
pressure boundary. The constant-pressure boundary can
be an aquifer or a water injection well. A sketch of the
reservoir model is shown in Fig. 3.2, where pe represents
the pressure at the constant-pressure boundary. Assuming
single-phase flow, the following theoretical relation can be
derived from Darcy’s law for an oil reservoir under the
steady-state flow condition due to a circular constant-
pressure boundary at distance re from wellbore:
q ¼
kh(pe À pwf )
141:2Bomo ln re
rw
þ S
, (3:5)
where ‘‘ln’’ denotes 2.718-based natural logarithm loge.
Derivation of Eq. (3.5) is left to readers for an exercise.
3.2.3 Pseudo–Steady-State Flow
‘‘Pseudo–steady-state’’ flow is defined as a flow regime
where the pressure at any point in the reservoir declines
at the same constant rate over time. This flow condition
prevails after the pressure funnel shown in Fig. 3.1 has
propagated to all no-flow boundaries. A no-flow bound-
ary can be a sealing fault, pinch-out of pay zone, or
boundaries of drainage areas of production wells. A sketch
of the reservoir model is shown in Fig. 3.3, where pe
represents the pressure at the no-flow boundary at time
t4. Assuming single-phase flow, the following theoretical
relation can be derived from Darcy’s law for an oil reser-
voir under pseudo–steady-state flow condition due to a
circular no-flow boundary at distance re from wellbore:
q ¼
kh(pe À pwf )
141:2Bomo ln re
rw
À 1
2 þ S
: (3:6)
The flow time required for the pressure funnel to reach the
circular boundary can be expressed as
tpss ¼ 1,200
fmoctr2
e
k
: (3:7)
Because the pe in Eq. (3.6) is not known at any given time,
the following expression using the average reservoir pres-
sure is more useful:
q ¼
kh(pp À pwf )
141:2Bomo ln re
rw
À 3
4 þ S
, (3:8)
where p¯ is the average reservoir pressure in psia. Deriv-
ations of Eqs. (3.6) and (3.8) are left to readers for exer-
cises.
If the no-flow boundaries delineate a drainage area of
noncircular shape, the following equation should be used
for analysis of pseudo–steady-state flow:
q ¼
kh(pp À pwf )
141:2Bomo
1
2 ln 4A
gCAr2
w
þ S
, (3:9)
where
A ¼ drainage area, ft2
g ¼ 1:78 ¼ Euler’s constant
CA ¼ drainage area shape factor, 31.6 for a circular
boundary.
The value of the shape factor CA can be found from
Fig. 3.4.
For a gas well located at the center of a circular drainage
area, the pseudo–steady-state solution is
qg ¼
kh[m(pp) À m(pwf )]
1,424T ln re
rw
À 3
4 þ S þ Dqg
, (3:10)
where
D ¼ non-Darcy flow coefficient, d/Mscf.
3.2.4 Horizontal Well
The transient flow, steady-state flow, and pseudo–steady-
state flow can also exist in reservoirs penetrated by horizon-
tal wells. Different mathematical models are available from
h p
rre
pe
pwf
rw
Figure 3.2 A sketch of a reservoir with a constant-pressure boundary.
p
h
rre
pe
pi
pwf
rw
t1
t2
t3
t4
Figure 3.3 A sketch of a reservoir with no-flow boundaries.
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![Well IPR curves are usually constructed using reservoir
inflow models, which can be from either a theoretical basis
or an empirical basis. It is essential to validate these
models with test points in field applications.
3.3.1 LPR for Single (Liquid)-Phase Reservoirs
All reservoir inflow models represented by Eqs. (3.1), (3.3),
(3.7), and (3.8) were derived on the basis of the assumption of
single-phase liquid flow. This assumption is valid for under-
saturated oil reservoirs, or reservoir portions where the pres-
sure is above the bubble-point pressure. These equations
define the productivity index (JÃ
) for flowing bottom-hole
pressures above the bubble-point pressure as follows:
JÃ
¼
q
(pi À pwf )
¼
kh
162:6Bomo log t þ log
k
fmoctr2
w
À 3:23 þ 0:87S
(3:15)
for radial transient flow around a vertical well,
JÃ
¼
q
(pe À pwf )
¼
kh
141:2Bomo ln re
rw
þ S
(3:16)
for radial steady-state flow around a vertical well,
JÃ
¼
q
(pp À pwf )
¼
kh
141:2Bomo
1
2 ln 4A
gCAr2
w
þ S
(3:17)
for pseudo–steady-state flow around a vertical well, and
JÃ
¼
q
(pe À pwf )
¼
kH h
141:2Bm ln
aþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2À(L=2)2
p
L=2
!
þ Ianih
L ln Iani h
rw(Ianiþ1)
h i '
(3:18)
for steady-state flow around a horizontal well.
Since the productivity index (JÃ
) above the bubble-point
pressure isindependent of productionrate, the IPR curve for a
single (liquid)-phase reservoir is simply a straight line drawn
from the reservoir pressure to the bubble-point pressure. If the
bubble-point pressure is 0 psig, the absolute open flow (AOF)
is the productivity index (JÃ
) times the reservoir pressure.
Example Problem 3.1 Construct IPR of a vertical well in
an oil reservoir. Consider (1) transient flow at 1 month, (2)
steady-state flow, and (3) pseudo–steady-state flow. The
following data are given:
Porosity: f ¼ 0:19
Effective horizontal permeability:k ¼ 8:2 md
Pay zone thickness: h ¼ 53 ft
Reservoir pressure: pe or pp ¼ 5,651 psia
Bubble-point pressure: pb ¼ 50 psia
Fluid formation volume factor:, Bo ¼ 1:1
Fluid viscosity: mo ¼ 1:7 cp
Total compressibility, ct ¼ 0:0000129 psiÀ1
Drainage area: A ¼ 640 acres
(re ¼ 2,980 ft)
Wellbore radius: rw ¼ 0:328 ft
Skin factor: S ¼ 0
Solution
1. For transient flow, calculated points are
JÃ
¼
kh
162:6Bm log t þ log fmctr2
w
À 3:23
¼
(8:2)(53)
162:6(1:1)(1:7) log [( (30)(24)] þ log (8:2)
(0:19)(1:7)(0:0000129)(0:328)2 À 3:23
¼ 0:2075 STB=d-psi
Transient IPR curve is plotted in Fig. 3.6.
2. For steady state flow:
JÃ
¼
kh
141:2Bm ln re
rw
þ S
¼
(8:2)(53)
141:2(1:1)(1:7) ln 2,980
0:328
¼ 0:1806 STB=d-psi
Calculated points are:
Steady state IPR curve is plotted in Fig. 3.7.
3. For pseudosteady state flow:
JÃ
¼
kh
141:2Bm ln re
rw
À 3
4 þ S
¼
(8:2)(53)
141:2(1:1)(1:7) ln 2,980
0:328 À 0:75
¼ 0:1968 STB=d-psi
0
1,000
2,000
3,000
4,000
5,000
6,000
0 200 400 600 800 1,000 1,200
qo (stb/day)
pwf(psia)
Figure 3.6 Transient IPR curve for Example Problem 3.1.
0
1,000
2,000
3,000
4,000
5,000
6,000
0 200 400 600 800 1,000 1,200
qo (stb/day)
pwf(psia)
Figure 3.7 Steady-state IPR curve for Example
Problem 3.1.
pwf (psi) qo(stb/day)
50 1,011
5,651 0
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 33 3.1.2007 8:30pm Compositor Name: SJoearun
RESERVOIR DELIVERABILITY 3/33](https://image.slidesharecdn.com/lvzgyxtjtgu380ahut9b-signature-b2ba6e1375509d785bab16b3cad4b5b4b504aa1845b2191a80963142ee8b462c-poli-140812083436-phpapp01/85/Petroleum-production-engineering-43-320.jpg)
![and
X1 ¼ log [(NL) þ 3]: (4:36)
Once the value of parameter (CNL) is determined, it is used
for calculating the value of the group
NvLp0:1
(CNL)
N0:575
vG p0:1
a ND
, where
p is the absolute pressure at the location where pressure
gradient is to be calculated, and pa is atmospheric pressure.
The value of this group is then used as an entry in the
second chart to determine parameter (yL=c). We have
found that the second chart can be represented by the
following correlation with good accuracy:
yL
c
¼ À0:10307 þ 0:61777[ log (X2) þ 6]
À 0:63295[ log (X2) þ 6]2
þ 0:29598[ log (X2)
þ 6]3
À 0:0401[ log (X2) þ 6]4
, (4:37)
where
X2 ¼
NvLp0:1
(CNL)
N0:575
vG p0:1
a ND
: (4:38)
According to Hagedorn and Brown (1965), the value of
parameter c can be determined from the third chart using
a value of group
NvGN0:38
L
N2:14
D
.
We have found that for
NvGN0:38
L
N2:14
D
0:01 the third chart
can be replaced by the following correlation with accept-
able accuracy:
c ¼ 0:91163 À 4:82176X3 þ 1,232:25X2
3
À 22,253:6X3
3 þ 116174:3X4
3 , (4:39)
where
X3 ¼
NvGN0:38
L
N2:14
D
: (4:40)
However, c ¼ 1:0 should be used for
NvGN0:38
L
N2:14
D
# 0:01.
Finally, the liquid holdup can be calculated by
yL ¼ c
yL
c
: (4:41)
The Fanning friction factor in Eq. (4.27) can be deter-
mined using either Chen’s correlation Eq. (4.5) or (4.16).
The Reynolds number for multiphase flow can be calcu-
lated by
NRe ¼
2:2 Â 10À2
mt
DmyL
L m(1ÀyL)
G
, (4:42)
where mt is mass flow rate. The modified mH-B method
uses the Griffith correlation for the bubble-flow regime.
The bubble-flow regime has been observed to exist when
lG LB, (4:43)
where
lG ¼
usG
um
(4:44)
and
LB ¼ 1:071 À 0:2218
u2
m
D
, (4:45)
which is valid for LB $ 0:13. When the LB value given by
Eq. (4.45) is less than 0.13, LB ¼ 0:13 should be used.
Neglecting the kinetic energy pressure drop term, the
Griffith correlation in U.S. field units can be expressed as
144
dp
dz
¼ rr þ
fF m2
L
7:413 Â 1010D5rLy2
L
, (4:46)
where mL is mass flow rate of liquid only. The liquid
holdup in Griffith correlation is given by the following
expression:
yL ¼ 1 À
1
2
1 þ
um
us
À
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ
um
us
2
À 4
usG
us
s2
4
3
5, (4:47)
where ms ¼ 0:8 ft=s. The Reynolds number used to obtain
the friction factor is based on the in situ average liquid
velocity, that is,
NRe ¼
2:2 Â 10À2
mL
DmL
: (4:48)
To speed up calculations, the Hagedorn–Brown cor-
relation has been coded in the spreadsheet program Hage-
dornBrownCorrelation.xls.
Example Problem 4.4 For the data given below, calculate
and plot pressure traverse in the tubing string:
Solution This example problem is solved with the
spreadsheet program HagedornBrownCorrelation.xls. The
result is shown in Table 4.3 and Fig. 4.4.
4.4 Single-Phase Gas Flow
The first law of thermodynamics (conservation of energy)
governs gas flow in tubing. The effect of kinetic energy
change is negligible because the variation in tubing diam-
eter is insignificant in most gas wells. With no shaft work
device installed along the tubing string, the first law of
thermodynamics yields the following mechanical balance
equation:
dP
r
þ
g
gc
dZ þ
fMn2
dL
2gcDi
¼ 0 (4:49)
Because dZ ¼ cos udL, r ¼
29ggP
ZRT , and n ¼ 4qsczPscT
pD2
i TscP
, Eq.
(4.49) can be rewritten as
zRT
29gg
dP
P
þ
g
gc
cos u þ
8fMQ2
scP2
sc
p2gcD5
i T2
sc
zT
P
!2
( )
dL ¼ 0, (4:50)
which is an ordinary differential equation governing
gas flow in tubing. Although the temperature T can be
approximately expressed as a linear function of length L
through geothermal gradient, the compressibility factor z
is a function of pressure P and temperature T. This makes
it difficult to solve the equation analytically. Fortunately,
the pressure P at length L is not a strong function of
temperature and compressibility factor. Approximate so-
lutions to Eq. (4.50) have been sought and used in the
natural gas industry.
Tubing shoe depth: 9,700 ft
Tubing inner diameter: 1.995 in.
Oil gravity: 40 8API
Oil viscosity: 5 cp
Production GLR: 75 scf/bbl
Gas-specific gravity: 0.7 air ¼ 1
Flowing tubing head pressure: 100 psia
Flowing tubing head temperature: 80 8F
Flowing temperature at tubing shoe: 180 8F
Liquid production rate: 758 stb/day
Water cut: 10 %
Interfacial tension: 30 dynes/cm
Specific gravity of water: 1.05 H2O ¼ 1
Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 53 22.12.2006 6:07pm
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![4.4.1 Average Temperature and Compressibility
Factor Method
If single average values of temperature and compressibility
factor over the entire tubing length can be assumed, Eq.
(4.50) becomes
zzRTT
29gg
dP
P
þ
g
gc
cos u þ
8fMQ2
csP2
sczz2 TT2
p2gcD5
i T2
scP2
'
dL ¼ 0: (4:51)
By separation of variables, Eq. (4.51) can be integrated
over the full length of tubing to yield
P2
wf ¼ Exp(s)P2
hf þ
8fM[Exp(s) À 1]Q2
scP2
sczz2 TT2
p2gcD5
i T2
sc cos u
, (4:52)
where
s ¼
58gggL cos u
gcRzzTT
: (4:53)
Equations (4.52) and (4.53) take the following forms when
U.S. field units (qsc in Mscf/d), are used (Katz et al., 1959):
Table 4.3 Result Given by HagedornBrownCorrelation.xls for Example Problem 4.4
HagedornBrownCorrelation.xls
Description: This spreadsheet calculates flowing pressures in tubing string based on tubing head pressure using the
Hagedorn–Brown correlation.
Instruction: (1) Select a unit system; (2) update parameter values in the Input data section;
(3) click ‘‘Solution’’ button; and (4) view result in the Solution section and charts.
Input data U.S. Field units SI units
Depth (D): 9,700 ft
Tubing inner diameter (dti): 1.995 in.
Oil gravity (API): 40 8API
Oil viscosity (mo): 5 cp
Production GLR (GLR): 75 scf/bbl
Gas-specific gravity (gg): 0.7 air ¼1
Flowing tubing head pressure (phf ): 100 psia
Flowing tubing head temperature (thf ): 80 8F
Flowing temperature at tubing shoe (twf ): 180 8F
Liquid production rate (qL): 758 stb/day
Water cut (WC): 10 %
Interfacial tension (s): 30 dynes/cm
Specific gravity of water (gw): 1.05 H2O ¼ 1
Solution
Depth Pressure
(ft) (m) (psia) (MPa)
0 0 100 0.68
334 102 183 1.24
669 204 269 1.83
1,003 306 358 2.43
1,338 408 449 3.06
1,672 510 543 3.69
2,007 612 638 4.34
2,341 714 736 5.01
2,676 816 835 5.68
3,010 918 936 6.37
3,345 1,020 1,038 7.06
3,679 1,122 1,141 7.76
4,014 1,224 1,246 8.48
4,348 1,326 1,352 9.20
4,683 1,428 1,459 9.93
5,017 1,530 1,567 10.66
5,352 1,632 1,676 11.40
5,686 1,734 1,786 12.15
6,021 1,836 1,897 12.90
6,355 1,938 2,008 13.66
6,690 2,040 2,121 14.43
7,024 2,142 2,234 15.19
7,359 2,243 2,347 15.97
7,693 2,345 2,461 16.74
8,028 2,447 2,576 17.52
8,362 2,549 2,691 18.31
8,697 2,651 2,807 19.10
9,031 2,753 2,923 19.89
9,366 2,855 3,040 20.68
9,700 2,957 3,157 21.48
Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 54 22.12.2006 6:07pm
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![p2
wf ¼ Exp(s)p2
hf
þ
6:67 Â 10À4
[Exp(s) À 1]fMq2
sczz2 TT2
d5
i cos u
(4:54)
and
s ¼
0:0375ggL cos u
zzTT
(4:55)
The Darcy–Wiesbach (Moody) friction factor fM can be
found in the conventional manner for a given tubing
diameter, wall roughness, and Reynolds number. How-
ever, if one assumes fully turbulent flow, which is the
case for most gas wells, then a simple empirical relation
may be used for typical tubing strings (Katz and Lee
1990):
fM ¼
0:01750
d0:224
i
for di# 4:277 in: (4:56)
fM ¼
0:01603
d0:164
i
for di 4:277 in: (4:57)
Guo (2001) used the following Nikuradse friction factor
correlation for fully turbulent flow in rough pipes:
fM ¼
1
1:74 À 2 log 2«
di
2
4
3
5
2
(4:58)
Because the average compressibility factor is a function of
pressure itself, a numerical technique such as Newton–
Raphson iteration is required to solve Eq. (4.54) for bot-
tom-hole pressure. This computation can be performed
automatically with the spreadsheet program Average
TZ.xls. Users need to input parameter values in the
Input data section and run Macro Solution to get results.
Example Problem 4.5 Suppose that a vertical well
produces 2 MMscf/d of 0.71 gas-specific gravity gas
through a 27
⁄8 in. tubing set to the top of a gas reservoir
at a depth of 10,000 ft. At tubing head, the pressure is
800 psia and the temperature is 150 8F; the bottom-hole
temperature is 200 8F. The relative roughness of tubing is
about 0.0006. Calculate the pressure profile along the
tubing length and plot the results.
Solution Example Problem 4.5 is solved with the
spreadsheet program AverageTZ.xls. Table 4.4 shows the
appearance of the spreadsheet for the Input data and Result
sections. The calculated pressure profile is plotted in
Fig. 4.5.
4.4.2 Cullender and Smith Method
Equation (4.50) can be solved for bottom-hole pressure
using a fast numerical algorithm originally developed by
Cullender and Smith (Katz et al., 1959). Equation (4.50)
can be rearranged as
0
2,000
4,000
6,000
8,000
10,000
12,000
Pressure (psia)
Depth(ft)
3,5000 500 1,000 1,500 2,000 2,500 3,000
Figure 4.4 Pressure traverse given by HagedornBrownCorrelation.xls for Example Problem 4.4.
Table 4.4 Spreadsheet AverageTZ.xls: the Input Data
and Result Sections
AverageTZ.xls
Description: This spreadsheet calculates tubing pressure
traverse for gas wells.
Instructions:
Step 1: Input your data in the Input data section.
Step 2: Click ‘‘Solution’’ button to get results.
Step 3: View results in table and in graph sheet ‘‘Profile’’.
Input data
gg ¼ 0.71
d ¼ 2.259 in.
«=d ¼ 0.0006
L ¼ 10.000 ft
u ¼ 0 degrees
phf ¼ 800 psia
Thf ¼ 150 8F
Twf ¼ 200 8F
qsc ¼ 2,000 Mscf/d
Solution
fM ¼ 0.017396984
Depth (ft) T (8R) p (psia) Zav
0 610 800 0.9028
1,000 615 827 0.9028
2,000 620 854 0.9027
3,000 625 881 0.9027
4,000 630 909 0.9026
5,000 635 937 0.9026
6,000 640 965 0.9026
7,000 645 994 0.9026
8,000 650 1023 0.9027
9,000 655 1053 0.9027
10,000 660 1082 0.9028
Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 55 22.12.2006 6:07pm
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![5.1 Introduction
Wellhead chokes are used to limit production rates for
regulations, protect surface equipment from slugging,
avoid sand problems due to high drawdown, and control
flow rate to avoid water or gas coning. Two types of well-
head chokes are used. They are (1) positive (fixed) chokes
and (2) adjustable chokes.
Placing a choke at the wellhead means fixing the well-
head pressure and, thus, the flowing bottom-hole pressure
and production rate. For a given wellhead pressure, by
calculating pressure loss in the tubing the flowing bottom-
hole pressure can be determined. If the reservoir pressure
and productivity index is known, the flow rate can then be
determined on the basis of inflow performance relation-
ship (IPR).
5.2 Sonic and Subsonic Flow
Pressure drop across well chokes is usually very significant.
There is no universal equation for predicting pressure drop
across the chokes for all types of production fluids. Differ-
ent choke flow models are available from the literature,
and they have to be chosen based on the gas fraction in the
fluid and flow regimes, that is, subsonic or sonic flow.
Both sound wave and pressure wave are mechanical
waves. When the fluid flow velocity in a choke reaches the
traveling velocity of sound in the fluid under the in situ
condition, the flow is called ‘‘sonic flow.’’ Under sonic
flow conditions, the pressure wave downstream of the
choke cannot go upstream through the choke because the
medium (fluid) is traveling in the opposite direction at the
same velocity. Therefore, a pressure discontinuity exists at
the choke, that is, the downstream pressure does not affect
the upstream pressure. Because of the pressure discontinu-
ity at the choke, any change in the downstream pressure
cannot be detected from the upstream pressure gauge. Of
course, any change in the upstream pressure cannot be
detected from the downstream pressure gauge either. This
sonic flow provides a unique choke feature that stabilizes
well production rate and separation operation conditions.
Whether a sonic flow exists at a choke depends on a
downstream-to-upstream pressure ratio. If this pressure
ratio is less than a critical pressure ratio, sonic (critical)
flow exists. If this pressure ratio is greater than or equal to
the critical pressure ratio, subsonic (subcritical) flow exists.
The critical pressure ratio through chokes is expressed as
poutlet
pup
c
¼
2
k þ 1
k
kÀ1
, (5:1)
where poutlet is the pressure at choke outlet, pup is the
upstream pressure, and k ¼ Cp=Cv is the specific heat
ratio. The value of the k is about 1.28 for natural gas.
Thus, the critical pressure ratio is about 0.55 for natural
gas. A similar constant is used for oil flow. A typical choke
performance curve is shown in Fig. 5.1.
5.3 Single-Phase Liquid Flow
When the pressure drop across a choke is due to kinetic
energy change, for single-phase liquid flow, the second
term in the right-hand side of Eq. (4.1) can be rearranged
as
q ¼ CDA
ffiffiffiffiffiffiffiffiffiffiffiffiffi
2gcDP
r
s
, (5:2)
where
q ¼ flow rate, ft3
=s
CD ¼ choke discharge coefficient
A ¼ choke area, ft2
gc ¼ unit conversion factor, 32.17 lbm-ft=lbf -s2
ÁP ¼ pressure drop, lbf =ft2
r ¼ fluid density, lbm=ft3
If U.S. field units are used, Eq. (5.2) is expressed as
q ¼ 8074CDd2
2
ffiffiffiffiffiffi
Dp
r
s
, (5:3)
where
q ¼ flow rate, bbl/d
d2 ¼ choke diameter, in.
Áp ¼ pressure drop, psi
The choke discharge coefficient CD can be determined
based on Reynolds number and choke/pipe diameter ratio
(Figs. 5.2 and 5.3). The following correlation has been
found to give reasonable accuracy for Reynolds numbers
between 104
and 106
for nozzle-type chokes (Guo and
Ghalambor, 2005):
CD ¼
d2
d1
þ
0:3167
d2
d1
0:6
þ 0:025[ log (NRe) À 4], (5:4)
where
d1 ¼ upstream pipe diameter, in.
d2 ¼ choke diameter, in.
NRe ¼ Reynolds number based on d2
5.4 Single-Phase Gas Flow
Pressure equations for gas flow through a choke are
derived based on an isentropic process. This is because
there is no time for heat to transfer (adiabatic) and the
friction loss is negligible (assuming reversible) at chokes.
In addition to the concern of pressure drop across the
chokes, temperature drop associated with choke flow is
also an important issue for gas wells, because hydrates
may form that may plug flow lines.
0
0 0.2 0.4 0.6 0.8 1 1.2
0.1
0.2
0.3
p2/p1
q
q
Critical
Sub-
critical
p1 p2
d1 d2
Figure 5.1 A typical choke performance curve.
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ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:3
(0:6)(75 þ 460)
2
1:3 þ 1
1:3þ1
1:3À1
s
qsc ¼ 12,743 Mscf=d
Check NRe:
m ¼ 0:01245 cp by the Carr–Kobayashi–Burrows cor-
relation.
NRe ¼
20qscgg
md2
¼
(20)(12,743)(0:6)
(0:01245)(1)
¼ 1:23 Â 107
106
(b)
Tdn ¼ Tup
zup
zoutlet
Poutlet
Pup
kÀ1
k
¼ (75 þ 460)(1)(0:5459)
1:3À1
1:3
¼ 465
R ¼ 5
F 32
F
Therefore, heating is needed to prevent icing.
(c)
Poutlet ¼ Pup
Poutlet
Pup
¼ (800)(0:5459) ¼ 437 psia
Example Problem 5.2 A 0.65 specific gravity natural gas
flows from a 2-in. pipe through a 1.5-in. nozzle-type
choke. The upstream pressure and temperature are
100 psia and 70 8F, respectively. The downstream
pressure is 80 psia (measured 2 ft from the nozzle). The
gas-specific heat ratio is 1.25. (a) What is the expected
daily flow rate? (b) Is icing a potential problem? (c) What
is the expected pressure at the nozzle outlet?
Solution (a)
Poutlet
Pup
c
¼
2
k þ 1
k
kÀ1
¼
2
1:25 þ 1
1:25
1:25À1
¼ 0:5549
Pdn
Pup
¼
80
100
¼ 0:8 0:5549 Subsonic flow exists:
d2
d1
¼
1:500
200
¼ 0:75
Assuming NRe 106
, Fig. 5.1 gives CD ¼ 1:2.
qsc ¼ 1,248CDAPup
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
(k À 1)ggTup
Pdn
Pup
2
k
À
Pdn
Pup
kþ1
k
#v
u
u
t
qsc ¼ (1,248)(1:2)[p(1:5)2
=4](100)
Â
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:25
(1:25 À 1)(0:65)(530)
80
100
2
1:25
À
80
100
1:25þ1
1:25
#v
u
u
t
qsc ¼ 5,572 Mscf=d
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![6.1 Introduction
Well deliverability is determined by the combination of
well inflow performance (see Chapter 3) and wellbore
flow performance (see Chapter 4). Whereas the former
describes the deliverability of the reservoir, the latter pre-
sents the resistance to flow of production string. This
chapter focuses on prediction of achievable fluid produc-
tion rates from reservoirs with specified production string
characteristics. The technique of analysis is called ‘‘Nodal
analysis’’ (a Schlumburger patent). Calculation examples
are illustrated with computer spreadsheets that are
provided with this book.
6.2 Nodal Analysis
Fluid properties change with the location-dependent pres-
sure and temperature in the oil and gas production system.
To simulate the fluid flow in the system, it is necessary to
‘‘break’’ the system into discrete nodes that separate sys-
tem elements (equipment sections). Fluid properties at the
elements are evaluated locally. The system analysis for
determination of fluid production rate and pressure at
a specified node is called ‘‘Nodal analysis’’ in petroleum
engineering. Nodal analysis is performed on the principle
of pressure continuity, that is, there is only one unique
pressure value at a given node regardless of whether the
pressure is evaluated from the performance of upstream
equipment or downstream equipment. The performance
curve (pressure–rate relation) of upstream equipment is
called ‘‘inflow performance curve’’; the performance
curve of downstream equipment is called ‘‘outflow per-
formance curve.’’ The intersection of the two performance
curves defines the operating point, that is, operating flow
rate and pressure, at the specified node. For the conveni-
ence of using pressure data measured normally at either
the bottom-hole or the wellhead, Nodal analysis is usually
conducted using the bottom-hole or wellhead as the solu-
tion node. This chapter illustrates the principle of Nodal
analysis with simplified tubing string geometries (i.e.,
single-diameter tubing strings).
6.2.1 Analysis with the Bottom-Hole Node
When the bottom-hole is used as a solution node in Nodal
analysis, the inflow performance is the well inflow per-
formance relationship (IPR) and the outflow performance
is the tubing performance relationship (TPR), if the tubing
shoe is set to the top of the pay zone. Well IPR can be
established with different methods presented in Chapter 3.
TPR can be modeled with various approaches as discussed
in Chapter 4.
Traditionally, Nodal analysis at the bottom-hole is car-
ried out by plotting the IPR and TPR curves and graph-
ically finding the solution at the intersection point of the
two curves. With modern computer technologies, the
solution can be computed quickly without plotting
the curves, although the curves are still plotted for visual
verification.
6.2.1.1 Gas Well
Consider the bottom-hole node of a gas well. If the IPR of
the well is defined by
qsc ¼ C( pp2
À p2
wf ) n
, (6:1)
and if the outflow performance relationship of the node
(i.e., the TPR) is defined by
p2
wf ¼ Exp(s)p2
hf
þ
6:67 Â 10À4
[Exp(s) À 1] fMq2
sczz2 TT2
d5
i cos u
, (6:2)
then the operating flow rate qsc and pressure pwf at the
bottom-hole node can be determined graphically by plot-
ting Eqs. (6.1) and (6.2) and finding the intersection point.
The operating point can also be solved analytically by
combining Eqs. (6.1) and (6.2). In fact, Eq. (6.1) can be
rearranged as
p2
wf ¼ pp2
À
qsc
C
1
n
: (6:3)
Substituting Eq. (6.3) into Eq. (6.2) yields
pp2
À
qsc
C
1
n
ÀExp(s)p2
hf
À
6:67 Â 10À4
[Exp(s) À 1] fMq2
sczz2 TT2
D5
i cos u
¼ 0; (6:4)
which can be solved with a numerical technique such as the
Newton–Raphson iteration for gas flow rate qsc. This
computation can be performed automatically with the
spreadsheet program BottomHoleNodalGas.xls.
Example Problem 6.1 Suppose that a vertical well
produces 0.71 specific gravity gas through a 27
⁄8 -in.
tubing set to the top of a gas reservoir at a depth of
10,000 ft. At tubing head, the pressure is 800 psia and
the temperature is 150 8F, whereas the bottom-hole
temperature is 200 8F. The relative roughness of tubing is
about 0.0006. Calculate the expected gas production rate
of the well using the following data for IPR:
Reservoir pressure: 2,000 psia
IPR model parameter C: 0.1 Mscf/d-psi2n
IPR model parameter n: 0.8
Solution Example Problem 6.1 is solved with the
spreadsheet program BottomHoleNodalGas.xls. Table 6.1
shows the appearance of the spreadsheet for the Input data
and Result sections. It indicates that the expected gas flow
rate is 1478 Mscf/d at a bottom-hole pressure of 1059 psia.
The inflow and outflow performance curves plotted in Fig.
6.1 confirm this operating point.
6.2.1.2 Oil Well
Consider the bottom-hole node of an oil well. As discussed
in Chapter 3, depending on reservoir pressure range, dif-
ferent IPR models can be used. For instance, if the reser-
voir pressure is above the bubble-point pressure, a straight-
line IPR can be used:
q ¼ JÃ
( pp À pwf ) (6:5)
The outflow performance relationship of the node (i.e., the
TPR) can be described by a different model. The simplest
model would be Poettmann–Carpenter model defined by
Eq. (4.8), that is,
pwf ¼ pwh þ rr þ
kk
rr
L
144
(6:6)
where pwh and L are tubing head pressure and well depth,
respectively, then the operating flow rate q and pressure
pwf at the bottom-hole node can be determined graphically
by plotting Eqs. (6.5) and (6.6) and finding the intersection
point.
The operating point can also be solved analytically by
combining Eqs. (6.5) and (6.6). In fact, substituting
Eq. (6.6) into Eq. (6.5) yields
q ¼ JÃ
pp À pwh þ rr þ
kk
rr
L
144
!
, (6:7)
which can be solved with a numerical technique such as the
Newton–Raphson iteration for liquid flow rate q. This
computation can be performed automatically with the
spreadsheet program BottomHoleNodalOil-PC.xls.
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![q. This computation can be performed automatically with
the spreadsheet program BottomHoleNodalOil-HB.xls.
Example Problem 6.4 For the data given in the following
table, predict the operating point:
Solution Example Problem 6.4 is solved with the spread-
sheet program BottomHoleNodalOil-HB.xls. Table 6.4
shows the appearance of the spreadsheet for the Input data
and Result sections. Figure 6.2 indicates that the expected
gas flow rate is 2200 stb/d at a bottom-hole pressure
of 3500 psia.
6.2.2 Analysis with Wellhead Node
When the wellhead is used as a solution node in Nodal
analysis, the inflow performance curve is the ‘‘wellhead
performance relationship’’ (WPR), which is obtained by
transforming the IPR to wellhead through the TPR.
The outflow performance curve is the wellhead choke
performance relationship (CPR). Some TPR models are
presented in Chapter 4. CPR models are discussed in
Chapter 5.
Nodal analysis with wellhead being a solution node
is carried out by plotting the WPR and CPR curves and
finding the solution at the intersection point of the two
curves. Again, with modern computer technologies, the solu-
tion can be computed quickly without plotting the curves,
although the curves are still plotted for verification.
6.2.2.1 Gas Well
If the IPR of a well is defined by Eq. (6.1) and the TPR is
represented by Eq. (6.2), substituting Eq. (6.2) into
Eq. (6.1) gives
qsc ¼ C pp2
À
Exp(s)p2
hf
þ
6:67 Â 10À4
[Exp(s) À 1] fMq2
sczz2 TT2
d5
i cos u
!n
, (6:12)
which defines a relationship between wellhead pressure phf
and gas production rate qsc, that is, WPR. If the CPR is
defined by Eq. (5.8), that is,
qsc ¼ 879CAphf
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
ggTup
!
2
k þ 1
kþ1
kÀ1
v
u
u
t , (6:13)
Depth: 9,850 ft
Tubing inner diameter: 1.995 in.
Oil gravity: 45 8API
Oil viscosity: 2 cp
Production GLR: 500 scf/bbl
Gas-specific gravity: 0.7 air ¼ 1
Flowing tubing head pressure: 450 psia
Flowing tubing head temperature: 80 8F
Flowing temperature at tubing shoe: 180 8F
Water cut: 10%
Reservoir pressure: 5,000 psia
Bubble-point pressure: 4,000 psia
Productivity index above bubble point: 1.5 stb/d-psi
Table 6.4 Solution Given by BottomHoleNodalOil-HB.xls
BottomHoleNodalOil-HB.xls
Description: This spreadsheet calculates operating point using the Hagedorn–Brown correlation.
Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution
button; and (4) view result in the Result section and charts.
Input data U.S. Field units SI units
Depth (D): 9,850 ft
Tubing inner diameter (dti): 1.995 in.
Oil gravity (API): 45 8API
Oil viscosity (mo): 2 cp
Production GLR (GLR): 500 scf/bbl
Gas-specific gravity (gg): 0.7 air ¼ 1
Flowing tubing head pressure (phf ): 450 psia
Flowing tubing head temperature (thf ): 80 8F
Flowing temperature at tubing shoe (twf ): 180 8F
Water cut: 10%
Reservoir pressure (pe): 5,000 psia
Bubble-point pressure (pb): 4,000 psia
Productivity index above bubble point (J*
): 1.5 stb/d-psi
Solution
US Field units :
qb ¼ 1,500
qmax ¼ 4,833
q (stb/d) pwf (psia)
IPR TPR
0 4,908
537 4,602 2,265
1,074 4,276 2,675
1,611 3,925 3,061
2,148 3,545 3,464
2,685 3,125 3,896
3,222 2,649 4,361
3,759 2,087 4,861
4,296 1,363 5,397
4,833 0 5,969
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![then the operating flow rate qsc and pressure phf at the
wellhead node can be determined graphically by plotting
Eqs. (6.12) and (6.13) and finding the intersection point.
The operating point can also be solved numerically by
combining Eqs. (6.12) and (6.13). In fact, Eq. (6.13) can be
rearranged as
phf ¼
qsc
879CA
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
ggTup
!
2
k þ 1
kþ1
kÀ1
v
u
u
t
: (6:14)
Substituting Eq. (6.14) into Eq. (6.12) gives
qsc ¼ C pp2
À Exp(s)
qsc
879CA
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
ggTup
2
kþ1
kþ1
kÀ1
r
0
B
B
@
1
C
C
A
2
0
B
B
B
@
2
6
6
6
4
þ
6:67 Â 10À4
[Exp(s) À 1]fMq2
sczz2 TT2
d5
i cos u
1
C
C
A
3
7
7
5
n
,
(6:15)
which can be solved numerically for gas flow rate qsc. This
computation can be performed automatically with the
spreadsheet program WellheadNodalGas-SonicFlow.xls.
Example Problem 6.5 Use the data given in the following
table to estimate gas production rate of a gas well:
Solution Example Problem 6.5 is solved with the
spreadsheet program WellheadNodalGas-SonicFlow.xls.
Table 6.5 shows the appearance of the spreadsheet for the
Input data and Result sections. It indicates that the expected
gas flow rate is 1,478 Mscf/d at a bottom-hole pressure of
1,050 psia. The inflow and outflow performance curves
plotted in Fig. 6.3 confirm this operating point.
6.2.2.2 Oil Well
As discussed in Chapter 3, depending on reservoir pressure
range, different IPR models can be used. For instance, if
the reservoir pressure is above the bubble-point pressure,
a straight-line IPR can be used:
q ¼ JÃ
pp À pwf
À Á
(6:16)
If the TPR is described by the Poettmann–Carpenter
model defined by Eq. (4.8), that is,
pwf ¼ pwh þ rr þ
kk
rr
L
144
(6:17)
substituting Eq. (6.17) into Eq. (6.16) gives
q ¼ JÃ
pp À pwh þ rr þ
kk
rr
L
144
!
, (6:18)
which describes inflow for the wellhead node and is called
the WPR. If the CPR is given by Eq. (5.12), that is,
pwh ¼
CRm
q
Sn
, (6:19)
the operating point can be solved analytically by combin-
ing Eqs. (6.18) and (6.19). In fact, substituting Eq. (6.19)
into Eq. (6.18) yields
q ¼ JÃ
pp À
CRm
q
Sn
þ rr þ
kk
rr
L
144
!
, (6:20)
which can be solved with a numerical technique. Because
the solution procedure involves loop-in-loop iterations, it
cannot be solved in MS Excel in an easy manner. A special
computer program is required. Therefore, a computer-
assisted graphical solution method is used in this text.
The operating flow rate q and pressure pwh at the well-
head node can be determined graphically by plotting
Eqs. (6.18) and (6.19) and finding the intersection point.
This computation can be performed automatically with
the spreadsheet program WellheadNodalOil-PC.xls.
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
Liquid Production Rate (bbl/d)
BottomHolePressure(psia)
IPR
TPR
0 5,0004,0003,0002,0001,000
Figure 6.2 Nodal analysis for Example Problem 6.4.
Gas-specific gravity: 0.71
Tubing inside diameter: 2.259 in.
Tubing wall relative roughness: 0.0006
Measured depth at tubing shoe: 10,000 ft
Inclination angle: 0 degrees
Wellhead choke size: 16 1
⁄64 4 in.
Flowline diameter: 2 in.
Gas-specific heat ratio: 1.3
Gas viscosity at wellhead: 0.01 cp
Wellhead temperature: 150 8F
Bottom-hole temperature: 200 8F
Reservoir pressure: 2,000 psia
C-constant in IPR model: 0.01 Mscf/ d-psi2n
n-exponent in IPR model: 0.8
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![Thus, the composite IPR,
q ¼ f phfn
À Á
, (6:31)
can be established implicitly.
It should be noted that the composite IPR model
described here is general. If the vertical section of the top
lateral is the production string (production through tubing
or/and casing), then phfn
will be the flowing wellhead pres-
sure. In this case, the relation expression (Eq. [6.31]) rep-
resents the WPR.
6.3.1 Gas well
For gas wells, Eq. (6.26) becomes
qgi
¼ Ci( pp2
i À p2
wfi
)ni
, (6:32)
where
Ci ¼ productivity coefficient of lateral i
ni ¼ productivity exponent of lateral i.
As described in Chapter 4, Eq. (6.27), in U.S. field units
(qgi in Mscf/d), can be approximated as (Katz et al., 1959)
p2
wfi
¼ eSi
p2
kfi
þ
6:67 Â 10À4
(eSi
À 1)fMiq2
gizz2
i T
2
i
d5
i cos (45)
, (6:33)
where
Si ¼
0:0375pggRi cos (45
)
2zziT
: (6:34)
The friction factor fMi can be found in the conventional
manner for a given tubing diameter, wall roughness, and
Reynolds number. However, if one assumes fully turbulent
flow, which is the case for most gas wells, then a simple
empirical relation may be used for typical tubing strings
(Katz and Lee, 1990):
fMi ¼
0:01750
d0:224
i
for di # 4:277 in: (6:35)
fMi ¼
0:01603
d0:164
i
for di 4:277 in: (6:36)
Guo (2001) used the following Nikuradse friction factor
correlation for fully turbulent flow in rough pipes:
fMi ¼
1
1:74 À 2 log
2«i
di
2
6
6
4
3
7
7
5
2
(6:37)
For gas wells, Eq. (6.28) can be expressed as (Katz et al.,
1959)
p2
hfi
¼ eSi
p2
hfi
þ
6:67 Â 10À4
(eSi
À 1)fMi
Pi
j¼1
qgi
!2
zz2
i T
2
i
d5
i
, (6:38)
where
Si ¼
0:0375ggHi
zziTi
: (6:39)
L1
L2
L3
Ln
R1
Rn
R3
R2
H1
H2
H3
Hn
Figure 6.6 Schematic of a multilateral well trajectory.
k3 h3 p3
k1 h1 p1
k2 h2 p2
kn hn pn
H3
Hn
H2
qn
qr
H1
q1
q2
q3
R3
R
2
R
n
R1
L3
L2
Ln
L1
Point 2
Point 1
Point 3
Pm2' q1+q2
Pm1' q1
Pm3' q1+q2+q3
pwf3
pwf 2
pwfn
pwfn
pwfn
pwf1
Figure 6.7 Nomenclature of a multilateral well.
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![Gi ¼
(43,560)(40)(78)(0:14)(1 À 0:27)
0:004236
¼ 3:28 Â 109
scf:
Assuming a circular drainage area, the equivalent radius of
the 40 acres is 745 ft. The time required for the pressure
wave to reach the reservoir boundary is estimated as
tpss % 1200
(0:14)(0:0251)(1:5 Â 10À4
)(745)2
0:17
¼ 2,065 hours ¼ 86 days:
The spreadsheet program PseudoPressure.xls gives
m( pi) ¼ m(4613) ¼ 1:27 Â 109
psi2
=cp
m( pwf ) ¼ m(1500) ¼ 1:85 Â 108
psi2
=cp
:
Substituting these and other given parameter values into
Eq. (7.15) yields
q ¼
(0:17)(78)[1:27 Â 109
À 1:85 Â 108
]
1638(180 þ 460) log (2065) þ log 0:17
(0:14)(0:0251)(1:5Â10À4)(0:328)2 À 3:23
¼ 2,092 Mscf=day:
Substituting q ¼ 2,092 Mscf=day into Eq. (7.16) gives
2,092 ¼
(0:17)(78)[m(pp) À 1:85 Â 108
]
1424(180 þ 460) ln 745
0:328 À 3
4 þ 0
À Á ,
which results in m(pp) ¼ 1:19 Â 109
psi2
=cp. The spread-
sheet program PseudoPressure.xls gives pp ¼ 4,409 psia at
the beginning of the pseudo–steady-state flow period.
If the flowing bottom-hole pressure is maintained at a level
of 1,500 psia during the pseudo–steady-state flowperiod (after
86 days of transient production), Eq. (7.16) is simplified as
q ¼
(0:17)(78)[m(pp) À 1:85 Â 108
]
1424(180 þ 460) ln 745
0:328 À 3
4 þ 0
À Á
or
q ¼ 2:09 Â 10À6
½m(pp) À 1:85 Â 108
Š,
which, combined with Eq. (7.17), gives the production fore-
cast shown in Table 7.6, where z-factors and real gas pseudo-
pressures were obtained using spreadsheet programs Hall-
Yarborogh-z.xls and PseudoPressure.xls, respectively. The
production forecast result is also plotted in Fig. 7.6.
Summary
This chapter illustrated how to perform production fore-
cast using the principle of Nodal analysis and material
balance. Accuracy of the forecast strongly depends on
the quality of fluid property data, especially for the two-
phase flow period. It is always recommended to use fluid
properties derived from PVT lab measurements in produc-
tion forecast calculations.
References
craft, b.c. and hawkins, m. Applied Petroleum Reservoir
Engineering, 2nd edition. Englewood Cliffs, NJ: Pren-
tice Hall, 1991.
Table 7.6 Result of Production Forecast for Example Problem 7.4
Reservoir
pressure (psia) z
Pseudo-
pressure
(108
psi2
=cp) Gp (MMscf) DGp (MMscf) q (Mscf/d) Dt (day) t (day)
4,409 1.074 11.90 130
4,200 1.067 11.14 260 130 1,942 67 67
4,000 1.060 10.28 385 125 1,762 71 138
3,800 1.054 9.50 514 129 1,598 81 218
3,600 1.048 8.73 645 131 1,437 91 309
3,400 1.042 7.96 777 132 1,277 103 413
3,200 1.037 7.20 913 136 1,118 122 534
3,000 1.032 6.47 1,050 137 966 142 676
2,800 1.027 5.75 1,188 139 815 170 846
2,600 1.022 5.06 1,328 140 671 209 1,055
2,400 1.018 4.39 1,471 143 531 269 1,324
2,200 1.014 3.76 1,615 144 399 361 1,686
2,000 1.011 3.16 1,762 147 274 536 2,222
0
500
1,000
1,500
2,000
2,500
Pseudosteady Production Time (days)
CumulativeProduction(MMscf)
ProductionRate(Mscf/day)
0
200
400
600
800
1,000
1,200
1,400
1,600
1,800
2,000
q (Mscf/d)
Gp (MMscf)
0 500 1,000 1,500 2,000 2,500
Figure 7.6 Result of production forecast for Example Problem 7.4.
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![decline data follow a harmonic decline model. If the plot of
Np versus log(q) shows a straight line (Fig. 8.4), according
to Eq. (8.34), the harmonic decline model should be used.
If no straight line is seen in these plots, the hyperbolic
decline model may be verified by plotting the relative
decline rate defined by Eq. (8.1). Figure 8.5 shows such a
plot. This work can be easily performed with computer
program UcomS.exe.
8.6 Determination of Model Parameters
Once a decline model is identified, the model parameters a
and b can be determined by fitting the data to the selected
model. For the exponential decline model, the b value can
be estimated on the basis of the slope of the straight line in
the plot of log(q) versus t (Eq. [8.23]). The b value can also
be determined based on the slope of the straight line in the
plot of q versus Np (Eq. [8.27]).
For the harmonic decline model, the b value can be
estimated on the basis of the slope of the straight line in
the plot of log(q) versus log(t) or Eq. (8.32):
b ¼
q0
q1
À 1
t1
: (8:40)
The b value can also be estimated based on the slope of the
straight line in the plot of Np versus log(q) (Eq. [8.34]).
For the hyperbolic decline model, determination of a
and b values is somewhat tedious. The procedure is shown
in Fig. 8.6.
Computer program UcomS.exe can be used for both
model identification and model parameter determination,
as well as production rate prediction.
8.7 Illustrative Examples
Example Problem 8.2 For the data given in Table 8.1,
identify a suitable decline model, determine model
parameters, and project production rate until a marginal
rate of 25 stb/day is reached.
Solution A plot of log(q) versus t is presented in Fig. 8.7,
which shows a straight line. According to Eq. (8.20), the
exponential decline model is applicable. This is further
evidenced by the relative decline rate shown in Fig. 8.8.
Select points on the trend line:
q
Np
Figure 8.4 A plot of Np versus log(q) indicating a har-
monic decline.
q
q∆t
∆q
−
Exponential decline
Harmonic decline
Hyperbolic decline
Figure 8.5 A plot of relative decline rate versus produc-
tion rate.
1. Select points (t1, q1)
and (t2, q2)
2. Read t3 at q3
3. Calculate
4. Find q0 at t = 0
5. Pick up any point (t*, q*)
6. Use
7. Finally
q
t
1
2
q1q2=
t
2
3− t1t2
t1 + t2 − 2t3
a
b
=
a
t *
a
b
q0
q * =
1+ 1+log
log
t *
a
b
q *
q0
a =
a
a
b
b =
q3
t3
(t*, q* )
Figure 8.6 Procedure for determining a- and b-values.
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 101 20.12.2006 10:36am
PRODUCTION DECLINE ANALYSIS 8/101](https://image.slidesharecdn.com/lvzgyxtjtgu380ahut9b-signature-b2ba6e1375509d785bab16b3cad4b5b4b504aa1845b2191a80963142ee8b462c-poli-140812083436-phpapp01/85/Petroleum-production-engineering-109-320.jpg)
![b ¼
EDo
2Rc
, (9:20)
where
sb ¼ bending stress, psi
E ¼ Young’s modulus, psi
Rc ¼ radius of hole curvature, in.
Do ¼ OD of tubing, in.
Because of the great variations in well operating condi-
tions, it is difficult to adopt a universal tubing design
criterion for all well situations. Probably the best design
practice is to consider the worst loading cases for collapse,
burst, and tension loads that are possible for the well to
experience during the life of the well. It is vitally important
to check the remaining strengths of tubing in a subject well
before any unexpected well treatment is carried out. Some
special considerations in well operations that affect tubing
string integrity are addressed in the sections that follow.
9.3.2 Buckling Prevention during Production
A completion fluid is in place in the annular space between
the tubing and the casing before a well is put into
production. The temperature at depth is T ¼ Tsf þ GT D,
where GT is geothermal gradient. When the oil is
produced, the temperature in the tubing will rise. This
will expand (thermal) the tubing length, and if there is
not sufficient landing tension, the tubing will buckle. The
temperature distribution in the tubing can be predicted on
the basis of the work of Ramey (1962), Hasan and Kabir
(2002), and Guo et al. (2005). The latter is described in
Chapter 11. A conservative approach to temperature
calculations is to assume the maximum possible tempera-
ture in the tubing string with no heat loss to formation
through annulus.
Example Problem 9.2 Consider a 27
⁄8 in. API, 6.40 lb/ft
Grade P-105 non-upset tubing anchored with a packer set
at 10,000 ft. The crude oil production through the tubing
from the bottom of the hole is 1,000 stb/day (no gas or
water production). A completion fluid is in place in the
annular space between the tubing and the casing (9.8 lb/
gal KCl water). Assuming surface temperature is 60 8F and
geothermal gradient of 0.01 8F/ft, determine the landing
tension to avoid buckling.
Solution The temperature of the fluid at the bottom of
the hole is estimated to be
T10,000 ¼ 60 þ 0:01(10,000) ¼ 160
F:
The average temperature of the tubing before oil produc-
tion is
Tav1 ¼
60 þ 160
2
¼ 110
F:
The maximum possible average temperature of the tubing
after oil production has started is
Tav2 ¼
160 þ 160
2
¼ 160
F:
This means that the approximate thermal expansion of the
tubing in length will be
DLT % b DTavg
À Á
L,
where b is the coefficient of thermal expansion (for steel,
this is bs ¼ 0:0000065 per 8F). Thus,
DLT % 0:0000065[160 À 110]10,000 ¼ 3:25 ft:
To counter the above thermal expansion, a landing tension
must be placed on the tubing string that is equivalent to the
above. Assumingthe tubingisasimpleuniaxialelement,then
A % pt(D À t) ¼ p(0:217)(2:875 À 0:217) ¼ 1:812 in:2
s ¼ E«
F
A
¼ E Á
DL
L
F ¼
AE DL
L
¼
(1:812)(30 Â 106
)(3:25)
10,000
¼ 17,667 lbf :
Thus,anadditionaltensionof17,667 lbf atthesurfacemustbe
placed on the tubing string to counter the thermal expansion.
It can be shown that turbulent flow will transfer
heat efficiently to the steel wall and then to the completion
fluid and then to the casing and out to the formation. While
laminar flow will not transfer heat very efficiently to the
steel then out to the formation. Thus, the laminar flow
situations are the most likely to have higher temperature
oil at the exit. Therefore, it is most likely the tubing will be
hotter via simple conduction. This effect has been consid-
ered in the work of Hasan and Kabir (2002). Obviously, in
the case of laminar flow, landing tension beyond the buoy-
ancy weight of the tubing may not be required, but in the
case of turbulent flow, the landing tension beyond the
buoyancy weight of the tubing is usually required to prevent
buckling of tubing string. In general, it is good practice to
calculate the buoyant force of the tubing and add approxi-
mately 4,000---5,000 lbf of additional tension when landing.
9.3.3 Considerations for Well Treatment and Stimulation
Tubing strings are designed to withstand the harsh
conditions during wellbore treatment and stimulation
operations such as hole cleaning, cement squeezing, gravel
packing, frac-packing, acidizing, and hydraulic fracturing.
Precautionary measures to take depend on tubing–packer
relation. If the tubing string is set through a non-restrain-
ing packer, the tubing is free to move. Then string buckling
and tubing–packer integrity will be major concerns. If the
tubing string is set on a restraining packer, the string is not
free to move and it will apply force to the packer.
The factors to be considered in tubing design include the
following:
. Tubing size, weight, and grade
. Well conditions
- Pressure effect
- Temperature effect
. Completion method
- Cased hole
- Open hole
- Multitubing
- Packer type (restraining, non-restraining)
9.3.3.1 Temperature Effect
As discussed in Example Problem 9.2, if the tubing string
is free to move, its thermal expansion is expressed as
DLT ¼ bLDTavg: (9:21)
If the tubing string is not free to move, its thermal expan-
sion will generate force. Since Hook’s Law gives
DLT ¼
LDF
AE
, (9:22)
substitution of Eq. (9.22) into Eq. (9.21) yields
DF ¼ AEbDTavg % 207ADTavg (9:23)
for steel tubing.
9.3.3.2 Pressure Effect
Pressures affect tubing string in different ways inclu-
ding piston effect, ballooning effect, and buckling effect.
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![Consider the tubing–pack relation shown in Fig. 9.3. The
total upward force acting on the tubing string from internal
and external pressures is expressed as
Fup ¼ pi(Ap À A
0
i) þ po(Ao À A
0
o), (9:24)
where
pi ¼ pressure in the tubing, psi
po ¼ pressure in the annulus, psi
Ap ¼ inner area of packer, in:2
A
0
i ¼ inner area of tubing sleeve, in:2
A
0
o ¼ outer area of tubing sleeve, in:2
The total downward force acting on the tubing string is
expressed as
Fdown ¼ pi(Ai À A
0
i) þ po(Ap À A
0
o), (9:25)
where Ai is the inner area of tubing. The net upward force
is then
F ¼ Fup À Fdown ¼ pi(Ap À Ai) À po(Ap À Ao): (9:26)
During a well treatment operation, the change (increase) in
the net upward force is expressed as
DF ¼ [Dpi(Ap À Ai) À Dpo(Ap À Ao)]: (9:27)
If the tubing string is anchored to a restraining packer, this
force will be transmitted to the packer, which may cause
packer failure. If the tubing string is free to move, this
force will cause the tubing string to shorten by
DLP ¼
LDF
AE
, (9:28)
which represents tubing string shrinkage due to piston
effect.
As shown in Fig. 9.4a, the ballooning effect is due to the
internal pressure being higher than the external pressure
during a well treatment. The change in tensile force can be
expressed as
DFB ¼ 0:6[Dpi avgAi À Dpo avgAo]: (9:29)
If the tubing string is set through a restraining packer, this
force will be transmitted to the packer, which may cause
packer failure. If the tubing string is free to move, this
force will cause the tubing string to shorten by
DLB ¼
2L
108
Dpi avg À R2
D po avg
R2 À 1
!
, (9:30)
where
Dpi avg ¼ the average pressure change in the tubing,
psi
Dpo avg ¼ the average pressure change in the annulus,
psi
R2
¼ Ao=Ai:
As illustrated in Fig. 9.4b, the buckling effect is caused by
the internal pressure being higher than the external pres-
sure during a well treatment. The tubing string buckles
when FBK ¼ Ap(pi À po) 0. If the tubing end is set
through a restraining packer, this force will be transmitted
to the packer, which may cause packer failure. If the
tubing string is not restrained at bottom, this force will
cause the tubing string to shorten by
DLBK ¼
r2
F2
BK
8EIW
, (9:31)
which holds true only if FBK is greater than 0, and
r ¼
Dci À Di
2
I ¼
p
64
(D4
o À D4
i )
W ¼ Wair þ Wfi À Wfo,
where
Dci ¼ inner diameter of casing, in.
Di ¼ inner diameter of tubing, in.
Do ¼ outer diameter of tubing, in.
Wair ¼ weight of tubing in air, lb/ft
Wft ¼ weight of fluid inside tubing, lb/ft
Wfo ¼ weight of fluid displaced by tubing, lb/ft.
9.3.3.3 Total Effect of Temperature and Pressure
The combination of Eqs. (9.22), (9.28), (9.30), and (9.31)
gives
DL ¼ DLT þ DLP þ DLB þ DLBK , (9:32)
which represents the tubing shortening with a non-
restraining packer. If a restraining packer is used, the
total tubing force acting on the packer is expressed as
Ac
Ap
AoЈ
AiЈ
Ai
Ao
pi po
Figure 9.3 Tubing–packer relation.
(a) (b)
Figure 9.4 Ballooning and buckling effects.
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 113 21.12.2006 2:16pm
WELL TUBING 9/113](https://image.slidesharecdn.com/lvzgyxtjtgu380ahut9b-signature-b2ba6e1375509d785bab16b3cad4b5b4b504aa1845b2191a80963142ee8b462c-poli-140812083436-phpapp01/85/Petroleum-production-engineering-119-320.jpg)
![DF ¼
AEDL
L
: (9:33)
Example Problem 9.3 The following data are given for a
cement squeezing job:
Tubing: 27
⁄8 in., 6.5 lb/ft (2.441-in. ID)
Casing: 7 in., 32 lb/ft (6.094-in. ID)
Packer: Bore size Dp ¼ 3.25 in., set at 10,000 ft
Initial condition: Tubing and casing are full of
30 API oil (S.G. ¼ 0.88)
Operation: Tubing is displaced with 15 ppg
cement with an injection pressure
5,000 psi and casing pressure
1,000 psi. The average temperature
drop is 20 8 F.
1. Calculate tubing movement if the tubing is not
restrained by the packer, and discuss solutions to the
possible operation problems.
2. Calculate the tubing force acting on a restraining packer.
Solution
Temperature Effect:
DlT ¼ bLDTavg ¼ (6:9 Â 10À6
)(10,000)(20) ¼ 1:38 ft
Piston Effect:
DPi ¼ (0:052)(10,000)[15 À (0:88)(8:33)] þ 5,000
¼ 8,988 psi
Dpo ¼ 0 psi
Ap ¼ 3:14(3:25)2
=4 ¼ 8:30 in:2
Ai ¼ 3:14(2:441)2
=4 ¼ 4:68 in:2
Ao ¼ 3:14(2:875)2
=4 ¼ 6:49 in:2
DF ¼ [Dpi(Ap À Ai) À Dpo(Ap À Ao)]
¼ [(8,988)(8:30 À 4:68) À (1,000)(8:30 À 6:49)]
¼ 30:727 lbf
DLP ¼
LDF
AE
¼
(10,000)(30,727)
(6:49 À 4:68)(30,000,000)
¼ 5:65 ft
Ballooning Effect:
DPi,avg ¼ (10,000=2)(0:052)[15 À (0:88)(8:33)] þ 5,000
¼ 6,994 psi
DPo,avg ¼ 1,000 psi
R2
¼ 6:49=4:68 ¼ 1:387
DLB ¼
2L
108
Dpi avg À R2
Dpo avg
R2 À 1
!
¼
2(10,000)
108
6,994 À 1:387(1,000)
1:387 À 1
!
¼ 2:898 ft
Since the tubing internal pressure is higher than the exter-
nal pressure during the cement squeezing, tubing string
buckling should occur.
pi ¼ 5,000 þ (0:052)(15)(10,000) ¼ 12,800 psi
po ¼ 1,000 þ (0:88)(0:433)(10,000) ¼ 4,810 psi
r ¼ (6:094 À 2:875)=2 ¼ 1:6095 in:
FBK ¼ Ap(pi À po) ¼ (8:30)(12,800 À 4,800) ¼ 66,317 lbf
I ¼
p
64
(2:875)4
À (2:441)4
À Á
¼ 1:61 in:4
Wair ¼ 6:5 lbf =ft
Wfi ¼ (15)(7:48)(4:68=144) ¼ 3:65 lbf =ft
Wfo ¼ (0:88)(62:4)(6:49=144) ¼ 2:48 lbf =ft
W ¼ 6:5 þ 3:65 À 2:48 ¼ 7:67 lbf =ft
DLBK ¼
r2
F2
BK
8EIW
¼
(1:6095)2
(66,317)2
(8)(30,000,000)(1:61)(7:67)
¼ 3:884 ft
1. Tubing is not restrained by the packer. The tubing
shortening is
DL ¼ DLT þ DLP þ DLB þ DLBK
¼ 1:38 þ 5:65 þ 2:898 þ 3:844 ¼ 13:77 ft:
Buckling point from bottom:
LBK ¼
FBK
W
¼
66,317
7:67
¼ 8,646 ft
To keep the tubing in the packer, one of the following
measures needs to be taken:
a. Use a sleeve longer than 13.77 ft
b. Use a restraining packer
c. Put some weight on the packer (slack-hook) before
treatment. Buckling due to slacking off needs to be
checked.
2. Tubing is restrained by the packer. The force acting on
the packer is
DF ¼
AE DL
L
¼
(6:49 À 4:68)(30,000,000)(13:77)
(10,000)
¼ 74,783 lbf :
Summary
This chapter presents strength of API tubing that can be
used for designing tubing strings for oil and gas wells.
Tubing design should consider operating conditions in
individual wells. Special care should be taken for tubing
strings before a well undergoes a treatment or stimulation.
References
guo, b., song, s., chacko, j., and ghalambor, a. Offshore
Pipelines. Burlington: Gulf Professional Publishing,
2005.
hasan, r. and kabir, c.s. Fluid Flow and Heat Transfer in
Wellbores, pp. 79–89. Richardson, TX: SPE, 2002.
ramey, h.j., jr. Wellbore heat transmission. Trans. AIME
April 1962;14:427.
Problems
9.1 Calculate the collapse resistance for a section of 3-in.
API 9.20 lb/ft, Grade J-55, non-upset tubing near the
surface of a 12,000-ft string suspended from the sur-
face in a well that is producing gas.
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 114 21.12.2006 2:16pm
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![where
P ¼ pressure, lb=ft2
L ¼ stroke length, ft
D ¼ piston diameter, ft.
Thus, for a triplex pump, the theoretical power is
Power ¼ 3P
pD2
4
LN
ft À lb
min
, (11:1)
where N is pumping speed in strokes per minute. The
theoretical horsepower is
HPth ¼
3P pD2
4
550(60)
LN (hp) (11:2)
or
HPth ¼
3P pD2
4
33,000
LN (hp): (11:3)
The input horsepower needed from the prime mover is
HPi ¼
3P pD2
4
33,000em
LN (hp), (11:4)
where em is the mechanical efficiency of the mechanical
system transferring power from the prime mover to
the fluid in the pump. Usually em is taken to be about
0.85.
The theoretical volume output from a triplex pump per
revolution is
Qth ¼ 3
pD2
4
LN
60
ft3
=sec
À Á
: (11:5)
The theoretical output in bbl/day is thus
qth ¼ 604LND2 bbl
day
: (11:6)
If we use inches (i.e., d [in.] and l [in.]), for D and L, then
qth ¼ 0:35lNd2 bbl
day
: (11:7)
The real output of the pump is dependent on how effici-
ently the pump can fill the chambers of the pistons. Using
the volumetric efficiency ev in Eq. 11.7 gives
qr ¼ 0:35evlNd2 bbl
day
(11:8)
or
qr ¼ 0:01evd2
lN (gal=min), (11:9)
where ev is usually taken to be 0.88–0.98.
As the above volumetric equation can be written in d
and l, then the horsepower equation can be written in d, l,
and p (psi). Thus,
HPi ¼
3p pd2
4
l
12 N
33,000em
(11:10)
reduces to
HPi ¼
pd2
lN
168,067em
: (11:11)
11.2.2 Duplex Pumps
The work per stroke cycle is expressed as
W1 ¼ P
pD2
1
4
L þ P
pD2
1
4
À
pD2
2
4
!
L(ft À lbs): (11:12)
The work per one rotation of crank is
W2 ¼ P
pD2
1
4
L þ P
pD2
1
4
À
pD2
2
4
!
L
'
(1)
ft À lbs
rotation
:
(11:13)
Thus, for a duplex pump, the theoretical power is
Power ¼ 2
 P
pD2
1
4
L þ P
pD2
1
4
À
pD2
2
4
!
L
'
N
ft À lbs
min
:
(11:14)
The theoretical horsepower is
HPth ¼
2 P
pD2
1
4
L þ P
pD2
1
4 À
pD2
2
4
h i
L
n o
N
550(60)
(hp)
or
HPth ¼
2 P p
4 D2
1
À Á
L þ P
pD2
1
4 À
pD2
2
4
h i
L
n o
N
33,000
: (11:15)
The input horsepower needed from the prime mover is
HPi ¼
2 P p
4 D2
1
À Á
L þ P
pD2
1
4 À
pD2
2
4
h i
L
n o
N
33,000em
(hp): (11:16)
The theoretical volume output from the double-acting
duplex pump per revolution is
Qth ¼ 2
pD2
1
4
L þ
pD2
1
4
À
pD2
2
4
!
L
'
N
60
ft3
=sec
À Á
: (11:17)
The theoretical output in gals/min is thus
qth ¼ 2
pD2
1
4
L þ
pD2
1
4
À
pD2
2
4
!
L
'
Â
N
0:1337
(gal= min): (11:18)
If we use inches (i.e., d [in.] and l [in.]), for D and L,
then
qth ¼ 2
pd2
1
4
l þ
pd2
1
4
À
pd2
2
4
!
l
'
N
231
(gal= min): (11:19)
The real output of the pump is
qr ¼ 2
pd2
1
4
l þ
pd2
1
4
À
pd2
2
4
!
l
'
N
231
ev(gal= min)
or
qr ¼ 0:0068 2d2
1 À d2
2
À Á
lNev (gal= min), (11:20)
that is,
qr ¼ 0:233 2d2
1 À d2
2
À Á
lNev (bbl=day): (11:21)
As in the volumetric output, the horsepower equation can
also be reduced to a form with p, d1, d2, and l
HPi ¼
p 2d2
1 À d2
2
À Á
lN
252,101em
: (11:22)
Returning to Eq. (11.16) for the duplex double-action
pump, let us derive a simplified pump equation. Rewriting
Eq. (11.16), we have
HPi ¼
2 P p
4 D2
1
À Á
L þ P
pD2
1
4 À
pD2
2
4
h i
L
n o
N
33,000em
: (11:23)
The flow rate is
Qth ¼ 2
pD2
1
4
L þ
pD2
1
4
À
pD2
2
4
!
L
'
N (ft3
= min ), (11:24)
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 135 3.1.2007 8:54pm Compositor Name: SJoearun
TRANSPORTATION SYSTEMS 11/135](https://image.slidesharecdn.com/lvzgyxtjtgu380ahut9b-signature-b2ba6e1375509d785bab16b3cad4b5b4b504aa1845b2191a80963142ee8b462c-poli-140812083436-phpapp01/85/Petroleum-production-engineering-140-320.jpg)
![so
HPi ¼
PQth
33,000em
: (11:25)
The usual form of this equation is in p (psi) and q (gal/
min):
HPi ¼
p(12)2
 Ã
[q(0:1337)]
33,000em
, (11:26)
that is,
HPi ¼
pq
1714em
: (11:27)
The other form of this equation is in p (psi) and qo (bbl/
day) for oil transportation:
HPi ¼
pqo
58,766em
: (11:28)
Equations (11.27) and (11.28) are valid for any type of
pump.
Example Problem 11.1 A pipeline transporting 5,000 bbl/
day of oil requires a pump with a minimum output
pressure of 1,000 psi. The available suction pressure is
300 psi. Select a triplex pump for this operation.
Solution Assuming a mechanical efficient of 0.85, the
horsepower requirement is
HPi ¼
pqo
58,766em
¼
(1,000)(5,000)
58,766(0:85)
¼ 100 hp:
According to a product sheet of the Oilwell Plunger
Pumps, the Model 336-ST Triplex with forged steel fluid
end has a rated brake horsepower of 160 hp at 320 rpm.
The maximum working pressure is 3,180 psi with the
minimum plunger (piston) size of 13
⁄4 in. It requires a
suction pressure of 275 psi. With 3-in. plungers, the
pump displacement is 0.5508 gal/rpm, and it can deliver
liquid flow rates in the range of 1,889 bbl/day (55.08 gpm)
at 100 rpm to 6,046 bbl/day (176.26 gpm) at 320 rpm,
allowing a maximum pressure of 1,420 psi. This pump
can be selected for the operation. The required operating
rpm is
RPM ¼
(5,000)(42)
(24)(60)(0:5508)
¼ 265 rpm:
11.3 Compressors
When natural gas does not have sufficient potential energy
to flow, a compressor station is needed. Five types of
compressor stations are generally used in the natural gas
production industry:
. Field gas-gathering stations to gather gas from wells in
which pressure is insufficient to produce at a desired rate
of flow into a transmission or distribution system. These
stations generally handle suction pressures from below
atmospheric pressure to 750 psig and volumes from a
few thousand to many million cubic feet per day.
. Relay or main-line stations to boost pressure in trans-
mission lines compress generally large volumes of gas at
a pressure range between 200 and 1,300 psig.
. Re-pressuring or recycling stations to provide gas pres-
sures as high as 6,000 psig for processing or secondary
oil recovery projects.
. Storage field stations to compress trunk line gas for
injection into storage wells at pressures up to 4,000 psig.
. Distribution plant stations to pump gas from holder
supply to medium- or high-pressure distribution lines
at about 20–100 psig, or pump into bottle storage up
to 2,500 psig.
11.3.1 Types of Compressors
The compressors used in today’s natural gas production
industry fall into two distinct types: reciprocating and
rotary compressors. Reciprocating compressors are most
commonly used in the natural gas industry. They are built
for practically all pressures and volumetric capacities. As
shown in Fig. 11.3, reciprocating compressors have more
moving parts and, therefore, lower mechanical efficiencies
than rotary compressors. Each cylinder assembly of a
reciprocation compressor consists of a piston, cylinder,
cylinder heads, suction and discharge valves, and other
parts necessary to convert rotary motion to reciprocation
motion. A reciprocating compressor is designed for a cer-
tain range of compression ratios through the selection of
proper piston displacement and clearance volume within
the cylinder. This clearance volume can be either fixed or
variable, depending on the extent of the operation range
and the percent of load variation desired. A typical recip-
rocating compressor can deliver a volumetric gas flow rate
up to 30,000 cubic feet per minute (cfm) at a discharge
pressure up to 10,000 psig.
Figure 11.3 Elements of a typical reciprocating compressor (courtesy of Petroleum Extension Services).
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![11.3.2 Reciprocating Compressors
Figure 11.5 shows a diagram volume relation during gas
compression. The shaft work put into the gas is expressed as
Ws ¼
V2
2
2g
À
V2
1
2g
þ P2v2 À
ð2
1
Pdv À P1v1
0
@
1
A, (11:29)
where
Ws ¼ mechanical shaft work into the system,
ft-lbs per lb of fluid
V1 ¼ inlet velocity of fluid to be compressed, ft/sec
V2 ¼ outlet velocity of compressed fluid, ft/sec
P1 ¼ inlet pressure, lb=ft2
abs
P2 ¼ outlet pressure, lb=ft2
abs
n1 ¼ specific volume at inlet, ft3
=lb
n2 ¼ specific volume at outlet, ft3
=lb.
Note that the mechanical kinetic energy term V2
2g is in
ft Á lb
lb
À Á
to get ft-lbs per lb.
Rewriting Eq. (11.29), we can get
Ws þ
V2
1
2g
À
V2
2
2g
þ P1v1 À P2v2 ¼ À
ð2
1
Pdv: (11:30)
An isentropic process is usually assumed for reciprocating
compression, that is, P1nk
1 ¼ P2nk
2 ¼ Pnk
¼ constant,
where k ¼
cp
cv
. Because P ¼
P1vk
1
vk , the right-hand side of Eq.
(11.30) is formulated as
À
ð2
1
Pdv ¼ À
ð2
1
P1vk
1
vk
dv ¼ ÀP1vk
1
ð2
1
dv
vk
¼ ÀP1vk
1
v1Àk
1 À k
!2
1
¼ À
P1vk
1
1 À k
[v1Àk
2 À v1Àk
1 ]
¼
P1vk
1
1 À k
v1Àk
1
v1Àk
1
[v1Àk
2 À v1Àk
1 ]
¼
P1v1
1 À k
v1Àk
2
v1Àk
1
À 1
!
¼
P1v1
1 À k
v1
v2
kÀ1
À1
#
: (11:31)
Using the ideal gas law
P
g
¼ RT, (11:32)
where g (lb=ft3
) is the specific weight of the gas and T (8R)
is the temperature and R ¼ 53:36 (lb-ft/lb-8R) is the gas
constant, and v ¼ 1
g, we can write Eq. (11.32) as
Pv ¼ RT (11:33)
or
P1v1 ¼ RT1: (11:34)
Using P1nk
1 ¼ P2nk
2 ¼ Pnk
¼ constant, which gives
v1
v2
k
¼
P2
P1
or
v1
v2
¼
P2
P1
1
k
: (11:35)
Substituting Eqs. (11.35) and (11.34) into Eq. (11.31) gives
À
ð2
1
Pdv ¼
RT1
k À 1
P2
P1
kÀ1
k
À1
#
: (11:36)
We multiply Eq. (11.33) by nkÀ1
, which gives
Pv vkÀ1
À Á
¼ RT vkÀ1
À Á
Pvk
¼ RTvkÀ1
¼ C1
Pvk
R
¼ TvkÀ1
¼
C1
R
¼ C
0
1
Thus,
TvkÀ1
¼ C
0
1: (11:37)
Also we can rise Pvk
¼ constant to the kÀ1
k
À Á
power. This is
(Pvk
)
kÀ1
k ¼ C1
0 kÀ1
k
P
kÀ1
k vkÀ1
¼ C
0kÀ1
k
1
or
nkÀ1
¼
C
0kÀ1
k
1
P
kÀ1
k
¼
C
00
1
P
kÀ1
k
: (11:38)
Substituting Eq. (3.38) into (3.37) gives
T
C
00
1
P
kÀ1
k
¼ C
0
1
or
T
P
kÀ1
k
¼
C
0
1
C
00
1
¼ C
000
1 ¼ constant: (11:39)
Thus, Eq. (11.39) can be written as
T1
P
kÀ1
k
1
¼
T2
P
kÀ1
k
2
: (11:40)
Thus, Eq. (11.40) is written
P2
P1
kÀ1
k
¼
T2
T1
: (11:41)
Substituting Eq. (11.41) into (11.36) gives
À
ð2
1
Pdv ¼
RT1
k À 1
T2
T1
À 1
!
À
ð2
1
Pdv ¼
R
k À 1
(T2 À T1):
(11:42)
Therefore, our original expression, Eq. (11.30), can be
written as
Ws þ
V2
1
2g
À
V2
2
2g
þ P1v1 À P2v2 ¼
R
k À 1
(T2 À T1)
orVolume
Pressure
m no
1
2
d c
b
a
Figure 11.5 Basic pressure–volume diagram.
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![Oil-specific gravity:
go ¼
141:5
131:5 þ 35
¼ 0:85
Reynolds number:
NRe ¼ 1,488
6
12
À Á
(1:66)(0:85)(62:4)
5
¼ 13,101
2,100 turbulent flow
Equation (11.89) gives
1
ffiffiffiffiffiffi
fM
p ¼ 1:14 À 2 log
0:0006
6
þ
21:25
(13,101)0:9
¼ 5:8759,
which gives
fM ¼ 0:02896:
Equation (11.93) gives
p1 ¼ 50 þ 0:433(0:85)(5)(5,280) sin (15
) þ 1:15 Â 10À5
Â
(0:02896)(0:85)(5,000)2
(5)(5,280)
(6)5
¼ 2,590 psi:
11.4.1.2 Gas Flow
Consider steady-state flow of dry gas in a constant-diam-
eter, horizontal pipeline. The mechanical energy equation,
Eq. (11.78), becomes
dp
dL
¼
fMru2
2gcD
¼
p(MW)a
zRT
fu2
2gcD
, (11:94)
which serves as a base for development of many pipeline
equations. The difference in these equations originated
from the methods used in handling the z-factor and fric-
tion factor. Integrating Eq. (11.94) gives
ð
dp ¼
(MW)a fMu2
2RgcD
ð
p
zT
dL: (11:95)
If temperature is assumed constant at average value in a
pipeline, T¯ , and gas deviation factor, z¯, is evaluated at
average temperature and average pressure, p¯, Eq. (11.95)
can be evaluated over a distance L between upstream
pressure, p1, and downstream pressure, p2:
p2
1 À p2
2 ¼
25ggQ2 TTzzfML
d5
, (11:96)
where
gg ¼ gas gravity (air ¼ 1)
Q ¼ gas flow rate, MMscfd (at 14.7 psia, 60 8F)
T¯ ¼ average temperature, 8R
z¯ ¼ gas deviation factor at T¯ and p¯
p¯ ¼ (p1 þ p2)/2
L ¼ pipe length, ft
d ¼ pipe internal diameter, in.
F ¼ Moody friction factor
Equation (11.96) may be written in terms of flow rate
measured at arbitrary base conditions (Tb and pb):
q ¼
CTb
pb
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
( p2
1 À p2
2)d5
gg
TTzzfML
s
, (11:97)
where C is a constant with a numerical value that depends
on the units used in the pipeline equation. If L is in miles
and q is in scfd, C ¼ 77:54.
The use of Eq. (11.97) involves an iterative procedure.
The gas deviation factor depends on pressure and the
friction factor depends on flow rate. This problem
prompted several investigators to develop pipeline flow
equations that are noniterative or explicit. This has in-
volved substitutions for the friction factor fM. The specific
substitution used may be diameter-dependent only
(Weymouth equation) or Reynolds number–dependent
only (Panhandle equations).
11.4.1.2.1 Weymouth Equation for Horizontal Flow
Equation (11.97) takes the following form when the unit of
scfh for gas flow rate is used:
qh ¼
3:23Tb
pb
ffiffiffiffiffiffi
1
fM
s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
( p2
1 À p2
2)d5
gg
TTzzL
s
, (11:98)
where
ffiffiffiffi
1
fM
q
is called the ‘‘transmission factor.’’ The friction
factor may be a function of flow rate and pipe roughness.
If flow conditions are in the fully turbulent region, Eq.
(11.89) degenerates to
fM ¼
1
[1:14 À 2 log (eD)]2
, (11:99)
where fM depends only on the relative roughness, eD.
When flow conditions are not completely turbulent, fM
depends on the Reynolds number also.
Therefore, use of Eq. (11.98) requires a trial-and-error
procedure to calculate qh. To eliminate the trial-and-error
procedure, Weymouth proposed that f vary as a function
of diameter as follows:
fM ¼
0:032
d1=3
(11:100)
With this simplification, Eq. (11.98) reduces to
qh ¼
18:062Tb
pb
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
( p2
1 À p2
2)D16=3
gg
TTzzL
s
, (11:101)
which is the form of the Weymouth equation commonly
used in the natural gas industry.
The use of the Weymouth equation for an existing
transmission line or for the design of a new transmission
line involves a few assumptions including no mechanical
work, steady flow, isothermal flow, constant compressibil-
ity factor, horizontal flow, and no kinetic energy change.
These assumptions can affect accuracy of calculation
results.
In the study of an existing pipeline, the pressure-mea-
suring stations should be placed so that no mechanical
energy is added to the system between stations. No me-
chanical work is done on the fluid between the points at
which the pressures are measured. Thus, the condition of
no mechanical work can be fulfilled.
Steady flow in pipeline operation seldom, if ever, exists
in actual practice because pulsations, liquid in the pipeline,
and variations in input or output gas volumes cause devi-
ations from steady-state conditions. Deviations from
steady-state flow are the major cause of difficulties experi-
enced in pipeline flow studies.
The heat of compression is usually dissipated into the
ground along a pipeline within a few miles downstream
from the compressor station. Otherwise, the temperature
of the gas is very near that of the containing pipe, and
because pipelines usually are buried, the temperature of
the flowing gas is not influenced appreciably by rapid
changes in atmospheric temperature. Therefore, the gas
flow can be considered isothermal at an average effective
temperature without causing significant error in long-
pipeline calculations.
The compressibility of the fluid can be considered con-
stant and an average effective gas deviation factor may be
used. When the two pressures p1 and p2 lie in a region
where z is essentially linear with pressure, it is accurate
enough to evaluate z¯ at the average pressure
pp ¼ ( p1 þ p2)=2. One can also use the arithmetic average
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![minimum yield strength. Equation (11.116) is valid for any
consistent units.
Most codes allow credit for external pressure. This credit
should be used whenever possible, although care should be
exercised for oil export lines to account for head of fluid and
for lines that traverse from deep to shallow water.
ASME B31.4 and DnV 1981 define Pi as the maximum
allowable operating pressure (MAOP) under normal condi-
tions, indicating that surge pressures up to 110% MAOP is
acceptable. In some cases, Pi is defined as wellhead shut-in
pressure (WSIP) for flowlines or specified by the operators.
In Eq. (11.116), the weld efficiency factor (Ew) is 1.0 for
seamless, ERW, and DSAW pipes. The temperature de-
rating factor (Ft) is equal to 1.0 for temperatures under
250 8F. The usage factor (h) is defined in Tables 11.2 and
11.3 for oil and gas lines, respectively.
The underthickness due to manufacturing tolerance is
taken into account in the design factor. There is no need to
add any allowance for fabrication to the wall thickness
calculated with Eq. (11.116).
11.4.2.1.3 Design for External Pressure Different
practices can be found in the industry using different
external pressure criteria. As a rule of thumb, or unless
qualified thereafter, it is recommended to use propagation
criterion for pipeline diameters under 16-in. and collapse
criterion for pipeline diameters more than or equal to 16-in.
Propagation Criterion: The propagation criterion is more
conservative and should be used where optimization of the
wall thickness is not required or for pipeline installation
methods not compatible with the use of buckle arrestors
such as reel and tow methods. It is generally economical to
design for propagation pressure for diameters less than
16-in. For greater diameters, the wall thickness penalty is
too high. When a pipeline is designed based on the collapse
criterion, buckle arrestors are recommended. The external
pressure criterion should be based on nominal wall thick-
ness, as the safety factors included below account for wall
variations.
Although a large number of empirical relationships have
been published, the recommended formula is the latest
given by AGA.PRC (AGA, 1990):
PP ¼ 33Sy
tNOM
D
2:46
, (11:117)
m1
n1
2
m
r
n
dr
dr
dr
dsr
+
sh
sh
2
sr
df
b
a
srb = −P
sra = − PO
P
m1
m
n1
n
PO
sr
Figure 11.12 Stresses generated by internal pressure p in a thick-wall pipe, D/t 20.
Table 11.2 Design and Hydrostatic Pressure Definitions and Usage Factors for Oil Lines
Parameter ASME B31.4, 1989 Edition Dnv (Veritas, 1981)
Design internal pressure Pa
d Pi À Pe[401:2:2] Pi À Pe[4.2.2.2]
Usage factor h 0.72 [402.3.1(a)] 0.72 [4.2.2.1]
Hydrotest pressure Ph 1:25 Pb
i [437.4.1(a)] 1:25Pd [8.8.4.3]
a
Credit can be taken for external pressure for gathering lines or flowlines when the MAOP (Pi) is applied at the wellhead
or at the seabed. For export lines, when Pi is applied on a platform deck, the head fluid shall be added to Pi for the
pipeline section on the seabed.
b
If hoop stress exceeds 90% of yield stress based on nominal wall thickness, special care should be taken to prevent
overstrain of the pipe.
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![which is valid for any consistent units. The nominal wall
thickness should be determined such that Pp 1:3 Pe. The
safety factor of 1.3 is recommended to account for uncer-
tainty in the envelope of data points used to derive Eq.
(11.117). It can be rewritten as
tNOM $ D
1:3PP
33Sy
1
2:46
: (11:118)
For the reel barge method, the preferred pipeline grade is
belowX-60.However, X-65 steel can beusedif the ductility is
kept high by selecting the proper steel chemistry and micro-
alloying. For deepwater pipelines, D/t ratios of less than 30
are recommended. It has been noted that bending loads have
no demonstrated influence on the propagation pressure.
Collapse Criterion: The mode of collapse is a function of
D/t ratio, pipeline imperfections, and load conditions. The
theoretical background is not given in this book. An em-
pirical general formulation that applies to all situations is
provided. It corresponds to the transition mode of collapse
under external pressure (Pe), axial tension (Ta), and bend-
ing strain (sb), as detailed elsewhere (Murphey and
Langner, 1985; AGA, 1990).
The nominal wall thickness should be determined such that
1:3PP
PC
þ
«b
«B
# gp, (11:119)
where 1.3 is the recommended safety factor on collapse,
«B is the bending strain of buckling failure due to pure
bending, and g is an imperfection parameter defined
below.
The safety factor on collapse is calculated for D/t ratios
along with the loads (Pe, «b, Ta) and initial pipeline out-of
roundness (do). The equations are
PC ¼
PelP
0
y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
P2
el þ P02
y
q , (11:120)
P
0
y ¼ Py
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 À 0:75
Ta
Ty
2
s
À
Ta
2Ty
2
4
3
5, (11:121)
Pel ¼
2E
1 À n2
t
D
3
, (11:122)
Py ¼ 2Sy
t
D
, (11:123)
Ty ¼ ASy, (11:124)
where gp is based on pipeline imperfections such as initial
out-of roundness (do), eccentricity (usually neglected), and
residual stress (usually neglected). Hence,
gp ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ p2
p2 À 1
f 2
p
,
v
u
u
t (11:125)
where
p ¼
P
0
y
Pel
, (11:126)
fp ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ do
D
t
2
s
À do
D
t
, (11:127)
«B ¼
t
2D
, (11:128)
and
do ¼
Dmax À Dmin
Dmax þ Dmin
: (11:129)
When a pipeline is designed using the collapse criterion, a
good knowledge of the loading conditions is required (Ta
and «b). An upper conservative limit is necessary and must
often be estimated.
Under high bending loads, care should be taken in esti-
mating «b using an appropriate moment-curvature
relationship. A Ramberg Osgood relationship can be used as
KÃ
¼ MÃ
þ AMÃB
, (11:130)
where KÃ
¼ K=Ky and MÃ
¼ M=My with Ky ¼ 2Sy=ED is
the yield curvature and My ¼ 2ISy=D is the yield moment.
The coefficients A and B are calculated from the two data
points on stress–strain curve generated during a tensile
test.
11.4.2.1.4 Corrosion Allowance To account for
corrosion when water is present in a fluid along with
contaminants such as oxygen, hydrogen sulfide (H2S),
and carbon dioxide (CO2), extra wall thickness is added.
A review of standards, rules, and codes of practices (Hill
and Warwick, 1986) shows that wall allowance is only one
of several methods available to prevent corrosion, and it is
often the least recommended.
For H2S and CO2 contaminants, corrosion is often
localized (pitting) and the rate of corrosion allowance
ineffective. Corrosion allowance is made to account for
damage during fabrication, transportation, and storage.
A value of 1
⁄16 in. may be appropriate. A thorough
assessment of the internal corrosion mechanism and rate
is necessary before any corrosion allowance is taken.
11.4.2.1.5 Check for Hydrotest Condition The min-
imum hydrotest pressure for oil and gas lines is given in
Tables 11.2 and 11.3, respectively, and is equal to 1.25
times the design pressure for pipelines. Codes do not
require that the pipeline be designed for hydrotest
conditions but sometimes give a tensile hoop stress limit
90% SMYS, which is always satisfied if credit has not been
taken for external pressure. For cases where the wall
thickness is based on Pd ¼ Pi À Pe, codes recommend
not to overstrain the pipe. Some of the codes are ASME
B31.4 (Clause 437.4.1), ASME B31.8 (no limit on hoop
stress during hydrotest), and DnV (Clause 8.8.4.3).
Table 11.3 Design and Hydrostatic Pressure Definitions and Usage Factors for Gas Lines
Parameter
ASME B31.8, 1989 Edition, 1990
Addendum DnV (Veritas, 1981)
Pa
d Pi À Pe[A842.221] Pi À Pe[4.2.2.2]
Usage factor h 0.72 [A842.221] 0.72 [4.2.2.1]
Hydrotest pressure Ph 1:25 Pb
i [A847.2] 1:25Pd [8.8.4.3]
a
Credit can be taken for external pressure for gathering lines or flowlines when the MAOP (Pi) is applied at wellhead or
at the seabed. For export lines, when Pi is applied on a platform deck, the head of fluid shall be added to Pi for the
pipeline section on the seabed (particularly for two-phase flow).
b
ASME B31.8 imposes Ph ¼ 1:4Pi for offshore risers but allows onshore testing of prefabricated portions.
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![where hi is film coefficient of pipeline inner surface; ho is
film coefficient of pipeline outer surface; Ai is area of pipe-
line inner surface; Ao is area of pipeline outer surface; rm is
radius of layer m; and km is thermal conductivity of layer m.
Similar equations exist for transient-heat flow, giving
an instantaneous rate for heat flow. Typically required
insulation performance, in terms of OHTC (U value) of
steel pipelines in water, is summarized in Table 11.5.
Pipeline insulation comes in two main types: dry insula-
tion and wet insulation. The dry insulations require an
outer barrier to prevent water ingress (PIP). The most
common types of this include the following:
. Closed-cell polyurethane foam
. Open-cell polyurethane foam
. Poly-isocyanurate foam
. Extruded polystyrene
. Fiber glass
. Mineral wool
. Vacuum-insulation panels
Under certain conditions, PIP systems may be considered
over conventional single-pipe systems. PIP insulation may
be required to produce fluids from high-pressure/high-
temperature (150 8C) reservoirs in deepwater (Carmi-
chael et al., 1999). The annulus between pipes can be filled
with different types of insulation materials such as foam,
granular particles, gel, and inert gas or vacuum.
A pipeline-bundled system—a special configuration of
PIP insulation—can be used to group individual flowlines
together to form a bundle (McKelvie, 2000); heat-up lines
can be included in the bundle, if necessary. The complete
bundle may be transported to site and installed with a
considerable cost savings relative to other methods. The
extra steel required for the carrier pipe and spacers can
sometimes be justified (Bai, 2001).
Wet-pipeline insulations are those materials that do not
need an exterior steel barrier to prevent water ingress, or the
water ingress is negligible and does not degrade the insulation
properties. The most common types of this are as follows:
. Polyurethane
. Polypropylene
. Syntactic polyurethane
. Syntactic polypropylene
. Multilayered
The main materials that have been used for deepwater insu-
lations have been polyurethane and polypropylene based.
Syntactic versions use plastic or glass matrix to improve
insulation with greater depth capabilities. Insulation coat-
ings with combinations of the two materials have also been
used. Guo et al. (2005) gives the properties of these wet
insulations. Because the insulation is buoyant, this effect
must be compensated by the steel pipe weight to obtain
lateral stability of the deepwater pipeline on the seabed.
11.4.2.2.2 Heat Transfer Models Heat transfer across
the insulation of pipelines presents a unique problem
affecting flow efficiency. Although sophisticated
computer packages are available for predicting fluid
temperatures, their accuracies suffer from numerical
treatments because long pipe segments have to be used to
save computing time. This is especially true for transient
fluid-flow analyses in which a very large number of
numerical iterations are performed.
Ramey (1962) was among the first investigators who stud-
ied radial-heat transfer across a well casing with no insula-
tion. He derived a mathematical heat-transfer model for an
outer medium that is infinitely large. Miller (1980) analyzed
heat transfer around a geothermal wellbore without ins-
ulation. Winterfeld (1989) and Almehaideb et al. (1989)
considered temperature effect on pressure-transient analyses
in well testing. Stone et al. (1989) developed a numerical
simulator to couple fluid flow and heat flow in a wellbore
and reservoir. More advanced studies on the wellbore heat-
transfer problem were conducted by Hasan and Kabir (1994,
2002), Hasan et al. (1997, 1998), and Kabir et al. (1996).
Although multilayers of materials have been considered in
these studies, the external temperature gradient in the longi-
tudinal direction has not been systematically taken into ac-
count. Traditionally, if the outer temperature changes with
length, the pipe must be divided into segments, with assumed
constant outer temperature in each segment, and numerical
algorithms are required for heat-transfer computation. The
accuracy of the computation depends on the number of
segments used. Fine segments can be employed to ensure
accuracy with computing time sacrificed.
Guo et al. (2006) presented three analytical heat-transfer
solutions. They are the transient-flow solution for startup
mode, steady-flow solution for normal operation mode,
and transient-flow solution for flow rate change mode
(shutting down is a special mode in which the flow rate
changes to zero).
Temperature and Heat Transfer for Steady Fluid Flow.
The internal temperature profile under steady fluid-flow
conditions is expressed as
T ¼
1
a2
b À abL À ag À eÀa(LþC)
 Ã
, (11:133)
where the constant groups are defined as
a ¼
2pRk
vrCpsA
, (11:134)
b ¼ aG cos (u), (11:135)
g ¼ ÀaT0, (11:136)
and
C ¼ À
1
a
ln (b À a2
Ts À ag), (11:137)
where T is temperature inside the pipe, L is longitudinal
distance from the fluid entry point, R is inner radius
of insulation layer, k is the thermal conductivity of the
insulation material, v is the average flow velocity of fluid in
the pipe, r is fluid density, Cp is heat capacity of fluid at
constant pressure, s is thickness of the insulation layer,
A is the inner cross-sectional area of pipe, G is principal
thermal-gradient outside the insulation, u is the angle be-
tween the principal thermal gradient and pipe orientation,
T0 is temperature of outer medium at the fluid entry
location, and Ts is temperature of fluid at the fluid entry
point.
The rate of heat transfer across the insulation layer over
the whole length of the pipeline is expressed as
q ¼ À
2pRk
s
 T0L G cos (u)
2
L2
À
1
a2
(b À ag)L À
ab
2
L2
þ
1
a
eÀa(LþC)
À eÀaC
 Ã
gÞ, (11:138)
where q is the rate of heat transfer (heat loss).
Transient Temperature During Startup. The internal
temperature profile after starting up a fluid flow is
expressed as follows:
T ¼
1
a2
{b À abL À ag À eÀa[Lþf (LÀvt)]
}, (11:139)
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 153 3.1.2007 8:54pm Compositor Name: SJoearun
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![where the function f is given by
f (L À vt) ¼ À(L À vt) À
1
a
ln {b À ab(L À vt)
À ag À a2
[Ts À G cos (u)(L À vt)]} (11:140)
and t is time.
Transient Temperature During Flow Rate Change.
Suppose that after increasing or decreasing the flow rate,
the fluid has a new velocity v’ in the pipe. The internal
temperature profile is expressed as follows:
T ¼
1
a02
{b0
À a0
b0
L À a0
g0
À eÀa0
[Lþf (LÀv0
t)]
}, (11:141)
where
a0
¼
2pRk
v0rCpsA
, (11:142)
b0
¼ a0
G cos (u), (11:143)
g0
¼ Àa0
T0, (11:144)
and the function f is given by
f (L À v0
t) ¼ À(L À v0
t) À
1
a0
ln (b0
À a0
b0
(L
À v0
t) À a0
g0
À
a0
a
2
{b À ab(L
À v0
t) À ag À eÀa[(LÀv0
t)þC]
}Þ: (11:145)
Example Problem 11.7 A design case is shown in this
example. Design base for a pipeline insulation is
presented in Table 11.6. The design criterion is to ensure
that the temperature at any point in the pipeline will not
drop to less than 25 8C, as required by flow assurance.
Insulation materials considered for the project were
polyethylene, polypropylene, and polyurethane.
Solution A polyethylene layer of 0.0254 M (1 in.) was
first considered as the insulation. Figure 11.13 shows the
temperature profiles calculated using Eqs. (11.133) and
(11.139). It indicates that at approximately 40 minutes
after startup, the transient-temperature profile in the
pipeline will approach the steady-flow temperature profile.
The temperature at the end of the pipeline will be slightly
lower than 20 8C under normal operating conditions.
Obviously, this insulation option does not meet design
criterion of 25 8C in the pipeline.
Figure 11.14 presents the steady-flow temperature pro-
files calculated using Eq. (11.133) with polyethylene layers
of four thicknesses. It shows that even a polyethylene layer
0.0635-M (2.5-in.) thick will still not give a pipeline tem-
perature higher than 25 8C; therefore, polyethylene should
not be considered in this project.
A polypropylene layer of 0.0254 M (1 in.) was then
considered as the insulation. Figure 11.15 illustrates the
temperature profiles calculated using Eq. (11.133) and
(11.139). It again indicates that at approximately 40 min-
utes after startup, the transient-temperature profile in
the pipe will approach the steady-flow temperature
profile. The temperature at the end of the pipeline will be
approximately 22.5 8C under normal operating conditions.
Obviously, this insulation option, again, does not meet
design criterion of 25 8C in the pipeline.
Figure 11.16 demonstrates the steady-flow temperature
profiles calculated using Eq. (11.133) with polypropylene
layers of four thicknesses. It shows that a polypropylene
layer of 0.0508 M (2.0 in.) or thicker will give a pipeline
temperature of higher than 25 8C.
A polyurethane layer of 0.0254 M (1 in.) was also
considered as the insulation. Figure 11.17 shows the tem-
perature profiles calculated using Eqs. (11.133) and
(11.139). It indicates that the temperature at the end
of pipeline will drop to slightly lower than 25 8C under
normal operating conditions. Figure 11.18 presents
the steady-flow temperature profiles calculated using
Eq. (11.133) with polyurethane layers of four thicknesses.
It shows that a polyurethane layer of 0.0381 M (1.5 in.)
0
5
10
15
20
25
30
0 2,000 4,000 6,000 8,000 10,000
Distance (M)
Temperature(ЊC)
0 minute
10 minutes
20 minutes
30 minutes
Steady flow
Figure 11.13 Calculated temperature profiles with a polyethylene layer of 0.0254 M (1 in.).
Table 11.6 Base Data for Pipeline Insulation Design
Length of pipeline: 8,047 M
Outer diameter of pipe: 0.2032 M
Wall thickness: 0.00635 M
Fluid density: 881 kg=M3
Fluid specific heat: 2,012 J/kg-8C
Average external temperature: 10 8C
Fluid temperature at entry point: 28 8C
Fluid flow rate: 7,950 M3
=day
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![12.1 Introduction
Sucker rod pumping is also referred to as ‘‘beam pump-
ing.’’ It provides mechanical energy to lift oil from bottom
hole to surface. It is efficient, simple, and easy for field
people to operate. It can pump a well down to very low
pressure to maximize oil production rate. It is applicable to
slim holes, multiple completions, and high-temperature
and viscous oils. The system is also easy to change to
other wells with minimum cost. The major disadvantages
of beam pumping include excessive friction in crooked/
deviated holes, solid-sensitive problems, low efficiency in
gassy wells, limited depth due to rod capacity, and bulky in
offshore operations. Beam pumping trends include
improved pump-off controllers, better gas separation, gas
handling pumps, and optimization using surface and
bottom-hole cards.
12.2 Pumping System
As shown in Fig. 12.1, a sucker rod pumping system
consists of a pumping unit at surface and a plunger
pump submerged in the production liquid in the well.
The prime mover is either an electric motor or an in-
ternal combustion engine. The modern method is to sup-
ply each well with its own motor or engine. Electric motors
are most desirable because they can easily be automated.
The power from the prime mover is transmitted to the
input shaft of a gear reducer by a V-belt drive. The output
shaft of the gear reducer drives the crank arm at a lower
speed ($4–40 revolutions per minute [rpm] depending on
well characteristics and fluid properties). The rotary mo-
tion of the crank arm is converted to an oscillatory motion
by means of the walking beam through a pitman arm. The
horse’s head and the hanger cable arrangement is used to
ensure that the upward pull on the sucker rod string is
vertical at all times (thus, no bending moment is applied to
the stuffing box). The polished rod and stuffing box com-
bine to maintain a good liquid seal at the surface and, thus,
force fluid to flow into the ‘‘T’’ connection just below the
stuffing box.
Conventional pumping units are available in a wide
range of sizes, with stroke lengths varying from 12 to
almost 200 in. The strokes for any pumping unit type are
available in increments (unit size). Within each unit size,
the stroke length can be varied within limits (about six
different lengths being possible). These different lengths
are achieved by varying the position of the pitman arm
connection on the crank arm.
Walking beam ratings are expressed in allowable pol-
ished rod loads (PRLs) and vary from approximately
3,000 to 35,000 lb. Counterbalance for conventional
pumping units is accomplished by placing weights directly
on the beam (in smaller units) or by attaching weights to
the rotating crank arm (or a combination of the two
methods for larger units). In more recent designs, the
rotary counterbalance can be adjusted by shifting the posi-
tion of the weight on the crank by a jackscrew or rack and
pinion mechanism.
There are two other major types of pumping units. These
are the Lufkin Mark II and the Air-Balanced Units
(Fig. 12.2). The pitman arm and horse’s head are in the
same side of the walking beam in these two types of units
(Class III lever system). Instead of using counter-weights in
Lufkin Mark II type units, air cylinders are used in the air-
balanced units to balance the torque on the crankshaft.
Pitman
Counter weight
Gear reducer
V-Belt
Prime
mover
Walking beamHorse head
Bridle
Polished rod Stuffing
box
Sampson
post
Crank
Casing
Tubing
Sucker rod
Downhole pump
Strokelength
Strokelength
Oil
Tee
Gas
Figure 12.1 A diagrammatic drawing of a sucker rod pumping system (Golan and Whitson, 1991).
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![passing through the open SV). The fluid continues to fill
the volume above the SV until the plunger reaches the top
of its stroke.
There are two basic types of plunger pumps: tubing
pump and rod pump (Fig. 12.4). For the tubing pump,
the working barrel or liner (with the SV) is made up (i.e.,
attached) to the bottom of the production tubing string
and must be run into the well with the tubing. The plunger
(with the TV) is run into the well (inside the tubing) on
the sucker rod string. Once the plunger is seated in the
working barrel, pumping can be initiated. A rod pump
(both working barrel and plunger) is run into the well on
the sucker rod string and is seated on a wedged type
seat that is fixed to the bottom joint of the production
tubing. Plunger diameters vary from 5
⁄8 to 45
⁄8 in. Plunger
area varies from 0:307 in:2
to 17:721 in:2
.
12.3 Polished Rod Motion
The theory of polished rod motion has been established
since 1950s (Nind, 1964). Figure 12.5 shows the cyclic
motion of a polished rod in its movements through the
stuffing box of the conventional pumping unit and the air-
balanced pumping unit.
Conventional Pumping Unit. For this type of unit, the
acceleration at the bottom of the stroke is somewhat
greater than true simple harmonic acceleration. At the
top of the stroke, it is less. This is a major drawback for
the conventional unit. Just at the time the TV is closing
and the fluid load is being transferred to the rods, the
acceleration for the rods is at its maximum. These two
factors combine to create a maximum stress on the rods
that becomes one of the limiting factors in designing an
installation. Table 12.1 shows dimensions of some API
conventional pumping units. Parameters are defined in
Fig. 12.6.
Air-Balanced Pumping Unit. For this type of unit, the
maximum acceleration occurs at the top of the stroke
(the acceleration at the bottom of the stroke is less than
simple harmonic motion). Thus, a lower maximum stress is
set up in the rod system during transfer of the fluid load to
the rods.
The following analyses of polished rod motion apply to
conventional units. Figure 12.7 illustrates an approximate
motion of the connection point between pitman arm and
walking beam.
If x denotes the distance of B below its top position C
and is measured from the instant at which the crank arm
and pitman arm are in the vertical position with the crank
arm vertically upward, the law of cosine gives
(AB)2
¼ (OA)2
þ (OB)2
À 2(OA)(OB) cos AOB,
that is,
h2
¼ c2
þ (h þ c À x)2
À 2c(h þ c À x) cos vt,
where v is the angular velocity of the crank. The equation
reduces to
x2
À 2x[h þ c(1 À cos vt)] þ 2c(h þ c)(1 À cos vt) ¼ 0
so that
x ¼ h þ c(1 À cos vt) Æ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2 cos2 vt þ (h2 À c2)
p
:
When vt is zero, x is also zero, which means that the
negative root sign must be taken. Therefore,
x ¼ h þ c(1 À cos vt) À
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2 cos2 vt þ (h2 þ c2)
p
:
Acceleration is
Polished rod
position
Polished
rod
Polished
rod
d1 d2
d2
d1
Conventional unit
Conventional
unit
Air-balanced unit
Crank angle, degrees
0
(b)
(a)
90 180 270 360
Air-balanced
unit
Simple harmonic
motion
Figure 12.5 Polished rod motion for (a) conventional pumping unit and (b) air-balanced unit (Nind, 1964).
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 165 4.1.2007 2:43pm Compositor Name: SJoearun
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![the rods. The minimum load is at or near the top of the
stroke. Neglecting the weight of the plunger and friction
term, the minimum PRL is
PRLmin ¼ ÀSf (62:4)
Wr
gs
þ Wr À WrF2,
which, for 50 8API oil, reduces to
PRLmin ¼ 0:9Wr À F2Wr ¼ (0:9 À F2)Wr, (12:22)
where for the conventional units
F2 ¼
SN2
(1 À c
h )
70,471:2
(12:23)
and for air-balanced units
F2 ¼
SN2
(1 þ c
h )
70,471:2
: (12:24)
12.4.3 Counterweights
To reduce the power requirements for the prime mover, a
counterbalance load is used on the walking beam (small
units) or the rotary crank. The ideal counterbalance load C
is the average PRL. Therefore,
C ¼
1
2
(PRLmax þ PRLmin):
Using Eqs. (12.19) and (12.22) in the above, we get
C ¼
1
2
Wf þ 0:9Wr þ
1
2
(F1 À F2)Wr (12:25)
or for conventional units
C ¼
1
2
Wf þ Wr 0:9 þ
SN2
70,471:2
c
h
(12:26)
and for air-balanced units
C ¼
1
2
Wf þ Wr 0:9 À
SN2
70,471:2
c
h
: (12:27)
The counterbalance load should be provided by structure
unbalance and counterweights placed at walking beam
(small units) or the rotary crank. The counterweights can
be selected from manufacturer’s catalog based on the cal-
culated C value. The relationship between the counterbal-
ance load C and the total weight of the counterweights is
C ¼ Cs þ Wc
r
c
d1
d2
,
where
Cs ¼ structure unbalance, lb
Wc ¼ total weight of counterweights, lb
r ¼ distance between the mass center of counter-
weights and the crank shaft center, in.
12.4.4 Peak Torque and Speed Limit
The peak torque exerted is usually calculated on the most
severe possible assumption, which is that the peak load
(polished rod less counterbalance) occurs when the ef-
fective crank length is also a maximum (when the crank
arm is horizontal). Thus, peak torque T is (Fig. 12.5)
T ¼ c C À (0:9 À F2)Wr½ Š
d2
d1
: (12:28)
Substituting Eq. (12.25) into Eq. (12.28) gives
T ¼
1
2
S C À (0:9 À F2)Wr½ Š (12:29)
or
T ¼
1
2
S
1
2
Wf þ
1
2
(F1 þ F2)Wr
!
or
T ¼
1
4
S Wf þ
2SN2
Wr
70,471:2
(in:-lb): (12-30)
Because the pumping unit itself is usually not perfectly
balanced (Cs 6¼ 0), the peak torque is also affected by
structure unbalance. Torque factors are used for correc-
tion:
T ¼
1
2 PRLmax(TF1) þ PRLmin(TF2)½ Š
0:93
, (12:31)
where
TF1 ¼ maximum upstroke torque factor
TF2 ¼ maximum downstroke torque factor
0.93 ¼ system efficiency.
For symmetrical conventional and air-balanced units,
TF ¼ TF1 ¼ TF2.
There is a limiting relationship between stroke length
and cycles per minute. As given earlier, the maximum
value of the downward acceleration (which occurs at the
top of the stroke) is equal to
amax = min ¼
SN2
g 1 Æ c
h
À Á
70,471:2
, (12:32)
(the + refers to conventional units or air-balanced units,
see Eqs. [12.9] and [12.12]). If this maximum acceleration
divided by g exceeds unity, the downward acceleration of
the hanger is greater than the free-fall acceleration of the
rods at the top of the stroke. This leads to severe pounding
when the polished rod shoulder falls onto the hanger
(leading to failure of the rod at the shoulder). Thus, a
limit of the above downward acceleration term divided
by g is limited to approximately 0.5 (or where L is deter-
mined by experience in a particular field). Thus,
SN2
1 Æ c
h
À Á
70,471:2
#L (12:33)
or
Nlimit ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
70,471:2L
S(1 Ç c
h )
s
: (12:34)
For L ¼ 0.5,
Nlimit ¼
187:7
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
S(1 Ç c
h )
p : (12:35)
The minus sign is for conventional units and the plus sign
for air-balanced units.
12.4.5 Tapered Rod Strings
For deep well applications, it is necessary to use a tapered
suckerrodstringstoreducethePRLatthesurface.Thelarger
diameter rod is placed at the top of the rod string, then the
next largest, and then the least largest. Usually these are in
sequences up to four different rod sizes. The tapered rod
strings are designated by 1/8-in. (in diameter) increments.
Taperedrodstringscanbeidentifiedbytheirnumberssuchas
a. No. 88 is a nontapered 8
⁄8 - or 1-in. diameter rod string
b. No. 76 is a tapered string with 7
⁄8 -in. diameter rod at
the top, then a 6
⁄8 -in. diameter rod at the bottom.
c. No. 75 is a three-way tapered string consisting of
7
⁄8 -in. diameter rod at top
6
⁄8 -in. diameter rod at middle
5
⁄8 -in. diameter rod at bottom
d. No. 107 is a four-way tapered string consisting of
10
⁄8 -in. (or 11
⁄4 -in.) diameter rod at top
9
⁄8 -in. (or 11
⁄8 -in.) diameter rod below 10
⁄8 -in. diameter rod
8
⁄8 -in. (or 1-in.) diameter rod below 9
⁄8 -in. diameter rod
7
⁄8 -in. diameter rod below 8
⁄8 -in. diameter rod
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![In a field-scale evaluation, if an unlimited amount of lift
gas is available for a given gas lift project, the injection rate
of gas to individual wells should be optimized to maximize
oil production of each well. If only a limited amount of gas
is available for the gas lift, the gas should be distributed to
individual wells based on predicted well lifting perfor-
mance, that is, the wells that will produce oil at higher
rates at a given amount of lift gas are preferably chosen to
receive more lift gas.
If an unlimited amount of gas lift gas is available for a
well, the well should receive a lift gas injection rate that
yields the optimum GLR in the tubing so that the flowing
bottom-hole pressure is minimized, and thus, oil produc-
tion is maximized. The optimum GLR is liquid flow rate
dependent and can be found from traditional gradient
curves such as those generated by Gilbert (Gilbert, 1954).
Similar curves can be generated with modern computer
programs using various multiphase correlations. The com-
puter program OptimumGLR.xls in the CD attached to
this book was developed based on modified Hagedorn and
Brown method (Brown, 1977) for multiphase flow calcu-
lations and the Chen method (1979) for friction factor
determination. It can be used for predicting the optimum
GLR in tubing at a given tubing head pressure and liquid
flow rate.
After the system analysis is completed with the
optimum GLRs in the tubing above the injection point,
the expected liquid production rate (well potential) is
known. The required injection GLR to the well can be
calculated by
GLRinj ¼ GLRopt,o À GLRfm, (13:1)
where
GLRinj ¼ injection GLR, scf/stb
GLRopt,o ¼ optimum GLR at operating
flow rate, scf/stb
GLRfm ¼ formation oil GLR, scf/stb.
Then the required gas injection rate to the well can be
calculated by
qg,inj ¼ GLRinjqo, (13:2)
where qo is the expected operating liquid flow rate.
If a limited amount of gas lift gas is available for a well,
the well potential should be estimated based on GLR
expressed as
GLR ¼ GLRfm þ
qg,inj
q
, (13:3)
where qg is the lift gas injection rate (scf/day) available to
the well.
Example Problem 13.1 An oil well has a pay zone around
the mid-perf depth of 5,200 ft. The formation oil has a
gravity of 26 8API and GLR of 300 scf/stb. Water cut
remains 0%. The IPR of the well is expressed as
q ¼ qmax 1 À 0:2
pwf
pp
À 0:8
pwf
pp
2
#
,
where
qmax ¼ 1500 stb=day
pp ¼ 2,000 psia.
A 21
⁄2 -in. tubing (2.259 in. inside diameter [ID]) can be
set with a packer at 200 ft above the mid-perf. What is the
maximum expected oil production rate from the well with
continuous gas lift at a wellhead pressure of 200 psia if
a. an unlimited amount of lift gas is available for the well?
b. only 1 MMscf/day of lift gas is available for the well?
Solution The maximum oil production rate is expected
when the gas injection point is set right above the packer.
Assuming that the pressure losses due to friction below
the injection point are negligible, the inflow-performance
curve for the gas injection point (inside tubing) can be
expressed as
pvf ¼ 0:125pp½
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
81 À 80 q=qmaxð Þ
p
À 1Š À GR D À Dvð Þ,
where pvf is the pressure at the gas injection point, GR is
the pressure gradient of the reservoir fluid, D is the pay
∆Pv
G
fb
PR
Flowing bottom
hole pressure
Pressure
Dv
0
0
Gfa
D-Dv
D
Depth
Tubing
pressure (Pt)
Casing
pressure (Pso)
Point of gas
injection
Point of
balance
Figure 13.4 Pressure relationship in a continuous gas lift.
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![pc,v ¼ pc,s 1 þ
Dv
40,000
, (13:9)
which gives
pc,s ¼
pc,v
1 þ
Dv
40,000
:
(13:10)
Neglecting the pressure losses between injection choke and
the casing head, the pressure downstream of the choke
( pdn) can be assumed to be the casing surface injection
pressure, that is,
pdn ¼ pc,s:
13.4.2.3 Pressure Upstream of the Choke
The pressure upstream of the injection choke depends on
flow condition at the choke, that is, sonic or subsonic flow.
Whether a sonic flow exists depends on a downstream-to-
upstream pressure ratio. If this pressure ratio is less than a
critical pressure ratio, sonic (critical) flow exists. If this
pressure ratio is greater than or equal to the critical pres-
sure ratio, subsonic (subcritical) flow exists. The critical
pressure ratio through chokes is expressed as
Rc ¼
2
k þ 1
k
kÀ1
, (13:11)
where k ¼ Cp=Cv is the gas-specific heat ratio. The value
of the k is about 1.28 for natural gas. Thus, the critical
pressure ratio is about 0.55.
Pressure equations for choke flow are derived based on
an isentropic process. This is because there is no time for
heat to transfer (adiabatic) and the friction loss is negli-
gible (assuming reversible) at choke.
13.4.2.3.1 Sonic Flow Under sonic flow conditions,
the gas passage rate reaches and remains its maximum
value. The gas passage rate is expressed in the following
equation for ideal gases:
qgM ¼ 879CcApup
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
ggTup
!
2
k þ 1
kþ1
kÀ1
v
u
u
t , (13:12)
where
qgM ¼ gas flow rate, Mscf/day
pup ¼ pressure upstream the choke, psia
A ¼ cross-sectional area of choke, in:2
Tup ¼ upstream temperature, 8R
gg ¼ gas specific gravity related to air
Cc ¼ choke flow coefficient.
The choke flow coefficient Cc can be determined using
charts in Figs. 5.2 and 5.3 (Chapter 5) for nozzle- and
orifice-type chokes, respectively. The following correlation
has been found to give reasonable accuracy for Reynolds
numbers between 104
and 106
for nozzle-type chokes (Guo
and Ghalambor, 2005):
C ¼
d
D
þ
0:3167
d
D
0:6
þ 0:025[ log (NRe) À 4], (13:13)
where
d ¼ choke diameter, inch
D ¼ pipe diameter, in.
NRe ¼ Reynolds number
and the Reynolds number is given by
NRe ¼
20qgMgg
md
, (13:14)
where
m ¼ gas viscosity at in situ temperature and pressure, cp.
Equation (13.12) indicates that the upstream pressure is
independent of downstream pressure under sonic flow
conditions. If it is desirable to make a choke work under
sonic flow conditions, the upstream pressure should meet
the following condition:
pup !
pdn
0:55
¼ 1:82pdn (13:15)
Once the pressure upstream of the choke/orifice is deter-
mined by Eq. (13.15), the required choke/orifice diameter
can be calculated with Eq. (13.12) using a trial-and-error
approach.
13.4.2.3.2 Subsonic Flow Under subsonic flow con-
ditions, gas passage through a choke can be expressed as
qgM ¼ 1,248CcApup
Â
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
k
(k À 1)ggTup
pdn
pup
2
k
À
pdn
pup
kþ1
k
#v
u
u
t : (13:16)
If it is desirable to make a choke work under subsonic flow
conditions, the upstream pressure should be determined
from Eq. (13.16) with a trial-and-error method.
13.4.2.4 Pressure of the Gas Distribution Line
The pressure at the inlet of gas distribution line can be
calculated using the Weymouth equation for horizontal
flow (Weymouth, 1912):
qgM ¼
0:433Tb
pb
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p2
L À p2
up
D16=3
gg
TTzzLg
v
u
u
t
, (13:17)
where
Tb ¼ base temperature, 8R
pb ¼ base pressure, psi
pL ¼ pressure at the inlet of gas distribution line, psia
Lg ¼ length of distribution line, mile
Equation (13.17) can be rearranged to solve for
pressure:
pL ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p2
up þ
qgMpb
0:433Tb
2
gg
TTzzLg
D16=3
s
(13:18)
Example Problem 13.2 Anoilfieldhas16oilwellstobegas
lifted. The gas lift gas at the central compressor station is
firstpumpedtotwoinjectionmanifoldswith4-in.ID,1-mile
lines and then is distributed to the wellheads with 4-in. ID,
0.2-mile lines. Given the following data, calculate the
required output pressure of compression station:
Gas-specific gravity (gg): 0.65
Valve depth (Dv): 5,000 ft
Maximum tubing pressure at valve
depth ( pt): 500 psia
Required lift gas injection rate per well: 2 MMscf/day
Pressure safety factor (Sf ): 1.1
Base temperature (Tb): 60 8F
Base pressure ( pb): 14.7 psia
Solution Using Dpv ¼ 100 psi, the injection pressure at
valve depth is then 600 psia. Equation (13.10) gives
pc,s ¼
pc,v
1 þ
Dv
40,000
¼
600
1 þ
5,000
40,000
¼ 533 psia:
Neglecting the pressure losses between the injection choke
and the casing head, pressure downstream of the choke
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![5. Starting from available kickoff surface pressure pk,s,
establish kickoff casing pressure line. This can be done
using Eq. (13.7) or Eq. (13.9).
6. Starting from pk À Dpkm at surface, where the kickoff
pressure margin Dpkm can be taken as 50 psi, establish a
design kickoff line parallel to the kickoff casing pres-
sure line. Pressures in this line, denoted pkd , represent
kickoff pressure after adjustment for kickoff pressure
margin. Set Dpkm ¼ 0 if kickoff casing pressure margin
is not required.
7. Calculate depth of the first valve. Based on the fact
that phf þ GsD1 ¼ pkd1, the depth of the top valve is
expressed as
D1 ¼
pkd1 À phf
Gs
, (13:47)
where
pkd1 ¼ kickoff pressure opposite the first valve (psia)
Gs ¼ static (dead liquid) gradient; psi/ft
Applying Eq. (13.9) gives
pkd1 ¼ pk,s À Dpkm
À Á
1 þ
D1
40,000
: (13:48)
Solving Eqs. (13.47) and (13.48) yields
D1 ¼
pk,s À Dpkm À phf
Gs À
pk À Dpkm
40,000
:
(13:49)
When the static liquid level is below the depth calculated
by use of Eq. (13.49), the first valve is placed at a depth
slightly deeper than the static level. If the static liquid
level is known, then
D1 ¼ Ds þ S1, (13:50)
where Ds is the static level and S1 is the submergence of
the valve below the static level.
8. Calculate the depths to other valves. Based on the fact
that phf þ Gfd D2 þ Gs(D2 À D1) ¼ pcd2, the depth of
valve 2 is expressed as
D2 ¼
pcd2 À Gfd D1 À phf
Gs
þ D1, (13:51)
where
pcd2 ¼ design injection pressure at valve 2, psig
Gfd ¼ design unloading gradient, psi/ft.
Applying Eq. (13.9) gives
pcd2 ¼ pc,s À Dpcm
À Á
1 þ
D2
40,000
: (13:52)
Solving Eqs. (13.51) and (13.52) yields
D2 ¼
pc,s À Dpcm À phf ,d þ Gs À Gfd
À Á
D1
Gs À
pc À Dpcm
40,000
:
(13:53)
Similarly, the depth to the third valve is
D3 ¼
pc,s À Dpcm À phf ,d þ Gs À Gfd
À Á
D2
Gs À
pc À Dpcm
40,000
:
(13:54)
Thus, a general equation for depth of valve i is
Di ¼
pc,s À Dpcm À phf ,d þ Gs À Gfd
À Á
DiÀ1
Gs À
pc À Dpcm
40,000
:
(13:55)
Depths of all valves can be calculated in a similar manner
until the minimum valve spacing ( $ 400 ft) is reached.
Example Problem 13.5 Only 1 MMscf/day of lift gas
is available for the well described in the Example
Problem 13.1. If 1,000 psia is available to kick off the
well and then a steady injection pressure of 800 psia is
maintained for gas lift operation against a wellhead
pressure of 130 psia, design locations of unloading
and operating valves. Assume a casing pressure margin
of 50 psi.
Solution The hydrostatic pressure of well fluid (26 8API
oil) is (0.39 psi/ft) (5,200 ft), or 2,028 psig, which is greater
than the given reservoir pressure of 2,000 psia. Therefore,
the well does not flow naturally. The static liquid level
depth is estimated to be
5,200 À (2,000 À 14:7)=(0:39) ¼ 110 ft:
Depth of the top valve is calculated with Eq. (13.49):
D1 ¼
1,000 À 50 À 130
0:39 À
1,000 À 50
40,000
¼ 2,245 ft 110 ft
Tubing pressure margin at surface is (0.25)(800), or
200 psi. The modified Hagedorn–Brown correlation gives
tubing pressure of 591 psia at depth of 5,000 ft. The design
tubing flowing gradient is Gfd ¼ [591 À (130 þ 200)]=
(5,000) or 0.052 psi/ft. Depth of the second valve is calcu-
lated with Eq. (13.53):
D2 ¼
1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(2,245)
0:39 À
1,000 À 50
40,000
¼ 3,004 ft
Similarly,
D3 ¼
1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(3,004)
0:39 À
1,000 À 50
40,000
¼ 3,676 ft
D4 ¼
1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(3,676)
0:39 À
1,000 À 50
40,000
¼ 4,269 ft
D5 ¼
1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(4,269)
0:39 À
1,000 À 50
40,000
¼ 4,792 ft,
which is the depth of the operating valve.
Similar problems can be quickly solved with the com-
puter spreadsheet GasLiftValveSpacing.xls.
Pressure
Operating Tubing
Pressure
Injection Operating
Pressure
Kick-off
Pressure
Depth
phf phf,d pc,s pk,s
∆ptm
∆pcm
∆pkm
G
s
G
s
G
s
G
s
G
s
G
s
G
s
Gf Gf,d
Figure 13.18 A flow diagram to illustrate procedure of
valve spacing.
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GAS LIFT 13/199](https://image.slidesharecdn.com/lvzgyxtjtgu380ahut9b-signature-b2ba6e1375509d785bab16b3cad4b5b4b504aa1845b2191a80963142ee8b462c-poli-140812083436-phpapp01/85/Petroleum-production-engineering-203-320.jpg)
![Centrilift-Hughes, Inc. Oilfield Centrilift-Hughes Submers-
ible Pump Handbook. Claremore, Oaklahoma, 1998.
cholet, h. Well Production Practical Handbook. Paris:
Editions TECHNIP, 2000.
foss, d.l. and gaul, r.b. Plunger lift performance criteria
with operating experience: Ventura Avenue field. Dril-
ling Production Practices API 1965:124–140.
guo, b. and ghalambor, a. Natural Gas Engineering
Handbook. Houston, TX: Gulf Publishing Co., 2005.
guo, b. and ghalambor, a. Gas Volume Requirements for
Underbalanced Drilling Deviated Holes. PennWell
Books Tulsa, Oaklahoma, 2002.
lea, j.f. Plunger lift vs velocity strings. Energy Sources
Technology Conference Exhibition (ETCE ’99), 1–2
February 1999, Houston Sheraton Astrodome Hotel in
Houston, Texas.
lea, j.f. Dynamic analysis of plunger lift operations. Pre-
sented at the 56th Annual Fall Technical Conference
and Exhibition, 5–7 October 1981, San Antonio,
Texas. Paper SPE 10253.
lebeaux, j.m. and sudduth, l.f. Theoretical and practical
aspects of free piston operation. JPT September
1955:33–35.
listiak, s.d. Plunger lift. In: Petroleum Engineering Hand-
book (Lake, L., ed.). Dallas: Society of Petroleum
Engineers, 2006.
mower, l.n., lea, j.f., beauregard, e., and ferguson, p.l.
Defining the characteristics and performance of gas-lift
plungers. Presented at the SPE Annual Technical Con-
ference and Exhibition held in 22–26 September 1985,
Las Vegas, Nevada. SPE Paper 14344.
rosina, l. A study of plunger lift dynamics [Masters
Thesis], University of Tulsa, 1983.
turner, r.g., hubbard, m.g., and dukler, a.e. Analysis
and prediction of minimum flow rate for the continu-
ous removal of liquids from gas wells. J. Petroleum
Technol. November 1969.
Problems
14.1 A 9,000-ft-deep well produces 26 8API oil with GOR
50 scf/stb and zero water cut through a 3-in. (2.992-
in. ID) tubing in a 7-in. casing. The oil has a forma-
tion volume factor of 1.20 and average viscosity of
8 cp. Gas-specific gravity is 0.75. The surface and
bottom-hole temperatures are 70 and 160 8F, re-
spectively. The IPR of the well can be described by
Vogel’s model with a reservoir pressure 4,050 psia
and AOF 12,000 stb/day. If the well is put in produc-
tion with an ESP to produce liquid at 7,000 stb/day
against a flowing well head pressure of 150 psia,
determine the required specifications for an ESP for
this application. Assume the minimum pump suction
pressure is 220 psia.
14.2 A 9,000-ft-deep well has a potential to produce
35 8API oil with GOR 120 scf/stb and 10% water
cut through a 2-in. (1.995-in. ID) tubing in a 7-in.
casing with a pump installation. The oil has a forma-
tion volume factor of 1.25 and average viscosity of
5 cp. Gas- and water-specific gravities are 0.75 and
1.05, respectively. The surface and bottom-hole tem-
peratures are 70 and 170 8F, respectively. The
IPR of the well can be described by Vogel’s model
with a reservoir pressure 2,000 psia and AOF
400 stb/day. If the well is to put in production with
a HPP at depth of 8,500 ft in an open power fluid
system to produce liquid at 210 stb/day against a
flowing well head pressure of 65 psia, determine the
required specifications for the HPP for this applica-
tion. Assume the overall efficiencies of the engine,
HHP, and surface pump to be 0.90, 0.80, and 0.85,
respectively.
14.3 Calculate required GLR, casing pressure, and plun-
ger lift operating range for the following given well
data:
Gas rate: 250 Mcfd expected
when unloaded
Liquid rate: 12 bbl/day expected
when unloaded
Liquid gradient: 0.40 psi/ft
Tubing, ID: 1.995 in.
Tubing, OD: 2.375 in.
Casing, ID: 4.56 in.
Depth to plunger: 7,000 ft
Line pressure: 120 psi
Available casing pressure: 850 psi
Reservoir pressure: 1250 psi
Average Z-factor: 0.99
Average temperature: 150 8F
Plunger weight: 10 lb
Plunger fall in gas: 750 fpm
Plunger fall in liquid: 150 fpm
Plunger rise velocity: 1,000 fpm
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 223 3.1.2007 9:10pm Compositor Name: SJoearun
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![Solution
Volume of CaCO3 to be removed:
Vm ¼ p r2
a À r2
w
À Á
1 À fð ÞCm
¼ p 1:3282
À 0:3282
À Á
1 À 0:2ð Þ(0:1)
¼ 0:42 ft3
CaCO3=ft pay zone
Initial pore volume:
VP ¼ p r2
a À r2
w
À Á
f
¼ p 1:3282
À 0:3282
À Á
(0:2) ¼ 1:05 ft3
=ft pay zone
Gravimetric dissolving power of the 15 wt% HCl solution:
b ¼ Ca
ymMWm
yaMWa
¼ (0:15)
(1)(100:1)
(2)(36:5)
¼ 0:21 lbm CaCO3=lbm 15 wt% HCl solution
Volumetric dissolving power of the 15 wt% HCl solution:
X ¼ b
ra
rm
¼ (0:21)
(1:07)(62:4)
(169)
¼ 0:082 ft3
CaCO3=ft3
15 wt% HCl solution
The required minimum HCl volume
Va ¼
Vm
X
þ VP þ Vm
¼
0:42
0:082
þ 1:05 þ 0:42
¼ 6:48 ft3
15 wt% HCl solution=ft pay zone
¼ (6:48)(7:48)
¼ 48 gal 15 wt% HCl solution=ft pay zone
The acid volume requirement for the main stage in a mud
acid treatment depends on mineralogy and acid type and
strength. Economides and Nolte (2000) provide a listing of
typical stage sequences and volumes for sandstone acidizing
treatments. For HCl acid, the volume requirement increases
from 50 to 200 gal/ft pay zone with HCl solubility of HF
changing from less than 5% to 20%. For HF acid, the volume
requirement is in the range of 75–100 gal/ft pay zone with
3.0–13.5% HCl and 0.5–3.0% HF depending on mineralogy.
Numerous efforts have been made to develop a rigorous
method for calculating the minimum required acid volume
in the past 2 decades. The most commonly used method is
the two-mineral model (Hekim et al., 1982; Hill et al., 1981;
Taha et al., 1989). This model requires a numerical tech-
nique to obtain a general solution. Schechter (1992) pre-
sented an approximate solution that is valid for Damkohler
number being greater than 10. This solution approximates
the HF fast-reacting mineral front as a sharp front. Readers
are referred to Schechter (1992) for more information.
Because mud acid treatments do not dissolve much of the
formation minerals but dissolve the materials clogging the
pore throats, Economides and Nolte (2000) suggest taking
the initial pour volume (Eq. [16.5]) within the radius of treat-
mentastheminimumrequiredacidvolumeforthemainstage
of acidizing treatment. Additional acid volume should be
considered for the losses in the injection tubing string.
16.3.3 Acid Injection Rate
Acid injection rate should be selected on the basis of
mineral dissolution and removal and depth of damaged
zone. Selecting an optimum injection rate is a difficult
process because the damaged zone is seldom known with
any accuracy and the competing effects of mineral disso-
lution and reaction product precipitation. Fortunately,
research results have shown that acidizing efficiency is
relatively insensitive to acid injection rate and that the
highest rate possible yields the best results. McLeod
(1984) recommends relatively low injection rates based
on the observation that acid contact time with the forma-
tion of 2–4 hours appears to give good results. da Motta
(1993) shows that with shallow damage, acid injection rate
has little effect on the residual skin after 100 gal/ft of
injection rate; and with deeper damage, the higher the
injection rate, the lower the residual skin. Paccaloni et al.
(1988) and Paccaloni and Tambini (1990) also report high
success rates in numerous field treatments using the high-
est injection rates possible.
There is always an upper limit on the acid injection rate
that is imposed by formation breakdown (fracture) pres-
sure pbd . Assuming pseudo–steady-state flow, the max-
imum injection rate limited by the breakdown pressure is
expressed as
qi,max ¼
4:917 Â 10À6
kh pbd À p À Dpsf
À Á
ma ln 0:472re
rw
þ S
, (16:6)
where
qi ¼ maximum injection rate, bbl/min
k ¼ permeability of undamaged formation, md
h ¼ thickness of pay zone to be treated, ft
pbd ¼ formation breakdown pressure, psia
p ¼ reservoir pressure, psia
Dpsf ¼ safety margin, 200 to 500 psi
ma ¼ viscosity of acid solution, cp
re ¼ drainage radius, ft
rw ¼ wellbore radius, ft
S ¼ skin factor, ft.
The acid injection rate can also be limited by surface
injection pressure at the pump available to the treatment.
This effect is described in the next section.
16.3.4 Acid Injection Pressure
In most acid treatment operations, only the surface tubing
pressure is monitored. It is necessary to predict the surface
injection pressure at the design stage for pump selection.
The surface tubing pressure is related to the bottom-hole
flowing pressure by
psi ¼ pwf À Dph þ Dpf , (16:7)
where
psi ¼ surface injection pressure, psia
pwf ¼ flowing bottom-hole pressure, psia
Dph ¼ hydrostatic pressure drop, psia
Dpf ¼ frictional pressure drop, psia.
The second and the third term in the right-hand side of
Eq. (16.7) can be calculated using Eq. (11.93). However, to
avert the procedure of friction factor determination,
the following approximation may be used for the frictional
pressure drop calculation (Economides and Nolte, 2000):
Dpf ¼
518r0:79
q1:79
m0:207
1,000D4:79
L, (16:8)
where
r ¼ density of fluid, g=cm3
q ¼ injection rate, bbl/min
m ¼ fluid viscosity, cp
D ¼ tubing diameter, in.
L ¼ tubing length, ft.
Equation (16.8) is relatively accurate for estimating fric-
tional pressures for newtonian fluids at flow rates less than
9 bbl/min.
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![Y ¼
L1
L
, X ¼
qt À q3
q3
: (18:44)
If, D1 ¼ D3, Eq. (18.43) can be rearranged as
Y ¼
1 À
1
1 þ Xð Þ2
1 À
1
1 þ R2:31
D
À Á2
, (18:45)
where RD is the ratio of the looping pipe diameter to the
original pipe diameter, that is, RD ¼ D2=D3. Equation
(18.45) can be rearranged to solve for X explicitly
X ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 À Y 1 À
1
1 þ R2:31
D
À Á2
!v
u
u
t
À 1: (18:46)
The effects of looped line on the increase of gas flow rate
for various pipe diameter ratios are shown in Fig. 18.11.
This figure indicates an interesting behavior of looping:
The increase in gas capacity is not directly proportional to
the fraction of looped pipeline. For example, looping of
40% of pipe with a new pipe of the same diameter will
increase only 20% of the gas flow capacity. It also shows
that the benefit of looping increases with the fraction of
looping. For example, looping of 80% of the pipe with a
new pipe of the same diameter will increase 60%, not 40%,
of gas flow capacity.
Example Problem 18.3 Consider a 4-in. pipeline that is 10
miles long. Assuming that the compression and delivery
pressures will maintain unchanged, calculate gas capacity
increases by using the following measures of improvement:
(a) replace 3 miles of the 4-in. pipeline by a 6-in. pipeline
segment; (b) place a 6-in. parallel pipeline to share gas
Table 18.1 Flash Calculation with Standing’s Method for ki Values
Flash calculation
nv ¼ 0:8791
Compound zi ki zi(ki À 1)=[nv(ki À 1) þ 1]
C1 0.6599 6.5255 0.6225
C2 0.0869 1.8938 0.0435
C3 0.0591 0.8552 À0:0098
i-C4 0.0239 0.4495 À0:0255
n-C4 0.0278 0.3656 À0:0399
i-C5 0.0157 0.1986 À0:0426
n-C5 0.0112 0.1703 À0:0343
C6 0.0181 0.0904 À0:0822
C7þ 0.0601 0.0089 À0:4626
N2 0.0194 30.4563 0.0212
CO2 0.0121 3.4070 0.0093
H2S 0.0058 1.0446 0.0002
Sum: 0.0000
nL ¼ 0.1209
Compound xi yi xiMWi yiMWi
C1 0.1127 0.7352 1.8071 11.7920
C2 0.0487 0.0922 1.4633 2.7712
C3 0.0677 0.0579 2.9865 2.5540
i-C4 0.0463 0.0208 2.6918 1.2099
n-C4 0.0629 0.0230 3.6530 1.3356
i-C5 0.0531 0.0106 3.8330 0.7614
n-C5 0.0414 0.0070 2.9863 0.5085
C6 0.0903 0.0082 7.7857 0.7036
C7þ 0.4668 0.0042 53.3193 0.4766
N2 0.0007 0.0220 0.0202 0.6156
CO2 0.0039 0.0132 0.1709 0.5823
H2S 0.0056 0.0058 0.1902 0.1987
Apparent molecular
weight of liquid phase:
23.51 80.91
Apparent molecular
weight of vapor phase:
0.76
Specific gravity
of liquid phase:
water ¼ 1
Specific gravity
of vapor phase:
0.81 air ¼ 1
Input vapor
phase z factor:
0.958
Density of liquid phase: 47.19 lbm=ft3
Density of vapor phase: 2.08 lbm=ft3
Volume of liquid phase: 0.04 bbl
Volume of vapor phase: 319.66 scf
GOR: 8,659 scf/bbl
API gravity of
liquid phase:
56
Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 274 4.1.2007 10:04pm Compositor Name: SJoearun
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Petroleum production engineering
- 1. • ISBN: 0750682701 • Publisher: Elsevier Science & Technology Books • Pub. Date: February 2007
- 2. Preface The advances in the digital computing technology in the last decade have revolutionized the petroleum industry. Using the modern computer technologies, today’s petro- leum production engineers work much more efficiently than ever before in their daily activities, including analyz- ing and optimizing the performance of their existing pro- duction systems and designing new production systems. During several years of teaching the production engineer- ing courses in academia and in the industry, the authors realized that there is a need for a textbook that reflects the current practice of what the modern production engineers do. Currently available books fail to provide adequate information about how the engineering principles are ap- plied to solving petroleum production engineering prob- lems with modern computer technologies. These facts motivated the authors to write this new book. This book is written primarily for production engineers and college students of senior level as well as graduate level. It is not authors’ intention to simply duplicate gen- eral information that can be found from other books. This book gathers authors’ experiences gained through years of teaching courses of petroleum production engineering in universities and in the petroleum industry. The mission of the book is to provide production engineers a handy guide- line to designing, analyzing, and optimizing petroleum production systems. The original manuscript of this book has been used as a textbook for college students of under- graduate and graduate levels in Petroleum Engineering. This book was intended to cover the full scope of pe- troleum production engineering. Following the sequence of oil and gas production process, this book presents its contents in eighteen chapters covered in four parts. Part I contains eight chapters covering petroleum pro- duction engineering fundamentals as the first course for the entry-level production engineers and undergraduate students. Chapter 1 presents an introduction to the petro- leum production system. Chapter 2 documents properties of oil and natural gases that are essential for designing and analysing oil and gas production systems. Chapters 3 through 6 cover in detail the performance of oil and gas wells. Chapter 7 presents techniques used to forecast well production for economics analysis. Chapter 8 describes empirical models for production decline analysis. Part II includes three chapters presenting principles and rules of designing and selecting the main components of petroleum production systems. These chapters are also written for entry-level production engineers and under- graduate students. Chapter 9 addresses tubing design. Chapter 10 presents rule of thumbs for selecting com- ponents in separation and dehydration systems. Chapter 11 details principles of selecting liquid pumps, gas com- pressors, and pipelines for oil and gas transportation. Part III consists of three chapters introducing artificial lift methods as the second course for the entry-level pro- duction engineers and undergraduate students. Chapter 12 presents an introduction to the sucker rod pumping system and its design procedure. Chapter 13 describes briefly gas lift method. Chapter 14 provides an over view of other artificial lift methods and design procedures. Part IV is composed of four chapters addressing pro- duction enhancement techniques. They are designed for production engineers with some experience and graduate students. Chapter 15 describes how to identify well prob- lems. Chapter 16 deals with designing acidizing jobs. Chapter 17 provides a guideline to hydraulic fracturing and job evaluation techniques. Chapter 18 presents some relevant information on production optimisation tech- niques. Since the substance of this book is virtually boundless in depth, knowing what to omit was the greatest difficulty with its editing. The authors believe that it requires many books to describe the foundation of knowledge in petro- leum production engineering. To counter any deficiency that might arise from the limitations of space, the book provides a reference list of books and papers at the end of each chapter so that readers should experience little diffi- culty in pursuing each topic beyond the presented scope. Regarding presentation, this book focuses on presen- ting and illustrating engineering principles used for designing and analyzing petroleum production systems rather than in-depth theories. Derivation of mathematical models is beyond the scope of this book, except for some special topics. Applications of the principles are illustrated by solving example problems. While the solutions to some simple problems not involving iterative procedures are demonstrated with stepwise calculations, compli- cated problems are solved with computer spreadsheet programs. The programs can be downloaded from the publisher’s website (http://books.elsevier.com/companions/ 9780750682701). The combination of the book and the computer programs provides a perfect tool kit to petrol- eum production engineers for performing their daily work in a most efficient manner. All the computer programs were written in spreadsheet form in MS Excel that is available in most computer platforms in the petroleum industry. These spreadsheets are accurate and very easy to use. Although the U.S. field units are used in the com- panion book, options of using U.S. field units and SI units are provided in the spreadsheet programs. This book is based on numerous documents including reports and papers accumulated through years of work in the University of Louisiana at Lafayette and the New Mexico Institute of Mining and Technology. The authors are grateful to the universities for permissions of publish- ing the materials. Special thanks go to the Chevron and American Petroleum Institute (API) for providing Chev- ron Professorship and API Professorship in Petroleum Engineering throughout editing of this book. Our thanks are due to Mr. Kai Sun of Baker Oil Tools, who made a thorough review and editing of this book. The authors also thank Malone Mitchell III of Riata Energy for he and his company’s continued support of our efforts to develop new petroleum engineering text and professional books for the continuing education and training of the industry’s vital engineers. On the basis of the collective experiences of authors and reviewer, we expect this book to be of value to the production engineers in the petrol- eum industry. Dr. Boyun Guo Chevron Endowed Professor in Petroleum Engineering University of Louisiana at Lafayette June 10, 2006 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page ix 29.12.2006 10:39am
- 3. List of Symbols A area, ft2 Ab total effective bellows area, in:2 Aeng net cross-sectional area of engine piston, in:2 Afb total firebox surface area, ft2 A 0 i inner area of tubing sleeve, in:2 A 0 o outer area of tubing sleeve, in:2 Ap valve seat area, gross plunger cross-sectional area, or inner area of packer, in:2 Apump net cross-sectional area of pump piston, in:2 Ar cross-sectional area of rods, in:2 At tubing inner cross-sectional area, in:2 o API API gravity of stock tank oil B formation volume factor of fluid, rb/stb b constant 1:5 Â 10À5 in SI units Bo formation volume factor of oil, rb/stb Bw formation volume factor of water, rb/bbl CA drainage area shape factor Ca weight fraction of acid in the acid solution Cc choke flow coefficient CD choke discharge coefficient Cg correction factor for gas-specific gravity Ci productivity coefficient of lateral i Cl clearance, fraction Cm mineral content, volume fraction Cs structure unbalance, lbs Ct correction factor for operating temperature ct total compressibility, psiÀ1 Cp specific heat of gas at constant pressure, lbf- ft/lbm-R CCp specific heat under constant pressure evaluated at cooler Cwi water content of inlet gas, lbm H2O=MMscf D outer diameter, in., or depth, ft, or non-Darcy flow coefficient, d/Mscf, or molecular diffusion coefficient, m2 =s d diameter, in. d1 upstream pipe diameter, in. d2 choke diameter, in. db barrel inside diameter, in. Dci inner diameter of casing, in. df fractal dimension constant 1.6 Dh hydraulic diameter, in. DH hydraulic diameter, ft Di inner diameter of tubing, in. Do outer diameter, in. dp plunger outside diameter, in. Dpump minimum pump depth, ft Dr length of rod string, ft E rotor/stator eccentricity, in., or Young’s modulus, psi Ev volumetric efficiency, fraction ev correction factor ep efficiency Fb axial load, lbf FCD fracture conductivity, dimensionless FF fanning friction factor Fgs modified Foss and Gaul slippage factor fhi flow performance function of the vertical section of lateral i fLi inflow performance function of the horizontal section of lateral i fM Darcy-Wiesbach (Moody) friction factor Fpump pump friction-induced pressure loss, psia fRi flow performance function of the curvic section of lateral i fsl slug factor, 0.5 to 0.6 G shear modulus, psia g gravitational acceleration, 32:17 ft=s2 Gb pressure gradient below the pump, psi/ft gc unit conversion factor, 32:17 lbmÀft=lbf Às2 Gfd design unloading gradient, psi/ft Gi initial gas-in-place, scf Gp cumulative gas production, scf G1 p cumulative gas production per stb of oil at the beginning of the interval, scf Gs static (dead liquid) gradient, psi/ft G2 mass flux at downstream, lbm=ft2 =sec GLRfm formation oil GLR, scf/stb GLRinj injection GLR, scf/stb GLRmin minimum required GLR for plunger lift, scf/ bbl GLRopt,o optimum GLR at operating flow rate, scf/stb GOR producing gas-oil ratio, scf/stb GWR glycol to water ratio, gal TEG=lbm H2O H depth to the average fluid level in the annulus, ft, or dimensionless head h reservoir thickness, ft, or pumping head, ft hf fracture height, ft HP required input power, hp HpMM required theoretical compression power, hp/ MMcfd Ht total heat load on reboiler, Btu/h Dh depth increment, ft DHpm mechanical power losses, hp rhi pressure gradient in the vertical section of lateral i, psi/ft J productivity of fractured well, stb/d-psi Ji productivity index of lateral i. Jo productivity of non-fractured well, stb/d-psi K empirical factor, or characteristic length for gas flow in tubing, ft k permeability of undamaged formation, md, or specific heat ratio kf fracture permeability, md kH the average horizontal permeability, md kh the average horizontal permeability, md ki liquid/vapor equilibrium ratio of compound i kp a constant kro the relative permeability to oil kV vertical permeability, md L length, ft , or tubing inner capacity, ft/bbl Lg length of gas distribution line, mile LN net lift, ft Lp length of plunger, in. M total mass associated with 1 stb of oil M2 mass flow rate at down stream, lbm/sec MWa molecular weight of acid MWm molecular weight of mineral N pump speed, spm, or rotary speed, rpm n number of layers, or polytropic exponent for gas NAc acid capillary number, dimensionless NCmax maximum number of cycles per day nG number of lb-mole of gas Ni initial oil in place in the well drainage area, stb ni productivity exponent of lateral i Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xi 29.12.2006 10:39am
- 4. nL number of mole of fluid in the liquid phase Nmax maximum pump speed, spm np number of pitches of stator N1 p cumulative oil production per stb of oil in place at the beginning of the interval Nf p,n forcasted annual cumulative production of fractured well for year n Nnf p,n predicted annual cumulative production of nonfractured well for year n Nno p,n predicted annual cumulative production of non-optimized well for year n Nop p,n forcasted annual cumulative production of optimized system for year n NRe Reunolds number Ns number of compression stages required Nst number of separation stages À1 nV number of mole of fluid in the vapor phase Nw number of wells DNp,n predicted annual incremental cumulative production for year n P pressure, lb=ft2 p pressure, psia pb base pressure, psia pbd formation breakdown pressure, psia Pc casing pressure, psig pc critical pressure, psia, or required casing pressure, psia, or the collapse pressure with no axial load, psia pcc the collapse pressure corrected for axial load, psia Pcd2 design injection pressure at valve 2, psig PCmin required minimum casing pressure, psia pc,s casing pressure at surface, psia pc,v casing pressure at valve depth, psia Pd pressure in the dome, psig pd final discharge pressure, psia peng,d engine discharge pressure, psia peng,i pressure at engine inlet, psia pf frictional pressure loss in the power fluid injection tubing, psi Ph hydraulic power, hp ph hydrostatic pressure of the power fluid at pump depth, psia phf wellhead flowing pressure, psia phfi flowing pressure at the top of lateral i, psia pL pressure at the inlet of gas distribution line, psia pi initial reservoir pressure, psia, or pressure in tubing, psia, or pressure at stage i, psia pkd1 kick-off pressure opposite the first valve, psia pkfi flowing pressure at the kick-out-point of lateral i, psia pL pressure at the inlet of the gas distribution line, psia Plf flowing liquid gradient, psi/bbl slug Plh hydrostatic liquid gradient, psi/bbl slug pLmax maximum line pressure, psia po pressure in the annulus, psia pout output pressure of the compression station, psia Pp Wp=At, psia pp pore pressure, psi ppc pseudocritical pressure, psia ppump,i pump intake pressure, psia ppump,d pump discharge pressure, psia Pr pitch length of rotor, ft pr pseudoreduced pressure Ps pitch length of stator, ft, or shaft power, ftÀlbf =sec ps surface operating pressure, psia, or suction pressure, psia, or stock-tank pressure, psia psc standard pressure, 14.7 psia psh slug hydrostatic pressure, psia psi surface injection pressure, psia psuction suction pressure of pump, psia Pt tubing pressure, psia ptf flowing tubing head pressure, psig pup pressure upstream the choke, psia Pvc valve closing pressure, psig Pvo valve opening pressure, psig pwh upstream (wellhead) pressure, psia pwf flowing bottom hole pressure, psia pwfi the average flowing bottom-lateral pressure in lateral i, psia pwfo dynamic bottom hole pressure because of cross-flow between, psia pc wf critical bottom hole pressure maintained during the production decline, psia pup upstream pressure at choke, psia P1 pressure at point 1 or inlet, lbf =ft2 P2 pressure at point 2 or outlet, lbf =ft2 p1 upstream/inlet/suction pressure, psia p2 downstream/outlet/discharge pressure, psia pp average reservoir pressure, psia ppf reservoir pressure in a future time, psia pp0 average reservoir pressure at decline time zero, psia ppt average reservoir pressure at decline time t, psia DP pressure drop, lbf =ft2 Dp pressure increment, psi dp head rating developed into an elementary cavity, psi Dpf frictional pressure drop, psia Dph hydrostatic pressure drop, psia Dpi avg the average pressure change in the tubing, psi Dpo avg the average pressure change in the annulus, psi Dpsf safety pressure margin, 200 to 500 psi Dpv pressure differential across the operating valve (orifice), psi Q volumetric flow rate q volumetric flow rate Qc pump displacement, bbl/day qeng flow rate of power fluid, bbl/day QG gas production rate, Mscf/day qG glycol circulation rate, gal/hr qg gas production rate, scf/d qg,inj the lift gas injection rate (scf/day) available to the well qgM gas flow rate, Mscf/d qg,total total output gas flow rate of the compression station, scf/day qh injection rate per unit thickness of formation, m3 =sec-m qi flow rate from/into layer i, or pumping rate, bpm qi,max maximum injection rate, bbl/min qL liquid capacity, bbl/day Qo oil production rate, bbl/day qo oil production rate, bbl/d qpump flow rate of the produced fluid in the pump, bbl/day Qs leak rate, bbl/day, or solid production rate, ft3 =day qs gas capacity of contactor for standard gas (0.7 specific gravity) at standard temperature (100 8F), MMscfd, or sand production rate, ft3 =day qsc gas flow rate, Mscf/d qst gas capacity at standard conditions, MMscfd qtotal total liquid flow rate, bbl/day Qw water production rate, bbl/day qw water production rate, bbl/d xii LIST OF SYMBOLS Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xii 29.12.2006 10:39am
- 5. qwh flow rate at wellhead, stb/day R producing gas-liquid ratio, Mcf/bbl, or dimensionless nozzle area, or area ratio Ap=Ab, or the radius of fracture, ft, or gas constant, 10:73 ft3 -psia=lbmol-R r distance between the mass center of counterweights and the crank shaft, ft or cylinder compression ratio ra radius of acid treatment, ft Rc radius of hole curvature, in. re drainage radius, ft reH radius of drainage area, ft Rp pressure ratio Rs solution gas oil ratio, scf/stb rw radius of wellbore, ft rwh desired radius of wormhole penetration, m R2 Ao=Ai rRi vertical pressure gradient in the curvic section of lateral i, psi/ft S skin factor, or choke size, 1 ⁄64 in. SA axial stress at any point in the tubing string, psi Sf specific gravity of fluid in tubing, water ¼ 1, or safety factor Sg specific gravity of gas, air ¼ 1 So specific gravity of produced oil, fresh water ¼ 1 Ss specific gravity of produced solid, fresh water ¼ 1 St equivalent pressure caused by spring tension, psig Sw specific gravity of produced water, fresh water ¼ 1 T temperature, 8R t temperature, 8F, or time, hour, or retention time, min Tav average temperature, 8R Tavg average temperature in tubing, 8F Tb base temperature, 8R, or boiling point, 8R Tc critical temperature, 8R Tci critical temperature of component i, 8R Td temperature at valve depth, 8R TF1 maximum upstroke torque factor TF2 maximum downstroke torque factor Tm mechanical resistant torque, lbf -ft tr retention time % 5:0 min Tsc standard temperature, 520 8R Tup upstream temperature, 8R Tv viscosity resistant torque, lbf -ft T1 suction temperature of the gas, 8R TT average temperature, 8R u fluid velocity, ft/s um mixture velocity, ft/s uSL superficial velocity of liquid phase, ft/s uSG superficial velocity of gas phase, ft/s V volume of the pipe segment, ft3 v superficial gas velocity based on total cross- sectional area A, ft/s Va the required minimum acid volume, ft3 Vfg plunger falling velocity in gas, ft/min Vfl plunger falling velocity in liquid, ft/min Vg required gas per cycle, Mscf Vgas gas volume in standard condition, scf VG1 gas specific volume at upstream, ft3 =lbm VG2 gas specific volume at downstream, ft3 =lbm Vh required acid volume per unit thickness of formation, m3 =m VL specific volume of liquid phase, ft3 =molÀlb, or volume of liquid phase in the pipe segment, ft3 , or liquid settling volume, bbl, or liquid specific volume at upstream, ft3 =lbm Vm volume of mixture associated with 1 stb of oil, ft3 , or volume of minerals to be removed, ft3 V0 pump displacement, ft3 VP initial pore volume, ft3 Vr plunger rising velocity, ft/min Vres oil volume in reservoir condition, rb Vs required settling volume in separator, gal Vslug slug volume, bbl Vst oil volume in stock tank condition, stb Vt At(D À VslugL), gas volume in tubing, Mcf VVsc specific volume of vapor phase under standard condition, scf/mol-lb V1 inlet velocity of fluid to be compressed, ft/sec V2 outlet velocity of compressed fluid, ft/sec n1 specific volume at inlet, ft3 =lb n2 specific volume at outlet, ft3 =lb w fracture width, ft, or theoretical shaft work required to compress the gas, ft-lbf =lbm Wair weight of tubing in air, lb/ft Wc total weight of counterweights, lbs Wf weight of fluid, lbs Wfi weight of fluid inside tubing, lb/ft Wfo weight of fluid displaced by tubing, lb/ft WOR producing water-oil ratio, bbl/stb Wp plunger weight, lbf Ws mechanical shaft work into the system, ft-lbs per lb of fluid ww fracture width at wellbore, in. ww average width, in. X volumetric dissolving power of acid solution, ft3 mineral/ ft3 solution xf fracture half-length, ft xi mole fraction of compound i in the liquid phase x1 free gas quality at upstream, mass fraction ya actual pressure ratio yc critical pressure ratio yi mole fraction of compound i in the vapor phase yL liquid hold up, fraction Z gas compressibility factor in average tubing condition z gas compressibility factor zb gas deviation factor at Tb and pb zd gas deviation factor at discharge of cylinder, or gas compressibility factor at valve depth condition zs gas deviation factor at suction of the cylinder z1 compressibility factor at suction conditions zz the average gas compressibility factor DZ elevation increase, ft Greek Symbols a Biot’s poroelastic constant, approximately 0.7 b gravimetric dissolving power of acid solution, lbm mineral=lbm solution «0 pipe wall roughness, in. f porosity, fraction h pump efficiency g 1.78 ¼ Euler’s constant ga acid specific gravity, water ¼ 1.0 gg gas-specific gravity, air ¼ 1 gL specific gravity of production fluid, water ¼ 1 gm mineral specific gravity, water ¼ 1.0 go oil specific gravity, water ¼ 1 goST specific gravity of stock-tank oil, water ¼ 1 gS specific weight of steel (490 lb=ft3 ) gs specific gravity of produced solid, water ¼ 1 gw specific gravity of produced water, fresh water ¼ 1 m viscosity ma viscosity of acid solution, cp mod viscosity of dead oil, cp LIST OF SYMBOLS xiii Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xiii 29.12.2006 10:39am
- 6. mf viscosity of the effluent at the inlet temperature, cp mG gas viscosity, cp mg gas viscosity at in-situ temperature and pressure, cp mL liquid viscosity, cp mo viscosity of oil, cp ms viscosity of the effluent at the surface temperature, cp n Poison’s ratio na stoichiometry number of acid nm stoichiometry number of mineral npf viscosity of power fluid, centistokes u inclination angle, deg., or dip angle from horizontal direction, deg. r fluid density lbm=ft3 r1 mixture density at top of tubing segment, lbf =ft3 r2 mixture density at bottom of segment, lbf =ft3 ra density of acid, lbm=ft3 rair density of air, lbm=ft3 rG in-situ gas density, lbm=ft3 rL liquid density, lbm=ft3 rm density of mineral, lbm=ft3 rm2 mixture density at downstream, lbm=ft3 ro,st density of stock tank oil, lbm=ft3 rw density of fresh water, 62:4 lbm=ft3 rwh density of fluid at wellhead, lbm=ft3 ri density of fluid from/into layer i, lbm=ft3 rr average mixture density (specific weight), lbf =ft3 s liquid-gas interfacial tension, dyne/cm s1 axial principal stress, psi, s2 tangential principal stress, psi s3 radial principal stress, psi sb bending stress, psi sv overburden stress, psi s 0 v effective vertical stress, psi xiv LIST OF SYMBOLS Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xiv 29.12.2006 10:39am
- 7. List of Tables Table 2.1: Result Given by the Spreadsheet Program OilProperties.xls Table 2.2: Results Given by the Spreadsheet Program MixingRule.xls Table 2.3: Results Given by the Spreadsheet Carr- Kobayashi-Burrows-GasViscosity.xls Table 2.4: Results Given by the Spreadsheet Program Brill.Beggs.Z.xls Table 2.5: Results Given by the Spreadsheet Program Hall.Yarborogh.z.xls Table 3.1: Summary of Test Points for Nine Oil Layers Table 3.2: Comparison of Commingled and Layer- Grouped Productions Table 4.1: Result Given by Poettmann-Carpenter BHP.xls for Example Problem 4.2 Table 4.2: Result Given by Guo.GhalamborBHP.xls for Example Problem 4.3 Table 4.3: Result Given by HagedornBrown Correlation.xls for Example Problem 4.4 Table 4.4: Spreadsheet Average TZ.xls for the Data Input and Results Sections Table 4.5: Appearance of the Spreadsheet Cullender. Smith.xls for the Data Input and Results Sections Table 5.1: Solution Given by the Spreadsheet Program GasUpChokePressure.xls Table 5.2: Solution Given by the Spreadsheet Program GasDownChokePressure.xls Table 5.3: A Summary of C, m and n Values Given by Different Researchers Table 5.4: An Example Calculation with Sachdeva’s Choke Model Table 6.1: Result Given by BottomHoleNodalGas.xls for Example Problem 6.1 Table 6.2: Result Given by BottomHoleNodalOil- PC.xls for Example Problem 6.2 Table 6.3: Result Given by BottomHoleNodaloil-GG. xls. for Example of Problem 6.2 Table 6.4: Solution Given by BottomHoleNodalOil- HB.xls Table 6.5: Solution Given by WellheadNodalGas- SonicFlow.xls. Table 6.6: Solution Given by WellheadNodalOil-PC.xls Table 6.7: Solution Given by WellheadNodalOil- GG.xls Table 6.8: Solution Given by WellheadNodalOil- HB.xls. Table 6.9: Solution Given by MultilateralGasWell Deliverability (Radial-Flow IPR).xls Table 6.10: Data Input and Result Sections of the Spreadsheet MultilateralOilWell Deliverability.xls Table 7.1: Sroduction Forecast Given by Transient ProductionForecast.xls Table 7.2: Production Forecast for Example Problem 7.2 Table 7.3: Oil Production Forecast for N ¼ 1 Table 7.4: Gas Production Forecast for N ¼ 1 Table 7.5: Production schedule forecast Table 7.6: Result of Production Forecast for Example Problem 7.4 Table 8.1: Production Data for Example Problem 8.2 Table 8.2: Production Data for Example Problem 8.3 Table 8.3: Production Data for Example Problem 8.4 Table 9.1: API Tubing Tensile Requirements Table 10.1: K-Values Used for Selecting Separators Table 10.2: Retention Time Required Under Various Separation Conditions Table 10.3: Settling Volumes of Standard Vertical High-Pressure Separators Table 10.4: Settling Volumes of Standard Vertical Low-Pressure Separators Table 10.5: Settling Volumes of Standard Horizontal High-Pressure Separators Table 10.6: Settling Volumes of Standard Horizontal Low-Pressure Separators Table 10.7: Settling Volumes of Standard Spherical High-Pressure Separators Table 10.8: Settling Volumes of Standard Spherical Low-Pressure Separators (125 psi) Table 10.9: Temperature Correction Factors for Trayed Glycol Contactors Table 10.10: Specific Gravity Correction Factors for Trayed Glycol Contactors Table 10.11: Temperature Correction Factors for Packed Glycol Contactors Table 10.12: Specific Gravity Correction Factors for Packed Glycol Contactors Table 11.1: Typical Values of Pipeline Efficiency Factors Table 11.2: Design and Hydrostatic Pressure Definitions and Usage Factors for Oil Lines Table 11.3: Design and Hydrostatic Pressure Definitions and Usage Factors for Gas Lines Table 11.4: Thermal Conductivities of Materials Used in Pipeline Insulation Table 11.5: Typical Performance of Insulated Pipelines Table 11.6: Base Data for Pipeline Insulation Design Table 11.7: Calculated Total Heat Losses for the Insulated Pipelines (kW) Table 12.1: Conventional Pumping Unit API Geometry Dimensions Table 12.2: Solution Given by Computer Program SuckerRodPumpingLoad.xls Table 12.3: Solution Given by SuckerRodPumping FlowratePower.xls Table 12.4: Design Data for API Sucker Rod Pumping Units Table 13.1: Result Given by Computer Program CompressorPressure.xls Table 13.2: Result Given by Computer Program ReciprocatingCompressorPower.xls for the First Stage Compression Table 13.3: Result Given by the Computer Program CentrifugalCompressorPower.xls Table 13.4: R Values for Otis Spreadmaster Valves Table 13.5: Summary of Results for Example Problem 13.7 Table 14.1: Result Given by the Computer Spreadsheet ESPdesign.xls Table 14.2: Solution Given by HydraulicPiston Pump.xls Table 14.3: Summary of Calculated Parameters Table 14.4: Solution Given by Spreadsheet Program PlungerLift.xls Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xv 29.12.2006 10:39am
- 8. Table 15.1: Basic Parameter Values for Example Problem 15.1 Table 15.2: Result Given by the Spreadsheet Program GasWellLoading.xls Table 16.1: Primary Chemical Reactions in Acid Treatments Table 16.2: Recommended Acid Type and Strength for Sandstone Acidizing Table 16.3: Recommended Acid Type and Strength for Carbonate Acidizing Table 17.1: Features of Fracture Geometry Models Table 17.2: Summary of Some Commercial Fracturing Models Table 17.3: Calculated Slurry Concentration Table 18.1: Flash Calculation with Standing’s Method for ki Values Table 18.2: Solution to Example Problem 18.3 Given by the Spreadsheet LoopedLines.xls Table 18.3: Gas Lift Performance Data for Well A and Well B Table 18.4: Assignments of Different Available Lift Gas Injection Rates to Well A and Well B xvi LIST OF TABLES Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xvi 29.12.2006 10:39am
- 9. List of Figures Figure 1.1: A sketch of a petroleum production system. Figure 1.2: A typical hydrocarbon phase diagram. Figure 1.3: A sketch of a water-drive reservoir. Figure 1.4: A sketch of a gas-cap drive reservoir. Figure 1.5: A sketch of a dissolved-gas drive reservoir. Figure 1.6: A sketch of a typical flowing oil well. Figure 1.7: A sketch of a wellhead. Figure 1.8: A sketch of a casing head. Figure 1.9: A sketch of a tubing head. Figure 1.10: A sketch of a ‘‘Christmas tree.’’ Figure 1.11: Sketch of a surface valve. Figure 1.12: A sketch of a wellhead choke. Figure 1.13: Conventional horizontal separator. Figure 1.14: Double action piston pump. Figure 1.15: Elements of a typical reciprocating compressor. Figure 1.16: Uses of offshore pipelines. Figure 1.17: Safety device symbols. Figure 1.18: Safety system designs for surface wellhead flowlines. Figure 1.19: Safety system designs for underwater wellhead flowlines. Figure 1.20: Safety system design for pressure vessel. Figure 1.21: Safety system design for pipeline pumps. Figure 1.22: Safety system design for other pumps. Figure 3.1: A sketch of a radial flow reservoir model: (a) lateral view, (b) top view. Figure 3.2: A sketch of a reservoir with a constant- pressure boundary. Figure 3.3: A sketch of a reservoir with no-flow boundaries. Figure 3.4: (a) Shape factors for various closed drainage areas with low-aspect ratios. (b) Shape factors for closed drainage areas with high-aspect ratios. Figure 3.5: A typical IPR curve for an oil well. Figure 3.6: Transient IPR curve for Example Problem 3.1. Figure 3.7: Steady-state IPR curve for Example Problem 3.1. Figure 3.8: Pseudo–steady-state IPR curve for Example Problem 3.1. Figure 3.9: IPR curve for Example Problem 3.2. Figure 3.10: Generalized Vogel IPR model for partial two-phase reservoirs. Figure 3.11: IPR curve for Example Problem 3.3. Figure 3.12: IPR curves for Example Problem 3.4, Well A. Figure 3.13: IPR curves for Example Problem 3.4, Well B Figure 3.14: IPR curves for Example Problem 3.5. Figure 3.15: IPR curves of individual layers. Figure 3.16: Composite IPR curve for all the layers open to flow. Figure 3.17: Composite IPR curve for Group 2 (Layers B4, C1, and C2). Figure 3.18: Composite IPR curve for Group 3 (Layers B1, A4, and A5). Figure 3.19: IPR curves for Example Problem 3.6. Figure 3.20: IPR curves for Example Problem 3.7. Figure 4.1: Flow along a tubing string. Figure 4.2: Darcy–Wiesbach friction factor diagram. Figure 4.3: Flow regimes in gas-liquid flow. Figure 4.4: Pressure traverse given by Hagedorn BrownCorreltion.xls for Example. Figure 4.5: Calculated tubing pressure profile for Example Problem 4.5. Figure 5.1: A typical choke performance curve. Figure 5.2: Choke flow coefficient for nozzle-type chokes. Figure 5.3: Choke flow coefficient for orifice-type chokes. Figure 6.1: Nodal analysis for Example Problem 6.1. Figure 6.2: Nodal analysis for Example Problem 6.4. Figure 6.3: Nodal analysis for Example Problem 6.5. Figure 6.4: Nodal analysis for Example Problem 6.6. Figure 6.5: Nodal analysis for Example Problem 6.8. Figure 6.6: Schematic of a multilateral well trajectory. Figure 6.7: Nomenclature of a multilateral well. Figure 7.1: Nodal analysis plot for Example Problem 7.1. Figure 7.2: Production forecast for Example Problem 7.2. Figure 7.3: Nodal analysis plot for Example Problem 7.2. Figure 7.4: Production forecast for Example Problem 7.2 Figure 7.3: Production forecast for Example Problem 7.3. Figure 7.4: Result of production forecast for Example Problem 7.4. Figure 8.1: A semilog plot of q versus t indicating an exponential decline. Figure 8.2: A plot of Np versus q indicating an exponential decline. Figure 8.3: A plot of log(q) versus log(t) indicating a harmonic decline. Figure 8.4: A plot of Np versus log(q) indicating a harmonic decline. Figure 8.5: A plot of relative decline rate versus production rate. Figure 8.6: Procedure for determining a- and b-values. Figure 8.7: A plot of log(q) versus t showing an exponential decline. Figure 8.8: Relative decline rate plot showing exponential decline. Figure 8.9: Projected production rate by an exponential decline model. Figure 8.10: Relative decline rate plot showing harmonic decline. Figure 8.11: Projected production rate by a harmonic decline model. Figure 8.12: Relative decline rate plot showing hyperbolic decline. Figure 8.13: Relative decline rate plot showing hyperbolic decline. Figure 8.14: Projected production rate by a hyperbolic decline model. Figure 9.1: A simple uniaxial test of a metal specimen. Figure 9.2: Effect of tension stress on tangential stress. Figure 9.3: Tubing–packer relation. Figure 9.4: Ballooning and buckling effects. Figure 10.1: A typical vertical separator. Figure 10.2: A typical horizontal separator. Figure 10.3: A typical horizontal double-tube separator. Figure 10.4: A typical horizontal three-phase separator. Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xvii 29.12.2006 10:39am
- 10. Figure 10.5: A typical spherical low-pressure separator. Figure 10.6: Water content of natural gases. Figure 10.7: Flow diagram of a typical solid desiccant dehydration plant. Figure 10.8: Flow diagram of a typical glycol dehydrator. Figure 10.9: Gas capacity of vertical inlet scrubbers based on 0.7-specific gravity at 100 8F. Figure 10.10: Gas capacity for trayed glycol contactors based on 0.7-specific gravity at 100 8F. Figure 10.11: Gas capacity for packed glycol contactors based on 0.7-specific gravity at 100 8F. Figure 10.12: The required minimum height of packing of a packed contactor, or the minimum number of trays of a trayed contactor. Figure 11.1: Double-action stroke in a duplex pump. Figure 11.2: Single-action stroke in a triplex pump. Figure 11.3: Elements of a typical reciprocating compressor. Figure 11.4: Cross-section of a centrifugal compressor. Figure 11.5: Basic pressure–volume diagram. Figure 11.6: Flow diagram of a two-stage compression unit. Figure 11.7: Fuel consumption of prime movers using three types of fuel. Figure 11.8: Fuel consumption of prime movers using natural gas as fuel. Figure 11.9: Effect of elevation on prime mover power. Figure 11.10: Darcy–Wiesbach friction factor chart. Figure 11.11: Stresses generated by internal pressure p in a thin-wall pipe, D=t 20. Figure 11.12: Stresses generated by internal pressure p in a thick-wall pipe, D=t 20. Figure 11.13: Calculated temperature profiles with a polyethylene layer of 0.0254 M (1 in.). Figure 11.14: Calculated steady-flow temperature profiles with polyethylene layers of various thicknesses. Figure 11.15: Calculated temperature profiles with a polypropylene layer of 0.0254 M (1 in.). Figure 11.16: Calculated steady-flow temperature profiles with polypropylene layers of various thicknesses. Figure 11.17: Calculated temperature profiles with a polyurethane layer of 0.0254 M (1 in.). Figure 11.18: Calculated steady-flow temperature profiles with polyurethane layers of four thicknesses. Figure 12.1: A diagrammatic drawing of a sucker rod pumping system. Figure 12.2: Sketch of three types of pumping units: (a) conventional unit; (b) Lufkin Mark II unit; (c) air-balanced unit. Figure 12.3: The pumping cycle: (a) plunger moving down, near the bottom of the stroke; (b) plunger moving up, near the bottom of the stroke; (c) plunger moving up, near the top of the stroke; (d) plunger moving down, near the top of the stroke. Figure 12.4: Two types of plunger pumps. Figure 12.5: Polished rod motion for (a) conventional pumping unit and (b) air-balanced unit. Figure 12.6: Definitions of conventional pumping unit API geometry dimensions. Figure 12.7: Approximate motion of connection point between pitman arm and walking beam. Figure 12.8: Sucker rod pumping unit selection chart. Figure 12.9: A sketch of pump dynagraph. Figure 12.10: Pump dynagraph cards: (a) ideal card, (b) gas compression on down-stroke, (c) gas expansion on upstroke, (d) fluid pound, (e) vibration due to fluid pound, (f) gas lock. Figure 12.11: Surface Dynamometer Card: (a) ideal card (stretch and contraction), (b) ideal card (acceleration), (c) three typical cards. Figure 12.12: Strain-gage–type dynamometer chart. Figure 12.13: Surface to down hole cards derived from surface dynamometer card. Figure 13.1: Configuration of a typical gas lift well. Figure 13.2: A simplified flow diagram of a closed rotary gas lift system for single intermittent well. Figure 13.3: A sketch of continuous gas lift. Figure 13.4: Pressure relationship in a continuous gas lift. Figure 13.5: System analysis plot given by GasLift Potential.xls for the unlimited gas injection case. Figure 13.6: System analysis plot given by GasLift Potential.xls for the limited gas injection case. Figure 13.7: Well unloading sequence. Figure 13.8: Flow characteristics of orifice-type valves. Figure 13.9: Unbalanced bellow valve at its closed condition. Figure 13.10: Unbalanced bellow valve at its open condition. Figure 13.11: Flow characteristics of unbalanced valves. Figure 13.12: A sketch of a balanced pressure valve. Figure 13.13: A sketch of a pilot valve. Figure 13.14: A sketch of a throttling pressure valve. Figure 13.15: A sketch of a fluid-operated valve. Figure 13.16: A sketch of a differential valve. Figure 13.17: A sketch of combination valve. Figure 13.18: A flow diagram to illustrate procedure of valve spacing. Figure 13.19: Illustrative plot of BHP of an intermittent flow. Figure 13.20: Intermittent flow gradient at mid-point of tubing. Figure 13.21: Example Problem 13.8 schematic and BHP build.up for slug flow. Figure 13.22: Three types of gas lift installations. Figure 13.23: Sketch of a standard two-packer chamber. Figure 13.24: A sketch of an insert chamber. Figure 13.25: A sketch of a reserve flow chamber. Figure 14.1: A sketch of an ESP installation. Figure 14.2: An internal schematic of centrifugal pump. Figure 14.3: A sketch of a multistage centrifugal pump. Figure 14.4: A typical ESP characteristic chart. Figure 14.5: A sketch of a hydraulic piston pump. Figure 14.6: Sketch of a PCP system. Figure 14.7: Rotor and stator geometry of PCP. Figure 14.8: Four flow regimes commonly encountered in gas wells. Figure 14.9: A sketch of a plunger lift system. Figure 14.10: Sketch of a hydraulic jet pump installation. Figure 14.11: Working principle of a hydraulic jet pump. Figure 14.12: Example jet pump performance chart. Figure 15.1: Temperature and spinner flowmeter- derived production profile. Figure 15.2: Notations for a horizontal wellbore. xviii LIST OF FIGURES Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xviii 29.12.2006 10:39am
- 11. Figure 15.3: Measured bottom-hole pressures and oil production rates during a pressure drawdown test. Figure 15.4: Log-log diagnostic plot of test data. Figure 15.5: Semi-log plot for vertical radial flow analysis. Figure 15.6: Square-root time plot for pseudo-linear flow analysis. Figure 15.7: Semi-log plot for horizontal pseudo- radial flow analysis. Figure 15.8: Match between measured and model calculated pressure data. Figure 15.9: Gas production due to channeling behind the casing. Figure 15.10: Gas production due to preferential flow through high-permeability zones. Figure 15.11: Gas production due to gas coning. Figure 15.12: Temperature and noise logs identifying gas channeling behind casing. Figure 15.13: Temperature and fluid density logs identifying a gas entry zone. Figure 15.14: Water production due to channeling behind the casing. Figure 15.15: Preferential water flow through high- permeability zones. Figure 15.16: Water production due to water coning. Figure 15.17: Prefracture and postfracture temperature logs identifying fracture height. Figure 15.18: Spinner flowmeter log identifying a watered zone at bottom. Figure 15.19: Calculated minimum flow rates with Turner et al.’s model and test flow rates. Figure 15.20: The minimum flow rates given by Guo et al.’s model and the test flow rates. Figure 16.1: Typical acid response curves. Figure 16.2: Wormholes created by acid dissolution of limestone. Figure 17.1: Schematic to show the equipment layout in hydraulic fracturing treatments of oil and gas wells. Figure 17.2: A schematic to show the procedure of hydraulic fracturing treatments of oil and gas wells. Figure 17.3: Overburden formation of a hydrocarbon reservoir. Figure 17.4: Concept of effective stress between grains. Figure 17.5: The KGD fracture geometry. Figure 17.6: The PKN fracture geometry. Figure 17.7: Relationship between fracture conductivity and equivalent skin factor. Figure 17.8: Relationship between fracture conductivity and equivalent skin factor. Figure 17.9: Effect of fracture closure stress on proppant pack permeability. Figure 17.10: Iteration procedure for injection time calculation. Figure 17.11: Calculated slurry concentration. Figure 17.12: Bottom-hole pressure match with three- dimensional fracturing model PropFRAC. Figure 17.13: Four flow regimes that can occur in hydraulically fractured reservoirs. Figure 18.1: Comparison of oil well inflow performance relationship (IPR) curves before and after stimulation. Figure 18.2: A typical tubing performance curve. Figure 18.3: A typical gas lift performance curve of a low-productivity well. Figure 18.4: Theoretical load cycle for elastic sucker rods. Figure 18.5: Actual load cycle of a normal sucker rod. Figure 18.6: Dimensional parameters of a dynamometer card. Figure 18.7: A dynamometer card indicating synchronous pumping speeds. Figure 18.8: A dynamometer card indicating gas lock. Figure 18.9: Sketch of (a) series pipeline and (b) parallel pipeline. Figure 18.10: Sketch of a looped pipeline. Figure 18.11: Effects of looped line and pipe diameter ratio on the increase of gas flow rate. Figure 18.12: A typical gas lift performance curve of a high-productivity well. Figure 18.13: Schematics of two hierarchical networks. Figure 18.14: An example of a nonhierarchical network. LIST OF FIGURES xix Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Guo-prelims Final Proof page xix 29.12.2006 10:39am
- 12. Table of Contents Preface List of Symbols List of Tables List of Figures Part I: Petroleum Production Engineering Fundamentals: Chapter 1: Petroleum Production System Chapter 2: Properties of Oil and Natural Gas Chapter 3: Reservoir Deliverability Chapter 4: Wellbore Performance Chapter 5: Choke Performance Chapter 6: Well Deliverability Chapter 7: Forecast of Well Production Chapter 8: Production Decline Analysis Part II: Equipment Design and Selection Chapter 9: Well Tubing Chapter 10: Separation Systems Chapter 11: Transportation Systems
- 13. Part III: Artificial Lift Methods Chapter 12: Sucker Rod Pumping Chapter 13: Gas Lift Chapter 14: Other Artificial Lift Methods Part IV: Production Enhancement Chapter 15: Well Problem Identification Chapter 16: Matrix Acidizing Chapter 17: Hydraulic Fracturing Chapter 18: Production Optimization Appendix A: Unit Conversion Factors Appendix B: The Minimum Performance Properties of API Tubing
- 14. Part I Petroleum Production Engineering Fundamentals The upstream of the petroleum industry involves itself in the business of oil and gas exploration and production (E P) activities. While the exploration activities find oil and gas reserves, the production activities deliver oil and gas to the downstream of the industry (i.e., processing plants). The petroleum production is definitely the heart of the petroleum industry. Petroleum production engineering is that part of petroleum engineering that attempts to maxi- mize oil and gas production in a cost-effective manner. To achieve this objective, production engineers need to have a thorough understanding of the petroleum production systems with which they work. To perform their job correctly, production engineers should have solid back- ground and sound knowledge about the properties of fluids they produce and working principles of all the major components of producing wells and surface facilities. This part of the book provides graduating production engineers with fundamentals of petroleum production engineering. Materials are presented in the following eight chapters: Chapter 1 Petroleum Production System 1/3 Chapter 2 Properties of Oil and Natural Gas 2/19 Chapter 3 Reservoir Deliverability 3/29 Chapter 4 Wellbore Performance 4/45 Chapter 5 Choke Performance 5/59 Chapter 6 Well Deliverability 6/69 Chapter 7 Forecast of Well Production 7/87 Chapter 8 Production Decline Analysis 8/97 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 1 4.1.2007 6:12pm Compositor Name: SJoearun
- 15. 1 Petroleum Production System Contents 1.1 Introduction 1/4 1.2 Reservoir 1/4 1.3 Well 1/5 1.4 Separator 1/8 1.5 Pump 1/9 1.6 Gas Compressor 1/10 1.7 Pipelines 1/11 1.8 Safety Control System 1/11 1.9 Unit Systems 1/17 Summary 1/17 References 1/17 Problems 1/17 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 3 4.1.2007 6:12pm Compositor Name: SJoearun
- 16. 1.1 Introduction The role of a production engineer is to maximize oil and gas production in a cost-effective manner. Familiarization and understanding of oil and gas production systems are essential to the engineers. This chapter provides graduat- ing production engineers with some basic knowledge about production systems. More engineering principles are discussed in the later chapters. As shown in Fig. 1.1, a complete oil or gas production system consists of a reservoir, well, flowline, separators, pumps, and transportation pipelines. The reservoir sup- plies wellbore with crude oil or gas. The well provides a path for the production fluid to flow from bottom hole to surface and offers a means to control the fluid production rate. The flowline leads the produced fluid to separators. The separators remove gas and water from the crude oil. Pumps and compressors are used to transport oil and gas through pipelines to sales points. 1.2 Reservoir Hydrocarbon accumulations in geological traps can be clas- sified as reservoir, field, and pool. A ‘‘reservoir’’ is a porous and permeable underground formation containing an indi- vidual bank of hydrocarbons confined by impermeable rock or water barriers and is characterized by a single natural pressure system. A ‘‘field’’ is an area that consists of one or more reservoirs all related to the same structural feature. A ‘‘pool’’ contains one or more reservoirs in isolated structures. Depending on the initial reservoir condition in the phase diagram (Fig. 1.2), hydrocarbon accumulations are classi- fied as oil, gas condensate, and gas reservoirs. An oil that is at a pressure above its bubble-point pressure is called an ‘‘undersaturated oil’’ because it can dissolve more gas at the given temperature. An oil that is at its bubble-point pressure is called a ‘‘saturated oil’’ because it can dissolve no more gas at the given temperature. Single (liquid)-phase flow prevails in an undersaturated oil reservoir, whereas two-phase (liquid oil and free gas) flow exists in a sat- urated oil reservoir. Wells in the same reservoir can fall into categories of oil, condensate, and gas wells depending on the producing gas–oilratio(GOR).GaswellsarewellswithproducingGOR being greater than 100,000 scf/stb; condensate wells are those with producing GOR being less than 100,000 scf/stb but greater than 5,000 scf/stb; and wells with producing GOR being less than 5,000 scf/stb are classified as oil wells. Oil reservoirs can be classified on the basis of boundary type, which determines driving mechanism, and which are as follows: . Water-drive reservoir . Gas-cap drive reservoir . Dissolved-gas drive reservoir In water-drive reservoirs, the oil zone is connected by a continuous path to the surface groundwater system (aqui- fer). The pressure caused by the ‘‘column’’ of water to the surface forces the oil (and gas) to the top of the reservoir against the impermeable barrier that restricts the oil and gas (the trap boundary). This pressure will force the oil and gas toward the wellbore. With the same oil production, reservoir pressure will be maintained longer (relative to other mech- anisms of drive) when there is an active water drive. Edge- water drive reservoir is the most preferable type of reservoir compared to bottom-water drive. The reservoir pressure can remain at its initial value above bubble-point pressure so that single-phase liquid flow exists in the reservoir for maximum well productivity. A steady-state flow condition can prevail in a edge-water drive reservoir for a long time before water breakthrough into the well. Bottom-water drive reservoir (Fig. 1.3) is less preferable because of water-coning problems that can affect oil production economics due to water treat- ment and disposal issues. Wellbore Reservoir Separator Wellhead Pwf PeP Gas Oil Water Figure 1.1 A sketch of a petroleum production system. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 4 4.1.2007 6:12pm Compositor Name: SJoearun 1/4 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 17. In a gas-cap drive reservoir, gas-cap drive is the drive mechanism where the gas in the reservoir has come out of solution and rises to the top of the reservoir to form a gas cap (Fig. 1.4). Thus, the oil below the gas cap can be produced. If the gas in the gas cap is taken out of the reservoir early in the production process, the reservoir pressure will decrease rapidly. Sometimes an oil reservoir is subjected to both water and gas-cap drive. A dissolved-gas drive reservoir (Fig. 1.5) is also called a ‘‘solution-gas drive reservoir’’ and ‘‘volumetric reservoir.’’ The oil reservoir has a fixed oil volume surrounded by no- flow boundaries (faults or pinch-outs). Dissolved-gas drive is the drive mechanism where the reservoir gas is held in solution in the oil (and water). The reservoir gas is actually in a liquid form in a dissolved solution with the liquids (at atmospheric conditions) from the reservoir. Compared to the water- and gas-drive reservoirs, expansion of solution (dissolved) gas in the oil provides a weak driving mech- anism in a volumetric reservoir. In the regions where the oil pressure drops to below the bubble-point pressure, gas escapes from the oil and oil–gas two-phase flow exists. To improve oil recovery in the solution-gas reservoir, early pressure maintenance is usually preferred. 1.3 Well Oil and gas wells are drilled like an upside-down telescope. The large-diameter borehole section is at the top of the well. Each section is cased to the surface, or a liner is placed in the well that laps over the last casing in the well. Each casing or liner is cemented into the well (usually up to at least where the cement overlaps the previous cement job). The last casing in the well is the production casing (or production liner). Once the production casing has been cemented into the well, the production tubing is run into the well. Usually a packer is used near the bottom of the tubing to isolate the annulus between the outside of the tubing and the inside of the casing. Thus, the produced fluids are forced to move out of the perforation into the bottom of the well and then into the inside of the tubing. Packers can be actuated by either mechanical or hydraulic mechanisms. The production tubing is often (particularly during initial well flow) provided with a bottom-hole choke to control the initial well flow (i.e., to restrict over- production and loss of reservoir pressure). Figure 1.6 shows a typical flowing oil well, defined as a well producing solely because of the natural pressure of the reservoir. It is composed of casings, tubing, packers, down-hole chokes (optional), wellhead, Christmas tree, and surface chokes. 0 50 150 200 250 300 350100 1,000 500 1,500 2,000 2,500 3,000 3,500 4,000 Reservoir Temperature (ЊF) ReservoirPressure(psia) Liquid Volume 40% 20% 10% 80% 5% 0% Bubble Point Gas Reservoirs Retrograde Condensate Reservoirs Critical Point pi, T ptf, Ttf pwf, Twf Dew Point Cricondentherm Point Figure 1.2 A typical hydrocarbon phase diagram. Water WOC Oil Figure 1.3 A sketch of a water-drive reservoir. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 5 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/5
- 18. Most wells produce oil through tubing strings, mainly because a tubing string provides good sealing performance and allows the use of gas expansion to lift oil. The Ameri- can Petroleum Institute (API) defines tubing size using nominal diameter and weight (per foot). The nominal diameter is based on the internal diameter of the tubing body. The weight of tubing determines the tubing outer diameter. Steel grades of tubing are designated H-40, J-55, C-75, L-80, N-80, C-90, and P-105, where the digits repre- sent the minimum yield strength in 1,000 psi. The min- imum performance properties of tubing are given in Chapter 9 and Appendix B. The ‘‘wellhead’’ is defined as the surface equipment set below the master valve. As we can see in Fig. 1.7, it includes casing heads and a tubing head. The casing head (lowermost) is threaded onto the surface casing. This can also be a flanged or studded connection. A ‘‘casing head’’ is a mechanical assembly used for hanging a casing string (Fig. 1.8). Depending on casing programs in well drilling, several casing heads can be installed during well construc- tion. The casing head has a bowl that supports the casing hanger. This casing hanger is threaded onto the top of the production casing (or uses friction grips to hold the cas- ing). As in the case of the production tubing, the produc- tion casing is landed in tension so that the casing hanger actually supports the production casing (down to the freeze point). In a similar manner, the intermediate cas- ing(s) are supported by their respective casing hangers (and bowls). All of these casing head arrangements are supported by the surface casing, which is in compression and cemented to the surface. A well completed with three casing strings has two casing heads. The uppermost casing head supports the production casing. The lowermost cas- ing head sits on the surface casing (threaded to the top of the surface casing). Most flowing wells are produced through a string of tubing run inside the production casing string. At the surface, the tubing is supported by the tubing head (i.e., the tubing head is used for hanging tubing string on the production casing head [Fig. 1.9]). The tubing head sup- ports the tubing string at the surface (this tubing is landed on the tubing head so that it is in tension all the way down to the packer). The equipment at the top of the producing wellhead is called a ‘‘Christmas tree’’ (Fig. 1.10) and it is used to control flow. The ‘‘Christmas tree’’ is installed above the tubing head. An ‘‘adaptor’’ is a piece of equipment used to join the two. The ‘‘Christmas tree’’ may have one flow outlet (a tee) or two flow outlets (a cross). The master valve is installed below the tee or cross. To replace a master valve, the tubing must be plugged. A Christmas tree consists of a main valve, wing valves, and a needle valve. These valves are used for closing the well when needed. At the top of the tee structure (on the top of the ‘‘Christmas tree’’), there is a pressure gauge that indicates the pressure in the tubing. Gas Cap Oil Figure 1.4 A sketch of a gas-cap drive reservoir. Oil and Gas Reservoir Figure 1.5 A sketch of a dissolved-gas drive reservoir. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 6 4.1.2007 6:12pm Compositor Name: SJoearun 1/6 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 19. The wing valves and their gauges allow access (for pressure measurements and gas or liquid flow) to the annulus spaces (Fig. 1.11). ‘‘Surface choke’’ (i.e., a restriction in the flowline) is a piece of equipment used to control the flow rate (Fig. 1.12). In most flowing wells, the oil production rate is altered by adjusting the choke size. The choke causes back-pressure in the line. The back-pressure (caused by the chokes or other restrictions in the flowline) increases the bottom- hole flowing pressure. Increasing the bottom-hole flowing pressure decreases the pressure drop from the reservoir to the wellbore (pressure drawdown). Thus, increasing the back-pressure in the wellbore decreases the flow rate from the reservoir. In some wells, chokes are installed in the lower section of tubing strings. This choke arrangement reduces well- head pressure and enhances oil production rate as a result of gas expansion in the tubing string. For gas wells, use of down-hole chokes minimizes the gas hydrate problem in the well stream. A major disadvantage of using down-hole chokes is that replacing a choke is costly. Certain procedures must be followed to open or close a well. Before opening, check all the surface equipment such as safety valves, fittings, and so on. The burner of a line heater must be lit before the well is opened. This is neces- sary because the pressure drop across a choke cools the fluid and may cause gas hydrates or paraffin to deposit out. A gas burner keeps the involved fluid (usually water) hot. Fluid from the well is carried through a coil of piping. The choke is installed in the heater. Well fluid is heated both before and after it flows through the choke. The upstream heating helps melt any solids that may be present in the producing fluid. The downstream heating prevents hydrates and paraffins from forming at the choke. Surface vessels should be open and clear before the well is allowed to flow. All valves that are in the master valve and other downstream valves are closed. Then follow the following procedure to open a well: 1. The operator barely opens the master valve (just a crack), and escaping fluid makes a hissing sound. When the fluid no longer hisses through the valve, the pressure has been equalized, and then the master valve is opened wide. 2. If there are no oil leaks, the operator cracks the next downstream valve that is closed. Usually, this will be Casing Perforation Wellhead Oil Reservoir Packer Bottom-hole Choke Tubing Annulus Production Casing Intermediate Casing Surface Casing Cement Wellbore Reservoir Figure 1.6 A sketch of a typical flowing oil well. Choke Wing Valve Master Valve Tubing Pressure Gauge Flow Fitting Tubing Intermediate Casing Surface CasingLowermost Casing Head Uppermost Casing Head Casing Valve Casing Pressure Gauge Production Casing Tubing Head Figure 1.7 A sketch of a wellhead. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 7 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/7
- 20. either the second (backup) master valve or a wing valve. Again, when the hissing sound stops, the valve is opened wide. 3. The operator opens the other downstream valves the same way. 4. To read the tubing pressure gauge, the operator must open the needle valve at the top of the Christmas tree. After reading and recording the pressure, the operator may close the valve again to protect the gauge. The procedure for ‘‘shutting-in’’ a well is the opposite of the procedure for opening a well. In shutting-in the well, the master valve is closed last. Valves are closed rather rapidly to avoid wearing of the valve (to prevent erosion). At least two valves must be closed. 1.4 Separator The fluids produced from oil wells are normally complex mixtures of hundreds of different compounds. A typical oil well stream is a high-velocity, turbulent, constantly expanding mixture of gases and hydrocarbon liquids, in- timately mixed with water vapor, free water, and some- times solids. The well stream should be processed as soon as possible after bringing them to the surface. Separators are used for the purpose. Three types of separators are generally available from manufacturers: horizontal, vertical, and spherical sep- arators. Horizontal separators are further classified into Bowl Production Casing Casing Head Surface Casing Casing Hanger Figure 1.8 A sketch of a casing head. Bowl Seal Tubing Hanger Tubing Head Figure 1.9 A sketch of a tubing head. Choke Wing Valve Wing Valve Choke Master Valve Tubing Head Adaptor Swabbing Valve Top Connection Gauge Valve Flow Fitting Figure 1.10 A sketch of a ‘‘Christmas tree.’’ Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 8 4.1.2007 6:12pm Compositor Name: SJoearun 1/8 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 21. two categories: single tube and double tube. Each type of separator has specific advantages and limitations. Selec- tion of separator type is based on several factors including characteristics of production steam to be treated, floor space availability at the facility site, transportation, and cost. Horizontal separators (Fig. 1.13) are usually the first choice because of their low costs. Horizontal separators are almost widely used for high-GOR well streams, foam- ing well streams, or liquid-from-liquid separation. They have much greater gas–liquid interface because of a large, long, baffled gas-separation section. Horizontal sep- arators are easier to skid-mount and service and require less piping for field connections. Individual separators can be stacked easily into stage-separation assemblies to min- imize space requirements. In horizontal separators, gas flows horizontally while liquid droplets fall toward the liquid surface. The moisture gas flows in the baffle surface and forms a liquid film that is drained away to the liquid section of the separator. The baffles need to be longer than the distance of liquid trajectory travel. The liquid-level control placement is more critical in a horizontal separator than in a vertical separator because of limited surge space. Vertical separators are often used to treat low to inter- mediate GOR well streams and streams with relatively large slugs of liquid. They handle greater slugs of liquid without carryover to the gas outlet, and the action of the liquid-level control is not as critical. Vertical separators occupy less floor space, which is important for facility sites such as those on offshore platforms where space is limited. Because of the large vertical distance between the liquid level and the gas outlet, the chance for liquid to re-vapor- ize into the gas phase is limited. However, because of the natural upward flow of gas in a vertical separator against the falling droplets of liquid, adequate separator diameter is required. Vertical separators are more costly to fabricate and ship in skid-mounted assemblies. Spherical separators offer an inexpensive and compact means of separation arrangement. Because of their com- pact configurations, these types of separators have a very limited surge space and liquid-settling section. Also, the placement and action of the liquid-level control in this type of separator is more critical. Chapter 10 provides more details on separators and dehydrators. 1.5 Pump After separation, oil is transported through pipelines to the sales points. Reciprocating piston pumps are used to provide mechanical energy required for the transportation. There are two types of piston strokes, the single-action Handwheel Packing Gate Port Figure 1.11 A sketch of a surface valve. Wellhead Choke Figure 1.12 A sketch of a wellhead choke. Figure 1.13 Conventional horizontal separator. (Courtesy Petroleum Extension Services.) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 9 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/9
- 22. piston stroke and the double-action piston stroke. The double-action stroke is used for duplex (two pistons) pumps. The single-action stroke is used for pumps with three pistons or greater (e.g., triplex pump). Figure 1.14 shows how a duplex pump works. More information about pumps is presented in Chapter 11. 1.6 Gas Compressor Compressors are used for providing gas pressure required to transport gas with pipelines and to lift oil in gas-lift operations. The compressors used in today’s natural gas production industry fall into two distinct types: reciprocat- ing and rotary compressors. Reciprocating compressors are most commonly used in the natural gas industry. They are built for practically all pressures and volumetric capacities. Asshown in Fig. 1.15, reciprocating compressors have more moving parts and, therefore, lower mechanical efficiencies than rotary compressors. Each cylinder assembly of a recip- rocating compressor consists of a piston, cylinder, cylinder heads, suction and discharge valves, and other parts neces- sary to convert rotary motion to reciprocation motion. A reciprocating compressor is designed for a certain range of compression ratios through the selection of proper piston displacement and clearance volume within the cylinder. This clearance volume can be either fixed or variable, depending on the extent of the operation range and the percent of load variation desired. A typical reciprocating compressor can deliver a volumetric gas flow rate up to 30,000 cubic feet per minute (cfm) at a discharge pressure up to 10,000 psig. Rotary compressors are divided into two classes: the centrifugal compressor and the rotary blower. A centrifu- Discharge Discharge P2 P1 P1 P2 Piston Rod Suction Suction dL dr Ls Piston Figure 1.14 Double-action piston pump. Piston Rod Wrist Pin Suction Valve Discharge Valve Cylinder Head Cylinder Crosshead Connecting Rod Crankshaft Piston Figure 1.15 Elements of a typical reciprocating compressor. (Courtesy Petroleum Extension Services.) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 10 4.1.2007 6:12pm Compositor Name: SJoearun 1/10 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 23. gal compressor consists of a housing with flow passages, a rotating shaft on which the impeller is mounted, bearings, and seals to prevent gas from escaping along the shaft. Centrifugal compressors have few moving parts because only the impeller and shaft rotate. Thus, its efficiency is high and lubrication oil consumption and maintenance costs are low. Cooling water is normally unnecessary be- cause of lower compression ratio and less friction loss. Compression rates of centrifugal compressors are lower because of the absence of positive displacement. Centrifu- gal compressors compress gas using centrifugal force. In this type of compressor, work is done on the gas by an impeller. Gas is then discharged at a high velocity into a diffuser where the velocity is reduced and its kinetic energy is converted to static pressure. Unlike reciprocating com- pressors, all this is done without confinement and physical squeezing. Centrifugal compressors with relatively unre- stricted passages and continuous flow are inherently high- capacity, low-pressure ratio machines that adapt easily to series arrangements within a station. In this way, each compressor is required to develop only part of the station compression ratio. Typically, the volume is more than 100,000 cfm and discharge pressure is up to 100 psig. More information about different types of compressors is provided in Chapter 11. 1.7 Pipelines The first pipeline was built in the United States in 1859 to transport crude oil (Wolbert, 1952). Through the one and half century of pipeline operating practice, the petro- leum industry has proven that pipelines are by far the most economical means of large-scale overland transpor- tation for crude oil, natural gas, and their products, clearly superior to rail and truck transportation over competing routes, given large quantities to be moved on a regular basis. Transporting petroleum fluids with pipelines is a continuous and reliable operation. Pipelines have demonstrated an ability to adapt to a wide variety of environments including remote areas and hostile environ- ments. With very minor exceptions, largely due to local peculiarities, most refineries are served by one or more pipelines, because of their superior flexibility to the alternatives. Figure 1.16 shows applications of pipelines in offshore operations. It indicates flowlines transporting oil and/or gas from satellite subsea wells to subsea manifolds, flow- lines transporting oil and/or gas from subsea manifolds to production facility platforms, infield flowlines transport- ing oil and/or gas from between production facility plat- forms, and export pipelines transporting oil and/or gas from production facility platforms to shore. The pipelines are sized to handle the expected pressure and fluid flow. To ensure desired flow rate of product, pipeline size varies significantly from project to project. To contain the pressures, wall thicknesses of the pipelines range from 3 ⁄8 inch to 11 ⁄2 inch. More information about pipelines is provided in Chapter 11. 1.8 Safety Control System The purpose of safety systems is to protect personnel, the environment, and the facility. The major objective of the safety system is to prevent the release of hydrocarbons Expansion Tie-in Spoolpiece Infield Flowline Riser Tie-in Subsea Manifold Flowlines (several can be bundled) Satellite Subsea Wells Flowlines Export Pipeline Existing Line Pipeline Crossing To Shore Figure 1.16 Uses of offshore pipelines. (Guo et al., 2005.) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 11 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/11
- 24. from the process and to minimize the adverse effects of such releases if they occur. This can be achieved by the following: 1. Preventing undesirable events 2. Shutting-in the process 3. Recovering released fluids 4. Preventing ignition The modes of safety system operation include 1. Automatic monitoring by sensors 2. Automatic protective action 3. Emergency shutdown Protection concepts and safety analysis are based on un- desirable events, which include A. Overpressure caused by 1. Increased input flow due to upstream flow-control device failure 2. Decreased output flow due to blockage 3. Heating of closed system B. Leak caused by 1. Corrosion 2. Erosion 3. Mechanical failure due to temperature change, overpressure and underpressure, and external im- pact force C. Liquid overflow caused by 1. Increased input flow due to upstream flow-control device failure 2. Decreased output flow due to blockage in the liquid discharge D. Gas blow-by caused by 1. Increased input flow due to upstream flow-control device failure 2. Decreased output flow due to blockage in the gas discharge E. Underpressure caused by 1. Outlet flow-control device (e.g., choke) failure 2. Inlet blockage 3. Cooling of closed system F. Excess temperature caused by 1. Overfueling of burner 2. External fire 3. Spark emission Figure 1.17 presents some symbols used in safety system design. Some API-recommended safety devices are shown in Figs. 1.18 through 1.22. Flow Safety Valve FSV Burner Safety Low BSL Flow Safety High FSH Flow Safety Low FSL Level Safety Low LSL Level Safety High Level Safety High Low LSHL Flow Safety High Low FSHL Pressure Safety High Low PSHL Pressure Safety Element PSE Pressure Safety Valve PSV PSV Temperature Safety High TSH Temperature Safety Low Temperature Safety High Low TSHL Temperature Safety Element TSE LSH TSL Figure 1.17 Safety device symbols. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 12 4.1.2007 6:12pm Compositor Name: SJoearun 1/12 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 25. Pressure Safety High PSH Pressure Safety Low PSL Surface Safety Valve SSV SSV Underwater Safety Valve USV USV Blow Down Valve BDV BDV Shut Down Valve SDV SDV Figure 1.17 (Continued) MAWP SITP 10’ (3M) TSE SSV PSHL FSV Outlet Option 1 MAWP SITP 10’ (3M) TSE SSV PSHL FSV Outlet PSL Option 2 Figure 1.18 Safety system designs for surface wellhead flowlines. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 13 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/13
- 26. MAWP SITP 10’ (3M) TSE SSV PSL FSV Outlet PSL PSHL Option 3 MAWP SITP TSE SSV PSHL FSV Outlet PSL 10’ (3M) MAWP SITP PSV Option 4 MAWP SITP SSV FSV Outlet PSHL TSE Option 5 Figure 1.18 (Continued) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 14 4.1.2007 6:12pm Compositor Name: SJoearun 1/14 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 27. MAWP SITP USV Outlet PSHL FSV MAWP SITP USV Outlet FSVPSHL Denotes Platform Limits PSL SDV Option 1 Option 2 Figure 1.19 Safety system designs for underwater wellhead flowlines. PSHL PSV Gas Outlet FSV Gas Makeup System TSE Inlet LSL LSH FSV Oil Outlet Pressure vessel Figure 1.20 Safety system design for pressure vessel. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 15 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/15
- 28. PSV TSE PSHL FSV SDV Discharge PumpFrom Storage Component Figure 1.21 Safety system design for pipeline pumps. PSV TSE PSHL FSV Discharge PumpSuction Figure 1.22 Safety system design for other pumps. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 16 4.1.2007 6:12pm Compositor Name: SJoearun 1/16 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 29. 1.9 Unit Systems This book uses U.S. oil field units in the text. However, the computer spreadsheet programs associated with this book were developed in both U.S. oil field units and S.I. units. Conversion factors between these two unit systems are presented in Appendix A. Summary This chapter provided a brief introduction to the compo- nentsin the petroleum production system. Working prin- ciples, especially flow performances, of the components are described in later chapters. References American Petroleum Institute. ‘‘Bulletin on Performance Properties of Casing, Tubing, and Drill Pipe,’’ 20th edition. Washington, DC: American Petroleum Insti- tute. API Bulletin 5C2, May 31, 1987. American Petroleum Institute. ‘‘Recommended Practice for Analysis, Design, Installation, and Testing of Basic Surface Safety Systems for Offshore Production Platforms,’’ 20th edition. Washington, DC: American Petroleum Institute. API Bulletin 14C, May 31, 1987. guo, b. and ghalambor a., Natural Gas Engineering Handbook. Houston: Gulf Publishing Company, 2005. guo, b., song, s., chacko, j., and ghalambor a., Offshore Pipelines. Amsterdam: Elsevier, 2005. sivalls, c.r. ‘‘Fundamentals of Oil and Gas Separation.’’ Proceedings of the Gas Conditioning Conference, University of Oklahoma, Norman, Oklahoma, 1977. wolbert, g.s., American Pipelines, University of Okla- homa Press, Norman (1952), p. 5. Problems 1.1 Explain why a water-drive oil reservoir is usually an unsaturated oil reservoir. 1.2 What are the benefits and disadvantages of using down-hole chokes over wellhead chokes? 1.3 What is the role of an oil production engineer? 1.4 Is the tubing nominal diameter closer to tubing outside diameter or tubing inside diameter? 1.5 What do the digits in the tubing specification repre- sent? 1.6 What is a wellhead choke used for? 1.7 What are the separators and pumps used for in the oil production operations? 1.8 Name three applications of pipelines. 1.9 What is the temperature safety element used for? Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap01 Final Proof page 17 4.1.2007 6:12pm Compositor Name: SJoearun PETROLEUM PRODUCTION SYSTEM 1/17
- 30. 2 Properties of Oil and Natural Gas Contents 2.1 Introduction 2/20 2.2 Properties of Oil 2/20 2.3 Properties of Natural Gas 2/21 Summary 2/26 References 2/26 Problems 2/26 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 19 22.12.2006 7:08pm
- 31. 2.1 Introduction Properties of crude oil and natural gas are fundamental for designing and analyzing oil and gas production systems in petroleum engineering. This chapter presents definitions of these fluid properties and some means of obtaining these property values other than experimental measurements. Applications of the fluid properties appear in subsequent chapters. 2.2 Properties of Oil Oil properties include solution gas–oil ratio (GOR), density, formation volume factor, viscosity, and compress- ibility. The latter four properties are interrelated through solution GOR. 2.2.1 Solution Gas–Oil Ratio ‘‘Solution GOR’’ is defined as the amount of gas (in standard condition) that will dissolve in unit volume of oil when both are taken down to the reservoir at the prevailing pressure and temperature; that is, Rs ¼ Vgas Voil , (2:1) where Rs ¼ solution GOR (in scf/stb) Vgas ¼ gas volume in standard condition (scf) Voil ¼ oil volume in stock tank condition (stb) The ‘‘standard condition’’ is defined as 14.7 psia and 60 8F in most states in the United States. At a given reservoir temperature, solution GOR remains constant at pressures above bubble-point pressure. It drops as pressure decreases in the pressure range below the bubble-point pressure. Solution GOR is measured in PTV laboratories. Empirical correlations are also available based on data from PVT labs. One of the correlations is, Rs ¼ gg p 18 100:0125( API) 100:00091t !1:2048 (2:2) where gg and 8API are defined in the latter sections, and p and t are pressure and temperature in psia and 8F, respectively. Solution GOR factor is often used for volumetric oil and gas calculations in reservoir engineering. It is also used as a base parameter for estimating other fluid properties such as density of oil. 2.2.2 Density of Oil ‘‘Density of oil’’ is defined as the mass of oil per unit volume, or lbm=ft3 in U.S. Field unit. It is widely used in hydraulics calculations (e.g., wellbore and pipeline per- formance calculations [see Chapters 4 and 11]). Because of gas content, density of oil is pressure depen- dent. The density of oil at standard condition (stock tank oil) is evaluated by API gravity. The relationship between the density of stock tank oil and API gravity is given through the following relations: API ¼ 141:5 go À 131:5 (2:3) and go ¼ ro,st rw , (2:4) where 8API ¼ API gravity of stock tank oil go ¼ specific gravity of stock tank oil, 1 for freshwater ro,st ¼ density of stock tank oil, lbm=ft3 rw ¼ density of freshwater, 62:4 lbm=ft3 The density of oil at elevated pressures and temperatures can be estimated on empirical correlations developed by a number of investigators. Ahmed (1989) gives a summary of correlations. Engineers should select and validate the correlations carefully with measurements before adopting any correlations. Standing (1981) proposed a correlation for estimating the oil formation volume factor as a function of solution GOR, specific gravity of stock tank oil, specific gravity of solution gas, and temperature. By coupling the mathemat- ical definition of the oil formation volume factor with Standing’s correlation, Ahmed (1989) presented the fol- lowing expression for the density of oil: ro ¼ 62:4go þ 0:0136Rsgg 0:972 þ 0:000147 Rs ffiffiffiffiffi gg go r þ 1:25t !1:175 , (2:5) where t ¼ temperature, 8F gg ¼ specific gravity of gas, 1 for air. 2.2.3 Formation Volume Factor of Oil ‘‘Formation volume factor of oil’’ is defined as the volume occupied in the reservoir at the prevailing pressure and temperature by volume of oil in stock tank, plus its dis- solved gas; that is, Bo ¼ Vres Vst , (2:6) where Bo ¼ formation volume factor of oil (rb/stb) Vres ¼ oil volume in reservoir condition (rb) Vst ¼ oil volume in stock tank condition (stb) Formation volume factor of oil is always greater than unity because oil dissolves more gas in reservoir condition than in stock tank condition. At a given reservoir tempera- ture, oil formation volume factor remains nearly constant at pressures above bubble-point pressure. It drops as pres- sure decreases in the pressure range below the bubble- point pressure. Formation volume factor of oil is measured in PTV labs. Numerous empirical correlations are available based on data from PVT labs. One of the correlations is Bo ¼ 0:9759 þ 0:00012 Rs ffiffiffiffiffi gg go r þ 1:25t !1:2 : (2:7) Formation volume factorof oil is oftenused for oil volumet- riccalculationsandwell-inflowcalculations.Itisalsousedas a base parameter for estimating other fluid properties. 2.2.4 Viscosity of Oil ‘‘Viscosity’’ is an empirical parameter used for describing the resistance to flow of fluid. The viscosity of oil is of interest in well-inflow and hydraulics calculations in oil production engineering. While the viscosity of oil can be measured in PVT labs, it is often estimated using empirical correlations developed by a number of investigators including Beal (1946), Beggs and Robinson (1975), Stand- ing (1981), Glaso (1985), Khan (1987), and Ahmed (1989). A summary of these correlations is given by Ahmed (1989). Engineers should select and validate a correlation with measurements before it is used. Standing’s (1981) correlation for dead oil is expressed as mod ¼ 0:32 þ 1:8 Â 107 API4:53 360 t þ 200 A , (2:8) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 20 22.12.2006 7:08pm 2/20 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 32. where A ¼ 10 0:43þ8:33 APIð Þ (2:9) and mod ¼ viscosity of dead oil (cp). Standing’s (1981) correlation for saturated crude oil is expressed as mob ¼ 10a mb od , (2:10) where mob ¼ viscosity of saturated crude oil in cp and a ¼ Rs(2:2 Â 10À7 Rs À 7:4 Â 10À4 ), (2:11) b ¼ 0:68 10c þ 0:25 10d þ 0:062 10e , (2:12) c ¼ 8:62 Â 10À5 Rs, (2:13) d ¼ 1:10 Â 10À3 Rs, (2:14) and e ¼ 3:74 Â 10À3 Rs, (2:15) Standing’s (1981) correlation for unsaturated crude oil is expressed as mo ¼ mob þ 0:001(p À pb)(0:024m1:6 ob þ 0:38m0:56 ob ): (2:16) 2.2.5 Oil Compressibility ‘‘Oil compressibility’’ is defined as co ¼ À 1 V @V @p T , (2:17) where T and V are temperature and volume, respectively. Oil compressibility is measured from PVT labs. It is often used in modeling well-inflow performance and reservoir simulation. Example Problem 2.1 The solution GOR of a crude oil is 600 scf/stb at 4,475 psia and 140 8F. Given the following PVT data, estimate density and viscosity of the crude oil at the pressure and temperature: Bubble-point pressure: 2,745 psia Oil gravity: 35 8API Gas-specific gravity: 0.77 air ¼ 1 Solution Example Problem 2.1 can be quickly solved using the spreadsheet program OilProperties.xls where Standing’s correlation for oil viscosity was coded. The input and output of the program is shown in Table 2.1. 2.3 Properties of Natural Gas Gas properties include gas-specific gravity, gas pseudo- critical pressure and temperature, gas viscosity, gas Table 2.1 Result Given by the Spreadsheet Program OilProperties.xls OilProperties.xls Description: This spreadsheet calculates density and viscosity of a crude oil. Instruction: (1) Click a unit-box to choose a unit system; (2) update parameter values in the Input data section; (3) view result in the Solution section and charts. Input data U.S. Field units SI units Pressure (p): 4,475 psia Temperature (t): 140 8F Bubble point pressure ( pb): 2,745 psia Stock tank oil gravity: 35 8API Solution gas oil ratio (Rs): 600 scf/stb Gas specific gravity (gg): 0.77 air ¼ 1 Solution go ¼ 141:5 API þ 131:5 ¼ 0.8498 H2O ¼ 1 ro ¼ 62:4go þ 0:0136Rsgg 0:972 þ 0:000147 Rs ffiffiffiffi gg go q þ 1:25t h i1:175 ¼ 44.90 lbm=ft3 A ¼ 10(0:43þ8:33=API) ¼ 4.6559 mod ¼ 0:32 þ 1:8 Â 107 API4:53 360 t þ 200 A ¼ 2.7956 cp a ¼ Rs(2:2 Â 10À7 Rs À 7:4 Â 10À4 ) ¼ À0.3648 c ¼ 8:62 Â 10À5 Rs ¼ 0.0517 d ¼ 1:10 Â 10À3 Rs ¼ 0.6600 e ¼ 3:74 Â 10À3 Rs ¼ 2.2440 b ¼ 0:68 10c þ 0:25 10d þ 0:062 10e ¼ 0.6587 mob ¼ 10a mb od ¼ 0.8498 cp 0.0008 Pa-s mo ¼ mob þ 0:001( p À pb)(0:024m1:6 ob þ 0:38m0:56 ob ) ¼ 1.4819 cp 0.0015 Pa-s Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 21 22.12.2006 7:08pm PROPERTIES OF OIL AND NATURAL GAS 2/21
- 33. compressibility factor, gas density, gas formation volume factor, and gas compressibility. The first two are com- position dependent.The latter four are pressure dependent. 2.3.1 Specific Gravity of Gas ‘‘Specific gravity gas’’ is defined as the ratio of the appar- ent molecular weight of the gas to that of air. The molecu- lar weight of air is usually taken as equal to 28.97 ($79% nitrogen and 21% oxygen). Therefore, the gas-specific gravity is gg ¼ MWa 28:97 , (2:18) where MWa is the apparent molecular weight of gas, which can be calculated on the basis of gas composition. Gas composition is usually determined in a laboratory and reported in mole fractions of components in the gas. Let yi be the mole fraction of component i, and the apparent molecular weight of the gas can be formulated using a mixing rule such as MWa ¼ XNc i¼1 yiMWi, (2:19) where MWi is the molecular weight of component i, and Nc is number of components. The molecular weights of compounds (MWi) can be found in textbooks on organic chemistry or petroleum fluids such as that by Ahmed (1989). Gas-specific gravity varies between 0.55 and 0.9. 2.3.2 Gas Pseudo-Critical Pressure and Temperature Similar to gas apparent molecular weight, the critical properties of a gas can be determined on the basis of the critical properties of compounds in the gas using the mix- ing rule. The gas critical properties determined in such a way are called ‘‘pseudo-critical properties.’’ Gas pseudo- critical pressure ( ppc) and pseudo-critical temperature (Tpc) are, respectively, expressed as ppc ¼ XNc i¼1 yipci (2:20) and Tpc ¼ XNc i¼1 yiTci, (2:21) where pci and Tci are critical pressure and critical tempera- ture of component i, respectively. Example Problem 2.2 For the gas composition given in the following text, determine apparent molecular weight, specific gravity, pseudo-critical pressure, and pseudo- critical temperature of the gas. Solution Example Problem 2.2 is solved with the spreadsheet program MixingRule.xls. Results are shown in Table 2.2. If the gas composition is not known but gas-specific gravity is given, the pseudo-critical pressure and tempera- ture can be determined from various charts or correlations developed based on the charts. One set of simple cor- relations is ppc ¼ 709:604 À 58:718gg (2:22) Tpc ¼ 170:491 þ 307:344gg, (2:23) which are valid for H2S 3%, N2 5%, and total content of inorganic compounds less than 7%. Corrections for impurities in sour gases are always necessary. The corrections can be made using either charts or correlations such as the Wichert and Aziz (1972) correction expressed as follows: A ¼ yH2S þ yCO2 (2:24) B ¼ yH2S (2:25) Table 2.2 Results Given by the Spreadsheet Program MixingRule.xls MixingRule.xls Description: This spreadsheet calculates gas apparent molecular weight, specific gravity, pseudo-critical pressure, and pseudo-critical temperature. Instruction: (1) Update gas composition data (yi); (2) read result. Compound yi MWi yiMWi pci (psia) yipci (psia) Tci, (8R) yiTci (8R) C1 0.775 16.04 12.43 673 521.58 344 266.60 C2 0.083 30.07 2.50 709 58.85 550 45.65 C3 0.021 44.10 0.93 618 12.98 666 13.99 i-C4 0.006 58.12 0.35 530 3.18 733 4.40 n-C4 0.002 58.12 0.12 551 1.10 766 1.53 i-C5 0.003 72.15 0.22 482 1.45 830 2.49 n-C5 0.008 72.15 0.58 485 3.88 847 6.78 C6 0.001 86.18 0.09 434 0.43 915 0.92 C7þ 0.001 114.23 0.11 361 0.36 1024 1.02 N2 0.050 28.02 1.40 227 11.35 492 24.60 CO2 0.030 44.01 1.32 1,073 32.19 548 16.44 H2S 0.020 34.08 0.68 672 13.45 1306 26.12 1.000 MWa ¼ 20.71 ppc ¼ 661 Tpc ¼ 411 gg ¼ 0.71 Component Mole Fraction C1 0.775 C2 0.083 C3 0.021 i-C4 0.006 n-C4 0.002 i-C5 0.003 n-C5 0.008 C6 0.001 C7þ 0.001 N2 0.050 CO2 0.030 H2S 0.020 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 22 22.12.2006 7:08pm 2/22 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 34. «3 ¼ 120(A0:9 À A1:6 ) þ 15(B0:5 À B4:0 ) (2:26) Tpc0 ¼ Tpc À «3(corrected Tpc) (2:27) Ppc0 ¼ PpcTpc0 Tpc þ B(1 À B)«3 (corrected ppc) (2:28) Correlations with impurity corrections for mixture pseudo-criticals are also available (Ahmed, 1989): ppc ¼ 678 À 50(gg À 0:5) À 206:7yN2 þ 440yCO2 þ 606:7yHsS (2:29) Tpc ¼ 326 þ 315:7(gg À 0:5) À 240yN2 À 83:3yCO2 þ 133:3yH2S: (2:30) Applications of the pseudo-critical pressure and temperature are normally found in petroleum engineer- ing through pseudo-reduced pressure and temperature defined as ppr ¼ p ppc (2:31) Tpr ¼ T Tpc : (2:32) 2.3.3 Viscosity of Gas Dynamic viscosity (mg) in centipoises (cp) is usually used in petroleum engineering. Kinematic viscosity (ng) is related to the dynamic viscosity through density (rg), ng ¼ mg rg : (2:33) Kinematic viscosity is not typically used in natural gas engineering. Direct measurements of gas viscosity are preferred for a new gas. If gas composition and viscosities of gas com- ponents are known, the mixing rule can be used to deter- mine the viscosity of the gas mixture: mg ¼ P (mgiyi ffiffiffiffiffiffiffiffiffiffiffi MWi p ) P (yi ffiffiffiffiffiffiffiffiffiffiffi MWi p ) (2:34) Viscosity of gas is very often estimated with charts or correlations developed based on the charts. Gas viscosity correlation of Carr et al. 1954 involves a two-step pro- cedure: The gas viscosity at temperature and atmospheric pressure is estimated first from gas-specific gravity and inorganic compound content. The atmospheric value is then adjusted to pressure conditions by means of a correc- tion factor on the basis of reduced temperature and pres- sure state of the gas. The atmospheric pressure viscosity (m1) can be expressed as m1 ¼ m1HC þ m1N2 þ m1CO2 þ m1H2S, (2:35) where m1HC ¼ 8:188 Â 10À3 À 6:15 Â 10À3 log (gg) þ (1:709 Â 10À5 À 2:062 Â 10À6 gg)T, (2:36) m1N2 ¼ [9:59 Â 10À3 þ 8:48 Â 10À3 log (gg)]yN2 , (2:37) m1CO2 ¼ [6:24 Â 10À3 þ 9:08 Â 10À3 log (gg)]yCO2 , (2:38) m1H2S ¼ [3:73 Â 10À3 þ 8:49 Â 10À3 log (gg)]yH2S, (2:39) Dempsey (1965) developed the following relation: mr ¼ ln mg m1 Tpr ¼ a0 þ a1ppr þ a2p2 pr þ a3p3 pr þ Tpr(a4 þ a5ppr þ a6p2 pr þ a7p3 pr) þ T2 pr(a8 þ a9ppr þ a10p2 pr þ a11p3 pr) þ T3 pr(a12 þ a13ppr þ a14p2 pr þ a15p3 pr), (2:40) where a0 ¼ À2:46211820 a1 ¼ 2:97054714 a2 ¼ À0:28626405 a3 ¼ 0:00805420 a4 ¼ 2:80860949 a5 ¼ À3:49803305 a6 ¼ 0:36037302 a7 ¼ À0:01044324 a8 ¼ À0:79338568 a9 ¼ 1:39643306 a10 ¼ À0:14914493 a11 ¼ 0:00441016 a12 ¼ 0:08393872 a13 ¼ À0:18640885 a14 ¼ 0:02033679 a15 ¼ À0:00060958 Thus, once the value of mr is determined from the right- hand side of this equation, gas viscosity at elevated pres- sure can be readily calculated using the following relation: mg ¼ m1 Tpr emr (2:41) Other correlations for gas viscosity include that of Dean and Stiel (1958) and Lee et al. (1966). Example Problem 2.3 A 0.65 specific–gravity natural gas contains 10% nitrogen, 8% carbon dioxide, and 2% hydrogen sulfide. Estimate viscosity of the gas at 10,000 psia and 1808F. Solution Example Problem 2.3 is solved with the spread- sheet Carr-Kobayashi-Burrows-GasViscosity.xls, which is attached to this book. The result is shown in Table 2.3. 2.3.4 Gas Compressibility Factor Gas compressibility factor is also called ‘‘deviation factor’’ or ‘‘z-factor.’’ Its value reflects how much the real gas deviates from the ideal gas at a given pressure and tem- perature. Definition of the compressibility factor is expressed as z ¼ Vactual Videal gas : (2:42) Introducing the z-factor to the gas law for ideal gas results in the gas law for real gas as pV ¼ nzRT, (2:43) where n is the number of moles of gas. When pressure p is entered in psia, volume V in ft3 , and temperature in 8R, the gas constant R is equal to10.73 psia À ft3 mole ÀR . Gas compressibility factor can be determined on the basis of measurements in PVT laboratories. For a given amount of gas, if temperature is kept constant and volume is mea- sured at 14.7 psia and an elevated pressure p1, z-factor can then be determined with the following formula: z ¼ p1 14:7 V1 V0 , (2:44) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 23 22.12.2006 7:08pm PROPERTIES OF OIL AND NATURAL GAS 2/23
- 35. where V0 and V1 are gas volumes measured at 14.7 psia and p1, respectively. Very often the z-factor is estimated with the chart devel- oped by Standing and Katz (1954). This chart has been set up for computer solution by a number of individuals. Brill and Beggs (1974) yield z-factor values accurate enough for many engineering calculations. Brill and Beggs’ z-factor correlation is expressed as follows: A ¼ 1:39(Tpr À 0:92)0:5 À 0:36Tpr À 0:10, (2:45) B ¼ (0:62 À 0:23Tpr)ppr þ 0:066 Tpr À 0:86 À 0:037 p2 pr þ 0:32 p6 pr 10E , (2:46) C ¼ 0:132 À 0:32 log (Tpr), (2:47) D ¼ 10F , (2:48) E ¼ 9(Tpr À 1), (2:49) F ¼ 0:3106 À 0:49Tpr þ 0:1824T2 pr, (2:50) and z ¼ A þ 1 À A eB þ CpD pr: (2:51) Example Problem 2.4 For the natural gas described in Example Problem 2.3, estimate z-factor at 5,000 psia and 180 8F. Solution Example Problem 2.4 is solved with the spreadsheet program Brill-Beggs-Z.xls. The result is shown in Table 2.4. Hall and Yarborough (1973) presented a more accurate correlation to estimate z-factor of natural gas. This cor- relation is summarized as follows: tr ¼ 1 Tpr (2:52) A ¼ 0:06125treÀ1:2(1Àtr)2 (2:53) B ¼ tr(14:76 À 9:76tr þ 4:58t2 r ) (2:54) C ¼ tr(90:7 À 242:2tr þ 42:4t2 r ) (2:55) D ¼ 2:18 þ 2:82tr (2:56) and z ¼ Appr Y , (2:57) where Y is the reduced density to be solved from f (Y) ¼ Y þ Y2 þ Y3 À Y4 (1 À Y)3 À Appr À BY2 þ CYD ¼ 0: (2:58) If the Newton and Raphson iteration method is used to solve Eq. (2.58) for Y, the following derivative is needed: df (Y) dY ¼ 1 þ 4Y þ 4Y2 À 4Y3 þ Y4 (1 À Y)4 À 2BY þ CDYDÀ1 (2:59) 2.3.5 Density of Gas Because gas is compressible, its density depends on pres- sure and temperature. Gas density can be calculated from gas law for real gas with good accuracy: rg ¼ m V ¼ MWap zRT , (2:60) where m is mass of gas and rg is gas density. Taking air molecular weight 29 and R ¼ 10:73 psia À ft3 mole À R , Eq. (2.60) is rearranged to yield rg ¼ 2:7ggp zT , (2:61) where the gas density is in lbm=ft3 . Table 2.3 Results Given by the Spreadsheet Carr-Kobayashi-Burrows-GasViscosity.xls Carr-Kobayashi-Burrows-GasViscosity.xls Description: This spreadsheet calculates gas viscosity with correlation of Carr et al. Instruction: (1) Select a unit system; (2) update data in the Input data section; (3) review result in the Solution section. Input data U.S. Field units SI units Pressure: 10,000 psia Temperature: 180 8F Gas-specific gravity: 0.65 air ¼ 1 Mole fraction of N2: 0.1 Mole fraction of CO2: 0.08 Mole fraction of H2S: 0.02 Solution Pseudo-critical pressure ¼ 697.164 psia Pseudo-critical temperature ¼ 345.357 8R Uncorrected gas viscosity at 14.7 psia ¼ 0.012174 cp N2 correction for gas viscosity at 14.7 psia ¼ 0.000800 cp CO2 correction for gas viscosity at 14.7 psia ¼ 0.000363 cp H2S correction for gas viscosity at 14.7 psia ¼ 0.000043 cp Corrected gas viscosity at 14.7 psia (m1) ¼ 0.013380 cp Pseudo-reduced pressure ¼ 14.34 Pseudo-reduced temperature ¼ 1.85 In(mg=m1 Ã Tpr) ¼ 1.602274 Gas viscosity ¼ 0.035843 cp Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 24 22.12.2006 7:08pm 2/24 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 36. Example Problem 2.5 A gas from oil has a specific gravity of 0.65, estimate z-factor and gas density at 5,000 psia and 180 8F. Solution Example Problem 2.5 is solved with the spreadsheet program Hall-Yarborogh-z.xls. The result is shown in Table 2.5. 2.3.6 Formation Volume Factor of Gas Gas formation volume factor is defined as the ratio of gas volume at reservoir condition to the gas volume at stan- dard condition, that is, Bg ¼ V Vsc ¼ psc p T Tsc z zsc ¼ 0:0283 zT p , (2:62) Table 2.4 Results Given by the Spreadsheet Program Brill-Beggs-Z.xls Brill-Beggs-Z.xls Description: This spreadsheet calculates gas compressibility factor based on the Brill and Beggs correlation. Instruction: (1) Select a unit system; (2) update data in the Input data section; (3) review result in the Solution section. Input data U.S. Field units SI units Pressure: 5,000 psia Temperature: 180 8F Gas specific gravity: 0.65 air ¼ 1 Mole fraction of N2: 0.1 Mole fraction of CO2: 0.08 Mole fraction of H2S: 0.02 Solution Pseudo-critical pressure ¼ 697 psia Pseudo-critical temperature ¼ 345 8R Pseudo-reduced pressure ¼ 7.17 Pseudo-reduced temperature ¼ 1.95 A ¼ 0.6063 B ¼ 2.4604 C ¼ 0.0395 D ¼ 1.1162 Gas compressibility factor z ¼ 0.9960 Table 2.5 Results Given by the Spreadsheet Program Hall-Yarborogh-z.xls Hall-Yarborogh-z.xls Description: This spreadsheet computes gas compressibility factor with the Hall–Yarborough method. Instruction: (1) Select a unit system; (2) update data in the Input data section; (3) click Solution button; (4) view result. Input data U.S. Field units SI units Temperature: 200 8F Pressure: 2,000 psia Gas-specific gravity: 0.7 air ¼ 1 Nitrogen mole fraction: 0.05 Carbon dioxide fraction: 0.05 Hydrogen sulfite fraction: 0.02 Solution Tpc ¼ 326 þ 315:7(gg À 0:5) À 240yN2 À 83:3yCO2 þ 133:3yH2S ¼ 375.641 8R ppc ¼ 678 À 50(gg À 0:5) À 206:7yN2 þ 440yCO2 þ 606:7yH2S ¼ 691.799 psia Tpr ¼ T Tpc ¼ 1.618967 tr ¼ 1 Tpr ¼ 0.617678 ppr ¼ p ppc ¼ 2.891013 A ¼ 0:06125treÀ1:2(1Àtr)2 ¼ 0.031746 B ¼ tr(14:76 À 9:76tr þ 4:58t2 r ) ¼ 6.472554 C ¼ tr(90:7 À 242:2tr þ 42:4t2 r ) ¼ À26.3902 D ¼ 2:18 þ 2:82tr ¼ 3.921851 Y ¼ ASSUMED ¼ 0.109759 f (Y) ¼ Y þ Y2 þ Y3 À Y4 (1 À Y)3 À Appr À BY2 þ CYD ¼ 0 ¼ 4.55E-06 z ¼ Appr Y ¼ 0.836184 rg ¼ 2:7ggp zT ¼ 6:849296 lbm=ft3 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 25 22.12.2006 7:08pm PROPERTIES OF OIL AND NATURAL GAS 2/25
- 37. where the unit of formation volume factor is ft3 =scf. If expressed in rb/scf, it takes the form Bg ¼ 0:00504 zT p : (2:63) Gas formation volume factor is frequently used in math- ematical modeling of gas well inflow performance relation- ship (IPR). Another way to express this parameter is to use gas expansion factor defined, in scf=ft3 , as E ¼ 1 Bg ¼ 35:3 P ZT (2:64) or E ¼ 198:32 p zT , (2:65) in scf/rb. It is normally used for estimating gas reserves. 2.3.7 Gas Compressibility Gas compressibility is defined as cg ¼ À 1 V @V @p T : (2:66) Because the gas law for real gas gives V ¼ nzRT p , @V @p ¼ nRT 1 p @z @p À z p2 : (2:67) Substituting Eq. (2.67) into Eq. (2.66) yields cg ¼ 1 p À 1 z @z @p : (2:68) Since the second term in the right-hand side is usually small, gas compressibility is approximately equal to the reciprocal of pressure. Summary This chapter presented definitions and properties of crude oil and natural gas. It also provided a few empirical cor- relations for determining the value of these properties. These correlations are coded in spreadsheet programs that are available with this book. Applications of these fluid properties are found in the later chapters. References ahmed, t. Hydrocarbon Phase Behavior. Houston: Gulf Publishing Company, 1989. American Petroleum Institute: ‘‘Bulletin on Performance Properties of Casing, Tubing, and Drill Pipe,’’ 20th edition. Washington, DC: American Petroleum Insti- tute. API Bulletin 5C2, May 31, 1987. beal, c. The viscosity of air, water, natural gas, crude oils and its associated gases at oil field temperatures and pressures. Trans. AIME 1946;165:94–112. beggs, h.d. and robinson, J.R. Estimating the viscosity of crude oil systems. J. Petroleum Technol. Sep. 1975:1140–1141. brill, j.p. and beggs, h.d. Two-phase flow in pipes. INTERCOMP Course, The Hague, 1974. carr, n.l., kobayashi, r., and burrows, d.b. Viscosity of hydrocarbon gases under pressure. Trans. AIME 1954;201:264–272. dean, d.e. and stiel, l.i. The viscosity of non-polar gas mixtures at moderate and high pressures. AIChE J. 1958;4:430–436. dempsey, j.r. Computer routine treats gas viscosity as a variable. Oil Gas J. Aug. 16, 1965:141. glaso, o. Generalized pressure-volume-temperature cor- relations. J. Petroleum Technol. May 1985:785–795. hall, k.r. and yarborough, l. A new equation of state for Z-factor calculations. Oil Gas J. June 18, 1973:82. khan, s.a. Viscosity correlations for Saudi Arabian crude oils. Presented at the 50th Middle East Conference and Exhibition held 7–10 March 1987, in Manama, Bah- rain. Paper SPE 15720. lee, a.l., gonzalez, m.h., and eakin, b.e. The viscosity of natural gases. J. Petroleum Technol. Aug. 1966:997– 1000. standing, m.b. Volume and Phase Behavior of Oil Field Hydrocarbon Systems, 9th edition. Dallas: Society of Petroleum Engineers, 1981. standing, m.b. and katz, d.l. Density of natural gases. Trans. AIME 1954;146:140–149. wichert, e. and aziz, k. Calculate Zs for sour gases. Hydrocarbon Processing 1972;51(May):119. Problems 2.1 Estimate the density of a 25-API gravity dead oil at 100 8F. 2.2 The solution gas–oil ratio of a crude oil is 800 scf/stb at 3,000 psia and 120 8F. Given the following PVT data: Bubble-point pressure: 2,500 psia Oil gravity: 35 8API Gas-specific gravity: 0.77 (air ¼ 1), estimate densities and viscosities of the crude oil at 120 8F, 2,500 psia, and 3,000 psia. 2.3 For the gas composition given below, determine apparent molecular weight, specific gravity, pseudo- critical pressure, and pseudo-critical temperature of the gas: 2.4 Estimate gas viscosities of a 0.70-specific gravity gas at 200 8F and 100 psia, 1,000 psia, 5,000 psia, and 10,000 psia. 2.5 Calculate gas compressibility factors and densities of a 0.65-specific gravity gas at 150 8F and 50 psia, 500 psia, and 5,000 psia with the Hall–Yarborough method. Compare the results with that given by the Brill and Beggs correlation. What is your conclusion? 2.6 For a 0.65-specific gravity gas at 250 8F, calculate and plot pseudo-pressures in a pressure range from 14.7 and 8,000 psia. Under what condition is the pseudo- pressure linearly proportional to pressure? 2.7 Estimate the density of a 0.8-specific gravity dead oil at 40 8C. Component Mole Fraction C1 0.765 C2 0.073 C3 0.021 i-C4 0.006 n-C4 0.002 i-C5 0.003 n-C5 0.008 C6 0.001 C7þ 0.001 N2 0.060 CO2 0.040 H2S 0.020 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 26 22.12.2006 7:08pm 2/26 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 38. 2.8 The solution gas–oil ratio of a crude oil is 4,000 sm3 =m3 at 20 MPa and 50 8C. Given the follow- ing PVT data: Bubble-point pressure: 15 MPa Oil-specific gravity: 0.8 water ¼ 1 Gas-specific gravity: 0.77 air ¼ 1, estimate densities and viscosities of the crude oil at 50 8C, 15 MPa, and 20 MPa. 2.9 For the gas composition given below, determine apparent molecular weight, specific gravity, pseudo- critical pressure, and pseudo-critical temperature of the gas. 2.10 Estimate gas viscosities of a 0.70-specific gravity gas at 90 8C and 1 MPa, 5 MPa, 10 MPa, and 50 MPa. 2.11 Calculate gas compressibility factors and densities of a 0.65-specific gravity gas at 80 8C and 1 MPa, 5 MPa, 10 MPa, and 50 MPa with the Hall–Yarbor- ough method. Compare the results with that given by the Brill and Beggs correlation. What is your conclu- sion? 2.12 For a 0.65-specific gravity gas at 110 8C, calculate and plot pseudo-pressures in a pressure range from 0.1 to 30 MPa. Under what condition is the pseudo- pressure linearly proportional to pressure? Component Mole Fraction C1 0.755 C2 0.073 C3 0.011 i-C4 0.006 n-C4 0.002 i-C5 0.003 n-C5 0.008 C6 0.001 C7þ 0.001 N2 0.070 CO2 0.050 H2S 0.020 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap02 Final Proof page 27 22.12.2006 7:08pm PROPERTIES OF OIL AND NATURAL GAS 2/27
- 39. 3 Reservoir Deliverability Contents 3.1 Introduction 3/30 3.2 Flow Regimes 3/30 3.3 Inflow Performance Relationship 3/32 3.4 Construction of IPR Curves Using Test Points 3/35 3.5 Composite IPR of Stratified Reservoirs 3/37 3.6 Future IPR 3/39 Summary 3/42 References 3/42 Problems 3/43 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 29 3.1.2007 8:30pm Compositor Name: SJoearun
- 40. 3.1 Introduction Reservoir deliverability is defined as the oil or gas produc- tion rate achievable from reservoir at a given bottom-hole pressure. It is a major factor affecting well deliverability. Reservoir deliverability determines types of completion and artificial lift methods to be used. A thorough knowl- edge of reservoir productivity is essential for production engineers. Reservoir deliverability depends on several factors in- cluding the following: . Reservoir pressure . Pay zone thickness and permeability . Reservoir boundary type and distance . Wellbore radius . Reservoir fluid properties . Near-wellbore condition . Reservoir relative permeabilities Reservoir deliverability can be mathematically modeled on the basis of flow regimes such as transient flow, steady state flow, and pseudo–steady state flow. An analytical relation between bottom-hole pressure and production rate can be formulated for a given flow regime. The relation is called ‘‘inflow performance relationship’’ (IPR). This chapter addresses the procedures used for establishing IPR of different types of reservoirs and well configurations. 3.2 Flow Regimes When a vertical well is open to produce oil at production rate q, it creates a pressure funnel of radius r around the wellbore, as illustrated by the dotted line in Fig. 3.1a. In this reservoir model, the h is the reservoir thickness, k is the effective horizontal reservoir permeability to oil, mo is viscosity of oil, Bo is oil formation volume factor, rw is wellbore radius, pwf is the flowing bottom hole pressure, and p is the pressure in the reservoir at the distance r from the wellbore center line. The flow stream lines in the cylindrical region form a horizontal radial flow pattern as depicted in Fig. 3.1b. 3.2.1 Transient Flow ‘‘Transient flow’’ is defined as a flow regime where/when the radius of pressure wave propagation from wellbore has not reached any boundaries of the reservoir. During tran- sient flow, the developing pressure funnel is small relative to the reservoir size. Therefore, the reservoir acts like an infinitively large reservoir from transient pressure analysis point of view. Assuming single-phase oil flow in the reservoir, several analytical solutions have been developed for describing the transient flow behavior. They are available from classic textbooks such as that of Dake (1978). A constant-rate solution expressed by Eq. (3.1) is frequently used in pro- duction engineering: pwf ¼ pi À 162:6qBomo kh  log t þ log k fmoctr2 w À 3:23 þ 0:87S , (3:1) where pwf ¼ flowing bottom-hole pressure, psia pi ¼ initial reservoir pressure, psia q ¼ oil production rate, stb/day mo ¼ viscosity of oil, cp k ¼ effective horizontal permeability to oil, md h ¼ reservoir thickness, ft t ¼ flow time, hour f ¼ porosity, fraction ct ¼ total compressibility, psiÀ1 rw ¼ wellbore radius to the sand face, ft S ¼ skin factor Log ¼ 10-based logarithm log10 Because oil production wells are normally operated at constant bottom-hole pressure because of constant well- head pressure imposed by constant choke size, a constant bottom-hole pressure solution is more desirable for well- inflow performance analysis. With an appropriate inner boundary condition arrangement, Earlougher (1977) developed a constant bottom-hole pressure solution, which is similar to Eq. (3.1): q ¼ kh( pi À pwf ) 162:6Bomo log t þ log k fmoctr2 w À 3:23 þ 0:87S , (3:2) which is used for transient well performance analysis in production engineering. Equation (3.2) indicates that oil rate decreases with flow time. This is because the radius of the pressure funnel, over which the pressure drawdown (pi À pwf ) acts, increases with time, that is, the overall pressure gradient in the reservoir drops with time. For gas wells, the transient solution is qg ¼ kh[m( pi) À m(pwf )] 1; 638T log t þ log k fmoctr2 w À 3:23 þ 0:87S , (3:3) where qg is production rate in Mscf/d, T is temperature in 8R, and m(p) is real gas pseudo-pressure defined as m( p) ¼ ðp pb 2p mz dp: (3:4) rw r pwf p mo Bok h q (a) k mo Bo p pwf r (b) rw Figure 3.1 A sketch of a radial flow reservoir model: (a) lateral view, (b) top view. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 30 3.1.2007 8:30pm Compositor Name: SJoearun 3/30 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 41. The real gas pseudo-pressure can be readily determined with the spreadsheet program PseudoPressure.xls. 3.2.2 Steady-State Flow ‘‘Steady-state flow’’ is defined as a flow regime where the pressure at any point in the reservoir remains constant over time. This flow condition prevails when the pressure funnel shown in Fig. 3.1 has propagated to a constant- pressure boundary. The constant-pressure boundary can be an aquifer or a water injection well. A sketch of the reservoir model is shown in Fig. 3.2, where pe represents the pressure at the constant-pressure boundary. Assuming single-phase flow, the following theoretical relation can be derived from Darcy’s law for an oil reservoir under the steady-state flow condition due to a circular constant- pressure boundary at distance re from wellbore: q ¼ kh(pe À pwf ) 141:2Bomo ln re rw þ S , (3:5) where ‘‘ln’’ denotes 2.718-based natural logarithm loge. Derivation of Eq. (3.5) is left to readers for an exercise. 3.2.3 Pseudo–Steady-State Flow ‘‘Pseudo–steady-state’’ flow is defined as a flow regime where the pressure at any point in the reservoir declines at the same constant rate over time. This flow condition prevails after the pressure funnel shown in Fig. 3.1 has propagated to all no-flow boundaries. A no-flow bound- ary can be a sealing fault, pinch-out of pay zone, or boundaries of drainage areas of production wells. A sketch of the reservoir model is shown in Fig. 3.3, where pe represents the pressure at the no-flow boundary at time t4. Assuming single-phase flow, the following theoretical relation can be derived from Darcy’s law for an oil reser- voir under pseudo–steady-state flow condition due to a circular no-flow boundary at distance re from wellbore: q ¼ kh(pe À pwf ) 141:2Bomo ln re rw À 1 2 þ S : (3:6) The flow time required for the pressure funnel to reach the circular boundary can be expressed as tpss ¼ 1,200 fmoctr2 e k : (3:7) Because the pe in Eq. (3.6) is not known at any given time, the following expression using the average reservoir pres- sure is more useful: q ¼ kh(pp À pwf ) 141:2Bomo ln re rw À 3 4 þ S , (3:8) where p¯ is the average reservoir pressure in psia. Deriv- ations of Eqs. (3.6) and (3.8) are left to readers for exer- cises. If the no-flow boundaries delineate a drainage area of noncircular shape, the following equation should be used for analysis of pseudo–steady-state flow: q ¼ kh(pp À pwf ) 141:2Bomo 1 2 ln 4A gCAr2 w þ S , (3:9) where A ¼ drainage area, ft2 g ¼ 1:78 ¼ Euler’s constant CA ¼ drainage area shape factor, 31.6 for a circular boundary. The value of the shape factor CA can be found from Fig. 3.4. For a gas well located at the center of a circular drainage area, the pseudo–steady-state solution is qg ¼ kh[m(pp) À m(pwf )] 1,424T ln re rw À 3 4 þ S þ Dqg , (3:10) where D ¼ non-Darcy flow coefficient, d/Mscf. 3.2.4 Horizontal Well The transient flow, steady-state flow, and pseudo–steady- state flow can also exist in reservoirs penetrated by horizon- tal wells. Different mathematical models are available from h p rre pe pwf rw Figure 3.2 A sketch of a reservoir with a constant-pressure boundary. p h rre pe pi pwf rw t1 t2 t3 t4 Figure 3.3 A sketch of a reservoir with no-flow boundaries. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 31 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/31
- 42. literature. Joshi (1988) presented the following relationship considering steady-state flow of oil in the horizontal plane and pseudo–steady-state flow in the vertical plane: q ¼ kH h(pe À pwf ) 141:2Bm ln aþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2À(L=2)2 p L=2 ! þ Ianih L ln Ianih rw(Iani þ 1) ! ' , (3:11) where a ¼ L 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 4 þ reH L=2 4 #v u u t v u u u t , (3:12) Iani ¼ ffiffiffiffiffiffi kH kV s , (3:13) and kH ¼ the average horizontal permeability, md kV ¼ vertical permeability, md reH ¼ radius of drainage area, ft L ¼ length of horizontal wellbore (L=2 0:9reH ), ft. 3.3 Inflow Performance Relationship IPR is used for evaluating reservoir deliverability in pro- duction engineering. The IPR curve is a graphical presen- tation of the relation between the flowing bottom-hole pressure and liquid production rate. A typical IPR curve is shown in Fig. 3.5. The magnitude of the slope of the IPR curve is called the ‘‘productivity index’’ (PI or J), that is, J ¼ q (pe À pwf ) , (3:14) where J is the productivity index. Apparently J is not a constant in the two-phase flow region. (a) Shape Factor CA 31.6 30.9 31.6 27.6 27.1 3.39 60 Reservoir Shape Well Location 1 1 1 1 1 1 1 4 4 1 Reservoir Shape Well Location 1/3 2 2 2 2 1 2 In water–drive reservoirs In reservoirs of unknown production character (b) 10.8 4.86 2.07 2.72 0.232 0.115 Shape Factor CA 3.13 0.607 0.111 0.098 5.38 2.36 Shape Factor CA 1 1 1 2 2 2 1 1 1 4 4 4 Reservoir Shape Well Location 2 1 2 1 2 1 3 4 1 4 1 5 Reservoir Shape Well Location 60 Shape Factor CA 21.9 22.6 12.9 4.5 19.1 25 Figure 3.4 (a) Shape factors for closed drainage areas with low-aspect ratios. (b) Shape factors for closed drainage areas with high-aspect ratios (Dietz, 1965). 0 1,000 2,000 3,000 4,000 5,000 6,000 qo (stb/day) pwf(psia) 6000 200 400 800 Figure 3.5 A typical IPR curve for an oil well. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 32 3.1.2007 8:30pm Compositor Name: SJoearun 3/32 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 43. Well IPR curves are usually constructed using reservoir inflow models, which can be from either a theoretical basis or an empirical basis. It is essential to validate these models with test points in field applications. 3.3.1 LPR for Single (Liquid)-Phase Reservoirs All reservoir inflow models represented by Eqs. (3.1), (3.3), (3.7), and (3.8) were derived on the basis of the assumption of single-phase liquid flow. This assumption is valid for under- saturated oil reservoirs, or reservoir portions where the pres- sure is above the bubble-point pressure. These equations define the productivity index (JÃ ) for flowing bottom-hole pressures above the bubble-point pressure as follows: JÃ ¼ q (pi À pwf ) ¼ kh 162:6Bomo log t þ log k fmoctr2 w À 3:23 þ 0:87S (3:15) for radial transient flow around a vertical well, JÃ ¼ q (pe À pwf ) ¼ kh 141:2Bomo ln re rw þ S (3:16) for radial steady-state flow around a vertical well, JÃ ¼ q (pp À pwf ) ¼ kh 141:2Bomo 1 2 ln 4A gCAr2 w þ S (3:17) for pseudo–steady-state flow around a vertical well, and JÃ ¼ q (pe À pwf ) ¼ kH h 141:2Bm ln aþ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2À(L=2)2 p L=2 ! þ Ianih L ln Iani h rw(Ianiþ1) h i ' (3:18) for steady-state flow around a horizontal well. Since the productivity index (JÃ ) above the bubble-point pressure isindependent of productionrate, the IPR curve for a single (liquid)-phase reservoir is simply a straight line drawn from the reservoir pressure to the bubble-point pressure. If the bubble-point pressure is 0 psig, the absolute open flow (AOF) is the productivity index (JÃ ) times the reservoir pressure. Example Problem 3.1 Construct IPR of a vertical well in an oil reservoir. Consider (1) transient flow at 1 month, (2) steady-state flow, and (3) pseudo–steady-state flow. The following data are given: Porosity: f ¼ 0:19 Effective horizontal permeability:k ¼ 8:2 md Pay zone thickness: h ¼ 53 ft Reservoir pressure: pe or pp ¼ 5,651 psia Bubble-point pressure: pb ¼ 50 psia Fluid formation volume factor:, Bo ¼ 1:1 Fluid viscosity: mo ¼ 1:7 cp Total compressibility, ct ¼ 0:0000129 psiÀ1 Drainage area: A ¼ 640 acres (re ¼ 2,980 ft) Wellbore radius: rw ¼ 0:328 ft Skin factor: S ¼ 0 Solution 1. For transient flow, calculated points are JÃ ¼ kh 162:6Bm log t þ log fmctr2 w À 3:23 ¼ (8:2)(53) 162:6(1:1)(1:7) log [( (30)(24)] þ log (8:2) (0:19)(1:7)(0:0000129)(0:328)2 À 3:23 ¼ 0:2075 STB=d-psi Transient IPR curve is plotted in Fig. 3.6. 2. For steady state flow: JÃ ¼ kh 141:2Bm ln re rw þ S ¼ (8:2)(53) 141:2(1:1)(1:7) ln 2,980 0:328 ¼ 0:1806 STB=d-psi Calculated points are: Steady state IPR curve is plotted in Fig. 3.7. 3. For pseudosteady state flow: JÃ ¼ kh 141:2Bm ln re rw À 3 4 þ S ¼ (8:2)(53) 141:2(1:1)(1:7) ln 2,980 0:328 À 0:75 ¼ 0:1968 STB=d-psi 0 1,000 2,000 3,000 4,000 5,000 6,000 0 200 400 600 800 1,000 1,200 qo (stb/day) pwf(psia) Figure 3.6 Transient IPR curve for Example Problem 3.1. 0 1,000 2,000 3,000 4,000 5,000 6,000 0 200 400 600 800 1,000 1,200 qo (stb/day) pwf(psia) Figure 3.7 Steady-state IPR curve for Example Problem 3.1. pwf (psi) qo(stb/day) 50 1,011 5,651 0 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 33 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/33
- 44. Calculated points are: Pseudo–steady-state IPR curve is plotted in Fig. 3.8. 3.3.2 LPR for Two-Phase Reservoirs The linear IPR model presented in the previous section is valid for pressure values as low as bubble-point pressure. Below the bubble-point pressure, the solution gas escapes from the oil and become free gas. The free gas occupies some portion of pore space, which reduces flow of oil. This effect is quantified by the reduced relative permeability. Also, oil viscosity in- creases as its solution gas content drops. The combination of the relative permeability effect and the viscosity effect results in lower oil production rate at a given bottom-hole pressure. This makes the IPR curve deviating from the linear trend below bubble-point pressure, as shown in Fig. 3.5. The lower the pressure, the larger the deviation. If the reservoir pressure is below the initial bubble-point pressure, oil and gas two- phase flow exists in the whole reservoir domain and the reservoir is referred as a ‘‘two-phase reservoir.’’ Only empirical equations are available for modeling IPR of two-phase reservoirs. These empirical equations include Vogel’s (1968) equation extended by Standing (1971), the Fetkovich (1973) equation, Bandakhlia and Aziz’s (1989) equation, Zhang’s (1992) equation, and Retnanto and Economides’ (1998) equation. Vogel’s equa- tion is still widely used in the industry. It is written as q ¼ qmax 1 À 0:2 pwf pp À 0:8 pwf pp 2 # (3:19) or pwf ¼ 0:125pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 q qmax s À 1 # , (3:20) where qmax is an empirical constant and its value represents the maximum possible value of reservoir deliverability, or AOF. The qmax can be theoretically estimated based on res- ervoir pressure and productivity index above the bubble- point pressure. The pseudo–steady-state flow follows that qmax ¼ JÃpp 1:8 : (3:21) Derivation of this relation is left to the reader for an exercise. Fetkovich’s equation is written as q ¼ qmax 1 À pwf pp 2 #n (3:22) or q ¼ C(pp2 À p2 wf )n , (3:23) where C and n are empirical constants and is related to qmax by C ¼ qmax=pp2n . As illustrated in Example Problem 3.5, the Fetkovich equation with two constants is more accurate than Vogel’s equation IPR modeling. Again, Eqs. (3.19) and (3.23) are valid for average reservoir pressure pp being at and below the initial bubble-point pres- sure. Equation (3.23) is often used for gas reservoirs. Example Problem 3.2 Construct IPR of a vertical well in a saturated oil reservoir using Vogel’s equation. The following data are given: Porosity: f ¼ 0:19 Effective horizontal permeability: k ¼ 8.2 md Pay zone thickness: h ¼ 53 ft Reservoir pressure: pp ¼ 5,651 psia Bubble point pressure: pb¼ 5,651 psia Fluid formation volume factor: Bo¼ 1:1 Fluid viscosity: mo¼ 1:7 cp Total compressibility: ct ¼ 0:0000129 psiÀ1 Drainage area: A ¼ 640 acres (re ¼ 2,980 ft) Wellbore radius: rw ¼ 0:328 ft Skin factor: S ¼ 0 Solution JÃ ¼ kh 141:2Bm ln re rw À 3 4 þ S ¼ (8:2)(53) 141:2(1:1)(1:7) ln 2,980 0:328 À 0:75 ¼ 0:1968 STB=d-psi qmax ¼ JÃpp 1:8 ¼ (0:1968)(5,651) 1:8 ¼ 618 stb=day Calculated points by Eq. (3.19) are The IPR curve is plotted in Fig. 3.9. 3.3.3 IPR for Partial Two-Phase Oil Reservoirs If the reservoir pressure is above the bubble-point pressure and the flowing bottom-hole pressure is below the bubble- point pressure, a generalized IPR model can be formu- lated. This can be done by combining the straight-line IPR model for single-phase flow with Vogel’s IPR model for two-phase flow. Figure 3.10 helps to understand the formulation. According to the linear IPR model, the flow rate at bubble-point pressure is qb ¼ JÃ (pp À pb), (3:24) 0 1,000 2,000 3,000 4,000 5,000 6,000 qo (stb/day) pwf(psia) 0 200 400 600 800 1,000 1,200 Figure 3.8 Pseudo–steady-state IPR curve for Example Problem 3.1. pwf (psi) qo(stb/day) 50 1,102 5,651 0 pwf (psi) qo (stb/day) 5,651 0 5,000 122 4,500 206 4,000 283 3,500 352 3,000 413 2,500 466 2,000 512 1,500 550 1,000 580 500 603 0 618 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 34 3.1.2007 8:30pm Compositor Name: SJoearun 3/34 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 45. Based on Vogel’s IPR model, the additional flow rate caused by a pressure below the bubble-point pressure is expressed as Dq ¼ qv 1 À 0:2 pwf pb À 0:8 pwf pb 2 # : (3:25) Thus, the flow rate at a given bottom-hole pressure that is below the bubble-point pressure is expressed as q ¼ qb þ qv 1 À 0:2 pwf pb À 0:8 pwf pb 2 # : (3:26) Because qv ¼ JÃ pb 1:8 , (3:27) Eq. (3.26) becomes q ¼ JÃ (pp À pb) þ JÃ pb 1:8 Â 1 À 0:2 pwf pb À 0:8 pwf pb 2 # : (3:28) Example Problem 3.3 Construct IPR of a vertical well in an undersaturated oil reservoir using the generalized Vogel equation. The following data are given: Porosity: f ¼ 0:19 Effective horizontal permeability: k ¼ 8.2 md Pay zone thickness: h ¼ 53 ft Reservoir pressure: pp ¼ 5,651 psia Bubble point pressure: pb¼ 3,000 psia Fluid formation volume factor: Bo¼ 1:1 Fluid viscosity: mo¼ 1:7 cp Total compressibility: ct ¼ 0:0000129 psiÀ1 Drainage area: A ¼ 640 acres (re¼ 2,980 ft) Wellbore radius: rw ¼ 0:328 ft Skin factor: S ¼ 0 Solution JÃ ¼ kh 141:2Bm ln re rw À 3 4 þ S ¼ (8:2)(53) 141:2(1:1)(1:7) ln 2,980 0:328 À 0:75 ¼ 0:1968 STB=d-psi qb ¼ JÃ (pp À pb) ¼ (0:1968)(5,651 À 3,000) ¼ 522 sbt=day qv ¼ JÃ pb 1:8 ¼ (0:1968)(3,000) 1:8 ¼ 328 stb=day Calculated points by Eq. (3.28) are The IPR curve is plotted in Fig. 3.11. 3.4 Construction of IPR Curves Using Test Points It has been shown in the previous section that well IPR curves can be constructed using reservoir parameters in- cluding formation permeability, fluid viscosity, drainage area, wellbore radius, and well skin factor. These param- eters determine the constants (e.g., productivity index) in the IPR model. However, the values of these parameters are not always available. Thus, test points (measured val- ues of production rate and flowing bottom-hole pressure) are frequently used for constructing IPR curves. Constructing IPR curves using test points involves back- ing-calculation of the constants in the IPR models. For a single-phase (unsaturated oil) reservoir, the model con- stant JÃ can be determined by JÃ ¼ q1 (pp À pwf 1) , (3:29) where q1 is the tested production rate at tested flowing bottom-hole pressure pwf 1. 0 1,000 2,000 3,000 4,000 5,000 6,000 0 100 200 300 400 500 600 700 q (stb/day) pwf(psia) Figure 3.9 IPR curve for Example Problem 3.2. pwf (psi) qo (stb/day) 0 850 565 828 1,130 788 1,695 729 2,260 651 2,826 555 3,000 522 5,651 0 pwf 0 AOF pi pb q qb qb = J ( p − pb )* 1.8 *J pb qV = Figure 3.10 Generalized Vogel IPR model for partial two-phase reservoirs. 0 1,000 2,000 3,000 4,000 5,000 6,000 0 200 400 600 800 qo (stb/day) pwf(psia) Figure 3.11 IPR curve for Example Problem 3.3. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 35 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/35
- 46. Forapartialtwo-phasereservoir,modelconstantJÃ inthe generalizedVogelequationmustbedeterminedbasedonthe range of tested flowing bottom-hole pressure. If the tested flowing bottom-hole pressure is greater than bubble-point pressure, the model constant JÃ should be determined by JÃ ¼ q1 (pp À pwf 1) : (3:30) If the tested flowing bottom-hole pressure is less than bubble-point pressure, the model constant JÃ should be determined using Eq. (3.28), that is, JÃ ¼ q1 (pp À pb) þ pb 1:8 1 À 0:2 pwf 1 pb À 0:8 pwf 1 pb 2 #! : (3:31) Example Problem 3.4 Construct IPR of two wells in an undersaturated oil reservoir using the generalized Vogel equation. The following data are given: Reservoir pressure: pp ¼ 5,000 psia Bubble point pressure: pb ¼ 3,000 psia Tested flowing bottom-hole pressure in Well A: pwf 1 ¼ 4,000 psia Tested production rate from Well A: q1 ¼ 300 stb=day Tested flowing bottom hole pressure in Well B: pwf 1 ¼ 2,000 psia Tested production rate from Well B: q1 ¼ 900 stb=day Solution Well A: JÃ ¼ q1 (pp À pwf 1) ¼ 300 (5,000 À 4,000) ¼ 0:3000 stb=day-psi Calculated points are The IPR curve is plotted in Fig. 3.12. Well B: JÃ ¼ q1 (pp À pb) þ pb 1:8 1 À 0:2 pwf 1 pb À 0:8 pwf 1 pb 2 ! ¼ 900 (5,000 À 3,000) þ 3,000 1:8 1 À 0:2 2,000 3,000 À 0:8 2,000 3,000 2 ! ¼ 0:3156 stb=day-psi Calculated points are The IPR curve is plotted in Fig. 3.13. For a two-phase (saturated oil) reservoir, if the Vogel equation, Eq. (3.20), is used for constructing the IPR curve, the model constant qmax can be determined by qmax ¼ q1 1 À 0:2 pwf 1 pp À 0:8 pwf 1 pp 2 : (3:32) The productivity index at and above bubble-point pres- sure, if desired, can then be estimated by JÃ ¼ 1:8qmax pp : (3:33) If Fetkovich’s equation, Eq. (3.22), is used, two test points are required for determining the values of the two model constant, that is, n ¼ log q1 q2 log pp2Àp2 wf 1 pp2Àp2 wf 2 (3:34) and C ¼ q1 (pp2 À p2 wf 1)n , (3:35) where q1 and q2 are the tested production rates at tested flowing bottom-hole pressures pwf 1 and pwf 1, respectively. Example Problem 3.5 Construct IPR of a well in a saturated oil reservoir using both Vogel’s equation and Fetkovich’s equation. The following data are given: pwf (psia) q (stb/day) 0 1,100 500 1,072 1,000 1,022 1,500 950 2,000 856 2,500 739 3,000 600 5,000 0 0 1,000 2,000 3,000 4,000 5,000 6,000 0 200 400 600 800 1,000 1,200 q (stb/day) pwf(psia) Figure 3.12 IPR curves for Example Problem 3.4, Well A. 0 1,000 2,000 3,000 4,000 5,000 6,000 0 200 400 600 800 1,000 1,200 q (stb/day) pwf(psia) Figure 3.13 IPR curves for Example Problem 3.4, Well B. pwf (psia) q (stb/day) 0 1,157 500 1,128 1,000 1,075 1,500 999 2,000 900 2,500 777 3,000 631 5,000 0 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 36 3.1.2007 8:30pm Compositor Name: SJoearun 3/36 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 47. Reservoir pressure, pp ¼ 3,000 psia Tested flowing bottom-hole pressure, pwf 1 ¼ 2,000 psia Tested production rate at pwf 1, q1 ¼ 500 stb=day Tested flowing bottom-hole pressure, pwf 2 ¼ 1,000 psia Tested production rate at pwf 2, q2 ¼ 800 stb=day Solution Vogel’s equation: qmax ¼ q1 1 À 0:2 pwf 1 pp À 0:8 pwf 1 pp 2 ¼ 500 1 À 0:2 2;000 3;000 À 0:8 2;000 3;000 2 ¼ 978 stb=day Calculated data points are Fetkovich’s equation: n ¼ log q1 q2 log pp2Àp2 wf 1 pp2Àp2 wf 2 ¼ log 500 800 log (3,000)2 À (2,000)2 (3,000)2 À (1,000)2 ! ¼ 1:0 C ¼ q1 (pp2 À p2 wf 1)n ¼ 500 ((3,000)2 À (2,000)2 )1:0 ¼ 0:0001 stb=day-psi2n Calculated data points are The IPR curves are plotted in Fig. 3.14, which indicates that Fetkovich’s equation with two constants catches more details than Vogel’s equation. 3.5 Composite IPR of Stratified Reservoirs Nearly all producing formations are stratified to some extent. This means that the vertical borehole in the production zone has different layers having different reservoir pressures, permeabilities, and producing fluids. If itisassumedthattherearenoothercommunicationbetween these formations (other than the wellbore), the production will come mainly from the higher permeability layers. As the well’s rate of production is gradually increased, the less consolidated layers will begin to produce one by one (at progressively lower GOR), and so the overall ratio of pro- ductionwill fallasthe rateisincreased.If, however,the most highly depleted layers themselves produce at high ratios because of high free gas saturations, the overall GOR will eventually start to rise as the rate is increased and this climb will be continued (after the most permeable zone has come onto production). Thus, it is to be expected that a well producing from a stratified formation will exhibit a minimum GOR as the rate of production is increased. One of the major concerns in a multiplayer system is that interlayer cross-flow may occur if reservoir fluids are produced from commingled layers that have unequal ini- tial pressures. This cross-flow greatly affects the composite IPR of the well, which may result in an optimistic estimate of production rate from the commingled layers. El-Banbi and Wattenbarger (1996, 1997) investigated productivity of commingled gas reservoirs based on history matching to production data. However, no information was given in the papers regarding generation of IPR curves. 3.5.1 Composite IPR Models The following assumptions are made in this section: 1. Pseudo–steady-state flow prevails in all the reservoir layers. 2. Fluids from/into all the layers have similar properties. 3. Pressure losses in the wellbore sections between layers are negligible (these pressure losses are considered in Chapter 6 where multilateral wells are addressed). 4. The IPR of individual layers is known. On the basis of Assumption 1, under steady-flow condi- tions, the principle of material balance dictates net mass flow rate from layers to the well ¼ mass flow rate at well head or Xn i¼1 riqi ¼ rwhqwh, (3:36) where ri ¼ density of fluid from/into layer i, qi ¼ flow rate from/into layer i, rwh ¼ density of fluid at wellhead, qwh ¼ flow rate at wellhead, and n ¼ number of layers. Fluid flow from wellbore to reservoir is indicated by negativeqi.UsingAssumption2andignoringdensitychange from bottom hole to well head, Eq. (3.36) degenerates to Xn i¼1 qi ¼ qwh (3:37) or Xn i¼1 Ji(ppi À pwf ) ¼ qwh, (3:38) where Ji is the productivity index of layer i. 3.5.1.1 Single-Phase Liquid Flow For reservoir layers containing undersaturated oils, if the flowing bottom-hole pressure is above the bubble-point pressures of oils in all the layers, single-phase flow in all the layers is expected. Then Eq. (3.38) becomes Xn i¼1 JÃ i (ppi À pwf ) ¼ qwh, (3:39) where JÃ i is the productivity index of layer i at and above the bubble-point pressure. Equations (3.39) represents a linear composite IPR of the well. A straight-line IPR can be pwf (psia) q (stb/day) 0 978 500 924 1,000 826 1,500 685 2,000 500 2,500 272 3,000 0 pwf (psia) q (stb/day) 0 900 500 875 1,000 800 1,500 675 2,000 500 2,500 275 3,000 0 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 37 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/37
- 48. drawn through two points at AOF and shut-in bottom- hole pressure (pwfo). It is apparent from Eq. (3.39) that AOF ¼ Xn i¼1 Jà i ppi ¼ Xn i¼1 AOFi (3:40) and pwfo ¼ Pn i¼1 Jà i ppi Pn i¼1 Jà i : (3:41) It should be borne in mind that pwfo is a dynamic bottom- hole pressure because of cross-flow between layers. 3.5.1.2 Two-Phase Flow For reservoir layers containing saturated oils, two-phase flow is expected. Then Eq. (3.38) takes a form of polyno- mial of order greater than 1. If Vogel’s IPR model is used, Eq. (3.38) becomes Xn i¼1 Jà i ppi 1:8 1 À 0:2 pwf ppi À 0:8 pwf ppi 2 # ¼ qwh, (3:42) which gives AOF ¼ Xn i¼1 Jà i ppi 1:8 ¼ Xn i¼1 AOFi (3:43) and pwfo ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 80 Pn i¼1 Jà i ppi Pn i¼1 Jà i ppi þ Pn i¼1 Jà i 2 s À Pn i¼1 Jà i 8 Pn i¼1 Jà i ppi : (3:44) Again, pwfo is a dynamic bottom-hole pressure because of cross-flow between layers. 3.5.1.3 Partial Two-Phase Flow The generalized Vogel IPR model can be used to describe well inflow from multilayer reservoirs where reservoir pressures are greater than oil bubble pressures and the wellbore pressure is below these bubble-point pressures. Equation (3.38) takes the form Xn i¼1 Jà i  (ppi À pbi) þ pbi 1:8 1 À 0:2 pwf pbi À 0:8 pwf pbi 2 #( ) ¼ qwh, (3:45) which gives AOF ¼ Xn i¼1 Jà i (ppi À 0:44pbi) ¼ Xn i¼1 AOFi (3:46) and pwfo ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 147 0:56 Pn i¼1 Jà i pbi þ Pn i¼1 Jà i (ppi À pbi) ! Pn i¼1 Jà i pbi þ Pn i¼1 Jà i 2 s À Pn i¼1 Jà i 8 Pn i¼1 Jà i pbi : (3:47) Again, pwfo is a dynamic bottom-hole pressure because of cross-flow between layers. 3.5.2 Applications The equations presented in the previous section can be readily used for generation of a composite IPR curve if all Jà i are known. Although numerous equations have been proposed to estimate Jà i for different types of wells, it is always recommended to determine Jà i based on flow tests of individual layers. If the tested rate (qi) was obtained at a wellbore pressure (pwfi) that is greater than the bubble- point pressure in layer i, the productivity index Jà i can be determined by Jà i ¼ qi ppi À pwfi : (3:48) If the tested rate (qi) was obtained at a wellbore pressure (pwfi) that is less than the bubble-point pressure in layer i, the productivity index Jà i should be determined by Jà i ¼ qi (ppi À pbi) þ pbi 1:8 1 À 0:2 pwfi pbi À 0:8 pwfi pbi 2 ! : (3:49) With Jà i , ppi, and pbi known, the composite IPR can be generated using Eq. (3.45). 0 500 1,000 1,500 2,000 2,500 3,000 3,500 0 200 400 600 800 1,000 1,200 q (stb/day) pwf(psai) Vogel's model Fetkovich's model Figure 3.14 IPR curves for Example Problem 3.5. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 38 3.1.2007 8:30pm Compositor Name: SJoearun 3/38 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 49. Case Study An exploration well in the south China Sea penetrated eight oil layers with unequal pressures within a short inter- val. These oil layers were tested in six groups. Layers B4 and C2 were tested together and Layers D3 and D4 were tested together. Test data and calculated productivity index (JÃ i ) are summarized in Table 3.1. The IPR curves of the individual layers are shown in Fig. 3.15. It is seen from this figure that productivities of Layers A4, A5, and B1 are significantly lower than those of other layers. It is expected that wellbore cross-flow should occur if the bottom pressure is above the lowest reservoir pressure of 2,254 psi. Layers B4, C1, and C2 should be the major thief zones because of their high injectivities (assum- ing to be equal to their productivities) and relatively low pressures. The composite IPR of these layers is shown in Fig. 3.16 where the net production rate from the well is plotted against bottom-hole pressure. It is seen from this figure that net oil production will not be available unless the bottom-hole pressure is reduced to below 2,658 psi. Figure 3.15 suggests that the eight oil layers be produced separately in three layer groups: Group 1: Layers D3 and D4 Group 2: Layers B4, C1, and C2 Group 3: Layers B1, A4 and A5 The composite IPR for Group 1 (D3 and D4) is the same as shown in Fig. 3.15 because these two layers were the commingle-tested. Composite IPRs of Group 2 and Group 3 are plotted in Figs. 3.17 and 3.18. Table 3.2 compares production rates read from Figs. 3.16, 3.17, and 3.18 at some pressures. This comparison indicates that significant production from Group 1 can be achieved at bottom-hole pressures higher than 2658 psi, while Group 2 and Group 3 are shut-in. A significant production from Group 1 and Group 2 can be achieved at bottom- hole pressures higher than 2,625 psi while Group 3 is shut- in. The grouped-layer production will remain beneficial until bottom-hole pressure is dropped to below 2,335 psi where Group 3 can be open for production. 3.6 Future IPR Reservoir deliverability declines with time. During transi- ent flow period in single-phase reservoirs, this decline is because the radius of the pressure funnel, over which the pressure drawdown (pi À pwf ) acts, increases with time, i.e., the overall pressure gradient in the reservoir drops with time. In two-phase reservoirs, as reservoir pressure depletes, reservoir deliverability drops due to reduced rela- tive permeability to oil and increased oil viscosity. Future IPR can be predicted by both Vogel’s method and Fetko- vich’s method. 3.6.1 Vogel’s Method Let JÃ p and JÃ f be the present productivity index and future productivity index, respectively. The following relation can be derived: Table 3.1 Summary of Test Points for Nine Oil Layers Layer no.: D3-D4 C1 B4-C2 B1 A5 A4 Layer pressure (psi) 3,030 2,648 2,606 2,467 2,302 2,254 Bubble point (psi) 26.3 4.1 4.1 56.5 31.2 33.8 Test rate (bopd) 3,200 3,500 3,510 227 173 122 Test pressure (psi) 2,936 2,607 2,571 2,422 2,288 2,216 JÃ (bopd=psi) 34 85.4 100.2 5.04 12.4 3.2 Oil Production Rate (stb/day) 0 500 1,000 1,500 2,000 2,500 3,000 3,500 300,000250,000200,000150,000100,00050,0000−50,000 BottomHolePressure(psi) D3-D4 C1 B4-C2 B1 A5 A4 Figure 3.15 IPR curves of individual layers. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 39 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/39
- 50. JÃ f JÃ p ¼ kro Bomo f kro Bomo p (3:50) or JÃ f ¼ JÃ p kro Bomo f kro Bomo p : (3:51) Thus, q ¼ JÃ f ppf 1:8 1 À 0:2 pwf ppf À 0:8 pwf ppf 2 # , (3:52) where ppf is the reservoir pressure in a future time. Example Problem 3.6 Determine the IPR for a well at the time when the average reservoir pressure will be 1,800 psig. The following data are obtained from laboratory tests of well fluid samples: Solution JÃ f ¼ JÃ p kro Bomo f kro Bomo p ¼ 1:01 0:685 3:59(1:150) 0:815 3:11(1:173) ¼ 0:75 stb=day-psi Vogel’s equation for future IPR: q ¼ JÃ f ppf 1:8 1 À 0:2 pwf ppf À 0:8 pwf ppf 2 # ¼ (0:75)(1,800) 1:8 1 À 0:2 pwf 1,800 À 0:8 pwf 1,800 2 # Calculated data points are as follows: Present and future IPR curves are plotted in Fig. 3.19. Table 3.2 Comparison of Commingled and Layer-Grouped Productions Production rate (stb/day) Grouped layers Bottom-hole pressure (psi) All layers commingled Group 1 Group 2 Group 3 Total 2,658 0 12,663 Shut-in Shut-in 12,663 2,625 7866 13,787 0 Shut-in 13,787 2,335 77,556 23,660 53,896 0 77,556 2,000 158,056 35,063 116,090 6,903 158,056 Figure 3.16 Composite IPR curve for all the layers open to flow. Composite IPR of layers B4, C1 and C2 Liquid Rate (stb/day) 100,0000−1E+05 0 500 1,000 1,500 2,000 2,500 3,000 3,500 200,000 300,000 400,000 500,000 BottomHolePressure(psi) Figure 3.17 Composite IPR curve for Group 2 (Layers B4, C1, and C2). Composite IPR of layers B1, A4 and A5 Liquid Rate (stb/day) 20,0000−2000 0 500 1,000 1,500 2,000 2,500 3,000 3,500 40,000 60,000 BottomHolePressure(psi) Figure 3.18 Composite IPR curve for Group 3 (Layers B1, A4, and A5). Reservoir properties Present Future Average pressure (psig) 2,250 1,800 Productivity index JÃ (stb/day-psi) 1.01 Oil viscosity (cp) 3.11 3.59 Oil formation volume factor (rb/stb) 1.173 1.150 Relative permeability to oil 0.815 0.685 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 40 3.1.2007 8:30pm Compositor Name: SJoearun 3/40 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 51. 3.6.2 Fetkovich’s Method The integral form of reservoir inflow relationship for multiphase flow is expressed as q¼ 0:007082kh ln re rw ðpe pwf f (p)dp, (3:53) where f(p) is a pressure function. The simplest two-phase flow case is that of constant pressure pe at the outer boundary (re), with pe less than the bubble-point pressure so that there is two-phase flow throughout the reservoir. Under these circumstances, f(p) takes on the value kro moBo , where kro is the relative permeability to oil at the satura- tion conditions in the formation corresponding to the pressure p. In this method, Fetkovich makes the key as- sumption that to a good degree of approximation, the expression kro moBo is a linear function of p, and is a straight line passing through the origin. If pi is the initial formation pressure (i.e., $ pe), then the straight-line assumption is kro moBo ¼ kro moBo p pi : (3:54) Substituting Eq. (3.54) into Eq. (3.53) and integrating the latter gives qo ¼ 0:007082kh ln re rw kro moBo i 1 2pi (p2 i À p2 wf ) (3:55) or qo ¼ J 0 i (p2 i À p2 wf ), (3:56) where J 0 i ¼ 0:007082kh ln re rw kro moBo i 1 2pi : (3:57) The derivative of Eq. (3.45) with respect to the flowing bottom-hole pressure is dqo dpwf ¼ À2J 0 i pwf : (3:58) This implies that the rate of change of q with respect to pwf is lower at the lower values of the inflow pressure. Next, we can modify Eq. (3.58) to take into account that in practice pe is not constant but decreases as cumulative production increases. The assumption made is that J 0 i will decrease in proportion to the decrease in average reservoir (drainage area) pressure. Thus, when the static pressure is pe( pi), the IPR equation is qo ¼ J 0 i pe pi (p2 e À p2 wf ) (3:59) or, alternatively, qo ¼ J0 (p2 e À p2 wf ), (3:60) where J0 ¼ J 0 i pe pi : (3:61) These equations may be used to extrapolate into the future. Example Problem 3.7 Using Fetkovich’s method plot the IPR curves for a well in which pi is 2,000 psia and J 0 i ¼ 5 Â 10À4 stb=day-psia2 . Predict the IPRs of the well at well shut-in static pressures of 1,500 and 1,000 psia. Reservoir pressure ¼ 2,250 psig Reservoir pressure ¼ 1,800 psig pwf (psig) q (stb/day) pwf (psig) q (stb/day) 2,250 0 1,800 0 2,025 217 1,620 129 1,800 414 1,440 246 1,575 591 1,260 351 1,350 747 1,080 444 1,125 884 900 525 900 1000 720 594 675 1096 540 651 450 1172 360 696 225 1227 180 729 0 1263 0 750 0 500 1,000 1,500 2,000 2,500 0 200 400 600 800 1,000 1,200 1,400 q (Stb/Day) pwf(psig) Reservoir pressure = 2,250 psig Reservoir pressure = 1,800 psig Figure 3.19 IPR curves for Example Problem 3.6. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 41 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/41
- 52. Solution The value of J 0 o at 1,500 psia is J 0 o ¼ 5 Â 10À4 1,500 2,000 ¼ 3:75 Â 10À4 stb=day (psia)2 , and the value of J 0 o at 1,000 psia is J 0 o ¼ 5 Â 10À4 1,000 2,000 ¼ 2:5 Â 10À4 stb=day(psia)2 : Using the above values for J 0 o and the accompanying pe in Eq. (3.46), the following data points are calculated: IPR curves are plotted in Fig. 3.20. Summary This chapter presented and illustrated various mathemat- ical models for estimating deliverability of oil and gas reservoirs. Production engineers should make selections of the models based on the best estimate of his/her reser- voir conditions, that is, flow regime and pressure level. The selected models should be validated with actual well pro- duction rate and bottom-hole pressure. At least one test point is required to validate a straight-line (single-liquid flow) IPR model. At least two test points are required to validate a curvic (single-gas flow or two-phase flow) IPR model. References bandakhlia, h. and aziz, k. Inflow performance relation- ship for solution-gas drive horizontal wells. Presented at the 64th SPE Annual Technical Conference and Exhibition held 8–11 October 1989, in San Antonio, Texas. Paper SPE 19823. chang, m. Analysis of inflow performance simulation of solution-gas drive for horizontal/slant vertical wells. Presented at the SPE Rocky Mountain Regional Meet- ing held 18–21 May 1992, in Casper, Wyoming. Paper SPE 24352. dietz, d.n. Determination of average reservoir pressure from build-up surveys. J. Pet. Tech. 1965; August. dake, l.p. Fundamentals of Reservoir Engineering. New York: Elsevier, 1978. earlougher, r.c. Advances in Well Test Analysis. Dallad: Society of Petroleum Engineers, 1977. el-banbi, a.h. and wattenbarger, r.a. Analysis of com- mingled tight gas reservoirs. Presented at the SPE Annual Technical Conference and Exhibition held 6– 9 October 1996, in Denver, Colorado. Paper SPE 36736. el-banbi, a.h. and wattenbarger, r.a. Analysis of com- mingled gas reservoirs with variable bottom-hole flow- ing pressure and non-Darcy flow. Presented at the SPE Annual Technical Conference and Exhibition held 5–8 October 1997, in San Antonio, Texas. Paper SPE 38866. fetkovich, m.j. The isochronal testing of oil wells. Pre- sented at the SPE Annual Technical Conference and Exhibition held 30 September–3 October 1973, Las Vegas, Nevada. Paper SPE 4529. joshi, s.d. Augmentation of well productivity with slant and horizontal wells. J. Petroleum Technol. 1988; June:729–739. retnanto, a. and economides, m. Inflow performance relationships of horizontal and multibranched wells in a solution gas drive reservoir. Presented at the 1998 SPE Annual Technical Conference and Exhibition held 27–30 September 1998, in New Orleans, Louisiana. Paper SPE 49054. pe ¼ 2,000 psig pe ¼ 1,500 psig pe ¼ 1,000 psig pwf ðpsigÞ q (stb/day) pwf ðpsigÞ q (stb/day) pwf ðpsigÞ q (stb/day) 2,000 0 1,500 0 1,000 0 1,800 380 1,350 160 900 48 1,600 720 1,200 304 800 90 1,400 1,020 1,050 430 700 128 1,200 1,280 900 540 600 160 1,000 1,500 750 633 500 188 800 1,680 600 709 400 210 600 1,820 450 768 300 228 400 1,920 300 810 200 240 200 1,980 150 835 100 248 0 2,000 0 844 0 250 0 500 1,000 1,500 2,000 2,500 0 500 1,000 1,500 2,000 2,500 q (stb/day) pwf(psig) Reservoir pressure = 2,000 psig Reservoir pressure = 1,500 psig Reservoir pressure = 1,000 psig Figure 3.20 IPR curves for Example Problem 3.7. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 42 3.1.2007 8:30pm Compositor Name: SJoearun 3/42 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 53. standing, m.b. Concerning the calculation of inflow per- formance of wells producing from solution gas drive reservoirs. J. Petroleum Technol. 1971; Sep.:1141–1142. vogel, j.v. Inflow performance relationships for solution- gas drive wells. J. Petroleum Technol. 1968; Jan.:83–92. Problems 3.1 Construct IPR of a vertical well in an oil reservoir. Consider (1) transient flow at 1 month, (2) steady-state flow, and (3) pseudo–steady-state flow. The following data are given: Porosity, f ¼ 0:25 Effective horizontal permeability, k ¼ 10 md Pay zone thickness, h ¼ 50 ft Reservoir pressure, pe or pp ¼ 5,000 psia Bubble point pressure, pb ¼ 100 psia Fluid formation volume factor, Bo ¼ 1:2 Fluid viscosity, mo ¼ 1:5 cp Total compressibility, ct ¼ 0:0000125 psiÀ1 Drainage area, A ¼ 640 acres (re ¼ 2,980 ft) Wellbore radius, rw ¼ 0:328 ft Skin factor, S ¼ 5 3.2 Construct IPR of a vertical well in a saturated oil reservoir using Vogel’s equation. The following data are given: Porosity, f ¼ 0:2 Effective horizontal permeability, k ¼ 80 md Pay zone thickness, h ¼ 55 ft Reservoir pressure, pp ¼ 4,500 psia Bubble point pressure, pb ¼ 4,500 psia Fluid formation volume factor, Bo ¼ 1:1 Fluid viscosity, mo ¼ 1:8 cp Total compressibility, ct ¼ 0:000013 psiÀ1 Drainage area, A ¼ 640 acres (re ¼ 2,980 ft) Wellbore radius, rw ¼ 0:328 ft Skin factor, S ¼ 2 3.3 Construct IPR of a vertical well in an unsaturated oil reservoir using generalized Vogel’s equation. The fol- lowing data are given: Porosity, f ¼ 0:25 Effective horizontal permeability, k ¼ 100 md Pay zone thickness, h ¼ 55 ft Reservoir pressure, pp ¼ 5,000 psia Bubble point pressure, pb ¼ 3,000 psia Fluid formation volume factor, Bo ¼ 1:2 Fluid viscosity, mo ¼ 1:8 cp Total compressibility, ct ¼ 0:000013 psiÀ1 Drainage area, A ¼ 640 acres (re ¼ 2,980 ft) Wellbore radius, rw ¼ 0:328 ft Skin factor, S ¼ 5.5 3.4 Construct IPR of two wells in an unsaturated oil reservoir using generalized Vogel’s equation. The fol- lowing data are given: Reservoir pressure, pp ¼ 5,500 psia Bubble point pressure, pb ¼ 3,500 psia Tested flowing bottom-hole pressure in Well A, pwf 1 ¼ 4,000 psia Tested production rate from Well A, q1 ¼ 400 stb=day Tested flowing bottom-hole pressure in Well B, pwf 1 ¼ 2,000 psia Tested production rate from Well B, q1 ¼ 1,000 stb=day 3.5 Construct IPR of a well in a saturated oil reservoir using both Vogel’s equation and Fetkovich’s equation. The following data are given: Reservoir pressure, pp ¼ 3,500 psia Tested flowing bottom-hole pressure, pwf 1 ¼ 2,500 psia Tested production rate at pwf 1,q1 ¼ 600 stb=day Tested flowing bottom-hole pressure, pwf 2 ¼ 1,500 psia Tested production rate at pwf 2,q2 ¼ 900 stb=day 3.6 Determine the IPR for a well at the time when the average reservoir pressure will be 1,500 psig. The fol- lowing data are obtained from laboratory tests of well fluid samples: 3.7 Using Fetkovich’s method, plot the IPR curve for a well in which pi is 3,000 psia and J 0 o ¼ 4 Â 10À4 stb=day-psia2 . Predict the IPRs of the well at well shut-in static pressures of 2,500 psia, 2,000 psia, 1,500 psia, and 1,000 psia. Reservoir properties Present Future Average pressure (psig) 2,200 1,500 Productivity index JÃ (stb/day-psi) 1.25 Oil viscosity (cp) 3.55 3.85 Oil formation volume factor (rb/stb) 1.20 1.15 Relative permeability to oil 0.82 0.65 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap03 Final Proof page 43 3.1.2007 8:30pm Compositor Name: SJoearun RESERVOIR DELIVERABILITY 3/43
- 54. 4 Wellbore Performance Contents 4.1 Introduction 4/46 4.2 Single-Phase Liquid Flow 4/46 4.3 Multiphase Flow in Oil Wells 4/48 4.4 Single-Phase Gas Flow 4/53 4.5 Mist Flow in Gas Wells 4/56 Summary 4/56 References 4/57 Problems 4/57 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 45 22.12.2006 6:07pm
- 55. 4.1 Introduction Chapter 3 described reservoir deliverability. However, the achievable oil production rate from a well is determined by wellhead pressure and the flow performance of production string, that is, tubing, casing, or both. The flow perform- ance of production string depends on geometries of the production string and properties of fluids being produced. The fluids in oil wells include oil, water, gas, and sand. Wellbore performance analysis involves establishing a re- lationship between tubular size, wellhead and bottom-hole pressure, fluid properties, and fluid production rate. Understanding wellbore flow performance is vitally im- portant to production engineers for designing oil well equipment and optimizing well production conditions. Oil can be produced through tubing, casing, or both in an oil well depending on which flow path has better per- formance. Producing oil through tubing is a better option in most cases to take the advantage of gas-lift effect. The traditional term tubing performance relationship (TPR) is used in this book (other terms such as vertical lift perform- ance have been used in the literature). However, the math- ematical models are also valid for casing flow and casing- tubing annular flow as long as hydraulic diameter is used. This chapter focuses on determination of TPR and pres- sure traverse along the well string. Both single-phase and multiphase fluids are considered. Calculation examples are illustrated with hand calculations and computer spread- sheets that are provided with this book. 4.2 Single-Phase Liquid Flow Single-phase liquid flow exists in an oil well only when the wellhead pressure is above the bubble-point pressure of the oil, which is usually not a reality. However, it is convenient to start from single-phase liquid for establishing the con- cept of fluid flow in oil wells where multiphase flow usually dominates. Consider a fluid flowing from point 1 to point 2 in a tubing string of length L and height Áz (Fig. 4.1). The first law of thermodynamics yields the following equation for pressure drop: DP ¼ P1 À P2 ¼ g gc rDz þ r 2gc Du2 þ 2fF ru2 L gcD (4:1) where ÁP ¼ pressure drop, lbf =ft2 P1 ¼ pressure at point 1, lbf =ft2 P2 ¼ pressure at point 2, lbf =ft2 g ¼ gravitational acceleration, 32:17 ft=s2 gc ¼ unit conversion factor, 32:17 lbm-ft=lbf -s2 r ¼ fluid density lbm=ft3 Áz ¼ elevation increase, ft u ¼ fluid velocity, ft/s fF ¼ Fanning friction factor L ¼ tubing length, ft D ¼ tubing inner diameter, ft The first, second, and third term in the right-hand side of the equation represent pressure drops due to changes in elevation, kinetic energy, and friction, respectively. The Fanning friction factor ( fF ) can be evaluated based on Reynolds number and relative roughness. Reynolds num- ber is defined as the ratio of inertial force to viscous force. The Reynolds number is expressed in consistent units as NRe ¼ Dur m (4:2) or in U.S. field units as NRe ¼ 1:48qr dm (4:3) where NRe ¼ Reynolds number q ¼ fluid flow rate, bbl/day r ¼ fluid density lbm=ft3 d ¼ tubing inner diameter, in. m ¼ fluid viscosity, cp For laminar flow where NRe 2,000, the Fanning friction factor is inversely proportional to the Reynolds number, or fF ¼ 16 NRe (4:4) For turbulent flow where NRe 2,100, the Fanning friction factor can be estimated using empirical cor- relations. Among numerous correlations developed by different investigators, Chen’s (1979) correlation has an explicit form and gives similar accuracy to the Cole- brook–White equation (Gregory and Fogarasi, 1985) that was used for generating the friction factor chart used in the petroleum industry. Chen’s correlation takes the following form: 1 ffiffiffiffiffi fF p ¼ À4 Â log « 3:7065 À 5:0452 NRe log «1:1098 2:8257 þ 7:149 NRe 0:8981 #( ) (4:5) where the relative roughness is defined as « ¼ d d, and d is the absolute roughness of pipe wall. The Fanning friction factor can also be obtained based on Darcy–Wiesbach friction factor shown in Fig. 4.2. The Darcy–Wiesbach friction factor is also referred to as the Moody friction factor ( fM) in some literatures. The rela- tion between the Moody and the Fanning friction factor is expressed as fF ¼ fM 4 : (4:6) Example Problem 4.1 Suppose that 1,000 bbl/day of 408API, 1.2 cp oil is being produced through 27 ⁄8-in., 8:6-lbm=ft tubing in a well that is 15 degrees from vertical. If the tubing wall relative roughness is 0.001, calculate the pressure drop over 1,000 ft of tubing. q ∆z L 1 2 Figure 4.1 Flow along a tubing string. Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 46 22.12.2006 6:07pm 4/46 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 56. Solution Oil-specific gravity: go ¼ 141:5 API þ 131:5 ¼ 141:5 40 þ 131:5 ¼ 0:825 Oil density: r ¼ 62:4go ¼ (62:5)(0:825) ¼ 51:57 lbm=ft3 Elevation increase: DZ ¼ cos (a)L ¼ cos (15)(1,000) ¼ 966 ft The 27 ⁄8-in., 8:6-lbm=ft tubing has an inner diameter of 2.259 in. Therefore, D ¼ 2:259 12 ¼ 0:188 ft: Fluid velocity can be calculated accordingly: u ¼ 4q pD2 ¼ 4(5:615)(1,000) p(0:188)2 (86,400) ¼ 2:34 ft=s: Reynolds number: NRe ¼ 1:48qr dm ¼ 1:48(1,000)(51:57) (2:259)(1:2) ¼ 28,115 2,100, turbulent flow Chen’s correlation gives 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 Reynolds Number FrictionFactor 0 0.000001 0.000005 0.00001 0.00005 0.0001 0.0002 0.0004 0.0006 0.001 0.002 0.004 0.006 0.01 0.015 0.02 0.03 0.04 0.05 Laminar Flow Relative Roughness Turbulent Flow 1.E+081.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 Figure 4.2 Darcy–Wiesbach friction factor diagram (used, with permission, from Moody, 1944). Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 47 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/47
- 57. 1 ffiffiffiffiffi fF p ¼ À4 log « 3:7065 À 5:0452 NRe log «1:1098 2:8257 þ 7:149 NRe 0:8981 #( ) ¼ 12:3255 fF ¼ 0:006583 If Fig. 4.2 is used, the chart gives a Moody friction factor of 0.0265. Thus, the Fanning friction factor is estimated as fF ¼ 0:0265 4 ¼ 0:006625 Finally, the pressure drop is calculated: DP ¼ g gc rDz þ r 2gc Du2 þ 2fF ru2 L gcD ¼ 32:17 32:17 (51:57)(966) þ 51:57 2(32:17) (0)2 þ 2(0:006625)(51:57)(2:34)2 (1000) (32:17)(0:188) ¼ 50,435 lbf =ft2 ¼ 350 psi 4.3 Multiphase Flow in Oil Wells In addition to oil, almost all oil wells produce a certain amount of water, gas, and sometimes sand. These wells are called multiphase-oil wells. The TPR equation for single- phase flow is not valid for multiphase oil wells. To analyze TPR of multiphase oil wells rigorously, a multiphase flow model is required. Multiphase flow is much more complicated than single- phase flow because of the variation of flow regime (or flow pattern). Fluid distribution changes greatly in different flow regimes, which significantly affects pressure gradient in the tubing. 4.3.1 Flow Regimes As shown in Fig. 4.3, at least four flow regimes have been identified in gas-liquid two-phase flow. They are bubble, slug, churn, and annular flow. These flow regimes occur as a progression with increasing gas flow rate for a given liquid flow rate. In bubble flow, gas phase is dispersed in the form of small bubbles in a continuous liquid phase. In slug flow, gas bubbles coalesce into larger bubbles that eventually fill the entire pipe cross-section. Between the large bubbles are slugs of liquid that contain smaller bub- bles of entrained gas. In churn flow, the larger gas bubbles become unstable and collapse, resulting in a highly turbu- lent flow pattern with both phases dispersed. In annular flow, gas becomes the continuous phase, with liquid flow- ing in an annulus, coating the surface of the pipe and with droplets entrained in the gas phase. 4.3.2 Liquid Holdup In multiphase flow, the amount of the pipe occupied by a phase is often different from its proportion of the total volumetric flow rate. This is due to density difference between phases. The density difference causes dense phase to slip down in an upward flow (i.e., the lighter phase moves faster than the denser phase). Because of this, the in situ volume fraction of the denser phase will be greater than the input volume fraction of the denser phase (i.e., the denser phase is ‘‘held up’’ in the pipe relative to the lighter phase). Thus, liquid ‘‘holdup’’ is defined as yL ¼ VL V , (4:7) where yL ¼ liquid holdup, fraction VL ¼ volume of liquid phase in the pipe segment, ft3 V ¼ volume of the pipe segment, ft3 Liquid holdup depends on flow regime, fluid proper- ties, and pipe size and configuration. Its value can be quantitatively determined only through experimental measurements. 4.3.3 TPR Models Numerous TPR models have been developed for analyzing multiphase flow in vertical pipes. Brown (1977) presents a thorough review of these models. TPR models for multi- phase flow wells fall into two categories: (1) homogeneous- flow models and (2) separated-flow models. Homogeneous models treat multiphase as a homogeneous mixture and do not consider the effects of liquid holdup (no-slip assump- tion). Therefore, these models are less accurate and are usually calibrated with local operating conditions in field applications. The major advantage of these models comes from their mechanistic nature. They can handle gas-oil- water three-phase and gas-oil-water-sand four-phase sys- tems. It is easy to code these mechanistic models in com- puter programs. Separated-flow models are more realistic than the homogeneous-flow models. They are usually given in the form of empirical correlations. The effects of liquid holdup (slip) and flow regime are considered. The major disad- vantage of the separated flow models is that it is difficult to code them in computer programs because most cor- relations are presented in graphic form. 4.3.3.1 Homogeneous-Flow Models Numerous homogeneous-flow models have been devel- oped for analyzing the TPR of multiphase wells since the pioneering works of Poettmann and Carpenter (1952). Poettmann–Carpenter’s model uses empirical two-phase friction factor for friction pressure loss calculations with- out considering the effect of liquid viscosity. The effect of liquid viscosity was considered by later researchers including Cicchitti (1960) and Dukler et al. (1964). A comprehensive review of these models was given by Hasan and Kabir (2002). Guo and Ghalambor (2005) presented work addressing gas-oil-water-sand four-phase flow. Assuming no slip of liquid phase, Poettmann and Car- penter (1952) presented a simplified gas-oil-water three- phase flow model to compute pressure losses in wellbores by estimating mixture density and friction factor. Accord- ing to Poettmann and Carpenter, the following equation can be used to calculate pressure traverse in a vertical tubing when the acceleration term is neglected: Dp ¼ rr þ kk rr Dh 144 (4:8) where Dp ¼ pressure increment, psi rr ¼ average mixture density (specific weight), lb=ft3 Dh ¼ depth increment, ft and kk ¼ f2F q2 oM2 7:4137 Â 1010D5 (4:9) where f2F ¼ Fanning friction factor for two-phase flow qo ¼ oil production rate, stb/day M ¼ total mass associated with 1 stb of oil D ¼ tubing inner diameter, ft The average mixture density rr can be calculated by rr ¼ r1 þ r2 2 (4:10) Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 48 22.12.2006 6:07pm 4/48 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 58. where r1 ¼ mixture density at top of tubing segment, lb=ft3 r2 ¼ mixture density at bottom of segment, lb=ft3 The mixture density at a given point can be calculated based on mass flow rate and volume flow rate: r ¼ M Vm (4:11) where M ¼ 350:17(go þ WOR gw) þ GORrairgg (4:12) Vm ¼ 5:615(Bo þ WOR Bw) þ (GOR À Rs) 14:7 p T 520 z 1:0 (4:13) and where go ¼ oil specific gravity, 1 for freshwater WOR ¼ producing water–oil ratio, bbl/stb gw ¼ water-specific gravity, 1 for freshwater GOR ¼ producing gas–oil ratio, scf/stb Flow Direction P OO 10 1.0 0.1 0.1 1.0 Superficial Gas Velocity, VSG, ft./sec. 10 100 H I J K L M N GFEDCBA R SuperficialWaterVelocity,VSL,ft./sec. Annular Mist (Water dispersed) Froth (Both phases dispersed) Slug 4 (Air dispersed) Bubble (Air dispersed) Figure 4.3 Flow regimes in gas-liquid flow (used, with permission, from Govier and Aziz, 1977). Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 49 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/49
- 59. rair ¼ density of air, lbm=ft3 gg ¼ gas-specific gravity, 1 for air Vm ¼ volume of mixture associated with 1 stb of oil, ft3 Bo ¼ formation volume factor of oil, rb/stb Bw ¼ formation volume factor of water, rb/bbl Rs ¼ solution gas–oil ratio, scf/stb p ¼ in situ pressure, psia T ¼ in situ temperature, 8R z ¼ gas compressibility factor at p and T. If data from direct measurements are not available, solution gas–oil ratio and formation volume factor of oil can be estimated using the following correlations: Rs ¼ gg p 18 100:0125API 100:00091t !1:2048 (4:14) Bo ¼ 0:9759 þ 0:00012 Rs gg go 0:5 þ1:25t #1:2 (4:15) where t is in situ temperature in 8F. The two-phase friction factor f2F can be estimated from a chart recommended by Poettmann and Carpenter (1952). For easy coding in com- puter programs, Guo and Ghalambor (2002) developed the following correlation to represent the chart: f2F ¼ 101:444À2:5 log (Drv) , (4:16) where (Drv) is the numerator of Reynolds number repre- senting inertial force and can be formulated as (Drv) ¼ 1:4737  10À5 Mqo D : (4:17) Because the Poettmann–Carpenter model takes a finite- difference form, this model is accurate for only short- depth incremental Áh. For deep wells, this model should be used in a piecewise manner to get accurate results (i.e., the tubing string should be ‘‘broken’’ into small segments and the model is applied to each segment). Because iterations are required to solve Eq. (4.8) for pressure, a computer spreadsheet program Poettmann- CarpenterBHP.xls has been developed. The program is available from the attached CD. Example Problem 4.2 For the following given data, calculate bottom-hole pressure: Tubing head pressure: 500 psia Tubing head temperature: 100 8F Tubing inner diameter: 1.66 in. Tubing shoe depth (near bottom hole): 5,000 ft Bottom hole temperature: 150 8F Liquid production rate: 2,000 stb/day Water cut: 25% Producing GLR: 1,000 scf/stb Oil gravity: 30 8API Water specific gravity: 1.05 1 for freshwater Gas specific gravity: 0.65 1 for air Solution This problem can be solved using the computer program Poettmann-CarpenterBHP.xls. The result is shown in Table 4.1. The gas-oil-water-sand four-phase flow model proposed by Guo and Ghalambor (2005) is similar to the gas-oil- water three-phase flow model presented by Poettmann and Carpenter (1952) in the sense that no slip of liquid phase was assumed. But the Guo–Ghalambor model takes a closed (integrated) form, which makes it easy to use. The Guo–Ghalambor model can be expressed as follows: 144b(p À phf ) þ 1 À 2bM 2 ln (144p þ M)2 þ N (144phf þ M)2 þ N À M þ b c N À bM2 ffiffiffiffiffi N p  tanÀ1 144p þ M ffiffiffiffiffi N p À tanÀ1 144phf þ M ffiffiffiffiffi N p ! ¼ a( cos u þ d2 e)L, (4:18) where the group parameters are defined as a ¼ 0:0765ggqg þ 350goqo þ 350gwqw þ 62:4gsqs 4:07Tavqg , (4:19) b ¼ 5:615qo þ 5:615qw þ qs 4:07TavQg , (4:20) c ¼ 0:00678 Tavqg A , (4:21) d ¼ 0:00166 A (5:615qo þ 5:615qw þ qs), (4:22) e ¼ fM 2gDH , (4:23) M ¼ cde cos u þ d2e , (4:24) N ¼ c2 e cos u ( cos u þ d2e)2 , (4:25) where A ¼ cross-sectional area of conduit, ft2 DH ¼ hydraulic diameter, ft fM ¼ Darcy–Wiesbach friction factor (Moody factor) g ¼ gravitational acceleration, 32:17 ft=s2 L ¼ conduit length, ft p ¼ pressure, psia phf ¼ wellhead flowing pressure, psia qg ¼ gas production rate, scf/d qo ¼ oil production rate, bbl/d qs ¼ sand production rate, ft3 =day qw ¼ water production rate, bbl/d Tav ¼ average temperature, 8R gg ¼ specific gravity of gas, air ¼ 1 go ¼ specific gravity of produced oil, freshwater ¼ 1 gs ¼ specific gravity of produced solid, fresh water ¼ 1 gw ¼ specific gravity of produced water, fresh water ¼ 1 The Darcy–Wiesbach friction factor (fM) can be obtained from diagram (Fig. 4.2) or based on Fanning friction factor (fF ) obtained from Eq. (4.16). The required relation is fM ¼ 4fF . Because iterations are required to solve Eq. (4.18) for pressure, a computer spreadsheet program Guo-Ghalam- borBHP.xls has been developed. Example Problem 4.3 For the following data, estimate bottom-hole pressure with the Guo–Ghalambor method: Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 50 22.12.2006 6:07pm 4/50 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 60. Solution This example problem is solved with the spreadsheet program Guo-GhalamborBHP.xls. The result is shown in Table 4.2. 4.3.3.2 Separated-Flow Models A number of separated-flow models are available for TPR calculations. Among many others are the Lockhart and Martinelli correlation (1949), the Duns and Ros correla- tion (1963), and the Hagedorn and Brown method (1965). Based on comprehensive comparisons of these models, Ansari et al. (1994) and Hasan and Kabir (2002) recom- mended the Hagedorn–Brown method with modifications for near-vertical flow. The modified Hagedorn–Brown (mH-B) method is an empirical correlation developed on the basis of the original work of Hagedorn and Brown (1965). The modifications include using the no-slip liquid holdup when the original correlation predicts a liquid holdup value less than the no- slip holdup and using the Griffith correlation (Griffith and Wallis, 1961) for the bubble flow regime. The original Hagedorn–Brown correlation takes the fol- lowing form: dP dz ¼ g gc rr þ 2fF rru2 m gcD þ rr D(u2 m) 2gcDz , (4:26) which can be expressed in U.S. field units as 144 dp dz ¼ rr þ fF M2 t 7:413 Â 1010D5 rr þ rr D(u2 m) 2gcDz , (4:27) where Mt ¼ total mass flow rate, lbm=d rr ¼ in situ average density, lbm=ft3 Total measured depth: 7,000 ft The average inclination angle: 20 deg Tubing inner diameter: 1.995 in. Gas production rate: 1 MMscfd Gas-specific gravity: 0.7 air ¼ 1 Oil production rate: 1,000 stb/d Oil-specific gravity: 0.85 H2O ¼ 1 Water production rate: 300 bbl/d Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom hole temperature: 224 8F Tubing head pressure: 300 psia Table 4.1 Result Given by Poettmann-CarpenterBHP.xls for Example Problem 4.2 Poettmann–CarpenterBHP.xls Description: This spreadsheet calculates flowing bottom-hole pressure based on tubing head pressure and tubing flow performance using the Poettmann–Carpenter method. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) Click ‘‘Solution’’ button; and (4) view result in the Solution section. Input data U.S. Field units Tubing ID: 1.66 in Wellhead pressure: 500 psia Liquid production rate: 2,000 stb/d Producing gas–liquid ratio (GLR): 1,000 scf/stb Water cut (WC): 25 % Oil gravity: 30 8API Water-specific gravity: 1.05 freshwater ¼1 Gas-specific gravity: 0.65 1 for air N2 content in gas: 0 mole fraction CO2 content in gas: 0 mole fraction H2S content in gas: 0 mole fraction Formation volume factor for water: 1.2 rb/stb Wellhead temperature: 100 8F Tubing shoe depth: 5,000 ft Bottom-hole temperature: 150 8F Solution Oil-specific gravity ¼ 0.88 freshwater ¼ 1 Mass associated with 1 stb of oil ¼ 495.66 lb Solution gas ratio at wellhead ¼ 78.42 scf/stb Oil formation volume factor at wellhead ¼ 1.04 rb/stb Volume associated with 1 stb oil @ wellhead ¼ 45.12 cf Fluid density at wellhead ¼ 10.99 lb/cf Solution gas–oil ratio at bottom hole ¼ 301.79 scf/stb Oil formation volume factor at bottom hole ¼ 1.16 rb/stb Volume associated with 1 stb oil @ bottom hole ¼ 17.66 cf Fluid density at bottom hole ¼ 28.07 lb/cf The average fluid density ¼ 19.53 lb/cf Inertial force (Drv) ¼ 79.21 lb/day-ft Friction factor ¼ 0.002 Friction term ¼ 293.12 (lb=cf)2 Error in depth ¼ 0.00 ft Bottom hole pressure ¼ 1,699 psia Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 51 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/51
- 61. um ¼ mixture velocity, ft/s and rr ¼ yLrL þ (1 À yL)rG, (4:28) um ¼ uSL þ uSG, (4:29) where rL ¼ liquid density, lbm=ft3 rG ¼ in situ gas density, lbm=ft3 uSL ¼ superficial velocity of liquid phase, ft/s uSG ¼ superficial velocity of gas phase, ft/s The superficial velocity of a given phase is defined as the volumetric flow rate of the phase divided by the pipe cross- sectional area for flow. The third term in the right-hand side of Eq. (4.27) represents pressure change due to kinetic energy change, which is in most instances negligible for oil wells. Obviously, determination of the value of liquid holdup yL is essential for pressure calculations. The mH-B cor- relation uses liquid holdup from three charts using the following dimensionless numbers: Liquid velocity number, NvL: NvL ¼ 1:938 uSL ffiffiffiffiffiffi rL s 4 r (4:30) Gas velocity number, NvG: NvG ¼ 1:938uSG ffiffiffiffiffiffi rL s 4 r (4:31) Pipe diameter number, ND: ND ¼ 120:872D ffiffiffiffiffiffi rL s r (4:32) Liquid viscosity number, NL: NL ¼ 0:15726 mL ffiffiffiffiffiffiffiffiffiffiffi 1 rLs3 4 s , (4:33) where D ¼ conduit inner diameter, ft s ¼ liquid–gas interfacial tension, dyne/cm mL ¼ liquid viscosity, cp mG ¼ gas viscosity, cp The first chart is used for determining parameter (CNL) based on NL. We have found that this chart can be re- placed by the following correlation with acceptable ac- curacy: (CNL) ¼ 10Y , (4:34) where Y ¼ À2:69851 þ 0:15841X1 À 0:55100X2 1 þ 0:54785X3 1 À 0:12195X4 1 (4:35) Table 4.2 Result Given by Guo-GhalamborBHP.xls for Example Problem 4.3 Guo-GhalamborBHP.xls Description: This spreadsheet calculates flowing bottom-hole pressure based on tubing head pressure and tubing flow performance using the Guo–Ghalambor Method. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click ‘‘Solution’’ button; and (4) view result in the Solution section. Input data U.S. Field units SI units Total measured depth: 7,000 ft Average inclination angle: 20 degrees Tubing inside diameter: 1.995 in. Gas production rate: 1,000,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil production rate: 1000 stb/d Oil-specific gravity: 0.85 H2O ¼ 1 Water production rate: 300 bbl/d Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 224 8F Tubing head pressure: 300 psia Solution A ¼ 3.1243196 in:2 D ¼ 0.16625 ft Tav ¼ 622 8R cos (u) ¼ 0.9397014 (Drv) ¼ 40.908853 fM ¼ 0.0415505 a ¼ 0.0001713 b ¼ 2.884E-06 c ¼ 1349785.1 d ¼ 3.8942921 e ¼ 0.0041337 M ¼ 20447.044 N ¼ 6.669Eþ09 Bottom-hole pressure, pwf ¼ 1,682 psia Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 52 22.12.2006 6:07pm 4/52 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 62. and X1 ¼ log [(NL) þ 3]: (4:36) Once the value of parameter (CNL) is determined, it is used for calculating the value of the group NvLp0:1 (CNL) N0:575 vG p0:1 a ND , where p is the absolute pressure at the location where pressure gradient is to be calculated, and pa is atmospheric pressure. The value of this group is then used as an entry in the second chart to determine parameter (yL=c). We have found that the second chart can be represented by the following correlation with good accuracy: yL c ¼ À0:10307 þ 0:61777[ log (X2) þ 6] À 0:63295[ log (X2) þ 6]2 þ 0:29598[ log (X2) þ 6]3 À 0:0401[ log (X2) þ 6]4 , (4:37) where X2 ¼ NvLp0:1 (CNL) N0:575 vG p0:1 a ND : (4:38) According to Hagedorn and Brown (1965), the value of parameter c can be determined from the third chart using a value of group NvGN0:38 L N2:14 D . We have found that for NvGN0:38 L N2:14 D 0:01 the third chart can be replaced by the following correlation with accept- able accuracy: c ¼ 0:91163 À 4:82176X3 þ 1,232:25X2 3 À 22,253:6X3 3 þ 116174:3X4 3 , (4:39) where X3 ¼ NvGN0:38 L N2:14 D : (4:40) However, c ¼ 1:0 should be used for NvGN0:38 L N2:14 D # 0:01. Finally, the liquid holdup can be calculated by yL ¼ c yL c : (4:41) The Fanning friction factor in Eq. (4.27) can be deter- mined using either Chen’s correlation Eq. (4.5) or (4.16). The Reynolds number for multiphase flow can be calcu- lated by NRe ¼ 2:2 Â 10À2 mt DmyL L m(1ÀyL) G , (4:42) where mt is mass flow rate. The modified mH-B method uses the Griffith correlation for the bubble-flow regime. The bubble-flow regime has been observed to exist when lG LB, (4:43) where lG ¼ usG um (4:44) and LB ¼ 1:071 À 0:2218 u2 m D , (4:45) which is valid for LB $ 0:13. When the LB value given by Eq. (4.45) is less than 0.13, LB ¼ 0:13 should be used. Neglecting the kinetic energy pressure drop term, the Griffith correlation in U.S. field units can be expressed as 144 dp dz ¼ rr þ fF m2 L 7:413 Â 1010D5rLy2 L , (4:46) where mL is mass flow rate of liquid only. The liquid holdup in Griffith correlation is given by the following expression: yL ¼ 1 À 1 2 1 þ um us À ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ um us 2 À 4 usG us s2 4 3 5, (4:47) where ms ¼ 0:8 ft=s. The Reynolds number used to obtain the friction factor is based on the in situ average liquid velocity, that is, NRe ¼ 2:2 Â 10À2 mL DmL : (4:48) To speed up calculations, the Hagedorn–Brown cor- relation has been coded in the spreadsheet program Hage- dornBrownCorrelation.xls. Example Problem 4.4 For the data given below, calculate and plot pressure traverse in the tubing string: Solution This example problem is solved with the spreadsheet program HagedornBrownCorrelation.xls. The result is shown in Table 4.3 and Fig. 4.4. 4.4 Single-Phase Gas Flow The first law of thermodynamics (conservation of energy) governs gas flow in tubing. The effect of kinetic energy change is negligible because the variation in tubing diam- eter is insignificant in most gas wells. With no shaft work device installed along the tubing string, the first law of thermodynamics yields the following mechanical balance equation: dP r þ g gc dZ þ fMn2 dL 2gcDi ¼ 0 (4:49) Because dZ ¼ cos udL, r ¼ 29ggP ZRT , and n ¼ 4qsczPscT pD2 i TscP , Eq. (4.49) can be rewritten as zRT 29gg dP P þ g gc cos u þ 8fMQ2 scP2 sc p2gcD5 i T2 sc zT P !2 ( ) dL ¼ 0, (4:50) which is an ordinary differential equation governing gas flow in tubing. Although the temperature T can be approximately expressed as a linear function of length L through geothermal gradient, the compressibility factor z is a function of pressure P and temperature T. This makes it difficult to solve the equation analytically. Fortunately, the pressure P at length L is not a strong function of temperature and compressibility factor. Approximate so- lutions to Eq. (4.50) have been sought and used in the natural gas industry. Tubing shoe depth: 9,700 ft Tubing inner diameter: 1.995 in. Oil gravity: 40 8API Oil viscosity: 5 cp Production GLR: 75 scf/bbl Gas-specific gravity: 0.7 air ¼ 1 Flowing tubing head pressure: 100 psia Flowing tubing head temperature: 80 8F Flowing temperature at tubing shoe: 180 8F Liquid production rate: 758 stb/day Water cut: 10 % Interfacial tension: 30 dynes/cm Specific gravity of water: 1.05 H2O ¼ 1 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 53 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/53
- 63. 4.4.1 Average Temperature and Compressibility Factor Method If single average values of temperature and compressibility factor over the entire tubing length can be assumed, Eq. (4.50) becomes zzRTT 29gg dP P þ g gc cos u þ 8fMQ2 csP2 sczz2 TT2 p2gcD5 i T2 scP2 ' dL ¼ 0: (4:51) By separation of variables, Eq. (4.51) can be integrated over the full length of tubing to yield P2 wf ¼ Exp(s)P2 hf þ 8fM[Exp(s) À 1]Q2 scP2 sczz2 TT2 p2gcD5 i T2 sc cos u , (4:52) where s ¼ 58gggL cos u gcRzzTT : (4:53) Equations (4.52) and (4.53) take the following forms when U.S. field units (qsc in Mscf/d), are used (Katz et al., 1959): Table 4.3 Result Given by HagedornBrownCorrelation.xls for Example Problem 4.4 HagedornBrownCorrelation.xls Description: This spreadsheet calculates flowing pressures in tubing string based on tubing head pressure using the Hagedorn–Brown correlation. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click ‘‘Solution’’ button; and (4) view result in the Solution section and charts. Input data U.S. Field units SI units Depth (D): 9,700 ft Tubing inner diameter (dti): 1.995 in. Oil gravity (API): 40 8API Oil viscosity (mo): 5 cp Production GLR (GLR): 75 scf/bbl Gas-specific gravity (gg): 0.7 air ¼1 Flowing tubing head pressure (phf ): 100 psia Flowing tubing head temperature (thf ): 80 8F Flowing temperature at tubing shoe (twf ): 180 8F Liquid production rate (qL): 758 stb/day Water cut (WC): 10 % Interfacial tension (s): 30 dynes/cm Specific gravity of water (gw): 1.05 H2O ¼ 1 Solution Depth Pressure (ft) (m) (psia) (MPa) 0 0 100 0.68 334 102 183 1.24 669 204 269 1.83 1,003 306 358 2.43 1,338 408 449 3.06 1,672 510 543 3.69 2,007 612 638 4.34 2,341 714 736 5.01 2,676 816 835 5.68 3,010 918 936 6.37 3,345 1,020 1,038 7.06 3,679 1,122 1,141 7.76 4,014 1,224 1,246 8.48 4,348 1,326 1,352 9.20 4,683 1,428 1,459 9.93 5,017 1,530 1,567 10.66 5,352 1,632 1,676 11.40 5,686 1,734 1,786 12.15 6,021 1,836 1,897 12.90 6,355 1,938 2,008 13.66 6,690 2,040 2,121 14.43 7,024 2,142 2,234 15.19 7,359 2,243 2,347 15.97 7,693 2,345 2,461 16.74 8,028 2,447 2,576 17.52 8,362 2,549 2,691 18.31 8,697 2,651 2,807 19.10 9,031 2,753 2,923 19.89 9,366 2,855 3,040 20.68 9,700 2,957 3,157 21.48 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 54 22.12.2006 6:07pm 4/54 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 64. p2 wf ¼ Exp(s)p2 hf þ 6:67 Â 10À4 [Exp(s) À 1]fMq2 sczz2 TT2 d5 i cos u (4:54) and s ¼ 0:0375ggL cos u zzTT (4:55) The Darcy–Wiesbach (Moody) friction factor fM can be found in the conventional manner for a given tubing diameter, wall roughness, and Reynolds number. How- ever, if one assumes fully turbulent flow, which is the case for most gas wells, then a simple empirical relation may be used for typical tubing strings (Katz and Lee 1990): fM ¼ 0:01750 d0:224 i for di# 4:277 in: (4:56) fM ¼ 0:01603 d0:164 i for di 4:277 in: (4:57) Guo (2001) used the following Nikuradse friction factor correlation for fully turbulent flow in rough pipes: fM ¼ 1 1:74 À 2 log 2« di 2 4 3 5 2 (4:58) Because the average compressibility factor is a function of pressure itself, a numerical technique such as Newton– Raphson iteration is required to solve Eq. (4.54) for bot- tom-hole pressure. This computation can be performed automatically with the spreadsheet program Average TZ.xls. Users need to input parameter values in the Input data section and run Macro Solution to get results. Example Problem 4.5 Suppose that a vertical well produces 2 MMscf/d of 0.71 gas-specific gravity gas through a 27 ⁄8 in. tubing set to the top of a gas reservoir at a depth of 10,000 ft. At tubing head, the pressure is 800 psia and the temperature is 150 8F; the bottom-hole temperature is 200 8F. The relative roughness of tubing is about 0.0006. Calculate the pressure profile along the tubing length and plot the results. Solution Example Problem 4.5 is solved with the spreadsheet program AverageTZ.xls. Table 4.4 shows the appearance of the spreadsheet for the Input data and Result sections. The calculated pressure profile is plotted in Fig. 4.5. 4.4.2 Cullender and Smith Method Equation (4.50) can be solved for bottom-hole pressure using a fast numerical algorithm originally developed by Cullender and Smith (Katz et al., 1959). Equation (4.50) can be rearranged as 0 2,000 4,000 6,000 8,000 10,000 12,000 Pressure (psia) Depth(ft) 3,5000 500 1,000 1,500 2,000 2,500 3,000 Figure 4.4 Pressure traverse given by HagedornBrownCorrelation.xls for Example Problem 4.4. Table 4.4 Spreadsheet AverageTZ.xls: the Input Data and Result Sections AverageTZ.xls Description: This spreadsheet calculates tubing pressure traverse for gas wells. Instructions: Step 1: Input your data in the Input data section. Step 2: Click ‘‘Solution’’ button to get results. Step 3: View results in table and in graph sheet ‘‘Profile’’. Input data gg ¼ 0.71 d ¼ 2.259 in. «=d ¼ 0.0006 L ¼ 10.000 ft u ¼ 0 degrees phf ¼ 800 psia Thf ¼ 150 8F Twf ¼ 200 8F qsc ¼ 2,000 Mscf/d Solution fM ¼ 0.017396984 Depth (ft) T (8R) p (psia) Zav 0 610 800 0.9028 1,000 615 827 0.9028 2,000 620 854 0.9027 3,000 625 881 0.9027 4,000 630 909 0.9026 5,000 635 937 0.9026 6,000 640 965 0.9026 7,000 645 994 0.9026 8,000 650 1023 0.9027 9,000 655 1053 0.9027 10,000 660 1082 0.9028 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 55 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/55
- 65. P zT dp g gc cos u P zT À Á2 þ 8fM Q2 scP2 sc p2gcD5 i T2 sc ¼ À 29gg R dL (4:59) that takes an integration form of ðPwf Phf P zT g gc cos u P zT À Á2 þ 8fM Q2 scP2 sc p2gcD5 i T2 sc 2 4 3 5dp ¼ 29ggL R : (4:60) In U.S. field units (qmsc in MMscf/d), Eq. (4.60) has the following form: ðpwf phf p zT 0:001 cos u p zT À Á2 þ 0:6666 fM q2 msc d5 i 2 4 3 5dp ¼ 18:75ggL (4:61) If the integrant is denoted with symbol I, that is, I ¼ p zT 0:001 cos u p zT À Á2 þ 0:6666 fM q2 sc d5 i , (4:62) Eq. (4.61) becomes ðpwf phf Idp ¼ 18:75ggL: (4:63) In the form of numerical integration, Eq. (4.63) can be expressed as (pmf À phf )(Imf þ Ihf ) 2 þ (pwf À pmf )(Iwf þ Imf ) 2 ¼ 18:75ggL, (4:64) where pmf is the pressure at the mid-depth. The Ihf , Imf , and Iwf are integrant Is evaluated at phf , pmf , and pwf , respectively. Assuming the first and second terms in the right-hand side of Eq. (4.64) each represents half of the integration, that is, (pmf À phf )(Imf þ Ihf ) 2 ¼ 18:75ggL 2 (4:65) (pwf À pmf )(Iwf þ Imf ) 2 ¼ 18:75ggL 2 , (4:66) the following expressions are obtained: pmf ¼ phf þ 18:75ggL Imf þ Ihf (4:67) pwf ¼ pmf þ 18:75ggL Iwf þ Imf (4:68) Because Imf is a function of pressure pmf itself, a numerical technique such as Newton–Raphson iteration is required to solve Eq. (4.67) for pmf . Once pmf is computed, pwf can be solved numerically from Eq. (4.68). These computa- tions can be performed automatically with the spreadsheet program Cullender-Smith.xls. Users need to input parameter values in the Input Data section and run Macro Solution to get results. Example Problem 4.6 Solve the problem in Example Problem 4.5 with the Cullender and Smith Method. Solution Example Problem 4.6 is solved with the spreadsheet program Cullender-Smith.xls. Table 4.5 shows the appearance of the spreadsheet for the Input data and Result sections. The pressures at depths of 5,000 ft and 10,000 ft are 937 psia and 1,082 psia, respectively. These results are exactly the same as that given by the Average Temperature and Compressibility Factor Method. 4.5 Mist Flow in Gas Wells In addition to gas, almost all gas wells produce certain amount of liquids. These liquids are formation water and/ or gas condensate (light oil). Depending on pressure and temperature, in some wells, gas condensate is not seen at surface, but it exists in the wellbore. Some gas wells pro- duce sand and coal particles. These wells are called multi- phase-gas wells. The four-phase flow model in Section 4.3.3.1 can be applied to mist flow in gas wells. Summary This chapter presented and illustrated different mathemat- ical models for describing wellbore/tubing performance. Among many models, the mH-B model has been found to give results with good accuracy. The industry practice is to conduct a flow gradient (FG) survey to measure the 0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 Pressure (psia) Depth(ft) 0 1,2001,000800600400200 Figure 4.5 Calculated tubing pressure profile for Example Problem 4.5. Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 56 22.12.2006 6:07pm 4/56 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 66. flowing pressures along the tubing string. The FG data are then employed to validate one of the models and tune the model if necessary before the model is used on a large scale. References ansari, a.m., sylvester, n.d., sarica, c., shoham, o., and brill, j.p. A comprehensive mechanistic model forupwardtwo-phaseflowinwellbores.SPEProduction and Facilities (May 1994) 143, Trans. AIME 1994; May:297. brown, k.e. The Technology of Artificial Lift Methods, Vol. 1. Tulsa, OK: PennWell Books, 1977, pp. 104– 158. chen, n.h. An explicit equation for friction factor in pipe. Ind. Eng. Chem. Fund. 1979;18:296. cicchitti, a. Two-phase cooling experiments—pressure drop, heat transfer and burnout measurements. Ener- gia Nucleare 1960;7(6):407. dukler, a.e., wicks, m., and cleveland, r.g. Frictional pressure drop in two-phase flow: a comparison of existing correlations for pressure loss and hold-up. AIChE J. 1964:38–42. duns, h. and ros, n.c.j. Vertical flow of gas and liquid mixtures in wells. Proceedings of the 6th World Petrol- eum Congress, Tokyo, 1963. goier, g.w. and aziz, k. The Flow of Complex Mixtures in Pipes. Huntington, NY: Robert E. Drieger Publishing Co., 1977. gregory, g.a. and fogarasi, m. Alternate to standard friction factor equation. Oil Gas J. 1985;April 1:120–127. griffith, p. and wallis, g.b. Two-phase slug flow. Trans. ASME 1961;83(Ser. C):307–320. guo, b. and ghalambor, a. Gas Volume Requirements for Underbalanced Drilling Deviated Holes. Tulsa, OK: PennWell Corporation, 2002, pp. 132–133. guo, b. and ghalambor, a. Natural Gas Enginee- ring Handbook. Houston: Gulf Publishing Company, 2005, pp. 59–61. hagedorn, a.r. and brown, k.e. Experimental study of pressure gradients occurring during continuous two- phase flow in small-diameter conduits. J. Petroleum Technol. 1965;475. hasan, a.r. and kabir, c.s. Fluid Flow and Heat Transfer in Wellbores. Richardson, TX: Society of Petroleum Engineers, 2002, pp. 10–15. katz, d.l., cornell, d., kobayashi, r., poettmann, f.h., vary, j.a., elenbaas, j.r., and weinaug, c.f. Handbook of Natural Gas Engineering. New York: McGraw-Hill Publishing Company, 1959. katz, d.l. and lee, r.l. Natural Gas Engineering—Produc- tion and Storage. New York: McGraw-Hill Publishing Company, 1990. lockhart, r.w. and martinelli, r.c. Proposed cor- relation of data for isothermal two-phase, two- component flow in pipes. Chem. Eng. Prog. 1949;39. poettmann, f.h. and carpenter, p.g. The multiphase flow of gas, oil, and water through vertical strings. API Dril. Prod. Prac. 1952:257–263. Problems 4.1 Suppose that 1,000 bbl/day of 16 8API, 5-cp oil is being produced through 27 ⁄8 -in., 8:6-lbm=ft tubing in a well that is 3 degrees from vertical. If the tubing wall relative roughness is 0.001, assuming no free gas in tubing string, calculate the pressure drop over 1,000 ft of tubing. 4.2 For the following given data, calculate bottom-hole pressure using the Poettmann–Carpenter method: Tubing head pressure: 300 psia Tubing head temperature: 100 8F Tubing inner diameter: 1.66 in. Tubing shoe depth (near bottom hole): 8,000 ft Bottom-hole temperature: 170 8F Liquid production rate: 2,000 stb/day Water cut: 30% Producing GLR: 800 scf/stb Oil gravity: 40 8API Water-specific gravity: 1.05 1 for freshwater Gas-specific gravity: 0.70 1 for air 4.3 For the data given below, estimate bottom-hole pres- sure with the Guo–Ghalambor method. Table 4.5. Spreadsheet Cullender-Smith.xls: the Input Data and Result Sections Cullender-SmithBHP.xls Description: This spreadsheet calculates bottom-hole pres- sure with the Cullender–Smith method. Instructions: Step 1: Input your data in the Input data section. Step 2: Click Solution button to get results. Input data gg ¼0.71 d ¼2.259 in. «=d ¼0.0006 L ¼10,000 ft u ¼0 degrees phf ¼800 psia Thf ¼150 8F Twf ¼200 8F qmsc ¼2 MMscf/d Solution fM ¼0.017397 Depth (ft) T (8R) p (psia) Z p/ZT I 0 610 800 0.9028 1.45263 501.137 5,000 635 937 0.9032 1.63324 472.581 10,000 660 1,082 0.9057 1.80971 445.349 Total measured depth: 8,000 ft The average inclination angle: 5 degrees Tubing inner diameter: 1.995 in. Gas production rate: 0.5 MMscfd Gas specific gravity: 0.75 air ¼ 1 Oil production rate: 2,000 stb/d Oil-specific gravity: 0.85 H2O ¼ 1 Water production rate: 500 bbl/d Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 4 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 170 8F Tubing head pressure: 500 psia (continued) Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 57 22.12.2006 6:07pm WELLBORE PERFORMANCE 4/57
- 67. 4.4 For the data given below, calculate and plot pressure traverse in the tubing string using the Hagedorn– Brown correlation: 4.5 Suppose 3 MMscf/d of 0.75 specific gravity gas is produced through a 31 ⁄2 -in. tubing string set to the top of a gas reservoir at a depth of 8,000 ft. At the tubing head, the pressure is 1,000 psia and the temperature is 120 8F; the bottom-hole temperature is 180 8F. The relative roughness of tubing is about 0.0006. Calculate the flowing bottom-hole pressure with three methods: (a) the average temperature and compressibility factor method; (b) the Cullender–Smith method; and (c) the four-phase flow method. Make comments on your re- sults. 4.6 Solve Problem 4.5 for gas production through a K-55, 17-lb/ft, 51 ⁄2-in casing. 4.7 Suppose 2 MMscf/d of 0.65 specific gravity gas is produced through a 27 ⁄8 -in. (2.259-in. inside diameter) tubing string set to the top of a gas reservoir at a depth of 5,000 ft. Tubing head pressure is 300 psia and the temperature is 100 8F; the bottom-hole temperature is 150 8F. The relative roughness of tubing is about 0.0006. Calculate the flowing bottom pressure with the average temperature and compressibility factor method. Tubing shoe depth: 6,000 ft Tubing inner diameter: 1.995 in. Oil gravity: 30 8API Oil viscosity: 2 cp Production GLR: 500 scf/bbl Gas-specific gravity: 0.65 air ¼ 1 Flowing tubing head pressure: 100 psia Flowing tubing head temperature: 80 8F Flowing temperature at tubing shoe: 140 8F Liquid production rate: 1,500 stb/day Water cut: 20% Interfacial tension: 30 dynes/cm Specific gravity of water: 1.05 H2O ¼ 1 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap04 Final Proof page 58 22.12.2006 6:07pm 4/58 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 68. 5 Choke Performance Contents 5.1 Introduction 5/60 5.2 Sonic and Subsonic Flow 5/60 5.3 Single-Phase Liquid Flow 5/60 5.4 Single-Phase Gas Flow 5/60 5.5 Multiphase Flow 5/63 Summary 5/66 References 5/66 Problems 5/66 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 59 21.12.2006 2:02pm
- 69. 5.1 Introduction Wellhead chokes are used to limit production rates for regulations, protect surface equipment from slugging, avoid sand problems due to high drawdown, and control flow rate to avoid water or gas coning. Two types of well- head chokes are used. They are (1) positive (fixed) chokes and (2) adjustable chokes. Placing a choke at the wellhead means fixing the well- head pressure and, thus, the flowing bottom-hole pressure and production rate. For a given wellhead pressure, by calculating pressure loss in the tubing the flowing bottom- hole pressure can be determined. If the reservoir pressure and productivity index is known, the flow rate can then be determined on the basis of inflow performance relation- ship (IPR). 5.2 Sonic and Subsonic Flow Pressure drop across well chokes is usually very significant. There is no universal equation for predicting pressure drop across the chokes for all types of production fluids. Differ- ent choke flow models are available from the literature, and they have to be chosen based on the gas fraction in the fluid and flow regimes, that is, subsonic or sonic flow. Both sound wave and pressure wave are mechanical waves. When the fluid flow velocity in a choke reaches the traveling velocity of sound in the fluid under the in situ condition, the flow is called ‘‘sonic flow.’’ Under sonic flow conditions, the pressure wave downstream of the choke cannot go upstream through the choke because the medium (fluid) is traveling in the opposite direction at the same velocity. Therefore, a pressure discontinuity exists at the choke, that is, the downstream pressure does not affect the upstream pressure. Because of the pressure discontinu- ity at the choke, any change in the downstream pressure cannot be detected from the upstream pressure gauge. Of course, any change in the upstream pressure cannot be detected from the downstream pressure gauge either. This sonic flow provides a unique choke feature that stabilizes well production rate and separation operation conditions. Whether a sonic flow exists at a choke depends on a downstream-to-upstream pressure ratio. If this pressure ratio is less than a critical pressure ratio, sonic (critical) flow exists. If this pressure ratio is greater than or equal to the critical pressure ratio, subsonic (subcritical) flow exists. The critical pressure ratio through chokes is expressed as poutlet pup c ¼ 2 k þ 1 k kÀ1 , (5:1) where poutlet is the pressure at choke outlet, pup is the upstream pressure, and k ¼ Cp=Cv is the specific heat ratio. The value of the k is about 1.28 for natural gas. Thus, the critical pressure ratio is about 0.55 for natural gas. A similar constant is used for oil flow. A typical choke performance curve is shown in Fig. 5.1. 5.3 Single-Phase Liquid Flow When the pressure drop across a choke is due to kinetic energy change, for single-phase liquid flow, the second term in the right-hand side of Eq. (4.1) can be rearranged as q ¼ CDA ffiffiffiffiffiffiffiffiffiffiffiffiffi 2gcDP r s , (5:2) where q ¼ flow rate, ft3 =s CD ¼ choke discharge coefficient A ¼ choke area, ft2 gc ¼ unit conversion factor, 32.17 lbm-ft=lbf -s2 ÁP ¼ pressure drop, lbf =ft2 r ¼ fluid density, lbm=ft3 If U.S. field units are used, Eq. (5.2) is expressed as q ¼ 8074CDd2 2 ffiffiffiffiffiffi Dp r s , (5:3) where q ¼ flow rate, bbl/d d2 ¼ choke diameter, in. Áp ¼ pressure drop, psi The choke discharge coefficient CD can be determined based on Reynolds number and choke/pipe diameter ratio (Figs. 5.2 and 5.3). The following correlation has been found to give reasonable accuracy for Reynolds numbers between 104 and 106 for nozzle-type chokes (Guo and Ghalambor, 2005): CD ¼ d2 d1 þ 0:3167 d2 d1 0:6 þ 0:025[ log (NRe) À 4], (5:4) where d1 ¼ upstream pipe diameter, in. d2 ¼ choke diameter, in. NRe ¼ Reynolds number based on d2 5.4 Single-Phase Gas Flow Pressure equations for gas flow through a choke are derived based on an isentropic process. This is because there is no time for heat to transfer (adiabatic) and the friction loss is negligible (assuming reversible) at chokes. In addition to the concern of pressure drop across the chokes, temperature drop associated with choke flow is also an important issue for gas wells, because hydrates may form that may plug flow lines. 0 0 0.2 0.4 0.6 0.8 1 1.2 0.1 0.2 0.3 p2/p1 q q Critical Sub- critical p1 p2 d1 d2 Figure 5.1 A typical choke performance curve. Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 60 21.12.2006 2:02pm 5/60 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 70. 5.4.1 Subsonic Flow Under subsonic flow conditions, gas passage through a choke can be expressed as qsc ¼ 1,248CDA2pup  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k (k À 1)ggTup pdn pup 2 k À pdn pup kþ1 k #v u u t , (5:5) where qsc ¼ gas flow rate, Mscf/d pup ¼ upstream pressure at choke, psia A2 ¼ cross-sectional area of choke, in:2 Tup ¼ upstream temperature, 8R g ¼ acceleration of gravity, 32:2 ft=s2 gg ¼ gas-specific gravity related to air The Reynolds number for determining CD is expressed as NRe ¼ 20qscgg md2 , (5:6) where m is gas viscosity in cp. Gas velocity under subsonic flow conditions is less than the sound velocity in the gas at the in situ conditions: n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 up þ 2gcCpTup 1 À zup zdn pdown pup kÀ1 k #v u u t , (5:7) where Cp ¼ specific heat of gas at constant pressure (187.7 lbf-ft/lbm-R for air). 0.9 0.95 1 1.05 1.1 1.15 1.2 1,000 10,000 100,000 1,000,000 10,000,000 Reynolds number CD 0.75 0.725 0.7 0.675 0.65 0.625 0.6 0.575 0.55 0.5 0.45 0.4 d2 /d1 Figure 5.2 Choke flow coefficient for nozzle-type chokes (data used, with permission, from Crane, 1957). 0.55 0.6 0.65 0.7 0.75 0.8 1,000 10,000 100,000 1,000,000 10,000,000 Reynolds number CD 0.75 0.725 0.7 0.65 0.6 0.5 0.45 0.4 0.3 0.2 d2 /d1 Figure 5.3 Choke flow coefficient for orifice-type chokes (data used, with permission, from Crane, 1957). Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 61 21.12.2006 2:02pm CHOKE PERFORMANCE 5/61
- 71. 5.4.2 Sonic Flow Under sonic flow conditions, the gas passage rate reaches its maximum value. The gas passage rate is expressed in the following equation for ideal gases: Qsc ¼ 879CDApup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup ! 2 k þ 1 kþ1 kÀ1 v u u t (5:8) The choke flow coefficient CD is not sensitive to the Rey- nolds number for Reynolds number values greater than 106 . Thus, the CD value at the Reynolds number of 106 can be assumed for CD values at higher Reynolds numbers. Gas velocity under sonic flow conditions is equal to sound velocity in the gas under the in situ conditions: n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n2 up þ 2gcCpTup 1 À zup zoutlet 2 k þ 1 !s (5:9) or n % 44:76 ffiffiffiffiffiffiffi Tup p (5:10) 5.4.3 Temperature at Choke Depending on the upstream-to-downstream pressure ratio, the temperature at choke can be much lower than expected. This low temperature is due to the Joule–Thomson cooling effect, that is, a sudden gas expansion below the nozzle causes a significant temperature drop. The temperature can easily drop to below ice point, resulting in ice-plugging if water exists. Even though the temperature still can be above ice point, hydrates can form and cause plugging problems. Assuming an isentropic process for an ideal gas flowing through chokes, the temperature at the choke downstream can be predicted using the following equation: Tdn ¼ Tup zup zoutlet poutlet pup kÀ1 k (5:11) The outlet pressure is equal to the downstream pressure in subsonic flow conditions. 5.4.4 Applications Equations (5.5) through (5.11) can be used for estimating . Downstream temperature . Gas passage rate at given upstream and downstream pressures . Upstream pressure at given downstream pressure and gas passage . Downstream pressure at given upstream pressure and gas passage To estimate the gas passage rate at given upstream and downstream pressures, the following procedure can be taken: Step 1: Calculate the critical pressure ratio with Eq. (5.1). Step 2: Calculate the downstream-to-upstream pressure ratio. Step 3: If the downstream-to-upstream pressure ratio is greater than the critical pressure ratio, use Eq. (5.5) to calculate gas passage. Otherwise, use Eq. (5.8) to calculate gas passage. Example Problem 5.1 A0.6 specific gravitygas flowsfrom a 2-in. pipe through a 1-in. orifice-type choke. The upstream pressure and temperature are 800 psia and 75 8F, respectively. The downstream pressure is 200 psia (measured 2 ft from the orifice). The gas-specific heat ratio is 1.3. (a) What is the expected daily flow rate? (b) Does heating need to be applied to ensure that the frost does not clog the orifice? (c) What is the expected pressure at the orifice outlet? Solution (a) Poutlet Pup c ¼ 2 k þ 1 k kÀ1 ¼ 2 1:3 þ 1 1:3 1:3À1 ¼ 0:5459 Pdn Pup ¼ 200 800 ¼ 0:25 0:5459 Sonic flow exists: d2 d1 ¼ 100 200 ¼ 0:5 Assuming NRe 106 , Fig. 5.2 gives CD ¼ 0:62. qsc ¼ 879CDAPup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup ! 2 k þ 1 kþ1 kÀ1 v u u t qsc ¼ (879)(0:62)[p(1)2 =4](800) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:3 (0:6)(75 þ 460) 2 1:3 þ 1 1:3þ1 1:3À1 s qsc ¼ 12,743 Mscf=d Check NRe: m ¼ 0:01245 cp by the Carr–Kobayashi–Burrows cor- relation. NRe ¼ 20qscgg md2 ¼ (20)(12,743)(0:6) (0:01245)(1) ¼ 1:23 Â 107 106 (b) Tdn ¼ Tup zup zoutlet Poutlet Pup kÀ1 k ¼ (75 þ 460)(1)(0:5459) 1:3À1 1:3 ¼ 465 R ¼ 5 F 32 F Therefore, heating is needed to prevent icing. (c) Poutlet ¼ Pup Poutlet Pup ¼ (800)(0:5459) ¼ 437 psia Example Problem 5.2 A 0.65 specific gravity natural gas flows from a 2-in. pipe through a 1.5-in. nozzle-type choke. The upstream pressure and temperature are 100 psia and 70 8F, respectively. The downstream pressure is 80 psia (measured 2 ft from the nozzle). The gas-specific heat ratio is 1.25. (a) What is the expected daily flow rate? (b) Is icing a potential problem? (c) What is the expected pressure at the nozzle outlet? Solution (a) Poutlet Pup c ¼ 2 k þ 1 k kÀ1 ¼ 2 1:25 þ 1 1:25 1:25À1 ¼ 0:5549 Pdn Pup ¼ 80 100 ¼ 0:8 0:5549 Subsonic flow exists: d2 d1 ¼ 1:500 200 ¼ 0:75 Assuming NRe 106 , Fig. 5.1 gives CD ¼ 1:2. qsc ¼ 1,248CDAPup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k (k À 1)ggTup Pdn Pup 2 k À Pdn Pup kþ1 k #v u u t qsc ¼ (1,248)(1:2)[p(1:5)2 =4](100) Â ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:25 (1:25 À 1)(0:65)(530) 80 100 2 1:25 À 80 100 1:25þ1 1:25 #v u u t qsc ¼ 5,572 Mscf=d Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 62 21.12.2006 2:02pm 5/62 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 72. Check NRe: m ¼ 0:0108 cp by the Carr–Kobayashi–Burrows cor- relation. NRe ¼ 20qscgg md ¼ (20)(5,572)(0:65) (0:0108)(1:5) ¼ 4:5 Â 106 106 (b) Tdn ¼ Tup zup zoutlet Poutlet Pup kÀ1 k ¼ (70 þ 460)(1)(0:8) 1:25À1 1:25 ¼ 507 R ¼ 47 F 32 F Heating may not be needed, but the hydrate curve may need to be checked. (c) Poutlet ¼ Pdn ¼ 80 psia for subcritical flow: To estimate upstream pressure at a given downstream pressure and gas passage, the following procedure can be taken: Step 1: Calculate the critical pressure ratio with Eq. (5.1). Step 2: Calculate the minimum upstream pressure re- quired for sonic flow by dividing the down- stream pressure by the critical pressure ratio. Step 3: Calculate gas flow rate at the minimum sonic flow condition with Eq. (5.8). Step 4: If the given gas passage is less than the calculated gas flow rate at the minimum sonic flow condi- tion, use Eq. (5.5) to solve upstream pressure numerically. Otherwise, Eq. (5.8) to calculate upstream pressure. Example Problem 5.3 For the following given data, estimate upstream pressure at choke: Solution Example Problem 5.3 is solved with the spreadsheet program GasUpChokePressure.xls. The result is shown in Table 5.1. Downstream pressure cannot be calculated on the basis of given upstream pressure and gas passage under sonic flow conditions, but it can be calculated under subsonic flow conditions. The following procedure can be followed: Step 1: Calculate the critical pressure ratio with Eq. (5.1). Step 2: Calculate the maximum downstream pressure for minimum sonic flow by multiplying the upstream pressure by the critical pressure ratio. Step 3: Calculate gas flow rate at the minimum sonic flow condition with Eq. (5.8). Step 4: If the given gas passage is less than the calculated gas flow rate at the minimum sonic flow condi- tion, use Eq. (5.5) to solve downstream pressure numerically. Otherwise, the downstream pressure cannot be calculated. The maximum possible downstream pressure for sonic flow can be esti- mated by multiplying the upstream pressure by the critical pressure ratio. Example Problem 5.4 For the following given data, estimate downstream pressure at choke: Solution Example Problem 5.4 is solved with the spreadsheet program GasDownChokePressure.xls. The result is shown in Table 5.2. 5.5 Multiphase Flow When the produced oil reaches the wellhead choke, the wellhead pressure is usually below the bubble-point pres- sure of the oil. This means that free gas exists in the fluid stream flowing through choke. Choke behaves differently depending on gas content and flow regime (sonic or subsonic flow). 5.5.1 Critical (Sonic) Flow Tangren et al. (1949) performed the first investigation on gas-liquid two-phase flowthrough restrictions. They pre- sented an analysis of the behavior of an expanding gas- liquid system. They showed that when gas bubbles are added to an incompressible fluid, above a critical flow velocity, the medium becomes incapable of transmitting pressure change upstream against the flow. Several Downstream pressure: 300 psia Choke size: 32 1/64 in. Flowline ID: 2 in. Gas production rate: 5,000 Mscf/d Gas-specific gravity: 0.75 1 for air Gas-specific heat ratio: 1.3 Upstream temperature: 110 8F Choke discharge coefficient: 0.99 Table 5.1 Solution Given by the Spreadsheet Program GasUpChokePressure.xls GasUpChokePressure.xls Description: This spreadsheet calculates upstream pressure at choke for dry gases. Instructions: (1) Update parameter values in blue; (2) click Solution button; (3) view results. Input data Downstream pressure: 300 psia Choke size: 32 1 ⁄64 in. Flowline ID: 2 in. Gas production rate: 5,000 Mscf/d Gas-specific gravity: 0.75 1 for air Gas-specific heat ratio (k): 1.3 Upstream temperature: 110 8F Choke discharge coefficient: 0.99 Solution Choke area: 0.19625 in:2 Critical pressure ratio: 0.5457 Minimum upstream pressure required for sonic flow: 549.72 psia Flow rate at the minimum sonic flow condition: 3,029.76 Mscf/d Flow regime (1 ¼ sonic flow; À1 ¼ subsonic flow): 1 Upstream pressure given by sonic flow equation: 907.21 psia Upstream pressure given by subsonic flow equation: 1,088.04 psia Estimated upstream pressure: 907.21 psia Upstream pressure: 600 psia Choke size: 32 1 ⁄64 in. Flowline ID: 2 in. Gas production rate: 2,500 Mscf/d Gas-specific gravity: 0.75 1 for air Gas-specific heat ratio: 1.3 Upstream temperature: 110 8F Choke discharge coefficient: 0.99 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 63 21.12.2006 2:02pm CHOKE PERFORMANCE 5/63
- 73. empirical choke flow models have been developed in the past half century. They generally take the following form for sonic flow: pwh ¼ CRm q Sn , (5:12) where pwh ¼ upstream (wellhead) pressure, psia q ¼ gross liquid rate, bbl/day R ¼ producing gas-liquid ratio, Scf/bbl S ¼ choke size, 1 ⁄64 in. and C, m, and n are empirical constants related to fluid properties. On the basis of the production data from Ten Section Field in California, Gilbert (1954) found the values for C, m, and n to be 10, 0.546, and 1.89, respectively. Other values for the constants were proposed by different researchers including Baxendell (1957), Ros (1960), Achong (1961), and Pilehvari (1980). A summary of these values is presented in Table 5.3. Poettmann and Beck (1963) extended the work of Ros (1960) to develop charts for different API crude oils. Omana (1969) derived dimen- sionless choke correlations for water-gas systems. 5.5.2 Subcritical (Subsonic) Flow Mathematical modeling of subsonic flow of multiphase fluid through choke has been controversial over decades. Fortunati (1972) was the first investigator who presented a model that can be used to calculate critical and subcritical two-phase flow through chokes. Ashford (1974) also developed a relation for two-phase critical flow based on the work of Ros (1960). Gould (1974) plotted the critical– subcritical boundary defined by Ashford, showing that different values of the polytropic exponents yield different boundaries. Ashford and Pierce (1975) derived an equa- tion to predict the critical pressure ratio. Their model assumes that the derivative of flow rate with respect to the downstream pressure is zero at critical conditions. One set of equations was recommended for both critical and subcritical flow conditions. Pilehvari (1980, 1981) also studied choke flow under subcritical conditions. Sachdeva (1986) extended the work of Ashford and Pierce (1975) and proposed a relationship to predict critical pressure ratio. He also derived an expression to find the boundary between critical and subcritical flow. Surbey et al. (1988, 1989) discussed the application of multiple orifice valve chokes for both critical and subcritical flow conditions. Empirical relations were developed for gas and water sys- tems. Al-Attar and Abdul-Majeed (1988) made a compari- son of existing choke flow models. The comparison was based on data from 155 well tests. They indicated that the best overall comparison was obtained with the Gilbert cor- relation, which predicted measured production rate within an average error of 6.19%. On the basis of energy equation, Perkins (1990) derived equations that describe isentropic flow of multiphase mixtures through chokes. Osman and Dokla (1990) applied the least-square method to field data to develop empirical correlations for gas condensate choke flow. Gilbert-type relationships were generated. Applica- tions of these choke flow models can be found elsewhere (Wallis, 1969; Perry, 1973; Brown and Beggs, 1977; Brill and Beggs, 1978; Ikoku, 1980; Nind, 1981; Bradley, 1987; Beggs, 1991; Rastion et al., 1992; Saberi, 1996). Sachdeva’s multiphase choke flow mode is representa- tive of most of these works and has been coded in some commercial network modeling software. This model uses the following equation to calculate the critical–subcritical boundary: yc ¼ k kÀ1 þ (1Àx1)VL(1Àyc) x1VG1 k kÀ1 þ n 2 þ n(1Àx1)VL x1VG2 þ n 2 (1Àx1)VL x1VG2 h i2 8 : 9 = ; k kÀ1 , (5:13) where yc ¼ critical pressure ratio k ¼ Cp=Cv, specific heat ratio n ¼ polytropic exponent for gas x1 ¼ free gas quality at upstream, mass fraction VL ¼ liquid specific volume at upstream, ft3 =lbm VG1 ¼ gas specific volume at upstream, ft3 =lbm VG2 ¼ gas specific volume at downstream, ft3 =lbm. The polytropic exponent for gas is calculated using n ¼ 1 þ x1(Cp À Cv) x1Cv þ (1 À x1)CL : (5:14) The gas-specific volume at upstream (VG1) can be deter- mined using the gas law based on upstream pressure and temperature. The gas-specific volume at downstream (VG2) is expressed as VG2 ¼ VG1y À1 k c : (5:15) The critical pressure ratio yc can be solved from Eq. (5.13) numerically. Table 5.2 Solution Given by the Spreadsheet Program GasDownChokePressure.xls GasDownChokePressure.xls Description: This spreadsheet calculates upstream pressure at choke for dry gases. Instructions: (1) Update values in the Input data section; (2) click Solution button; (3) view results. Input data Upstream pressure: 700 psia Choke size: 32 1 ⁄64 in. Flowline ID: 2 in. Gas production rate: 2,500 Mscf/d Gas-specific gravity: 0.75 1 for air Gas-specific heat ratio (k): 1.3 Upstream temperature: 110 8F Choke discharge coefficient: 0.99 Solution Choke area: 0.19625 in:2 Critical pressure ratio: 0.5457 Minimum downstream pressure for minimum sonic flow: 382 psia Flow rate at the minimum sonic flow condition: 3,857 Mscf/d Flow regime (1 ¼ sonic flow; À1 ¼ subsonic flow): À1 The maximum possible downstream pressure in sonic flow: 382 psia Downstream pressure given by subsonic flow equation: 626 psia Estimated downstream pressure: 626 psia Table 5.3 A Summary of C, m, and n Values Given by Different Researchers Correlation C m n Gilbert 10 0.546 1.89 Ros 17.4 0.5 2 Baxendell 9.56 0.546 1.93 Achong 3.82 0.65 1.88 Pilehvari 46.67 0.313 2.11 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 64 21.12.2006 2:02pm 5/64 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 74. The actual pressure ratio can be calculated by ya ¼ p2 p1 , (5:16) where ya ¼ actual pressure ratio p1 ¼ upstream pressure, psia p2 ¼ downstream pressure, psia If ya yc, critical flow exists, and the yc should be used (y ¼ yc). Otherwise, subcritical flow exists, and ya should be used (y ¼ ya). The total mass flux can be calculated using the following equation: G2 ¼ CD 288gcp1r2 m2 (1 À x1)(1 À y) rL þ x1k k À 1 (VG1 À yVG2) ! '0:5 , (5:17) where G2 ¼ mass flux at downstream, lbm=ft2 =s CD ¼ discharge coefficient, 0.62–0.90 rm2 ¼ mixture density at downstream, lbm=ft3 rL ¼ liquid density, lbm=ft3 The mixture density at downstream (rm2) can be calcu- lated using the following equation: 1 rm2 ¼ x1VG1yÀ1 k þ (1 À x1)VL (5:18) Once the mass flux is determined from Eq. (5.17), mass flow rate can be calculated using the following equation: M2 ¼ G2A2, (5:19) where A2 ¼ choke cross-sectional area, ft2 M2 ¼ mass flow rate at down stream, lbm/s Liquid mass flow rate is determined by ML2 ¼ (1 À x2)M2: (5:20) At typical velocities of mixtures of 50–150 ft/s flowing through chokes, there is virtually no time for mass transfer between phases at the throat. Thus, x2 ¼ x1 can be as- sumed. Liquid volumetric flow rate can then be deter- mined based on liquid density. Gas mass flow rate is determined by MG2 ¼ x2M2: (5:21) Gas volumetric flow rate at choke downstream can then be determined using gas law based on downstream pressure and temperature. The major drawback of Sachdeva’s multiphase choke flow model is that it requires free gas quality as an input parameter to determine flow regime and flow rates, and this parameter is usually not known before flow rates are known. A trial-and-error approach is, therefore, needed in flow rate computations. Table 5.4 shows an example cal- culation with Sachdeva’s choke model. Guo et al. (2002) investigated the applicability of Sachdeva’s choke flow model in southwest Louisiana gas condensate wells. A total of 512 data sets from wells in southwest Louisiana were gathered for this study. Out of these data sets, 239 sets were collected from oil wells and 273 from condensate wells. Each of the data sets includes choke size, gas rate, oil rate, condensate rate, water rate, gas–liquid ratio, up- stream and downstream pressures, oil API gravity, and gas deviation factor (z-factor). Liquid and gas flow rates from these wells were also calculated using Sachdeva’s choke model. The overall performance of the model was studied in predicting the gas flow rate from both oil and gas con- densate wells. Out of the 512 data sets, 48 sets failed to comply with the model. Mathematical errors occurred in finding square roots of negative numbers. These data sets were from the condensate wells where liquid densities ranged from 46.7 to 55:1 lb=ft3 and recorded pressure dif- ferential across the choke less than 1,100 psi. Therefore, only 239 data sets from oil wells and 235 sets from conden- sate wells were used. The total number of data sets is 474. Different values of discharge coefficient CD were used to improve the model performance. Based on the cases stud- ied, Guo et al. (2002) draw the following conclusions: Table 5.4 An Example Calculation with Sachdeva’s Choke Model Input data Choke diameter (d2): 241 ⁄64 in. Discharge coefficient (CD): 0.75 Downstream pressure (p2): 50 psia Upstream pressure (p1): 80 psia Upstream temperature (T1): 100 8F Downstream temperature (T2): 20 8F Free gas quality (x1): 0.001 mass fraction Liquid-specific gravity: 0.9 water ¼ 1 Gas-specific gravity: 0.7 air ¼ 1 Specific heat of gas at constant pressure (Cp): 0.24 Specific heat of gas at constant volume (Cv): 0.171429 Specific heat of liquid (CL): 0.8 Precalculations Gas-specific heat ratio (k ¼ Cp=Cv): 1.4 Liquid-specific volume (VL): 0.017806 ft3 =lbm Liquid density (rL): 56.16 lb=ft3 Upstream gas density (rG1): 0.27 lb=ft3 Downstream gas density (rG2): 0.01 lb=ft4 Upstream gas-specific volume (VG1): 3.70 ft3 =lbm Polytropic exponent of gas (n): 1.000086 Critical pressure ratio computation k/(k-1) ¼ 3.5 (1 À x1)=x1 ¼ 999 n/2 ¼ 0.500043 VL=VG1 ¼ 0.004811 Critical pressure ratio (yc): 0.353134 VG2 ¼ 7.785109 ft3 =lbm VL=VG2 ¼ 0.002287 Equation residue (goal seek 0 by changing yc): 0.000263 Flow rate calculations Pressure ratio (yactual): 0.625 Critical flow index: À1 Subcritical flow index: 1 Pressure ratio to use (y): 0.625 Downstream mixture density (rm2): 43.54 lb=ft3 Downstream gas-specific volume (VG2): 5.178032 Choke area (A2) ¼ 0.000767 ft2 Mass flux (G2) ¼ 1432.362 lbm=ft2 =s Mass flow rate (M) ¼ 1.098051 lbm/s Liquid mass flow rate (ML) ¼ 1.096953 lbm/s Liquid glow rate ¼ 300.5557 bbl/d Gas mass flow rate (MG) ¼ 0.001098 lbm/s Gas flow rate ¼ 0.001772 MMscfd Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 65 21.12.2006 2:02pm CHOKE PERFORMANCE 5/65
- 75. 1. The accuracy of Sachdeva’s choke model can be im- proved by using different discharge coefficients for dif- ferent fluid types and well types. 2. For predicting liquid rates of oil wells and gas rates of gas condensate wells, a discharge coefficient of CD ¼ 1:08 should be used. 3. A discharge coefficient CD ¼ 0:78 should be used for predicting gas rates of oil wells. 4. A discharge coefficient CD ¼ 1:53 should be used for predicting liquid rates of gas condensate wells. Summary This chapter presented and illustrated different mathemat- ical models for describing choke performance. While the choke models for gas flow have been well established with fairly good accuracy in general, the models for two-phase flow are subject to tuning to local oil properties. It is essential to validate two-phase flow choke models before they are used on a large scale. References achong, i.b. ‘‘Revised Bean and Performance Formula for Lake Maracaibo Wells,’’ Shell Internal Report, Octo- ber 1961. al-attar, h.h. and abdul-majeed, g. Revised bean per- formance equation for east Baghdad oil wells. SPE Production Eng. 1988;February:127–131. ashford, f.e. An evaluation of critical multiphase flow performance through wellhead chokes. J. Petroleum Technol. 1974;26(August):843–848. ashford, f.e. and pierce, p.e. Determining multiphase pressure drop and flow capabilities in down hole safety valves. J. Petroleum Technol. 1975;27(Septem- ber):1145–1152. baxendell, p.b. Bean performance-lake wells. Shell Internal Report, October 1957. beggs, h.d. Production Optimization Using Nodal Analysis. Tulsa, OK: OGTC Publications, 1991. bradley, h.b. Petroleum Engineering Handbook. Richard- son, TX: Society of Petroleum Engineers, 1987. brill, j.p. and beggs, h.d. Two-Phase Flow in Pipes. Tulsa, OK: The University of Tulsa Press, 1978. brown, k.e. and beggs, h.d. The Technology of Artificial Lift Methods. Tulsa, OK: PennWell Books, 1977. Crane, Co. ‘‘Flow of Fluids through Valves, Fittings, and Pipe. Technical paper No. 410. Chicago, 1957. fortunati, f. Two-phase flow through wellhead chokes. Presented at the SPE European Spring Meeting held 16–18 May 1972 in Amsterdam, the Netherlands. SPE paper 3742. gilbert, w.e. Flowing and gas-lift well performance. API Drilling Production Practice 1954;20:126–157. gould, t.l. Discussion of an evaluation of critical multi- phase flow performance through wellhead chokes. J. Petroleum Technol. 1974;26(August):849–850. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston, TX: Gulf Publishing Company, 2005. guo, b., al-bemani, a., and ghalambor, a. Applicability of Sachdeva’s choke flow model in southwest Louisi- ana gas condensate wells. Presented at the SPE Gas technology Symposium held 30 April–2 May 2002 in Calgary, Canada. Paper SPE 75507. ikoku, c.u. Natural Gas Engineering. Tulsa, OK: Penn- Well Books, 1980. nind, t.e.w. Principles of Oil Well Production, 2nd edition. New York: McGraw-Hill Book Co., 1981. omana, r., houssiere, c., jr., brown, k.e., brill, j.p., and thompson, r.e. Multiphase flow through chokes. Pre- sented at the SPE 44th Annual Meeting held 28–31 September 1969 in Denver, Colorado. SPE paper 2682. osman, m.e. and dokla, m.e. Has condensate flow through chokes. Presented at 23 April 1990. SPE paper 20988. perkins, t.k. Critical and subcritical flow of multiphase mixtures through chokes. Presented at the SPE 65th Annual Technical Conference and Exhibition held 23– 26 September 1990 in New Orleans, Louisiana. SPE paper 20633. perry, r.h. Chemical Engineers’ Handbook, 5th edition. New York: McGraw-Hill Book Co., 1973. pilehvari, a.a. Experimental study of subcritical two- phase flow through wellhead chokes. Tulsa, OK: Uni- versity of Tulsa Fluid Flow Projects Report, Septem- ber 1980. pilehvari, a.a. Experimental study of critical two-phase flow through wellhead chokes. Tulsa, OK: University of Tulsa Fluid Flow Projects Report, June 1981. poettmann, f.h. and beck, r.l. New charts developed to predict gas-liquid flow through chokes. World Oil 1963;March:95–101. poettmann, f.h., beck, r.l., and beck, r.l. A review of multiphase flow through chokes. Paper presented at the ASME Winter Annual Meeting held 8–13 Novem- ber 1992, Anaheim, California, pp. 51–62. ros, n.c.j. An analysis of critical simultaneous gas/ liquid flow through a restriction and its application to flow metering. Applied Sci. Res. 1960; Section A(9):374–389. saberi, m. A study on flow through wellhead chokes and choke size selection, MS thesis, University of South- western Louisiana, Lafayette, 1996, pp. 78–89. sachdeva, r., schmidt, z., brill, j.p., and blais, r.m. Two-phase flow through chokes. Paper presented at the SPE 61st Annual Technical Conference and Ex- hibition held 5–8 October 1986 in New Orleans, Louisiana. SPE paper 15657. secen, j.a. Surface-choke measurement equation improved by field testing and analysis. Oil Gas J. 1976;30(August):65–68. surbey, d.w., kelkar, b.g., and brill, j.p. Study of sub- critical flow through multiple orifice valves. SPE Pro- duction Eng. 1988;February:103–108. surbey, d.w., kelkar, b.g., and brill, j.p. Study of multi- phase critical flow through wellhead chokes. SPE Pro- duction Eng. 1989;May:142–146. tangren, r.f., dodge, c.h., and seifert, h.s. Compress- ibility effects in two-phase flow. J. Applied Physics 1947;20:637–645. wallis, g.b. One Dimensional Two-Phase Flow. New York: McGraw-Hill Book Co., 1969. Problems 5.1 A well is producing 40 8API oil at 200 stb/d and no gas. If the beam size is 1 in., pipe size is 2 in., tem- perature is 100 8F, estimate pressure drop across a nozzle-type choke. 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- 76. 5.2 A well is producing at 200 stb/d of liquid along with a 900 scf/stb of gas. If the beam size is ½ in., assuming sonic flow, calculate the flowing wellhead pressure using Gilbert’s formula. 5.3 A 0.65 specific gravity gas flows from a 2-in. pipe through a 1-in. orifice-type choke. The upstream pressure and temperature are 850 psia and 85 8F, respectively. The downstream pressure is 210 psia (measured 2 ft from the orifice). The gas-specific heat ratio is 1.3. (a) What is the expected daily flow rate? (b) Does heating need to be applied to ensure that the frost does not clog the orifice? (c) What is the expected pressure at the orifice outlet? 5.4 A 0.70 specific gravity natural gas flows from a 2-in. pipe through a 1.5-in. nozzle-type choke. The up- stream pressure and temperature are 120 psia and 75 8F, respectively. The downstream pressure is 90 psia (measured 2 ft from the nozzle). The gas- specific heat ratio is 1.25. (a) What is the expected daily flow rate? (b) Is icing a potential problem? (c) What is the expected pressure at the nozzle outlet? 5.5 For the following given data, estimate upstream gas pressure at choke: 5.6 For the following given data, estimate downstream gas pressure at choke: 5.7 For the following given data, assuming subsonic flow, estimate liquid and gas production rate: Downstream pressure: 350 psia Choke size: 32 1 ⁄64 in. Flowline ID: 2 in. Gas production rate: 4,000 Mscf/d Gas-specific gravity: 0.70 1 for air Gas-specific heat ratio: 1.25 Upstream temperature: 100 8F Choke discharge coefficient: 0.95 Upstream pressure: 620 psia Choke size: 32 1 ⁄64 in. Flowline ID: 2 in. Gas production rate: 2,200 Mscf/d Gas-specific gravity: 0.65 1 for air Gas-specific heat ratio: 1.3 Upstream temperature: 120 8F Choke discharge coefficient: 0.96 Choke diameter: 32 1 ⁄64 in. Discharge coefficient: 0.85 Downstream pressure: 60 psia Upstream pressure: 90 psia Upstream temperature: 120 8F Downstream temperature: 30 8F Free gas quality: 0.001 mass fraction Liquid-specific gravity: 0.85 water ¼ 1 Gas-specific gravity: 0.75 air ¼ 1 Specific heat of gas at constant pressure: 0.24 Specific heat of gas at constant volume: 0.171429 Specific heat of liquid: 0.8 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_chap05 Final Proof page 67 21.12.2006 2:02pm CHOKE PERFORMANCE 5/67
- 77. 6 Well Deliverability Contents 6.1 Introduction 6/70 6.2 Nodal Analysis 6/70 6.3 Deliverability of Multilateral Well 6/79 Summary 6/84 References 6/85 Problems 6/85 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 69 3.1.2007 8:40pm Compositor Name: SJoearun
- 78. 6.1 Introduction Well deliverability is determined by the combination of well inflow performance (see Chapter 3) and wellbore flow performance (see Chapter 4). Whereas the former describes the deliverability of the reservoir, the latter pre- sents the resistance to flow of production string. This chapter focuses on prediction of achievable fluid produc- tion rates from reservoirs with specified production string characteristics. The technique of analysis is called ‘‘Nodal analysis’’ (a Schlumburger patent). Calculation examples are illustrated with computer spreadsheets that are provided with this book. 6.2 Nodal Analysis Fluid properties change with the location-dependent pres- sure and temperature in the oil and gas production system. To simulate the fluid flow in the system, it is necessary to ‘‘break’’ the system into discrete nodes that separate sys- tem elements (equipment sections). Fluid properties at the elements are evaluated locally. The system analysis for determination of fluid production rate and pressure at a specified node is called ‘‘Nodal analysis’’ in petroleum engineering. Nodal analysis is performed on the principle of pressure continuity, that is, there is only one unique pressure value at a given node regardless of whether the pressure is evaluated from the performance of upstream equipment or downstream equipment. The performance curve (pressure–rate relation) of upstream equipment is called ‘‘inflow performance curve’’; the performance curve of downstream equipment is called ‘‘outflow per- formance curve.’’ The intersection of the two performance curves defines the operating point, that is, operating flow rate and pressure, at the specified node. For the conveni- ence of using pressure data measured normally at either the bottom-hole or the wellhead, Nodal analysis is usually conducted using the bottom-hole or wellhead as the solu- tion node. This chapter illustrates the principle of Nodal analysis with simplified tubing string geometries (i.e., single-diameter tubing strings). 6.2.1 Analysis with the Bottom-Hole Node When the bottom-hole is used as a solution node in Nodal analysis, the inflow performance is the well inflow per- formance relationship (IPR) and the outflow performance is the tubing performance relationship (TPR), if the tubing shoe is set to the top of the pay zone. Well IPR can be established with different methods presented in Chapter 3. TPR can be modeled with various approaches as discussed in Chapter 4. Traditionally, Nodal analysis at the bottom-hole is car- ried out by plotting the IPR and TPR curves and graph- ically finding the solution at the intersection point of the two curves. With modern computer technologies, the solution can be computed quickly without plotting the curves, although the curves are still plotted for visual verification. 6.2.1.1 Gas Well Consider the bottom-hole node of a gas well. If the IPR of the well is defined by qsc ¼ C( pp2 À p2 wf ) n , (6:1) and if the outflow performance relationship of the node (i.e., the TPR) is defined by p2 wf ¼ Exp(s)p2 hf þ 6:67 Â 10À4 [Exp(s) À 1] fMq2 sczz2 TT2 d5 i cos u , (6:2) then the operating flow rate qsc and pressure pwf at the bottom-hole node can be determined graphically by plot- ting Eqs. (6.1) and (6.2) and finding the intersection point. The operating point can also be solved analytically by combining Eqs. (6.1) and (6.2). In fact, Eq. (6.1) can be rearranged as p2 wf ¼ pp2 À qsc C 1 n : (6:3) Substituting Eq. (6.3) into Eq. (6.2) yields pp2 À qsc C 1 n ÀExp(s)p2 hf À 6:67 Â 10À4 [Exp(s) À 1] fMq2 sczz2 TT2 D5 i cos u ¼ 0; (6:4) which can be solved with a numerical technique such as the Newton–Raphson iteration for gas flow rate qsc. This computation can be performed automatically with the spreadsheet program BottomHoleNodalGas.xls. Example Problem 6.1 Suppose that a vertical well produces 0.71 specific gravity gas through a 27 ⁄8 -in. tubing set to the top of a gas reservoir at a depth of 10,000 ft. At tubing head, the pressure is 800 psia and the temperature is 150 8F, whereas the bottom-hole temperature is 200 8F. The relative roughness of tubing is about 0.0006. Calculate the expected gas production rate of the well using the following data for IPR: Reservoir pressure: 2,000 psia IPR model parameter C: 0.1 Mscf/d-psi2n IPR model parameter n: 0.8 Solution Example Problem 6.1 is solved with the spreadsheet program BottomHoleNodalGas.xls. Table 6.1 shows the appearance of the spreadsheet for the Input data and Result sections. It indicates that the expected gas flow rate is 1478 Mscf/d at a bottom-hole pressure of 1059 psia. The inflow and outflow performance curves plotted in Fig. 6.1 confirm this operating point. 6.2.1.2 Oil Well Consider the bottom-hole node of an oil well. As discussed in Chapter 3, depending on reservoir pressure range, dif- ferent IPR models can be used. For instance, if the reser- voir pressure is above the bubble-point pressure, a straight- line IPR can be used: q ¼ JÃ ( pp À pwf ) (6:5) The outflow performance relationship of the node (i.e., the TPR) can be described by a different model. The simplest model would be Poettmann–Carpenter model defined by Eq. (4.8), that is, pwf ¼ pwh þ rr þ kk rr L 144 (6:6) where pwh and L are tubing head pressure and well depth, respectively, then the operating flow rate q and pressure pwf at the bottom-hole node can be determined graphically by plotting Eqs. (6.5) and (6.6) and finding the intersection point. The operating point can also be solved analytically by combining Eqs. (6.5) and (6.6). In fact, substituting Eq. (6.6) into Eq. (6.5) yields q ¼ JÃ pp À pwh þ rr þ kk rr L 144 ! , (6:7) which can be solved with a numerical technique such as the Newton–Raphson iteration for liquid flow rate q. This computation can be performed automatically with the spreadsheet program BottomHoleNodalOil-PC.xls. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 70 3.1.2007 8:40pm Compositor Name: SJoearun 6/70 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 79. 0 500 1,000 1,500 2,000 2,500 Gas Production Rate (Mscf/d) BottomHolePressure(psia) IPR TPR 2,0000 200 400 600 800 1,000 1,200 1,400 1,600 1,800 Figure 6.1 Nodal analysis for Example Problem 6.1. Table 6.1 Result Given by BottomHoleNodalGas.xls for Example Problem 6.1 BottomHoleNodalGas.xls Description: This spreadsheet calculates gas well deliverability with bottom-hole node. Instructions: (1) Input your data in the Input data section; (2) click Solution button; (3) view results in table and in graph sheet ‘‘Plot.’’ Input data Gas-specific gravity (gg): 0.71 Tubing inside diameter (D): 2.259 in. Tubing relative roughness (e/D): 0.0006 Measured depth at tubing shoe (L): 10,000 ft Inclination angle (Q): 0 degrees Wellhead pressure (phf ): 800 psia Wellhead temperature (Thf ): 150 8F Bottom-hole temperature (Twf ): 200 8F Reservoir pressure (p $): 2000 psia C-constant in back-pressure IPR model: 0:01 Mscf=d-psi2n n-exponent in back-pressure IPR model: 0.8 Solution Tav ¼ 635 8R Zav ¼ 0.8626 s ¼ 0.486062358 es ¼ 1.62590138 fM ¼ 0.017396984 AOF ¼ 1912.705 Mscf/d qsc (Mscf/d) IPR TPR 0 2,000 1,020 191 1,943 1,021 383 1,861 1,023 574 1,764 1,026 765 1,652 1,031 956 1,523 1,037 1,148 1,374 1,044 1,339 1,200 1,052 1,530 987 1,062 1,721 703 1,073 1,817 498 1,078 1,865 353 1,081 1,889 250 1,083 1,913 0 1,084 Operating flow rate ¼ 1,470 Mscf/d Residual of objective function ¼ À0.000940747 Operating pressure ¼ 1,059 psia Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 71 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/71
- 80. Example Problem 6.2 For the data given in the following table, predict the operating point: Solution Example Problem 6.2 is solved with the spreadsheet program BottomHoleNodalOil-PC.xls. Table 6.2 shows the appearance of the spreadsheet for the Input data and Result sections. It indicates that the expected oil flow rate is 1127 stb/d at a bottom-hole pressure of 1,873 psia. If the reservoir pressure is below the bubble-point pressure, Vogel’s IPR can be used q ¼ qmax 1 À 0:2 pwf pp À 0:8 pwf pp 2 # (6:8) or pwf ¼ 0:125pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 q qmax s À 1 # (6:9) If the outflow performance relationship of the node (i.e., the TPR) is described by the Guo–Ghalambor model defined by Eq. (4.18), that is, Reservoir pressure: 3,000 psia Tubing ID: 1.66 in. Wellhead pressure: 500 psia Productivity index above bubble point: 1 stb/d-psi Producing gas–liquid ratio (GLR): 1,000 scf/stb Water cut (WC): 25 % Oil gravity: 30 8API Water-specific gravity: 1.05 1 for fresh-water Gas-specific gravity: 0.65 1 for air N2 content in gas: 0 mole fraction CO2 content in gas: 0 mole fraction H2S content in gas: 0 mole fraction Formation volume factor of oil: 1.2 rb/stb Wellhead temperature: 100 8F Tubing shoe depth: 5,000 ft Bottom-hole temperature: 150 8F Table 6.2 Result Given by BottomHoleNodalOil-PC.xls for Example Problem 6.2 BottomHoleNodalOil-PC.xls Description: This spreadsheet calculates the operating point using the Poettmann–Carpenter method with bottom-hole node. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution button; and (4) view result in the Solution section. Input data U.S. Field units SI units Reservoir pressure: 3,000 psia Tubing ID: 1.66 in. Wellhead pressure: 500 psia Productivity index above bubble point: 1 stb/d-psi Producing gas–liquid ratio (GLR): 1,000 scf/stb Water cut: 25 % Oil gravity: 30 8API Water-specific gravity: 1.05, 1 for water Gas-specific gravity: 0.65, 1 for air N2 content in gas: 0 mole fraction CO2 content in gas: 0 mole fraction H2S content in gas: 0 mole fraction Formation volume factor of oil: 1.2 rb/stb Wellhead temperature: 100 8F Tubing shoe depth: 5,000 ft Bottom-hole temperature: 150 8F Solution Oil-specific gravity ¼ 0.88, 1 for water Mass associated with 1 stb of oil ¼ 495.66 lb Solution–gas ratio at wellhead ¼ 78.42 scf/stb Oil formation volume factor at wellhead ¼ 1.04 rb/stb Volume associated with 1 stb of oil at wellhead ¼ 45.12 cf Fluid density at wellhead ¼ 10.99 lb/cf Solution gas–oil ratio at bottom-hole ¼ 339.39 scf/stb Oil formation volume factor at bottom-hole ¼ 1.18 rb/stb Volume associated with 1 stb of oil at bottom-hole ¼ 16.56 cf Fluid density at bottom-hole ¼ 29.94 lb/cf The average fluid density ¼ 20.46 lb/cf Inertial force (Drv) ¼ 44.63 lb/day-ft Friction factor ¼ 0.0084 Friction term ¼ 390.50 (lb=cf)2 Error in liquid rate ¼ 0.00 stb/d Bottom-hole pressure ¼ 1,873 psia Liquid production rate: 1,127 stb/d Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 72 3.1.2007 8:40pm Compositor Name: SJoearun 6/72 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 81. 144b( pwf À phf ) þ 1 À 2bM 2 ln (144pwf þ M)2 þ N (144phf þ M)2 þ N À M þ b c N À bM2 ffiffiffiffiffi N p tanÀ1 144pwf þ M ffiffiffiffiffi N p À tanÀ1 144phf þ M ffiffiffiffiffi N p ! ¼ a( cos u þ d2 e)L, (6:10) substituting Eq. (6.9) into Eq. (6.10) will give an equation to solve for liquid production rate q. The equation can be solved with a numerical technique such as the Newton–Raphson iteration. This computation is performed automatically with the spreadsheet program BottomHoleNodalOil-GG.xls. Example Problem 6.3 For the data given in the following table, predict the operating point: Solution Example Problem 6.3 is solved with the spreadsheet program BottomHoleNodalOil-GG.xls. Table 6.3 shows the appearance of the spreadsheet for the Input data and Result sections. It indicates that the expected oil flow rate is 1,268 stb/d at a bottom-hole pressure of 1,688 psia. If the reservoir pressure is above the bubble-point pres- sure, but the flowing bottom-hole pressure is in the range of below bubble-point pressure, the generalized Vogel’s IPR can be used: q ¼ qb þ qv 1 À 0:2 pwf pb À 0:8 pwf pb 2 # (6:11) Iftheoutflowperformancerelationshipofthenode(i.e.,TPR) is described by Hagedorn-Brown correlation, Eq. (4.27) can be used for generating the TPR curve. Combining Eqs. (6.11) and (4.27) can be solved with a numerical technique such as the Newton–Raphson iteration for liquid flow rate Reservoir pressure: 3,000 psia Total measured depth: 7,000 ft Average inclination angle: 20 degree Tubing ID: 1.995 in. Gas production rate: 1,000,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.85 H2O ¼ 1 Water cut: 30 % Table 6.3 Result Given by BottomHoleNodalOil-GG.xls for Example Problem 6.2 BottomHoleNodalOil-GG.xls Description: This spreadsheet calculates flowing bottom-hole pressure based on tubing head pressure and tubing flow performance using the Guo–Ghalambor method. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Result button; and (4) view result in the Result section. Input data U.S. Field units SI units Reservoir pressure: 3,000 psia Total measured depth: 7,000 ft Average inclination angle: 20 degrees Tubing ID: 1.995 in. Gas production rate: 1,000,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.85 H2O ¼ 1 Water cut: 30% Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 160 8F Tubing head pressure: 300 psia Absolute open flow (AOF): 2000 bbl/d Solution A ¼ 3:1243196 in:2 D ¼ 0.16625 ft Tav ¼ 622 8R cos (u) ¼ 0.9397014 (Drv) ¼ 40.576594 fM ¼ 0.0424064 a ¼ 0.0001699 b ¼ 2.814E-06 c ¼ 1,349,785.1 d ¼ 3.7998147 e ¼ 0.0042189 M ¼ 20,395.996 N ¼ 6.829Eþ09 Liquid production rate, q ¼ 1,268 bbl/d Bottom hole pressure, pwf ¼ 1,688 psia Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 160 8F Tubing head pressure: 300 psia Absolute open flow (AOF): 2,000 bbl/d Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 73 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/73
- 82. q. This computation can be performed automatically with the spreadsheet program BottomHoleNodalOil-HB.xls. Example Problem 6.4 For the data given in the following table, predict the operating point: Solution Example Problem 6.4 is solved with the spread- sheet program BottomHoleNodalOil-HB.xls. Table 6.4 shows the appearance of the spreadsheet for the Input data and Result sections. Figure 6.2 indicates that the expected gas flow rate is 2200 stb/d at a bottom-hole pressure of 3500 psia. 6.2.2 Analysis with Wellhead Node When the wellhead is used as a solution node in Nodal analysis, the inflow performance curve is the ‘‘wellhead performance relationship’’ (WPR), which is obtained by transforming the IPR to wellhead through the TPR. The outflow performance curve is the wellhead choke performance relationship (CPR). Some TPR models are presented in Chapter 4. CPR models are discussed in Chapter 5. Nodal analysis with wellhead being a solution node is carried out by plotting the WPR and CPR curves and finding the solution at the intersection point of the two curves. Again, with modern computer technologies, the solu- tion can be computed quickly without plotting the curves, although the curves are still plotted for verification. 6.2.2.1 Gas Well If the IPR of a well is defined by Eq. (6.1) and the TPR is represented by Eq. (6.2), substituting Eq. (6.2) into Eq. (6.1) gives qsc ¼ C pp2 À Exp(s)p2 hf þ 6:67 Â 10À4 [Exp(s) À 1] fMq2 sczz2 TT2 d5 i cos u !n , (6:12) which defines a relationship between wellhead pressure phf and gas production rate qsc, that is, WPR. If the CPR is defined by Eq. (5.8), that is, qsc ¼ 879CAphf ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup ! 2 k þ 1 kþ1 kÀ1 v u u t , (6:13) Depth: 9,850 ft Tubing inner diameter: 1.995 in. Oil gravity: 45 8API Oil viscosity: 2 cp Production GLR: 500 scf/bbl Gas-specific gravity: 0.7 air ¼ 1 Flowing tubing head pressure: 450 psia Flowing tubing head temperature: 80 8F Flowing temperature at tubing shoe: 180 8F Water cut: 10% Reservoir pressure: 5,000 psia Bubble-point pressure: 4,000 psia Productivity index above bubble point: 1.5 stb/d-psi Table 6.4 Solution Given by BottomHoleNodalOil-HB.xls BottomHoleNodalOil-HB.xls Description: This spreadsheet calculates operating point using the Hagedorn–Brown correlation. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution button; and (4) view result in the Result section and charts. Input data U.S. Field units SI units Depth (D): 9,850 ft Tubing inner diameter (dti): 1.995 in. Oil gravity (API): 45 8API Oil viscosity (mo): 2 cp Production GLR (GLR): 500 scf/bbl Gas-specific gravity (gg): 0.7 air ¼ 1 Flowing tubing head pressure (phf ): 450 psia Flowing tubing head temperature (thf ): 80 8F Flowing temperature at tubing shoe (twf ): 180 8F Water cut: 10% Reservoir pressure (pe): 5,000 psia Bubble-point pressure (pb): 4,000 psia Productivity index above bubble point (J* ): 1.5 stb/d-psi Solution US Field units : qb ¼ 1,500 qmax ¼ 4,833 q (stb/d) pwf (psia) IPR TPR 0 4,908 537 4,602 2,265 1,074 4,276 2,675 1,611 3,925 3,061 2,148 3,545 3,464 2,685 3,125 3,896 3,222 2,649 4,361 3,759 2,087 4,861 4,296 1,363 5,397 4,833 0 5,969 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 74 3.1.2007 8:40pm Compositor Name: SJoearun 6/74 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 83. then the operating flow rate qsc and pressure phf at the wellhead node can be determined graphically by plotting Eqs. (6.12) and (6.13) and finding the intersection point. The operating point can also be solved numerically by combining Eqs. (6.12) and (6.13). In fact, Eq. (6.13) can be rearranged as phf ¼ qsc 879CA ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup ! 2 k þ 1 kþ1 kÀ1 v u u t : (6:14) Substituting Eq. (6.14) into Eq. (6.12) gives qsc ¼ C pp2 À Exp(s) qsc 879CA ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup 2 kþ1 kþ1 kÀ1 r 0 B B @ 1 C C A 2 0 B B B @ 2 6 6 6 4 þ 6:67 Â 10À4 [Exp(s) À 1]fMq2 sczz2 TT2 d5 i cos u 1 C C A 3 7 7 5 n , (6:15) which can be solved numerically for gas flow rate qsc. This computation can be performed automatically with the spreadsheet program WellheadNodalGas-SonicFlow.xls. Example Problem 6.5 Use the data given in the following table to estimate gas production rate of a gas well: Solution Example Problem 6.5 is solved with the spreadsheet program WellheadNodalGas-SonicFlow.xls. Table 6.5 shows the appearance of the spreadsheet for the Input data and Result sections. It indicates that the expected gas flow rate is 1,478 Mscf/d at a bottom-hole pressure of 1,050 psia. The inflow and outflow performance curves plotted in Fig. 6.3 confirm this operating point. 6.2.2.2 Oil Well As discussed in Chapter 3, depending on reservoir pressure range, different IPR models can be used. For instance, if the reservoir pressure is above the bubble-point pressure, a straight-line IPR can be used: q ¼ JÃ pp À pwf À Á (6:16) If the TPR is described by the Poettmann–Carpenter model defined by Eq. (4.8), that is, pwf ¼ pwh þ rr þ kk rr L 144 (6:17) substituting Eq. (6.17) into Eq. (6.16) gives q ¼ JÃ pp À pwh þ rr þ kk rr L 144 ! , (6:18) which describes inflow for the wellhead node and is called the WPR. If the CPR is given by Eq. (5.12), that is, pwh ¼ CRm q Sn , (6:19) the operating point can be solved analytically by combin- ing Eqs. (6.18) and (6.19). In fact, substituting Eq. (6.19) into Eq. (6.18) yields q ¼ JÃ pp À CRm q Sn þ rr þ kk rr L 144 ! , (6:20) which can be solved with a numerical technique. Because the solution procedure involves loop-in-loop iterations, it cannot be solved in MS Excel in an easy manner. A special computer program is required. Therefore, a computer- assisted graphical solution method is used in this text. The operating flow rate q and pressure pwh at the well- head node can be determined graphically by plotting Eqs. (6.18) and (6.19) and finding the intersection point. This computation can be performed automatically with the spreadsheet program WellheadNodalOil-PC.xls. 0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 Liquid Production Rate (bbl/d) BottomHolePressure(psia) IPR TPR 0 5,0004,0003,0002,0001,000 Figure 6.2 Nodal analysis for Example Problem 6.4. Gas-specific gravity: 0.71 Tubing inside diameter: 2.259 in. Tubing wall relative roughness: 0.0006 Measured depth at tubing shoe: 10,000 ft Inclination angle: 0 degrees Wellhead choke size: 16 1 ⁄64 4 in. Flowline diameter: 2 in. Gas-specific heat ratio: 1.3 Gas viscosity at wellhead: 0.01 cp Wellhead temperature: 150 8F Bottom-hole temperature: 200 8F Reservoir pressure: 2,000 psia C-constant in IPR model: 0.01 Mscf/ d-psi2n n-exponent in IPR model: 0.8 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 75 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/75
- 84. Example Problem 6.6 Use the following data to estimate the liquid production rate of an oil well: Solution Example Problem 6.6 is solved with the spreadsheet program WellheadNodalOil-PC.xls. Table 6.6 shows the appearance of the spreadsheet for the Input data and Result sections. The inflow and outflow performance curves are plotted in Fig. 6.4, which indicates that the expected oil flow rate is 3280 stb/d at a wellhead pressure of 550 psia. If the reservoir pressure is below the bubble-point pressure, Vogel’s IPR can be rearranged to be pwf ¼ 0:125pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 q qmax s À 1 # (6:21) Table 6.5 Solution Given by WellheadNodalGas-SonicFlow.xls WellheadNodalGas-SonicFlow.xls Description: This spreadsheet calculates well deliverability with wellhead node. Instructions: Step 1: Input your data in the Input data section. Step 2: Click Solution button to get results. Step 3: View results in table and in the plot graph sheet. Input data Gas-specific gravity (gg): 0.71 Tubing inside diameter (D): 2.259 in. Tubing relative roughness ˙ («=D): 0.0006 Measured depth at tubing shoe (L): 10,000 ft Inclination angle (u): 0 degrees Wellhead choke size (Dck): 16 1/64 in. Flowline diameter (Dfl): 2 in. Gas-specific heat ratio (k): 1.3 Gas viscosity at wellhead (mg): 0.01 cp Wellhead temperature (Thf ): 120 8F Bottom-hole temperature (Twf ): 180 8F Reservoir pressure (p $): 2,000 psia C-constant in back-pressure IPR model: 0:01 Mscf=d-psi2n n-exponent in back-pressure IPR model: 0.8 Solution Tav ¼ 610 8R Zav ¼ 0.8786 s ¼ 0.4968 es ¼ 1.6434 fm ¼ 0.0174 AOF ¼ 1,913 Mscf/d Dck=Dfl ¼ 0.125 Re = 8,348,517 Cck ¼ 1:3009 in:2 Ack ¼ 0.0490625 qsc (Mscf/d) WPR CPR 0 1,600 0 191 1,554 104 383 1,489 207 574 1,411 311 765 1,321 415 956 1,218 518 1,148 1,099 622 1,339 960 726 1,530 789 830 1,721 562 933 1,817 399 985 1,865 282 1,011 1,889 200 1,024 1,913 1 1,037 Operating flow rate ¼ 1,470 Mscf/d Operating pressure ¼ 797 psia Reservoir pressure: 6,000 psia Tubing ID: 3.5 in. Choke size: 64 1 ⁄64 in. Productivity index above bubble point: 1 stb/d-psi Producing gas–liquid ratio (GLR): 1000 scf/stb Water cut: 25% Oil gravity: 30 8API Water-specific gravity: 1.05 1 for fresh- water Gas-specific gravity: 0.65 1 for air Choke constant: 10 Choke GLR exponent: 0.546 Choke-size exponent: 1.89 Formation volume factor of oil: 1 rb/stb Wellhead temperature: 100 8F Tubing shoe depth: 12,000 ft Bottom-hole temperature: 150 8F Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 76 3.1.2007 8:40pm Compositor Name: SJoearun 6/76 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 85. If the TPR is described by the Guo–Ghalambor model defined by Eq. (4.18), that is, 144b pwf À phf À Á þ 1 À 2bM 2 ln 144pwf þ M À Á2 þN 144phf þ M À Á2 þN À M þ b c N À bM2 ffiffiffiffiffi N p tanÀ1 144pwf þ M ffiffiffiffiffi N p À tanÀ1 144phf þ M ffiffiffiffiffi N p ! ¼ a( cos u þ d2 e)L, (6:22) and the CPR is given by Eq. (5.12), that is, phf ¼ CRm q Sn , (6:23) solving Eqs. (6.21), (6.22), and (6.23) simultaneously will give production rate q and wellhead pressure phf : The solution procedure has been coded in the spreadsheet pro- gram WellheadNodalOil-GG.xls. Example Problem 6.7 Use the following data to estimate the liquid production rate of an oil well: 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 Gas Production Rate (Mscf/d) WellheadPressure(psia) WPR CPR 2,0000 200 400 600 800 1,000 1,200 1,400 1,600 1,800 Figure 6.3 Nodal analysis for Example Problem 6.5. 0 500 1,000 1,500 2,000 2,500 0 1,000 2,000 3,000 4,000 5,000 6,000 Liquid Production Rate (bbl/d) WellheadPressure(psia) WPR CPR Figure 6.4 Nodal analysis for Example Problem 6.6. Choke size: 64 1/64 in. Reservoir pressure: 3,000 psia Total measured depth: 7,000 ft Average inclination angle: 20 degrees Tubing ID: 1.995 in. Gas production rate: 1,000,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.85 H2O ¼ 1 Water cut: 30% Water specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 160 8F Absolute openflow (AOF): 2,000 bbl/d Choke flow constant: 10 Choke GLR exponent: 0.546 Choke-size exponent: 1.89 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 77 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/77
- 86. Solution Example Problem 6.7 is solved with the spreadsheet program WellheadNodalOil-GG.xls. Table 6.7 shows the appearance of the spreadsheet for the Data Input and Result sections. It indicates that the expected oil flow rate is 1,289 stb/d at a wellhead pressure of 188 psia. If the reservoir pressure is above the bubble-point pres- sure, but the flowing bottom-hole pressure is in the range of below bubble-point pressure, the generalized Vogel’s IPR can be used: q ¼ qb þ qv 1 À 0:2 pwf pb À 0:8 pwf pb 2 # (6:24) Hagedorn–Brown correlation, Eq. (4.27), can be used for translating the IPR to the WPR. Again, if the CPR is given by Eq. (5.12), that is, phf ¼ CRm q Sn , (6:25) solving Eqs. (6.24), (4.27), and (6.25) simultaneously will give production rate q and wellhead pressure phf . Because the solution procedure involves loop-in-loop iter- ations, it cannot be solved in MS Excel in an easy manner. A special computer program is required. Therefore, a computer-assisted graphical solution method is used in this text. The operating flow rate q and pressure phf at the well- head node can be determined graphically. This computa- tion can be performed automatically with the spreadsheet program WellheadNodalOil-HB.xls. Example Problem 6.8 For the following data, predict the operating point: Solution Example Problem 6.8 is solved with the spreadsheet program WellheadNodalOil-HB.xls. Table 6.8 shows the appearance of the spreadsheet for the Input data and Result sections. Figure 6.5 indicates that the expected oil flow rate is 4,200 stb/d at a wellhead pressure of 1,800 psia. Table 6.6 Solution Given by WellheadNodalOil-PC.xls WellheadNodalOil-PC.xls Description: This spreadsheet calculates operating point using the Poettmann–Carpenter method with wellhead node. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution button; and (4) view result in the Solution section and charts. Input data U.S. Field Units SI Units Reservoir pressure: 6,000 psia Tubing ID: 3.5 in. Choke size: 64 1 ⁄64 in. Productivity index above bubble point: 1 stb/d-psi Producing gas–liquid ratio: 1,000 scf/stb Water cut: 25% Oil gravity: 30 8API Water-specific gravity: 1.05 1 for fresh- water Gas-specific gravity: 0.65 1 for air Choke constant: 10 Choke gas–liquid ratio exponent: 0.546 Choke-size exponent: 1.89 Formation volume factor for water: 1 rb/stb Wellhead temperature: 100 8F Tubing shoe depth: 12,000 ft Bottom-hole temperature: 150 8F Solution: q (stb/d) pwf ðpsiaÞ pwh (psia) WPR CPR 0 6,000 0 600 5,400 2,003 101 1,200 4,800 1,630 201 1,800 4,200 1,277 302 2,400 3,600 957 402 3,000 3,000 674 503 3,600 2,400 429 603 4,200 1,800 220 704 4,800 1,200 39 805 Depth: 7,000 ft Tubing inner diameter: 3.5 in. Oil gravity: 45 8API Oil viscosity: 0.5 cp Production gas–liquid ratio (GLR): 500 scf/bbl Gas-specific gravity: 0.7 air ¼ 1 Choke size: 32 1/64 in. Flowing tubing head temperature: 80 8F Flowing temperature at tubing shoe: 150 8F Water cut: 10 % Reservoir pressure: 4,000 psia Bubble-point pressure: 3,800 psia Productivity index above bubble point: 5 stb/d-psi Choke flow constant: 10.00 Choke GLR exponent: 0.546 Choke-size exponent: 1.89 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 78 3.1.2007 8:40pm Compositor Name: SJoearun 6/78 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 87. 6.3 Deliverability of Multilateral Well Following the work of Pernadi et al. (1996), several math- ematical models have been proposed to predict the deliver- ability of multilateral wells. Some of these models are found from Salas et al. (1996), Larsen (1996), and Chen et al. (2000). Some of these models are oversimplified and some others are too complex to use. Consider a multilateral well trajectory depicted in Fig. 6.6. Nomenclatures are illustrated in Fig. 6.7. Suppose the well has n laterals and each lateral consists of three sections: horizontal, curvic, and vertical. Let Li, Ri, and Hi denote the length of the horizontal section, radius of curvature of the curvic section, and length of the vertical section of lateral i, respectively. Assuming the pressure losses in the horizontal sections are negligible, pseudo–steady IPR of the laterals can be expressed as follows: qi ¼ fLi pwfi À Á i ¼ 1, 2, . . . , n, (6:26) where qi ¼ production rate from lateral i fLi ¼ inflow performance function of the horizontal section of lateral i pwfi ¼ the average flowing bottom-lateral pressure in lateral i. The fluid flow in the curvic sections can be described by pwfi ¼ fRi pkfi ,qi À Á i ¼ 1, 2, . . . , n, (6:27) where fRi ¼ flow performance function of the curvic section of lateral i pkfi ¼ flowing pressure at the kick-out-point of lateral i. The fluid flow in the vertical sections may be described by pkfi ¼ fhi phfi , Xi j¼1 qj ! i ¼ 1, 2, . . . , n, (6:28) where fhi ¼ flow performance function of the vertical section of lateral i phfi ¼ flowing pressure at the top of lateral i. The following relation holds true at the junction points: pkfi ¼ phfiÀ1 i ¼ 1, 2, . . . , n (6:29) Table 6.7 Solution Given by WellheadNodalOil-GG.xls WellheadNodalOil-GG.xls Description: This spreadsheet calculates operating point based on CPR and Guo–Ghalambor TPR. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution button; and (4) view result in the Solution section. Input data U.S. Field units SI units Choke size: 64 1/64 in. Reservoir pressure: 3,000 psia Total measured depth: 7,000 ft Average inclination angle: 20 degrees Tubing ID: 1.995 in. Gas production rate: 1,000,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.85 H2O ¼ 1 Water cut: 30% Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 1 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 100 8F Bottom-hole temperature: 160 8F Absolute open flow (AOF): 2,000 bbl/d Choke flow constant: 10 Choke GLR exponent: 0.546 Choke-size exponent: 1.89 Solution A ¼ 3:1243196 in:2 D ¼ 0.16625 ft Tav ¼ 622 8R cos(u) ¼ 0.9397014 (Drv) ¼ 41.163012 fM ¼ 0.0409121 a ¼ 0.0001724 b ¼ 2.86EÀ06 c ¼ 1349785.1 d ¼ 3.8619968 e ¼ 0.0040702 M ¼ 20003.24 N ¼ 6.591Eþ09 Liquid production rate, q ¼ 1,289 bbl/d 205 m3 =d Bottom hole pressure, pwf ¼ 1,659 psia 11.29 MPa Wellhead pressure, phf ¼ 188 psia 1.28 MPa Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 79 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/79
- 88. Equations (6.26) through (6.29) contain (4n À 1) equa- tions. For a given flowing pressure phfn at the top of lateral n, the following (4n À1) unknowns can be solved from the (4n À1) equations: q1, q2, . . . qn pwf1 , pwf2 , . . . pwfn pkf1 , pkf2 , . . . pkfn phf1 , phf2 , . . . phfnÀ1 Then the production rate of the multilateral well can be determined by q ¼ Xn i¼1 qi: (6:30) Table 6.8 Solution Given by WellheadNodalOil-HB.xls WellheadNodalOil-HB.xls Description: This spreadsheet calculates operating point using Hagedorn–Brown correlation. Instruction: (1) Select a unit system; (2) update parameter values in the Input data section; (3) click Solution button; and (4) view result in the Solution section and charts. Input data U.S. Field units SI units Depth (D): 7,000 ft Tubing inner diameter (dti): 3.5 in. Oil gravity (API): 45 8API Oil viscosity (mo): 0.5 cp Production gas–liquid ratio: 500 scf/bbl Gas-specific gravity (gg): 0.7 air ¼ 1 Choke size (S): 32 1/64 in. Flowing tubing head temperature (thf ): 80 8F Flowing temperature at tubing shoe (twf ): 150 8F Water cut: 10% Reservoir pressure (pe): 4,000 psia Bubble-point pressure (pb): 3,800 psia Productivity above bubble point (J*): 5 stb/d-psi Choke flow constant (C): 10.00 Choke gas–liquid ratio exponent (m): 0.546 Choke-size exponent (n): 1.89 Solution q (stb/d) pwf (psia) phf (psia) WPR CPR 0 3,996 0 1,284 3,743 2,726 546 2,568 3,474 2,314 1,093 3,852 3,185 1,908 1,639 5,136 2,872 1,482 2,185 6,420 2,526 1,023 2,732 7,704 2,135 514 3,278 8,988 1,674 0 3,824 0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 Liquid Production Rate (bbl/d) WellheadPressure(psia) WPR CPR 0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 Figure 6.5 Nodal analysis for Example Problem 6.8. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 80 3.1.2007 8:40pm Compositor Name: SJoearun 6/80 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 89. Thus, the composite IPR, q ¼ f phfn À Á , (6:31) can be established implicitly. It should be noted that the composite IPR model described here is general. If the vertical section of the top lateral is the production string (production through tubing or/and casing), then phfn will be the flowing wellhead pres- sure. In this case, the relation expression (Eq. [6.31]) rep- resents the WPR. 6.3.1 Gas well For gas wells, Eq. (6.26) becomes qgi ¼ Ci( pp2 i À p2 wfi )ni , (6:32) where Ci ¼ productivity coefficient of lateral i ni ¼ productivity exponent of lateral i. As described in Chapter 4, Eq. (6.27), in U.S. field units (qgi in Mscf/d), can be approximated as (Katz et al., 1959) p2 wfi ¼ eSi p2 kfi þ 6:67 Â 10À4 (eSi À 1)fMiq2 gizz2 i T 2 i d5 i cos (45) , (6:33) where Si ¼ 0:0375pggRi cos (45 ) 2zziT : (6:34) The friction factor fMi can be found in the conventional manner for a given tubing diameter, wall roughness, and Reynolds number. However, if one assumes fully turbulent flow, which is the case for most gas wells, then a simple empirical relation may be used for typical tubing strings (Katz and Lee, 1990): fMi ¼ 0:01750 d0:224 i for di # 4:277 in: (6:35) fMi ¼ 0:01603 d0:164 i for di 4:277 in: (6:36) Guo (2001) used the following Nikuradse friction factor correlation for fully turbulent flow in rough pipes: fMi ¼ 1 1:74 À 2 log 2«i di 2 6 6 4 3 7 7 5 2 (6:37) For gas wells, Eq. (6.28) can be expressed as (Katz et al., 1959) p2 hfi ¼ eSi p2 hfi þ 6:67 Â 10À4 (eSi À 1)fMi Pi j¼1 qgi !2 zz2 i T 2 i d5 i , (6:38) where Si ¼ 0:0375ggHi zziTi : (6:39) L1 L2 L3 Ln R1 Rn R3 R2 H1 H2 H3 Hn Figure 6.6 Schematic of a multilateral well trajectory. k3 h3 p3 k1 h1 p1 k2 h2 p2 kn hn pn H3 Hn H2 qn qr H1 q1 q2 q3 R3 R 2 R n R1 L3 L2 Ln L1 Point 2 Point 1 Point 3 Pm2' q1+q2 Pm1' q1 Pm3' q1+q2+q3 pwf3 pwf 2 pwfn pwfn pwfn pwf1 Figure 6.7 Nomenclature of a multilateral well. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 81 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/81
- 90. At the junction points, pkfi ¼ phfiÀ1 : (6:40) Equations (6.32), (6.33), (6.38), and (6.40) contain (4n À1) equations. For a given flowing pressure phfn at the top of lateral n, the following (4n À1) unknowns can be solved from the (4n À1) equations: qg1, qg2, . . . qgn pwf1 , pwf2 , . . . pwfn pkf1 , pkf2 , . . . pkfn phf1 , phf2 , . . . phfnÀ1 Then the gas production rate of the multilateral well can be determined by qg ¼ Xn i¼1 qgi: (6:41) Thus, the composite IPR, qg ¼ f phfn À Á , (6:42) can be established implicitly. The solution procedure has been coded in the spreadsheet program MultilateralGas WellDeliverability(C-nIPR).xls. It has been found that the program does not allow cross-flow to be computed because of difficulty of computing roof of negative number with Eq. (6.32). Therefore, another spreadsheet was dev- eloped to solve the problem. The second spreadsheet is MultilateralGasWellDeliverability(Radial-FlowIPR).xls and it employs the following IPR model for individual laterals: qg ¼ khh( pp2 À p2 wf ) 1424mZT 1 ln 0:472reh L=4 2 6 6 4 3 7 7 5 (6:43) Example Problem 6.9 For the data given in the following table, predict gas production rate against 1,000 psia wellhead pressure and 100 8F wellhead temperature: Solution Example Problem 6.9 is solved with the spreadsheet program MultilateralGasWellDeliverability (Radial-FlowIPR).xls. Table 6.9 shows the appearance of the spreadsheet for the Input data and Result sections. It indicates that the expected total gas flow rate is 4,280 Mscf/d from the four laterals. Lateral 3 will steals 6,305 Mscf/d. 6.3.2 Oil well The inflow performance function for oil wells can be ex- pressed as qoi ¼ Ji( ppi À pwfi ), (6:44) where Ji ¼ productivity index of lateral i. The fluid flow in the curvic sections can be approxi- mated as pwfi ¼ pkfi þ rRi Ri, (6:45) where rRi ¼ vertical pressure gradient in the curvic sec- tion of lateral i. The pressure gradient rRi may be estimated by the Poettmann–Carpenter method: rRi ¼ rr2 i þ kki 144rri , (6:46) where rri ¼ MF Vmi , (6:47) MF ¼ 350:17 go þ WOR gwð Þ þ 0:0765GOR gg, (6:48) Vmi ¼ 5:615 Bo þ WOR Bwð Þ þ zz GOR À R sð Þ Â 29:4 pwfi þ pxfi Ti 520 , (6:49) kki ¼ f2F q2 oiMF 7:4137 Â 1010d5 i : (6:50) The fluid flow in the vertical sections may be expressed as pkfi ¼ phfi þ rhi Hi i ¼ 1 , 2, . . . , n, (6:51) where rhi ¼ pressure gradient in the vertical section of lateral i. Horizontal sections Lateral no.: 1 2 3 4 Length of horizontal section (L) 500 600 700 400 ft Horizontal permeability (k) 1 2 3 4 md Net pay thickness (h) 20 20 20 20 ft Reservoir pressure (p-bar) 3700 3500 1,800 2,800 psia Radius of drainage (reh) 2,000 2,500 1,700 2,100 ft Gas viscosity (mg) 0.02 0.02 0.02 0.02 cp Wellbore diameter (Di) 8.00 8.00 8.00 8.00 in. Bottom-hole temperature (T) 270 260 250 230 8F Gas compressibility factor (z) 0.85 0.90 0.95 0.98 Gas-specific gravity (gg) 0.85 0.83 0.80 0.75 air ¼ 1 Curvic sections Lateral no.: 1 2 3 4 Radius of curve (R) 250 300 200 270 ft Average inclination angle (u) 45 45 45 45 8F Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 4 Interval length (H) 250 300 200 8,000 ft Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 82 3.1.2007 8:40pm Compositor Name: SJoearun 6/82 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 91. Based on the Poettmann–Carpenter method, the pres- sure gradient rhi may be estimated by the follow equation: rhi ¼ rr2 i þ kki 144rri , (6:52) where rri ¼ MF Vmi , (6:53) MF ¼ 350:17 go þ WOR gwð Þ þ 0:0765GOR gg, (6:54) Vmi ¼ 5:615 Bo þ WOR Bwð Þ þ zz GOR À Rsð Þ Â 29:4 pxfi þ phfi Ti 520 , (6:55) kki ¼ f2F Pi j¼1 qoi !2 MF 7:4137 Â 1010d5 i : (6:56) Table 6.9 Solution Given by MultilateralGasWellDeliverability(Radial-FlowIPR).xls Horizontal sections Lateral no.: 1 2 3 4 Length of horizontal section (L) 500 600 700 400 ft Bottom-hole pressure (pwf ) 2,701 2,686 2,645 2,625 psia Horizontal permeability (k) 1 2 3 4 md Net pay thickness (h) 20 20 20 20 ft Reservoir pressure (p-bar) 3,700 3,500 1,800 2,800 psia Radius of drainage (reh) 2,000 2,500 1,700 2,100 ft Gas viscosity (mg) 0.02 0.02 0.02 0.02 cp Wellbore diameter (Di) 8.00 8.00 8.00 8.00 in. Bottom-hole temperature (T) 270 260 250 230 8F Gas compressibility factor (z) 0.85 0.90 0.95 0.98 Gas-specific gravity (gg) 0.85 0.83 0.80 0.75 air ¼ 1 Curvic sections Lateral no.: 1 2 3 4 Radius of curve (R) 250 300 200 270 ft Average inclination angle (u) 45 45 45 45 degrees Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 4 Interval length () 250 300 200 8,000 ft Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Kick-off points 1 2 3 4 Flow rate (q) 3,579 8,870 2,564 4,280 Mscf/d Pressure (p) 2,682 2,665 2,631 2,609 psia Temperature (T) 265 250 240 230 8F Total Production rate (q) ¼ 3,579 5,290 (6,305) 1,716 4,280 Mscf/d Horizontal sections Lateral no.: 1 2 3 4 Reservoir pressure (p-bar) 3,700 3,500 3,300 2,800 psia Oil formation factor (Bo) 1.20 1.15 1.10 1.1 stb/rb Water formation factor (Bw) 1.00 1.00 1.00 1.00 stb/rb Bottom-hole temperature (T) 270 260 250 230 8F Gas compressibility factor (z) 0.85 0.90 0.95 0.98 Gas-specific gravity (gg) 0.85 0.83 0.80 0.75 air ¼ 1 Oil-specific gravity (go) 0.80 0.78 0.87 0.85 water ¼ 1 Water-specific gravity (gw) 1.07 1.06 1.05 1.04 water ¼ 1 Water–oil ratio (WOR) 0.10 0.40 0.20 0.30 stb/stb Gas–oil ratio (GOR) 1000 1,500 2,000 2,500 scf/stb Solution–gas–oil ratio (Rs) 800 1,200 1,500 2,000 scf/stb Productivity index (J) 1 0.8 0.7 0.6 stb/d/psi Curvic sections Lateral no.: 1 2 3 4 Radius of curve (R) 200 200 200 200 ft Average inclination angle (u) 45 45 45 45 degrees Tubing diameter (di) 5 5 5 5 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 4 Interval length (H) 500 400 300 3,000 ft Tubing diameter (di) 5 5 5 5 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 83 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/83
- 92. At the junction points, pkfi ¼ phfiÀ1 : (6:57) Equations (6.44), (4.45), (6.51), and (6.57) contain (4n À1) equations. For a given flowing pressure phfn at the top of lateral n, the following (4n À1) unknowns can be solved from the (4n À1) equations: qo1 , qo2 , . . . qon pwf1 , pwf2 , . . . pwfn pkf1 , pkf2 , . . . pkfn phf1 , phf2 , . . . phfnÀ1 Then the oil production rate of the multilateral well can be determined by qo ¼ Xn i¼1 qoi: (6:58) Thus, the composite IPR, qo ¼ f phfn À Á , (6:59) can be established implicitly. The solution procedure has been coded in spreadsheet program MultilateralOilWell Deliverability.xls. Example Problem 6.10 For the data given in the last page, predict the oil production rate against 1,800 psia wellhead pressure and 100 8F wellhead temperature. Solution Example Problem 6.10 is solved with the spreadsheet program MultilateralOilWellDeliverability.xls. Table 6.10 shows the appearance of the spreadsheet for the data Input and Result sections. It indicates that the expected total oil production rate is 973 stb/d. Lateral 4 would steal 39 stb/d. Summary This chapter illustrated the principle of system analysis (Nodal analysis) with simplified well configurations. In the industry, the principle is applied with a piecewise approach to handle local flow path dimension, fluid prop- erties, and heat transfer to improve accuracy. It is vitally important to validate IPR and TPR models before performing Nodal analysis on a large scale. A Nodal analysis model is not considered to be reliable before it can match well production rates at two bottom- hole pressures. Table 6.10 Data Input and Result Sections of the Spreadsheet MultilateralOilWellDeliverability.xls MultilateralOilWellDeliverability.xls Instruction: (1) Update parameter values in the Input data section; (2) click Calculate button; and (3) view result. Input data Top node Pressure (pwh) 1,800 psia Temperature (Twh) 100 8F Calculate Horizontal sections Lateral no.: 1 2 3 4 Initial guess for pwf 3,249 3,095 2,961 2,865 psia Reservoir pressure (p-bar) 3,700 3,500 3,300 2,800 psia Oil formation factor (Bo) 1.20 1.15 1.10 1.1 stb/rb Water formation factor (Bw) 1.00 1.00 1.00 1.00 stb/rb Bottom-hole temperature (T) 270 260 250 230 8F Gas compressibility factor (z) 0.85 0.90 0.95 0.98 Gas-specific gravity (gg) 0.85 0.83 0.80 0.75 air ¼ 1 Oil-specific gravity (go) 0.80 0.78 0.87 0.85 water ¼ 1 Water-specific gravity (gw) 1.07 1.06 1.05 1.04 water ¼ 1 Water–oil ratio (WOR) 0.10 0.40 0.20 0.30 stb/stb Gas–oil ratio (GOR) 1,000 1,500 2,000 2,500 scf/stb Solution–gas–oil ratio (Rs) 800 1,200 1,500 2,000 scf/stb Productivity index (J) 1 0.8 0.7 0.6 stb/d/psi Curvic sections Lateral no.: 1 2 3 4 Radius of curve (R) 200 200 200 200 ft Average inclination angle (u) 45 45 45 45 8F Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 4 Interval length (H) 500 400 300 3,000 ft Tubing diameter (di) 3 3 3 3 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Kick off points 1 2 3 4 Flow rate (q) 451 775 1,012 973 stb/d Pressure (p) 3,185 3,027 2,895 2,797 psia Temperature (T) 265 250 240 230 8F Total: 973 451 451 237 (39) stb/d Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 84 3.1.2007 8:40pm Compositor Name: SJoearun 6/84 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 93. References chen, w., zhu, d., and hill, a.d. A comprehensive model of multilateral well deliverability. Presented at the SPE International Oil and Gas Conference and Exhibition held 7–10 November 2000 in Beijing, China. Paper SPE 64751. greene, w.r. Analyzing the performance of gas wells. J. Petroleum Technol. 1983:31–39. larsen, l. Productivity computations for multilateral, branched and other generalized and extended well con- cepts. Presented at the SPE Annual Technical Confer- ence and Exhibition held 6–9 October 1996 in Denver, Colorado. Paper SPE 36753. nind, t.e.w. Principles of Oil Well Production, 2nd edition. New York: McGraw-Hill, 1981. pernadi, p., wibowo, w., and permadi, a.k. Inflow per- formance of stacked multilateral well. Presented at the SPE Asia Pacific Conference on Integrated Modeling for Asset Management held 23–24 March 1996 in Kuala Lumpur, Malaysia. Paper SPE 39750. russell, d.g., goodrich, j.h., perry, g.e., and bruskot- ter, j.f. Methods for predicting gas well performance. J. Petroleum Technol. January 1966:50–57. salas, j.r., clifford, p.j., and hill, a.d. Multilateral well performance prediction. Presented at the SPE Western Regional Meeting held 22–24 May 1996 in Anchorage, Alaska. Paper SPE 35711. Problems 6.1 Suppose that a vertical well produces 0.65 specific gravity gas through a 27 ⁄8 -in. tubing set to the top of a gas reservoir at a depth of 8,000 ft. At tubing head, the pressure is 600 psia and the temperature is 120 8F, and the bottom-hole temperature is 180 8F. The rela- tive roughness of tubing is about 0.0006. Calculate the expected gas production rate of the well using the following data for IPR: Reservoir pressure: 1,800 psia IPR model parameter C: 0:15 Mscf=d-psi2n IPR model parameter n: 0.82 6.2 For the data given in the following table, predict the operating point using the bottom-hole as a solution node: 6.3 For the data given in the following table, predict the operating point using the bottom-hole as the solution node: 6.4 For the data given in the following table, predict the operating point using the bottom-hole as the solution node: 6.5 Use the following data to estimate the gas production rate of a gas well: 6.6 Use the following data to estimate liquid production rate of an oil well: Reservoir pressure: 3,200 psia Tubing ID: 1.66 in. Wellhead pressure: 600 psia Productivity index above bubble point: 1.5 stb/d-psi Producing gas–liquid ratio (GLR): 800 scf/stb Water cut (WC): 30% Oil gravity: 408API Water-specific gravity: 1.05 1 for freshwater Gas-specific gravity: 0.75 1 for air N2 content in gas: 0.05 mole fraction CO2 content in gas: 0.03 mole fraction H2S content in gas: 0.02 mole fraction Formation volume factor for water: 1.25 rb/stb Wellhead temperature: 110 8F Tubing shoe depth: 6,000 ft Bottom-hole temperature: 140 8F Reservoir pressure: 3,500 psia Total measured depth: 8,000 ft Average inclination angle: 10 degrees Tubing ID: 1.995 in. Gas production rate: 500,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.82 H2O ¼ 1 Water cut: 20% Water-specific gravity: 1.07 H2O ¼ 1 Solid production rate: 2ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 120 8F Bottom-hole temperature: 160 8F Tubing head pressure: 400 psia Absolute open flow (AOF): 2,200 bbl/d Depth: 9,500 ft Tubing inner diameter: 1.995 in. Oil gravity: 40 8API Oil viscosity: 3 cp Production gas–liquid ratio: 600 scf/bbl Gas-specific gravity: 0.75 air ¼ 1 Flowing tubing head pressure: 500 psia Flowing tubing head temperature: 90 8F Flowing temperature at tubing shoe: 190 8F Water cut: 20% Reservoir pressure: 5,250 psia Bubble-point pressure: 4,200 psia Productivity above bubble point: 1.2 stb/d-psi Gas-specific gravity: 0.75 Tubing inside diameter: 2.259 in. Tubing wall relative roughness: 0.0006 Measured depth at tubing shoe: 8,000 ft Inclination angle: 0 degrees Wellhead choke size: 24 1 ⁄64 in. Flowline diameter: 2 in. Gas-specific heat ratio: 1.3 Gas viscosity at wellhead: 0.01 cp Wellhead temperature: 140 8F Bottom-hole temperature: 180 8F Reservoir pressure: 2,200 psia C-constant in backpressure IPR model: 0:01 Mscf d-psi2n n-exponent in backpressure IPR model: 0.84 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 85 3.1.2007 8:40pm Compositor Name: SJoearun WELL DELIVERABILITY 6/85
- 94. 6.7 Use the following data to estimate the liquid produc- tion rate of an oil well: 6.8 For the following data, predict the oil production rate: 6.9 For the following data, predict the gas production rate against 1,200 psia wellhead pressure and 90 8F wellhead temperature: 6.10 For the following data, predict the gas production rate against 2,000 psia wellhead pressure and 80 8F wellhead temperature: Reservoir pressure: 6,500 psia Tubing ID: 3.5 in Choke size: 64 1 ⁄64 in. Productivity index above bubble point: 1.2 stb/d-psi Producing gas–liquid ratio: 800 scf/stb Water cut: 35 % Oil gravity: 40 8API Water-specific gravity: 1.05 1 for freshwater Gas-specific gravity: 0.75 1 for air Choke constant: 10 Choke gas–liquid ratio exponent: 0.546 Choke-size exponent: 1.89 Formation volume factor for water: 1 rb/stb Wellhead temperature: 110 8F Tubing shoe depth: 10,000 ft Bottom-hole temperature: 200 8F Choke size: 48 1 ⁄64 in. Reservoir pressure: 3,200 psia Total measured depth: 7,000 ft Average inclination angle: 10 degrees Tubing ID: 1.995 in. Gas production rate: 600,000 scfd Gas-specific gravity: 0.7 air ¼ 1 Oil-specific gravity: 0.85 H2O ¼ 1 Water cut: 20% Water-specific gravity: 1.05 H2O ¼ 1 Solid production rate: 0:5 ft3 =d Solid-specific gravity: 2.65 H2O ¼ 1 Tubing head temperature: 120 8F Bottom-hole temperature: 180 8F Absolute open flow (AOF): 2,200 bbl/d Choke flow constant: 10 Choke gas–liquid ratio exponent: 0.546 Choke size exponent: 1.89 Depth: 7,500 ft Tubing inner diameter: 3.5 in. Oil gravity: 40 8API Oil viscosity: 0.8 cp Production GLR: 700 scf/bbl Gas-specific gravity: 0.7 air ¼ 1 Choke size: 48 1 ⁄64 in. Flowing tubing head temperature: 90 8F Flowing temperature at tubing shoe: 160 8F Water cut: 20% Reservoir pressure: 4,200 psia Bubble-point pressure: 4,000 psia Productivity above bubble point: 4 stb/d-psi Choke flow constant: 10 Choke gas–liquid ratio exponent: 0.546 Choke-size exponent: 1.89 Horizontal sections Lateral no.: 1 2 3 Length of horizontal section (L) 1,000 1,100 1,200 ft Horizontal permeability (k) 8 5 4 md Net pay thickness (h) 40 50 30 ft Reservoir pressure (p-bar) 3,500 3,450 3,400 psia Radius of drainage area (reh) 2,000 2,200 2,400 ft Gas viscosity (mg) 0.02 0.02 0.02 cp Wellbore diameter (Di) 6.00 6.00 6.00 in. Bottom-hole temperature (T) 150 140 130 8F Gas compressibility factor (z) 0.95 0.95 0.95 Gas specific gravity (gg) 0.80 0.80 0.80 air ¼ 1 Curvic sections Lateral no.: 1 2 3 Radius of curve (R) 333 400 500 ft Average inclination angle (u) 45 45 45 degrees Tubing diameter (di) 1.995 1.995 1.995 in. Pipe roughness (e) 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 Interval length (H) 500 500 6,000 ft Tubing diameter (di) 1.995 1.995 1.995 in. Pipe roughness (e) 0.0018 0.0018 0.0018 in. Horizontal sections Lateral no.: 1 2 3 4 Reservoir pressure (p-bar) 3,500 3,300 3,100 2,900 psia Oil formation factor (Bo) 1.25 1.18 1.19 1.16 stb/rb Water formation factor (Bw) 1.00 1.00 1.00 1.00 stb/rb Bottom-hole temperature (T) 170 160 150 130 8F Gas compressibility factor (z) 0.9 0.90 0.90 0.90 Gas-specific gravity (gg) 0.75 0.73 0.70 0.75 air ¼ 1 Oil-specific gravity (go) 0.85 0.88 0.87 0.8 6 water ¼ 1 Water-specific gravity (gw) 1.07 1.06 1.05 1.0 4 water ¼ 1 Water–oil ratio (WOR) 0.30 0.20 0.10 0.1 0 stb/stb Gas–oil ratio (GOR) 1,000 1,200 1,500 2,000 scf/stb Solution–gas–oil ratio (Rs) 600 1,000 1,200 1,800 scf/stb Productivity index (J) 2 1.8 1.7 1.6 stb/d/psi Curvic sections Lateral no.: 1 2 3 4 Radius of curve (R) 400 400 400 400 ft Average inclination angle (u) 45 45 45 45 degrees Tubing diameter (di) 2.441 2.441 2.441 2.441 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Vertical sections Lateral no.: 1 2 3 4 Interval length (H) 100 100 100 5,000 ft Tubing diameter (di) 2.441 2.441 2.441 2.441 in. Pipe roughness (e) 0.0018 0.0018 0.0018 0.0018 in. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap06 Final Proof page 86 3.1.2007 8:40pm Compositor Name: SJoearun 6/86 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 95. 7 Forecast of Well Production Contents 7.1 Introduction 7/88 7.2 Oil Production during Transient Flow Period 7/88 7.3 Oil Production during Pseudo–Steady Flow Period 7/88 7.4 Gas Production during Transient Flow Period 7/92 7.5 Gas Production during Pseudo–Steady-State Flow Period 7/92 Summary 7/94 References 7/94 Problems 7/95 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 87 3.1.2007 8:47pm Compositor Name: SJoearun
- 96. 7.1 Introduction With the knowledge of Nodal analysis, it is possible to forecast well production, that is, future production rate and cumulative production of oil and gas. Combined with information of oil and gas prices, the results of a produc- tion forecast can be used for field economics analyses. A production forecast is performed on the basis of principle of material balance. The remaining oil and gas in the reservoir determine future inflow performance relation- ship (IPR) and, therefore, production rates of wells. Production rates are predicted using IPR (see Chapter 3) and tubing performance relationship (TPR) (see Chapter 4) in the future times. Cumulative productions are predicted by integrations of future production rates. A complete production forecast should be carried out in different flow periods identified on the basis of flow regimes and drive mechanisms. For a volumetric oil reservoir, these periods include the following: . Transient flow period . Pseudo–steady one-phase flow period . Pseudo–steady two-phase flow period 7.2 Oil Production during Transient Flow Period The production rate during the transient flow period can be predicted by Nodal analysis using transient IPR and steady flow TPR. IPR model for oil wells is given by Eq. (3.2), that is, q ¼ kh( pi À pwf ) 162:6Bomo log t þ log k fmoctr2 w À 3:23 þ 0:87S : (7:1) Equation 7.1 can be used for generating IPR curves for future time t before any reservoir boundary is reached by the pressure wave from the wellbore. After all reservoir boundaries are reached, either pseudo–steady-state flow or steady-state flow should prevail depending on the types of reservoir boundaries. The time required for the pressure wave to reach a circular reservoir boundary can be with tpss % 1,200 fmctr2 e k . The same TPR is usually used in the transient flow period assuming fluid properties remain the same in the well over the period. Depending on the producing gas–liquid ratio (GLR), the TPR model can be chosen from simple ones such as Poettmann–Carpenter and sophisticated ones such as the modified Hagedorn–Brown. It is essential to validate the selected TPR model based on measured data such as flow gradient survey from local wells. Example Problem 7.1 Suppose a reservoir can produce oil under transient flow for the next 6 months. Predict oil production rate and cumulative oil production over the 6 months using the following data: Solution To solve Example Problem 7.1, the spreadsheet program TransientProductionForecast.xls was used to perform Nodal analysis for each month. Operating points are shown in Fig. 7.1. The production forecast result is shown in Table 7.1, which also includes calculated cumulative production at the end of each month. The data in Table 7.1 are plotted in Fig. 7.2. 7.3 Oil Production during Pseudo–Steady Flow Period It is generally believed that oil production during a pseudo– steady-state flow period is due to fluid expansion in under- saturated oil reservoirs and solution-gas drive in saturated oil reservoirs. An undersaturated oil reservoir becomes a saturated oil reservoir when the reservoir pressure drops to below the oil bubble-point pressure. Single-phase flow dominates in undersaturated oil reservoirs and two-phase flow prevails in saturated oil reservoirs. Different math- ematical models have been used for time projection in production forecast for these two types of reservoirs, or the same reservoir at different stages of development based on reservoir pressure. IPR changes over time due to the changes in gas saturation and fluid properties. 7.3.1 Oil Production During Single-Phase Flow Period Following a transient flow period and a transition time, oil reservoirs continue to deliver oil through single-phase flow under a pseudo–steady-state flow condition. The IPR changes with time because of the decline in reservoir pres- sure, while the TPR may be considered constant because fluid properties do not significantly vary above the bubble- point pressure. The TPR model can be chosen from simple ones such as Poettmann–Carpenter and sophisticated ones such as the modified Hagedorn–Brown. The IPR model is given by Eq. (3.7), in Chapter 3, that is, q ¼ kh( pp À pwf ) 141:2Bomo 1 2 ln 4A gCAr2 w þ S : (7:2) The driving mechanism above the bubble-point pressure is essentially the oil expansion because oil is slightly compressible. The isothermal compressibility is defined as c ¼ À 1 V @V @p , (7:3) where V is the volume of reservoir fluid and p is pressure. The isothermal compressibility c is small and essentially constant for a given oil reservoir. The value of c can be measured experimentally. By separating variables, integra- tion of Eq. (7.3) from the initial reservoir pressure pi to the current average-reservoir pressure pp results in V Vi ¼ ec(piÀpp) , (7:4) Reservoir porosity (f): 0.2 Effective horizontal permeability (k): 10 md Pay zone thickness (h): 50 ft Reservoir pressure ( pi): 5,500 psia Oil formation volume factor (Bo): 1.2 rb/stb Total reservoir compressibility (ct): 0.000013 psiÀ1 Wellbore radius (rw): 0.328 ft Skin factor (S ): 0 Well depth (H): 10,000 ft Tubing inner diameter (d ): 2.441 Oil gravity (API): 30 API Oil viscosity (mo): 1.5 cp Producing GLR (GLR): 300 scf/bbl Gas-specific gravity (gg): 0.7 air ¼ 1 Flowing tubing head pressure (phf ): 800 psia Flowing tubing head temperature (Thf ): 150 8F Flowing temperature at tubing shoe (Twf ): 180 8F Water cut: 10% Interfacial tension (s): 30 dynes/cm Specific gravity of water (gw): 1.05 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 88 3.1.2007 8:47pm Compositor Name: SJoearun 7/88 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 97. where Vi is the reservoir volume occupied by the reservoir fluid. The fluid volume V at lower pressure pp includes the volume of fluid that remains in the reservoir (still Vi) and the volume of fluid that has been produced, that is, V ¼ Vi þ Vp: (7:5) Substituting Eq. (7.5) into Eq. (7.4) and rearranging the latter give r ¼ Vp Vi ¼ ec(piÀpp) À 1, (7:6) where r is the recovery ratio. If the original oil in place N is known, the cumulative recovery (cumulative production) is simply expressed as Np ¼ rN. For the case of an undersaturated oil reservoir, forma- tion water and rock also expand as reservoir pressure drops. Therefore, the compressibility c should be the total compressibility ct, that is, ct ¼ coSo þ cwSw þ cf , (7:7) where co, cw, and cf are the compressibilities of oil, water, and rock, respectively, and So and Sw are oil and water saturations, respectively. The following procedure is taken to perform the production forecast during the single-phase flow period: 0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,000 100 300 500 700 900 1,100 1,300 Production Rate (stb/day) FlowingBottomHolePressure(psia) 1-month IPR 2-month IPR 3-month IPR 4-month IPR 5-month IPR 6-month IPR TPR Figure 7.1 Nodal analysis plot for Example Problem 7.1. Table 7.1 Production Forecast Given by TransientProductionForecast.xls Time (mo) Production rate (stb/d) Cumulative production (stb) 1 639 19,170 2 618 37,710 3 604 55,830 4 595 73,680 5 588 91,320 6 583 108,795 570 580 590 600 610 620 630 640 650 Time (month) ProductionRate(stb/d) 0 20,000 40,000 60,000 80,000 100,000 120,000 CumulativeProduction(stb) Production Rate (stb/d) Cumulative Production (stb) 0 7654321 Figure 7.2 Production forecast for Example Problem 7.1. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 89 3.1.2007 8:47pm Compositor Name: SJoearun FORECAST OF WELL PRODUCTION 7/89
- 98. 1. Assume a series of average-reservoir pressure p¯ values between the initial reservoir pressure pi and oil bubble- point pressure pb. Perform Nodal analyses to estimate production rate q at each average-reservoir pressure and obtain the average production rate q¯ over the pressure interval. 2. Calculate recovery ratio r, cumulative production Np at each average-reservoir pressure, and the incremental cumulative production DNp within each average-reser- voir pressure interval. 3. Calculate production time Dt for each average-reservoir pressure interval by Dt ¼ DNp=qq and the cumulative production time by t ¼ P Dt. Example Problem 7.2 Suppose the reservoir described in Example Problem 7.1 begins to produce oil under pseudo– steady-state flow conditions immediately after the 6-month transient flow. If the bubble-point pressure is 4,500 psia, predict the oil production rate and cumulative oil production over the time interval before the reservoir pressure declines to bubble-point pressure. Solution Based on the transient flow IPR, Eq. (7.1), the productivity index will drop to 0.2195 stb/d-psi and production rate will drop to 583 stb/d at the end of the 6 months. If a pseudo–steady-state flow condition assumes immediately after the 6-month transient flow, the same production rate should be given by the pseudo–steady-state flow IPR, Eq. (7.2). These conditions require that the average-reservoir pressure be 5,426 psia by pp ¼ pÀ35:3 e q kh and drainage be 1458 acres by Eq. (3.9). Assuming an initial water saturation of 0.35, the original oil in place (OOIP) in the drainage area is estimated to be 87,656,581 stb. Using these additional data, Nodal analyses were per- formed with spreadsheet program Pseudo-Steady-1Phase ProductionForecast.xls at 10 average-reservoir pressures from 5,426 to bubble-point pressure of 4,500 psia. Operating points are shown in Fig. 7.3. The production forecast result is shown in Table 7.2. The production rate and cumulative production data in Table 7.2 are plotted in Fig. 7.4. 7.3.2 Oil Production during Two-Phase Flow Period Upon the average-reservoir pressure drops to bubble-point pressure, a significant amount of solution gas becomes free gas in the reservoir, and solution-gas drive becomes a dominating mechanism of fluid production. The gas–oil two-phase pseudo–steady-state flow begins to prevail the reservoir. Both IPR and TPR change with time because of the significant variations of fluid properties, relative per- meabilities, and gas–liquid ratio (GLR). The Hagedorn– Brown correlation should be used to model the TPR. The IPR can be described with Vogel’s model by Eq. (3.19), in Chapter 3, that is, 0 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,000 Production Rate (stb/day) FlowingBottomHolePressure(psia) IPR for reservoir pressure 5,426 psia IPR for reservoir pressure 5,300 psia IPR for reservoir pressure 5,200 psia IPR for reservoir pressure 5,100 psia IPR for reservoir pressure 5,000 psia IPR for reservoir pressure 4,900 psia IPR for reservoir pressure 4,800 psia IPR for reservoir pressure 4,700 psia IPR for reservoir pressure 4,600 psia IPR for reservoir pressure 4,500 psia TPR 1,300100 300 500 700 900 1,100 Figure 7.3 Nodal analysis plot for Example Problem 7.2. Table 7.2 Production Forecast for Example Problem 7.2 Reservoir pressure (psia) Production rate (stb/d) Recovery ratio Cumulative production (stb) Incremental production (stb) Incremental production time (days) Pseudo– steady-state production time (days) 5,426 583 0.0010 84,366 0 5,300 563 0.0026 228,204 143,837 251 251 5,200 543 0.0039 342,528 114,325 207 458 5,100 523 0.0052 457,001 114,473 215 673 5,000 503 0.0065 571,624 114,622 223 896 4,900 483 0.0078 686,395 114,771 233 1,129 4,800 463 0.0091 801,315 114,921 243 1,372 4,700 443 0.0105 916,385 115,070 254 1,626 4,600 423 0.0118 1,031,605 115,220 266 1,892 4,500 403 0.0131 1,146,975 115,370 279 2,171 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 90 3.1.2007 8:47pm Compositor Name: SJoearun 7/90 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 99. q ¼ JÃpp 1:8 1 À 0:2 pwf pp À 0:8 pwf pp 2 # : (7:8) To perform production forecast for solution-gas drive reservoirs, material balance models are used for establish- ing the relation of the cumulative production to time. The commonly used material balance model is found in Craft and Hawkins (1991), which was based on the original work of Tarner (1944). The following procedure is taken to carry out a produc- tion forecast during the two-phase flow period: Step 1: Assume a series of average-reservoir pressure p¯ values between the bubble-point pressure pb and abandonment reservoir pressure pa. Step 2: Estimate fluid properties at each average-reservoir pressure, and calculate incremental cumulative production DNp and cumulative production Np within each average-reservoir pressure interval. Step 3: Perform Nodal analyses to estimate production rate q at each average-reservoir pressure. Step 4: Calculate production time Dt for each average- reservoir pressure interval by Dt ¼ DNp=q and the cumulative production time by t ¼ P Dt. Step 2 is further described in the following procedure: 1. Calculate coefficients Fn and Fg for the two pressure values that define the pressure interval, and obtain average values FFn and FFg in the interval. The Fn and Fg are calculated using Fn ¼ Bo À RsBg (Bo À Boi) þ (Rsi À Rs)Bg , (7:9) Fg ¼ Bg (Bo À Boi) þ (Rsi À Rs)Bg , (7:10) where Bg should be in rb/scf if Rs is in scf/stb. 2. Assume an average gas–oil ratio R¯ in the interval, and calculate incremental oil and gas production per stb of oil in place by DN1 p ¼ 1 À FFnN1 p À FFgG1 p FFn þ RR FFg , (7:11) DG1 p ¼ DN1 p RR, (7:12) where N1 p and G 1 p are the cumulative oil and gas pro- duction per stb of oil in place at the beginning of the interval. 3. Calculate cumulative oil and gas production at the end of the interval by adding DN1 p and DG1 p to N1 p and G1 p, respectively. 4. Calculate oil saturation by So ¼ Bo Boi (1 À Sw)(1 À N1 p ): (7:13) 5. Obtain the relative permeabilities krg and kro based on So. 6. Calculate the average gas–oil ratio by RR ¼ Rs þ krgmoBo kromgBg , (7:14) where again Bg should be in rb/scf if Rs is in scf/stb. 7. Compare the calculated RR with the value assumed in Step 2. Repeat Steps 2 through 6 until RR converges. Example Problem 7.3 For the oil reservoir described in Example Problem 7.2, predict the oil production rate and cumulative oil production over the time interval during which reservoir pressure declines from bubble-point pressure to abandonment reservoir pressure of 2,500. The following additional data are given: 0 100 200 300 400 500 600 Pseudosteady State Production Time (days) ProductionRate(stb/day) 0.E+00 2.E+05 4.E+05 6.E+05 8.E+05 1.E+06 1.E+06 1.E+06 CumulativeProduction(stb) Production Rate Cumulative Production 0 500 1,000 1,500 2,000 2,500 Figure 7.4 Production forecast for Example Problem 7.2. Reservoir pressure (psia) Bo (rb /stb) Bg (rb /scf) Rs (rb /scf) mg (cp) 4,500 1.200 6.90EÀ04 840 0.01 4,300 1.195 7.10EÀ04 820 0.01 4,100 1.190 7.40EÀ04 770 0.01 3,900 1.185 7.80EÀ04 730 0.01 3,700 1.180 8.10EÀ04 680 0.01 3,500 1.175 8.50EÀ04 640 0.01 3,300 1.170 8.90EÀ04 600 0.01 3,100 1.165 9.30EÀ04 560 0.01 2,900 1.160 9.80EÀ04 520 0.01 2,700 1.155 1.00EÀ03 480 0.01 2,500 1.150 1.10EÀ03 440 0.01 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 91 3.1.2007 8:47pm Compositor Name: SJoearun FORECAST OF WELL PRODUCTION 7/91
- 100. kro ¼ 10À(4:8455Sgþ0:301) krg ¼ 0:730678S1:892 g Solution Example Problem 7.3 is solved using spreadsheets Pseudo-Steady-2PhaseProductionForecast.xls and Pseudo- steady2PhaseForecastPlot.xls. The former computes operating points and the latter performs material balance calculations. The results are shown in Tables 7.3, 7.4, and 7.5. Production forecast curves are given in Fig. 7.5. 7.4 Gas Production during Transient Flow Period Similar to oil production, the gas production rate during a transient flow period can be predicted by Nodal analysis using transient IPR and steady-state flow TPR. The IPR model for gas wells is described in Chapter 3, that is, q ¼ kh½m(pi) À m(pwf )Š 1638T log t þ log k fmoctr2 w À 3:23 þ 0:87S : (7:15) Equation (7.15) can be used for generating IPR curves for future time t before any reservoir boundary is ‘‘felt.’’ After all reservoir boundaries are reached, a pseudo–steady-state flow should prevail for a volumetric gas reservoir. For a circular reservoir, the time required for the pressure wave to reach the reservoir boundary can be estimated with tpss % 1200 fmctr2 e k . The same TPR is usually used in the transient flow period assuming fluid properties remain the same in the well over the period. The average temperature– average z-factor method can be used for constructing TPR. 7.5 Gas Production during Pseudo–Steady-State Flow Period Gas production during pseudo–steady-state flow period is due to gas expansion. The IPR changes over time due to the change in reservoir pressure. An IPR model is described in Chapter 3, that is, q ¼ kh½m(pp) À m(pwf )Š 1424T ln re rw À 3 4 þ S þ Dq : (7:16) Table 7.3 Oil Production Forecast for N ¼ 1 p-bar (psia) Bo (rb=stb) Bg (rb=scf) Rs (rb=scf) Fn Fg Rav (rb=scf) DN1 p (stb) N1 p (stb) 4,500 1.200 6.9EÀ04 840 1.195 7.1EÀ04 820 66.61 0.077 859 7.52E-03 7.52EÀ03 4,300 7.52EÀ03 1.190 7.4EÀ04 770 14.84 0.018 1,176 2.17E-02 2.92EÀ02 4,100 2.92EÀ02 1.185 7.8EÀ04 730 8.69 0.011 1,666 1.45E-02 4.38EÀ02 3,900 4.38EÀ02 1.180 8.1EÀ04 680 5.74 0.007 2,411 1.41E-02 5.79EÀ02 3,700 5.79EÀ02 1.175 8.5EÀ04 640 4.35 0.006 3,122 9.65E-03 6.76EÀ02 3,500 6.76EÀ02 1.170 8.9EÀ04 600 3.46 0.005 3,877 8.18E-03 7.57EÀ02 3,300 7.57EÀ02 1.165 9.3EÀ04 560 2.86 0.004 4,658 7.05E-03 8.28EÀ02 3,100 8.28EÀ02 1.160 9.8EÀ04 520 2.38 0.004 5,436 6.43E-03 8.92EÀ02 2,900 8.92EÀ02 1.155 1.0EÀ03 480 2.07 0.003 6,246 5.47E-03 9.47EÀ02 2,700 9.47EÀ02 1.150 1.1EÀ03 440 1.83 0.003 7,066 4.88E-03 9.96EÀ02 2,500 9.96EÀ02 Table 7.4 Gas Production Forecast for N ¼ 1 p-bar (psia) DG1 p (scf) G1 p (scf) So Sg kro krg Rav (rb=scf) 4,500 6.46Eþ00 6.46Eþ00 0.642421 0.007579 0.459492 7.11066EÀ05 859 4,300 6.46Eþ00 2.55Eþ01 3.20Eþ01 0.625744 0.024256 0.381476 0.000642398 1,176 4,100 3.20Eþ01 2.42Eþ01 5.62Eþ01 0.61378 0.03622 0.333809 0.001371669 1,666 3,900 5.62Eþ01 3.41Eþ01 9.03Eþ01 0.602152 0.047848 0.293192 0.002322907 2,411 3,700 9.03Eþ01 3.01Eþ01 1.20Eþ02 0.593462 0.056538 0.266099 0.003185377 3,122 3,500 1.20Eþ02 3.17Eþ01 1.52Eþ02 0.585749 0.064251 0.244159 0.004057252 3,877 3,300 1.52Eþ02 3.28Eþ01 1.85Eþ02 0.578796 0.071204 0.225934 0.004927904 4,658 3,100 1.85Eþ02 3.50Eþ01 2.20Eþ02 0.572272 0.077728 0.210073 0.005816961 5,436 2,900 2.20Eþ02 3.41Eþ01 2.54Eþ02 0.566386 0.083614 0.19672 0.006678504 6,246 2,700 2.54Eþ02 3.45Eþ01 2.89Eþ02 0.560892 0.089108 0.185024 0.007532998 7,066 2,500 2.89Eþ02 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 92 3.1.2007 8:47pm Compositor Name: SJoearun 7/92 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 101. Constant TPR is usually assumed if liquid loading is not a problem and the wellhead pressure is kept constant over time. The gas production schedule can be established through the material balance equation, Gp ¼ Gi 1 À pp z pi zi ! , (7:17) where Gp and Gi are the cumulative gas production and initial ‘‘gas in place,’’ respectively. If the gas production rate is predicted by Nodal analysis at a given reservoir pressure level and the cumulative gas production is estimated with Eq. (7.17) at the same reservoir pressure level, the corresponding production time can be calculated and, thus, production forecast can be carried out. Example Problem 7.4 Use the following data and develop a forecast of a well production after transient flow until the average reservoir pressure declines to 2,000 psia: Reservoir depth: 10,000 ft Initial reservoir pressure: 4,613 psia Reservoir temperature: 180 8F Pay zone thickness: 78 ft Formation permeability: 0.17 md Formation porosity: 0.14 Water saturation: 0.27 Gas-specific gravity: 0:7 air ¼ 1 Total compressibility: 1:5 Â 10À4 psiÀ1 Darcy skin factor: 0 Non-Darcy flow coefficient: 0 Drainage area: 40 acres Wellbore radius: 0.328 ft Tubing inner diameter: 2.441 in. Desired flowing bottom-hole pressure: 1,500 psia Solution The spreadsheet program Carr-Kobayashi- Burrows-GasViscosity.xls gives a gas viscosity value of 0.0251 cp at the initial reservoir pressure of 4,613 psia and temperature of 180 8F for the 0.7 specific gravity gas. The spreadsheet program Hall-Yarborogh-z.xls gives a z-factor value of 1.079 at the same conditions. Formation volume factor at the initial reservoir pressure is calculated with Eq. (2.62): Bgi ¼ 0:0283 (1:079)(180 þ 460) 4,613 ¼ 0:004236 ft3 =scf The initial ‘‘gas in place’’ within the 40 acres is Table 7.5 Production Schedule Forecast p-bar (psia) qo (stb=d) DNp (stb) Np (stb) DGp (scf) Gp (scf) Dt (d) t (d) 4,500 393 2.8Eþ04 2.37Eþ07 70 4,300 27,601 2.37Eþ07 7.02Eþ01 363 8.0Eþ04 9.36Eþ07 219 4,100 107,217 1.17Eþ08 2.90Eþ02 336 5.3Eþ04 8.89Eþ07 159 3,900 160,565 2.06Eþ08 4.48Eþ02 305 5.2Eþ04 1.25Eþ08 170 3,700 212,442 3.31Eþ08 6.18Eþ02 276 3.5Eþ04 1.10Eþ08 128 3,500 247,824 4.42Eþ08 7.47Eþ02 248 3.0Eþ04 1.16Eþ08 121 3,300 277,848 5.58Eþ08 8.68Eþ02 217 2.6Eþ04 1.21Eþ08 119 3,100 303,716 6.79Eþ08 9.87Eþ02 187 2.4Eþ04 1.28Eþ08 126 2,900 327,302 8.07Eþ08 1.11Eþ03 155 2.0Eþ04 1.25Eþ08 129 2,700 347,354 9.32Eþ08 1.24Eþ03 120 1.8Eþ04 1.27Eþ08 149 2,500 365,268 1.06Eþ09 1.39Eþ03 0 50 100 150 200 250 300 350 400 450 0 10 20 30 40 50 Two-Phase Production Time (months) ProductionRate(stb/d) 0.0E+00 5.0E+04 1.0E+05 1.5E+05 2.0E+05 2.5E+05 3.0E+05 3.5E+05 4.0E+05 CumulativeProduction(stb) Production Rate Cumulative Production Figure 7.5 Production forecast for Example Problem 7.3. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 93 3.1.2007 8:47pm Compositor Name: SJoearun FORECAST OF WELL PRODUCTION 7/93
- 102. Gi ¼ (43,560)(40)(78)(0:14)(1 À 0:27) 0:004236 ¼ 3:28 Â 109 scf: Assuming a circular drainage area, the equivalent radius of the 40 acres is 745 ft. The time required for the pressure wave to reach the reservoir boundary is estimated as tpss % 1200 (0:14)(0:0251)(1:5 Â 10À4 )(745)2 0:17 ¼ 2,065 hours ¼ 86 days: The spreadsheet program PseudoPressure.xls gives m( pi) ¼ m(4613) ¼ 1:27 Â 109 psi2 =cp m( pwf ) ¼ m(1500) ¼ 1:85 Â 108 psi2 =cp : Substituting these and other given parameter values into Eq. (7.15) yields q ¼ (0:17)(78)[1:27 Â 109 À 1:85 Â 108 ] 1638(180 þ 460) log (2065) þ log 0:17 (0:14)(0:0251)(1:5Â10À4)(0:328)2 À 3:23 ¼ 2,092 Mscf=day: Substituting q ¼ 2,092 Mscf=day into Eq. (7.16) gives 2,092 ¼ (0:17)(78)[m(pp) À 1:85 Â 108 ] 1424(180 þ 460) ln 745 0:328 À 3 4 þ 0 À Á , which results in m(pp) ¼ 1:19 Â 109 psi2 =cp. The spread- sheet program PseudoPressure.xls gives pp ¼ 4,409 psia at the beginning of the pseudo–steady-state flow period. If the flowing bottom-hole pressure is maintained at a level of 1,500 psia during the pseudo–steady-state flowperiod (after 86 days of transient production), Eq. (7.16) is simplified as q ¼ (0:17)(78)[m(pp) À 1:85 Â 108 ] 1424(180 þ 460) ln 745 0:328 À 3 4 þ 0 À Á or q ¼ 2:09 Â 10À6 ½m(pp) À 1:85 Â 108 Š, which, combined with Eq. (7.17), gives the production fore- cast shown in Table 7.6, where z-factors and real gas pseudo- pressures were obtained using spreadsheet programs Hall- Yarborogh-z.xls and PseudoPressure.xls, respectively. The production forecast result is also plotted in Fig. 7.6. Summary This chapter illustrated how to perform production fore- cast using the principle of Nodal analysis and material balance. Accuracy of the forecast strongly depends on the quality of fluid property data, especially for the two- phase flow period. It is always recommended to use fluid properties derived from PVT lab measurements in produc- tion forecast calculations. References craft, b.c. and hawkins, m. Applied Petroleum Reservoir Engineering, 2nd edition. Englewood Cliffs, NJ: Pren- tice Hall, 1991. Table 7.6 Result of Production Forecast for Example Problem 7.4 Reservoir pressure (psia) z Pseudo- pressure (108 psi2 =cp) Gp (MMscf) DGp (MMscf) q (Mscf/d) Dt (day) t (day) 4,409 1.074 11.90 130 4,200 1.067 11.14 260 130 1,942 67 67 4,000 1.060 10.28 385 125 1,762 71 138 3,800 1.054 9.50 514 129 1,598 81 218 3,600 1.048 8.73 645 131 1,437 91 309 3,400 1.042 7.96 777 132 1,277 103 413 3,200 1.037 7.20 913 136 1,118 122 534 3,000 1.032 6.47 1,050 137 966 142 676 2,800 1.027 5.75 1,188 139 815 170 846 2,600 1.022 5.06 1,328 140 671 209 1,055 2,400 1.018 4.39 1,471 143 531 269 1,324 2,200 1.014 3.76 1,615 144 399 361 1,686 2,000 1.011 3.16 1,762 147 274 536 2,222 0 500 1,000 1,500 2,000 2,500 Pseudosteady Production Time (days) CumulativeProduction(MMscf) ProductionRate(Mscf/day) 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2,000 q (Mscf/d) Gp (MMscf) 0 500 1,000 1,500 2,000 2,500 Figure 7.6 Result of production forecast for Example Problem 7.4. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 94 3.1.2007 8:47pm Compositor Name: SJoearun 7/94 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 103. tarner, j. How different size gas caps and pressure main- tenance programs affect amount of recoverable oil. Oil Weekly June 12, 1944;144:32–34. Problems 7.1 Suppose an oil reservoir can produce under transient flow for the next 1 month. Predict oil production rate and cumulative oil production over the 1 month using the following data: 7.2 Suppose the reservoir described in Problem 7.1 begins to produce oil under a pseudo–steady-state flow con- dition immediately after the 1-month transient flow. If the bubble-point pressure is 4,000 psia, predict oil production rate and cumulative oil production over the time interval before reservoir pressure declines to bubble-point pressure. 7.3 For the oil reservoir described in Problem 7.2, predict oil production rate and cumulative oil production over the time interval during which reservoir pressure declines from bubble-point pressure to abandonment reservoir pressure of 2,000. The following additional data are given: kro ¼ 10À(4:5Sgþ0:3) krg ¼ 0:75S1:8 g 7.4 Assume that a 0.328-ft radius well in a gas reservoir drains gas from an area of 40 acres at depth 8,000 ft through a 2.441 inside diameter (ID) tubing against a wellhead pressure 500 psia. The reservoir has a net pay of 78 ft, porosity of 0.14, permeability of 0.17 md, and water saturation of 0.27. The initial reservoir pressure is 4,613 psia. Reservoir temperature is 180 8F. Gas- specific gravity is 0.65. The total system compressibility is 0:00015 psiÀ1 . Both Darcy and non-Darcy skin are negligible. Considering both transient and pseudo– steady-state flow periods, generate a gas production forecast until the reservoir pressure drops to 3,600 psia. 7.5 Use the following data and develop a forecast of a gas well production during the transient flow period: Reservoir depth: 9,000 ft Initial reservoir pressure: 4,400 psia Reservoir temperature: 1708F Pay zone thickness: 60 ft Formation permeability: 0.25 md Formation porosity: 0.15 Water saturation: 0.30 Gas-specific gravity: 0.7 air ¼ 1 Total compressibility: 1:6 Â 10À4 psiÀ1 Darcy skin factor: 0 Non-Darcy flow coefficient: 0 Drainage area: 40 acres Wellbore radius: 0.328 ft Tubing inner diameter: 2.441 in. Desired flowing bottom-hole pressure: 1,100 psia 7.6 Use the following data and develop a forecast of a gas well production after transient flow until the average reservoir pressure declines to 2,000 psia: Reservoir depth: 8,000 ft Initial reservoir pressure: 4,300 psia Reservoir temperature: 1608F Pay zone thickness: 50 ft Formation permeability: 0.20 md Formation porosity: 0.15 Water saturation: 0.30 Gas-specific gravity: 0.7 air ¼ 1 Total compressibility: 1:6 Â 10À4 psiÀ1 Darcy skin factor: 0 Non-Darcy flow coefficient: 0 Drainage area: 160 acres Wellbore radius: 0.328 ft Tubing inner diameter: 1.995 in. Desired flowing bottom-hole pressure: 1,200 psia 7.7 Use the following data and develop a forecast of a gas well production after transient flow until the average reservoir pressure declines to 2,000 psia: Reservoir depth: 8,000 ft Initial reservoir pressure: 4,300 psia Reservoir temperature: 1608F Pay zone thickness: 50 ft Formation permeability: 0.20 md Reservoir porosity (f): 0.25 Effective horizontal permeability (k): 50 md Pay zone thickness (h): 75 ft Reservoir pressure (pi): 5000 psia Oil formation volume factor (Bo): 1.3 rb/stb Total reservoir compressibility (ct): 0.000012 psiÀ1 Wellbore radius (rw): 0.328 ft Skin factor (S): 0 Well depth (H): 8,000 ft Tubing inner diameter (d): 2.041 Oil gravity (API): 35 API Oil viscosity (mo): 1.3 cp Producing gas–liquid ratio: 400 scf/bbl Gas specific gravity (gg): 0.7 air ¼ 1 Flowing tubing head pressure (phf ): 500 psia Flowing tubing head temperature (Thf ): 120 8F Flowing temperature at tubing shoe (Twf ): 160 8F Water cut: 10% Interfacial tension (s): 30 dynes/cm Specific gravity of water (gw): 1.05 Reservoir pressure (psia) Bo(rb/stb) Bg (rb/scf) Rs (rb/scf) mg (cp) 4,000 1.300 6.80EÀ04 940 0.015 3,800 1.275 7.00EÀ04 920 0.015 3,600 1.250 7.20EÀ04 870 0.015 3,400 1.225 7.40EÀ04 830 0.015 3,200 1.200 8.00EÀ04 780 0.015 3,000 1.175 8.20EÀ04 740 0.015 2,800 1.150 8.50EÀ04 700 0.015 2,600 1.125 9.00EÀ04 660 0.015 2,400 1.120 9.50EÀ04 620 0.015 2,200 1.115 1.00EÀ03 580 0.015 2,000 1.110 1.10EÀ03 540 0.015 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 95 3.1.2007 8:47pm Compositor Name: SJoearun FORECAST OF WELL PRODUCTION 7/95
- 104. Formation porosity: 0.15 Water saturation: 0.30 Gas-specific gravity: 0.7 air ¼ 1 Total compressibility: 1:6 Â 10À4 psiÀ1 Darcy skin factor: 0 Non-Darcy flow coefficient: 0 Drainage area: 160 acres Wellbore radius: 0.328 ft Tubing inner diameter: 1.995 in. Desired flowing wellhead pressure: 800 psia Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap07 Final Proof page 96 3.1.2007 8:47pm Compositor Name: SJoearun 7/96 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 105. 8 Production Decline Analysis Contents 8.1 Introduction 8/98 8.2 Exponential Decline 8/98 8.3 Harmonic Decline 8/100 8.4 Hyperbolic Decline 8/100 8.5 Model Identification 8/100 8.6 Determination of Model Parameters 8/101 8.7 Illustrative Examples 8/101 Summary 8/104 References 8/104 Problems 8/104 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 97 20.12.2006 10:36am
- 106. 8.1 Introduction Production decline analysis is a traditional means of identifying well production problems and predicting well performance and life based on real production data. It uses empirical decline models that have little fundamental justifications. These models include the following: . Exponential decline (constant fractional decline) . Harmonic decline . Hyperbolic decline Although the hyperbolic decline model is more general, the other two models are degenerations of the hyperbolic decline model. These three models are related through the following relative decline rate equation (Arps, 1945): 1 q dq dt ¼ Àbqd , (8:1) where b and d are empirical constants to be deter- mined based on production data. When d ¼ 0, Eq. (8.1) degenerates to an exponential decline model, and when d ¼ 1, Eq. (8.1) yields a harmonic decline model. When 0 d 1, Eq. (8.1) derives a hyperbolic decline model. The decline models are applicable to both oil and gas wells. 8.2 Exponential Decline The relative decline rate and production rate decline equa- tions for the exponential decline model can be derived from volumetric reservoir model. Cumulative production expression is obtained by integrating the production rate decline equation. 8.2.1 Relative Decline Rate Consider an oil well drilled in a volumetric oil reservoir. Suppose the well’s production rate starts to decline when a critical (lowest permissible) bottom-hole pressure is reached. Under the pseudo–steady-state flow condition, the production rate at a given decline time t can be expressed as q ¼ kh( ppt À pc wf ) 141:2B0m ln 0:472re rw þ s h i , (8:2) where ppt ¼ average reservoir pressure at decline time t, pc wf ¼ the critical bottom-hole pressure maintained during the production decline. The cumulative oil production of the well after the production decline time t can be expressed as Np ¼ ðt 0 kh( ppt À pc wf ) 141:2Bom ln 0:472re rw þ s h i dt: (8:3) The cumulative oil production after the production de- cline upon decline time t can also be evaluated based on the total reservoir compressibility: Np ¼ ctNi Bo ( pp0 À ppt), (8:4) where ct ¼ total reservoir compressibility, Ni ¼ initial oil in place in the well drainage area, pp0 ¼ average reservoir pressure at decline time zero. Substituting Eq. (8.3) into Eq. (8.4) yields ðt 0 kh( ppt À pc wf ) 141:2Bom ln 0:472re rw þ s h i dt ¼ ctNi Bo ( pp0 À ppt): (8:5) Taking derivative on both sides of this equation with respect to time t gives the differential equation for reser- voir pressure: kh( ppt À pc wf ) 141:2m ln 0:472re rw þ s h i ¼ ÀctNi dppt dt (8:6) Because the left-hand side of this equation is q and Eq. (8.2) gives dq dt ¼ kh 141:2B0m ln 0:472re rw þ s h i dppt dt , (8:7) Eq. (8.6) becomes q ¼ À141:2ctNim ln 0:472re rw þ s h i kh dq dt (8:8) or the relative decline rate equation of 1 q dq dt ¼ Àb, (8:9) where b ¼ kh 141:2mctNi ln 0:472re rw þ s h i : (8:10) 8.2.2 Production rate decline Equation (8.6) can be expressed as Àb( ppt À pc wf ) ¼ dppt dt : (8:11) By separation of variables, Eq. (8.11) can be integrated, À ðt 0 bdt ¼ ðppt pp0 dppt ( ppt À pc wf ) , (8:12) to yield an equation for reservoir pressure decline: ppt ¼ pc wf þ pp0 À pc wf eÀbt (8:13) Substituting Eq. (8.13) into Eq. (8.2) gives the well pro- duction rate decline equation: q ¼ kh( pp0 À pc wf ) 141:2Bom ln 0:472re rw þ s h i eÀbt (8:14) or q ¼ bctNi Bo ( pp0 À pc wf ) eÀbt , (8:15) which is the exponential decline model commonly used for production decline analysis of solution-gas-drive reser- voirs. In practice, the following form of Eq. (8.15) is used: q ¼ qieÀbt , (8:16) where qi is the production rate at t ¼ 0. It can be shown that q2 q1 ¼ q3 q2 ¼ . . . . . . ¼ qn qnÀ1 ¼ eÀb . That is, the fractional decline is constant for exponential decline. As an exercise, this is left to the reader to prove. 8.2.3 Cumulative production Integration of Eq. (8.16) over time gives an expression for the cumulative oil production since decline of Np ¼ ðt 0 qdt ¼ ðt 0 qieÀbt dt, (8:17) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 98 20.12.2006 10:36am 8/98 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 107. that is, Np ¼ qi b 1 À eÀbt À Á : (8:18) Since q ¼ qieÀbt , Eq. (8.18) becomes Np ¼ 1 b qi À qð Þ: (8:19) 8.2.4 Determination of decline rate The constant b is called the continuous decline rate. Its value can be determined from production history data. If production rate and time data are available, the b value can be obtained based on the slope of the straight line on a semi-log plot. In fact, taking logarithm of Eq. (8.16) gives ln (q) ¼ ln (qi) À bt, (8:20) which implies that the data should form a straight line with a slope of Àb on the log(q) versus t plot, if exponential decline is the right model. Picking up any two points, (t1, q1) and (t2, q2), on the straight line will allow analyt- ical determination of b value because ln (q1) ¼ ln (qi) À bt1 (8:21) and ln (q2) ¼ ln (qi) À bt2 (8:22) give b ¼ 1 (t2 À t1) ln q1 q2 : (8:23) If production rate and cumulative production data are available, the b value can be obtained based on the slope of the straight line on an Np versus q plot. In fact, rearranging Eq. (8.19) yields q ¼ qi À bNp: (8:24) Picking up any two points, (Np1, q1) and (Np2, q2), on the straight line will allow analytical determination of the b value because q1 ¼ qi À bNp1 (8:25) and q2 ¼ qi À bNp2 (8:26) give b ¼ q1 À q2 Np2 À Np1 : (8:27) Depending on the unit of time t, the b can have different units such as monthÀ1 and yearÀ1 . The following relation can be derived: ba ¼ 12bm ¼ 365bd , (8:28) where ba, bm, and bd are annual, monthly, and daily decline rates, respectively. 8.2.5 Effective decline rate Because the exponential function is not easy to use in hand calculations, traditionally the effective decline rate has been used. Since eÀx % 1 À x for small x-values based on Taylor’s expansion, eÀb % 1 À b holds true for small values of b. The b is substituted by b’, the effective decline rate, in field applications. Thus, Eq. (8.16) becomes q ¼ qi(1 À b0 )t : (8:29) Again, it can be shown that q2 q1 ¼ q3 q2 ¼ . . . . . . ¼ qn qnÀ1 ¼ 1 À b0 . Depending on the unit of time t, the b’ can have different units such as monthÀ1 and yearÀ1 . The following relation can be derived: (1 À b 0 a) ¼ (1 À b 0 m)12 ¼ (1 À b 0 d )365 , (8:30) where b 0 a, b 0 m, and b 0 d are annual, monthly, and daily effective decline rates, respectively. Example Problem 8.1 Given that a well has declined from 100 stb/day to 96 stb/day during a 1-month period, use the exponential decline model to perform the following tasks: 1. Predict the production rate after 11 more months 2. Calculate the amount of oil produced during the first year 3. Project the yearly production for the well for the next 5 years Solution 1. Production rate after 11 more months: bm ¼ 1 (t1m À t0m) ln q0m q1m ¼ 1 1 ln 100 96 ¼ 0:04082=month Rate at end of 1 year: q1m ¼ q0meÀbmt ¼ 100eÀ0:04082(12) ¼ 61:27 stb=day If the effective decline rate b’ is used, b 0 m ¼ q0m À q1m q0m ¼ 100 À 96 100 ¼ 0:04=month: From 1 À b 0 y ¼ (1 À b 0 m)12 ¼ (1 À 0:04)12 , one gets b 0 y ¼ 0:3875=yr Rate at end of 1 year: q1 ¼ q0(1 À b 0 y) ¼ 100(1 À 0:3875) ¼ 61:27 stb=day 2. The amount of oil produced during the first year: by ¼ 0:04082(12) ¼ 0:48986=year Np,1 ¼ q0 À q1 by ¼ 100 À 61:27 0:48986 365 ¼ 28,858 stb or bd ¼ ln 100 96 ! 1 30:42 ¼ 0:001342 1 day Np,1 ¼ 100 0:001342 (1 À eÀ0:001342(365) ) ¼ 28,858 stb 3. Yearly production for the next 5 years: Np,2 ¼ 61:27 0:001342 (1 À eÀ0:001342(365) ) ¼ 17; 681 stb q2 ¼ qieÀbt ¼ 100eÀ0:04082(12)(2) ¼ 37:54 stb=day Np,3 ¼ 37:54 0:001342 (1 À eÀ0:001342(365) ) ¼ 10,834 stb q3 ¼ qieÀbt ¼ 100eÀ0:04082(12)(3) ¼ 23:00 stb=day Np,4 ¼ 23:00 0:001342 (1 À eÀ0:001342(365) ) ¼ 6639 stb q4 ¼ qieÀbt ¼ 100eÀ0:04082(12)(4) ¼ 14:09 stb=day Np,5 ¼ 14:09 0:001342 (1 À eÀ0:001342(365) ) ¼ 4061 stb In summary, Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 99 20.12.2006 10:36am PRODUCTION DECLINE ANALYSIS 8/99
- 108. 8.3 Harmonic Decline When d ¼ 1, Eq. (8.1) yields differential equation for a harmonic decline model: 1 q dq dt ¼ Àbq, (8:31) which can be integrated as q ¼ q0 1 þ bt , (8:32) where q0 is the production rate at t ¼ 0. Expression for the cumulative production is obtained by integration: Np ¼ ðt 0 qdt, which gives Np ¼ q0 b ln (1 þ bt): (8:33) Combining Eqs. (8.32) and (8.33) gives Np ¼ q0 b ln (q0) À ln (q)½ Š: (8:34) 8.4 Hyperbolic Decline When 0 d 1, integration of Eq. (8.1) gives ðq q0 dq q1þd ¼ À ðt 0 bdt, (8:35) which results in q ¼ q0 (1 þ dbt)1=d (8:36) or q ¼ q0 1 þ b a t À Áa , (8:37) where a ¼ 1=d. Expression for the cumulative production is obtained by integration: Np ¼ ðt 0 qdt, which gives Np ¼ aq0 b(a À 1) 1 À 1 þ b a t 1Àa # : (8:38) Combining Eqs. (8.37) and (8.38) gives Np ¼ a b(a À 1) q0 À q 1 þ b a t ! : (8:39) 8.5 Model Identification Production data can be plotted in different ways to iden- tify a representative decline model. If the plot of log(q) versus t shows a straight line (Fig. 8.1), according to Eq. (8.20), the decline data follow an exponential decline model. If the plot of q versus Np shows a straight line (Fig. 8.2), according to Eq. (8.24), an exponential decline model should be adopted. If the plot of log(q) versus log(t) shows a straight line (Fig. 8.3), according to Eq. (8.32), the Year Rate at End of Year (stb/day) Yearly Production (stb) 0 100.00 — 1 61.27 28,858 2 37.54 17,681 3 23.00 10,834 4 14.09 6,639 5 8.64 4,061 68,073 q t Figure 8.1 A semilog plot of q versus t indicating an exponential decline. q Np Figure 8.2 A plot of Np versus q indicating an exponen- tial decline. q t Figure 8.3 A plot of log(q) versus log(t) indicating a harmonic decline. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 100 20.12.2006 10:36am 8/100 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 109. decline data follow a harmonic decline model. If the plot of Np versus log(q) shows a straight line (Fig. 8.4), according to Eq. (8.34), the harmonic decline model should be used. If no straight line is seen in these plots, the hyperbolic decline model may be verified by plotting the relative decline rate defined by Eq. (8.1). Figure 8.5 shows such a plot. This work can be easily performed with computer program UcomS.exe. 8.6 Determination of Model Parameters Once a decline model is identified, the model parameters a and b can be determined by fitting the data to the selected model. For the exponential decline model, the b value can be estimated on the basis of the slope of the straight line in the plot of log(q) versus t (Eq. [8.23]). The b value can also be determined based on the slope of the straight line in the plot of q versus Np (Eq. [8.27]). For the harmonic decline model, the b value can be estimated on the basis of the slope of the straight line in the plot of log(q) versus log(t) or Eq. (8.32): b ¼ q0 q1 À 1 t1 : (8:40) The b value can also be estimated based on the slope of the straight line in the plot of Np versus log(q) (Eq. [8.34]). For the hyperbolic decline model, determination of a and b values is somewhat tedious. The procedure is shown in Fig. 8.6. Computer program UcomS.exe can be used for both model identification and model parameter determination, as well as production rate prediction. 8.7 Illustrative Examples Example Problem 8.2 For the data given in Table 8.1, identify a suitable decline model, determine model parameters, and project production rate until a marginal rate of 25 stb/day is reached. Solution A plot of log(q) versus t is presented in Fig. 8.7, which shows a straight line. According to Eq. (8.20), the exponential decline model is applicable. This is further evidenced by the relative decline rate shown in Fig. 8.8. Select points on the trend line: q Np Figure 8.4 A plot of Np versus log(q) indicating a har- monic decline. q q∆t ∆q − Exponential decline Harmonic decline Hyperbolic decline Figure 8.5 A plot of relative decline rate versus produc- tion rate. 1. Select points (t1, q1) and (t2, q2) 2. Read t3 at q3 3. Calculate 4. Find q0 at t = 0 5. Pick up any point (t*, q*) 6. Use 7. Finally q t 1 2 q1q2= t 2 3− t1t2 t1 + t2 − 2t3 a b = a t * a b q0 q * = 1+ 1+log log t * a b q * q0 a = a a b b = q3 t3 (t*, q* ) Figure 8.6 Procedure for determining a- and b-values. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 101 20.12.2006 10:36am PRODUCTION DECLINE ANALYSIS 8/101
- 110. t1 ¼ 5 months, q1 ¼ 607 stb=day t2 ¼ 20 months, q2 ¼ 135 stb=day Decline rate is calculated with Eq. (8.23): b ¼ 1 (5 À 20) ln 135 607 ¼ 0:11=month Projected production rate profile is shown in Fig. 8.9. Example Problem 8.3 For the data given in Table 8.2, identify a suitable decline model, determine model parameters, and project production rate until the end of the fifth year. Solution A plot of relative decline rate is shown in Fig. 8.10, which clearly indicates a harmonic decline model. On the trend line, select q0 ¼ 10,000 stb=day at t ¼ 0 q1 ¼ 5,680 stb=day at t ¼ 2 years: Therefore, Eq. (8.40) gives Table 8.1 Production Data for Example Problem 8.2 t (mo) q (stb/day) t (mo) q (stb/day) 1.00 904.84 13.00 272.53 2.00 818.73 14.00 246.60 3.00 740.82 15.00 223.13 4.00 670.32 16.00 201.90 5.00 606.53 17.00 182.68 6.00 548.81 18.00 165.30 7.00 496.59 19.00 149.57 8.00 449.33 20.00 135.34 9.00 406.57 21.00 122.46 10.00 367.88 22.00 110.80 11.00 332.87 23.00 100.26 12.00 301.19 24.00 90.720 1 10 100 1,000 10,000 0 10 15 20 25 30 t (month) q(STB/D) 5 Figure 8.7 A plot of log(q) versus t showing an expo- nential decline. 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 3 203 403 603 803 1,003 q (stb/d) −∆q/∆t/q(month−1) Figure 8.8 Relative decline rate plot showing exponen- tial decline. 0 100 200 300 400 500 600 700 800 900 1,000 0 10 20 30 40 t (month) q(stb/d) Figure 8.9 Projected production rate by a exponential decline model. 0.15 0.20 0.25 0.30 0.35 0.10 0.40 4.00 5.00 6.00 7.00 8.00 9.003.00 10.00 q (1000 stb/d) −∆q/∆t/q(year−1) Figure 8.10 Relative decline rate plot showing har- monic decline. Table 8.2 Production Data for Example Problem 8.3 t (yr) q (1,000 stb/day) t (yr) q (1,000 stb/day) 0.20 9.29 2.10 5.56 0.30 8.98 2.20 5.45 0.40 8.68 2.30 5.34 0.50 8.40 2.40 5.23 0.60 8.14 2.50 5.13 0.70 7.90 2.60 5.03 0.80 7.67 2.70 4.94 0.90 7.45 2.80 4.84 1.00 7.25 2.90 4.76 1.10 7.05 3.00 4.67 1.20 6.87 3.10 4.59 1.30 6.69 3.20 4.51 1.40 6.53 3.30 4.44 1.50 6.37 3.40 4.36 1.60 6.22 3.50 4.29 1.70 6.08 3.60 4.22 1.80 5.94 3.70 4.16 1.90 5.81 3.80 4.09 2.00 5.68 3.90 4.03 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 102 20.12.2006 10:36am 8/102 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 111. b ¼ 10;000 5;680 À 1 2 ¼ 0:38 1=yr: Projected production rate profile is shown in Fig. 8.11. Example Problem 8.4 For the data given in Table 8.3, identify a suitable decline model, determine model parameters, and project production rate until the end of the fifth year. Solution A plot of relative decline rate is shown in Fig. 8.12, which clearly indicates a hyperbolic decline model. Select points t1 ¼ 0:2 year, q1 ¼ 9,280 stb=day t2 ¼ 3:8 years, q2 ¼ 3,490 stb=day q3 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (9,280)(3,490) p ¼ 5,670 stb=day b a ¼ 0:2 þ 3:8 À 2(1:75) (1:75)2 À (0:2)(3:8) ¼ 0:217 Read from decline curve (Fig. 8.13) t3 ¼ 1:75 years at q3 ¼ 5,670 stb=day. Read from decline curve (Fig. 8.13) q0 ¼ 10,000 stb=day at t0 ¼ 0. Pick up point (tà ¼ 1:4 years, qà ¼ 6,280 stb=day). a ¼ log 10,000 6,280 log (1 þ (0:217)(1:4) ) ¼ 1:75 b ¼ (0:217)(1:758) ¼ 0:38 Projected production rate profile is shown in Fig. 8.14. Table 8.3 Production Data for Example Problem 8.4 t (yr) q (1,000 stb/day) t (yr) q (1,000 stb/day) 0.10 9.63 2.10 5.18 0.20 9.28 2.20 5.05 0.30 8.95 2.30 4.92 0.40 8.64 2.40 4.80 0.50 8.35 2.50 4.68 0.60 8.07 2.60 4.57 0.70 7.81 2.70 4.46 0.80 7.55 2.80 4.35 0.90 7.32 2.90 4.25 1.00 7.09 3.00 4.15 1.10 6.87 3.10 4.06 1.20 6.67 3.20 3.97 1.30 6.47 3.30 3.88 1.40 6.28 3.40 3.80 1.50 6.10 3.50 3.71 1.60 5.93 3.60 3.64 1.70 5.77 3.70 3.56 1.80 5.61 3.80 3.49 1.90 5.46 3.90 3.41 2.00 5.32 4.00 3.34 0 2 4 6 8 10 12 0.0 1.0 2.0 3.0 4.0 5.0 6.0 t (year) q(1,000Stb/d) Figure 8.11 Projected production rate by a harmonic decline model. 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.20 0.38 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 q (1,000 stb/d) −∆q/∆t/q(year−1) Figure 8.12 Relative decline rate plot showing hyper- bolic decline. 2 4 6 8 10 0 12 1.0 2.0 3.0 4.00.0 5.0 t (year) q(1,000stb/d) Figure 8.13 Relative decline rate shot showing hyper- bolic decline. 2 4 6 8 10 0 12 1.0 2.0 0.3 4.0 5.00.0 6.0 t (year) q(1,000stb/d) Figure 8.14 Projected production rate by a hyperbolic decline model. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 103 20.12.2006 10:36am PRODUCTION DECLINE ANALYSIS 8/103
- 112. Summary This chapter presents empirical models and procedure for using the models to perform production decline data ana- lyses. Computer program UcomS.exe can be used for model identification, model parameter determination, and production rate prediction. References arps, j.j. Analysis of decline curves. Trans. AIME 1945;160:228–247. golan, m. and whitson, c.m. Well Performance, pp. 122– 125. Upper Saddle River, NJ: International Human Resource Development Corp., 1986. economides, m.j., hill, a.d., and ehlig-economides, c. Petroleum Production Systems, pp. 516–519. Upper Saddle River, NJ: Prentice Hall PTR, 1994. Problems 8.1 For the data given in the following table, identify a suitable decline model, determine model parameters, and project production rate until the end of the tenth year. Predict yearly oil productions: 8.2 For the data given in the following table, identify a suit- able decline model, determine model parameters, predict the time when the production rate will decline to a mar- ginal value of 500 stb/day, and the reverses to be recov- ered before the marginal production rate is reached: 8.3 For the data given in the following table, identify a suit- able decline model, determine model parameters, predict the time when the production rate will decline to a marginal value of 50 Mscf/day, and the reverses to be recovered before the marginal production rate is reached: Time (yr) Production Rate (1,000 stb/day) 0.1 9.63 0.2 9.29 0.3 8.98 0.4 8.68 0.5 8.4 0.6 8.14 0.7 7.9 0.8 7.67 0.9 7.45 1 7.25 1.1 7.05 1.2 6.87 1.3 6.69 1.4 6.53 1.5 6.37 1.6 6.22 1.7 6.08 1.8 5.94 1.9 5.81 2 5.68 2.1 5.56 2.2 5.45 2.3 5.34 2.4 5.23 2.5 5.13 2.6 5.03 2.7 4.94 2.8 4.84 2.9 4.76 3 4.67 3.1 4.59 3.2 4.51 3.3 4.44 3.4 4.36 Time (yr) Production Rate (stb/day) 0.1 9.63 0.2 9.28 0.3 8.95 0.4 8.64 0.5 8.35 0.6 8.07 0.7 7.81 0.8 7.55 0.9 7.32 1 7.09 1.1 6.87 1.2 6.67 1.3 6.47 1.4 6.28 1.5 6.1 1.6 5.93 1.7 5.77 1.8 5.61 1.9 5.46 2 5.32 2.1 5.18 2.2 5.05 2.3 4.92 2.4 4.8 2.5 4.68 2.6 4.57 2.7 4.46 2.8 4.35 2.9 4.25 3 4.15 3.1 4.06 3.2 3.97 3.3 3.88 3.4 3.8 Time (mo) Production Rate (Mscf/day) 1 904.84 2 818.73 3 740.82 4 670.32 5 606.53 6 548.81 7 496.59 8 449.33 9 406.57 10 367.88 11 332.87 12 301.19 13 272.53 14 246.6 15 223.13 16 201.9 17 182.68 18 165.3 19 149.57 20 135.34 21 122.46 22 110.8 23 100.26 24 90.72 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 104 20.12.2006 10:36am 8/104 PETROLEUM PRODUCTION ENGINEERING FUNDAMENTALS
- 113. 8.4 For the data given in the following table, identify a suitable decline model, determine model parameters, predict the time when the production rate will decline to a marginal value of 50 stb/day, and yearly oil pro- ductions: Time (mo) Production Rate (stb/day) 1 1,810 2 1,637 3 1,482 4 1,341 5 1,213 6 1,098 7 993 8 899 9 813 10 736 11 666 12 602 13 545 14 493 15 446 16 404 17 365 18 331 19 299 20 271 21 245 22 222 23 201 24 181 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap08 Final Proof page 105 20.12.2006 10:36am PRODUCTION DECLINE ANALYSIS 8/105
- 114. Part II Equipment Design and Selection The role of a petroleum production engineer is to maximize oil and gas production in a cost- effective manner. Design and selection of the right equipment for production systems is essential for a production engineer to achieve his or her job objective. To perform their design work correctly, production engineers should have thorough knowledge of the principles and rules used in the industry for equipment design and selection. This part of the book provides graduating production engineers with principles and rules used in the petroleum production engineering practice. Materials are presented in the following three chapters: Chapter 9 Well Tubing 9/109 Chapter 10 Separation Systems 10/117 Chapter 11 Transportation Systems 11/133 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 107 21.12.2006 2:16pm
- 115. 9 Well Tubing Contents 9.1 Introduction 9/110 9.2 Strength of Tubing 9/110 9.3 Tubing Design 9/111 Summary 9/114 References 9/114 Problems 9/114 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 109 21.12.2006 2:16pm
- 116. 9.1 Introduction Most oil wells produce reservoir fluids through tubing strings. This is mainly because tubing strings provide good sealing performance and allow the use of gas expan- sion to lift oil. Gas wells produce gas through tubing strings to reduce liquid loading problems. Tubing strings are designed considering tension, col- lapse, and burst loads under various well operating condi- tions to prevent loss of tubing string integrity including mechanical failure and deformation due to excessive stresses and buckling. This chapter presents properties of the American Petroleum Institute (API) tubing and special considerations in designing tubing strings. 9.2 Strength of Tubing The API defines ‘‘tubing size’’ using nominal diameter and weight (per foot). The nominal diameter is based on the internal diameter of tubing body. The weight of tubing determines the tubing outer diameter. Steel grades of tub- ing are designated to H-40, J-55, C-75, L-80, N-80, C-90, and P-105, where the digits represent the minimum yield strength in 1,000 psi. Table 9.1 gives the tensile requirements of API tubing. The minimum performance properties of API tubing are listed in Appendix B of this book. The tubing collapse strength data listed in Appendix B do not reflect the effect of biaxial stress. The effect of tension of the collapse resistance is analyzed as follows. Consider a simple uniaxial test of a metal specimen as shown in Fig. 9.1, Hooke’s Law applies to the elastic portion before yield point: s ¼ E«, (9:1) where s, «, and E are stress, strain, and Young’s modulus, respectively. The energy in the elastic portion of the test is Uu ¼ 1 2 s« ¼ 1 2 P A Dl L ¼ 1 2 (P Á Dl) V Uu ¼ 1 2 W V , (9:2) where P, A, L, V, and Dl are force, area, length, volume, and length change, respectively. However, using Hooke’s Law, we have Uu ¼ 1 2 s« ¼ 1 2 s s E ¼ 1 2 s2 E : (9:3) To assess whether a material is going to fail, we use various material failure criteria. One of the most import- ant is the Distortion Energy Criteria. This is for 3D and is U ¼ 1 2 1þv 3E ½ s1 Às2ð Þ2 þ s2 Às3ð Þ2 þ s3 Às1ð Þ2 Š, (9:4) where n ¼ Poison’s ratio s1 ¼ axial principal stress, psi s2 ¼ tangential principal stress, psi s3 ¼ radial principal stress, psi. For our case of the uniaxial test, we would have s1 ¼ s s2 ¼ 0 s3 ¼ 0 : (9:5) Then from Eq. (9.4), we would get U ¼ 1 2 1 þ v 3E s2 þ s2 Â Ã U ¼ 1 þ v 3E s2 : (9:6) If the failure of a material is taken to be when the material is at the yield point, then Eq. (9.6) is written Uf ¼ 1 þ v 3E s 2 y, (9:7) where sy is yield stress. The definition of an ‘‘equivalent stress’’ is the energy level in 3D, which is equivalent to the criteria energy level. Thus, 1 þ v 3E s2 e ¼ 1 þ v 3E s2 y and se ¼ sy, (9:8) where se is the equivalent stress. The collapse pressure is expressed as pc ¼ 2sy D t À Á À 1 D t À Á2 # , (9:9) where D is the tubing outer diameter (OD) and t is wall thickness. For the 3D case, we can consider U ¼ 1 þ v 3E s2 e , (9:10) where se is the equivalent stress for the 3D case of 1 þ v 3E s2 e ¼ 1 2 1 þ v 3E Â (s1 À s2)2 þ (s2 À s3)2 þ (s3 À s1)2 Ã ; (9:11) thus, s2 e ¼ 1 2 s1 À s2ð Þ2 þ s2 À s3ð Þ2 þ ðs3 À s1Þ2 n o : (9:12) Table 9.1 API Tubing Tensile Requirements Tubing grade Yield strength (psi) Minimum tensile strength (psi)Minimum Maximum H-40 40,000 80,000 60,000 J-55 55,000 80,000 75,000 C-75 75,000 90,000 95,000 L-80 80,000 95,000 95,000 N-80 80,000 110,000 100,000 C-90 90,000 105,000 100,000 P-105 105,000 135,000 120,000 Strain (e) Stress(s) Figure 9.1 A simple uniaxial test of a metal specimen. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 110 21.12.2006 2:16pm 9/110 EQUIPMENT DESIGN AND SELECTION
- 117. Consider the case in which we have only tensile axial loads, and compressive pressure on the outside of the tubing, then Eq. (9.12) reduces to s2 e ¼ 1 2 s1 À s2ð Þ2 þ s2ð Þ2 þ Às1ð Þ2 n o (9:13) or s2 e ¼ s2 1 À s1s2 þ s2 2: (9:14) Further, we can define s1 ¼ W A s2 Ym ¼ À pcc pc ; (9:15) where Ym ¼ minimum yield stress pcc ¼ the collapse pressure corrected for axial load pc ¼ the collapse pressure with no axial load. se ¼ ÀYm Thus, Eq. (9.14) becomes Y2 m ¼ W A 2 þ W A pcc pc Á Ym þ pcc pc 2 Y2 m (9:16) pcc pc 2 þ W AYm Á pcc pc þ W AYm 2 À1 ¼ 0: (9:17) We can solve Eq. (9.17) for the term pcc pc . This yields pcc pc ¼ À W AYm Æ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi W AYm 2 À 4 W AYm 2 þ 4 r 2 (9:18) pcc ¼ pc ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À 0:75 SA Ym 2 s À 0:5 SA Ym 8 : 9 = ; , (9:19) where SA ¼ W A is axial stress at any point in the tubing string. In Eq. (9.19), it can be seen that as W (or SA) increases, the corrected collapse pressure resistance decreases (from the nonaxial load case). In general, there are four cases, as shown in Fig. 9.2: Case 1: Axial tension stress (s1 0) and collapse pressure (s2 0) Case 2: Axial tension stress (s1 0) and burst pressure (s2 0) Case 3: Axial compression stress (s1 0) and collapse pressure (s2 0) Case 4: Axial compression stress (s1 0) and burst pres- sure (s2 0) Example Problem 9.1 Calculate the collapse resistance for a section of 27 ⁄8 in. API 6.40 lb/ft, Grade J-55, non- upset tubing near the surface of a 10,000-ft string suspended from the surface in a well that is producing gas. Solution Appendix B shows an inner diameter of tubing of 2.441 in., therefore, t ¼ (2:875 À 2:441)=2 ¼ 0:217 in: D t ¼ 2:875 0:217 ¼ 13:25 pc ¼ 2(55; 000) 13:25 À 1 (13:25)2 ! ¼ 7,675:3 psi, which is consistent with the rounded value of 7,680 psi listed in Appendix B. A ¼ pt(D À t) ¼ p(0:217)(2:875 À 0:217) ¼ 1:812 in:2 SA ¼ 6:40(10,000) 1:812 ¼ 35,320 psi: Using Eq. (9.19), we get pcc ¼ 7675:3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À 0:75 35,320 55,000 2 s À 0:5 35; 320 55,000 8 : 9 = ; ¼ 3,914:5 psi: 9.3 Tubing Design Tubing design should consider tubing failure due to tension, collapse, and burst loads under various well operating conditions. Forces affecting tubing strings in- clude the following: 1. Axial tension due to weight of tubing and compression due to buoyancy 2. External pressure (completion fluids, oil, gas, forma- tion water) 3. Internal pressure (oil, gas, formation water) 4. Bending forces in deviated portion of well 5. Forces due to lateral rock pressure 6. Other forces due to thermal gradient or dynamics 9.3.1 Tension, Collapse, and Burst Design The last three columns of the tables in Appendix B present tubing collapse resistance, internal yield pressure, and joint yield strength. These are the limiting strengths for a given tubing joint without considering the biaxial effect shown in Fig. 9.2. At any point should the net external pressure, net internal pressure, and buoyant tensile load not be allowed to exceed tubing’s axial load-corrected collapse resistance, internal yield pressure, and joint yield strength, respectively. Tubing strings should be designed to have strengths higher than the maximum expected loads with safety factors greater than unity. In addition, bending stress should be considered in tension design for deviated and horizontal wells. The tensile stress due to bending is expressed as Case 4 Case 2 Case 3 Case 1 s2 s1 Figure 9.2 Effect of tension stress on tangential stress. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 111 21.12.2006 2:16pm WELL TUBING 9/111
- 118. b ¼ EDo 2Rc , (9:20) where sb ¼ bending stress, psi E ¼ Young’s modulus, psi Rc ¼ radius of hole curvature, in. Do ¼ OD of tubing, in. Because of the great variations in well operating condi- tions, it is difficult to adopt a universal tubing design criterion for all well situations. Probably the best design practice is to consider the worst loading cases for collapse, burst, and tension loads that are possible for the well to experience during the life of the well. It is vitally important to check the remaining strengths of tubing in a subject well before any unexpected well treatment is carried out. Some special considerations in well operations that affect tubing string integrity are addressed in the sections that follow. 9.3.2 Buckling Prevention during Production A completion fluid is in place in the annular space between the tubing and the casing before a well is put into production. The temperature at depth is T ¼ Tsf þ GT D, where GT is geothermal gradient. When the oil is produced, the temperature in the tubing will rise. This will expand (thermal) the tubing length, and if there is not sufficient landing tension, the tubing will buckle. The temperature distribution in the tubing can be predicted on the basis of the work of Ramey (1962), Hasan and Kabir (2002), and Guo et al. (2005). The latter is described in Chapter 11. A conservative approach to temperature calculations is to assume the maximum possible tempera- ture in the tubing string with no heat loss to formation through annulus. Example Problem 9.2 Consider a 27 ⁄8 in. API, 6.40 lb/ft Grade P-105 non-upset tubing anchored with a packer set at 10,000 ft. The crude oil production through the tubing from the bottom of the hole is 1,000 stb/day (no gas or water production). A completion fluid is in place in the annular space between the tubing and the casing (9.8 lb/ gal KCl water). Assuming surface temperature is 60 8F and geothermal gradient of 0.01 8F/ft, determine the landing tension to avoid buckling. Solution The temperature of the fluid at the bottom of the hole is estimated to be T10,000 ¼ 60 þ 0:01(10,000) ¼ 160 F: The average temperature of the tubing before oil produc- tion is Tav1 ¼ 60 þ 160 2 ¼ 110 F: The maximum possible average temperature of the tubing after oil production has started is Tav2 ¼ 160 þ 160 2 ¼ 160 F: This means that the approximate thermal expansion of the tubing in length will be DLT % b DTavg À Á L, where b is the coefficient of thermal expansion (for steel, this is bs ¼ 0:0000065 per 8F). Thus, DLT % 0:0000065[160 À 110]10,000 ¼ 3:25 ft: To counter the above thermal expansion, a landing tension must be placed on the tubing string that is equivalent to the above. Assumingthe tubingisasimpleuniaxialelement,then A % pt(D À t) ¼ p(0:217)(2:875 À 0:217) ¼ 1:812 in:2 s ¼ E« F A ¼ E Á DL L F ¼ AE DL L ¼ (1:812)(30 Â 106 )(3:25) 10,000 ¼ 17,667 lbf : Thus,anadditionaltensionof17,667 lbf atthesurfacemustbe placed on the tubing string to counter the thermal expansion. It can be shown that turbulent flow will transfer heat efficiently to the steel wall and then to the completion fluid and then to the casing and out to the formation. While laminar flow will not transfer heat very efficiently to the steel then out to the formation. Thus, the laminar flow situations are the most likely to have higher temperature oil at the exit. Therefore, it is most likely the tubing will be hotter via simple conduction. This effect has been consid- ered in the work of Hasan and Kabir (2002). Obviously, in the case of laminar flow, landing tension beyond the buoy- ancy weight of the tubing may not be required, but in the case of turbulent flow, the landing tension beyond the buoyancy weight of the tubing is usually required to prevent buckling of tubing string. In general, it is good practice to calculate the buoyant force of the tubing and add approxi- mately 4,000---5,000 lbf of additional tension when landing. 9.3.3 Considerations for Well Treatment and Stimulation Tubing strings are designed to withstand the harsh conditions during wellbore treatment and stimulation operations such as hole cleaning, cement squeezing, gravel packing, frac-packing, acidizing, and hydraulic fracturing. Precautionary measures to take depend on tubing–packer relation. If the tubing string is set through a non-restrain- ing packer, the tubing is free to move. Then string buckling and tubing–packer integrity will be major concerns. If the tubing string is set on a restraining packer, the string is not free to move and it will apply force to the packer. The factors to be considered in tubing design include the following: . Tubing size, weight, and grade . Well conditions - Pressure effect - Temperature effect . Completion method - Cased hole - Open hole - Multitubing - Packer type (restraining, non-restraining) 9.3.3.1 Temperature Effect As discussed in Example Problem 9.2, if the tubing string is free to move, its thermal expansion is expressed as DLT ¼ bLDTavg: (9:21) If the tubing string is not free to move, its thermal expan- sion will generate force. Since Hook’s Law gives DLT ¼ LDF AE , (9:22) substitution of Eq. (9.22) into Eq. (9.21) yields DF ¼ AEbDTavg % 207ADTavg (9:23) for steel tubing. 9.3.3.2 Pressure Effect Pressures affect tubing string in different ways inclu- ding piston effect, ballooning effect, and buckling effect. 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- 119. Consider the tubing–pack relation shown in Fig. 9.3. The total upward force acting on the tubing string from internal and external pressures is expressed as Fup ¼ pi(Ap À A 0 i) þ po(Ao À A 0 o), (9:24) where pi ¼ pressure in the tubing, psi po ¼ pressure in the annulus, psi Ap ¼ inner area of packer, in:2 A 0 i ¼ inner area of tubing sleeve, in:2 A 0 o ¼ outer area of tubing sleeve, in:2 The total downward force acting on the tubing string is expressed as Fdown ¼ pi(Ai À A 0 i) þ po(Ap À A 0 o), (9:25) where Ai is the inner area of tubing. The net upward force is then F ¼ Fup À Fdown ¼ pi(Ap À Ai) À po(Ap À Ao): (9:26) During a well treatment operation, the change (increase) in the net upward force is expressed as DF ¼ [Dpi(Ap À Ai) À Dpo(Ap À Ao)]: (9:27) If the tubing string is anchored to a restraining packer, this force will be transmitted to the packer, which may cause packer failure. If the tubing string is free to move, this force will cause the tubing string to shorten by DLP ¼ LDF AE , (9:28) which represents tubing string shrinkage due to piston effect. As shown in Fig. 9.4a, the ballooning effect is due to the internal pressure being higher than the external pressure during a well treatment. The change in tensile force can be expressed as DFB ¼ 0:6[Dpi avgAi À Dpo avgAo]: (9:29) If the tubing string is set through a restraining packer, this force will be transmitted to the packer, which may cause packer failure. If the tubing string is free to move, this force will cause the tubing string to shorten by DLB ¼ 2L 108 Dpi avg À R2 D po avg R2 À 1 ! , (9:30) where Dpi avg ¼ the average pressure change in the tubing, psi Dpo avg ¼ the average pressure change in the annulus, psi R2 ¼ Ao=Ai: As illustrated in Fig. 9.4b, the buckling effect is caused by the internal pressure being higher than the external pres- sure during a well treatment. The tubing string buckles when FBK ¼ Ap(pi À po) 0. If the tubing end is set through a restraining packer, this force will be transmitted to the packer, which may cause packer failure. If the tubing string is not restrained at bottom, this force will cause the tubing string to shorten by DLBK ¼ r2 F2 BK 8EIW , (9:31) which holds true only if FBK is greater than 0, and r ¼ Dci À Di 2 I ¼ p 64 (D4 o À D4 i ) W ¼ Wair þ Wfi À Wfo, where Dci ¼ inner diameter of casing, in. Di ¼ inner diameter of tubing, in. Do ¼ outer diameter of tubing, in. Wair ¼ weight of tubing in air, lb/ft Wft ¼ weight of fluid inside tubing, lb/ft Wfo ¼ weight of fluid displaced by tubing, lb/ft. 9.3.3.3 Total Effect of Temperature and Pressure The combination of Eqs. (9.22), (9.28), (9.30), and (9.31) gives DL ¼ DLT þ DLP þ DLB þ DLBK , (9:32) which represents the tubing shortening with a non- restraining packer. If a restraining packer is used, the total tubing force acting on the packer is expressed as Ac Ap AoЈ AiЈ Ai Ao pi po Figure 9.3 Tubing–packer relation. (a) (b) Figure 9.4 Ballooning and buckling effects. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 113 21.12.2006 2:16pm WELL TUBING 9/113
- 120. DF ¼ AEDL L : (9:33) Example Problem 9.3 The following data are given for a cement squeezing job: Tubing: 27 ⁄8 in., 6.5 lb/ft (2.441-in. ID) Casing: 7 in., 32 lb/ft (6.094-in. ID) Packer: Bore size Dp ¼ 3.25 in., set at 10,000 ft Initial condition: Tubing and casing are full of 30 API oil (S.G. ¼ 0.88) Operation: Tubing is displaced with 15 ppg cement with an injection pressure 5,000 psi and casing pressure 1,000 psi. The average temperature drop is 20 8 F. 1. Calculate tubing movement if the tubing is not restrained by the packer, and discuss solutions to the possible operation problems. 2. Calculate the tubing force acting on a restraining packer. Solution Temperature Effect: DlT ¼ bLDTavg ¼ (6:9 Â 10À6 )(10,000)(20) ¼ 1:38 ft Piston Effect: DPi ¼ (0:052)(10,000)[15 À (0:88)(8:33)] þ 5,000 ¼ 8,988 psi Dpo ¼ 0 psi Ap ¼ 3:14(3:25)2 =4 ¼ 8:30 in:2 Ai ¼ 3:14(2:441)2 =4 ¼ 4:68 in:2 Ao ¼ 3:14(2:875)2 =4 ¼ 6:49 in:2 DF ¼ [Dpi(Ap À Ai) À Dpo(Ap À Ao)] ¼ [(8,988)(8:30 À 4:68) À (1,000)(8:30 À 6:49)] ¼ 30:727 lbf DLP ¼ LDF AE ¼ (10,000)(30,727) (6:49 À 4:68)(30,000,000) ¼ 5:65 ft Ballooning Effect: DPi,avg ¼ (10,000=2)(0:052)[15 À (0:88)(8:33)] þ 5,000 ¼ 6,994 psi DPo,avg ¼ 1,000 psi R2 ¼ 6:49=4:68 ¼ 1:387 DLB ¼ 2L 108 Dpi avg À R2 Dpo avg R2 À 1 ! ¼ 2(10,000) 108 6,994 À 1:387(1,000) 1:387 À 1 ! ¼ 2:898 ft Since the tubing internal pressure is higher than the exter- nal pressure during the cement squeezing, tubing string buckling should occur. pi ¼ 5,000 þ (0:052)(15)(10,000) ¼ 12,800 psi po ¼ 1,000 þ (0:88)(0:433)(10,000) ¼ 4,810 psi r ¼ (6:094 À 2:875)=2 ¼ 1:6095 in: FBK ¼ Ap(pi À po) ¼ (8:30)(12,800 À 4,800) ¼ 66,317 lbf I ¼ p 64 (2:875)4 À (2:441)4 À Á ¼ 1:61 in:4 Wair ¼ 6:5 lbf =ft Wfi ¼ (15)(7:48)(4:68=144) ¼ 3:65 lbf =ft Wfo ¼ (0:88)(62:4)(6:49=144) ¼ 2:48 lbf =ft W ¼ 6:5 þ 3:65 À 2:48 ¼ 7:67 lbf =ft DLBK ¼ r2 F2 BK 8EIW ¼ (1:6095)2 (66,317)2 (8)(30,000,000)(1:61)(7:67) ¼ 3:884 ft 1. Tubing is not restrained by the packer. The tubing shortening is DL ¼ DLT þ DLP þ DLB þ DLBK ¼ 1:38 þ 5:65 þ 2:898 þ 3:844 ¼ 13:77 ft: Buckling point from bottom: LBK ¼ FBK W ¼ 66,317 7:67 ¼ 8,646 ft To keep the tubing in the packer, one of the following measures needs to be taken: a. Use a sleeve longer than 13.77 ft b. Use a restraining packer c. Put some weight on the packer (slack-hook) before treatment. Buckling due to slacking off needs to be checked. 2. Tubing is restrained by the packer. The force acting on the packer is DF ¼ AE DL L ¼ (6:49 À 4:68)(30,000,000)(13:77) (10,000) ¼ 74,783 lbf : Summary This chapter presents strength of API tubing that can be used for designing tubing strings for oil and gas wells. Tubing design should consider operating conditions in individual wells. Special care should be taken for tubing strings before a well undergoes a treatment or stimulation. References guo, b., song, s., chacko, j., and ghalambor, a. Offshore Pipelines. Burlington: Gulf Professional Publishing, 2005. hasan, r. and kabir, c.s. Fluid Flow and Heat Transfer in Wellbores, pp. 79–89. Richardson, TX: SPE, 2002. ramey, h.j., jr. Wellbore heat transmission. Trans. AIME April 1962;14:427. Problems 9.1 Calculate the collapse resistance for a section of 3-in. API 9.20 lb/ft, Grade J-55, non-upset tubing near the surface of a 12,000-ft string suspended from the sur- face in a well that is producing gas. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 114 21.12.2006 2:16pm 9/114 EQUIPMENT DESIGN AND SELECTION
- 121. 9.2 Consider a 27 ⁄8 -in. API, 6.40 lb/ft Grade J-55 non- upset tubing anchored with a packer set at 8,000 ft. The crude oil production through the tubing from the bottom of the hole is 1,500 stb/day (no gas or water production). A completion fluid is in place in the annular space between the tubing and the casing (9.6 lb/gal KCl water). Assuming surface temperature is 80 8F and geothermal gradient of 0.01 8F/ft, deter- mine the landing tension to avoid buckling. 9.3 The following data are given for a frac-packing job: Tubing: 27 ⁄8 in., 6.5 lb/ft (2.441 in. ID) Casing: 7 in., 32 lb/ft (6.094 in. ID) Packer: Bore size Dp ¼ 3:25 in., set at 8,000 ft Initial condition: Tubing and casing are full of 30 API oil (S.G. ¼ 0.88) Operation: Tubing is displaced with 12 ppg cement with an injection pressure 4,500 psi and casing pressure 1,200 psi. The average temperature drop is 30 8F. a. Calculate tubing movement if the tubing is not restrained by the packer, and discuss solutions to the possible operation problems. b. Calculate the tubing force acting on a restraining packer. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap09 Final Proof page 115 21.12.2006 2:16pm WELL TUBING 9/115
- 122. 10 Separation Systems Contents 10.1 Introduction 10/118 10.2 Separation System 10/118 10.3 Dehydration System 10/125 Summary 10/132 References 10/132 Problems 10/132 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 117 4.1.2007 8:26pm Compositor Name: SJoearun
- 123. 10.1 Introduction Oil and gas produced from wells are normally complex mixtures of hundreds of different compounds. A typical well stream is a turbulent mixture of oil, gas, water, and sometimes solid particles. The well stream should be pro- cessed as soon as possible after bringing it to the surface. Field separation processes fall into two categories: (1) separation of oil, water, and gas; and (2) dehydration that removes condensable water vapor and other undesirable compounds, such as hydrogen sulfide or carbon dioxide. This chapter focuses on the principles of separation and dehydration and selection of required separators and dehydrators. 10.2 Separation System Separation of well stream gas from free liquids is the first and most critical stage of field-processing operations. Composition of the fluid mixture and pressure determine what type and size of separator are required. Separators are also used in other locations such as upstream and downstream of compressors, dehydration units, and gas sweetening units. At these locations, separators are referred to as scrubbers, knockouts, and free liquid knock- outs. All these vessels are used for the same purpose: to separate free liquids from the gas stream. 10.2.1 Principles of Separation Separators work on the basis of gravity segregation and/or centrifugal segregation. A separator is normally con- structed in such a way that it has the following features: 1. It has a centrifugal inlet device where the primary sep- aration of the liquid and gas is made. 2. It provides a large settling section of sufficient height or length to allow liquid droplets to settle out of the gas stream with adequate surge room for slugs of liquid. 3. It is equipped with a mist extractor or eliminator near the gas outlet to coalesce small particles of liquid that do not settle out by gravity. 4. It allows adequate controls consisting of level control, liquid dump valve, gas backpressure valve, safety relief valve, pressure gauge, gauge glass, instrument gas regulator, and piping. The centrifugal inlet device makes the incoming stream spin around. Depending on the mixture flow rate, the reac- tion force from the separator wall can generate a centripetal acceleration of up to 500 times the gravitational acceler- ation. This action forces the liquid droplets together where they fall to the bottom of the separator into the settling section. The settling section in a separator allows the tur- bulence of the fluid stream to subside and the liquid drop- lets to fall to the bottom of the vessel due to gravity segregation. A large open space in the vessel is required for this purpose. Use of internal baffling or plates may produce more liquid to be discharged from the separator. However, the product may not be stable because of the light ends entrained in it. Sufficient surge room is essential in the settling section to handle slugs of liquid without carryover to the gas outlet. This can be achieved by placing the liquid level control in the separator, which in turn determines the liquid level. The amount of surge room required depends on the surge level of the production steam and the separator size used for a particular application. Small liquid droplets that do not settle out of the gas stream due to little gravity difference between them and the gas phase tend to be entrained and pass out of the separator with the gas. A mist eliminator or extractor near the gas outlet allows this to be almost eliminated. The small liquid droplets will hit the eliminator or extractor surfaces, coalesce, and collect to form larger droplets that will then drain back to the liquid section in the bottom of the sep- arator. A stainless steel woven-wire mesh mist eliminator can remove up to 99.9% of the entrained liquids from the gas stream. Cane mist eliminators can be used in areas where there is entrained solid material in the gas phase that may collect and plug a wire mesh mist eliminator. 10.2.2 Types of Separators Three types of separators are generally available from manufacturers: vertical, horizontal, and spherical separ- ators. Horizontal separators are further classified into two categories: single tube and double tube. Each type of separator has specific advantages and limitations. Selec- tion of separator type is based on several factors including characteristics of production steam to be treated, floor space availability at the facility site, transportation, and cost. 10.2.2.1 Vertical Separators Figure 10.1 shows a vertical separator. The inlet diverter baffle is a centrifugal inlet device making the incoming stream spin around. This action forces the liquid droplets to stay together and fall to the bottom of the separator along the separator wall due to gravity. Sufficient surge room is available in the settling section of the vertical separator to handle slugs of liquid without carryover to the gas outlet. A mist eliminator or extractor near the gas outlet allows the entrained liquid in the gas to be almost eliminated. Vertical separators are often used to treat low to inter- mediate gas–oil ratio well streams and streams with rela- tively large slugs of liquid. They handle greater slugs of liquid without carryover to the gas outlet, and the action of the liquid level control is not as critical. Vertical sep- arators occupy less floor space, which is important for facility sites such as those on offshore platforms where space is limited. Because of the large vertical distance between the liquid level and the gas outlet, the chance for liquid to revaporize into the gas phase is limited. However, because of the natural upward flow of gas in a vertical separator against the falling droplets of liquid, adequate separator diameter is required. Vertical separators are more costly to fabricate and ship in skid-mounted assemblies. 10.2.2.2 Horizontal Separators Figure 10.2 presents a sketch of a horizontal separator. In horizontal separators, gas flows horizontally while liquid droplets fall toward the liquid surface. The moisture gas flows in the baffle surface and forms a liquid film that is drained away to the liquid section of the separator. The baffles need to be longer than the distance of liquid trajec- tory travel. The liquid-level control placement is more critical in a horizontal separator than in a vertical sep- arator because of limited surge space. Horizontal separators are usually the first choice because of their low costs. They are almost widely used for high gas–oil ratio well streams, foaming well streams, or liquid-from-liquid separation. They have much greater gas–liquid interface because of a large, long, baffled gas- separation section. Horizontal separators are easier to skid-mount and service and require less piping for field connections. Individual separators can be stacked easily into stage-separation assemblies to minimize space re- quirements. Figure 10.3 demonstrates a horizontal double-tube sep- arator consisting of two tube sections. The upper tube section is filled with baffles, gas flows straight through and at higher velocities, and the incoming free liquid is immediately drained away from the upper tube Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 118 4.1.2007 8:26pm Compositor Name: SJoearun 10/118 EQUIPMENT DESIGN AND SELECTION
- 124. section into the lower tube section. Horizontal double- tube separators have all the advantages of normal horizontal single-tube separators, plus much higher liquid capacities. Figure 10.4 illustrates a horizontal oil–gas–water three- phase separator. This type of separator is commonly used for well testing and in instances where free water readily separates from the oil or condensate. Three-phase sep- aration can be accomplished in any type of separator. This can be achieved by installing either special internal baffling to construct a water leg or water siphon arrange- ment. It can also be achieved by using an interface liquid- level control. In three-phase operations, two liquid dump valves are required. Gas outlet Final centrifugal gas−liquid separation section Well/Stream inlet Liquid quieting baffle Drain connection Liquid outlet Liquid discharge valve Liquid−level control Gas equalizer pipe Inlet diverter baffle Figure 10.1 A typical vertical separator (courtesy Petroleum Extension Services). Figure 10.2 A typical horizontal separator (courtesy Petroleum Extension Services). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 119 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/119
- 125. 10.2.2.3 Spherical Separators A spherical separator is shown in Fig. 10.5. Spherical separators offer an inexpensive and compact means of separation arrangement. Because of their compact config- urations, this type of separator has a very limited surge space and liquid settling section. Also, the placement and action of the liquid-level control in this type of separator is very critical. 10.2.3 Factors Affecting Separation Separation efficiency is dominated by separator size. For a given separator, factors that affect separation of liquid and gas phases include separator operating pressure, separator operating temperature, and fluid stream composition. Changes in any of these factors will change the amount of gas and liquid leaving the separator. An increase in operating pressure or a decrease in operating temperature generally increases the liquid covered in a separator. How- ever, this is often not true for gas condensate systems in which an optimum pressure may exist that yields the max- imum volume of liquid phase. Computer simulation (flash vaporization calculation) of phase behavior of the well stream allows the designer to find the optimum pressure and temperature at which a separator should operate to give maximum liquid recovery (see Chapter 18). However, it is often not practical to operate at the optimum point. This is because storage system vapor losses may become too great under these optimum conditions. In field separation facilities, operators tend to determine the optimum conditions for them to maximize revenue. As the liquid hydrocarbon product is generally worth more than the gas, high liquid recovery is often desirable, provided that it can be handled in the available storage system. The operator can control operating pressure to some extent by use of backpressure valves. However, pipe- line requirements for Btu content of the gas should also be considered as a factor affecting separator operation. It is usually unfeasible to try to lower the operating temperature of a separator without adding expensive mechanical refrigeration equipment. However, an indirect heater can be used to heat the gas before pressure reduc- tion to pipeline pressure in a choke. This is mostly applied to high-pressure wells. By carefully operating this indirect heater, the operator can prevent overheating the gas stream ahead of the choke. This adversely affects the temperature of the downstream separator. 10.2.4 Selection of Separators Petroleum engineers normally do not perform detailed designing of separators but carry out selection of sepa- rators suitable for their operations from manufacturers’ product catalogs. This section addresses how to determine Figure 10.3 A typical horizontal double-tube separator (courtesy Petroleum Extension Services). Figure 10.4 A typical horizontal three-phase separator (courtesy Petroleum Extension Services). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 120 4.1.2007 8:26pm Compositor Name: SJoearun 10/120 EQUIPMENT DESIGN AND SELECTION
- 126. separator specifications based on well stream conditions. The specifications are used for separator selections. 10.2.4.1 Gas Capacity The following empirical equations proposed by Souders– Brown are widely used for calculating gas capacity of oil/gas separators: v ¼ K ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rL À rg rg s (10:1) and q ¼ Av, (10:2) where A ¼ total cross-sectional area of separator, ft2 v ¼ superficial gas velocity based on total cross-sectional area A, ft/sec q ¼ gas flow rate at operating conditions, ft3 =sec rL ¼ density of liquid at operating conditions, lbm=ft3 rg ¼ density of gas at operating conditions, lbm=ft3 K ¼ empirical factor Table 10.1 presents K values for various types of sep- arators. Also listed in the table are K values used for other designs such as mist eliminators and trayed towers in dehydration or gas sweetening units. Substituting Eq. (10.1) into Eq. (10.2) and applying real gas law gives qst ¼ 2:4D2 Kp z(T þ 460) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rL À rg rg s , (10:3) where qst ¼ gas capacity at standard conditions, MMscfd D ¼ internal diameter of vessel, ft p ¼ operation pressure, psia T ¼ operating temperature, 8F z ¼ gas compressibility factor It should be noted that Eq. (10.3) is empirical. Height differences in vertical separators and length differences in horizontal separators are not considered. Field experience has indicated that additional gas capacity can be obtained by increasing height of vertical separators and length of horizontal separators. The separator charts (Sivalls, 1977; Ikoku, 1984) give more realistic values for the gas capacity of separators. In addition, for single-tube horizontal ves- sels, corrections must be made for the amount of liquid in the bottom of the separator. Although one-half full of liquid is more or less standard for most single-tube hori- zontal separators, lowering the liquid level to increase the available gas space within the vessel can increase the gas capacity. Figure 10.5 A typical spherical low-pressure separator (Sivalls, 1977). Table 10.1 K Values Used for Selecting Separators Separator type K Remarks Vertical separators 0.06–0.35 Horizontal separators 0.40–0.50 Wire mesh mist eliminators 0.35 Bubble cap trayed columns 0.16 24-in. spacing Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 121 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/121
- 127. 10.2.4.2 Liquid Capacity Retention time of the liquid within the vessel determines liquid capacity of a separator. Adequate separation re- quires sufficient time to obtain an equilibrium condition between the liquid and gas phase at the temperature and pressure of separation. The liquid capacity of a separator relates to the retention time through the settling volume: qL ¼ 1,440VL t (10:4) where qL ¼ liquid capacity, bbl/day VL ¼ liquid settling volume, bbl t ¼ retention time, min Table10.2presentstvaluesforvarioustypesofseparators tested in fields. It is shown that temperature has a strong impact on three-phase separations at low pressures. Tables 10.3 through 10.8 present liquid-settling volumes with the conventional placement of liquid-level controls for typical oil/gas separators. Proper sizing of a separator requires the use of both Eq. (10.3) for gas capacity and Eq. (10.4) for liquid capacity. Experience shows that for high-pressure separators used for treating high gas/oil ratio well streams, the gas capacity is usually the controlling factor for separator selection. However, the reverse may be true for low-pressure sep- arators used on well streams with low gas/oil ratios. Example Problem 10.1 Calculate the minimum required size of a standard oil/gas separator for the following conditions. Consider both vertical and horizontal separators. Gas flow rate: 5.0 MMscfd Gas-specific gravity: 0.7 Condensate flow rate: 20 bbl/MMscf Condensate gravity: 608API Operating pressure: 800 psia Operating temperature: 808F Solution The total required liquid flow capacity is (5)(20) ¼ 100 bbl/day. Assuming a 20-in. Â 71 ⁄2 -ft vertical separator, Table 10.1 suggests an average K value of 0.205. The spreadsheet program Hall-Yarborogh-z.xls gives z ¼ 0.8427 and rg ¼ 3:38 lbm=ft3 at 800 psig and 808F. Liquid density is calculated as Table 10.2 Retention Time Required Under Various Separation Conditions Separation condition T (8F) t (min) Oil/gas separation 1 High-pressure oil/gas/water separation 2–5 Low-pressure oil/gas/water separation 100 5–10 90 10–15 80 15–20 70 20–25 60 25–30 Table 10.3 Settling Volumes of Standard Vertical High-Pressure Separators (230–2,000 psi working pressure) VL (bbl) Size (D Â H) Oil/Gas separators Oil/Gas/Water separators 16 00 Â 50 0.27 0.44 16 00 Â 71 ⁄2 0 0.41 0.72 16 00 Â 100 0.51 0.94 20 00 Â 50 0.44 0.71 20 00 Â 71 ⁄2 0 0.65 1.15 20 00 Â 100 0.82 1.48 24 00 Â 50 0.66 1.05 24 00 Â 71 ⁄2 0 0.97 1.68 24 00 Â 100 1.21 2.15 30 00 Â 50 1.13 1.76 30 00 Â 71 ⁄2 0 1.64 2.78 30 00 Â 100 2.02 3.54 36 00 Â 71 ⁄2 0 2.47 4.13 36 00 Â 100 3.02 5.24 36 00 Â 150 4.13 7.45 42 00 Â 71 ⁄2 0 3.53 5.80 42 00 Â 100 4.29 7.32 42 00 Â 150 5.80 10.36 48 00 Â 71 ⁄2 0 4.81 7.79 48 00 Â 100 5.80 9.78 48 00 Â 150 7.79 13.76 54 00 Â 71 ⁄2 6.33 10.12 54 00 Â 100 7.60 12.65 54 00 Â 150 10.12 17.70 60 00 Â 71 ⁄2 0 8.08 12.73 60 00 Â 100 9.63 15.83 60 00 Â 150 12.73 22.03 60 00 Â 200 15.31 27.20 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 122 4.1.2007 8:26pm Compositor Name: SJoearun 10/122 EQUIPMENT DESIGN AND SELECTION
- 128. Table 10.4 Settling Volumes of Standard Vertical Low-Pressure Separators (125 psi working pressure) VL (bbl) Size (D Â H) Oil/Gas separators Oil/Gas/Water separators 24 00 Â 50 0.65 1.10 24 00 Â 71 ⁄2 0 1.01 1.82 30 00 Â 100 2.06 3.75 36 00 Â 50 1.61 2.63 36 00 Â 71 ⁄2 0 2.43 4.26 36 00 Â 100 3.04 5.48 48 00 Â 100 5.67 10.06 48 00 Â 150 7.86 14.44 60 00 Â 100 9.23 16.08 60 00 Â 150 12.65 12.93 60 00 Â 200 15.51 18.64 Table 10.5 Settling Volumes of Standard Horizontal High-Pressure Separators (230–2,000 psi working pressure) VL(bbl) Size (D Â L) 1 ⁄2 Full 1 ⁄3 Full 1 ⁄4 Full 123 ⁄4 00 Â50 0.38 0.22 0.15 123 ⁄4 00 Â71 ⁄2 0 0.55 0.32 0.21 123 ⁄4 00 Â100 0.72 0.42 0.28 16 00 Â 50 0.61 0.35 0.24 16 00 Â 71 ⁄2 0 0.88 0.50 0.34 16 00 Â 100 1.14 0.66 0.44 20 00 Â 50 0.98 0.55 0.38 20 00 Â 71 ⁄2 0 1.39 0.79 0.54 20 00 Â 100 1.80 1.03 0.70 24 00 Â 50 1.45 0.83 0.55 24 00 Â 71 ⁄2 0 2.04 1.18 0.78 24 00 Â 100 2.63 1.52 1.01 24 00 Â 150 3.81 2.21 1.47 30 00 Â 50 2.43 1.39 0.91 30 00 Â 71 ⁄2 0 3.40 1.96 1.29 30 00 Â 100 4.37 2.52 1.67 30 00 Â 150 6.30 3.65 2.42 36 00 Â 71 ⁄2 4.99 2.87 1.90 36 00 Â 100 6.38 3.68 2.45 36 00 Â 150 9.17 5.30 3.54 36 00 Â 200 11.96 6.92 4.63 42 00 Â 71 ⁄2 0 6.93 3.98 2.61 42 00 Â 100 8.83 5.09 3.35 42 00 Â 150 12.62 7.30 4.83 42 00 Â 200 16.41 9.51 6.32 48 00 Â 71 ⁄2 0 9.28 5.32 3.51 48 00 Â 100 11.77 6.77 4.49 48 00 Â 150 16.74 9.67 6.43 48 00 Â 200 21.71 12.57 8.38 54 00 Â 71 ⁄2 0 12.02 6.87 4.49 54 00 Â 100 15.17 8.71 5.73 54 00 Â 150 12.49 12.40 8.20 54 00 Â 200 27.81 16.08 10.68 60 00 Â 71 ⁄2 0 15.05 8.60 5.66 60 00 Â 100 18.93 10.86 7.17 60 00 Â 150 26.68 15.38 10.21 60 00 Â 200 34.44 19.90 13.24 Table 10.6 Settling Volumes of Standard Horizontal Low-Pressure Separators (125 psi working pressure) VL (bbl) Size (D Â L) 1 ⁄2 Full 1 ⁄3 Full 1 ⁄4 Full 24 00 Â 50 1.55 0.89 0.59 24 00 Â 71 ⁄2 0 2.22 1.28 0.86 24 00 Â 100 2.89 1.67 1.12 (Continued ) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 123 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/123
- 129. rL ¼ 62:4 141:5 131:5 þ 60 ¼ 46:11 lbm=ft3 : Equation (10.3) gives qst ¼ (2:4)(20=12)2 (0:205)(800) (0:8427)(80 þ 460) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 46:11 À 3:38 3:38 r ¼ 8:70 MMscfd: Sivalls’s chart gives 5.4 MMscfd. From Table 10.3, a 20-in. Â 71 ⁄2 -ft separator will handle the following liquid capacity: qL ¼ 1440(0:65) 1:0 ¼ 936 bbl=day, which is much higher than the liquid load of 100 bbl/day. Consider a 16-in. Â 5-ft horizontal separator and Eq. (10.3) gives qst ¼ (2:4)(16=12)2 (0:45)(800) (0:8427)(80 þ 460) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 46:11 --- 3:38 3:38 r ¼ 12:22 MMscfd: If the separator is one-half full of liquid, it can still treat 6.11 MMscfd of gas. Sivalls’s chart indicates that a 16-in. Â 5-ft horizontal separator will handle 5.1 MMscfd. From Table 10.5, a half-full, 16-in. Â 5-ft horizontal separator will handle qL ¼ 1440(0:61) 1:0 ¼ 878 bbl=day, which again is much higher than the liquid load of 100 bbl/day. This example illustrates a case of high gas/oil ratio well streams where the gas capacity is the controlling factor for separator selection. It suggests that a smaller horizontal separator would be required and would be more econom- ical. The selected separator should have at least a 1,000 psig working pressure. 10.2.5 Stage separation Stage separation is a process in which hydrocarbon mix- tures are separated into vapor and liquid phases by mul- tiple equilibrium flashes at consecutively lower pressures. A two-stage separation requires one separator and a stor- age tank, and a three-stage separation requires two sep- arators and a storage tank. The storage tank is always counted as the final stage of vapor/liquid separation. Stage separation reduces the pressure a little at a time, in steps or stages, resulting in a more stable stock-tank liquid. Usually a stable stock-tank liquid can be obtained by a stage separation of not more than four stages. In high-pressure gas-condensate separation systems, a stepwise reduction of the pressure on the liquid condensate can significantly increase the recovery of stock-tank liquids. Prediction of the performance of the various sep- arators in a multistage separation system can be carried out with compositional computer models using the initial well stream composition and the operating temperatures and pressures of the various stages. Although three to four stages of separation theoretically increase the liquid recovery over a two-stage separation, the incremental liquid recovery rarely pays out the cost of the additional separators. It has been generally recognized that two stages of separation plus the stock tank are practically optimum. The increase in liquid recovery for two-stage separation over single-stage separation usually varies from 2 to 12%, although 20 to 25% increases in liquid recoveries have been reported. The first-stage separator operating pressure is generally determined by the flowline pressure and operating charac- teristics of the well. The pressure usually ranges from 600 to 1,200 psi. In situations in which the flowline pres- sure is greater than 600 psi, it is practical to let the first- stage separator ride the line or operate at the flowline pressure. Pressures at low-stage separations can be deter- mined based on equal pressure ratios between the stages (Campbell, 1976): Rp ¼ p1 ps 1 Nst , (10:5) Table 10.6 Settling Volumes of Standard Horizontal Low-Pressure Separators (125 psi working pressure)(Continued ) VL (bbl) Size (D Â L) 1 ⁄2 Full 1 ⁄3 Full 1 ⁄4 Full 30 00 Â 50 2.48 1.43 0.94 30 00 Â 71=20 3.54 2.04 1.36 30 00 Â 100 4.59 2.66 1.77 36 00 Â 100 6.71 3.88 2.59 36 00 Â 150 9.76 5.66 3.79 48 00 Â 100 12.24 7.07 4.71 48 00 Â 150 17.72 10.26 6.85 60 00 Â 100 19.50 11.24 7.47 60 00 Â 150 28.06 16.23 10.82 60 00 Â 200 36.63 21.21 14.16 Table 10.7 Settling Volumes of Standard Spherical High-Pressure Separators (230–3,000 psi working pressure) Size (OD) VL (bbl) 24’’ 0.15 30’’ 0.30 36’’ 0.54 42’’ 0.88 48’’ 1.33 60’’ 2.20 Table 10.8 Settling Volumes of Standard Spherical Low-Pressure Separators (125 psi) Size (OD) VL (bbl) 41’’ 0.77 46’’ 1.02 54’’ 1.60 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 124 4.1.2007 8:26pm Compositor Name: SJoearun 10/124 EQUIPMENT DESIGN AND SELECTION
- 130. where Rp ¼ pressure ratio Nst ¼ number of stages À1 p1 ¼ first-stage or high-pressure separator pressure, psia ps ¼ stock-tank pressure, psia Pressures at the intermediate stages can then be designed with the following formula: pi ¼ piÀ1 Rp , (10:6) where pi ¼ pressure at stage i, psia. 10.3 Dehydration System All natural gas downstream from the separators still con- tain water vapor to some degree. Water vapor is probably the most common undesirable impurity found in the untreated natural gas. The main reason for removing water vapor from natural gas is that water vapor becomes liquid water under low-temperature and/or high-pressure conditions. Specifically, water content can affect long- distance transmission of natural gas because of the follow- ing facts: 1. Liquid water and natural gas can form hydrates that may plug the pipeline and other equipment. 2. Natural gas containing CO2 and/or H2S is corrosive when liquid water is present. 3. Liquid water in a natural gas pipeline potentially causes slugging flow conditions resulting in lower flow effi- ciency of the pipeline. 4. Water content decreases the heating value of natural gas being transported. Dehydration systems are designed for further separating water vapor from natural gas before the gas is transported by pipeline. 10.3.1 Water Content of Natural Gas Streams Solubility of water in natural gas increases with tempera- ture and decreases with pressure. The presence of salt in the liquid water reduces the water content of the gas. Water content of untreated natural gases is normally in the magnitude of a few hundred pounds of water per million standard cubic foot of gas (lbm=MMscf); while gas pipelines normally require water content to be in the range of 6---8 lbm=MMscf and even lower for offshore pipelines. The water content of natural gas is indirectly indicated by the ‘‘dew point,’’ defined as the temperature at which the natural gas is saturated with water vapor at a given pressure. At the dew point, natural gas is in equilibrium with liquid water; any decrease in temperature or increase in pressure will cause the water vapor to begin condensing. The difference between the dew point temperature of a water-saturated gas stream and the same stream after it has been dehydrated is called ‘‘dew-point depression.’’ It is essential to accurately estimate the saturated water vapor content of natural gas in the design and operation of dehydration equipment. Several methods are available for this purpose including the correlations of McCarthy et al. (1950) and McKetta and Wehe (1958). Dalton’s law of partial pressures is valid for estimating water vapor con- tent of gas at near-atmospheric pressures. Readings from the chart by McKetta and Wehe (1958) were re-plotted in Fig. 10.6 by Guo and Ghalambor (2005). Example Problem 10.2 Estimate water content of a natural gas at a pressure of 3,000 psia and temperature of 150 8F. Solution The chart in Fig. 10.6 gives water contents of Cw140F ¼ 84 lbm=MMcf Cw160F ¼ 130 lbm=MMcf Linear interpolation yields: Cw150F ¼ 107 lbm=MMcf 10.3.2 Methods for Dehydration Dehydration techniques used in the petroleum industry fall into four categories in principle: (a) direct cooling, (b) compression followed by cooling, (c) absorption, and (d) adsorption. Dehydration in the first two methods does 1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 Pressure (psia) WaterContent(lbm/MMcf@60ЊFand14.7psia) 280 240 200 180 160 140 120 100 80 60 40 20 0 −20 −40 −60 Temperature (ЊF) 1 10 100 1,000 10,000 Figure 10.6 Water content of natural gases (Guo and Ghalambor, 2005). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 125 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/125
- 131. not result in sufficiently low water contents to permit injection into a pipeline. Further dehydration by absorp- tion or adsorption is often required. 10.3.2.1 Dehydration by Cooling The ability of natural gas to contain water vapor decreases as the temperature is lowered at constant pressure. During the cooling process, the excess water in the vapor state becomes liquid and is removed from the system. Natural gas containing less water vapor at low temperature is output from the cooling unit. The gas dehydrated by cooling is still at its water dew point unless the temperature is raised again or the pressure is decreased. Cooling for the purpose of gas dehydration is sometimes economical if the gas temperature is unusually high. It is often a good practice that cooling is used in conjunction with other dehydration processes. Gas compressors can be used partially as dehydrators. Because the saturation water content of gases decreases at higher pressure, some water is condensed and removed from gas at compressor stations by the compressor dis- charge coolers. Modern lean oil absorption gas plants use mechanical refrigeration to chill the inlet gas stream. Ethylene glycol is usually injected into the gas chilling section of the plant, which simultaneously dehydrates the gas and recovers liquid hydrocarbons, in a manner similar to the low-temperature separators. 10.3.2.2 Dehydration by Adsorption ‘‘Adsorption’’ is defined as the ability of a substance to hold gases or liquids on its surface. In adsorption dehy- dration, the water vapor from the gas is concentrated and held at the surface of the solid desiccant by forces caused by residual valiancy. Solid desiccants have very large sur- face areas per unit weight to take advantage of these surface forces. The most common solid adsorbents used today are silica, alumina, and certain silicates known as molecular sieves. Dehydration plants can remove practic- ally all water from natural gas using solid desiccants. Because of their great drying ability, solid desiccants are employed where higher efficiencies are required. Depicted in Fig. 10.7 is a typical solid desiccant dehy- dration plant. The incoming wet gas should be cleaned preferably by a filter separator to remove solid and liquid contaminants in the gas. The filtered gas flows downward during dehydration through one adsorber containing a desiccant bed. The down-flow arrangement reduces dis- turbance of the bed caused by the high gas velocity during the adsorption. While one adsorber is dehydrating, the other adsorber is being regenerated by a hot stream of inlet gas from the regeneration gas heater. A direct-fired heater, hot oil, steam, or an indirect heater can supply the necessary regeneration heat. The regeneration gas usually flows upward through the bed to ensure thorough regen- eration of the bottom of the bed, which is the last area contacted by the gas being dehydrated. The hot regener- ated bed is cooled by shutting off or bypassing the heater. The cooling gas then flows downward through the bed so that any water adsorbed from the cooling gas will be at the top of the bed and will not be desorbed into the gas during the dehydration step. The still-hot regeneration gas and the cooling gas flow through the regeneration gas cooler to condense the desorbed water. Power-operated valves acti- vated by a timing device switch the adsorbers between the dehydration, regeneration, and cooling steps. Under normal operating conditions, the usable life of a desiccant ranges from 1 to 4 years. Solid desiccants become less effective in normal use because of loss of effective surface area as they age. Abnormally fast degradation occurs through blockage of the small pores and capillary openings lubricating oils, amines, glycols, corrosion inhibi- tors, and other contaminants, which cannot be removed during the regeneration cycle. Hydrogen sulfide can also damage the desiccant and reduce its capacity. The advantages of solid-desiccant dehydration include the following: . Lower dew point, essentially dry gas (water content 1.0 lb/MMcf) can be produced . Higher contact temperatures can be tolerated with some adsorbents . Higher tolerance to sudden load changes, especially on startup . Quick start up after a shutdown . High adaptability for recovery of certain liquid hydro- carbons in addition to dehydration functions Operating problems with the solid-desiccant dehydration include the following: Figure 10.7 Flow diagram of a typical solid desiccant dehydration plant (Guenther, 1979). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 126 4.1.2007 8:26pm Compositor Name: SJoearun 10/126 EQUIPMENT DESIGN AND SELECTION
- 132. . Space adsorbents degenerate with use and require replacement Dehydrating tower must be regenerated and cooled for operation before another tower approaches exhaustion. The maximum allowable time on dehydration gradually shortens because desiccant loses capacity with use. Although this type of dehydrator has high adaptability to sudden load changes, sudden pressure surges should be avoided because they may upset the desiccant bed and channel the gas stream resulting in poor dehydration. If a plant is operated above its rated capacity, high-pressure loss may introduce some attrition to occur. Attrition causes fines, which may in turn cause excessive pressure loss and result in loss of capacity. Replacing the desiccant should be scheduled and com- pleted ahead of the operating season. To maintain con- tinuous operation, this may require discarding the desiccant before its normal operating life is reached. To cut operating costs, the inlet part of the tower can be recharged and the remainder of the desiccant retained because it may still possess some useful life. Additional service life of the desiccant may be obtained if the direction of gas flow is reversed at a time when the tower would normally be recharged. 10.3.2.3 Dehydration by Absorption Water vapor is removed from the gas by intimate contact with a hygroscopic liquid desiccant in absorption dehydra- tion. The contacting is usually achieved in packed or trayed towers. Glycols have been widely used as effective liquid desiccants. Dehydration by absorption with glycol is usually economically more attractive than dehydration by solid desiccant when both processes are capable of meeting the required dew point. Glycols used for dehydrating natural gas are ethylene glycol (EG), diethylene glycol (DEG), triethylene glycol (TEG), and tetraethylene glycol (T4EG). Normally a single type of pure glycol is used in a dehydrator, but sometimes a glycol blend is economically attractive. TEG has gained nearly universal acceptance as the most cost effective of the glycols because of its superior dew-point depression, operating cost, and operation reliability. TEG has been successfully used to dehydrate sweet and sour natural gases over wide ranges of operating conditions. Dew-point depression of 40–1408F can be achieved at a gas pressure ranging from 25 to 2,500 psig and gas tem- perature between 40 and 1608F. The dew-point depression obtained depends on the equilibrium dew-point tempera- ture for a given TEG concentration and contact tempera- ture. Increased glycol viscosity may cause problems at lower contact temperature. Thus, heating of the natural gas may be desirable. Very hot gas streams are often cooled before dehydration to prevent vaporization of TEG. The feeding-in gas must be cleaned to remove all liquid water and hydrocarbons, wax, sand, drilling muds, and other impurities. These substances can cause severe foam- ing, flooding, higher glycol losses, poor efficiency, and increased maintenance in the dehydration tower or ab- sorber. These impurities can be removed using an efficient scrubber, separator, or even a filter separator for very contaminated gases. Methanol, injected at the wellhead as hydrate inhibitor, can cause several problems for glycol dehydration plants. It increases the heat requirements of the glycol regeneration system. Slugs of liquid methanol can cause flooding in the absorber. Methanol vapor vented to the atmosphere with the water vapor from the regener- ation system is hazardous and should be recovered or vented at nonhazardous concentrations. 10.3.2.3.1 Glycol Dehydration Process Illustrated in Fig. 10.8 shows the process and flow through a typical glycol dehydrator. The dehydration process can be described as follows: 1. The feeding-in gas stream first enters the unit through an inlet gas scrubber to remove liquid accumulations. A two-phase inlet scrubber is normally required. 2. The wet gas is then introduced to the bottom of the glycol-gas contactor and allowed to flow upward through the trays, while glycol flows downward through the column. The gas contacts the glycol on Glycol−gas contactor Inlet scrubber Gas inlet Distillate outlet Glycol cooler Glycol filter Gas outlet Glycol pump Glycol strainer Heat exchanger surge tank Flash separator Fuel gas scrubber Vent gas Fuel gas Stripping gas Resoiler Stripping still Water vapor outlet Figure 10.8 Flow diagram of a typical glycol dehydrator (Sivalls, 1977). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 127 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/127
- 133. each tray and the glycol absorbs the water vapor from the gas steam. 3. The gas then flows down through a vertical glycol cooler, usually fabricated in the form of a concentric pipe heat exchanger, where the outlet dry gas aids in cooling the hot regenerated glycol before it enters the contactor. The dry gas then leaves the unit from the bottom of the glycol cooler. 4. The dry glycol enters the top of the glycol-gas contactor from the glycol cooler and is injected onto the top tray. The glycol flows across each tray and down through a downcomer pipe onto the next tray. The bottom tray downcomer is fitted with a seal pot to hold a liquid seal on the trays. 5. The wet glycol, which has now absorbed the water vapor from the gas stream, leaves the bottom of the glycol-gas contactor column, passes through a high-pressure glycol filter, which removes any foreign solid particles that may have been picked up from the gas stream, and enters the power side of the glycol pump. 6. In the glycol pump, the wet high-pressure glycol from the contactor column pumps the dry regenerated glycol into the column. The wet glycol stream flows from the glycol pump to the flash separator, which allows for the release of the entrained solution gas. 7. The gas separated in the flash separator leaves the top of the flash separator vessel and can be used to supple- ment the fuel gas required for the reboiler. Any excess vent gas is discharged through a backpressure valve. The flash separator is equipped with a liquid level control and diaphragm motor valve that discharges the wet glycol stream through a heat exchange coil in the surge tank to preheat the wet glycol stream. 8. The wet glycol stream leaves the heat exchange coil in the surge tank and enters the stripping still mounted on top of the reboiler at the feed point in the still. The stripping still is packed with a ceramic intalox saddle- type packing, and the glycol flows downward through the column and enters the reboiler. The wet glycol passing downward through the still is contacted by hot rising glycol and water vapors passing upward through the column. The water vapors released in the reboiler and stripped from the glycol in the stripping still pass upward through the still column through anatmospheric reflux condenser that provides a partial reflux for the column. The water vapor then leaves the top of the stripping still column and is released to the atmosphere. 9. The glycol flows through the reboiler in essentially a horizontal path from the stripping still column to the opposite end. In the reboiler, the glycol is heated to approximately 350–4008F to remove enough water vapor to re-concentrate it to 99.5% or higher. In field dehydration units, the reboiler is generally equipped with a direct-fired firebox, using a portion of the natural gas stream for fuel. 10. The re-concentrated glycol leaves the reboiler through an overflow pipe and passes into the shell side of the heat exchanger/surge tank. In the surge tank, the hot re-concentrated glycol is cooled by exchanging heat with the wet glycol stream passing through the coil. The surge tank also acts as a liquid accumulator for feed for the glycol pump. The re-concentrated glycol flows from the surge tank through a strainer and into the glycol pump. From the pump, it passes into the shell side of the glycol cooler mounted on the glycol- gas contactor. It then flows upward through the glycol cooler where it is further cooled and enters the column on the top tray. 10.3.2.3.2 Advantages and Limitations Glycol dehy- drators have several advantages including the following: . Low initial equipment cost . Low pressure drop across absorption towers . Continuous operation . Makeup requirements may be added readily . Recharging of towers presents no problems . Plant may be used satisfactorily in the presence of materials that would cause fouling of some solid adsorbents Glycol dehydrators also present several operating prob- lems including the following: . Suspended matter, such as dirt, scale, and iron oxide, may contaminate glycol solutions. 0 10 20 30 40 50 60 70 Operating Pressure (psia) GasCapacity(MMscfd) 16 20 24 30 36 42 48 54 60 OD, in. 0 200 400 600 800 1,000 1,200 Figure 10.9 Gas capacity of vertical inlet scrubbers based on 0.7-specific gravity at 100 8F (Guo and Ghalambor, 2005). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 128 4.1.2007 8:26pm Compositor Name: SJoearun 10/128 EQUIPMENT DESIGN AND SELECTION
- 134. . Overheating of solution may produce both low and high boiling decomposition products. . The resultant sludge may collect on heating surfaces, causing some loss in efficiency, or in severe cases, com- plete flow stoppage. . When both oxygen and hydrogen sulfide are present, corrosion may become a problem because of the forma- tion of acid material in glycol solution. . Liquids (e.g., water, light hydrocarbons, or lubrication oils) in inlet gas may require installation of an efficient separator ahead of the absorber. Highly mineralized water entering the system with inlet gas may, over long periods, crystallize and fill the reboiler with solid salts. . Foaming of solution may occur with a resultant carry over of liquid. The addition of a small quantity of anti- foam compound usually remedies this problem. . Some leakage around the packing glands of pumps may be permitted because excessive tightening of packing may result in the scouring of rods. This leakage is col- lected and periodically returned to the system. . Highly concentrated glycol solutions tend to become viscous at low temperatures and, therefore, are hard to pump. Glycol lines may solidify completely at low tem- peratures when the plant is not operating. In cold wea- ther, continuous circulation of part of the solution through the heater may be advisable. This practice can also prevent freezing in water coolers. . To start a plant, all absorber trays must be filled with glycol before good contact of gas and liquid can be expected. This may also become a problem at low cir- culation rates because weep holes on trays may drain solution as rapidly as it is introduced. . Sudden surges should be avoided in starting and shut- ting down a plant. Otherwise, large carryover losses of solution may occur. 10.3.2.3.3 Sizing Glycol Dehydrator Unit Dehydrators with TEG in trays or packed-column contactors can be sized from standard models by using the following information: . Gas flow rate . Specific gravity of gas . Operating pressure . Maximum working pressure of contact . Gas inlet temperature . Outlet gas water content required One of the following two design criteria can be employed: 1. Glycol/water ratio (GWR): A value of 2–6 gal TEG=lbm H2O removed is adequate for most glycol dehydration requirements. Very often 2.5–4.0 gal TEG=lbm H2O is used for field dehydrators. 2. Lean TEG concentration from re-concentrator. Most glycol re-concentrators can output 99.0–99.9% lean TEG. A value of 99.5% lean TEG is used in most designs. Inlet Scrubber. It is essential to have a good inlet scrubber for efficient operation of a glycol dehydrator unit. Two-phase inlet scrubbers are generally constructed with 71 ⁄2 -ft shell heights. The required minimum diameter of a vertical inlet scrubber can be determined based on the operating pressure and required gas capacity using Fig. 10.9, which was prepared by Guo and Ghalambor (2005) based on Sivalls’s data (1977). Glycol-Gas Contactor. Glycol contactors are generally con- structed with a standard height of 71 ⁄2 ft. The minimum required diameter of the contactor can be determined based on the gas capacity of the contactor for standard gas of 0.7 specific gravity at standard temperature 100 8F. If the gas is not the standard gas and/or the operating temperature is different from the standard temperature, a correction should be first made using the following relation: qs ¼ q CtCg , (10:7) where q ¼ gas capacity of contactor at operating conditions, MMscfd qs ¼ gas capacity of contactor for standard gas (0.7 specific gravity) at standard temperature (100 8F), MMscfd Ct ¼ correction factor for operating temperature Cg ¼ correction factor for gas-specific gravity The temperature and gas-specific gravity correction fac- tors for trayed glycol contactors are given in Tables 10.9 and 10.10, respectively. The temperature and specific grav- ity factors for packed glycol contactors are contained in Tables 10.11 and 10.12, respectively. Once the gas capacity of the contactor for standard gas at standard temperature is calculated, the required minimum diameter of a trayed glycol contactor can be calculated using Fig. 10.10. The required minimum diam- eter of a packed glycol contactor can be determined based on Fig. 10.11. Table 10.9 Temperature Correction Factors for Trayed Glycol Contactors Operating temperature (8F) Correction factor (Ct) 40 1.07 50 1.06 60 1.05 70 1.04 80 1.02 90 1.01 100 1.00 110 0.99 120 0.98 Source: Used, with permission, from Sivalls, 1977. Table 10.10 Specific Gravity Correction Factors for Trayed Glycol Contactors Gas-specific gravity (air ¼ 1) Correction factor (Cg) 0.55 1.14 0.60 1.08 0.65 1.04 0.70 1.00 0.75 0.97 0.80 0.93 0.85 0.90 0.90 0.88 Source: Used, with permission, from Sivalls, 1977. Table 10.11 Temperature Correction Factors for Packed Glycol Contactors Operating temperature (8F) Correction factor (Ct) 50 0.93 60 0.94 70 0.96 80 0.97 90 0.99 100 1.00 110 1.01 120 1.02 Source: Used, with permission, from Sivalls, 1977. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 129 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/129
- 135. The required minimum height of packing of a packed contactor, or the minimum number of trays of a trayed contactor, can be determined based on Fig. 10.12. Example Problem 10.2 Size a trayed-type glycol contactor for a field installation to meet the following requirements: Gas flow rate: 12 MMscfd Gas specific gravity: 0.75 Operating line pressure: 900 psig Maximum working pressure of contactor: 1,440 psig Gas inlet temperature: 90 8F Outlet gas water content: 6 lb H2O=MMscf Design criteria: GWR ¼ 3 gal TEG=lbm H2O with 99.5% TEG Solution Because the given gas is not a standard gas and the inlet temperature is not the standard temperature, corrections need to be made. Tables 10.9 and 10.10 give Ct ¼ 1:01 and Cg ¼ 0:97. The gas capacity of contactor is calculated with Eq. (10.7): qs ¼ 12 (1:01)(0:97) ¼ 12:25 MMscfd: Figure 10.10 gives contactor diameter DC ¼ 30 in. Figure 10.6 gives water content of inlet gas: Cwi ¼ 50 lbm=MMscf. The required water content of outlet gas determines the dew-point temperature of the outlet gas through Fig. 10.6: tdo ¼ 28 F. Therefore, the dew-point depression is Dtd ¼ 90 À 28 ¼ 62 F. Based on GWR ¼ 3 gal TEG=lbm H2O and Dtd ¼ 62 F, Fig. 10.12 gives the number of trays rounded off to be four. Glycol Re-concentrator: Sizing the various components of a glycol re-concentrator starts from calculating the required glycol circulation rate: qG ¼ (GWR)Cwiq 24 , (10:8) where qG ¼ glycol circulation rate, gal/hr GWR ¼ GWR, gal TEG=lbm H2O Cwi ¼ water content of inlet gas, lbm H2O=MMscf q ¼ gas flow rate, MMscfd Reboiler: The required heat load for the reboiler can be approximately estimated from the following equation: Ht ¼ 2,000qG, (10:9) where Table 10.12 Specific Gravity Correction Factors for Packed Glycol Contactors Gas-specific gravity (air ¼1) Correction Factor (Cg) 0.55 1.13 0.60 1.08 0.65 1.04 0.70 1.00 0.75 0.97 0.80 0.94 0.85 0.91 0.90 0.88 Source: Used, with permission, from Sivalls, 1977. 0 20 40 60 80 100 120 140 Operating Pressure (psia) GasCapacity(MMscfd) 12 15 18 20 24 30 36 42 48 54 60 66 72 OD, in. 0 200 400 600 800 1,000 1,200 Figure 10.10 Gas capacity for trayed glycol contactors based on 0.7-specific gravity at 100 8F (Sivalls, 1977). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 130 4.1.2007 8:26pm Compositor Name: SJoearun 10/130 EQUIPMENT DESIGN AND SELECTION
- 136. Ht ¼ total heat load on reboiler, Btu/hr Equation (10.9) is accurate enough for most high-pres- sure glycol dehydrator sizing. A more detailed procedure for determination of the required reboiler heat load can be found from Ikoku (1984). The general overall size of the reboiler can be determined as follows: Afb ¼ Ht 7,000 , (10:10) where Afb is the total firebox surface area in squared feet. Glycol Circulating Pump: The glycol circulating pump can be sized using the glycol circulation rate and the maximum operatingpressureof the contactor.Commonlyusedglycol- powered pumps use the rich glycol from the bottom of the contactortopowerthepumpandpumptheleanglycoltothe top of the contactor. The manufacturers of these pumps should be consulted to meet the specific needs of the glycol dehydrator. Glycol Flash Separator: A glycol flash separator is usually installed downstream from the glycol pump to remove any entrained hydrocarbons from the rich glycol. A small 125-psi vertical two-phase separator is usually adequate for this purpose. The separator should be sized based on a liquid retention time in the vessel of at least 5 minutes. Vs ¼ qGtr 60 , (10:11) where Vs ¼ required settling volume in separator, gal qG ¼ glycol circulation rate, gph tr ¼ retention time approximately 5 minute Liquid hydrocarbon is not allowed to enter the glycol-gas contactor. If this is a problem, a three-phase glycol flash separator should be used to keep these liquid hydrocarbons out of the reboiler and stripping still. Three-phase flash 0 2 4 6 8 10 12 14 Operating Pressure (psia) GasCapacity(MMscfd) 10 3/4 12 3/4 14 16 18 20 24 OD, in. 0 200 400 600 800 1,000 1,200 1,400 1,600 Figure 10.11 Gas capacity for packed glycol contactors based on 0.7-specific gravity at 100 8F (Sivalls, 1977). 0 2 4 6 8 10 12 14 0 1 2 3 4 5 6 7 8 9 Glycol to Water Ratio (gal TEG/lbm H2O) NumberofValveTraysorFeetofPacking 55 65 75 85 95 td , 8F ∇ Figure 10.12 The required minimum height of packing of a packed contactor, or the minimum number of trays of a trayed contactor (Sivalls, 1977). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 131 4.1.2007 8:26pm Compositor Name: SJoearun SEPARATION SYSTEMS 10/131
- 137. separators should be sized with a liquid retention time of 20– 30 minutes. The hydrocarbon gas released from the flash separator can be piped to the reboiler to use as fuel gas and stripping gas. Based on the glycol circulation rate and the operating pressure of the contactor, the amount of gas avail- able from the glycol pump can be determined. Stripping Still: The size of the packed stripping still for the glycol re-concentrator can be determined based on the glycol-to-water circulation rate (gas TEG=lbm H2O) and the glycol circulation rate (gph). The required diameter for the stripping still is normally based on the required diam- eter at the base of the still using the vapor and liquid loading conditions at the base point. The vapor load con- sists of the water vapor and stripping gas flowing up through the still. The liquid load consists of the rich glycol stream and reflux flowing downward through the still column. One tray is normally sufficient for most stripping still requirements for TEG dehydration units. The amount of stripping gas required to re-concentrate the glycol is approximately 2---10 ft3 per gal of glycol circulated. Summary Thischaptergivesabriefintroductiontofluidseparationand gas dehydration systems. A guideline to selection of system componentsisalso presented.Operators needtoconsultwith equipment providers in designing their separation systems. References ahmed, t. Hydrocarbon Phase Behavior. Houston: Gulf Publishing Company, 1989. campbell, j.m. Gas Conditioning and Processing. Norman, OK: Campbell Petroleum Services, 1976. guenther, j.d. Natural gas dehydration. Paper presented at the Seminar on Process Equipment and Systems on Treatment Platforms, April 26, 1979, Taastrup, Denmark. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston: Gulf Publishing Company, 2005. ikoku, c.u. Natural Gas Production Engineering. New York: John Wiley Sons, 1984. mccarthy, e.l., boyd, w.l., and reid, l.s. The water vapor content of essentially nitrogen-free natural gas saturated at various conditions of temperature and pressure. Trans. AIME 1950;189:241–243. mcketta, j.j. and wehe, w.l. Use this chart for water content of natural gases. Petroleum Refinery 1958;37:153–154. sivalls, c.r. Fundamentals of oil and gas separation. Proceedings of the Gas Conditioning Conference, Uni- versity of Oklahoma, Norman, Oklahoma, 1977. Problems 10.1 Calculate the minimum required size of a standard oil/gas separator for the following conditions (con- sider vertical, horizontal, and spherical separators): Gas flow rate: 4.0 MMscfd Gas-specific gravity: 0.7 Condensate-gas ratio (CGR): 15 bbl/MMscf Condensate gravity: 65 8API Operating pressure: 600 psig Operating temperature: 70 8F 10.2 A three-stage separation is proposed to treat a well stream at a flowline pressure of 1,000 psia. Calculate pressures at each stage of separation. 10.3 Estimate water contents of a natural gas at a pressure of 2,000 psia and temperatures of 40, 80, 120, 160, 200, and 240 8F. 10.4 Design a glycol contactor for a field dehydration installation to meet the following requirements. Con- sider both trayed-type and packed-type contactors. Gas flow rate: 10 MMscfd Gas-specific gravity: 0.65 Operating line pressure: 1,000 psig Maximum working pressure of contactor: 1,440 psig Gas inlet temperature: 90 8F Outlet gas water content: 7 lb H2O=MMscf Design criteria with 99.5% TEG: GWR ¼ 3 gal TEG=lbm H2O Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap10 Final Proof page 132 4.1.2007 8:26pm Compositor Name: SJoearun 10/132 EQUIPMENT DESIGN AND SELECTION
- 138. 11 Transportation Systems Contents 11.1 Introduction 11/134 11.2 Pumps 11/134 11.3 Compressors 11/136 11.4 Pipelines 11/143 Summary 11/156 References 11/157 Problems 11/157 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 133 3.1.2007 8:54pm Compositor Name: SJoearun
- 139. 11.1 Introduction Crude oil and natural gas are transmitted over short and long distances mainly through pipelines. Pumps and compressors are used for providing pressures required for the transportation. This chapter presents principles of pumps and compressors and techniques that are used for selecting these equipments. Pipeline design criteria and fluid flow in pipelines are also discussed. Flow assurance issues are addressed. 11.2 Pumps Reciprocating piston pumps (also called ‘‘slush pumps’’ or ‘‘power pumps’’) are widely used for transporting crude oil through pipelines. There are two types of piston strokes: the single-action piston stroke and the double-action piston stroke. These are graphically shown in Figs. 11.1 and 11.2. The double-action stroke is used for duplex (two pistons) pumps. The single-action stroke is used for pumps with three or more pistons (e.g., triplex pump). Normally, duplex pumps can handle higher flow rate and triplex pumps can provide higher pressure. 11.2.1 Triplex Pumps The work per stroke for a single piston is expressed as W1 ¼ P pD2 4 L (ft À lbs): The work per one rotation of crank is W2 ¼ P pD2 4 L(1) (ft À lbs)=rotation, Figure 11.1 Double-action stroke in a duplex pump. Figure 11.2 Single-action stroke in a triplex pump. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 134 3.1.2007 8:54pm Compositor Name: SJoearun 11/134 EQUIPMENT DESIGN AND SELECTION
- 140. where P ¼ pressure, lb=ft2 L ¼ stroke length, ft D ¼ piston diameter, ft. Thus, for a triplex pump, the theoretical power is Power ¼ 3P pD2 4 LN ft À lb min , (11:1) where N is pumping speed in strokes per minute. The theoretical horsepower is HPth ¼ 3P pD2 4 550(60) LN (hp) (11:2) or HPth ¼ 3P pD2 4 33,000 LN (hp): (11:3) The input horsepower needed from the prime mover is HPi ¼ 3P pD2 4 33,000em LN (hp), (11:4) where em is the mechanical efficiency of the mechanical system transferring power from the prime mover to the fluid in the pump. Usually em is taken to be about 0.85. The theoretical volume output from a triplex pump per revolution is Qth ¼ 3 pD2 4 LN 60 ft3 =sec À Á : (11:5) The theoretical output in bbl/day is thus qth ¼ 604LND2 bbl day : (11:6) If we use inches (i.e., d [in.] and l [in.]), for D and L, then qth ¼ 0:35lNd2 bbl day : (11:7) The real output of the pump is dependent on how effici- ently the pump can fill the chambers of the pistons. Using the volumetric efficiency ev in Eq. 11.7 gives qr ¼ 0:35evlNd2 bbl day (11:8) or qr ¼ 0:01evd2 lN (gal=min), (11:9) where ev is usually taken to be 0.88–0.98. As the above volumetric equation can be written in d and l, then the horsepower equation can be written in d, l, and p (psi). Thus, HPi ¼ 3p pd2 4 l 12 N 33,000em (11:10) reduces to HPi ¼ pd2 lN 168,067em : (11:11) 11.2.2 Duplex Pumps The work per stroke cycle is expressed as W1 ¼ P pD2 1 4 L þ P pD2 1 4 À pD2 2 4 ! L(ft À lbs): (11:12) The work per one rotation of crank is W2 ¼ P pD2 1 4 L þ P pD2 1 4 À pD2 2 4 ! L ' (1) ft À lbs rotation : (11:13) Thus, for a duplex pump, the theoretical power is Power ¼ 2 Â P pD2 1 4 L þ P pD2 1 4 À pD2 2 4 ! L ' N ft À lbs min : (11:14) The theoretical horsepower is HPth ¼ 2 P pD2 1 4 L þ P pD2 1 4 À pD2 2 4 h i L n o N 550(60) (hp) or HPth ¼ 2 P p 4 D2 1 À Á L þ P pD2 1 4 À pD2 2 4 h i L n o N 33,000 : (11:15) The input horsepower needed from the prime mover is HPi ¼ 2 P p 4 D2 1 À Á L þ P pD2 1 4 À pD2 2 4 h i L n o N 33,000em (hp): (11:16) The theoretical volume output from the double-acting duplex pump per revolution is Qth ¼ 2 pD2 1 4 L þ pD2 1 4 À pD2 2 4 ! L ' N 60 ft3 =sec À Á : (11:17) The theoretical output in gals/min is thus qth ¼ 2 pD2 1 4 L þ pD2 1 4 À pD2 2 4 ! L ' Â N 0:1337 (gal= min): (11:18) If we use inches (i.e., d [in.] and l [in.]), for D and L, then qth ¼ 2 pd2 1 4 l þ pd2 1 4 À pd2 2 4 ! l ' N 231 (gal= min): (11:19) The real output of the pump is qr ¼ 2 pd2 1 4 l þ pd2 1 4 À pd2 2 4 ! l ' N 231 ev(gal= min) or qr ¼ 0:0068 2d2 1 À d2 2 À Á lNev (gal= min), (11:20) that is, qr ¼ 0:233 2d2 1 À d2 2 À Á lNev (bbl=day): (11:21) As in the volumetric output, the horsepower equation can also be reduced to a form with p, d1, d2, and l HPi ¼ p 2d2 1 À d2 2 À Á lN 252,101em : (11:22) Returning to Eq. (11.16) for the duplex double-action pump, let us derive a simplified pump equation. Rewriting Eq. (11.16), we have HPi ¼ 2 P p 4 D2 1 À Á L þ P pD2 1 4 À pD2 2 4 h i L n o N 33,000em : (11:23) The flow rate is Qth ¼ 2 pD2 1 4 L þ pD2 1 4 À pD2 2 4 ! L ' N (ft3 = min ), (11:24) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 135 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/135
- 141. so HPi ¼ PQth 33,000em : (11:25) The usual form of this equation is in p (psi) and q (gal/ min): HPi ¼ p(12)2 Â Ã [q(0:1337)] 33,000em , (11:26) that is, HPi ¼ pq 1714em : (11:27) The other form of this equation is in p (psi) and qo (bbl/ day) for oil transportation: HPi ¼ pqo 58,766em : (11:28) Equations (11.27) and (11.28) are valid for any type of pump. Example Problem 11.1 A pipeline transporting 5,000 bbl/ day of oil requires a pump with a minimum output pressure of 1,000 psi. The available suction pressure is 300 psi. Select a triplex pump for this operation. Solution Assuming a mechanical efficient of 0.85, the horsepower requirement is HPi ¼ pqo 58,766em ¼ (1,000)(5,000) 58,766(0:85) ¼ 100 hp: According to a product sheet of the Oilwell Plunger Pumps, the Model 336-ST Triplex with forged steel fluid end has a rated brake horsepower of 160 hp at 320 rpm. The maximum working pressure is 3,180 psi with the minimum plunger (piston) size of 13 ⁄4 in. It requires a suction pressure of 275 psi. With 3-in. plungers, the pump displacement is 0.5508 gal/rpm, and it can deliver liquid flow rates in the range of 1,889 bbl/day (55.08 gpm) at 100 rpm to 6,046 bbl/day (176.26 gpm) at 320 rpm, allowing a maximum pressure of 1,420 psi. This pump can be selected for the operation. The required operating rpm is RPM ¼ (5,000)(42) (24)(60)(0:5508) ¼ 265 rpm: 11.3 Compressors When natural gas does not have sufficient potential energy to flow, a compressor station is needed. Five types of compressor stations are generally used in the natural gas production industry: . Field gas-gathering stations to gather gas from wells in which pressure is insufficient to produce at a desired rate of flow into a transmission or distribution system. These stations generally handle suction pressures from below atmospheric pressure to 750 psig and volumes from a few thousand to many million cubic feet per day. . Relay or main-line stations to boost pressure in trans- mission lines compress generally large volumes of gas at a pressure range between 200 and 1,300 psig. . Re-pressuring or recycling stations to provide gas pres- sures as high as 6,000 psig for processing or secondary oil recovery projects. . Storage field stations to compress trunk line gas for injection into storage wells at pressures up to 4,000 psig. . Distribution plant stations to pump gas from holder supply to medium- or high-pressure distribution lines at about 20–100 psig, or pump into bottle storage up to 2,500 psig. 11.3.1 Types of Compressors The compressors used in today’s natural gas production industry fall into two distinct types: reciprocating and rotary compressors. Reciprocating compressors are most commonly used in the natural gas industry. They are built for practically all pressures and volumetric capacities. As shown in Fig. 11.3, reciprocating compressors have more moving parts and, therefore, lower mechanical efficiencies than rotary compressors. Each cylinder assembly of a reciprocation compressor consists of a piston, cylinder, cylinder heads, suction and discharge valves, and other parts necessary to convert rotary motion to reciprocation motion. A reciprocating compressor is designed for a cer- tain range of compression ratios through the selection of proper piston displacement and clearance volume within the cylinder. This clearance volume can be either fixed or variable, depending on the extent of the operation range and the percent of load variation desired. A typical recip- rocating compressor can deliver a volumetric gas flow rate up to 30,000 cubic feet per minute (cfm) at a discharge pressure up to 10,000 psig. Figure 11.3 Elements of a typical reciprocating compressor (courtesy of Petroleum Extension Services). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 136 3.1.2007 8:54pm Compositor Name: SJoearun 11/136 EQUIPMENT DESIGN AND SELECTION
- 142. Rotary compressors are divided into two classes: the centrifugal compressor and the rotary blower. A centrifu- gal compressor (Fig. 11.4) consists of a housing with flow passages, a rotating shaft on which the impeller is mounted, bearings, and seals to prevent gas from escaping along the shaft. Centrifugal compressors have few moving parts because only the impeller and shaft rotate. Thus, its efficiency is high and lubrication oil consumption and maintenance costs are low. Cooling water is normally unnecessary because of lower compression ratio and lower friction loss. Compression rates of centrifugal com- pressors are lower because of the absence of positive dis- placement. Centrifugal compressors compress gas using centrifugal force. In this type of compressor, work is done on the gas by an impeller. Gas is then discharged at a high velocity into a diffuser where the velocity is reduced and its kinetic energy is converted to static pressure. Unlike reciprocating compressors, all this is done without confinement and physical squeezing. Centrifugal compres- sors with relatively unrestricted passages and continuous flow are inherently high-capacity, low-pressure ratio ma- chines that adapt easily to series arrangements within a station. In this way, each compressor is required to develop only part ofthe station compression ratio. Typically, the volume is more than 100,000 cfm and discharge pressure is up to 100 psig. A rotary blower is built of a casing in which one or more impellers rotate in opposite directions. Rotary blowers are primarily used in distribution systems where the pressure differential between suction and discharge is less than 15 psi. They are also used for refrigeration and closed regeneration of adsorption plants. The rotary blower has several advantages: large quantities of low-pressure gas can be handled at comparatively low horsepower, it has small initial cost and low maintenance cost, it is simple to install and easy to operate and attend, it requires minimum floor space for the quantity of gas removed, and it has almost pulsation-less flow. As its disadvantages, it cannot withstand high pressures, it has noisy operation because of gear noise and clattering impellers, it improperly seals the clearance between the impellers and the casing, and it overheats if operated above safe pressures. Typically, rotary blowers deliver a volumetric gas flow rate of up to 17,000 cfm and have a maximum intake pressure of 10 psig and a differential pressure of 10 psi. When selecting a compressor, the pressure–volume characteristics and the type of driver must be considered. Small rotary compressors (vane or impeller type) are gen- erally driven by electric motors. Large-volume positive compressors operate at lower speeds and are usually driven by steam or gas engines. They may be driven through reduction gearing by steam turbines or an electric motor. Reciprocation compressors driven by steam tur- bines or electric motors are most widely used in the natural gas industry as the conventional high-speed compression machine. Selection of compressors requires considerations of volumetric gas deliverability, pressure, compression ratio, and horsepower. The following are important characteristics of the two types of compressors: . Reciprocating piston compressors can adjust pressure output to backpressure. . Reciprocating compressors can vary their volumetric flow-rate output (within certain limits). . Reciprocating compressors have a volumetric efficiency, which is related to the relative clearance volume of the compressor design. . Rotary compressors have a fixed pressure ratio, so they have a constant pressure output. . Rotary compressors can vary their volumetric flow-rate output (within certain limits). Impeller Guide vane Suction bearing and seal housing Stub shaft Moving parts Discharge bearing and seal housing Rotor spacer Exit guide vane Outer spacer Interstage diaphragm Inlet diaphragm Stationary parts Figure 11.4 Cross-section of a centrifugal compressor (courtesy of Petroleum Extension Services). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 137 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/137
- 143. 11.3.2 Reciprocating Compressors Figure 11.5 shows a diagram volume relation during gas compression. The shaft work put into the gas is expressed as Ws ¼ V2 2 2g À V2 1 2g þ P2v2 À ð2 1 Pdv À P1v1 0 @ 1 A, (11:29) where Ws ¼ mechanical shaft work into the system, ft-lbs per lb of fluid V1 ¼ inlet velocity of fluid to be compressed, ft/sec V2 ¼ outlet velocity of compressed fluid, ft/sec P1 ¼ inlet pressure, lb=ft2 abs P2 ¼ outlet pressure, lb=ft2 abs n1 ¼ specific volume at inlet, ft3 =lb n2 ¼ specific volume at outlet, ft3 =lb. Note that the mechanical kinetic energy term V2 2g is in ft Á lb lb À Á to get ft-lbs per lb. Rewriting Eq. (11.29), we can get Ws þ V2 1 2g À V2 2 2g þ P1v1 À P2v2 ¼ À ð2 1 Pdv: (11:30) An isentropic process is usually assumed for reciprocating compression, that is, P1nk 1 ¼ P2nk 2 ¼ Pnk ¼ constant, where k ¼ cp cv . Because P ¼ P1vk 1 vk , the right-hand side of Eq. (11.30) is formulated as À ð2 1 Pdv ¼ À ð2 1 P1vk 1 vk dv ¼ ÀP1vk 1 ð2 1 dv vk ¼ ÀP1vk 1 v1Àk 1 À k !2 1 ¼ À P1vk 1 1 À k [v1Àk 2 À v1Àk 1 ] ¼ P1vk 1 1 À k v1Àk 1 v1Àk 1 [v1Àk 2 À v1Àk 1 ] ¼ P1v1 1 À k v1Àk 2 v1Àk 1 À 1 ! ¼ P1v1 1 À k v1 v2 kÀ1 À1 # : (11:31) Using the ideal gas law P g ¼ RT, (11:32) where g (lb=ft3 ) is the specific weight of the gas and T (8R) is the temperature and R ¼ 53:36 (lb-ft/lb-8R) is the gas constant, and v ¼ 1 g, we can write Eq. (11.32) as Pv ¼ RT (11:33) or P1v1 ¼ RT1: (11:34) Using P1nk 1 ¼ P2nk 2 ¼ Pnk ¼ constant, which gives v1 v2 k ¼ P2 P1 or v1 v2 ¼ P2 P1 1 k : (11:35) Substituting Eqs. (11.35) and (11.34) into Eq. (11.31) gives À ð2 1 Pdv ¼ RT1 k À 1 P2 P1 kÀ1 k À1 # : (11:36) We multiply Eq. (11.33) by nkÀ1 , which gives Pv vkÀ1 À Á ¼ RT vkÀ1 À Á Pvk ¼ RTvkÀ1 ¼ C1 Pvk R ¼ TvkÀ1 ¼ C1 R ¼ C 0 1 Thus, TvkÀ1 ¼ C 0 1: (11:37) Also we can rise Pvk ¼ constant to the kÀ1 k À Á power. This is (Pvk ) kÀ1 k ¼ C1 0 kÀ1 k P kÀ1 k vkÀ1 ¼ C 0kÀ1 k 1 or nkÀ1 ¼ C 0kÀ1 k 1 P kÀ1 k ¼ C 00 1 P kÀ1 k : (11:38) Substituting Eq. (3.38) into (3.37) gives T C 00 1 P kÀ1 k ¼ C 0 1 or T P kÀ1 k ¼ C 0 1 C 00 1 ¼ C 000 1 ¼ constant: (11:39) Thus, Eq. (11.39) can be written as T1 P kÀ1 k 1 ¼ T2 P kÀ1 k 2 : (11:40) Thus, Eq. (11.40) is written P2 P1 kÀ1 k ¼ T2 T1 : (11:41) Substituting Eq. (11.41) into (11.36) gives À ð2 1 Pdv ¼ RT1 k À 1 T2 T1 À 1 ! À ð2 1 Pdv ¼ R k À 1 (T2 À T1): (11:42) Therefore, our original expression, Eq. (11.30), can be written as Ws þ V2 1 2g À V2 2 2g þ P1v1 À P2v2 ¼ R k À 1 (T2 À T1) orVolume Pressure m no 1 2 d c b a Figure 11.5 Basic pressure–volume diagram. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 138 3.1.2007 8:54pm Compositor Name: SJoearun 11/138 EQUIPMENT DESIGN AND SELECTION
- 144. Ws ¼ R k À 1 (T2 À T1) þ P2v2 À P1v1 þ (V2 2 À V2 1 ) 2g : (11:43) And because P1v1 ¼ RT1 (11:44) and P2v2 ¼ RT2, (11:45) Eq. (11.43) becomes Ws ¼ R k À 1 (T2 À T1) þ R(T2 À T1) þ (V2 2 À V2 1 ) 2g , (11:46) but rearranging Eq. (11.46) gives Ws ¼ k k À 1 RT1 T2 T1 À 1 þ (V2 2 À V2 1 ) 2g : Substituting Eq. (11.41) and (11.44) into the above gives Ws ¼ k k À 1 P1v1 P2 P1 kÀ1 k À1 # þ (V2 2 À V2 1 ) 2g : (11:47) Neglecting the kinetic energy term, we arrive at Ws ¼ k k À 1 P1v1 P2 P1 kÀ1 k À1 # , (11:48) where Ws is ft-lb/lb, that is, work done per lb. It is convenient to obtain an expression for power under conditions of steady state gas flow. Substituting Eq. (11.44) into (11.48) yields Ws ¼ k k À 1 RT1 P2 P1 kÀ1 k À1 # : (11:49) If we multiply both sides of Eq. (11.49) by the weight rate of flow, wt (lb/sec), through the system, we get Ps ¼ k k À 1 wtRT1 P2 P1 kÀ1 k À1 # , (11:50) where Ps ¼ Wswt ftÀlb sec and is shaft power. However, the term wt is wt ¼ g1Q1 ¼ g2Q2, (11:51) where Q1 (ft3 =sec) is the volumetric flow rate into the compressor and Q2 (ft3 =sec) would be the compressed volumetric flow rate out of the compressor. Substituting Eq. (11.32) and (11.51) into (11.50) yields Ps ¼ k k À 1 P1Q1 P2 P1 kÀ1 k À1 # : (11:52) If we use more conventional field terms such as P1 ¼ p1(144) where p1 is in psia P2 ¼ p2(144) where p2 is in psia and Q1 ¼ q1 60 where q1 is in cfm, and knowing that 1 horsepower ¼ 550 ft-lb/sec, then Eq. (11.52) becomes HP ¼ k (k À 1) p1(144)q1 550(60) p2 p1 kÀ1 k À1 # , which yields HP ¼ k (k À 1) p1q1 229:2 p2 p1 kÀ1 k À1 # : (11:53) If the gas flow rate is given in QMM (MMscf/day) in a standard base condition at base pressure pb (e.g., 14.7 psia) and base temperature Tb (e.g., 520 8R), since q1 ¼ pbT1QMM(1,000,000) p1Tb(24) , (11:54) Eq. (11.53) becomes HP ¼ 181:79pbT1QMM Tb k (k À 1) p2 p1 kÀ1 k À1 # : (11:55) It will beshown later thatthe efficiency ofcompressiondrops with increased compression ratio p2=p1. Most field applica- tions require multistage compressors (two, three, and some- times four stages) to reduce compression ratio in each stage. Figure 11.6 shows a two-stage compression unit. Using com- pressorstageswithperfectintercoolingbetweenstagesgivesa theoretical minimum power for gas compression. To obtain this minimum power, the compression ratio in each stage must be the same and the cooling between each stage must bring the gas entering each stage to the same temperature. 1 1 4 7 Knockout drums (to remove condensed liquids) Compressors (first and second stages) Interstage cooler/intercooler (air−type) Aftercooler (air−type) 5 3 2 6 2 3 4 5 6 7 Figure 11.6 Flow diagram of a two-stage compression unit. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 139 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/139
- 145. The compression ratio in each stage should be less than six to increase compression efficiency. The equation to calculate stage-compression ratio is rs ¼ Pdis Pin 1=ns , (11:56) where Pdis, Pin, and ns are final discharge pressure, inlet pressure, and number of stages, respectively. For a two-stage compression, the compression ratio for each stage should be rs ¼ ffiffiffiffiffiffiffiffi Pdis Pin r : (11:57) Using Eq. (11.50), we can write the total power require- ment for the two-stage compressor as Ptotal ¼ k k À 1 wtRTin1 Pdis1 Pin1 kÀ1 k À1 # þ k k À 1 wtRTin2 Pdis2 Pin2 kÀ1 k À1 # : (11:58) The ideal intercooler will cool the gas flow stage one to stage two to the temperature entering the compressor. Thus, we have Tin1 ¼ Tin2. Also, the pressure Pin2 ¼ Pdis1. Equation (11.58) may be written as Ptotal ¼ k k À 1 wtRTin1 Pdis1 Pin1 kÀ1 k À1 # þ k k À 1 wtRTin1 Pdis2 Pdis1 kÀ1 k À1 # : (11:59) We can find the value of Pdis1 that will minimize the power required, Ptotal. We take the derivative of Eq. (11.59) with respect to Pdis1 and set this equal to zero and solve for Pdis1. This gives Pdis1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Pin1Pdis2 p , which proves Eq. (11.57). For the two-stage compressor, Eq. (11.59) can be rewritten as Ptotal ¼ 2 k k À 1 wtRT1 Pdis2 Pin1 kÀ1 2k À1 # : (11:60) The ideal intercooling does not extend to the gas exiting the compressor. Gas exiting the compressor is governed by Eq. (11.41). Usually there is an adjustable after-cooler on a compressor that allows the operators to control the temperature of the exiting flow of gas. For greater number of stages, Eq. (11.60) can be written in field units as HPt ¼ nsp1q1 229:2 k (k À 1) p2 p1 kÀ1 nsk À1 # (11:61) or HPt ¼ 181:79nspbT1QMM Tb k (k À 1) p2 p1 kÀ1 nsk À1 # : (11:62) In the above, p1 (psia) is the intake pressure of the gas and p2 (psia) is the outlet pressure of the compressor after the final stage, q1 is the actual cfm of gas into the compressor, HPt is the theoretical horsepower needed to compress the gas. This HPt value has to be matched with a prime mover motor. The proceeding equations have been coded in the spreadsheet ReciprocatingCompressorPower.xls for quick calculations. Reciprocating compressors have a clearance at the end of the piston. This clearance produces a volumetric effi- ciency ev. The relation is given by ev ¼ 0:96 1 À « r 1 k s À 1 h in o , (11:63) where « is the clearance ratio defined as the clearance volume at the end of the piston stroke divided by the entire volume of the chamber (volume contacted by the gas in the cylinder). In addition, there is a mechanical efficiency em of the compressor and its prime mover. This results in two separate expressions for calculating the required HPt for reciprocating compressors and rotary compressors. The required minimum input prime mover motor to practically operate the compressor (either reciprocating or rotary) is HPin ¼ HPt evem , (11:64) where ev % 0:80 À 0:99 and em % 0:80 to 0:95 for recipro- cating compressors, and ev ¼ 1:0 and em % 0:70 to 0:75 for rotary compressors. Equation (11.64) stands for the input power required by the compressor, which is the minimum power to be provided by the prime mover. The prime movers usually have fixed power HPp under normal operating conditions. The usable prime mover power ratio is PR ¼ HPin HPp : (11:65) If the prime mover is not fully loaded by the compressor, its rotary speed increases and fuel consumption thus increases. Figure 11.7 shows fuel consumption curves for prime movers using gasoline, propane/butane, and diesel as fuel. Figure 11.8 presents fuel consumption curve for prime movers using nat- ural gas as fuel. It is also important to know that the prime mover power drops with surface location elevation (Fig. 11.9). ExampleProblem11.2 Considerathree-stagereciprocating compressor that is rated at q ¼ 900 scfm and a maximum pressure capability of pmax ¼ 240 psig (standard conditions at sea level). The diesel prime mover is a diesel motor (naturally aspirated) rated at 300 horsepower (at sea-level conditions). The reciprocating compressor has a clearance ratio of « ¼ 0:06 and em % 0:90. Determine the gallons/hr of fuel consumption if the working backpressure is 150 psig, and do for 1. operating at sea level 2. operating at 6,000 ft. Solution 1. Operating at sea level: rs ¼ ffiffiffiffiffiffiffi pdis pin 3 r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 150 þ 14:7 14:7 3 r ¼ ffiffiffiffiffiffiffiffiffiffiffi 164:7 14:7 3 r ¼ 2:24 ev ¼ 0:96 1 À 0:06 (2:24) 1 1:4 À 1 h in o ¼ 0:9151 Required theoretical power to compress the gas: HPt ¼ (3) 14:7(900) 229:2 1:4 0:4 164:7 14:7 0:4 3(1:4) À1 # ¼ 156:8 hp Required input power to the compressor: HPr ¼ HPt emev ¼ 156:8 0:90(0:9151) ¼ 190:3 hp Since the available power from the prime mover is 300 hp, which is greater than HPr, the prime mover is okay. The power ratio is PR ¼ 190:3 300:0 ¼ 0:634 or 63:4%: From Fig. 11.7, fuel usage is approximately 0.56 lb/hp-hr. The weight of fuel requirement is, therefore, Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 140 3.1.2007 8:54pm Compositor Name: SJoearun 11/140 EQUIPMENT DESIGN AND SELECTION
- 146. wf (lb=hr) % 0:56(190:3) ¼ 106:6 lb=hr: The volumetric fuel requirement is qf (gallons=hr) % 106:6 6:9 ¼ 15:4 gallons=hr: 2. Operating at 6,000 ft, the atmospheric pressure at an elevation of 6,000 is about 11.8 psia (Lyons et al., 2001). Figure 11.9 shows a power reduction of 22%. rs ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 150 þ 11:8 11:8 3 r ¼ 2:39 ev ¼ 0:96 1 À 0:06 (2:39)0:714 À 1 Â ÃÈ É ¼ 0:9013 HPt ¼ (3) 11:8(900) 229:2 1:4 0:4 161:8 11:8 0:0952 À1 # ¼ 137:7 hp HPr ¼ 137:7 emev ¼ 137:7 0:90(0:9103) ¼ 168:1 hp HPin ¼ 300(1 À 0:22) ¼ 234 hp 168:1 hp, so okay: PR ¼ 161:8 234 ¼ 0:718 or 71:8% Figure 11.7 shows that a fuel usage of 0.54 lb/hp-hr at 71.8% power ratio. Thus, wf (lbs=hr) % 0:54(168:1) ¼ 90:8 lbs=hr qf (gallons=hr) % 90:8 6:9 ¼ 13:2 gallons=hr : 0.50 0.55 0.60 0.65 0.70 0.75 0.80 50 60 70 80 90 100 % Load on Prime Mover FuelConsumption(lbs/hp-hr) Gasoline Propane/butane Diesel Figure 11.7 Fuel consumption of prime movers using three types of fuel. 9.0 9.5 10.0 10.5 11.0 11.5 12.0 50 60 70 80 90 100 % Load on Prime Mover FuelConsumption(lbs/hp-h) Figure 11.8 Fuel consumption of prime movers using natural gas as fuel. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 141 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/141
- 147. 11.3.3 Centrifugal Compressors Although the adiabatic compression process can be assumed in centrifugal compression, polytropic compression process is commonly considered as the basis for comparing centrifu- gal compressor performance. The process is expressed as pVn ¼ constant, (11:66) where n denotes the polytropic exponent. The isentropic exponent k applies to the ideal frictionless adiabatic pro- cess, while the polytropic exponent n applies to the actual process with heat transfer and friction. The n is related to k through polytropic efficiency Ep: n À 1 n ¼ k À 1 k  1 Ep : (11:67) The polytropic efficiency of centrifugal compressors is nearly proportional to the logarithm of gas flow rate in the range of efficiency between 0.7 and 0.75. The polytro- pic efficiency chart presented by Rollins (1973) can be represented by the following correlation: Ep ¼ 0:61 þ 0:03 log (q1), (11:68) where q1 ¼ gas capacity at the inlet condition, cfm. There is a lower limit of gas flow rate, below which severe gas surge occurs in the compressor. This limit is called ‘‘surge limit.’’ The upper limit of gas flow rate is called ‘‘stone-wall limit,’’ which is controlled by compressor horsepower. The procedure of preliminary calculations for selection of centrifugal compressors is summarized as follows: 1. Calculate compression ratio based on the inlet and discharge pressures: r ¼ p2 p1 (11:69) 2. Based on the required gas flow rate under standard condition (q), estimate the gas capacity at inlet condi- tion (q1) by ideal gas law: q1 ¼ pb p1 T1 Tb q (11:70) 3. Find a value for the polytropic efficiency Ep from the manufacturer’s manual based on q1. 4. Calculate polytropic ratio (n-1)/n using Eq. (11.67): Rp ¼ n À 1 n ¼ k À 1 k  1 Ep (11:71) 5. Calculate discharge temperature by T2 ¼ T1rRp : (11:72) 6. Estimate gas compressibility factor values at inlet and discharge conditions. 7. Calculate gas capacity at the inlet condition (q1) by real gas law: q1 ¼ z1pb z2p1 T1 Tb q (11:73) 8. Repeat Steps 2–7 until the value of q1 converges within an acceptable deviation. 9. Calculate gas horsepower by Hpg ¼ q1p1 229Ep z1 þ z2 2z1 rRp À 1 Rp : (11:74) Some manufacturers present compressor specifications using polytropic head in lbf -ft=lbm defined as Hg ¼ RT1 z1 þ z2 2 rRp À 1 Rp , (11:75) where R is the gas constant given by 1,544=MWa in psia-ft3 =lbm- R. The polytropic head relates to the gas horsepower by Hpg ¼ mtHg 33,000Ep , (11:76) where mt is mass flow rate in lbm=min. 10. Calculate gas horsepower by: Hpb ¼ Hpg þ DHpm, (11:77) 0 10 20 30 40 Surface Location Elevation (1,000 ft) PercentReductioninInputHorsepower Aspirated Input Motor Turbocharged Input Motor 0 108642 Figure 11.9 Effect of elevation on prime mover power. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 142 3.1.2007 8:54pm Compositor Name: SJoearun 11/142 EQUIPMENT DESIGN AND SELECTION
- 148. where DHpm is mechanical power losses, which is usually taken as 20 horsepower for bearing and 30 horsepower for seals. The proceeding equations have been coded in the spreadsheet CentrifugalCompressorPower.xls for quick calculations. Example Problem 11.3 Size a centrifugal compressor for the following given data: Gas-specific gravity: 0.68 Gas-specific heat ratio: 1.24 Gas flow rate: 144 MMscfd at 14.7 psia and 60 8F Inlet pressure: 250 psia Inlet temperature: 100 8F Discharge pressure: 600 psia Polytropic efficiency: Ep ¼ 0:61 þ 0:03 log (q1) Solution Calculate compression ratio based on the inlet and discharge pressures: r ¼ 600 250 ¼ 2:4 Calculate gas flow rate in scfm: q ¼ 144,000,000 (24)(60) ¼ 100,000 scfm Based on the required gas flow rate under standard condi- tion (q), estimate the gas capacity at inlet condition (q1) by ideal gas law: q1 ¼ (14:7) (250) (560) (520) (100,000) ¼ 6,332 cfm Find a value for the polytropic efficiency based on q1: Ep ¼ 0:61 þ 0:03 log (6,332) ¼ 0:724 Calculate polytropic ratio (n–1)/n: Rp ¼ n À 1 n ¼ 1:24 À 1 1:24 Â 1 0:724 ¼ 0:2673 Calculate discharge temperature: T2 ¼ (560) (2:4)0:2673 ¼ 707:7 R ¼ 247:7 F Estimate gas compressibility factor values at inlet and discharge conditions (spreadsheet program Hall- Yaborough-z.xls can be used): z1 ¼ 0:97 at 250 psia and 100 F z2 ¼ 0:77 at 600 psia and 247:7 F Calculate gas capacity at the inlet condition (q1) by real gas law: q1 ¼ (0:97)(14:7) (0:77)(250) (560) (520) (100,000) ¼ 7,977 cfm Use the new value of q1 to calculate Ep: Ep ¼ 0:61 þ 0:03 log (7,977) ¼ 0:727 Calculate the new polytropic ratio (n–1)/n: Rp ¼ n À 1 n ¼ 1:24 À 1 1:24 Â 1 0:727 ¼ 0:2662 Calculate the new discharge temperature: T2 ¼ (560) (2:4)0:2662 ¼ 707 R ¼ 247 F Estimate the new gas compressibility factor value: z2 ¼ 0:77 at 600 psia and 247 F Because z2 did not change, q1 remains the same value of 7,977 cfm. Calculate gas horsepower: Hpg ¼ (7,977)(250) (229)(0:727) 0:97 þ 0:77 2(0:97) 2:40:2662 À 1 0:2662 ¼ 10,592 hp Calculate gas apparent molecular weight: MWa ¼ (0:68)(29) ¼ 19:72 Calculated gas constant: R ¼ 1,544 19:72 ¼ 78:3 psia-ft3 =lbm- R Calculate polytropic head: Hg ¼ (78:3)(560) 0:97 þ 0:77 2 2:40:2662 À 1 0:2662 ¼ 37,610 lbf -ft=lbm Calculate gas horsepower requirement: Hpb ¼ 10,592 þ 50 ¼ 10,642 hp: 11.4 Pipelines Transporting petroleum fluids with pipelines is a continu- ous and reliable operation. Pipelines have demonstrated an ability to adapt to a wide variety of environments including remote areas and hostile environments. With very minor exceptions, largely due to local peculiarities, most refineries are served by one or more pipelines, because of their superior flexibility to the alternatives. Pipelinescan bedivided intodifferentcategories,including the following: . Flowlines transporting oil and/or gas from satellite wells to manifolds . Flowlines transporting oil and/or gas from manifolds to production facility . Infield flowlines transporting oil and/or gas from between production facilities . Export pipelines transporting oil and/or gas from production facilities to refineries/users The pipelines are sized to handle the expected pressure and fluid flow on the basis of flow assurance analysis. This section covers the following topics: 1. Flow in oil and gas pipelines 2. Design of pipelines 3. Operation of pipelines. 11.4.1 Flow in Pipelines Designing a long-distance pipeline for transportation of crude oil and natural gas requires knowledge of flow formulas for calculating capacity and pressure require- ments. Based on the first law of thermal dynamics, the total pressure gradient is made up of three distinct components: dP dL ¼ g gc r sin u þ fMru2 2gcD þ rudu gcdL , (11:78) where g gc r sin u ¼ pressure gradient due to elevation or potential energy change fM ru2 2gcD ¼ pressure gradient due to frictional losses Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 143 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/143
- 149. rudu gcdL ¼ pressure gradient due to acceleration or kinetic energy change P ¼ pressure, lbf=ft2 L ¼ pipe length, ft g ¼ gravitational acceleration, ft=sec2 gc ¼ 32:17, ft-lbm=lbf-sec2 r ¼ density lbm=ft3 u ¼ dip angle from horizontal direction, degrees fM ¼ Darcy–Wiesbach (Moody) friction factor u ¼ flow velocity, ft/sec D ¼ pipe inner diameter, ft The elevation component is pipe-angle dependent. It is zero for horizontal flow. The friction loss component applies to any type of flow at any pipe angle and causes a pressure drop in the direction of flow. The acceleration component causes a pressure drop in the direction of velocity increase in any flow condition in which velocity changes occurs. It is zero for constant-area, incompressible flow. This term is normally negligible for both oil and gas pipelines. The friction factor fM in Eq. (11.78) can be determined based on flow regimes, that is, laminar flow or turbulent flow. Reynolds number (NRe) is used as a parameter to distinguish between laminar and turbulent fluid flow. Reynolds number is defined as the ratio of fluid momen- tum force to viscous shear force. The Reynolds number can be expressed as a dimensionless group defined as NRe ¼ Dur m , (11:79) where D ¼ pipe ID, ft u ¼ fluid velocity, f/sec r ¼ fluid density, lbm=ft3 m ¼ fluid viscosity, lbm=ft-sec. The change from laminar to turbulent flow is usually assumed to occur at a Reynolds number of 2,100 for flow in a circular pipe. If U.S. field units of ft for diameter, ft/ sec for velocity, lbm=ft3 for density and centipoises for viscosity are used, the Reynolds number equation becomes NRe ¼ 1,488 Dur m : (11:80) For a gas with specific gravity gg and viscosity mg (cp) flowing in a pipe with an inner diameter D (in.) at flow rate q (Mcfd) measured at base conditions of Tb (8R) and pb (psia), the Reynolds number can be expressed as NRe ¼ 711pbqgg TbDmg : (11:81) The Reynolds number usually takes values greater than 10,000 in gas pipelines. As Tb is 520 8R and pb varies only from 14.4 to 15.025 psia in the United States, the value of 711pb/Tb varies between 19.69 and 20.54. For all practical purposes, the Reynolds number for natural gas flow prob- lems may be expressed as NRe ¼ 20qgg mgd , (11:82) where q ¼ gas flow rate at 60 8F and 14.73 psia, Mcfd gg ¼ gas-specific gravity (air ¼ 1) mg ¼ gas viscosity at in-situ temperature and pressure, cp d ¼ pipe diameter, in. The coefficient 20 becomes 0.48 if q is in scfh. Figure 11.10 is a friction factor chart covering the full range of flow conditions. It is a log-log graph of (log fM) versus (log NRe). Because of the complex nature of the curves, the equation for the friction factor in terms of the Reynolds number and relative roughness varies in different regions. In the laminar flow region, the friction factor can be determined analytically. The Hagen–Poiseuille equation for laminar flow is 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 Reynolds Number FrictionFactor 0 0.000001 0.000005 0.00001 0.00005 0.0001 0.0002 0.0004 0.0006 0.001 0.002 0.004 0.006 0.01 0.015 0.02 0.03 0.04 0.05 Laminar Flow Relative roughness Turbulent Flow Figure 11.10 Darcy–Wiesbach friction factor chart (Moody, 1944). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 144 3.1.2007 8:54pm Compositor Name: SJoearun 11/144 EQUIPMENT DESIGN AND SELECTION
- 150. dp dL f ¼ 32mu gcD2 : (11:83) Equating the frictional pressure gradients given by Eqs. (11.78) and (11.83) gives fMru2 2gcD ¼ 32mu gcD2 , (11:84) which yields fM ¼ 64m dur ¼ 64 NRe : (11:85) In the turbulent flow region, a number of empirical cor- relations for friction factors are available. Only the most accurate ones are presented in this section. For smooth wall pipes in the turbulent flow region, Drew et al. (1930) presented the most commonly used correlation: fM ¼ 0:0056 þ 0:5 N 0:32 Re , (11:86) which is valid over a wide range of Reynolds numbers, 3 Â 103 NRe 3 Â 106 . For rough wall pipes in the turbulent flow region, the effect of wall roughness on friction factor depends on the relative roughness and Reynolds number. The Nikur- adse (1933) friction factor correlation is still the best one available for fully developed turbulent flow in rough pipes: 1 ffiffiffiffiffiffi fM p ¼ 1:74 À 2 log (2eD) (11:87) This equation is valid for large values of the Reynolds number where the effect of relative roughness is dominant. The correlation that is used as the basis for modern friction factor charts was proposed by Colebrook (1938): 1 ffiffiffiffiffiffi fM p ¼ 1:74 À 2 log 2eD þ 18:7 NRe ffiffiffiffiffiffi fM p ! , (11:88) which is applicable to smooth pipes and to flow in transition and fully rough zones of turbulent flow. It degenerates to the Nikuradse correlation at large values of the Reynolds number. Equation (11.88) is not explicit in fM. However, values of fM can be obtained by a numer- ical procedure such as Newton–Raphson iteration. An explicit correlation for friction factor was presented by Jain (1976): 1 ffiffiffiffiffiffi fM p ¼ 1:14 À 2 log eD þ 21:25 N 0:9 Re : (11:89) This correlation is comparable to the Colebrook correlation. For relative roughness between 10À6 and 10À2 and the Reynolds number between 5 Â 103 and 108 , the errors were reported to be within + 1% when compared with the Cole- brookcorrelation.Therefore,Eq.(11.89)isrecommendedfor all calculations requiring friction factor determination of turbulent flow. The wall roughness is a function of pipe material, method of manufacturing, and the environment to which it has been exposed. From a microscopic sense, wall roughness is not uniform, and thus, the distance from the peaks to valleys on the wall surface will vary greatly. The absolute roughness, «, of a pipe wall is defined as the mean protruding height of relatively uniformly distributed and sized, tightly packed sand grains that would give the same pressure gradient behavior as the actual pipe wall. Analysis has suggested that the effect of roughness is not due to its absolute dimensions, but to its dimensions relative to the inside diameter of the pipe. Relative rough- ness, eD, is defined as the ratio of the absolute roughness to the pipe internal diameter: eD ¼ « D , (11:90) where « and D have the same unit. The absolute roughness is not a directly measurable property for a pipe, which makes the selection of value of pipe wall roughness difficult. The way to evaluate the absolute roughness is to compare the pressure gradients obtained from the pipe of interest with a pipe that is sand- roughened. If measured pressure gradients are available, the friction factor and Reynolds number can be calculated and an effective eD obtained from the Moody diagram. This value of eD should then be used for future predictions until updated. If no information is available on roughness, a value of « ¼ 0:0006 in. is recommended for tubing and line pipes. 11.4.1.1 Oil Flow This section addresses flow of crude oil in pipelines. Flow of multiphase fluids is discussed in other literatures such as that of Guo et al. (2005). Crude oil can be treated as an incompressible fluid. The relation between flow velocity and driving pressure differ- ential for a given pipeline geometry and fluid properties is readily obtained by integration of Eq. (11.78) when the kinetic energy term is neglected: P1 À P2 ¼ g gc r sin u þ fMru2 2gcD L, (11:91) which can be written in flow rate as P1 À P2 ¼ g gc r sin u þ fMrq2 2gcDA2 L, (11:92) where q ¼ liquid flow rate, ft3 =sec A ¼ inner cross-sectional area, ft2 When changed to U.S. field units, Eq. (11.92) becomes p1 À p2 ¼ 0:433goL sin u þ 1:15 Â 10À5 Â fMgoQ2 L d5 , (11:93) where p1 ¼ inlet pressure, psi p2 ¼ outlet pressure, psi go ¼ oil specific gravity, water ¼ 1.0 Q ¼ oil flow rate, bbl/day d ¼ pipe inner diameter, in. Example Problem 11.4 A 35 API gravity, 5 cp, oil is transported through a 6-in. (I.D.) pipeline with an uphill angle of 15 degrees across a distance of 5 miles at a flow rate of 5,000 bbl/day. Estimate the minimum required pump pressure to deliver oil at 50 psi pressure at the outlet. Assume e ¼ 0.0006 in. Solution Pipe inner area: A ¼ p 4 6 12 2 ¼ 0:1963 ft2 The average oil velocity in pipe: u ¼ (5,000)(5:615) (24)(60)(60)(0:1963) ¼ 1:66 ft=sec Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 145 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/145
- 151. Oil-specific gravity: go ¼ 141:5 131:5 þ 35 ¼ 0:85 Reynolds number: NRe ¼ 1,488 6 12 À Á (1:66)(0:85)(62:4) 5 ¼ 13,101 2,100 turbulent flow Equation (11.89) gives 1 ffiffiffiffiffiffi fM p ¼ 1:14 À 2 log 0:0006 6 þ 21:25 (13,101)0:9 ¼ 5:8759, which gives fM ¼ 0:02896: Equation (11.93) gives p1 ¼ 50 þ 0:433(0:85)(5)(5,280) sin (15 ) þ 1:15 Â 10À5 Â (0:02896)(0:85)(5,000)2 (5)(5,280) (6)5 ¼ 2,590 psi: 11.4.1.2 Gas Flow Consider steady-state flow of dry gas in a constant-diam- eter, horizontal pipeline. The mechanical energy equation, Eq. (11.78), becomes dp dL ¼ fMru2 2gcD ¼ p(MW)a zRT fu2 2gcD , (11:94) which serves as a base for development of many pipeline equations. The difference in these equations originated from the methods used in handling the z-factor and fric- tion factor. Integrating Eq. (11.94) gives ð dp ¼ (MW)a fMu2 2RgcD ð p zT dL: (11:95) If temperature is assumed constant at average value in a pipeline, T¯ , and gas deviation factor, z¯, is evaluated at average temperature and average pressure, p¯, Eq. (11.95) can be evaluated over a distance L between upstream pressure, p1, and downstream pressure, p2: p2 1 À p2 2 ¼ 25ggQ2 TTzzfML d5 , (11:96) where gg ¼ gas gravity (air ¼ 1) Q ¼ gas flow rate, MMscfd (at 14.7 psia, 60 8F) T¯ ¼ average temperature, 8R z¯ ¼ gas deviation factor at T¯ and p¯ p¯ ¼ (p1 þ p2)/2 L ¼ pipe length, ft d ¼ pipe internal diameter, in. F ¼ Moody friction factor Equation (11.96) may be written in terms of flow rate measured at arbitrary base conditions (Tb and pb): q ¼ CTb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)d5 gg TTzzfML s , (11:97) where C is a constant with a numerical value that depends on the units used in the pipeline equation. If L is in miles and q is in scfd, C ¼ 77:54. The use of Eq. (11.97) involves an iterative procedure. The gas deviation factor depends on pressure and the friction factor depends on flow rate. This problem prompted several investigators to develop pipeline flow equations that are noniterative or explicit. This has in- volved substitutions for the friction factor fM. The specific substitution used may be diameter-dependent only (Weymouth equation) or Reynolds number–dependent only (Panhandle equations). 11.4.1.2.1 Weymouth Equation for Horizontal Flow Equation (11.97) takes the following form when the unit of scfh for gas flow rate is used: qh ¼ 3:23Tb pb ffiffiffiffiffiffi 1 fM s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)d5 gg TTzzL s , (11:98) where ffiffiffiffi 1 fM q is called the ‘‘transmission factor.’’ The friction factor may be a function of flow rate and pipe roughness. If flow conditions are in the fully turbulent region, Eq. (11.89) degenerates to fM ¼ 1 [1:14 À 2 log (eD)]2 , (11:99) where fM depends only on the relative roughness, eD. When flow conditions are not completely turbulent, fM depends on the Reynolds number also. Therefore, use of Eq. (11.98) requires a trial-and-error procedure to calculate qh. To eliminate the trial-and-error procedure, Weymouth proposed that f vary as a function of diameter as follows: fM ¼ 0:032 d1=3 (11:100) With this simplification, Eq. (11.98) reduces to qh ¼ 18:062Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)D16=3 gg TTzzL s , (11:101) which is the form of the Weymouth equation commonly used in the natural gas industry. The use of the Weymouth equation for an existing transmission line or for the design of a new transmission line involves a few assumptions including no mechanical work, steady flow, isothermal flow, constant compressibil- ity factor, horizontal flow, and no kinetic energy change. These assumptions can affect accuracy of calculation results. In the study of an existing pipeline, the pressure-mea- suring stations should be placed so that no mechanical energy is added to the system between stations. No me- chanical work is done on the fluid between the points at which the pressures are measured. Thus, the condition of no mechanical work can be fulfilled. Steady flow in pipeline operation seldom, if ever, exists in actual practice because pulsations, liquid in the pipeline, and variations in input or output gas volumes cause devi- ations from steady-state conditions. Deviations from steady-state flow are the major cause of difficulties experi- enced in pipeline flow studies. The heat of compression is usually dissipated into the ground along a pipeline within a few miles downstream from the compressor station. Otherwise, the temperature of the gas is very near that of the containing pipe, and because pipelines usually are buried, the temperature of the flowing gas is not influenced appreciably by rapid changes in atmospheric temperature. Therefore, the gas flow can be considered isothermal at an average effective temperature without causing significant error in long- pipeline calculations. The compressibility of the fluid can be considered con- stant and an average effective gas deviation factor may be used. When the two pressures p1 and p2 lie in a region where z is essentially linear with pressure, it is accurate enough to evaluate z¯ at the average pressure pp ¼ ( p1 þ p2)=2. One can also use the arithmetic average Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 146 3.1.2007 8:54pm Compositor Name: SJoearun 11/146 EQUIPMENT DESIGN AND SELECTION
- 152. of the z’s with zz ¼ (z1 þ z2)=2, where z1 and z2 are obtained at p1 and p2, respectively. On the other hand, should p1 and p2 lie in the range where z is not linear with pressure (double-hatched lines), the proper average would result from determining the area under the z-curve and dividing it by the difference in pressure: zz ¼ Ðp2 p1 zdp ( p1 À p2) , (11:102) where the numerator can be evaluated numerically. Also, z¯ can be evaluated at an average pressure given by pp ¼ 2 3 p3 1 À p3 2 p2 1 À p2 2 : (11:103) Regarding the assumption of horizontal pipeline, in actual practice, transmission lines seldom, if ever, are horizontal, so that factors are needed in Eq. (11.101) to compensate for changes in elevation. With the trend to higher operat- ing pressures in transmission lines, the need for these factors is greater than is generally realized. This issue of correction for change in elevation is addressed in the next section. If the pipeline is long enough, the changes in the kinetic- energy term can be neglected. The assumption is justified for work with commercial transmission lines. Example Problem 11.5 For the following data given for a horizontal pipeline, predict gas flow rate in ft3 =hr through the pipeline. Solve the problem using Eq. (11.101) with the trial-and-error method for friction factor and the Weymouth equation without the Reynolds number– dependent friction factor: d ¼ 12.09 in. L ¼ 200 mi e ¼ 0.0006 in. T ¼ 80 8F gg ¼ 0:70 Tb ¼ 520 R pb ¼ 14:7 psia p1 ¼ 600 psia p2 ¼ 200 psia Solution The average pressure is pp ¼ (200 þ 600)=2 ¼ 400 psia: With p¯ ¼ 400 psia, T ¼ 540 8R and gg ¼ 0:70, Brill-Beggs- Z.xls gives zz ¼ 0:9188: With p¯ ¼ 400 psia, T ¼ 540 8R and gg ¼ 0:70, Carr- Kobayashi-BurrowsViscosity.xls gives m ¼ 0:0099 cp: Relative roughness: eD ¼ 0:0006=12:09 ¼ 0:00005 A. Trial-and-error calculation: First trial: qh ¼ 500,000 scfh NRe ¼ 0:48(500,000)(0:7) (0:0099)(12:09) ¼ 1,403,733 1 ffiffiffiffiffiffi fM p ¼ 1:14 À 2 log 0:00005 þ 21:25 (1,403,733)0:9 fM ¼ 0:01223 qh ¼ 3:23(520) 14:7 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0:01223 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (6002 À 2002)(12:09)5 (0:7)(540)(0:9188)(200) s ¼ 1,148,450 scfh Second trial: qh ¼ 1,148,450 cfh NRe ¼ 0:48(1,148,450)(0:7) (0:0099)(12:09) ¼ 3,224,234 1 ffiffiffiffiffiffi fM p ¼ 1:14 À 2 log 0:00005 þ 21:25 (3,224,234)0:9 fM ¼ 0:01145 qh ¼ 3:23(520) 14:7 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0:01145 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6002 À 2002ð Þ(12:09)5 (0:7)(540)(0:9188)(200) s ¼ 1,186,759 scfh Third trial: qh ¼ 1,186,759 scfh NRe ¼ 0:48(1,186,759)(0:7) (0:0099)(12:09) ¼ 3,331,786 1 ffiffiffiffiffiffi fM p ¼ 1:14 À 2 log 0:00005 þ 21:25 (3,331,786)0:9 fM ¼ 0:01143 qh ¼ 3:23(520) 14:7 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0:01143 r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6002 À 2002ð Þ(12:09)5 (0:7)(540)(0:9188)(200) s ¼ 1,187,962 scfh, which is close to the assumed 1,186,759 scfh. B. Using the Weymouth equation: qh ¼ 18:062(520) 14:7 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (6002 À 2002)(12:09)16=3 (0:7)(540)(0:9188)(200) s ¼ 1,076,035 scfh Problems similar to this one can be quickly solved with the spreadsheet program PipeCapacity.xls. 11.4.1.2.2 Weymouth Equation for Non-horizontal Flow Gas transmission pipelines are often nonhorizontal. Account should be taken of substantial pipeline elevation changes. Considering gas flow from point 1 to point 2 in a nonhorizontal pipe, the first law of thermal dynamics gives ð2 1 vdP þ g gc Dz þ ð2 1 fMu2 2gcD dL ¼ 0: (11:104) Based on the pressure gradient due to the weight of gas column, dP dz ¼ rg 144 , (11:105) and real gas law, rg ¼ p(MW)a zRT ¼ 29ggp zRT , Weymouth (1912) developed the following equation: qh ¼ 3:23Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À esp2 2)d5 fMgg TTzzL s , (11:106) where Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 147 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/147
- 153. e ¼ 2.718 and s ¼ 0:0375ggDz TTzz , (11:107) and Dz is equal to outlet elevation minus inlet elevation (note that Dz is positive when outlet is higher than inlet). A general and more rigorous form of the Weymouth equa- tion with compensation for elevation is qh ¼ 3:23Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À esp2 2)d5 fMgg TTzzLe s , (11:108) where Le is the effective length of the pipeline. For a uniform slope, Le is defined as Le ¼ (es À1)L s . For a non-uniform slope (where elevation change cannot be simplified to a single section of constant gradient), an approach in steps to any number of sections, n, will yield Le ¼ (es1 À 1) s1 L1 þ es1 (es2 À 1) s2 L2 þ es1þs2 (es3 À 1) s3 L3 þ . . . . . . : : þ Xn i¼1  e PiÀ1 j¼1 sj (esi À 1) si Li, (11:109) where si ¼ 0:0375ggDzi TTzz : (11:110) 11.4.1.2.3 Panhandle-A Equation for Horizontal Flow The Panhandle-A pipeline flow equation assumes the following Reynolds number–dependent friction factor: fM ¼ 0:085 N0:147 Re (11:111) The resultant pipeline flow equation is, thus, q ¼ 435:87 d2:6182 g0:4604 g Tb pb 1:07881 ( p2 1 À p2 2) TTzzL !0:5394 , (11:112) where q is the gas flow rate in scfd measured at Tb and pb, and other terms are the same as in the Weymouth equa- tion. 11.4.1.2.4 Panhandle-B Equation for Horizontal Flow (Modified Panhandle) The Panhandle-B equation is the most widely used equation for long transmission and delivery lines. It assumes that fM varies as fM ¼ 0:015 N0:0392 Re , (11:113) and it takes the following resultant form: q ¼ 737d2:530 Tb pb 1:02 ( p2 1 À p2 2) TTzzLg0:961 g #0:510 (11:114) 11.4.1.2.5 Clinedinst Equation for Horizontal Flow The Clinedinst equation rigorously considers the deviation of natural gas from ideal gas through integration. It takes the following form: q ¼ 3973:0 zbpbppc pb  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d5 TTfMLgg ðpr1 0 pr z dpr À ðpr2 0 pr z dpr 0 @ 1 A v u u u t , (11:115) where q ¼ volumetric flow rate, Mcfd ppc ¼ pseudocritical pressure, psia d ¼ pipe internal diameter, in. L ¼ pipe length, ft pr ¼ pseudo-reduced pressure T¯ ¼ average flowing temperature, 8R gg ¼ gas gravity, air ¼ 1.0 zb ¼ gas deviation factor at Tb and pb, normally accepted as 1.0. Based on Eqs. (2.29), (2.30), and (2.51), Guo and Ghalam- bor (2005) generated curves of the integral function Ðpr 0 pr z dpr for various gas-specific gravity values. 11.4.1.2.6 Pipeline Efficiency All pipeline flow equ- ations were developed for perfectly clean lines filled with gas. In actual pipelines, water, condensates, sometimes crude oil accumulates in low spots in the line. There are often scales and even ‘‘junk’’ left in the line. The net result is that the flow rates calculated for the 100% efficient cases are often modified by multiplying them by an efficiency factor E. The efficiency factor expresses the actual flow capacity as a fraction of the theoretical flow rate. An efficiency factor ranging from 0.85 to 0.95 would represent a ‘‘clean’’ line. Table 11.1 presents typical values of efficiency factors. 11.4.2 Design of Pipelines Pipeline design includes determination of material, diam- eter, wall thickness, insulation, and corrosion protection measure. For offshore pipelines, it also includes weight coating and trenching for stability control. Bai (2001) provides a detailed description on the analysis–analysis- based approach to designing offshore pipelines. Guo et al. (2005) presents a simplified approach to the pipeline design. The diameter of pipeline should be determined based on flow capacity calculations presented in the previous sec- tion. This section focuses on the calculations to design wall thickness and insulation. 11.4.2.1 Wall Thickness Design Wall thickness design for steel pipelines is governed by U.S. Code ASME/ANSI B32.8. Other codes such as Z187 (Canada), DnV (Norway), and IP6 (UK) have es- sentially the same requirements but should be checked by the readers. Except for large-diameter pipes (30 in.), material grade is usually taken as X-60 or X-65 (414 or 448 MPa) for high-pressure pipelines or on deepwater. Higher grades can be selected in special cases. Lower grades such as X-42, X-52, or X-56 can be selected in shallow water or for low- pressure, large-diameter pipelines to reduce material cost or in cases in which high ductility is required for improved impact resistance. Pipe types include . Seamless . Submerged arc welded (SAW or DSAW) Table 11.1 Typical Values of Pipeline Efficiency Factors Type of line Liquid content (gal/MMcf) Efficiency E Dry-gas field 0.1 0.92 Casing-head gas 7.2 0.77 Gas and condensate 800 0.6 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 148 3.1.2007 8:54pm Compositor Name: SJoearun 11/148 EQUIPMENT DESIGN AND SELECTION
- 154. . Electric resistance welded (ERW) . Spiral weld. Except in specific cases, only seamless or SAW pipes are to be used, with seamless being the preference for diameters of 12 in. or less. If ERW pipe is used, special inspection provisions such as full-body ultrasonic testing are re- quired. Spiral weld pipe is very unusual for oil/gas pipe- lines and should be used only for low-pressure water or outfall lines. 11.4.2.1.1 Design Procedure Determination of pipeline wall thickness is based on the design internal pressure or the external hydrostatic pressure. Maximum longitudinal stresses and combined stresses are sometimes limited by applicable codes and must be checked for installation and operation. However, these criteria are not normally used for wall thickness determination. Increasing wall thickness can sometimes ensure hydrodynamic stability in lieu of other stabilization methods (such as weight coating). This is not normally economical, except in deepwater where the presence of concrete may interfere with the preferred installation method. We recommend the following procedure for designing pipeline wall thickness: Step 1: Calculate the minimum wall thickness required for the design internal pressure. Step 2: Calculate the minimum wall thickness required to withstand external pressure. Step 3: Add wall thickness allowance for corrosion if ap- plicable to the maximum of the above. Step 4: Select next highest nominal wall thickness. Step 5: Check selected wall thickness for hydrotest condi- tion. Step 6: Check for handling practice, that is, pipeline han- dling is difficult for D/t larger than 50; welding of wall thickness less than 0.3 in (7.6 mm) requires special provisions. Note that in certain cases, it may be desirable to order a nonstandard wall. This can be done for large orders. Pipelines are sized on the basis of the maximum expected stresses in the pipeline under operating condi- tions. The stress calculation methods are different for thin-wall and thick-wall pipes. A thin-wall pipe is defined as a pipe with D/t greater than or equal to 20. Figure 11.11 shows stresses in a thin-wall pipe. A pipe with D/t less than 20 is considered a thick-wall pipe. Figure 11.12 illustrates stresses in a thick-wall pipe. 11.4.2.1.2 Design for Internal Pressure Three pipe- line codes typically used for design are ASME B31.4 (ASME, 1989), ASME B31.8 (ASME, 1990), and DnV 1981 (DnV, 1981). ASME B31.4 is for all oil lines in North America. ASME B31.8 is for all gas lines and two- phase flow pipelines in North America. DnV 1981 is for oil, gas, and two-phase flow pipelines in North Sea. All these codes can be used in other areas when no other code is available. The nominal pipeline wall thickness (tNOM) can be calculated as follows: tNOM ¼ Pd D 2EwhsyFt þ ta, (11:116) where Pd is the design internal pressure defined as the difference between the internal pressure (Pi) and external pressure (Pe), D is nominal outside diameter, ta is thick- ness allowance for corrosion, and sy is the specified t t r PD sL 4t = rdf dφ t sh = D P = ppr2 sL 2t PD sh f f P = ppr2 df Figure 11.11 Stresses generated by internal pressure p in a thin-wall pipe, D/t 20. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 149 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/149
- 155. minimum yield strength. Equation (11.116) is valid for any consistent units. Most codes allow credit for external pressure. This credit should be used whenever possible, although care should be exercised for oil export lines to account for head of fluid and for lines that traverse from deep to shallow water. ASME B31.4 and DnV 1981 define Pi as the maximum allowable operating pressure (MAOP) under normal condi- tions, indicating that surge pressures up to 110% MAOP is acceptable. In some cases, Pi is defined as wellhead shut-in pressure (WSIP) for flowlines or specified by the operators. In Eq. (11.116), the weld efficiency factor (Ew) is 1.0 for seamless, ERW, and DSAW pipes. The temperature de- rating factor (Ft) is equal to 1.0 for temperatures under 250 8F. The usage factor (h) is defined in Tables 11.2 and 11.3 for oil and gas lines, respectively. The underthickness due to manufacturing tolerance is taken into account in the design factor. There is no need to add any allowance for fabrication to the wall thickness calculated with Eq. (11.116). 11.4.2.1.3 Design for External Pressure Different practices can be found in the industry using different external pressure criteria. As a rule of thumb, or unless qualified thereafter, it is recommended to use propagation criterion for pipeline diameters under 16-in. and collapse criterion for pipeline diameters more than or equal to 16-in. Propagation Criterion: The propagation criterion is more conservative and should be used where optimization of the wall thickness is not required or for pipeline installation methods not compatible with the use of buckle arrestors such as reel and tow methods. It is generally economical to design for propagation pressure for diameters less than 16-in. For greater diameters, the wall thickness penalty is too high. When a pipeline is designed based on the collapse criterion, buckle arrestors are recommended. The external pressure criterion should be based on nominal wall thick- ness, as the safety factors included below account for wall variations. Although a large number of empirical relationships have been published, the recommended formula is the latest given by AGA.PRC (AGA, 1990): PP ¼ 33Sy tNOM D 2:46 , (11:117) m1 n1 2 m r n dr dr dr dsr + sh sh 2 sr df b a srb = −P sra = − PO P m1 m n1 n PO sr Figure 11.12 Stresses generated by internal pressure p in a thick-wall pipe, D/t 20. Table 11.2 Design and Hydrostatic Pressure Definitions and Usage Factors for Oil Lines Parameter ASME B31.4, 1989 Edition Dnv (Veritas, 1981) Design internal pressure Pa d Pi À Pe[401:2:2] Pi À Pe[4.2.2.2] Usage factor h 0.72 [402.3.1(a)] 0.72 [4.2.2.1] Hydrotest pressure Ph 1:25 Pb i [437.4.1(a)] 1:25Pd [8.8.4.3] a Credit can be taken for external pressure for gathering lines or flowlines when the MAOP (Pi) is applied at the wellhead or at the seabed. For export lines, when Pi is applied on a platform deck, the head fluid shall be added to Pi for the pipeline section on the seabed. b If hoop stress exceeds 90% of yield stress based on nominal wall thickness, special care should be taken to prevent overstrain of the pipe. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 150 3.1.2007 8:54pm Compositor Name: SJoearun 11/150 EQUIPMENT DESIGN AND SELECTION
- 156. which is valid for any consistent units. The nominal wall thickness should be determined such that Pp 1:3 Pe. The safety factor of 1.3 is recommended to account for uncer- tainty in the envelope of data points used to derive Eq. (11.117). It can be rewritten as tNOM $ D 1:3PP 33Sy 1 2:46 : (11:118) For the reel barge method, the preferred pipeline grade is belowX-60.However, X-65 steel can beusedif the ductility is kept high by selecting the proper steel chemistry and micro- alloying. For deepwater pipelines, D/t ratios of less than 30 are recommended. It has been noted that bending loads have no demonstrated influence on the propagation pressure. Collapse Criterion: The mode of collapse is a function of D/t ratio, pipeline imperfections, and load conditions. The theoretical background is not given in this book. An em- pirical general formulation that applies to all situations is provided. It corresponds to the transition mode of collapse under external pressure (Pe), axial tension (Ta), and bend- ing strain (sb), as detailed elsewhere (Murphey and Langner, 1985; AGA, 1990). The nominal wall thickness should be determined such that 1:3PP PC þ «b «B # gp, (11:119) where 1.3 is the recommended safety factor on collapse, «B is the bending strain of buckling failure due to pure bending, and g is an imperfection parameter defined below. The safety factor on collapse is calculated for D/t ratios along with the loads (Pe, «b, Ta) and initial pipeline out-of roundness (do). The equations are PC ¼ PelP 0 y ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P2 el þ P02 y q , (11:120) P 0 y ¼ Py ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À 0:75 Ta Ty 2 s À Ta 2Ty 2 4 3 5, (11:121) Pel ¼ 2E 1 À n2 t D 3 , (11:122) Py ¼ 2Sy t D , (11:123) Ty ¼ ASy, (11:124) where gp is based on pipeline imperfections such as initial out-of roundness (do), eccentricity (usually neglected), and residual stress (usually neglected). Hence, gp ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ p2 p2 À 1 f 2 p , v u u t (11:125) where p ¼ P 0 y Pel , (11:126) fp ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ do D t 2 s À do D t , (11:127) «B ¼ t 2D , (11:128) and do ¼ Dmax À Dmin Dmax þ Dmin : (11:129) When a pipeline is designed using the collapse criterion, a good knowledge of the loading conditions is required (Ta and «b). An upper conservative limit is necessary and must often be estimated. Under high bending loads, care should be taken in esti- mating «b using an appropriate moment-curvature relationship. A Ramberg Osgood relationship can be used as KÃ ¼ MÃ þ AMÃB , (11:130) where KÃ ¼ K=Ky and MÃ ¼ M=My with Ky ¼ 2Sy=ED is the yield curvature and My ¼ 2ISy=D is the yield moment. The coefficients A and B are calculated from the two data points on stress–strain curve generated during a tensile test. 11.4.2.1.4 Corrosion Allowance To account for corrosion when water is present in a fluid along with contaminants such as oxygen, hydrogen sulfide (H2S), and carbon dioxide (CO2), extra wall thickness is added. A review of standards, rules, and codes of practices (Hill and Warwick, 1986) shows that wall allowance is only one of several methods available to prevent corrosion, and it is often the least recommended. For H2S and CO2 contaminants, corrosion is often localized (pitting) and the rate of corrosion allowance ineffective. Corrosion allowance is made to account for damage during fabrication, transportation, and storage. A value of 1 ⁄16 in. may be appropriate. A thorough assessment of the internal corrosion mechanism and rate is necessary before any corrosion allowance is taken. 11.4.2.1.5 Check for Hydrotest Condition The min- imum hydrotest pressure for oil and gas lines is given in Tables 11.2 and 11.3, respectively, and is equal to 1.25 times the design pressure for pipelines. Codes do not require that the pipeline be designed for hydrotest conditions but sometimes give a tensile hoop stress limit 90% SMYS, which is always satisfied if credit has not been taken for external pressure. For cases where the wall thickness is based on Pd ¼ Pi À Pe, codes recommend not to overstrain the pipe. Some of the codes are ASME B31.4 (Clause 437.4.1), ASME B31.8 (no limit on hoop stress during hydrotest), and DnV (Clause 8.8.4.3). Table 11.3 Design and Hydrostatic Pressure Definitions and Usage Factors for Gas Lines Parameter ASME B31.8, 1989 Edition, 1990 Addendum DnV (Veritas, 1981) Pa d Pi À Pe[A842.221] Pi À Pe[4.2.2.2] Usage factor h 0.72 [A842.221] 0.72 [4.2.2.1] Hydrotest pressure Ph 1:25 Pb i [A847.2] 1:25Pd [8.8.4.3] a Credit can be taken for external pressure for gathering lines or flowlines when the MAOP (Pi) is applied at wellhead or at the seabed. For export lines, when Pi is applied on a platform deck, the head of fluid shall be added to Pi for the pipeline section on the seabed (particularly for two-phase flow). b ASME B31.8 imposes Ph ¼ 1:4Pi for offshore risers but allows onshore testing of prefabricated portions. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 151 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/151
- 157. For design purposes, condition sh # sy should be con- firmed, and increasing wall thickness or reducing test pres- sure should be considered in other cases. For offshore pipelines connected to riser sections requiring Ph ¼ 1:4Pi, it is recommended to consider testing the riser separately (for prefabricated sections) or to determine the hydrotest pressure based on the actual internal pressure experienced by the pipeline section. It is important to note that most pressure testing of subsea pipelines is done with water, but on occasion, nitrogen or air has been used. For low D/t ratios (20), the actual hoop stress in a pipeline tested from the surface is overestimated when using the thin wall equations provided in this chapter. Credit for this effect is allowed by DnV Clause 4.2.2.2 but is not normally taken into account. ExampleProblem11.6 Calculatetherequiredwallthickness forthepipelineinExampleProblem11.4assumingaseamless still pipe of X-60 grade and onshore gas field (external pressure Pe ¼ 14:65 psia). Solution The wall thickness can be designed based on the hoop stress generated by the internal pressure Pi ¼ 2,590 psia. The design pressure is Pd ¼ Pi À Pe ¼ 2,590 À 14:65 ¼ 2,575:35 psi: The weld efficiency factor is Ew ¼ 1:0. The temperature de-rating factor Ft ¼ 1:0. Table 11.3 gives h ¼ 0:72. The yield stress is sy ¼ 60,000 psi. A corrosion allowance 1 ⁄16 in. is considered. The nominal pipeline wall thickness can be calculated using Eq. (11.116) as tNOM ¼ (2,574:3)(6) 2(1:0)(0:72)(60,000)(1:0) þ 1 16 ¼ 0:2413 in: Considering that welding of wall thickness less than 0.3 in. requires special provisions, the minimum wall thickness is taken, 0.3 in. 11.4.2.2 Insulation Design Oil and gas field pipelines are insulated mainly to conserve heat. The need to keep the product fluids in the pipeline at a temperature higher than the ambient temperature could exist, for reasons including the following: . Preventing formation of gas hydrates . Preventing formation of wax or asphaltenes . Enhancing product flow properties . Increasing cool-down time after shutting down In liquefied gas pipelines, such as liquefied natural gas, insulation is required to maintain the cold temperature of the gas to keep it in a liquid state. Designing pipeline insulation requires thorough knowl- edge of insulation materials and heat transfer mechanisms across the insulation. Accurate predictions of heat loss and temperature profile in oil- and gas-production pipelines are essential to designing and evaluating pipeline oper- ations. 11.4.2.2.1 Insulation Materials Polypropylene, poly- ethylene, and polyurethane are three base materials widely used in the petroleum industry for pipeline insulation. Their thermal conductivities are given in Table 11.4 (Carter et al., 2002). Depending on applications, these base materials are used in different forms, resulting in different overall conductivities. A three-layer polypropylene applied to pipe surface has a conductivity of 0.225 W/M-8C (0.13 btu/hr-ft- 8F), while a four-layer polypropylene has a conductivity of 0.173 W/M-8C (0.10 btu/hr-ft-8F). Solid polypropylene has higher conductivity than polypropylene foam. Polymer syntactic polyurethane has a conductivity of 0.121 W/M-8C (0.07 btu/hr-ft-8F), while glass syntactic polyurethane has a conductivity of 0.156 W/M-8C (0.09 btu/hr-ft-8F). These materials have lower conductivities in dry conditions such as that in pipe-in-pipe (PIP) applications. Because of their low thermal conductivities, more and more polyurethane foams are used in deepwater pipeline applications. Physical properties of polyurethane foams include density, compressive strength, thermal conductiv- ity, closed-cell content, leachable halides, flammability, tensile strength, tensile modulus, and water absorption. Typical values of these properties are available elsewhere (Guo et al., 2005). In steady-state flow conditions in an insulated pipeline segment, the heat flow through the pipe wall is given by Qr ¼ UArDT, (11:131) where Qr is heat-transfer rate; U is overall heat-transfer coefficient (OHTC) at the reference radius; Ar is area of the pipeline at the reference radius; DT is the difference in temperature between the pipeline product and the ambient temperature outside. The OHTC, U, for a system is the sum of the thermal resistances and is given by (Holman, 1981): U ¼ 1 Ar 1 Aihi þ Pn m¼1 ln (rmþ1=rm) 2pLkm þ 1 Aoho ! , (11:132) Table 11.4 Thermal Conductivities of Materials Used in Pipeline Insulation Thermal conductivity Material name W/M-8C Btu/hr-ft-8F Polyethylene 0.35 0.20 Polypropylene 0.22 0.13 Polyurethane 0.12 0.07 Table 11.5 Typical Performance of Insulated Pipelines U-Value Water depth (M) Insulation type (Btu=hr À ft2 À F) W=M2 À K Field proven Potential Solid polypropylene 0.50 2.84 1,600 4,000 Polypropylene foam 0.28 1.59 700 2,000 Syntactic polyurethane 0.32 1.81 1,200 3,300 Syntactic polyurethane foam 0.30 1.70 2,000 3,300 Pipe-in-pipe syntactic polyurethane foam 0.17 0.96 3,100 4,000 Composite 0.12 0.68 1,000 3,000 Pipe-in-pipe high efficiency 0.05 0.28 1,700 3,000 Glass syntactic polyurethane 0.03 0.17 2,300 3,000 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 152 3.1.2007 8:54pm Compositor Name: SJoearun 11/152 EQUIPMENT DESIGN AND SELECTION
- 158. where hi is film coefficient of pipeline inner surface; ho is film coefficient of pipeline outer surface; Ai is area of pipe- line inner surface; Ao is area of pipeline outer surface; rm is radius of layer m; and km is thermal conductivity of layer m. Similar equations exist for transient-heat flow, giving an instantaneous rate for heat flow. Typically required insulation performance, in terms of OHTC (U value) of steel pipelines in water, is summarized in Table 11.5. Pipeline insulation comes in two main types: dry insula- tion and wet insulation. The dry insulations require an outer barrier to prevent water ingress (PIP). The most common types of this include the following: . Closed-cell polyurethane foam . Open-cell polyurethane foam . Poly-isocyanurate foam . Extruded polystyrene . Fiber glass . Mineral wool . Vacuum-insulation panels Under certain conditions, PIP systems may be considered over conventional single-pipe systems. PIP insulation may be required to produce fluids from high-pressure/high- temperature (150 8C) reservoirs in deepwater (Carmi- chael et al., 1999). The annulus between pipes can be filled with different types of insulation materials such as foam, granular particles, gel, and inert gas or vacuum. A pipeline-bundled system—a special configuration of PIP insulation—can be used to group individual flowlines together to form a bundle (McKelvie, 2000); heat-up lines can be included in the bundle, if necessary. The complete bundle may be transported to site and installed with a considerable cost savings relative to other methods. The extra steel required for the carrier pipe and spacers can sometimes be justified (Bai, 2001). Wet-pipeline insulations are those materials that do not need an exterior steel barrier to prevent water ingress, or the water ingress is negligible and does not degrade the insulation properties. The most common types of this are as follows: . Polyurethane . Polypropylene . Syntactic polyurethane . Syntactic polypropylene . Multilayered The main materials that have been used for deepwater insu- lations have been polyurethane and polypropylene based. Syntactic versions use plastic or glass matrix to improve insulation with greater depth capabilities. Insulation coat- ings with combinations of the two materials have also been used. Guo et al. (2005) gives the properties of these wet insulations. Because the insulation is buoyant, this effect must be compensated by the steel pipe weight to obtain lateral stability of the deepwater pipeline on the seabed. 11.4.2.2.2 Heat Transfer Models Heat transfer across the insulation of pipelines presents a unique problem affecting flow efficiency. Although sophisticated computer packages are available for predicting fluid temperatures, their accuracies suffer from numerical treatments because long pipe segments have to be used to save computing time. This is especially true for transient fluid-flow analyses in which a very large number of numerical iterations are performed. Ramey (1962) was among the first investigators who stud- ied radial-heat transfer across a well casing with no insula- tion. He derived a mathematical heat-transfer model for an outer medium that is infinitely large. Miller (1980) analyzed heat transfer around a geothermal wellbore without ins- ulation. Winterfeld (1989) and Almehaideb et al. (1989) considered temperature effect on pressure-transient analyses in well testing. Stone et al. (1989) developed a numerical simulator to couple fluid flow and heat flow in a wellbore and reservoir. More advanced studies on the wellbore heat- transfer problem were conducted by Hasan and Kabir (1994, 2002), Hasan et al. (1997, 1998), and Kabir et al. (1996). Although multilayers of materials have been considered in these studies, the external temperature gradient in the longi- tudinal direction has not been systematically taken into ac- count. Traditionally, if the outer temperature changes with length, the pipe must be divided into segments, with assumed constant outer temperature in each segment, and numerical algorithms are required for heat-transfer computation. The accuracy of the computation depends on the number of segments used. Fine segments can be employed to ensure accuracy with computing time sacrificed. Guo et al. (2006) presented three analytical heat-transfer solutions. They are the transient-flow solution for startup mode, steady-flow solution for normal operation mode, and transient-flow solution for flow rate change mode (shutting down is a special mode in which the flow rate changes to zero). Temperature and Heat Transfer for Steady Fluid Flow. The internal temperature profile under steady fluid-flow conditions is expressed as T ¼ 1 a2 b À abL À ag À eÀa(LþC)  à , (11:133) where the constant groups are defined as a ¼ 2pRk vrCpsA , (11:134) b ¼ aG cos (u), (11:135) g ¼ ÀaT0, (11:136) and C ¼ À 1 a ln (b À a2 Ts À ag), (11:137) where T is temperature inside the pipe, L is longitudinal distance from the fluid entry point, R is inner radius of insulation layer, k is the thermal conductivity of the insulation material, v is the average flow velocity of fluid in the pipe, r is fluid density, Cp is heat capacity of fluid at constant pressure, s is thickness of the insulation layer, A is the inner cross-sectional area of pipe, G is principal thermal-gradient outside the insulation, u is the angle be- tween the principal thermal gradient and pipe orientation, T0 is temperature of outer medium at the fluid entry location, and Ts is temperature of fluid at the fluid entry point. The rate of heat transfer across the insulation layer over the whole length of the pipeline is expressed as q ¼ À 2pRk s  T0L À G cos (u) 2 L2 À 1 a2 (b À ag)L À ab 2 L2 þ 1 a eÀa(LþC) À eÀaC  à gÞ, (11:138) where q is the rate of heat transfer (heat loss). Transient Temperature During Startup. The internal temperature profile after starting up a fluid flow is expressed as follows: T ¼ 1 a2 {b À abL À ag À eÀa[Lþf (LÀvt)] }, (11:139) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 153 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/153
- 159. where the function f is given by f (L À vt) ¼ À(L À vt) À 1 a ln {b À ab(L À vt) À ag À a2 [Ts À G cos (u)(L À vt)]} (11:140) and t is time. Transient Temperature During Flow Rate Change. Suppose that after increasing or decreasing the flow rate, the fluid has a new velocity v’ in the pipe. The internal temperature profile is expressed as follows: T ¼ 1 a02 {b0 À a0 b0 L À a0 g0 À eÀa0 [Lþf (LÀv0 t)] }, (11:141) where a0 ¼ 2pRk v0rCpsA , (11:142) b0 ¼ a0 G cos (u), (11:143) g0 ¼ Àa0 T0, (11:144) and the function f is given by f (L À v0 t) ¼ À(L À v0 t) À 1 a0 ln (b0 À a0 b0 (L À v0 t) À a0 g0 À a0 a 2 {b À ab(L À v0 t) À ag À eÀa[(LÀv0 t)þC] }Þ: (11:145) Example Problem 11.7 A design case is shown in this example. Design base for a pipeline insulation is presented in Table 11.6. The design criterion is to ensure that the temperature at any point in the pipeline will not drop to less than 25 8C, as required by flow assurance. Insulation materials considered for the project were polyethylene, polypropylene, and polyurethane. Solution A polyethylene layer of 0.0254 M (1 in.) was first considered as the insulation. Figure 11.13 shows the temperature profiles calculated using Eqs. (11.133) and (11.139). It indicates that at approximately 40 minutes after startup, the transient-temperature profile in the pipeline will approach the steady-flow temperature profile. The temperature at the end of the pipeline will be slightly lower than 20 8C under normal operating conditions. Obviously, this insulation option does not meet design criterion of 25 8C in the pipeline. Figure 11.14 presents the steady-flow temperature pro- files calculated using Eq. (11.133) with polyethylene layers of four thicknesses. It shows that even a polyethylene layer 0.0635-M (2.5-in.) thick will still not give a pipeline tem- perature higher than 25 8C; therefore, polyethylene should not be considered in this project. A polypropylene layer of 0.0254 M (1 in.) was then considered as the insulation. Figure 11.15 illustrates the temperature profiles calculated using Eq. (11.133) and (11.139). It again indicates that at approximately 40 min- utes after startup, the transient-temperature profile in the pipe will approach the steady-flow temperature profile. The temperature at the end of the pipeline will be approximately 22.5 8C under normal operating conditions. Obviously, this insulation option, again, does not meet design criterion of 25 8C in the pipeline. Figure 11.16 demonstrates the steady-flow temperature profiles calculated using Eq. (11.133) with polypropylene layers of four thicknesses. It shows that a polypropylene layer of 0.0508 M (2.0 in.) or thicker will give a pipeline temperature of higher than 25 8C. A polyurethane layer of 0.0254 M (1 in.) was also considered as the insulation. Figure 11.17 shows the tem- perature profiles calculated using Eqs. (11.133) and (11.139). It indicates that the temperature at the end of pipeline will drop to slightly lower than 25 8C under normal operating conditions. Figure 11.18 presents the steady-flow temperature profiles calculated using Eq. (11.133) with polyurethane layers of four thicknesses. It shows that a polyurethane layer of 0.0381 M (1.5 in.) 0 5 10 15 20 25 30 0 2,000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) 0 minute 10 minutes 20 minutes 30 minutes Steady flow Figure 11.13 Calculated temperature profiles with a polyethylene layer of 0.0254 M (1 in.). Table 11.6 Base Data for Pipeline Insulation Design Length of pipeline: 8,047 M Outer diameter of pipe: 0.2032 M Wall thickness: 0.00635 M Fluid density: 881 kg=M3 Fluid specific heat: 2,012 J/kg-8C Average external temperature: 10 8C Fluid temperature at entry point: 28 8C Fluid flow rate: 7,950 M3 =day Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 154 3.1.2007 8:54pm Compositor Name: SJoearun 11/154 EQUIPMENT DESIGN AND SELECTION
- 160. 19 20 21 22 23 24 25 26 27 28 29 2,0000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) s = 1.0 in s = 1.5 in s = 2.0 in s = 2.5 in Figure 11.14 Calculated steady-flow temperature profiles with polyethylene layers of various thicknesses. 0 5 10 15 20 25 30 0 minute 10 minutes 20 minutes 30 minutes Steady flow 2,0000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) Figure 11.15 Calculated temperature profiles with a polypropylene layer of 0.0254 M (1 in.). 19 20 21 22 23 24 25 26 27 28 29 2,0000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) s = 1.0 in s = 1.5 in s = 2.0 in s = 2.5 in Figure 11.16 Calculated steady-flow temperature profiles with polypropylene layers of various thicknesses. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 155 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/155
- 161. is required to keep pipeline temperatures higher than 25 8C under normal operating conditions. Therefore, either a polypropylene layer of 0.0508 M (2.0 in.) or a polyurethane layer of 0.0381 M (1.5 in.) should be chosen for insulation of the pipeline. Cost analyses can justify one of the options, which is beyond the scope of this example. The total heat losses for all the steady-flow cases were calculated with Eq. (11.138). The results are summarized in Table 11.7. These data may be used for sizing heaters for the pipeline if heating of the product fluid is necessary. Summary This chapter described oil and gas transportation systems. The procedure for selection of pumps and gas compressors were presented and demonstrated. Theory and applica- tions of pipeline design were illustrated. 0 minute 10 minutes 20 minutes 30 minutes Steady flow 0 5 10 15 20 25 30 2,0000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) Figure 11.17 Calculated temperature profiles with a polyurethane layer of 0.0254 M (1 in.). s = 1.0 in s = 1.5 in s = 2.0 in s = 2.5 in 19 20 21 22 23 24 25 26 27 28 29 2,0000 4,000 6,000 8,000 10,000 Distance (M) Temperature(ЊC) Figure 11.18 Calculated steady-flow temperature profiles with polyurethane layers of four thicknesses. Table 11.7 Calculated Total Heat Losses for the Insulated Pipelines (kW) Insulation thickness Material name (M) 0.0254 0.0381 0.0508 0.0635 Polyethylene 1,430 1,011 781 636 Polypropylene 989 685 524 424 Polyurethane 562 383 290 234 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 156 3.1.2007 8:54pm Compositor Name: SJoearun 11/156 EQUIPMENT DESIGN AND SELECTION
- 162. References almehaideb, r.a., aziz, k., and pedrosa, o.a., jr. A reser- voir/wellbore model for multiphase injection and pres- sure transient analysis. Presented at the 1989 SPE Middle East Oil Show, Bahrain, 11–14 March. Paper SPE 17941. American Gas Association. Collapse of Offshore Pipe- lines. Pipeline Research Committee, Seminar held 20 February, 1990, in Houston, Texas. American Society of Mechanical Engineers. Liquid transpor- tation systems for hydrocarbons, liquid petroleum gas, anhydrous ammonia and alcohols. ASME B31.4 1989. Washington. American Society of Mechanical Engineers. Gas Trans- mission and Distribution Piping Systems,’’ ASME Code for Pressure Piping, B31.8–(1989 Edition and 1990 Addendum). Washington. bai, y. Pipelines and Risers, Vol. 3, Ocean Engineering Book Series. Amsterdam: Elsevier, 2001. brown, g.g. A series of enthalpy-entropy charts for nat- ural gases. Trans. AIME 1945;60:65. carmichael, r., fang, j., and tam, c. Pipe-in-pipe systems for deepwater developments. Proceedings of the Deep- water Pipeline Technology Conference, in New Or- leans, 1999. carter, b., gray, c., and cai, j. 2002 survey of offshore non-chemical flow assurance solutions. Poster pub- lished by Offshore Magazine, Houston, 2003. Det norske Veritas. Rules for Submarine Pipeline Systems. 1981. guo, b. et al. Offshore Pipelines. Burlington: Gulf Profes- sional Publishing, 2005. guo, b., duan s., and ghalambor, a. A simple model for predicting heat loss and temperature profiles in insu- lated pipelines. SPE Prod. Operations J. February 2006. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston, TX: Gulf Publishing Company, 2005. hasan, a.r. and kabir, c.s. Aspects of wellbore heat trans- fer during two-phase flow. SPEPF 1994;9(2):211–218. hasan, a.r., kabir, c.s., and Wang, x. Wellbore two- phase flow and heat transfer during transient testing. SPEJ 1998:174–181. hasan, a.r., kabir, c.s., and wang, x. Development and application of a wellbore/reservoir simulator for testing oil wells. SPEFE 1997;12(3)182–189. hasan, r. and kabir, c.s. Fluid Flow and Heat Transfer in Wellbores. Richardson, TX: SPE, 2002. hill, r.t. and warwick, p.c. Internal Corrosion Allow- ance for Marine Pipelines: A Question of Validity. OTC paper No. 5268, 1986. holman, j.p. Heat Transfer. New York: McGraw-Hill Book Co., 1981. ikoku, c.u. Natural Gas Production Engineering. New York: John Wiley Sons, 1984. kabir, c.s. et al. A wellbore/reservoir simulator for testing gas well in high-temperature reservoirs. SPEFE 1996;11(2):128–135. katz, d.l. and lee, r.l. Natural Gas Engineering— Production and Storage. New York: McGraw-Hill Publishing Co., 1990. lyons, w.c. Air and Gas Drilling Manual. New York: McGraw-Hill, 2001:4–5. mckelvie, m. ‘‘Bundles—Design and Construction,’’ Inte- grated Graduate Development Scheme, Heriot-Watt U., 2000. miller, c.w. Wellbore storage effect in geothermal wells. SPEJ 1980:555. murphey, c.e. and langner, c.g. Ultimate pipe strength under bending, collapse, and fatigue. Proceedings of the OMAE Conference, 1985. ramey, h.j., jr. Wellbore heat transmission. JPT April 1962;427, Trans. AIME 14. rollins, j.p. Compressed Air and Gas Handbook. New York: Compressed Air and Gas Institute, 1973. stone, t.w., edmunds, n.r., and kristoff, b.j. A compre- hensive wellbore/reservoir simulator. Presented at the 1989 SPE Reservoir Simulation Symposium, 6–8 Feb- ruary, in Houston. Paper SPE 18419. winterfeld, p.h. Simulation of pressure buildup in a multiphase wellbore/reservoir system. SPEFE 1989; 4(2):247–252. Problems 11.1 A pipeline transporting 10,000 bbl/day of oil requires a pump with a minimum output pressure of 500 psi. The available suction pressure is 300 psi. Select a triplex pump for this operation. 11.2 A pipeline transporting 8,000 bbl/day of oil requires a pump with a minimum output pressure of 400 psi. The available suction pressure is 300 psi. Select a duplex pump for this operation. 11.3 For a reciprocating compressor, calculate the the- oretical and brake horsepower required to compress 30 MMcfd of a 0.65 specific gravity natural gas from 100 psia and 70 8F to 2,000 psia. If intercoolers and end-coolers cool the gas to 90 8F, what is the heat load on the coolers? Assuming the overall efficiency is 0.80. 11.4 For a centrifugal compressor, use the following data to calculate required input horsepower and polytro- pic head: Gas-specific gravity: 0.70 Gas-specific heat ratio: 1.30 Gas flow rate: 50 MMscfd at 14.7 psia and 60 8F Inlet pressure: 200 psia Inlet tempera- ture: 70 8F Discharge pressure: 500 psia Polytropic efficiency: Ep ¼ 061 þ 003 log (q1) 11.5 For the data given in Problem 11.4, calculate the required brake horsepower if a reciprocating com- pressor is used. 11.6 A 40-API gravity, 3-cp oil is transported through an 8-in. (I.D.) pipeline with a downhill angle of 5 de- grees across a distance of 10 miles at a flow rate of 5,000 bbl/day. Estimate the minimum required pump pressure to deliver oil at 100 psi pressure at the out- let. Assume e ¼ 0.0006 in. 11.7 For the following data given for a horizontal pipe- line, predict gas flow rate in cubic feet per hour through the pipeline. Solve the problem using Eq. (11.101) with the trial-and-error method for friction factor and the Weymouth equation without the Rey- nolds number–dependent friction factor: d ¼ 6 in. L ¼ 100 mi e ¼ 0.0006 in. T ¼ 70 8F gg ¼ 0:70 Tb ¼ 520 R Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 157 3.1.2007 8:54pm Compositor Name: SJoearun TRANSPORTATION SYSTEMS 11/157
- 163. pb ¼ 14:65 psia p1 ¼ 800 psia p2 ¼ 200 psia 11.8 Solve Problem 11.7 using a. Panhandle-A Equation b. Panhandle-B Equation 11.9 Assuming a 10-degree uphill angle, solve Problem 11.7 using the Weymouth equation. 11.10 Calculate the required wall thickness for a pipeline using the following data: Water depth 2,000 ft offshore oil field Water temperature 45 8F 12.09 in. pipe inner diameter Seamless still pipe of X-65 grade Maximum pipeline pressure 3,000 psia 11.11 Design insulation for a pipeline with the following given data: Length of pipeline: 7,000 M Outer diameter of pipe: 0.254 M Wall thickness: 0.0127M Fluid density: 800 kg=M3 Fluid specific heat: 2,000 J/kg- 8C Average external temperature: 15 8C Fluid temperature at entry point: 30 8C Fluid flow rate: 5,000 M3 =day Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap11 Final Proof page 158 3.1.2007 8:54pm Compositor Name: SJoearun 11/158 EQUIPMENT DESIGN AND SELECTION
- 164. Part III Artificial Lift Methods Most oil reservoirs are of the volumetric type where the driving mechanism is the expansion of solution gas when reservoir pressure declines because of fluid production. Oil reservoirs will eventually not be able to produce fluids at economical rates unless natural driving mechanisms (e.g., aquifer and/or gas cap) or pressure maintenance mechanisms (e.g., water flooding or gas injection) are present to maintain reservoir energy. The only way to obtain a high production rate of a well is to increase production pressure drawdown by reducing the bottom-hole pressure with artificial lift methods. Approximately 50% of wells worldwide need artificial lift systems. The commonly used artificial lift methods include the following: . Sucker rod pumping . Gas lift . Electrical submersible pumping . Hydraulic piston pumping . Hydraulic jet pumping . Plunger lift . Progressing cavity pumping Each method has applications for which it is the optimum installation. Proper selection of an artificial lift method for a given production system (reservoir and fluid properties, wellbore configuration, and surface facility restraints) requires a thorough understanding of the system. Economics analysis is always performed. Relative advantages and disadvantages of artificial lift systems are discussed in the beginning of each chapter in this part of this book. The chapters in this part provide production engineers with fundamentals of sucker rod pumping and gas lifts, as well as an introduction to other artificial lift systems. The following three chapters are included in this part of the book: Chapter 12: Sucker Rod Pumping Chapter 13: Gas Lift Chapter 14: Other Artificial Lift Methods Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 159 4.1.2007 2:43pm Compositor Name: SJoearun
- 165. 12 Sucker Rod Pumping Contents 12.1 Introduction 12/162 12.2 Pumping System 12/162 12.3 Polished Rod Motion 12/165 12.4 Load to the Pumping Unit 12/168 12.5 Pump Deliverability and Power Requirements 12/170 12.6 Procedure for Pumping Unit Selection 12/172 12.7 Principles of Pump Performance Analysis 12/174 Summary 12/179 References 12/179 Problems 12/179 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 161 4.1.2007 2:43pm Compositor Name: SJoearun
- 166. 12.1 Introduction Sucker rod pumping is also referred to as ‘‘beam pump- ing.’’ It provides mechanical energy to lift oil from bottom hole to surface. It is efficient, simple, and easy for field people to operate. It can pump a well down to very low pressure to maximize oil production rate. It is applicable to slim holes, multiple completions, and high-temperature and viscous oils. The system is also easy to change to other wells with minimum cost. The major disadvantages of beam pumping include excessive friction in crooked/ deviated holes, solid-sensitive problems, low efficiency in gassy wells, limited depth due to rod capacity, and bulky in offshore operations. Beam pumping trends include improved pump-off controllers, better gas separation, gas handling pumps, and optimization using surface and bottom-hole cards. 12.2 Pumping System As shown in Fig. 12.1, a sucker rod pumping system consists of a pumping unit at surface and a plunger pump submerged in the production liquid in the well. The prime mover is either an electric motor or an in- ternal combustion engine. The modern method is to sup- ply each well with its own motor or engine. Electric motors are most desirable because they can easily be automated. The power from the prime mover is transmitted to the input shaft of a gear reducer by a V-belt drive. The output shaft of the gear reducer drives the crank arm at a lower speed ($4–40 revolutions per minute [rpm] depending on well characteristics and fluid properties). The rotary mo- tion of the crank arm is converted to an oscillatory motion by means of the walking beam through a pitman arm. The horse’s head and the hanger cable arrangement is used to ensure that the upward pull on the sucker rod string is vertical at all times (thus, no bending moment is applied to the stuffing box). The polished rod and stuffing box com- bine to maintain a good liquid seal at the surface and, thus, force fluid to flow into the ‘‘T’’ connection just below the stuffing box. Conventional pumping units are available in a wide range of sizes, with stroke lengths varying from 12 to almost 200 in. The strokes for any pumping unit type are available in increments (unit size). Within each unit size, the stroke length can be varied within limits (about six different lengths being possible). These different lengths are achieved by varying the position of the pitman arm connection on the crank arm. Walking beam ratings are expressed in allowable pol- ished rod loads (PRLs) and vary from approximately 3,000 to 35,000 lb. Counterbalance for conventional pumping units is accomplished by placing weights directly on the beam (in smaller units) or by attaching weights to the rotating crank arm (or a combination of the two methods for larger units). In more recent designs, the rotary counterbalance can be adjusted by shifting the posi- tion of the weight on the crank by a jackscrew or rack and pinion mechanism. There are two other major types of pumping units. These are the Lufkin Mark II and the Air-Balanced Units (Fig. 12.2). The pitman arm and horse’s head are in the same side of the walking beam in these two types of units (Class III lever system). Instead of using counter-weights in Lufkin Mark II type units, air cylinders are used in the air- balanced units to balance the torque on the crankshaft. Pitman Counter weight Gear reducer V-Belt Prime mover Walking beamHorse head Bridle Polished rod Stuffing box Sampson post Crank Casing Tubing Sucker rod Downhole pump Strokelength Strokelength Oil Tee Gas Figure 12.1 A diagrammatic drawing of a sucker rod pumping system (Golan and Whitson, 1991). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 162 4.1.2007 2:43pm Compositor Name: SJoearun 12/162 ARTIFICIAL LIFT METHODS
- 167. The American Petroleum Institute (API) has established designations for sucker rod pumping units using a string of characters containing four fields. For example, C---228D---200---74: The first field is the code for type of pumping unit. C is for conventional units, A is for air-balanced units, B is for beam counterbalance units, and M is for Mark II units. The second field is the code for peak torque rating in thousands of inch-pounds and gear reducer. D stands for double-reduction gear reducer. The third field is the code for PRL rating in hundreds of pounds. The last field is the code for stroke length in inches. Pitman Well load Fulcrum Force Counter balance Walking beam Walking beam (a) (b) (c) Fulcrum Pitman Force Counter balance Well load Fulcrum Pitman Force Counterbalance Walking beam Well load Figure 12.2 Sketch of three types of pumping units: (a) conventional unit; (b) Lufkin Mark II Unit; (c) air-balanced unit. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 163 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/163
- 168. Figure 12.3 illustrates the working principle of a plunger pump. The pump is installed in the tubing string below the dynamic liquid level. It consists of a working barrel and liner, standing valve (SV), and traveling valve (TV) at the bottom of the plunger, which is connected to sucker rods. As the plunger is moved downward by the sucker rod string, the TV is open, which allows the fluid to pass through the valve, which lets the plunger move to a po- sition just above the SV. During this downward motion of the plunger, the SV is closed; thus, the fluid is forced to pass through the TV. When the plunger is at the bottom of the stroke and starts an upward stroke, the TV closes and the SV opens. As upward motion continues, the fluid in the well below the SV is drawn into the volume above the SV (fluid Tubing Sucker rods Working barrel and liner Traveling valve plunger Standing valve (a) (b) (c) (d) Figure 12.3 The pumping cycle: (a) plunger moving down, near the bottom of the stroke; (b) plunger moving up, near the bottom of the stroke; (c) plunger moving up, near the top of the stroke; (d) plunger moving down, near the top of the stroke (Nind, 1964). Tubing pump Rod pump(a) (b) Figure 12.4 Two types of plunger pumps (Nind, 1964). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 164 4.1.2007 2:43pm Compositor Name: SJoearun 12/164 ARTIFICIAL LIFT METHODS
- 169. passing through the open SV). The fluid continues to fill the volume above the SV until the plunger reaches the top of its stroke. There are two basic types of plunger pumps: tubing pump and rod pump (Fig. 12.4). For the tubing pump, the working barrel or liner (with the SV) is made up (i.e., attached) to the bottom of the production tubing string and must be run into the well with the tubing. The plunger (with the TV) is run into the well (inside the tubing) on the sucker rod string. Once the plunger is seated in the working barrel, pumping can be initiated. A rod pump (both working barrel and plunger) is run into the well on the sucker rod string and is seated on a wedged type seat that is fixed to the bottom joint of the production tubing. Plunger diameters vary from 5 ⁄8 to 45 ⁄8 in. Plunger area varies from 0:307 in:2 to 17:721 in:2 . 12.3 Polished Rod Motion The theory of polished rod motion has been established since 1950s (Nind, 1964). Figure 12.5 shows the cyclic motion of a polished rod in its movements through the stuffing box of the conventional pumping unit and the air- balanced pumping unit. Conventional Pumping Unit. For this type of unit, the acceleration at the bottom of the stroke is somewhat greater than true simple harmonic acceleration. At the top of the stroke, it is less. This is a major drawback for the conventional unit. Just at the time the TV is closing and the fluid load is being transferred to the rods, the acceleration for the rods is at its maximum. These two factors combine to create a maximum stress on the rods that becomes one of the limiting factors in designing an installation. Table 12.1 shows dimensions of some API conventional pumping units. Parameters are defined in Fig. 12.6. Air-Balanced Pumping Unit. For this type of unit, the maximum acceleration occurs at the top of the stroke (the acceleration at the bottom of the stroke is less than simple harmonic motion). Thus, a lower maximum stress is set up in the rod system during transfer of the fluid load to the rods. The following analyses of polished rod motion apply to conventional units. Figure 12.7 illustrates an approximate motion of the connection point between pitman arm and walking beam. If x denotes the distance of B below its top position C and is measured from the instant at which the crank arm and pitman arm are in the vertical position with the crank arm vertically upward, the law of cosine gives (AB)2 ¼ (OA)2 þ (OB)2 À 2(OA)(OB) cos AOB, that is, h2 ¼ c2 þ (h þ c À x)2 À 2c(h þ c À x) cos vt, where v is the angular velocity of the crank. The equation reduces to x2 À 2x[h þ c(1 À cos vt)] þ 2c(h þ c)(1 À cos vt) ¼ 0 so that x ¼ h þ c(1 À cos vt) Æ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 cos2 vt þ (h2 À c2) p : When vt is zero, x is also zero, which means that the negative root sign must be taken. Therefore, x ¼ h þ c(1 À cos vt) À ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 cos2 vt þ (h2 þ c2) p : Acceleration is Polished rod position Polished rod Polished rod d1 d2 d2 d1 Conventional unit Conventional unit Air-balanced unit Crank angle, degrees 0 (b) (a) 90 180 270 360 Air-balanced unit Simple harmonic motion Figure 12.5 Polished rod motion for (a) conventional pumping unit and (b) air-balanced unit (Nind, 1964). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 165 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/165
- 170. G I HR P A C Figure 12.6 Definitions of conventional pumping unit API geometry dimensions. Table 12.1 Conventional Pumping Unit API Geometry Dimensions API Unit designation A (in.) C (in.) I (in.) P (in.) H (in.) G (in.) R1, R2, R3 (in.) Cs (lb) Torque factor C-912D-365-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 À1,500 80.32 C-912D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 À1,500 80.32 C-640D-365-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 À1,500 80.32 C-640D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 À1,500 80.32 C-456D-305-168 210 120.03 120 148.5 237.88 86.88 47, 41, 35 À1,500 80.32 C-912D-427-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 À650 68.82 C-912D-365-144 180 120.03 120 148.5 237.88 86.88 47, 41, 35 À650 68.82 C-640D-365-144 180 120.03 120 148.5 238.88 89.88 47, 41, 35 À650 68.82 C-640D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 À520 68.45 C-456D-305-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 À520 68.45 C-640D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 À400 68.45 C-456D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 À400 68.45 C-320D-256-144 180 120.08 120 144.5 238.88 89.88 47, 41, 35 À400 68.45 C-456D-365-120 152 120.03 120 148.5 238.88 89.88 47, 41, 35 570 58.12 C-640D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 À120 57.02 C-456D-305-120 155 111.09 111 133.5 213 75 42, 36, 30 À120 57.02 C-320D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05 C-456D-256-120 155 111.07 111 132 211 75 42, 36, 30 55 57.05 C-456D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05 C-320D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05 C-228D-213-120 155 111.07 111 132 211 75 42, 36, 30 0 57.05 C-456D-265-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48 C-320D-265-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48 C-320D-305-100 129 111.07 111 132 211 75 42, 36, 30 550 47.48 C-228D-213-100 129 96.08 96 113 180 63 37, 32, 27 0 48.37 C-228D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37 C-160D-173-100 129 96.05 96 114 180 63 37, 32, 27 0 48.37 C-320D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96 C-228D-246-86 111 111.04 111 133 211 75 42, 36, 30 800 40.96 C-320D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-228D-213-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-160D-173-86 111 96.05 96 114 180 63 37, 32, 27 450 41.61 C-114D-119-86 111 84.05 84 93.75 150.13 53.38 32, 27, 22 115 40.98 C-320D-245-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-228D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-160D-200-74 96 96.05 96 114 180 63 37, 32, 27 800 35.99 C-228D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49 C-160D-173-74 96 84.05 84 96 152.38 53.38 32, 27, 22 450 35.49 C-160D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49 C-114D-143-74 96 84.05 84 93.75 150.13 53.38 32, 27, 22 300 35.49 C-160D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02 C-114D-173-64 84 84.05 84 93.75 150.13 53.38 32, 27, 22 550 31.02 C-160D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59 C-114D-143-64 84 72.06 72 84 132 45 27, 22, 17 360 30.59 C-80D-119-64 84 64 64 74.5 116 41 24, 20, 16 0 30.85 C-160D-173-54 72 72.06 72 84 132 45 27, 22, 17 500 26.22 (Continued) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 166 4.1.2007 2:43pm Compositor Name: SJoearun 12/166 ARTIFICIAL LIFT METHODS
- 171. a ¼ d2 x dt2 : Carrying out the differentiation for acceleration, it is found that the maximum acceleration occurs when vt is equal to zero (or an even multiple of p radians) and that this maximum value is amax ¼ v2 c(1 þ c h ): (12:1) It also appears that the minimum value of acceleration is amin ¼ v2 c(1 À c h ): (12:2) If N is the number of pumping strokes per minute, then Table 12.1 Conventional Pumping Unit API Geometry Dimensions (Continued) API Unit designation A (in.) C (in.) I (in.) P (in.) H (in.) G (in.) R1, R2, R3 (in.) Cs (lb) Torque factor C-114D-133-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45 C-80D-133-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45 C-80D-119-54 72 64 64 74.5 116 41 24, 20, 16 330 26.45 C-P57D-76-54 64 51 51 64 103 39 21, 16, 11 105 25.8 C-P57D-89-54 64 51 51 64 103 39 21, 16, 11 105 25.8 C-80D-133-48 64 64 64 74.5 116 41 24, 20, 16 440 23.51 C-80D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3 C-57D-109-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3 C-57D-95-48 64 56.05 56 65.63 105 37 21, 16, 11 320 23.3 C-P57D-109-48 57 51 51 64 103 39 21, 16, 11 180 22.98 C-P57D-95-48 57 51 51 64 103 39 21, 16, 11 180 22.98 C-40D-76-48 64 48.17 48 57.5 98.5 37 18, 14, 10 0 23.1 C-P40D-76-48 61 47 47 56 95 39 18, 14, 10 190 22.92 C-P57D-89-42 51 51 51 64 103 39 21, 16, 11 280 20.56 C-P57D-76-42 51 51 51 64 103 39 21, 16, 11 280 20.56 C-P40D-89-42 53 47 47 56 95 39 18, 14, 10 280 19.92 C-P40D-76-42 53 47 47 56 95 39 18, 14, 10 280 19.92 C-57D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-57D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-40D-89-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-40D-76-42 56 48.17 48 57.5 98.5 37 18, 14, 10 150 20.27 C-40D-89-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37 C-P40D-89-36 47 47 47 56 95 39 18, 14, 10 375 17.66 C-25D-67-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37 C-25D-56-36 48 48.17 48 57.5 98.5 37 18, 14, 10 275 17.37 C-25D-67-30 45 36.22 36 49.5 84.5 31 12, 8 150 14.53 C-25D-53-30 45 36.22 36 49.5 84.5 31 12, 9 150 14.53 C B x h c O Pitman arm Crank arm AB = length of pitman arm (h) OA = length of crank arm (c) OB = distance from center O to pitman arm-walking beam connection at B A wt w Figure 12.7 Approximate motion of connection point between pitman arm and walking beam (Nind, 1964). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 167 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/167
- 172. v ¼ 2pN 60 (rad=sec): (12:3) The maximum downward acceleration of point B (which occurs when the crank arm is vertically upward) is amax ¼ cN2 91:2 1 þ c h (ft=sec2 ) (12:4) or amax ¼ cN2 g 2936:3 1 þ c h (ft=sec2 ): (12:5) Likewise the minimum upward (amin) acceleration of point B (which occurs when the crank arm is vertically down- ward) is amin ¼ cN2 g 2936:3 1 À c h (ft=sec2 ): (12:6) It follows that in a conventional pumping unit, the max- imum upward acceleration of the horse’s head occurs at the bottom of the stroke (polished rod) and is equal to amax ¼ d1 d2 cN2 g 2936:3 1 þ c h (ft=sec2 ), (12:7) where d1 and d2 are shown in Fig. 12.5. However, 2cd2 d1 ¼ S, where S is the polished rod stroke length. So if S is mea- sured in inches, then 2cd2 d1 ¼ S 12 or cd2 d1 ¼ S 24 : (12:8) So substituting Eq. (12.8) into Eq. (12.7) yields amax ¼ SN2 g 70471:2 1 þ c h (ft=sec2 ), (12:9) or we can write Eq. (12.9) as amax ¼ SN2 g 70,471:2 M(ft=sec2 ), (12:10) where M is the machinery factor and is defined as M ¼ 1 þ c h : (12:11) Similarly, amin ¼ SN2 g 70471:2 1 À c h (ft=sec2 ): (12:12) For air-balanced units, because of the arrangements of the levers, the acceleration defined in Eq. (12.12) occurs at the bottom of the stroke, and the acceleration defined in Eq. (12.9) occurs at the top. With the lever system of an air- balanced unit, the polished rod is at the top of its stroke when the crank arm is vertically upward (Fig. 12.5b). 12.4 Load to the Pumping Unit The load exerted to the pumping unit depends on well depth, rod size, fluid properties, and system dynamics. The maximum PRL and peak torque are major concerns for pumping unit. 12.4.1 Maximum PRL The PRL is the sum of weight of fluid being lifted, weight of plunger, weight of sucker rods string, dynamic load due to acceleration, friction force, and the up-thrust from below on plunger. In practice, no force attributable to fluid acceleration is required, so the acceleration term involves only acceleration of the rods. Also, the friction term and the weight of the plunger are neglected. We ignore the reflective forces, which will tend to underestimate the maximum PRL. To compensate for this, we set the up- thrust force to zero. Also, we assume the TV is closed at the instant at which the acceleration term reaches its maxi- mum. With these assumptions, the PRLmax becomes PRLmax ¼ Sf (62:4)D( Ap À Ar) 144 þ gsDAr 144 þ gsDAr 144 SN2 M 70,471:2 , (12:13) where Sf ¼ specific gravity of fluid in tubing D ¼ length of sucker rod string (ft) Ap ¼ gross plunger cross-sectional area (in:2 ) Ar ¼ sucker rod cross-sectional area (in:2 ) gs ¼ specific weight of steel (490 lb=ft3 ) M ¼ Eq. (12.11). Note that for the air-balanced unit, M in Eq. (12.13) is replaced by 1-c/h. Equation (12.13) can be rewritten as PRLmax ¼ Sf (62:4) DAp 144 À Sf (62:4) DAr 144 þ gsDAr 144 þ gsDAr 144 SN2 M 70,471:2 : (12:14) If the weight of the rod string in air is Wr ¼ gsDAr 144 , (12:15) which can be solved for Ar, which is Ar ¼ 144Wr gsD : (12:16) Substituting Eq. (12.16) into Eq. (12.14) yields PRLmax ¼ Sf (62:4) DAp 144 À Sf (62:4) Wr gs þ Wr þ Wr SN2 M 70,471:2 : (12:17) The above equation is often further reduced by taking the fluid in the second term (the subtractive term) as an 50 8API with Sf ¼ 0.78. Thus, Eq. (12.17) becomes (where gs ¼ 490) PRLmax ¼ Sf (62:4) DAp 144 À 0:1Wr þ Wr þ Wr SN2 M 70,471:2 or PRLmax ¼ Wf þ 0:9Wr þ Wr SN2 M 70,471:2 , (12:18) where Wf ¼ Sf (62:4) DAp 144 and is called the fluid load (not to be confused with the actual fluid weight on the rod string). Thus, Eq. (12.18) can be rewritten as PRLmax ¼ Wf þ (0:9 þ F1)Wr, (12:19) where for conventional units F1 ¼ SN2 (1 þ c h ) 70,471:2 (12:20) and for air-balanced units F1 ¼ SN2 (1 À c h ) 70,471:2 : (12:21) 12.4.2 Minimum PRL The minimum PRL occurs while the TV is open so that the fluid column weight is carried by the tubing and not Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 168 4.1.2007 2:43pm Compositor Name: SJoearun 12/168 ARTIFICIAL LIFT METHODS
- 173. the rods. The minimum load is at or near the top of the stroke. Neglecting the weight of the plunger and friction term, the minimum PRL is PRLmin ¼ ÀSf (62:4) Wr gs þ Wr À WrF2, which, for 50 8API oil, reduces to PRLmin ¼ 0:9Wr À F2Wr ¼ (0:9 À F2)Wr, (12:22) where for the conventional units F2 ¼ SN2 (1 À c h ) 70,471:2 (12:23) and for air-balanced units F2 ¼ SN2 (1 þ c h ) 70,471:2 : (12:24) 12.4.3 Counterweights To reduce the power requirements for the prime mover, a counterbalance load is used on the walking beam (small units) or the rotary crank. The ideal counterbalance load C is the average PRL. Therefore, C ¼ 1 2 (PRLmax þ PRLmin): Using Eqs. (12.19) and (12.22) in the above, we get C ¼ 1 2 Wf þ 0:9Wr þ 1 2 (F1 À F2)Wr (12:25) or for conventional units C ¼ 1 2 Wf þ Wr 0:9 þ SN2 70,471:2 c h (12:26) and for air-balanced units C ¼ 1 2 Wf þ Wr 0:9 À SN2 70,471:2 c h : (12:27) The counterbalance load should be provided by structure unbalance and counterweights placed at walking beam (small units) or the rotary crank. The counterweights can be selected from manufacturer’s catalog based on the cal- culated C value. The relationship between the counterbal- ance load C and the total weight of the counterweights is C ¼ Cs þ Wc r c d1 d2 , where Cs ¼ structure unbalance, lb Wc ¼ total weight of counterweights, lb r ¼ distance between the mass center of counter- weights and the crank shaft center, in. 12.4.4 Peak Torque and Speed Limit The peak torque exerted is usually calculated on the most severe possible assumption, which is that the peak load (polished rod less counterbalance) occurs when the ef- fective crank length is also a maximum (when the crank arm is horizontal). Thus, peak torque T is (Fig. 12.5) T ¼ c C À (0:9 À F2)Wr½ Š d2 d1 : (12:28) Substituting Eq. (12.25) into Eq. (12.28) gives T ¼ 1 2 S C À (0:9 À F2)Wr½ Š (12:29) or T ¼ 1 2 S 1 2 Wf þ 1 2 (F1 þ F2)Wr ! or T ¼ 1 4 S Wf þ 2SN2 Wr 70,471:2 (in:-lb): (12-30) Because the pumping unit itself is usually not perfectly balanced (Cs 6¼ 0), the peak torque is also affected by structure unbalance. Torque factors are used for correc- tion: T ¼ 1 2 PRLmax(TF1) þ PRLmin(TF2)½ Š 0:93 , (12:31) where TF1 ¼ maximum upstroke torque factor TF2 ¼ maximum downstroke torque factor 0.93 ¼ system efficiency. For symmetrical conventional and air-balanced units, TF ¼ TF1 ¼ TF2. There is a limiting relationship between stroke length and cycles per minute. As given earlier, the maximum value of the downward acceleration (which occurs at the top of the stroke) is equal to amax = min ¼ SN2 g 1 Æ c h À Á 70,471:2 , (12:32) (the + refers to conventional units or air-balanced units, see Eqs. [12.9] and [12.12]). If this maximum acceleration divided by g exceeds unity, the downward acceleration of the hanger is greater than the free-fall acceleration of the rods at the top of the stroke. This leads to severe pounding when the polished rod shoulder falls onto the hanger (leading to failure of the rod at the shoulder). Thus, a limit of the above downward acceleration term divided by g is limited to approximately 0.5 (or where L is deter- mined by experience in a particular field). Thus, SN2 1 Æ c h À Á 70,471:2 #L (12:33) or Nlimit ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 70,471:2L S(1 Ç c h ) s : (12:34) For L ¼ 0.5, Nlimit ¼ 187:7 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S(1 Ç c h ) p : (12:35) The minus sign is for conventional units and the plus sign for air-balanced units. 12.4.5 Tapered Rod Strings For deep well applications, it is necessary to use a tapered suckerrodstringstoreducethePRLatthesurface.Thelarger diameter rod is placed at the top of the rod string, then the next largest, and then the least largest. Usually these are in sequences up to four different rod sizes. The tapered rod strings are designated by 1/8-in. (in diameter) increments. Taperedrodstringscanbeidentifiedbytheirnumberssuchas a. No. 88 is a nontapered 8 ⁄8 - or 1-in. diameter rod string b. No. 76 is a tapered string with 7 ⁄8 -in. diameter rod at the top, then a 6 ⁄8 -in. diameter rod at the bottom. c. No. 75 is a three-way tapered string consisting of 7 ⁄8 -in. diameter rod at top 6 ⁄8 -in. diameter rod at middle 5 ⁄8 -in. diameter rod at bottom d. No. 107 is a four-way tapered string consisting of 10 ⁄8 -in. (or 11 ⁄4 -in.) diameter rod at top 9 ⁄8 -in. (or 11 ⁄8 -in.) diameter rod below 10 ⁄8 -in. diameter rod 8 ⁄8 -in. (or 1-in.) diameter rod below 9 ⁄8 -in. diameter rod 7 ⁄8 -in. diameter rod below 8 ⁄8 -in. diameter rod Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 169 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/169
- 174. Tapered rod strings are designed for static (quasi-static) lads with a sufficient factor of safety to allow for random low level dynamic loads. Two criteria are used in the design of tapered rod strings: 1. Stress at the top rod of each rod size is the same throughout the string. 2. Stress in the top rod of the smallest (deepest) set of rods should be the highest ($30,000 psi) and the stress pro- gressively decreases in the top rods of the higher sets of rods. The reason for the second criterion is that it is preferable that any rod breaks occur near the bottom of the string (otherwise macaroni). Example Problem 12.1 The following geometric dim- ensions are for the pumping unit C–320D–213–86: d1 ¼ 96:05 in. d2 ¼ 111 in. c ¼ 37 in. c/h ¼ 0.33. If this unit is used with a 21 ⁄2 -in. plunger and 7 ⁄8 -in. rods to lift 25 8API gravity crude (formation volume factor 1.2 rb/stb) at depth of 3,000 ft, answer the following questions: a. What is the maximum allowable pumping speed if L ¼ 0.4 is used? b. What is the expected maximum polished rod load? c. What is the expected peak torque? d. What is the desired counterbalance weight to be placed at the maximum position on the crank? Solution The pumping unit C–320D–213–86 has a peak torque of gearbox rating of 320,000 in.-lb, a polished rod rating of 21,300 lb, and a maximum polished rod stroke of 86 in. a. Based on the configuration for conventional unit shown in Fig. 12.5a and Table 12.1, the polished rod stroke length can be estimated as S ¼ 2c d2 d1 ¼ (2)(37) 111 96:05 ¼ 85:52 in: The maximum allowable pumping speed is N ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 70,471:2L S(1 À c h ) s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (70,471:2)(0:4) (85:52)(1 À 0:33) s ¼ 22 SPM: b. The maximum PRL can be calculated with Eq. (12.17). The 25 8API gravity has an Sf ¼ 0:9042. The area of the 21 ⁄2 -in. plunger is Ap ¼ 4:91 in:2 . The area of the 7 ⁄8 -in. rod is Ar ¼ 0:60 in:2 . Then Wf ¼ Sf (62:4) DAp 144 ¼ (0:9042)(62:4) (3,000)(4:91) 144 ¼ 5,770 lbs Wr ¼ gsDAr 144 ¼ (490)(3,000)(0:60) 144 ¼ 6,138 lbs F1 ¼ SN2 1 þ c h À Á 70,471:2 ¼ (85:52)(22)2 (1 þ 0:33) 70,471:2 ¼ 0:7940: Then the expected maximum PRL is PRLmax ¼ Wf À Sf (62:4) Wr gs þ Wr þ WrF1 ¼ 5,770 À (0:9042)(62:4)(6,138)=(490) þ 6,138 þ (6,138)(0:794) ¼ 16,076 lbs 21,300 lb : c. The peak torque is calculated by Eq. (12.30): T ¼ 1 4 S Wf þ 2SN2 Wr 70,471:2 ¼ 1 4 (85:52) 5,770 þ 2(85:52)(22)2 (6,138) 70,471:2 ! ¼ 280,056 lb-in: 320,000 lb-in: d. Accurate calculation of counterbalance load requires the minimum PRL: F2 ¼ SN2 (1 À c h ) 70,471:2 ¼ (85:52)(22)2 (1 À 0:33) 70,471:2 ¼ 0:4 PRLmin ¼ ÀSf (62:4) Wr gs þ Wr À WrF2 ¼ À(0:9042)(62:4) 6,138 490 þ 6,138 À (6,138)(0:4) ¼ 2,976 lb C ¼ 1 2 (PRLmax þ PRLmin) ¼ 1 2 (16,076 þ 2,976) ¼ 9,526 lb: A product catalog of LUFKIN Industries indicates that the structure unbalance is 450 lb and 4 No. 5ARO coun- terweights placed at the maximum position (c in this case) on the crank will produce an effective counterbalance load of 10,160 lb, that is, Wc (37) (37) (96:05) (111) þ 450 ¼ 10,160, which gives Wc ¼ 11,221 lb. To generate the ideal counter- balance load of C ¼ 9,526 lb, the counterweights should be placed on the crank at r ¼ (9,526)(111) (11,221)(96:05) (37) ¼ 36:30 in: The computer program SuckerRodPumpingLoad.xls can be used for quickly seeking solutions to similar problems. It is available from the publisher with this book. The solution is shown in Table 12.2. 12.5 Pump Deliverability and Power Requirements Liquid flow rate delivered by the plunger pump can be expressed as q ¼ Ap 144 N Sp 12 Ev Bo (24)(60) 5:615 (bbl=day) or q ¼ 0:1484 ApNSpEv Bo (stb=day), where Sp is the effective plunger stroke length (in.), Ev is the volumetric efficiency of the plunger, and Bo formation volume factor of the fluid. 12.5.1 Effective Plunger Stroke Length The motion of the plunger at the pump-setting depth and the motion of the polished rod do not coincide in time and in magnitude because sucker rods and tubing strings are elastic. Plunger motion depends on a number of factors including polished rod motion, sucker rod stretch, and Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 170 4.1.2007 2:43pm Compositor Name: SJoearun 12/170 ARTIFICIAL LIFT METHODS
- 175. tubing stretch. The theory in this subject has been well established (Nind, 1964). Two major sources of difference in the motion of the pol- ishedrodandtheplungerareelasticstretch(elongation)ofthe rod string and overtravel. Stretch is caused by the periodic transfer of the fluid load from the SV to the TV and back again. The result is a function of the stretch of the rod string and the tubing string. Rod string stretch is caused by the weight of the fluid column in the tubing coming on to the rodstringatthebottomofthestrokewhentheTVcloses(this load is removed from the rod string at the top of the stroke whentheTVopens).Itisapparentthattheplungerstrokewill be less than the polished rod stroke length S by an amount equal to the rod stretch. The magnitude of the rod stretch is dlr ¼ Wf Dr ArE , (12:36) where Wf ¼ weight of fluid (lb) Dr ¼ length of rod string (ft) Ar ¼ cross-sectional area of rods (in:2 ) E ¼ modulus of elasticity of steel (30 Â 106 lb=in:2 ). Tubing stretch can be expressed by a similar equation: dlt ¼ Wf Dt AtE (12:37) But because the tubing cross-sectional area At is greater than the rod cross-sectional area Ar, the stretch of the tubing is small and is usually neglected. However, the tub- ing stretch can cause problems with wear on the casing. Thus, for this reason a tubing anchor is almost always used. Plunger overtravel at the bottom of the stroke is a result of the upward acceleration imposed on the downward- moving sucker rod elastic system. An approximation to the extent of the overtravel may be obtained by consider- ing a sucker rod string being accelerated vertically upward at a rate n times the acceleration of gravity. The vertical force required to supply this acceleration is nWr. The magnitude of the rod stretch due to this force is dlo ¼ n WrDr ArE (ft): (12:38) But the maximum acceleration term n can be written as n ¼ SN2 1 Æ c h À Á 70,471:2 so that Eq. (12.38) becomes dlo ¼ WrDr ArE SN2 1 Æ c h À Á 70,471:2 (ft), (12:39) where again the plus sign applies to conventional units and the minus sign to air-balanced units. Table 12.2 Solution Given by Computer Program SuckerRodPumpingLoad.xls SuckerRodPumpingLoad.xls Description: This spreadsheet calculates the maximum allowable pumping speed, the maximum PRL, the minimum PRL, peak torque, and counterbalance load. Instruction: (1) Update parameter values in the Input section; and (2) view result in the Solution section. Input data Pump setting depth (D): 3,000 ft Plunger diameter (dp): 2.5 in. Rod section 1, diameter (dr1): 1 in. Length (L1): 0 ft Rod section 2, diameter (dr2): 0.875 in. Length (L2): 3,000 ft Rod section 3, diameter (dr3): 0.75 in. Length (L3): 0 ft Rod section 4, diameter (dr4): 0.5 in. Length (L4): 0 ft Type of pumping unit (1 ¼ conventional; À1 ¼ Mark II or Air-balanced): 1 Beam dimension 1 (d1) 96.05 in. Beam dimension 2 (d2) 111 in. Crank length (c): 37 in. Crank to pitman ratio (c/h): 0.33 Oil gravity (API): 25 8API Maximum allowable acceleration factor (L): 0.4 Solution S ¼ 2c d2 d1 ¼ 85.52 in. N ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 70471:2L S(1Àc h) q ¼ 22 SPM Ap ¼ pd2 p 4 ¼ 4.91 in:2 Ar ¼ pd2 r 4 ¼ 0.60 in. Wf ¼ Sf (62:4) DAp 144 ¼ 5,770 lb Wr ¼ gsDAr 144 ¼ 6,138 lb F1 ¼ SN2 (1Æc h) 70,471:2 ¼ 0.7940 8 PRLmax ¼ Wf À Sf (62:4) Wr gs þ Wr þ WrF1 ¼ 16,076 lb T ¼ 1 4 S Wf þ 2SN2Wr 70,471:2 ¼ 280,056 lb F2 ¼ SN2 (1Çc h) 70,471:2 ¼ 0.40 PRLmin ¼ ÀSf (62:4) Wr gs þ Wr À WrF2 ¼ 2,976 lb C ¼ 1 2 (PRLmax þ PRLmin) ¼ 9,526 lb Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 171 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/171
- 176. Let us restrict our discussion to conventional units. Then Eq. (12.39) becomes dlo ¼ WrDr ArE SN2 M 70,471:2 (ft): (12:40) Equation (12.40) can be rewritten to yield dlo in inches. Wr is Wr ¼ gsArDr and gS ¼ 490 lb=ft3 with E ¼ 30 Â 106 lb=m2 . Eq. (12.40) becomes dlo ¼ 1:93 Â 10À11 D2 r SN2 M(in:), (12:41) which is the familiar Coberly expression for overtravel (Coberly, 1938). Plunger stroke is approximated using the above expres- sions as Sp ¼ S À dlr À dlt þ dlo or Sp ¼ S À 12D E Â Wf 1 Ar þ 1 At À SN2 M 70,471:2 Wr Ar ! (in:): (12:42) If pumping is carried out at the maximum permissible speed limited by Eq. (12.34), the plunger stroke becomes Sp ¼ S À 12D E Â Wf 1 Ar þ 1 At À 1 þ c h 1 À c h LWr Ar ! (in:): (12:43) For the air-balanced unit, the term 1þc h 1Àc h is replaced by its reciprocal. 12.5.2 Volumetric Efficiency Volumetric efficiency of the plunger mainly depends on the rate of slippage of oil past the pump plunger and the solution–gas ratio under pump condition. Metal-to-metal plungers are commonly available with plunger-to-barrel clearance on the diameter of À0.001, À0.002, À0.003, À0.004, and À0.005 in. Such fits are re- ferred to as À1, À2, À3, À4, and À5, meaning the plunger outside diameter is 0.001 in. smaller than the barrel inside diameter. In selecting a plunger, one must consider the viscosity of the oil to be pumped. A loose fit may be acceptable for a well with high viscosity oil (low 8API gravity). But such a loose fit in a well with low viscosity oil may be very inefficient. Guidelines are as follows: a. Low-viscosity oils (1–20 cps) can be pumped with a plunger to barrel fit of À0.001 in. b. High-viscosity oils (7,400 cps) will probably carry sand in suspension so a plunger-to-barrel fit or approxi- mately 0.005 in. can be used. An empirical formula has been developed that can be used to calculate the slippage rate, qs (bbl/day), through the annulus between the plunger and the barrel: qs ¼ kp m db À dp À Á2:9 db þ dp À Á d0:1 b Dp Lp , (12:44) where kp ¼ a constant dp ¼ plunger outside diameter (in.) db ¼ barrel inside diameter (in.) Dp ¼ differential pressure drop across plunger (psi) Lp ¼ length of plunger (in.) m ¼ viscosity of oil (cp). The value of kp is 2:77 Â 106 to 6:36 Â 106 depending on field conditions. An average value is 4:17 Â 106 . The value of Dp may be estimated on the basis of well productivity index and production rate. A reasonable estimate may be a value that is twice the production drawdown. Volumetric efficiency can decrease significantly due to the presence of free gas below the plunger. As the fluid is elevated and gas breaks out of solution, there is a significant difference between the volumetric displace- ment of the bottom-hole pump and the volume of the fluid delivered to the surface. This effect is denoted by the shrinkage factor greater than 1.0, indicating that the bottom-hole pump must displace more fluid by some additional percentage than the volume delivered to the surface (Brown, 1980). The effect of gas on volumetric efficiency depends on solution–gas ratio and bottom-hole pressure. Down-hole devices, called ‘‘gas anchors,’’ are usually installed on pumps to separate the gas from the liquid. In summary, volumetric efficiency is mainly affected by the slippage of oil and free gas volume below plunger. Both effects are difficult to quantify. Pump efficiency can vary over a wide range but are commonly 70–80%. 12.5.3 Power Requirements The prime mover should be properly sized to provide adequate power to lift the production fluid, to overcome friction loss in the pump, in the rod string and polished rod, and in the pumping unit. The power required for lifting fluid is called ‘‘hydraulic power.’’ It is usually ex- pressed in terms of net lift: Ph ¼ 7:36 Â 10À6 qglLN , (12:45) where Ph ¼ hydraulic power, hp q ¼ liquid production rate, bbl/day gl ¼ liquid specific gravity, water ¼ 1 LN ¼ net lift, ft, and LN ¼ H þ ptf 0:433gl , (12:46) where H ¼ depth to the average fluid level in the annulus, ft ptf ¼ flowing tubing head pressure, psig. The power required to overcome friction losses can be empirically estimated as Pf ¼ 6:31 Â 10À7 WrSN: (12:47) Thus, the required prime mover power can be expressed as Ppm ¼ Fs(Ph þ Pf ), (12:48) where Fs is a safety factor of 1.25–1.50. Example Problem 12.2 A well is pumped off (fluid level is the pump depth) with a rod pump described in Example Problem 12.1. A 3-in. tubing string (3.5-in. OD, 2.995 ID) in the well is not anchored. Calculate (a) expected liquid production rate (use pump volumetric efficiency 0.8), and (b) required prime mover power (use safety factor 1.35). Solution This problem can be quickly solved using the program SuckerRodPumpingFlowratePower.xls. The solution is shown in Table 12.3. 12.6 Procedure for Pumping Unit Selection The following procedure can be used for selecting a pump- ing unit: Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 172 4.1.2007 2:43pm Compositor Name: SJoearun 12/172 ARTIFICIAL LIFT METHODS
- 177. 1. From the maximum anticipated fluid production (based on IPR) and estimated volumetric efficiency, calculate required pump displacement. 2. Based on well depth and pump displacement, determine API rating and stroke length of the pumping unit to be used.ThiscanbedoneusingeitherFig.12.8orTable12.4. 3. Select tubing size, plunger size, rod sizes, and pumping speed from Table 12.4. 4. Calculate the fractional length of each section of the rod string. 5. Calculate the length of each section of the rod string to the nearest 25 ft. 6. Calculate the acceleration factor. 7. Determine the effective plunger stroke length. 8. Using the estimated volumetric efficiency, determine the probable production rate and check it against the desired production rate. 9. Calculate the dead weight of the rod string. 10. Calculate the fluid load. 11. Determine peak polished rod load and check it against the maximum beam load for the unit selected. 12. Calculate the maximum stress at the top of each rod size and check it against the maximum permissible working stress for the rods to be used. 13. Calculate the ideal counterbalance effect and check it againstthecounterbalanceavailablefortheunitselected. 14. From the manufacturer’s literature, determine the position of the counterweight to obtain the ideal coun- terbalance effect. 15. On the assumption that the unit will be no more than 5% out of counterbalance, calculate the peak torque on the gear reducer and check it against the API rating of the unit selected. 16. Calculate hydraulic horsepower, friction horsepower, and brake horsepower of the prime mover. Select the prime mover. 17. From the manufacturer’s literature, obtain the gear reduction ratio and unit sheave size for the unit selected, and the speed of the prime mover. From this, determine the engine sheave size to obtain the desired pumping speed. Example Problem 12.3 A well is to be put on a sucker rod pump. The proposed pump setting depth is 3,500 ft. The anticipated production rate is 600 bbl/day oil of 0.8 specific gravity against wellhead pressure 100 psig. It is assumed that the working liquid level is low, and a sucker rod string having a working stress of 30,000 psi is Table 12.3 Solution Given by SuckerRodPumpingFlowratePower.xls SuckerRodPumpingFlowRatePower.xls Description: This spreadsheet calculates expected deliverability and required prime mover power for a given sucker rod pumping system. Instruction: (1) Update parameter values in the Input section; and (2) view result in the Solution section. Input data Pump setting depth (D): 4,000 ft Depth to the liquid level in annulus (H): 4,000 ft Flowing tubing head pressure (ptf ): 100 ft Tubing outer diameter (dto): 3.5 in. Tubing inner diameter (dti): 2.995 in. Tubing anchor (1 ¼ yes; 0 ¼ no): 0 Plunger diameter (dp): 2.5 in. Rod section 1, diameter (dr1): 1 in. Length (L1): 0 ft Rod section 2, diameter (dr2): 0.875 in. Length (L2): 0 ft Rod section 3, diameter (dr3): 0.75 in. Length (L3): 4,000 ft Rod section 4, diameter (dr4): 0.5 in. Length (L4): 0 ft Type of pumping unit (1 ¼ conventional; À1 ¼ Mark II or Air-balanced): 1 Polished rod stroke length (S) 86 in. Pumping speed (N) 22 spm Crank to pitman ratio (c/h): 0.33 8 Oil gravity (API): 25 8API Fluid formation volume factor (Bo): 1.2 rb/stb Pump volumetric efficiency (Ev): 0.8 Safety factor to prime mover power (Fs): 1.35 Solution At ¼ pd2 t 4 ¼ 2.58 in:2 Ap ¼ pd2 p 4 ¼ 4.91 in:2 Ar ¼ pd2 r 4 ¼ 0.44 in. Wf ¼ Sf (62:4) DAp 144 ¼ 7,693 lb Wr ¼ gsDAr 144 ¼ 6,013 lb M ¼ 1 Æ c h ¼ 1.33 Sp ¼ S À 12D E Wf 1 Ar þ 1 At À SN2 M 70471:2 Wr Ar h i ¼ 70 in. q ¼ 0:1484 ApNSpEv Bo ¼ 753 sbt/day LN ¼ H þ ptf 0:433gl ¼ 4,255 ft Ph ¼ 7:36 Â 10À6 qglLN ¼ 25.58 hp Pf ¼ 6:31 Â 10À7 WrSN ¼ 7.2 hp Ppm ¼ Fs(Ph þ Pf ) ¼ 44.2 hp Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 173 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/173
- 178. to be used. Select surface and subsurface equipment for the installation. Use a safety factor of 1.35 for the prime mover power. Solution 1. Assuming volumetric efficiency of 0.8, the required pump displacement is (600)=(0:8) ¼ 750 bbl=day: 2. Based on well depth 3,500 ft and pump displacement 750 bbl/day, Fig. 12.8 suggests API pump size 320 unit with 84 in. stroke, that is, a pump is selected with the following designation: C---320D---213---86 3. Table 12.4 g suggests the following: Tubing size: 3 in. OD, 2.992 in. ID Plunger size: 21 ⁄2 in. Rod size: 7 ⁄8 in. Pumping speed: 18 spm 4. Table 12.1 gives d1 ¼ 96:05 in., d2 ¼ 111 in., c ¼ 37 in., and h ¼ 114 in., thus c/h ¼ 0.3246. The spreadsheet program SuckerRodPumpingFlowRatePower.xls gives qo ¼ 687 bbl=day 600 bbl/day Ppm ¼ 30:2 hp 5. The spreadsheet program SuckerRodPumpingLoad.xls gives PRLmax ¼ 16,121 lb PRLmin ¼ 4,533 lb T ¼ 247,755 lb 320,000 in.-lb C ¼ 10,327 lb 6. The cross-sectional area of the 7 ⁄8 -in. rod is 0.60 in.2 . Thus, the maximum possible stress in the sucker rod is smax ¼ (16,121)=(0:60) ¼ 26,809 psi 30,000 psi: Therefore, the selected pumping unit and rod meet well load and volume requirements. 7. If a LUFKIN Industries C–320D–213–86 unit is chosen, the structure unbalance is 450 lb and 4 No. 5 ARO counterweights placed at the maximum position (c in this case) on the crank will produce an effective counterbalance load of 12,630 lb, that is, Wc (37) (37) (96:05) (111) þ 450 ¼ 12,630 lb, which gives Wc ¼ 14,075 lb. To generate the ideal counter- balance load of C ¼ 10,327 lb, the counterweights should be placed on the crank at r ¼ (10,327)(111) (14,076)(96:05) (37) ¼ 31:4 in: 8. The LUFKIN Industries C–320D–213–86 unit has a gear ratio of 30.12 and unit sheave sizes of 24, 30, and 44 in. are available. If a 24-in. unit sheave and a 750- rpm electric motor are chosen, the diameter of the motor sheave is de ¼ (18)(30:12)(24) (750) ¼ 17:3 in: 12.7 Principles of Pump Performance Analysis The efficiency of sucker rod pumping units is usually ana- lyzed using the information from pump dynagraph and polisher rod dynamometer cards. Figure 12.9 shows a sche- matic of a pump dynagraph. This instrument is installed immediately above the plunger to record the plunger stroke and the loads carried by the plunger during the pump cycle. The relative motion between the cover tube (which is attached to the pump barrel and hence anchored to the tubing) and the calibrated rod (which is an integral part of the sucker rod string) is recorded as a horizontal line on the recording tube. This is achieved by having the record- ing tube mounted on a winged nut threaded onto the calibrated rod and prevented from rotating by means of 0 500 1,000 1,500 2,000 2,500 0 2,000 4,000 6,000 8,000 10,000 12,000 Pump Setting Depth (ft) PumpDisplacement(bbl/day) A B C D E F G H Curve API size Stroke 34 42 48 54 64 74 84 144 40 57 80 114 160 228 320 640 Figure 12.8 Sucker rod pumping unit selection chart (Kelley and Willis, 1954). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 174 4.1.2007 2:43pm Compositor Name: SJoearun 12/174 ARTIFICIAL LIFT METHODS
- 179. Table 12.4 Design Data for API Sucker Rod Pumping Units (a) Size 40 unit with 34-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 1,000–1,100 23 ⁄4 3 7 ⁄8 24–19 1,100–1,250 21 ⁄2 3 7 ⁄8 24–19 1,250–1,650 21 ⁄4 21 ⁄2 3 ⁄4 24–19 1,650–1,900 2 21 ⁄2 3 ⁄4 24–19 1,900–2,150 13 ⁄4 21 ⁄2 3 ⁄4 24–19 2,150–3,000 11 ⁄2 2 5 ⁄8 –3 ⁄4 24–19 3,000–3,700 11 ⁄4 2 5 ⁄8 –3 ⁄5 22–18 3,700–4,000 1 2 5 ⁄8 –3 ⁄6 21–18 (b) Size 57 unit with 42-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 1,150–1,300 23 ⁄4 3 7 ⁄8 24–19 1,300–1,450 21 ⁄2 3 7 ⁄8 24–19 1,450–1,850 21 ⁄4 21 ⁄2 3 ⁄4 24–19 1,850–2,200 2 21 ⁄2 3 ⁄4 24–19 2,200–2,500 13 ⁄4 21 ⁄2 3 ⁄4 24–19 2,500–3,400 11 ⁄2 2 5 ⁄8 –3 ⁄4 23–18 3,400–4,200 11 ⁄4 2 5 ⁄8 –3 ⁄5 22–17 4,200–5,000 1 2 5 ⁄8 –3 ⁄6 21–17 (c) Size 80 unit with 48-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 1,400–1,500 23 ⁄4 3 7 ⁄8 24–19 1,550–1,700 21 ⁄2 3 7 ⁄8 24–19 1,700–2,200 21 ⁄4 21 ⁄2 3 ⁄4 24–19 2,200–2,600 2 21 ⁄2 3 ⁄4 24–19 2,600–3,000 13 ⁄4 21 ⁄2 3 ⁄4 23–18 3,000–4,100 11 ⁄2 2 5 ⁄8 –3 ⁄4 23–19 4,100–5,000 11 ⁄4 2 5 ⁄8 –3 ⁄5 21–17 5,000–6,000 1 2 5 ⁄8 –3 ⁄6 19–17 (d) Size 114 unit with 54-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 1,700–1,900 23 ⁄4 3 7 ⁄8 24–19 1,900–2,100 21 ⁄2 3 7 ⁄8 24–19 2,100–2,700 21 ⁄4 21 ⁄2 3 ⁄4 24–19 2,700–3,300 2 21 ⁄2 3 ⁄4 23–18 3,300–3,900 13 ⁄4 21 ⁄2 3 ⁄4 22–17 3,900–5,100 11 ⁄2 2 5 ⁄8 –3 ⁄4 21–17 5,100–6,300 11 ⁄4 2 5 ⁄8 –3 ⁄5 19–16 6,300–7,000 1 2 5 ⁄8 –3 ⁄6 17–16 (e) Size 160 unit with 64-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 2,000–2,200 23 ⁄4 3 7 ⁄8 24–19 2,200–2,400 21 ⁄2 3 7 ⁄8 24–19 2,400–3,000 21 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 24–19 3,000–3,600 2 21 ⁄2 3 ⁄4 –7 ⁄8 23–18 3,600–4,200 13 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 22–17 4,200–5,400 11 ⁄2 2 5 ⁄8 –3 ⁄4 –7 ⁄8 21–17 5,400–6,700 11 ⁄4 2 5 ⁄8 –3 ⁄4 –7 ⁄8 19–15 6,700–7,700 1 2 5 ⁄8 –3 ⁄4 –7 ⁄8 17–15 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 175 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/175
- 180. (f) Size 228 unit with 74-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 2,400–2,600 23 ⁄4 3 7 ⁄8 24–20 2,600–3,000 21 ⁄2 3 7 ⁄8 23–18 3,000–3,700 21 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 22–17 3,700–4,500 2 21 ⁄2 3 ⁄4 –7 ⁄8 21–16 4,500–5,200 13 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 19–15 5,200–6,800 11 ⁄2 2 5=8---3 ⁄4 –7 ⁄8 18–14 6,800–8,000 11 ⁄4 2 5=8---3 ⁄4 –7 ⁄8 16–13 8,000–8,500 11=16 2 5=8---3 ⁄4 –7 ⁄8 14–13 (g) Size 320 unit with 84-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 2,800–3,200 23 ⁄4 3 7 ⁄8 23–18 3,200–3,600 21 ⁄2 3 7 ⁄8 21–17 3,600–4,100 21 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 21–17 4,100–4,800 2 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 20–16 4,800–5,600 13 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 19–16 5,600–6,700 11 ⁄2 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 18–15 6,700–8,000 11 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 17–13 8,000–9,500 11=16 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 14–11 (h) Size 640 unit with 144-in. stroke Pump depth (ft) Plunger size (in.) Tubing size (in.) Rod sizes (in.) Pumping speed (stroke/min) 3,200–3,500 23 ⁄4 3 7 ⁄8 –1 18–14 3,500–4,000 21 ⁄2 3 7 ⁄8 –1 17–13 4,000–4,700 21 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 16–13 4,700–5,700 2 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 15–12 5,700–6,600 13 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 14–12 6,600–8,000 11 ⁄2 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 14–11 8,000–9,600 11 ⁄4 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 13–10 9,600–11,000 11=16 21 ⁄2 3 ⁄4 –7 ⁄8 ---1 12–10 Sucker rod string Self-aligning bear Lugs Rotating tube with spiral grooves Cover tube with vertical grooves Recording tube Winged nut Stylus Lugs Tubing Calibrated rod Plunger assembly Pump liner Figure 12.9 A sketch of pump dynagraph (Nind, 1964). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 176 4.1.2007 2:43pm Compositor Name: SJoearun 12/176 ARTIFICIAL LIFT METHODS
- 181. two lugs, which are attached to the winged nut, which run in vertical grooves in the cover tube. The stylus is mounted on a third tube, which is free to rotate and is connected by a self-aligning bearing to the upper end of the calibrated rod. Lugs attached to the cover tube run in spiral grooves cut in the outer surface of the rotating tube. Consequently, vertical motion of the plunger assembly relative to the barrel results in rotation of the third tube, and the stylus cuts a horizontal line on a recording tube. Any change in plunger loading causes a change in length of the section of the calibrated rod between the winged nut supporting the recording tube and the self-aligning bearing supporting the rotating tube (so that a vertical line is cut on the recording tube by the stylus). When the pump is in operation, the stylus traces a series of cards, one on top of the other. To obtain a new series of cards, the polished rod at the well head is rotated. This rotation is transmitted to the plunger in a few pump strokes. Because the recording tube is prevented from rotating by the winged nut lugs that run in the cover tube grooves, the rotation of the sucker rod string causes the winged nut to travel—upward or downward depending on the direction of rotation—on the threaded calibrated rod. Upon the completion of a series of tests, the recording tube (which is 36 in. long) is removed. It is important to note that although the bottom-hole dynagraph records the plunger stroke and variations in plunger loading, no zero line is obtained. Thus, quantita- tive interpretation of the cards becomes somewhat specu- lative unless a pressure element is run with the dynagraph. Figure 12.10 shows some typical dynagraph card results. Card (a) shows an ideal case where instantaneous valve actions at the top and bottom of the stroke are indicated. In general, however, some free gas is drawn into the pump on the upstroke, so a period of gas compression can occur on the down-stroke before the TV opens. This is shown in card (b). Card (c) shows gas expansion during the upstroke giving a rounding of the card just as the upstroke begins. Card (d) shows fluid pounding that occurs when the well is almost pumped off (the pump displacement rate is higher than the formation of potential liquid production rate). This fluid pounding results in a rapid fall off in stress in the rod string and the sudden imposed shock to the system. Card (e) shows that the fluid pounding has progressed so that the mechanical shock causes oscillations in the sys- tem. Card (f) shows that the pump is operating at a very low volumetric efficiency where almost all the pump stroke is being lost in gas compression and expansion (no liquid is being pumped). This results in no valve action and the area between the card nearly disappears (thus, is gas locked). Usually, this gas-locked condition is only temporary, and as liquid leaks past the plunger, the volume of liquid in the pump barrel increases until the TV opens and pumping recommences. The use of the pump dynagraph involves pulling the rods and pump from the well bath to install the instrument and to recover the recording tube. Also, the dynagraph cannot be used in a well equipped with a tubing pump. Thus, the dynagraph is more a research instrument than an operational device. Once there is knowledge from a dyna- graph, surface dynamometer cards can be interpreted. The surface, or polished rod, dynamometer is a device that records the motion of (and its history) the polished rod during the pumping cycle. The rod string is forced by the pumping unit to follow a regular time versus position pattern. However, the polished rod reacts with the load- ings (on the rod string) that are imposed by the well. The surface dynamometer cards record the history of the variations in loading on the polished rod during a cycle. The cards have three principal uses: a. To obtain information that can be used to determine load, torque, and horsepower changes required of the pump equipment b. To improve pump operating conditions such as pump speed and stroke length c. To check well conditions after installation of equipment to prevent or diagnose various operating problems (like pounding, etc.) Surface instruments can be mechanical, hydraulic, and electrical. One of the most common mechanical instru- ments is a ring dynamometer installed between the hanger bar and the polished rod clamp in such a manner as the ring may carry the entire well load. The deflection of the ring is proportional to the load, and this deflection is amplified and transmitted to the recording arm by a series of levers. A stylus on the recording arm traces a record of the imposed loads on a waxed (or via an ink pen) paper card located on a drum. The loads are obtained in terms of polished rod displacements by having the drum oscillate back and forth to reflect the polished rod motion. Correct interpretation of surface dynamometer card leads to esti- mate of various parameter values. . Maximum and minimum PRLs can be read directly from the surface card (with the use of instrument cali- bration). These data then allow for the determination of the torque, counterbalance, and horsepower require- ments for the surface unit. . Rod stretch and contraction is shown on the surface dynamometer card. This phenomenon is reflected in the surface unit dynamometer card and is shown in Fig. 12.11a for an ideal case. . Acceleration forces cause the ideal card to rotate clock- wise. The PRL is higher at the bottom of the stroke and lower at the top of the stroke. Thus, in Fig. 12.11b, Point A is at the bottom of the stroke. Figure 12.10 Pump dynagraph cards: (a) ideal card, (b) gas compression on down-stroke, (c) gas expansion on upstroke, (d) fluid pound, (e) vibration due to fluid pound, (f) gas lock (Nind, 1964). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 177 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/177
- 182. . Rod vibration causes a serious complication in the in- terpretation of the surface card. This is result of the closing of the TV and the ‘‘pickup’’ of the fluid load by the rod string. This is, of course, the fluid pounding. This phenomenon sets up damped oscillation (longitu- dinal and bending) in the rod string. These oscillations result in waves moving from one end of the rod string to the other. Because the polished rod moves slower near the top and bottom of the strokes, these stress (or load) fluctuations due to vibrations tend to show up more prominently at those locations on the cards. Figure 12.11c shows typical dynamometer card with vibrations of the rod string. Figure 12.12 presents a typical chart from a strain-gage type of dynamometer measured for a conventional unit operated with a 74-in. stroke at 15.4 strokes per minute. It shows the history of the load on the polished rod as a function of time (this is for a well 825 ft in depth with a No. 86 three-tapered rod string). Figure 12.13 reproduces the data in Fig. 12.12 in a load versus displacement dia- gram. In the surface chart, we can see the peak load of 22,649 lb (which is 28,800 psi at the top of the 1-in. rod) in Fig. 12.13a. In Fig. 12.13b, we see the peak load of 17,800 lb (which is 29,600 psi at the top of the 7 ⁄8 -in. rod). In Fig. 12.13c, we see the peak load of 13,400 lb (which is 30,300 psi at the top of the 3 ⁄4 -in. rod). In Fig. 12.13d is Rod stretch (a) (b) (c) Rod contraction Zero line Zero line Zero line Zero lineZero line F A Wr-Wrb+Wf Wr-Wrb Wr-Wrb+Wf Wr-Wrb B C Figure 12.11 Surface dynamometer card: (a) ideal card (stretch and contraction), (b) ideal card (acceleration), (c) three typical cards (Nind, 1964). 0 3,000 9,000 6,000 15,000 12,000 21,000 18,000 24,000 PolishedRodLoad(lb) 0 1 2 3 4 5 6 7 8 9 8 6 4 2 0 PolishedRod Displacement(ft) Time (sec) Figure 12.12 Strain-gage–type dynamometer chart. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 178 4.1.2007 2:43pm Compositor Name: SJoearun 12/178 ARTIFICIAL LIFT METHODS
- 183. the dynagraph card at the plunger itself. This card indicates gross pump stroke of 7.1 ft, a net liquid stroke of 4.6 ft, and a fluid load of Wf ¼ 3,200 lb. The shape of the pump card, Fig. 12.13d, indicates some down-hole gas compression. The shape also indicates that the tubing anchor is holding properly. A liquid displacement rate of 200 bbl/day is cal- culated and, compared to the surface measured production of 184 bbl/day, indicated no serious tubing flowing leak. The negative in Fig. 12.13d is the buoyancy of the rod string. The information derived from the dynamometer card (dynagraph) can be used for evaluation of pump perfor- mance and troubleshooting of pumping systems. This sub- ject is thoroughly addressed by Brown (1980). Summary This chapter presents the principles of sucker rod pumping systems and illustrates a procedure for selecting components of rod pumping systems. Major tasks include calculations of polished rod load, peak torque, stresses in the rod string, pump deliverability, and counterweight placement. Opti- mization of existing pumping systems is left to Chapter 18. References brown, k.e. The Technology of Artificial Lift Methods, Vol. 2a. Tulsa, OK: Petroleum Publishing Co., 1980. coberly, c.j. Problems in modern deep-well pumping. Oil Gas J. May 12, 1938. golan, m. and whitson, c.h. Well Performance, 2nd edi- tion. Englewood Cliffs: Prentice Hall, 1991. nind, t.e.w. Principles of Oil Well Production. New York: McGraw-Hill Book Co., New York, 1964. Problems 12.1 If the dimensions d1, d2, and c take the same values for both conventional unit (Class I lever system) and air-balanced unit (Class III lever system), how differ- ent will their polished rod strokes length be? 12.2 What are the advantages of the Lufkin Mark II and air-balanced units in comparison with conventional units? 12.3 Use your knowledge of kinematics to prove that for Class I lever systems, a. the polished rod will travel faster in down stroke than in upstroke if the distance between crank- shaft and the center of Sampson post is less than dimension d1. b. the polished rod will travel faster in up stroke than in down stroke if the distance between crankshaft and the center of Sampson post is greater than dimension d1. 12.4 Derive a formula for calculating the effective di- ameter of a tapered rod string. 12.5 Derive formulas for calculating length fractions of equal-top-rod-stress tapered rod strings for (a) two- sized rod strings, (b) three-sized rod strings, and (c) four-sized rod strings. Plot size fractions for each case as a function of plunger area. 12.6 A tapered rod string consists of sections of 5 ⁄8 - and 1 ⁄2 - in. rods and a 2-in. plunger. Use the formulas from Problem 12.5 to calculate length fraction of each size of rod. 12.7 A tapered rod string consists of sections of 3 ⁄4 -, 5 ⁄8 -, and 1 ⁄2 -in. rods and a 13 ⁄4 -in. plunger. Use the for- mulas from Problem 12.5 to calculate length fraction of each size of rod. 12.8 The following geometry dimensions are for the pumping unit C–80D–133–48: d1 ¼ 64 in. d2 ¼ 64 in. c ¼ 24 in. h ¼ 74.5 in. Can this unit be used with a 2-in. plunger and 3 ⁄4 -in. rods to lift 30 8API gravity crude (formation volume factor 1.25 rb/stb) at depth of 2,000 ft? If yes, what is the required counter-balance load? 12.9 The following geometry dimensions are for the pumping unit C–320D–256–120: d1 ¼ 111:07 in. d2 ¼ 155 in. c ¼ 42 in. h ¼ 132 in. Can this unit be used with a 21 ⁄2 -in. plunger and 3 ⁄4 -, 7 ⁄8 -, 1-in. tapered rod string to lift 22 8API gravity crude (formation volume factor 1.22 rb/stb) at a −4,000 0 8,000 4,000 16,000 12,000 24,000 20,000 28,000 Load(lb) (d) 8 7 6 5 4 3 2 1 0 −1 −2 −3 −4 (a) (b) (c) Displacement (ft) Figure 12.13 Surface to down hole cards derived from surface dynamometer card. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 179 4.1.2007 2:43pm Compositor Name: SJoearun SUCKER ROD PUMPING 12/179
- 184. depth of 3,000 ft? If yes, what is the required coun- ter-balance load? 12.10 A well is pumped off with a rod pump described in Problem 12.8. A 21 ⁄2 -in. tubing string (2.875-in. OD, 2.441 ID) in the well is not anchored. Calculate (a) expected liquid production rate (use pump volu- metric efficiency 0.80) and (b) required prime mover power (use safety factor 1.3). 12.11 A well is pumped with a rod pump described in Problem 12.9 to a liquid level of 2,800 ft. A 3-in. tubing string (31 ⁄2 -in. OD, 2.995-in. ID) in the well is anchored. Calculate (a) expected liquid production rate (use pump volumetric efficiency 0.85) and (b) required prime mover power (use safety factor 1.4). 12.12 A well is to be put on a sucker rod pump. The proposed pump setting depth is 4,500 ft. The antici- pated production rate is 500 bbl/day oil of 40 8API gravity against wellhead pressure 150 psig. It is as- sumed that the working liquid level is low, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsur- face equipment for the installation. Use a safety factor of 1.40 for prime mover power. 12.13 A well is to be put on a sucker rod pump. The proposed pump setting depth is 4,000 ft. The antici- pated production rate is 550 bbl/day oil of 35 8API gravity against wellhead pressure 120 psig. It is as- sumed that working liquid level will be about 3,000 ft, and a sucker rod string having a working stress of 30,000 psi is to be used. Select surface and subsurface equipment for the installation. Use a safety factor of 1.30 for prime mover power. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap12 Final Proof page 180 4.1.2007 2:43pm Compositor Name: SJoearun 12/180 ARTIFICIAL LIFT METHODS
- 185. 13 Gas Lift Contents 13.1 Introduction 13/182 13.2 Gas Lift System 13/182 13.3 Evaluation of Gas Lift Potential 13/183 13.4 Gas Lift Gas Compression Requirements 13/185 13.5 Selection of Gas Lift Valves 13/192 13.6 Special Issues in Intermittent-Flow Gas Lift 13/201 13.7 Design of Gas Lift Installations 13/203 Summary 13/205 References 13/205 Problems 13/205 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 181 3.1.2007 9:07pm Compositor Name: SJoearun
- 186. 13.1 Introduction Gas lift technology increases oil production rate by injec- tion of compressed gas into the lower section of tubing through the casing–tubing annulus and an orifice installed in the tubing string. Upon entering the tubing, the com- pressed gas affects liquid flow in two ways: (a) the energy of expansion propels (pushes) the oil to the surface and (b) the gas aerates the oil so that the effective density of the fluid is less and, thus, easier to get to the surface. There are four categories of wells in which a gas lift can be considered: 1. High productivity index (PI), high bottom-hole pres- sure wells 2. High PI, low bottom-hole pressure wells 3. Low PI, high bottom-hole pressure wells 4. Low PI, low bottom-hole pressure wells Wells having a PI of 0.50 or less are classified as low productivity wells. Wells having a PI greater than 0.50 are classified as high productivity wells. High bottom-hole pressures will support a fluid column equal to 70% of the well depth. Low bottom-hole pressures will support a fluid column less than 40% of the well depth. Gas lift technology has been widely used in the oil fields that produce sandy and gassy oils. Crooked/deviated holes present no problem. Well depth is not a limitation. It is also applicable to offshore operations. Lifting costs for a large number of wells are generally very low. However, it requires lift gas within or near the oil fields. It is usually not efficient in lifting small fields with a small number of wells if gas compression equipment is required. Gas lift advancements in pressure control and automation systems have enabled the optimization of individual wells and gas lift systems. 13.2 Gas Lift System A complete gas lift system consists of a gas compression station, a gas injection manifold with injection chokes and time cycle surface controllers, a tubing string with instal- lations of unloading valves and operating valve, and a down-hole chamber. Figure 13.1 depicts a configuration of a gas-lifted well with installations of unloading valves and operating valve on the tubing string. There are four principal advantages to be gained by the use of multiple valves in a well: 1. Deeper gas injection depths can be achieved by using valves for wells with fixed surface injection pressures. 2. Variation in the well’s productivity can be obtained by selectively injecting gas valves set at depths ‘‘higher’’ or ‘‘lower’’ in the tubing string. 3. Gas volumes injected into the well can be ‘‘metered’’ into the well by the valves. 4. Intermittent gas injection at progressively deeper set valves can be carried out to ‘‘kick off’’ a well to either continuous or intermittent flow. A continuous gas lift operation is a steady-state flow of the aerated fluid from the bottom (or near bottom) of the well to the surface. Intermittent gas lift operation is char- acterized by a start-and-stop flow from the bottom (or near bottom) of the well to the surface. This is unsteady state flow. In continuous gas lift, a small volume of high-pressure gas is introduced into the tubing to aerate or lighten the fluid column. This allows the flowing bottom-hole pres- sure with the aid of the expanding injection gas to deliver liquid to the surface. To accomplish this efficiently, it is desirable to design a system that will permit injection through a single valve at the greatest depth possible with the available injection pressure. Continuous gas lift method is used in wells with a high PI ( !0:5 stb=day=psi) and a reasonably high reser- voir pressure relative to well depth. Intermittent gas lift method is suitable to wells with (1) high PI and low reservoir pressure or (2) low PI and low reservoir pressure. The type of gas lift operation used, continuous or intermittent, is also governed by the volume of fluids to be produced, the available lift gas as to both volume and pressure, and the well reservoir’s conditions such as the case when the high instantaneous BHP drawdown encountered with intermittent flow would cause exces- sive sand production, or coning, and/or gas into the wellbore. Figure 13.2 illustrates a simplified flow diagram of a closed rotary gas lift system for a single well in an intermittent gas lift operation. The time cycle surface controller regulates the start-and-stop injection of lift gas to the well. For proper selection, installation, and operations of gas lift systems, the operator must know the equipment and the fundamentals of gas lift technology. The basic equip- ment for gas lift technology includes the following: a. Main operating valves b. Wire-line adaptations c. Check valves d. Mandrels e. Surface control equipment f. Compressors This chapter covers basic system engineering design fun- damentals for gas lift operations. Relevant topics include the following: 1. Liquid flow analysis for evaluation of gas lift potential 2. Gas flow analysis for determination of lift gas compres- sion requirements 3. Unloading process analysis for spacing subsurface valves 4. Valve characteristics analysis for subsurface valve selection 5. Installation design for continuous and intermittent lift systems. Operating valve Unloading valves Gas inlet Production Figure 13.1 Configuration of a typical gas lift well. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 182 3.1.2007 9:07pm Compositor Name: SJoearun 13/182 ARTIFICIAL LIFT METHODS
- 187. 13.3 Evaluation of Gas Lift Potential Continuous gas lift can be satisfactorily applied to most wells having a reasonable degree of bottom-hole maintenance and a PI of approximately 0.5 bbl/day/psi or greater. A PI as low as 0.2 bbl/day/psi can be used for a continuous gas lift operation if injection gas is available at a sufficiently high pressure. An intermittent gas lift is usually applied to wells having a PI less than 0.5 bbl/day/psi. Continuous gas lift wells are changed to intermittent gas lift wells after reservoir pressures drop to below a certain level. Therefore, intermittent gas lift wells usually give lower production rates than continuous gas lift wells. The decision of whether to use gas lift technology for oil well production starts from evaluating gas lift potential with continuous gas injection. Evaluation of gas lift potential requires system analyses to determine well operating points for various lift gas availabilities. The principle is based on the fact that there is only one pressure at a given point (node) in any system; no matter, the pressure is estimated based on the informa- tion from upstream (inflow) or downstream (outflow). The node of analysis is usually chosen to be the gas injection point inside the tubing, although bottom hole is often used as a solution node. The potential of gas lift wells is controlled by gas injec- tion rate or gas liquid ratio (GLR). Four gas injection rates are significant in the operation of gas lift installations: 1. Injection rates of gas that result in no liquid (oil or water) flow up the tubing. The gas amount is insuffi- cient to lift the liquid. If the gas enters the tubing at an extremely low rate, it will rise to the surface in small semi-spheres (bubbly flow). 2. Injection rates of maximum efficiency where a min- imum volume of gas is required to lift a given amount of liquid. 3. Injection rate for maximum liquid flow rate at the ‘‘optimum GLR.’’ 4. Injection rate of no liquid flow because of excessive gas injection. This occurs when the friction (pipe) produced by the gas prevents liquid from entering the tubing. Figure 13.3 depicts a continuous gas lift operation. The tubing is filled with reservoir fluid below the injection point and with the mixture of reservoir fluid and injected gas above the injection point. The pressure relationship is shown in Fig. 13.4. The inflow performance curve for the node at the gas injection point inside the tubing is well IPR curve minus the pressure drop from bottom hole to the node. The outflow performance curve is the vertical lift performance curve, with total GLR being the sum of formation GLR and injected GLR. Intersection of the two curves defines the operation point, that is, the well production potential. Compressor station Suction regulator High pressure system By-pass regulator Stock tank Intermittent gas lift well Flowline Separator Vent or sales line regulator Low pressure system Make-up gas line for charging system Injection gas line Time cycle surface controller Figure 13.2 A simplified flow diagram of a closed rotary gas lift system for single intermittent well. GR Gu Injection gas Pcs PWH Injection choke Unloading valves Point of gas injection (operation valve) Additional valve or valves Packer Point of balance Kill fluid Figure 13.3 A sketch of continuous gas lift. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 183 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/183
- 188. In a field-scale evaluation, if an unlimited amount of lift gas is available for a given gas lift project, the injection rate of gas to individual wells should be optimized to maximize oil production of each well. If only a limited amount of gas is available for the gas lift, the gas should be distributed to individual wells based on predicted well lifting perfor- mance, that is, the wells that will produce oil at higher rates at a given amount of lift gas are preferably chosen to receive more lift gas. If an unlimited amount of gas lift gas is available for a well, the well should receive a lift gas injection rate that yields the optimum GLR in the tubing so that the flowing bottom-hole pressure is minimized, and thus, oil produc- tion is maximized. The optimum GLR is liquid flow rate dependent and can be found from traditional gradient curves such as those generated by Gilbert (Gilbert, 1954). Similar curves can be generated with modern computer programs using various multiphase correlations. The com- puter program OptimumGLR.xls in the CD attached to this book was developed based on modified Hagedorn and Brown method (Brown, 1977) for multiphase flow calcu- lations and the Chen method (1979) for friction factor determination. It can be used for predicting the optimum GLR in tubing at a given tubing head pressure and liquid flow rate. After the system analysis is completed with the optimum GLRs in the tubing above the injection point, the expected liquid production rate (well potential) is known. The required injection GLR to the well can be calculated by GLRinj ¼ GLRopt,o À GLRfm, (13:1) where GLRinj ¼ injection GLR, scf/stb GLRopt,o ¼ optimum GLR at operating flow rate, scf/stb GLRfm ¼ formation oil GLR, scf/stb. Then the required gas injection rate to the well can be calculated by qg,inj ¼ GLRinjqo, (13:2) where qo is the expected operating liquid flow rate. If a limited amount of gas lift gas is available for a well, the well potential should be estimated based on GLR expressed as GLR ¼ GLRfm þ qg,inj q , (13:3) where qg is the lift gas injection rate (scf/day) available to the well. Example Problem 13.1 An oil well has a pay zone around the mid-perf depth of 5,200 ft. The formation oil has a gravity of 26 8API and GLR of 300 scf/stb. Water cut remains 0%. The IPR of the well is expressed as q ¼ qmax 1 À 0:2 pwf pp À 0:8 pwf pp 2 # , where qmax ¼ 1500 stb=day pp ¼ 2,000 psia. A 21 ⁄2 -in. tubing (2.259 in. inside diameter [ID]) can be set with a packer at 200 ft above the mid-perf. What is the maximum expected oil production rate from the well with continuous gas lift at a wellhead pressure of 200 psia if a. an unlimited amount of lift gas is available for the well? b. only 1 MMscf/day of lift gas is available for the well? Solution The maximum oil production rate is expected when the gas injection point is set right above the packer. Assuming that the pressure losses due to friction below the injection point are negligible, the inflow-performance curve for the gas injection point (inside tubing) can be expressed as pvf ¼ 0:125pp½ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 q=qmaxð Þ p À 1Š À GR D À Dvð Þ, where pvf is the pressure at the gas injection point, GR is the pressure gradient of the reservoir fluid, D is the pay ∆Pv G fb PR Flowing bottom hole pressure Pressure Dv 0 0 Gfa D-Dv D Depth Tubing pressure (Pt) Casing pressure (Pso) Point of gas injection Point of balance Figure 13.4 Pressure relationship in a continuous gas lift. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 184 3.1.2007 9:07pm Compositor Name: SJoearun 13/184 ARTIFICIAL LIFT METHODS
- 189. zone depth, and Dv is the depth of the gas injection point. Based on the oil gravity of 26 8API, GR is calculated to be 0.39 psi/ft. D and Dv are equal to 5,200 ft and 5,000 ft, respectively in this problem. The outflow performance curve for the gas injection point can be determined based on 2.259-in. tubing ID, 200 psia wellhead pressure, and the GLRs. a. Spreadsheet OptimumGLR.xls gives the following: Using these data to run computer program HagedornBrown- Correlation.xls (on the CD attached to this book) gives Figure 13.5 shows the system analysis plot given by the computer program GasLiftPotential.xls. It indicates an operating point of q ¼ 632 stb=day and pt,v ¼ 698 psia tubing pressure at the depth of injection. The optimum GLR at the operating point is calculated with interpolation as GLRopt,o ¼ 2,400 þ 3,200 À 2,400 800 À 600 800 À 632ð Þ ¼ 3,072 scf=stb: The injection GLR is GLRinj ¼ 3,072 À 300 ¼ 2,772 scf=stb: Then the required gas injection rate to the well can be calculated: qg,inj ¼ (2,772)(632) ¼ 1,720,000 scf=day b. For a given amount of lift gas 1 MMscf/day, the GLR can be calculated with Eq. (13.3) as Using these data to run computer program Hagedorn- BrownCorrelation.xls gives Figure 13.6 shows the system analysis plot given by the computer program GasLiftPotential.xls. It indicates an operating point of q ¼ 620 stb=day and pt ¼ 702 psia tubing pressure at the depth of injection. This example shows that increasing the gas injection rate from 1 MMscf/day to 1.58 MMscf/day will not make a significant difference in the oil production rate. 13.4 Gas Lift Gas Compression Requirements The gas compression station should be designed to provide an adequate gas lift gas flow rate at sufficiently high pressure. These gas flow rates and output pressures deter- mine the required power of the compression station. 13.4.1 Gas Flow Rate Requirement The total gas flow rate of the compression station should be designed on the basis of gas lift at peak operating condition for all the wells with a safety factor for system leak consideration, that is, qg,total ¼ Sf XNw i¼1 qg,inj À Á i , (13:4) where qg ¼ total output gas flow rate of the compression station, scf/day Sf ¼ safety factor, 1.05 or higher Nw ¼ number of wells. The procedure for determination of lift gas injection rate qg,inj to each well has been illustrated in Example Prob- lem 13.1. 13.4.2 Output Gas Pressure Requirement Kickoff of a dead well (non-natural flowing) requires much higher compressor output pressures than the ulti- mate goal of steady production (either by continuous gas lift or by intermittent gas lift operations). Mobil compressor trailers are used for the kickoff operations. The output pressure of the compression station should be designed on the basis of the gas distribution pressure under normal flow conditions, not the kickoff conditions. It can be expressed as pout ¼ Sf pL, (13:5) where pout ¼ output pressure of the compression station, psia Sf ¼ safety factor pL ¼ pressure at the inlet of the gas distribution line, psia. Starting from the tubing pressure at the valve ( pt,v), the pressure at the inlet of the gas distribution line can be estimated based on the relationships of pressures along the injection path. These relationships are discussed in the following subsections. 13.4.2.1 Injection Pressure at Valve Depth The injection pressure at valve depth in the casing side can be expressed as pc,v ¼ pt,v þ Dpv, (13:6) where pc,v ¼ casing pressure at valve depth, psia Dpv ¼ pressure differential across the operating valve (orifice). It is a common practice to use Dpv ¼ 100 psi. The required size of the orifice can be determined using the choke-flow equations presented in Subsection 13.4.2.3. 13.4.2.2 Injection Pressure at Surface Accurate determination of the surface injection pressure pc,s requires rigorous methods such as the Cullender and q (stb/d) GLRopt (scf/stb) 400 4,500 600 3,200 800 2,400 q (stb/day) pt (psia) 400 603 600 676 800 752 q (stb/day) GLR (scf/stb) 400 2,800 600 1,967 800 1,550 q (stb/day) pt (psia) 400 614 600 694 800 774 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 185 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/185
- 190. Smith method (Katz et al., 1959). The average temperature and compressibility factor method also gives results with acceptable accuracy. In both methods, the frictional pressure losses in the annulus are considered. However, because of the large cross-sectional area of the annular space, the frictional pressure losses are often negligible. Then the average temperature and compressibility factor model degenerates to (Economides et al., 1994) pc,v ¼ pc,s e0:01875 ggDv zzTT , (13:7) where pc,v ¼ casing pressure at valve depth, psia pc,s ¼ casing pressure at surface, psia gg ¼ gas specific gravity, air ¼ 1:0 zz ¼ the average gas compressibility factor TT ¼ the average temperature, 8R. Equation (13.7) can be rearranged to be pc,s ¼ pc,veÀ0:01875 ggDv zzTT : (13:8) Since the z factor also depends on pc,s, this equation can be solved for pc,s with a trial-and-error approach. Because Eq. (13.8) involves exponential function that is difficult to handle without a calculator, an approximation to the equation has been used traditionally. In fact, when Eq. (13.7) is expended as a Taylor series, and if common fluid properties for a natural gas and reservoir are consid- ered such as gg ¼ 0:7, zz ¼ 0:9, and TT ¼ 600 8R, it can be approximated as −500 0 500 1,000 1,500 2,000 2,500 0 100 200 300 400 500 600 700 800 900 Liquid Flow Rate (stb/day) TubingPressureattheInjectionDepth(psia) Inflow Pressure (psia) Outflow Pressure (psia) Figure 13.5 System analysis plot given by GasLiftPotential.xls for the unlimited gas injection case. −500 0 500 1,000 1,500 2,000 2,500 0 100 200 300 400 500 600 700 800 900 Liquid Flow Rate (stb/day) TubingPressureattheInjectionDepth(psia) Inflow Pressure (psia) Outflow Pressure (psia) Figure 13.6 System analysis plot given by GasLiftPotential.xls for the limited gas injection case. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 186 3.1.2007 9:07pm Compositor Name: SJoearun 13/186 ARTIFICIAL LIFT METHODS
- 191. pc,v ¼ pc,s 1 þ Dv 40,000 , (13:9) which gives pc,s ¼ pc,v 1 þ Dv 40,000 : (13:10) Neglecting the pressure losses between injection choke and the casing head, the pressure downstream of the choke ( pdn) can be assumed to be the casing surface injection pressure, that is, pdn ¼ pc,s: 13.4.2.3 Pressure Upstream of the Choke The pressure upstream of the injection choke depends on flow condition at the choke, that is, sonic or subsonic flow. Whether a sonic flow exists depends on a downstream-to- upstream pressure ratio. If this pressure ratio is less than a critical pressure ratio, sonic (critical) flow exists. If this pressure ratio is greater than or equal to the critical pres- sure ratio, subsonic (subcritical) flow exists. The critical pressure ratio through chokes is expressed as Rc ¼ 2 k þ 1 k kÀ1 , (13:11) where k ¼ Cp=Cv is the gas-specific heat ratio. The value of the k is about 1.28 for natural gas. Thus, the critical pressure ratio is about 0.55. Pressure equations for choke flow are derived based on an isentropic process. This is because there is no time for heat to transfer (adiabatic) and the friction loss is negli- gible (assuming reversible) at choke. 13.4.2.3.1 Sonic Flow Under sonic flow conditions, the gas passage rate reaches and remains its maximum value. The gas passage rate is expressed in the following equation for ideal gases: qgM ¼ 879CcApup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ggTup ! 2 k þ 1 kþ1 kÀ1 v u u t , (13:12) where qgM ¼ gas flow rate, Mscf/day pup ¼ pressure upstream the choke, psia A ¼ cross-sectional area of choke, in:2 Tup ¼ upstream temperature, 8R gg ¼ gas specific gravity related to air Cc ¼ choke flow coefficient. The choke flow coefficient Cc can be determined using charts in Figs. 5.2 and 5.3 (Chapter 5) for nozzle- and orifice-type chokes, respectively. The following correlation has been found to give reasonable accuracy for Reynolds numbers between 104 and 106 for nozzle-type chokes (Guo and Ghalambor, 2005): C ¼ d D þ 0:3167 d D 0:6 þ 0:025[ log (NRe) À 4], (13:13) where d ¼ choke diameter, inch D ¼ pipe diameter, in. NRe ¼ Reynolds number and the Reynolds number is given by NRe ¼ 20qgMgg md , (13:14) where m ¼ gas viscosity at in situ temperature and pressure, cp. Equation (13.12) indicates that the upstream pressure is independent of downstream pressure under sonic flow conditions. If it is desirable to make a choke work under sonic flow conditions, the upstream pressure should meet the following condition: pup ! pdn 0:55 ¼ 1:82pdn (13:15) Once the pressure upstream of the choke/orifice is deter- mined by Eq. (13.15), the required choke/orifice diameter can be calculated with Eq. (13.12) using a trial-and-error approach. 13.4.2.3.2 Subsonic Flow Under subsonic flow con- ditions, gas passage through a choke can be expressed as qgM ¼ 1,248CcApup  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k (k À 1)ggTup pdn pup 2 k À pdn pup kþ1 k #v u u t : (13:16) If it is desirable to make a choke work under subsonic flow conditions, the upstream pressure should be determined from Eq. (13.16) with a trial-and-error method. 13.4.2.4 Pressure of the Gas Distribution Line The pressure at the inlet of gas distribution line can be calculated using the Weymouth equation for horizontal flow (Weymouth, 1912): qgM ¼ 0:433Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 L À p2 up D16=3 gg TTzzLg v u u t , (13:17) where Tb ¼ base temperature, 8R pb ¼ base pressure, psi pL ¼ pressure at the inlet of gas distribution line, psia Lg ¼ length of distribution line, mile Equation (13.17) can be rearranged to solve for pressure: pL ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 up þ qgMpb 0:433Tb 2 gg TTzzLg D16=3 s (13:18) Example Problem 13.2 Anoilfieldhas16oilwellstobegas lifted. The gas lift gas at the central compressor station is firstpumpedtotwoinjectionmanifoldswith4-in.ID,1-mile lines and then is distributed to the wellheads with 4-in. ID, 0.2-mile lines. Given the following data, calculate the required output pressure of compression station: Gas-specific gravity (gg): 0.65 Valve depth (Dv): 5,000 ft Maximum tubing pressure at valve depth ( pt): 500 psia Required lift gas injection rate per well: 2 MMscf/day Pressure safety factor (Sf ): 1.1 Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia Solution Using Dpv ¼ 100 psi, the injection pressure at valve depth is then 600 psia. Equation (13.10) gives pc,s ¼ pc,v 1 þ Dv 40,000 ¼ 600 1 þ 5,000 40,000 ¼ 533 psia: Neglecting the pressure losses between the injection choke and the casing head, pressure downstream of the choke Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 187 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/187
- 192. ( pdn) can be assumed to be the surface injection pressure, that is, pdn ¼ pc,s ¼ 533 psia: Assuming minimum sonic flow at the injection choke, the pressure upstream of the choke is calculated as pup ! pdn 0:55 ¼ 1:82pdn ¼ (1:82)(533) ¼ 972 psia: The gas flow rate in each of the two gas distribution lines is (2)(16)/(2), or 16 MMscf/day. Using the trial-and-error method, Eq. (13.18) gives pL ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (972)2 þ (16,000)(14:7) 0:433(60 þ 460) 2 (0:65)(530)(0:79)(1) (4)16=3 s ¼ 1,056 psia: The required output pressure of the compressor is deter- mined to be pout ¼ Sf pL ¼ (1:1)(1,056) ¼ 1,162 psia: The computer program CompressorPressure.xls can be used for solving similar problems. The solution given by the program to this example problem is shown in Table 13.1. 13.4.3 Compression Power Requirement The compressors used in the petroleum industry fall into two distinct categories: reciprocating and rotary compres- sors. Reciprocating compressors are built for practically all pressures and volumetric capacities. Reciprocating compressors have more moving parts and, therefore, lower mechanical efficiencies than rotary compressors. Each cylinder assembly of a reciprocation compressor consists of a piston, cylinder, cylinder heads, suction and discharge valves, and other parts necessary to convert rotary motion to reciprocation motion. A reciprocating compressor is designed for a certain range of compression ratios through the selection of proper piston displacement and clearance volume within the cylinder. This clearance volume can be either fixed or variable, depending on the extent of the operation range and the percent of load variation desired. A typical reciprocating compressor can deliver a volumetric gas flow rate up to 30,000 cubic feet per minute (cfm) at a discharge pressure up to 10,000 psig. Rotary compressors are divided into two classes: the centrifugal compressor and the rotary blower. A centri- fugal compressor consists of a housing with flow passages, a rotating shaft on which the impeller is mounted, bearings, and seals to prevent gas from escaping along the shaft. Centrifugal compressors have few moving parts because only the impeller and shaft rotate. Thus, its efficiency is high and lubrication oil consumption and maintenance costs are low. Cooling water is normally unnecessary because of lower compression ratio and lower friction loss. Compression rates of centrifugal compressors are lower because of the absence of positive displacement. Centrifugal compressors compress gas using centrifugal force. Work is done on the gas by an impeller. Gas is then discharged at a high velocity into a diffuser where the velocity is reduced and its kinetic energy is converted to static pressure. Unlike reciprocating compressors, all this is done without confinement and physical squeezing. Centri- fugal compressors with relatively unrestricted passages and continuous flow are inherently high-capacity, low-pressure ratio machines that adapt easily to series arrangements within a station. In this way, each compressor is required to develop only part of the station compression ratio. Typically, the volume is more than 100,000 cfm and dis- charge pressure is up to 100 psig. When selecting a compressor, the pressure-volume char- acteristics and the type of driver must be considered. Small Table 13.1 Result Given by Computer Program CompressorPressure.xls CompressorPressure.xls Description: This spreadsheet calculates required pressure from compressor. Instruction: (1) Select a unit system; (2) click ‘‘Solution’’ button; and (3) view result. Input data U.S. units SI units 1 Depth of operating valve (Dv): 5,000 ft Length of the main distribution line (Lg): 1 mi ID of the main distribution line (D): 4.00 in. Gas flow rate in main distribution line (qg,l): 16 MMscf/day Surface temperature (Ts): 70 8F Temperature at valve depth (Tv): 120 8F Gas-specific gravity (gg): 0:65 (air ¼ 1) Gas-specific heat ratio (k): 1.25 Tubing pressure at valve depth ( pt): 500 psia Valve pressure differential (Dpv): 100 psia Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia Pressure safety factor (Sf ): 1.1 Solution pc,v ¼ pt,v þ Dpv 600 psia Average z-factor in annulus: 0.9189? pc,s À pc,veÀ0:01875 ggDv zzTT ¼ 0 gives pc,s 532 psia pdn ¼ pc,s 532 psia pup ! pdn 0:55 ¼ 1:82pdn 969 psia Average z-factor at surface: 0.8278 pL À ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 up þ qgMpb 0:433Tb 2 gg TTzzLg D16=3 s ¼ 0 gives pL 1,063 psia pout ¼ Sf pL 1,170 psia Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 188 3.1.2007 9:07pm Compositor Name: SJoearun 13/188 ARTIFICIAL LIFT METHODS
- 193. rotary compressors (vane or impeller type) are generally driven by electric motors. Large-volume positive compres- sors operate at lower speeds and are usually driven by steam or gas engines. They may be driven through reduc- tion gearing by steam turbines or an electric motor. Re- ciprocation compressors driven by steam turbines or electric motors are most widely used in the petroleum industry as the conventional high-speed compression ma- chine. Selection of compressors requires considerations of volumetric gas deliverability, pressure, compression ratio, and horsepower. 13.4.3.1 Reciprocating Compressors Two basic approaches are used to calculate the horse- power theoretically required to compress natural gas. One is to use analytical expressions. In the case of adiabatic compression, the relationships are complicated and are usually based on the ideal-gas equation. When used for real gases where deviation from ideal-gas law is appre- ciable, they are empirically modified to take into consid- eration the gas deviation factor. The second approach is the enthalpy-entropy or Mollier diagram for real gases. This diagram provides a simple, direct, and rigorous pro- cedure for determining the horsepower theoretically neces- sary to compress the gas. Even though in practice the cylinders in the reciprocating compressors may be water-cooled, it is customary to con- sider the compression process as fundamentally adiabatic— that is, to idealize the compression as one in which there is no cooling of the gas. Furthermore, the process is usually considered to be essentially a perfectly reversible adiabatic, that is, an isentropic process. Thus, in analyzing the performance of a typical reciprocating compressor, one may look upon the compression path following the general law pVk ¼ a constant: (13:19) For real natural gases in the gravity range 0:55 gg 1, the following relationship can be used at approximately 150 8F: k150 F % 2:738 À log gg 2:328 (13:20) When a real gas is compressed in a single-stage compres- sion, the compression is polytropic tending to approach adiabatic or constant-entropy conditions. Adiabatic com- pression calculations give the maximum theoretical work or horsepower necessary to compress a gas between any two pressure limits, whereas isothermal compression cal- culations give the minimum theoretical work or horse- power necessary to compress a gas. Adiabatic and isothermal work of compression, thus, give the upper and lower limits, respectively, of work or horsepower requirements to compress a gas. One purpose of intercool- ers between multistage compressors is to reduce the horse- power necessary to compress the gas. The more intercoolers and stages, the closer the horsepower require- ment approaches the isothermal value. 13.4.3.1.1 Volumetric Efficiency The volumetric effi- ciency represents the efficiency of a compressor cylinder to compress gas. It may be defined as the ratio of the volume of gas actually delivered to the piston displacement, corrected to suction temperature and pressure. The principal reasons that the cylinder will not deliver the piston displacement capacity are wire-drawing, a throttling effect on the valves; heating of the gas during admission to the cylinder; leakage past valves and piston rings; and re- expansion of the gas trapped in the clearance-volume space from the previous stroke. Re-expansion has by far the greatest effect on volumetric efficiency. The theoretical formula for volumetric efficiency is Ev ¼ 1 À (r1=k À 1) Cl, (13:21) where Ev ¼ volumetric efficiency, fraction r ¼ cylinder compression ratio Cl ¼ clearance, fraction. In practice, adjustments are made to the theoretical formula in computing compressor performance: Ev ¼ 0:97 À zs zd r1=k À 1 ! Cl À ev, (13:22) where zs ¼ gas deviation factor at suction of the cylinder zd ¼ gas deviation factor at discharge of the cylinder ev ¼ correction factor. In this equation, the constant 0.97 is a reduction of 1 to correct for minor inefficiencies such as incomplete filling of the cylinder during the intake stroke. The correction factor ev is to correct for the conditions in a particular application that affect the volumetric efficiency and for which the theoretical formula is inadequate. 13.4.3.1.2 Stage Compression The ratio of the dis- charge pressure to the inlet pressure is called the pressure ratio.Thevolumetricefficiencybecomesless,andmechanical stress limitation becomes more, pronounced as pressure ratio increases. Natural gas is usually compressed in stages, with the pressure ratio per stage being less than 6. In field practice, the pressure ratio seldom exceeds 4 when boosting gas from low pressure for processing or sale. When the total compression ratio is greater than this, more stages of compression are used to reach high pressures. The total power requirement is a minimum when the pressure ratio in each stage is the same. This may be expressed in equation form as r ¼ pd ps 1=Ns , (13:23) where pd ¼ final discharge pressure, absolute ps ¼ suction pressure, absolute Ns ¼ number of stages required. As large compression ratios result in gas being heated to undesirably high temperatures, it is common practice to cool the gas between stages and, if possible, after the final stage of compression. 13.4.3.1.3 Isentropic Horsepower The computation is basedontheassumptionthattheprocessisidealisentropicor perfectly reversible adiabatic. The total ideal horsepower for a given compression is the sum of the ideal work computed for each stage of compression. The ideal isentropic work can be determined for each stage of compression in a number of ways.Onewaytosolveacompressionproblemisbyusingthe Mollier diagram. This method is not used in this book because it is not easily computerized. Another approach commonly used is to calculate the horsepower for each stage from the isentropic work formula: w ¼ k k À 1 53:241T1 gg p2 p1 (kÀ1)=k À1 # , (13:24) where w ¼ theoretical shaft work required to compress the gas, ft-lbf =lbm T1 ¼ suction temperature of the gas, 8R Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 189 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/189
- 194. gg ¼ gas-specific gravity, air ¼ 1 p1 ¼ suction pressure of the gas, psia p2 ¼ pressure of the gas at discharge point, psia. When the deviation from ideal gas behavior is appre- ciable, Eq. (13.24) is empirically modified. One such modi- fication is w ¼ k k À 1 53:241T1 gg p2 p1 Z1(kÀ1)=k À1 # (13:25) or, in terms of power, HpMM ¼ k k À 1 3:027pb Tb T1 p2 p1 Z1(kÀ1)=k À1 # , (13:26) where HpMM ¼ required theoretical compression power, hp/ MMcfd z1 ¼ compressibility factor at suction conditions. The theoretical adiabatic horsepower obtained by the proceedingequationscan beconverted to brakehorsepower (Hpb) required at the end of prime mover of the compressor using an overall efficiency factor, Eo. The brake horsepower is the horsepower input into the compressor. The efficiency factor Eo consists of two components: compression efficiency (compressor-valve losses) and the mechanical efficiency of the compressor. The overall efficiency of a compressor depends on a number of factors, including design details of the compressor, suction pressure, speed of the compressor, compression ratio, loading, and general mechanical condition of the unit. In most modern compres- sors, the compression efficiency ranges from 83 to 93%. The mechanical efficiency of most modern compressors ranges from 88 to 95%. Thus, most modern compressors have an overall efficiency ranging from 75 to 85%, based on the ideal isentropic compression process as a standard. The actual efficiency curves can be obtained from the manufacturer. Applying these factors to the theoretical horsepower gives Hpb ¼ qMMHpMM Eo , (13:27) where qMM is the gas flow rate in MMscfd. The discharge temperature for real gases can be calcu- lated by T2 ¼ T1 p2 p1 z1(kÀ1)=k : (13:28) Calculation of the heat removed by intercoolers and after- coolers can be accomplished using constant pressure- specific heat data: DH ¼ nGCpDT, (13:29) where nG ¼ number of lb-mole of gas Cp ¼ specific heat under constant pressure evaluated at cooler operating pressure and the average tem- perature, btu/lb-mol-8F. Example Problem 13.3 For data given in Example Problem 13.2, assuming the overall efficiency is 0.80, calculate the theoretical and brake horsepower required to compress the 32 MMcfd of a 0.65-specific gravity natural gas from 100 psia and 70 8F to 1,165 psia. If intercoolers cool the gas to 70 8F, what is the heat load on the intercoolers and what is the final gas temperature? Solution The overall compression ratio is rov ¼ 1,165 100 ¼ 11:65: Because this is greater than 6, more than one-stage com- pression is required. Using two stages of compression gives r ¼ 1,165 100 1=2 ¼ 3:41: The gas is compressed from 100 to 341 psia in the first stage, and from 341 to 1,165 psia in the second stage. Based on gas-specific gravity, the following gas property data can be obtained: Tc ¼ 358 R pc ¼ 671 psia Tr ¼ 1:42 pr,1 ¼ 0:149 at 100 psia pr,2 ¼ 0:595 at 341 psia z1 ¼ 0:97 at 70 F and 100 psia z2 ¼ 0:95 at 70 F and 341 psia: First stage: HpMM ¼ 1:25 0:25 3:027 Â 14:7 520 530 (3:41)0:97(0:25=1:25) À 1 h i ¼ 61 hp=MMcfd Second stage: HpMM ¼ 1:25 0:25 3:027 Â 14:7 520 530 (3:41)0:95(0:25=1:25) À 1 h i ¼ 59 hp=MMcfd Total theoretical compression work ¼ 61 þ 59 ¼ 120 hp= MMcfd. Required brake horsepower is Hpb ¼ (32)(120) (0:8) ¼ 4,800 hp: Number of moles of gas is nG ¼ 1,000,000 378:6 (32) ¼ 2:640 Â 103 (32) ¼ 84 Â 106 lb-mole=day: Gas temperature after the first stage of compression is T2 ¼ (530)(3:41)0:97(0:25=1:25) ¼ 670 R ¼ 210 F: The average cooler temperature is 210 þ 70 2 ¼ 140 F. Cp at 140 F and 341 psia ¼ 9:5 btu lb À mol F : Intercooler load ¼ 2:640 Â 103 (32)(9:5)(210 À 70) ¼ 55:67 Â 106 btu=day: Final gas temperature: Td ¼ (530)(3:41)0:95(0:25=1:25) ¼ 669 R ¼ 209 F It can be shown that the results obtained using the analyt- ical expressions compare very well to those obtained from the Mollier diagram. ThecomputerprogramReciprocatingCompressorPower.xls can be used for computing power requirement of each stage of compression. The solution given by the program for the first stage of compression in this example problem is shown in Table 13.2. 13.4.3.2 Centrifugal Compressors Although the adiabatic compression process can be assumed in centrifugal compression, polytropic compression process is commonly considered as the basis for comparing centrifu- gal compressor performance. The process is expressed as Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 190 3.1.2007 9:07pm Compositor Name: SJoearun 13/190 ARTIFICIAL LIFT METHODS
- 195. pVn ¼ constant, (13:30) where n denotes the polytropic exponent. The isentropic exponent k applies to the ideal frictionless adiabatic pro- cess, while the polytropic exponent n applies to the actual process with heat transfer and friction. The n is related to k through polytropic efficiency Ep: n À 1 n ¼ k À 1 k  1 Ep (13:31) The polytropic efficiency of centrifugal compressors is nearly proportional to the logarithm of gas flow rate in the range of efficiency between 0.7 and 0.75. The polytro- pic efficiency chart presented by Rollins (1973) can be represented by the following correlation (Guo and Ghalambor, 2005): Ep ¼ 0:61 þ 0:03 log (q1), (13:32) where q1 ¼ gas capacity at the inlet condition, cfm. There is a lower limit of gas flow rate below which severe gas surge occurs in the compressor. This limit is called surge limit. The upper limit of gas flow rate is called stone-wall limit, which is controlled by compressor horsepower. The procedure of preliminary calculations for selection of centrifugal compressors is summarized as follows: 1. Calculate compression ratio based on the inlet and discharge pressures: r ¼ p2 p1 (13:33) 2. Based on the required gas flow rate under standard condition (q), estimate the gas capacity at inlet condi- tion (q1) by ideal gas law: q1 ¼ pb p1 T1 Tb q (13:34) 3. Find a value for the polytropic efficiency Ep from the manufacturer’s manual based on q1. 4. Calculate polytropic ratio (n À 1)=n: Rp ¼ n À 1 n ¼ k À 1 k  1 Ep (13:35) 5. Calculate discharge temperature by T2 ¼ T1 rRp : (13:36) 6. Estimate gas compressibility factor values at inlet and discharge conditions. 7. Calculate gas capacity at the inlet condition (q1) by real gas law: q1 ¼ z1pb z2p1 T1 Tb q (13:37) 8. Repeat steps 2 through 7 until the value of q1 converges within an acceptable deviation. 9. Calculate gas horsepower by Hpg ¼ q1p1 229Ep z1 þ z2 2z1 rRp À 1 Rp : (13:38) Some manufacturers present compressor specifications using polytropic head in lbf -ft=lbm defined as Hg ¼ RT1 z1 þ z2 2 rRp À 1 Rp , (13:39) where R is the gas constant given by 1,544=MWa in psia-ft3 =lbm-8R. The polytropic head relates to the gas horsepower by Hpg ¼ MF Hg 33,000Ep , (13:40) where MF is mass flow rate in lbm= min. 10. Calculate gas horsepower by Hpb ¼ Hpg þ DHpm, (13:41) Table 13.2 Result Given by Computer Program ReciprocatingCompressorPower.xls for the First- Stage Compression ReciprocatingCompressorPower.xls Description: This spreadsheet calculates stage power of reciprocating compressor. Instruction: (1) Update parameter valves in the ‘‘Input data’’ in blue; (2) click ‘‘Solution’’ button; (3) view result in the Solution section. Input data Gas flow rate (qg): 32 MMscf/day Stage inlet temperature (T1): 70 8F Stage inlet pressure ( p1): 100 psia Gas-specific gravity (gg): 0:65(air ¼ 1) Stage outlet pressure ( p2): 341 psia Gas-specific heat ratio (k): 1.25 Overall efficiency (Eo): 0.8 Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia Solution z ¼ Hall–Yarborogh Method ¼ 0:9574 r ¼ p2 p1 ¼ 3:41 HpMM ¼ k k À 1 3:027pb Tb T1 p2 p1 Z1(kÀ1)=k À1 # ¼ 60 hp Hpb ¼ qMMHpMM Eo ¼ 2,401 hp T2 ¼ T1 p2 p1 z1(kÀ1)=k ¼ 210:33 F Tavg ¼ T1 þ T2 2 ¼ 140:16 F Cp ¼ 9:50 btu=lbm-mol F Cooler load ¼ 2:640  103 qMM CCp Tavg À T1 À Á ¼ 56,319,606 btu=day Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 191 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/191
- 196. where DHpm is mechanical power losses, which is usually taken as 20 horsepower for bearing and 30 horsepower for seals. The proceeding equations have been coded in the com- puter program CnetriComp.xls (on the CD attached to this book) for quick calculation. Example Problem 13.4 Assuming two centrifugal com- pressors in series are used to compress gas for a gas lift operation. Size the first compressor using the formation given in Example Problem 13.3. Solution Calculate compression ratio based on the inlet and discharge pressures: r ¼ ffiffiffiffiffiffiffiffiffiffiffi 1,165 100 r ¼ 3:41 Calculate gas flow rate in scfm: q ¼ 32,000,000 (24)(60) ¼ 22,222 scfm Based on the required gas flow rate under standard condi- tion (q), estimate the gas capacity at inlet condition (q1) by ideal gas law: q1 ¼ (14:7) (250) (560) (520) (22,222) ¼ 3,329 cfm Find a value for the polytropic efficiency based on q1: Ep ¼ 0:61 þ 0:03 log (3,329) ¼ 0:719 Calculate polytropic ratio (n À 1)=n: Rp ¼ 1:25 À 1 1:25 Â 1 0:719 ¼ 0:278 Calculate discharge temperature by T2 ¼ (530)(3:41)0:278 ¼ 745 R ¼ 285 F: Estimate gas compressibility factor values at inlet and discharge conditions: z1 ¼ 1:09 at 100 psia and 70 8F z2 ¼ 0:99 at 341 psia and 590 8F Calculate gas capacity at the inlet condition (q1) by real gas law: q1 ¼ (1:09)(14:7) (0:99)(100) (530) (520) (22,222) ¼ 3,674 cfm Use the new value of q1 to calculate Ep: Ep ¼ 0:61 þ 0:03 log (3,674) ¼ 0:721 Calculate the new polytropic ratio (n À 1)=n: Rp ¼ 1:25 À 1 1:25 Â 1 0:721 ¼ 0:277 Calculate the new discharge temperature: T2 ¼ (530)(3:41)0:277 ¼ 746 R ¼ 286 F Estimate the new gas compressibility factor value: z2 ¼ 0:99 at 341 psia and 286 8F Because z2 did not change, q1 remains the same value of 3,674 cfm. Calculate gas horsepower: Hpg ¼ (3,674)(100) (229)(0:721) 1:09 þ 0:99 2(1:09) 3:410:277 À 1 0:277 ¼ 3,100 hp Calculate gas apparent molecular weight: MWa ¼ (0:65)(29) ¼ 18:85 Calculated gas constant: R ¼ 1,544 18:85 ¼ 81:91 psia-ft3 =lbm- R Calculate polytropic head: Hg ¼ (81:91)(530) 1:09 þ 0:99 2 3:410:277 À 1 0:277 ¼ 65,850 lbf -ft=lbm Calculate gas horsepower: Hpb ¼ 3,100 þ 50 ¼ 3,150 hp ThecomputerprogramCentrifugalCompressorPower.xlscan be used for solving similar problems. The solution given by the program to this example problem is shown in Table 13.3. 13.5 Selection of Gas Lift Valves Kickoff of a dead well requires a much higher gas pressure than the ultimate operating pressure. Because of the kickoff problem, gas lift valves have been developed and are run as part of the overall tubing string. These valves permit the introduction of gas (which is usually injected down the annu- lus) into the fluid column in tubing at intermediate depths to unload the well and initiate well flow. Proper design of these valve depths to unsure unloading requires a thorough under- standing of the unloading process and valve characteristics. 13.5.1 Unloading Sequence Figure 13.7 shows a well unloading process. Usually all valves are open at the initial condition, as depicted in Fig. 13.7a, due to high tubing pressures. The fluid in tubing has a pressure gradient Gs of static liquid column. When the gas enters the first (top) valve as shown in Fig. 13.7b, it creates a slug of liquid–gas mixture of less- density in the tubing above the valve depth. Expansion of the slug pushes the liquid column above it to flow to the surface. It can also cause the liquid in the bottom hole to flow back to reservoir if no check valve is installed at the end of the tubing string. However, as the length of the light slug grows due to gas injection, the bottom-hole pressure will eventually decrease to below reservoir pressure, which causes inflow of reservoir fluid. When the tubing pressure at the depth of the first valve is low enough, the first valve should begin to close and the gas should be forced to the second valve as shown in Fig. 13.7c. Gas injection to the second valve will gasify the liquid in the tubing between the first and the second valve. This will further reduce bottom-hole pressure and cause more inflow. By the time the slug reaches the depth of the first valve, the first valve should be closed, allowing more gas to be injected to the second valve. The same process should occur until the gas enters the main valve (Fig. 13.7d). The main valve (sometimes called the master valve or operating valve) is usually the lower most valve in the tubing string. It is an orifice type of valve that never closes. In continuous gas lift operations, once the well is fully unloaded and a steady-state flow is established, the main valve is the only valve open and in operation (Fig. 13.7e). 13.5.2 Valve Characteristics Equations (13.12) and (13.16) describing choke flow are also applicable to the main valve of orifice type. Flow characteristics of this type of valve are depicted in Fig. 13.8. Under sonic flow conditions, the gas passage is independent of tubing pressure but not casing pressure. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 192 3.1.2007 9:07pm Compositor Name: SJoearun 13/192 ARTIFICIAL LIFT METHODS
- 197. There are different types of unloading valves, namely casing pressure-operated valve (usually called a pressure valve), throt- tling pressure valve (also called a proportional valve or continu- ous flow valve), fluid-operated valve (also called a fluid valve), and combination valve (also called a fluid open-pressure closed valve). Different gas lift design methods have been developed and used in the oil industry for applications of these valves. 13.5.2.1 Pressure Valve Pressure valves are further classified as unbalanced bellow valves, balanced pressure valves, and pilot valves. Tubing pressure affects the opening action of the unbalanced valves, but it does not affect the opening or closing of balanced valves. Pilot valves were developed for intermit- tent gas lift with large ports. 13.5.2.1.1 Unbalanced Bellow Valve As shown in Fig. 13.9, an unbalanced bellow valve has a pressure- charged nitrogen dome and an optional spring loading element. While the forces from the dome pressure and spring act to cause closing of the valve, the forces due to casing and tubing pressures act to cause opening of the valve. Detailed discussions of valve mechanics can be found in Brown (1980). When a valve is at its closed condition (as shown in Fig. 13.9), the minimum casing pressure required to open the valve is called the valve opening pressure and is expressed as Pvo ¼ 1 1 À R Pd þ St À R 1 À R Pt, (13:42) where Pvo ¼ valve opening pressure, psig Pd ¼ pressure in the dome, psig St ¼ equivalent pressure caused by spring tension, psig Pt ¼ tubing pressure at valve depth when the valve opens, psi R ¼ area ratio Ap=Ab Ap ¼ valve seat area, in:2 Ab ¼ total effective bellows area, in:2 . The term R 1ÀRPt is called tubing effect (T.E.) and R 1ÀR is called tubing effect factor (T.E.F.). With other parameters given, Eq. (13.42) is used for determining the required dome pressure at depth, that is, Pd ¼ (1 À R)Pvo À St þ RPt, in valve selection. When a valve is at its open condition (as shown in Fig. 13.10), the maximum pressure under the ball (assumed to be casing pressure) required to close the valve is called the valve closing pressure and is expressed as Pvc ¼ Pd þ St 1 À Rð Þ, (13:43) where Pvc ¼ valve closing pressure, psig. The difference between the valve opening and closing pressures,Pvo À Pvc,iscalledspread.Spreadcanbeimportant Table 13.3 Result Given by the Computer Program CentrifugalCompressorPower.xls CentrifugalCompressorPower.xls Description: This spreadsheet calculates stage power of reciprocating compressor. Instruction: (1) Update parameter valves in the ‘‘Input data’’ in blue; (2) click ‘‘Solution’’ button; (3) view result in the Solution section. Input data Gas flow rate (qg): 32 MMscf/day Inlet temperature (T1): 70 8F Inlet pressure ( p1): 100 psia Gas-specific gravity (gg): 0:65 (air ¼ 1) Discharge pressure ( p2): 341 psia Gas-specific heat ratio (k): 1.25 Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia Solution r ¼ p2 p1 ¼ 3:41 q ¼ qMM (24)(60) ¼ 22,222 scfm q1 ¼ pb p1 T1 Tb q ¼ 3,329 scfm Ep ¼ 0:61 þ 0:03 log (q1) ¼ 0:7192 Rp ¼ n À 1 n ¼ k À 1 k  1 Ep ¼ 0:2781 T2 ¼ T1rRp ¼ 285 F z1 by Hall–Yarborogh Method ¼ 1:0891 z2 by Hall–Yarborogh Method ¼ 0:9869 q1 ¼ z1pb z2p1 T1 Tb q ¼ 3,674 Ep ¼ 0:61 þ 0:03 log (q1) ¼ 0:7205 Rp ¼ n À 1 n ¼ k À 1 k  1 Ep ¼ 0:2776 T2 ¼ T1rRp ¼ 285 F Hpg ¼ q1p1 229Ep z1 þ z2 2z1 rRp À 1 Rp ¼ 3,102 hp Hpb ¼ Hpg þ 50 ¼ 3,152 hp MWa ¼ 29gg ¼ 18:85 R ¼ 1,544 MWa ¼ 81:91 Hg ¼ RT1 z1þz2 2 À Á rRp À1 Rp ¼ 65,853 lbf-ft=lbm Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 193 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/193
- 198. in continuous flow installations but is particularly important in intermittent gas lift installations where unbalanced valves are used. The spread controls the minimum amount of gas used for each cycle. As the spread increases, the amount of gas injected during the cycle increases. Gas passage of unbalanced valves are tubing-pressure dependent due to partial travel of the valve stem. Figure 13.11 illustrates flow characteristics of unbalanced valves. 13.5.2.1.2 Balanced Pressure Valve Figure 13.12 depicts a balanced pressure valve. Tubing pressure does not influence valve status when in the closed or open condition. The valve opens and closes at the same pressure—dome pressure. Balanced pressure valves act as expanding orifice regulators, opening to pass any amount of gas injected from the surface and partial closing to control the lower gas flow rate. 13.5.2.1.3 Pilot Valve Figure 13.13 shows a sketch of a pilot valve used for intermittent gas lift where a large port Initial condition(a) (b) (c) (d) (e) Gas enters the first valve Gas enters the second valve GsAll valves open Ptbg All valves open GsGf Gf The first valve begins to close Ptbg Ptbg Gas enters the last valve Unloaded condition Valves closed Valve begins to close All unloading valves closed Gf Gf Ptbg Ptbg Figure 13.7 Well unloading sequence. Tubing Pressure, (psi) Pc Critical flow ratio qg,scf/day Figure 13.8 Flow characteristics of orifice-type valves. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 194 3.1.2007 9:07pm Compositor Name: SJoearun 13/194 ARTIFICIAL LIFT METHODS
- 199. for gas passage and a close control over the spread characteristics are desirable. It has two ports. The smaller port (control port) is used for opening calculations and the large port (power port) is used for gas passage calculations. The equations derived from unbalanced valves are also valid for pilot valves. 13.5.2.2 Throttling Pressure Valve Throttling pressure valves are also called continuous flow valves. As shown in Fig. 13.14, the basic elements of a throttling valve are the same as the pressure-operated valve except that the entrance port of the valve is choked to drop the casing pressure to tubing pressure by using a tapered stem or seat, which allows the port area to sense tubing pressure when the valve is open. Unlike pressure-operated valves where the casing pressure must drop to a pressure set by dome pressure and spring for the valve to close, a throttling pressure valve will close on a reduction in tubing pressure with the casing pressure held constant. The equations derived from pressure-operated Pd AbSt Pc Ap Pt Figure 13.9 Unbalanced bellow valve at its closed condition. AbSt Pc Ap Pt Pd Figure 13.10 Unbalanced bellow valve at its open condition. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 195 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/195
- 200. valves are also to be applied to throttling valves for opening pressure calculations. 13.5.2.3 Fluid-Operated Valve As shown in Fig. 13.15, the basic elements of a fluid-oper- ated valve are identical to those in a pressure-operated valve except that tubing pressure now acts on the larger area of the bellows and casing pressure acts on the area of the port. This configuration makes the valve mostly sensitive to the tubing fluid pressure. Therefore, the opening pressure is defined as the tubing pressure required to open the valve under actual operating conditions. Force balance gives Pvo ¼ 1 1 À R Pd þ St À R 1 À R Pc, (13:44) where Pc ¼ casing pressure, psig. The term R 1ÀR Pc is called the C.E. and R 1ÀR is called T.E.F. for fluid valves. With other parameters given, Eq. (13.44) is used for determining required dome pressure at depth, that is, Pd ¼ (1 À R)Pvo À St þ RPc, in valve selection. When a fluid valve is in its open position under operat- ing conditions, the maximum pressure under the ball (as- sumed to be tubing pressure) required to close the valve is called the valve closing pressure and is expressed as Pvc ¼ Pd þ St 1 À Rð Þ, (13:45) which is identical to that for a pressure-operated valve. The first generation of fluid valves is a differential valve. As illustrated in Fig. 13.16, a differential valve relies on the difference between the casing pressure and the spring pressure effect to open and close. The opening and closing pressures are the same tubing pressure defined as Pvo ¼ Pvc ¼ Pc À St: (13:46) 13.5.2.4 Combination Valves Figure 13.17 shows that a combination valve consists of two portions. The upper portion is essentially the same as that found in pressure-operated valves, and the lower portion is a fluid pilot, or a differential pressure device incorporating a stem and a spring. Holes in the pilot housing allow the casing pressure to act on the area of the stem at the upper end. The spring acts to hold the stem qg Gas Flow Rate Pt TubingPressure Casing pressure Throttlingrange Maximum flow rate Figure 13.11 Flow characteristics of unbalanced valves. Pt Pt Pd Ab Dome As Stem Seal Port Ap Pc Tubing Mandrel Pt Pc Piston (Bellows) Figure 13.12 A sketch of a balanced pressure valve. Dome Pc Pt Ab Ap Pd Pc Pt Piston (Bellows) Pilot Port Piston Piston Bleed Port Seal Main Port Tubing Mandrel Figure 13.13 A sketch of a pilot valve. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 196 3.1.2007 9:07pm Compositor Name: SJoearun 13/196 ARTIFICIAL LIFT METHODS
- 201. in the upward position. This is the open position for the pilot. The casing pressure acts to move the stem to the closed position. The fluid pilot will only open when tubing pressure acting on the pilot area is sufficient to overcome the casing pressure force and move the stem up to the open position. At the instant of opening, the pilot opens completely, providing instantaneous operation for inter- mittent lift. 13.5.3 Valve Spacing Various methods are being used in the industry for design- ing depths of valves of different types. They are the universal design method, the API-recommended method, the fallback method, and the percent load method. However, the basic objective should be the same: 1. To be able to open unloading valves with kickoff and injection operating pressures 2. To ensure single-point injection during unloading and normal operating conditions 3. To inject gas as deep as possible No matter which method is used, the following principles apply: . The design tubing pressure at valve depth is between gas injection pressure (loaded condition) and the minimum tubing pressure (fully unloaded condition). . Depth of the first valve is designed on the basis of kickoff pressure from a special compressor for well kickoff oper- ations. . Depths of other valves are designed on the basis of injection operating pressure. . Kickoff casing pressure margin, injection operating cas- ing pressure margin, and tubing transfer pressure mar- gin are used to consider the following effects: 8 Pressure drop across the valve 8 Tubing pressure effect of the upper valve 8 Nonlinearity of the tubing flow gradient curve. The universal design method explained in this section is valid for all types of continuous-flow gas lift valves. Still, different procedures are used with the universal design method, including the following: a. Design procedure using constant surface opening pres- sure for pressure-operated valves. b. Design procedure using 10- to 20-psi drop in surface closing pressures between valves for pressure-operated valves. Tubing Choke Ap Ab Pd Pt Tubing Entry Port Gas Charged Dome Bellows Pc Valve Entry Ports Stem and Seat Valve Entry Ports Figure 13.14 A sketch of a throttling pressure valve. Tubing Pressure Pt Area of Port Ap Dome (loading element) Bellows (response element) Ab, Area of Bellows St, Spring Tension Pc, Casing Pressure Figure 13.15 A sketch of a fluid-operated valve. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 197 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/197
- 202. c. Design procedure for fluid-operated valves. d. Design procedure for combination of pressure-closed fluid-opened values. Detailed descriptions of these procedures are given by Brown (1980). Only the design procedure using constant surface opening pressure for pressure-operated valves is illustrated in this section. Figure 13.18 illustrates a graphical solution procedure of valve spacing using constant surface opening pressure for pressure-operated valves. The arrows in the figure depict the sequence of line drawing. For a continuous-flow gas lift, the analytical solution procedure is outlined as follows: 1. Starting from a desired wellhead pressure phf at surface, compute a flowing tubing-pressure traverse under fully unloaded condition. This can be done using various two- phase flow correlations such as the modified Hagedorn– Brown correlation (HagedornBrownCorrelation.xls). 2. Starting from a design wellhead pressure phf ,d ¼ phf þ Dphf ,d at surface, where Dphf can be taken as 0:25pc,s establish a design tubing line meeting the flow- ing tubing-pressure traverse at tubing shoe. Pressures in this line, denoted by ptd , represent tubing pressure after adjustment for tubing pressure margin. Gradient of this line is denoted by Gfd . Set Dphf ¼ 0 if tubing pressure margin is not required. 3. Starting from a desired injection operating pressure pc at surface, compute a injection operating pressure line. This can be done using Eq. (13.7) or Eq. (13.9). 4. Starting from pcs À Dpcm at surface, where the casing pressure margin Dpcm can be taken as 50 psi, establish a design casing line parallel to the injection operating pressure line. Pressures in this line, denoted by pcd , represent injection pressure after adjustment for casing pressure margin. Set Dpcm ¼ 0 if the casing pressure margin is not required as in the case of using the universal design method. Small orifice Pt Pc St Pc Ap Ap Figure 13.16 A sketch of a differential valve. Pd Ab Pt Pt Pc Pd Ab Pc Pt Pc Pd Ab Pc Pc Both bellows and pilot valves closed Bellows valve open and pilot valve closed Both bellows and pilot valves open Figure 13.17 A sketch of combination valve. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 198 3.1.2007 9:07pm Compositor Name: SJoearun 13/198 ARTIFICIAL LIFT METHODS
- 203. 5. Starting from available kickoff surface pressure pk,s, establish kickoff casing pressure line. This can be done using Eq. (13.7) or Eq. (13.9). 6. Starting from pk À Dpkm at surface, where the kickoff pressure margin Dpkm can be taken as 50 psi, establish a design kickoff line parallel to the kickoff casing pres- sure line. Pressures in this line, denoted pkd , represent kickoff pressure after adjustment for kickoff pressure margin. Set Dpkm ¼ 0 if kickoff casing pressure margin is not required. 7. Calculate depth of the first valve. Based on the fact that phf þ GsD1 ¼ pkd1, the depth of the top valve is expressed as D1 ¼ pkd1 À phf Gs , (13:47) where pkd1 ¼ kickoff pressure opposite the first valve (psia) Gs ¼ static (dead liquid) gradient; psi/ft Applying Eq. (13.9) gives pkd1 ¼ pk,s À Dpkm À Á 1 þ D1 40,000 : (13:48) Solving Eqs. (13.47) and (13.48) yields D1 ¼ pk,s À Dpkm À phf Gs À pk À Dpkm 40,000 : (13:49) When the static liquid level is below the depth calculated by use of Eq. (13.49), the first valve is placed at a depth slightly deeper than the static level. If the static liquid level is known, then D1 ¼ Ds þ S1, (13:50) where Ds is the static level and S1 is the submergence of the valve below the static level. 8. Calculate the depths to other valves. Based on the fact that phf þ Gfd D2 þ Gs(D2 À D1) ¼ pcd2, the depth of valve 2 is expressed as D2 ¼ pcd2 À Gfd D1 À phf Gs þ D1, (13:51) where pcd2 ¼ design injection pressure at valve 2, psig Gfd ¼ design unloading gradient, psi/ft. Applying Eq. (13.9) gives pcd2 ¼ pc,s À Dpcm À Á 1 þ D2 40,000 : (13:52) Solving Eqs. (13.51) and (13.52) yields D2 ¼ pc,s À Dpcm À phf ,d þ Gs À Gfd À Á D1 Gs À pc À Dpcm 40,000 : (13:53) Similarly, the depth to the third valve is D3 ¼ pc,s À Dpcm À phf ,d þ Gs À Gfd À Á D2 Gs À pc À Dpcm 40,000 : (13:54) Thus, a general equation for depth of valve i is Di ¼ pc,s À Dpcm À phf ,d þ Gs À Gfd À Á DiÀ1 Gs À pc À Dpcm 40,000 : (13:55) Depths of all valves can be calculated in a similar manner until the minimum valve spacing ( $ 400 ft) is reached. Example Problem 13.5 Only 1 MMscf/day of lift gas is available for the well described in the Example Problem 13.1. If 1,000 psia is available to kick off the well and then a steady injection pressure of 800 psia is maintained for gas lift operation against a wellhead pressure of 130 psia, design locations of unloading and operating valves. Assume a casing pressure margin of 50 psi. Solution The hydrostatic pressure of well fluid (26 8API oil) is (0.39 psi/ft) (5,200 ft), or 2,028 psig, which is greater than the given reservoir pressure of 2,000 psia. Therefore, the well does not flow naturally. The static liquid level depth is estimated to be 5,200 À (2,000 À 14:7)=(0:39) ¼ 110 ft: Depth of the top valve is calculated with Eq. (13.49): D1 ¼ 1,000 À 50 À 130 0:39 À 1,000 À 50 40,000 ¼ 2,245 ft 110 ft Tubing pressure margin at surface is (0.25)(800), or 200 psi. The modified Hagedorn–Brown correlation gives tubing pressure of 591 psia at depth of 5,000 ft. The design tubing flowing gradient is Gfd ¼ [591 À (130 þ 200)]= (5,000) or 0.052 psi/ft. Depth of the second valve is calcu- lated with Eq. (13.53): D2 ¼ 1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(2,245) 0:39 À 1,000 À 50 40,000 ¼ 3,004 ft Similarly, D3 ¼ 1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(3,004) 0:39 À 1,000 À 50 40,000 ¼ 3,676 ft D4 ¼ 1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(3,676) 0:39 À 1,000 À 50 40,000 ¼ 4,269 ft D5 ¼ 1,000 À 50 À 330 þ 0:39 À 0:052ð Þ(4,269) 0:39 À 1,000 À 50 40,000 ¼ 4,792 ft, which is the depth of the operating valve. Similar problems can be quickly solved with the com- puter spreadsheet GasLiftValveSpacing.xls. Pressure Operating Tubing Pressure Injection Operating Pressure Kick-off Pressure Depth phf phf,d pc,s pk,s ∆ptm ∆pcm ∆pkm G s G s G s G s G s G s G s Gf Gf,d Figure 13.18 A flow diagram to illustrate procedure of valve spacing. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 199 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/199
- 204. 13.5.4 Valve selection and testing Valve selection starts from sizing of valves to determine required proper port size Ap and area ratio R. Valve test- ing sets dome pressure Pd and/or string load St. Both of the processes are valve-type dependent. 13.5.4.1 Valve Sizing Gas lift valves are sized on the basis of required gas passage through the valve. All the equations presented in Section 13.4.2.3 for choke flow are applicable to valve port area calculations. Unloading and operating valves (ori- fices) are sized on the basis of subcritical (subsonic flow) that occurs when the pressure ratio Pt=Pc is greater than the critical pressure ratio defined in the right-hand side of Eq. (13.11). The value of the k is about 1.28 for natural gas. Thus, the critical pressure ratio is about 0.55. Re- arranging Eq. (13.12) gives Ap ¼ qgM 1,248Cpup ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k (kÀ1)ggTup pdn pup 2 k À pdn pup kþ1 k !s : (13:56) Since the flow coefficient C is port-diameter dependent, a trial-and-error method is required to get a solution. A conservative C value is 0.6 for orifice-type valve ports. Once the required port area is determined, the port diam- eter can then be calculated by dp ¼ 1:1284 ffiffiffiffiffiffi Ap p and up-rounded off to the nearest 1 ⁄16 in. The values of the port area to bellows area ratio R are fixed for given valve sizes and port diameters by valve manufacturers. Table 13.4 presents R values for Otis Spreadmaster Valves. Example Problem 13.6 Size port for the data given below: Upstream pressure: 900 psia Downstream pressure for subsonic flow: 600 psia Tubing ID: 2.259 in. Gas rate: 2,500 Mscf/day Gas-specific gravity: 0.75 (1 for air) Gas-specific heat ratio: 1.3 Upstream temperature: 110 8F Gas viscosity: 0.02 cp Choke discharge coefficient: 0.6 Use Otis Spreadmaster Valve Solution Ap ¼ 2,500 1,248(0:6)(900) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:3 (1:3À1)(0:75)(110þ460) 600 900 À Á2 k À 600 900 À Á1:3þ1 1:3 h ir Ap ¼ 0:1684 in:2 dp ¼ 1:1284 ffiffiffiffiffiffiffiffiffiffiffi 1:684 p ¼ 0:4631 in: Table 13.1 shows that an Otis 11 ⁄2 -in. outside diameter (OD) valve with 1 ⁄2 -in. diameter seat will meet the require- ment. It has an R value of 0.2562. 13.5.4.2 Valve Testing Before sending to field for installation, every gas lift valve should be set and tested at an opening pressure in the shop that corresponds to the desired opening pressure in the well. The pressure is called test rack opening pressure (Ptro). The test is run with zero tubing pressure for pres- sure-operated valves and zero casing pressure for fluid- operated valves at a standard temperature (60 8F in the U.S. petroleum industry). For pressure-operated unbal- anced bellow valves at zero tubing pressure, Eq. (13.42) becomes Ptro ¼ Pd at 60 F 1 À R þ St: (13:57) For fluid-operated valves at zero casing pressure, Eq. (13.44) also reduces to Eq. (13.57) at zero casing pressure and 60 8F. To set Pd at 60 8F to a value representing Pd at valve depth condition, real gas law must be used for cor- rection: Pd at 60 F ¼ 520z60 FPd Td zd , (13:58) where Td ¼ temperature at valve depth, 8R zd ¼ gas compressibility factor at valve depth condition. The z factors in Eq. (13.58) can be determined using the Hall–Yarborogh correlation. Computer spreadsheet Hall- Yarborogh-z.xls is for this purpose. Table 13.4 R Values for Otis Spreadmaster Valves Port Diameter (in.) 9 ⁄16 -in. OD Valves 1-in. OD Valves 11 ⁄2 -in. OD Valves R 1 À R T.E.F. R 1 À R T.E.F. R 1 À R T.E.F. (1 ⁄8 ) 0.1250 0.1016 0.8984 0.1130 0.0383 0.9617 0.0398 0.1520 0.1508 0.8429 0.1775 0.1730 0.1958 0.8042 0.2434 (3 ⁄16 ) 0.1875 0.0863 0.9137 0.0945 0.0359 0.9641 0.0372 0.1960 0.2508 0.7492 0.3347 (13 ⁄64 ) 0.2031 0.1013 0.8987 0.1127 0.2130 0.2966 0.7034 0.4216 0.2460 0.3958 0.6042 0.6550 (1 ⁄4 ) 0.2500 0.1534 0.8466 0.1812 0.0638 0.9362 0.0681 (9 ⁄32 ) 0.2812 0.1942 0.8058 0.2410 (5 ⁄16 ) 0.3125 0.2397 0.7603 0.3153 0.0996 0.9004 0.1106 (11 ⁄32 ) 0.3437 0.2900 0.7100 0.4085 (3 ⁄8 ) 0.3750 0.3450 0.6550 0.5267 0.1434 0.8566 0.1674 (7 ⁄16 ) 0.4375 0.4697 0.5303 0.8857 0.1952 0.8048 0.2425 (1 ⁄2 ) 0.5000 0.2562 0.7438 0.3444 (9 ⁄16 ) 0.5625 0.3227 0.6773 0.4765 (5 ⁄8 ) 0.6250 0.3984 0.6016 0.6622 (3 ⁄4 ) 0.7500 0.5738 0.4262 1.3463 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 200 3.1.2007 9:07pm Compositor Name: SJoearun 13/200 ARTIFICIAL LIFT METHODS
- 205. Equation (13.57) indicates that the Ptro also depends on the optional string load St for double-element valves. The St value can be determined on the basis of manufacturer’s literature. The procedure for setting and testing valves in a shop is as follows: . Install valve in test rack. . Adjust spring setting until the valve opens with St psig applied pressure. This sets St value in the valve. . Pressureupthedomewithnitrogengas.Coolvalveto60 8F. . Bleed pressure off of dome until valve opens with Ptro psig applied pressure. Example Problem 13.7 Design gas lift valves using the following data: Pay zone depth: 6,500 ft Casing size and weight: 7 in., 23 lb. Tubing 23 ⁄8 in., 4.7 lb. (1.995 in. ID) Liquid level surface: Kill fluid gradient: 0.4 psi/ft Gas gravity: 0.75 Bottom-hole temperature: 170 8F Temperature surface flowing: 100 8F Injection depth: 6,300 ft Minimum tubing pressure at injection point: 600 psi Pressure kickoff: 1,000 psi Pressure surface operating: 900 psi Pressure of wellhead: 120 psi Tubing pressure margin at surface: 200 psi Casing pressure margin: 0 psi Valve specifications given by Example Problem 13.6 Solution Design tubing pressure at surface ( phf ,d ): 120 þ 200 ¼ 320 psia Design tubing pressure gradient (Gfd ): (600 À 320)=6,300 ¼ 0:044 psi=ft Temperature gradient (Gt): (170 À 100)=6,300 ¼ 0:011 F=ft 1 À R 1:0 À 0:2562 ¼ 0:7438 T:E:F: ¼ R=(1 À R) 0:2562=0:7438 ¼ 0:3444 Depth of the top valve is calculated with Eq. (13.49): D1 ¼ 1,000 À 0 À 120 0:40 À 1,000 À 0 40,000 ¼ 2,347 ft Temperature at the top valve: 100 þ (0:011) (2,347) ¼ 126 F Design tubing pressure at the top valve: 320 þ (0:044) (2,347) ¼ 424 psia For constant surface opening pressure of 900 psia, the valve opening pressure is calculated with Eq. (13.9): pvo1 ¼ (900) 1 þ 2,347 40,000 ¼ 953 psia The dome pressure at the valve depth is calculated on the basis of Eq. (13.42): Pd ¼ 0:7438(953) À 0 þ (0:2562)(424 ) ¼ 817 psia The valve closing pressure at the valve depth is calculated with Eq. (13.43): Pvc ¼ 817 þ (0) 0:7438ð Þ ¼ 817 psia The dome pressure at 60 8F can be calculated with a trial- and-error method. The first estimate is given by idea gas law: Pd at 60 F ¼ 520Pd Td ¼ (520)(817) (126 þ 460) ¼ 725 psia Spreadsheet programs give z60F ¼ 0:80 at 725 psia and 60 8F. The same spreadsheet gives zd ¼ 0:85 at 817 psia and 126 8F. Then Eq. (13.58) gives Pd at 60 F ¼ (520)(0:80)Pd (126 þ 460)(0:85) (817) ¼ 683 psia: Test rack opening pressure is given by Eq. (13.57) as Ptro ¼ 683 0:7438 þ 0 ¼ 918 psia: Following the same procedure, parameters for other valves are calculated. The results are summarized in Table 13.5. The spreadsheet program GasLiftValveDesign.xls can be used to seek solutions of similar problems. 13.6 Special Issues in Intermittent-Flow Gas Lift The intermittent-flow mechanism is very different from that of the continuous-flow gas lift. It is normally applicable in either high-BHP–low PI or low-BHP–low PI reservoirs. In these two reservoir cases, an excessive high drawdown is needed, which results in a prohibitively high GLR to pro- duce the desired quantity of oil (liquid) by continuous gas lift. In many instances, the reservoir simply is not capable of giving up the desired liquid regardless of drawdown. The flow from a well using intermittent gas lift techniques is called ‘‘ballistic’’ or ‘‘slug’’ flow. Two major factors that define the intermittent-gas lift process must be understood: 1. Complex flowing gradient of the gas lifted liquids from the well. 2. Contribution of the PI of the well to the actual deliver- ability of liquid to the surface. Figure 13.19 shows the BHP of a well being produced by intermittent-flow gas lift. The BHP at the instant the valve opens is indicated by Point A. The pressure impulse results in an instantaneous pressure buildup at Point B, which reaches a maximum at C after the initial acceleration of the oil column. Figure 13.20 shows the intermittent-flowing gradient, which is a summation of the gradient of gas above the slug, the gradient of the slug, and the gradient of the lift gas and entrained liquids below the slug. Table 13.5 Summary of Results for Example Problem 13.7 Valve no. Valve depth (ft) Temperature (8F) Design tubing pressure (psia) Surface opening pressure (psia) Valve opening pressure (psia) Dome pressure at depth (psia) Valve closing pressure (psia) Dome pressure at 60 8F (psia) Test rack opening (psia) 1 2,347 126 424 900 953 817 817 683 918 2 3,747 142 487 900 984 857 857 707 950 3 5,065 156 545 900 1,014 894 894 702 944 4 6,300 170 600 900 1,042 929 929 708 952 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 201 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/201
- 206. Example Problem 13.8 Determine the depth to the oper- ating (master) valve and the minimum GLR ratio for the following well data: Depth ¼ 8,000 ft pso ¼ 800 psig 23 ⁄8 -in. tubing ¼ 1:995 in: ID 51 ⁄2 -in., 20 lb/ft casing No water production go ¼ 0:8762, 30 8API BHP (SI) ¼ 2,000 psig PI ¼ 0:10 bbl=day=psi ptf ¼ 50 psig tav ¼ 127 F Cycle time: 45 minutes Desired production: 100 bbl/day gg ¼ 0:80 Solution The static gradient is Gs ¼ 0:8762(0:433) ¼ 0:379 psi=ft: Thus, the average flowing BHP is Pbhfave ¼ 2,000 À 1,000 ¼ 1,000 psig: The depth to the static fluid level with the ptf ¼ 50 psig, is Ds ¼ 8,000 À 2,000 À 50 0:379 ¼ 2,855 ft: The hydrostatic head after a 1,000 psi drawdown is Ddds ¼ 1,000 0:379 ¼ 2,639 ft: Thus, the depth to the working fluid level is WFL ¼ Ds þ Ddds ¼ 2,855 þ 2,639 ¼ 5,494 ft: Shut-in pressure build-up curve Bottom-holepressure PWS T1 T2 T3 Time Valve open Valve closed Valve open Valve closed A B C D E Figure 13.19 Illustrative plot of BHP of an intermittent flow. Pressure PSP Slug Depth Operating valve Static G radient PWS Gradient of gas and entrained liquid below slug Gradient of slug Gradient in gas Above slug Figure 13.20 Intermittent flow gradient at midpoint of tubing. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 202 3.1.2007 9:07pm Compositor Name: SJoearun 13/202 ARTIFICIAL LIFT METHODS
- 207. Figure 13.21 shows the example well and the WFL. The number of cycles per day is approximately 24 60ð Þ 45 ¼ 32 cycles/day. The number of bbls per cycle is 100 32 % 3 bbls=cycle. Intermittent-gas lift operating experience shows that depending on depth, 30–60% of the total liquid slug is lost due to slippage or fallback. If a 40% loss of starting slug is assumed, the volume of the starting slug is 3 0:60 % 5:0 bbl=cycle. Because the capacity of our tubing is 0.00387 bbl/ft, the length of the starting slug is 5:0 0:00387 % 1,292 ft. This means that the operating valve should be located 1,292 2 ¼ 646 ft below the working fluid level. Therefore, the depth to the operating valve is 5,494 þ 646 ¼ 6,140 ft. The pressure in the tubing opposite the operating valve with the 50 psig surface back-pressure (neglecting the weight of the gas column) is pt ¼ 50 þ 1,292ð Þ 0:379ð Þ ¼ 540 psig: For minimum slippage and fallback, a minimum velocity of the slug up the tubing should be 100 ft/min. This is accomplished by having the pressure in the casing opposite the operating valve at the instant the valve opens to be at least 50% greater than the tubing pressure with a minimum differential of 200 psi. Therefore, for a tubing pressure at the valve depth of 540 psig, at the instant the valve opens, the minimum casing pressure at 6,140 ft is pmin c ¼ 540 þ 540=2 ¼ 810 psig: Equation (13.10) gives a pso ¼ 707 psig. The minimum volume of gas required to lift the slug to the surface will be that required to fill the tubing from injection depth to surface, less the volume occupied by the slug. Thus, this volume is 6,140 þ 1,292ð Þ 0:00387 ¼ 18:8 bbls, which converts to 105:5 ft3 . The approximate pressure in the tubing immediately under a liquid slug at the instant the slug surfaces is equal to the pressure due to the slug length plus the tubing backpressure. This is pts ¼ 50 þ 3:0 0:00387 ! 0:379ð Þ ¼ 344 psig: Thus, the average pressure in the tubing is ptave ¼ 810 þ 344 2 ¼ 577 psig ¼ 591:7 psia: The average temperature in the tubing is 127 8F or 587 8R. This gives z ¼ 0:886. The volume of gas at standard condi- tions (API 60 8F, 14.695 psia) is Vsc ¼ 105:5 591:7 14:695 520 587 1 0:886 ¼ 4,246 scf=cycle: 13.7 Design of Gas Lift Installations Different types of gas lift installations are used in the in- dustry depending on well conditions. They fall into four categories: (1) open installation, (2) semiclosed installation, (3) closed installation, and (4) chamber installation. As shown in Fig. 13.22a, no packer is set in open installa- tions. This type of installation is suitable for continuous flow gas lift in wells with good fluid seal. Although this type of installation is simple, it exposes all gas lift valves beneath the pointofgasinjectiontoseverefluiderosionduetothedynamic changing of liquid level in the annulus. Open installation is not recommended unless setting packer is not an option. Time 8000’ Depthtooperatingvalve=6140’ 646’WFL=5494’ SFL=28551292’ P=490PSI 1993’ P=755PSI SFL Max working BHP=1245PSIG MIN Working BHP=755PSIG AVG Working BHP=1000PSIG WFL Bottomhole pressure BHPS = 2000PSIG Pressure build-up curve Valve opens Slug surfacing 245PSI 490PSI 755PSI 45MIN AVGDD=1000PSI 45MIN Figure 13.21 Example Problem 13.8 schematic and BHP buildup for slug flow. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 203 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/203
- 208. Figure 13.22b demonstrates a semiclosed installation. It is identical to the open installation except that a packer is set between the tubing and casing. This type of installation can be used for both continuous- and intermittent-flow gas lift operations. It avoids all the problems associated with the open installations. However, it still does not prevent flow of well fluids back to formation during unloading processes, which is especially important for intermittent operating. Illustrated in Fig. 13.22c is a closed installation where a standing valve is placed in the tubing string or below the bottom gas lift valve. The standing valve effectively pre- vents the gas pressure from acting on the formation, which increases the daily production rate from a well of the intermittent type. Chamber installations are used for accumulating liquid volume at bottom hole of intermittent-flow gas lift wells. A chamber is an ideal installation for a low BHP and high PI well. The chambers can be configured in various ways including using two packers, insert chamber, and reverse flow chamber. Figure 13.23 shows a standard two-packer chamber. This type of chamber is installed to ensure a large storage volume of liquids with a minimum amount of backpressure on the formation so that the liquid pro- duction rate is not hindered. a Production Out Gas In Open Production Out Gas In Semi-Closed Production Out Gas In Closed cb Continuous Flow Applications Intermitting Lift Applications Figure 13.22 Three types of gas lift installations. a Flow Unloading Gas lift Valves Bottom unloading Gas lift Valves Operating Chamber Gas lift Valves By-pass Packer Perforated sub Bottom Packer Standing Valve Gas Bleed Valve Standing Valve b Standing Valve modified for handing Sand Figure 13.23 Sketch of a standard two-packer chamber. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 204 3.1.2007 9:07pm Compositor Name: SJoearun 13/204 ARTIFICIAL LIFT METHODS
- 209. Figure 13.24 illustrates an insert chamber. It is normally used in a long open hole or perforated interval where squeezing of fluids back to formation by gas pressure is a concern. It takes the advantage of existing bottom-hole pressure. The disadvantage of the installation is that the chamber size is limited by casing diameter. Shown in Fig. 13.25 is a reverse flow chamber. It ensures venting of all formation gas into the tubing string to empty the chamber for liquid accumulation. For wells with high- formation GLR, this option appears to be an excellent choice. Summary This chapter presents the principles of gas lift systems and illustrates a procedure for designing gas lift operations. Major tasks include calculations of well deliverability, pressure and horsepower requirements for gas lift gas compression, gas lift valve selection and spacing, and selection of installation methods. Optimization of existing gas lift systems is left to Chapter 18. References brown, k.e. The Technology of Artificial Lift Methods, Vol. 1. Tulsa, OK: PennWell Books, 1977. brown, k.e. The Technology of Artificial Lift Methods, Vol. 2a. Tulsa, OK: Petroleum Publishing Co., 1980. economides, m.j., hill, a.d., and ehig-economides, c. Petroleum Production Systems. New Jersey: Prentice Hall PTR, 1994. gilbert, w.e. Flowing and gas-lift well performance. API Drill. Prod. Practice 1954. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston, TX: Gulf Publishing Co., 2005. katz, d.l., cornell, d., kobayashi, r., poettmann, f.h., vary, j.a., elenbaas, j.r., and weinaug, c.f. Handbook of Natural Gas Engineering. New York: McGraw-Hill Publishing Company, 1959. weymouth, t.r. Problems in Natural Gas Engineering. Trans. ASME 1912;34:185. Problems 13.1 An oil well has a pay zone around the mid-perf depth of 5,200 ft. The formation oil has a gravity of 30 8API and GLR of 500 scf/stb. Water cut remains 10%. The IPR of the well is expressed as q ¼ Jbpp À pwf c, where J ¼ 0:5 stb=day=psi pp ¼ 2,000 psia. A 2-in. tubing (1.995-in. ID) can be set with a packerat200 ftabovethemid-perf.Whatisthemaximum expectedoilproductionratefromthewellwithcontinuous gas lift at a wellhead pressure of 200 psia if a. unlimited amount of lift gas is available for the well? b. only 1.2 MMscf/day of lift gas is available for the well? 13.2 An oil well has a pay zone around the mid-perf depth of 6,200 ft. The formation oil has a gravity of 30 8API and GLR of 500 scf/stb. Water cut remains 10%. The IPR of the well is expressed as q ¼ qmax 1 À 0:2 pwf pp À 0:80:2 pwf pp 2 # , where qmax ¼ 2,000 stb=day pp ¼ 2,500 psia. A21 ⁄2 -in.tubing(2.259-in.ID)canbesetwithapacker at 200 ft above the mid-perf. What is the maximum expected oil production rate from the well with con- tinuous gas lift at a wellhead pressure of 150 psia if a. unlimited amount of lift gas is available for the well? b. only 1.0 MMscf/day of lift gas is available for the well? 13.3 An oil field has 24 oil wells defined in Problem 13.1. The gas lift gas at the central compressor station is first pumped to three injection manifolds with 6-in. ID, 2-mile lines and then distributed to the well heads with 4 in. ID, 0.5-mile lines. Given the following Gas Flow Unloading Gas lift Valve Bottom unloading Gas lift Valve Hanger Nipple for Dip tube Operating Chamber Gas lift Valve Packer Bleed Port or Valve Standing Valve Figure 13.24 A sketch of an insert chamber. Unloading Valves Operating Valve Stinger tube Top Packer Perforated Nipple Stinger Receiver Perforated Nipple Standing Valve Packer Liquids Gas Gas Figure 13.25 A sketch of a reserve flow chamber. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 205 3.1.2007 9:07pm Compositor Name: SJoearun GAS LIFT 13/205
- 210. data, calculate the required output pressure of the compression station: Gas-specific gravity (gg): 0.75 Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia. 13.4 An oil field has 32 oil wells defined in Problem 13.2. The gas lift gas at the central compressor station is first pumped to four injection manifolds with 4-in. ID, 1.5-mile lines and then distributed to the well- heads with 4-in. ID, 0.4-mile lines. Given the fol- lowing data, calculate the required output pressure of compression station: Gas-specific gravity (gg): 0.70 Base temperature (Tb): 60 8F Base pressure ( pb): 14.7 psia 13.5 For a reciprocating compressor, calculate the theo- retical and brake horsepower required to compress 50 MMcfd of a 0.7-gravity natural gas from 200 psia and 70 8F to 2,500 psia. If intercoolers cool the gas to 90 8F, what is the heat load on the intercoolers and what is the final gas temperature? Assuming the overall efficiency is 0.75. 13.6 For a reciprocating compressor, calculate the theo- retical and brake horsepower required to compress 30 MMcfd of a 0.65-gravity natural gas from 100 psia and 70 8F to 2,000 psia. If intercoolers and endcoolers cool the gas to 90 8F, what is the heat load on the coolers? Assuming the overall efficiency is 0.80. 13.7 For a centrifugal compressor, use the following data to calculate required input horsepower and polytro- pic head: Gas-specific gravity: 0.70 Gas-specific heat ratio: 1.30 Gas flow rate: 50 MMscfd at 14.7 psia and 60 8F Inlet pressure: 200 psia Inlet temperature: 70 8F Discharge pressure: 500 psia Polytropicefficiency: Ep ¼ 0:61 þ 0:03 log (q1) 13.8 For the data given in Problem 13.7, calculate the required brake horsepower if a reciprocating com- pressor is used. 13.9 Only 1 MMscf/day of lift gas is available for the well described in Problem 13.3. If 1,000 psia is available to kick off the well and then a steady injection pressure of 800 psia is maintained for gas lift oper- ation against a wellhead pressure of 130 psia, design locations of unloading and operating valves. As- sume a casing pressure margin of 0 psi. 13.10 An unlimited amount of lift gas is available for the well described in Problem 13.4. If 1,100 psia is avail- able to kick off the well and then a steady injection pressure of 900 psia is maintained for gas lift oper- ation against a wellhead pressure of 150 psia, design locations of unloading and operating valves. Assume a casing pressure margin of 50 psi. 13.11 Size port for the data given below: Upstream pressure: 950 psia Downstream pressure for subsonic flow: 650 psia Tubing ID: 2.259 in. Gas rate: 2,000 Mscf/day Gas-specific gravity: 0.70 (1 for air) Gas-specific heat ratio: 1.3 Upstream temperature: 100 8F Gas viscosity: 0.02 cp Choke discharge coefficient: 0.6 Use Otis Spreadmaster Valve 13.12 Size port for the data given below: Upstream pressure: 950 psia Downstream pressure for subsonic flow: 550 psia Tubing ID: 1.995 in. Gas rate: 1,500 Mscf/day Gas specific gravity: 0.70 (1 for air) Gas specific heat ratio: 1.3 Upstream temperature: 80 8F Gas viscosity: 0.03 cp Choke discharge coefficient: 0.6 Use Otis Spreadmaster Valve 13.13 Design gas lift valves using the following data: Pay zone depth: 5,500 ft Casing size and weight: 7 in., 23 lb Tubing 23 ⁄8 in., 4.7 lb (1.995-in. ID): Liquid level surface: Kill fluid gradient: 0.4 psi/ft Gas gravity: 0.65 Bottom-hole temperature: 150 8F Temperature surface flowing: 80 8F Injection depth: 5,300 ft The minimum tubing pressure at injection point: 550 psi Pressure kickoff: 950 psi Pressure surface operating: 900 psi Pressure of wellhead: 150 psi Tubing pressure margin at surface: 200 psi Casing pressure margin: 0 psi Otis 11 ⁄2 -in. OD valve with 1 ⁄2 -in. diameter seat: R ¼ 0:2562 13.14 Design gas lift valves using the following data: Pay zone depth: 7,500 ft Casing size and weight: 7 in., 23 lb Tubing 23 ⁄8 -in., 4.7 lb (1.995 in. ID): Liquid level surface: Kill fluid gradient: 0.4 psi/ft Gas gravity: 0.70 Bottom-hole temperature: 160 8F Temperature surface flowing: 90 8F Injection depth: 7,300 ft The minimum tubing pressure at injection point: 650 psi Pressure kickoff: 1,050 psi Pressure surface operating: 950 psi Pressure of wellhead: 150 psi Tubing pressure margin at surface: 200 psi Casing pressure margin: 10 psi Otis 1-in. OD valve with 1 ⁄2 -in. diameter seat: R = 0.1942 13.15 Determine the gas lift gas requirement for the following well data: Depth ¼ 7,500 ft pso ¼ 800 psig 23 ⁄8 -in. tubing ¼ 1:995 in: ID 51 ⁄2 -in., 20-lb/ft casing No water production go ¼ 0:8762,30 API BHP (SI) ¼ 1,800 psig PI ¼ 0:125 bbl=day=psi ptf ¼ 50 psig tav ¼ 120 F Cycle time: 45 minutes Desired production: 150 bbl/day gg ¼ 0:70 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap13 Final Proof page 206 3.1.2007 9:07pm Compositor Name: SJoearun 13/206 ARTIFICIAL LIFT METHODS
- 211. 14 Other Artificial Lift Methods Contents 14.1 Introduction 14/208 14.2 Electrical Submersible Pump 14/208 14.3 Hydraulic Piston Pumping 14/211 14.4 Progressive Cavity Pumping 14/213 14.5 Plunger Lift 14/215 14.6 Hydraulic Jet Pumping 14/220 Summary 14/222 References 14/222 Problems 14/223 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 207 3.1.2007 9:10pm Compositor Name: SJoearun
- 212. 14.1 Introduction In addition to beam pumping and gas lift systems, other artificial lift systems are used in the oil industry. They are electrical submersible pumping, hydraulic piston pumping, hydraulic jet pumping, progressive cavity pumping, and plunger lift systems. All these systems are continuous pumping systems except the plunger lift, which is very similar to intermittent gas lift systems. 14.2 Electrical Submersible Pump Electrical submersible pumps (ESPs) are easy to install and operate. They can lift extremely high volumes from highly productive oil reservoirs. Crooked/deviated holes present no problem. ESPs are applicable to offshore operations. Lifting costs for high volumes are generally very low. Limitations to ESP applications include high- voltage electricity availability, not applicable to multiple completions, not suitable to deep and high-temperature oil reservoirs, gas and solids production is troublesome, and costly to install and repair. ESP systems have higher horsepower, operate in hotter applications, are used in dual installations and as spare down-hole units, and include down-hole oil/water separation. Sand and gas problems have led to new products. Automation of the systems includes monitoring, analysis, and control. The ESP is a relatively efficient artificial lift. Under certain conditions, it is even more efficient than sucker rod beam pumping. As shown in Fig. 14.1, an ESP consists of subsurface and surface components. a. Subsurface components - Pump - Motor - Seal electric cable - Gas separator b. Surface components - Motor controller (or variable speed controller) - Transformer - Surface electric cable The overall ESP system operates like any electric pump commonly used in other industrial applications. In ESP operations, electric energy is transported to the down-hole electric motor via the electric cables. These electric cables are run on the side of (and are attached to) the production tubing. The electric cable provides the electrical energy needed to actuate the down-hole electric motor. The elec- tric motor drives the pump and the pump imparts energy to the fluid in the form of hydraulic power, which lifts the fluid to the surface. 14.2.1 Principle ESPs are pumps made of dynamic pump stages or centri- fugal pump stages. Figure 14.2 gives the internal schematic of a single-stage centrifugal pump. Figure 14.3 shows a cutaway of a multistage centrifugal pump. The electric motor connects directly to the centrifugal pump module in an ESP. This means that the electric motor shaft connects directly to the pump shaft. Thus, the pump rotates at the same speed as the electric motor. Switchboard AMP meter Transformers Well head Drain valve Check valve Cable-round Splice Motor flat Pump Intake Seal section Motor Surface cable Vent box Tubing Casing Figure 14.1 A sketch of an ESP installation (Centrilift-Hughes, Inc., 1998). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 208 3.1.2007 9:10pm Compositor Name: SJoearun 14/208 ARTIFICIAL LIFT METHODS
- 213. Like most down-hole tools in the oil field, ESPs are clas- sified by their outside diameter (from 3.5 to 10.0 in.). The number of stages to be used in a particular outside diam- eter sized pump is determined by the volumetric flow rate and the lift (height) required. Thus, the length of a pump module can be 40–344 in. in length. Electric motors are three-phase (AC), squirrel cage, induction type. They can vary from 10 to 750 hp at 60 Hz or 50 Hz (and range from 33 ⁄4 to 71 ⁄4 in. in diameter). Their voltage requirements vary from 420–4,200 V. The seal system (the protector) separates the well fluids from the electric motor lubrication fluids and the electrical wiring. The electric controller (surface) serves to energize the ESP, sensing such conditions as overload, well pump- off, short in cable, and so on. It also shuts down or starts up in response to down-hole pressure switches, tank levels, or remote commands. These controllers are available in conventional electromechanical or solid-state devices. Conventional electromechanical controllers give a fixed- speed, fixed flow rate pumping. To overcome this limita- tion, the variable speed controller has been developed (solid state). These controllers allow the frequency of the electric current to vary. This results in a variation in speed (rpm) and, thus, flow rate. Such a device allows changes to be made (on the fly) whenever a well changes volume (static level), pressure, GLR, or WOR. It also allows flexi- bility for operations in wells where the PI is not well known. The transformer (at surface) changes the voltage of the distribution system to a voltage required by the ESP system. Unlike positive-displacement pumps, centrifugal pumps do not displace a fixed amount of fluid but create a rela- tively constant amount of pressure increase to the flow system. The output flow rate depends on backpressure. The pressure increase is usually expressed as pumping head, the equivalent height of freshwater that the pressure differential can support (pumps are tested with freshwater by the manufacturer). In U.S. field units, the pumping head is expressed as h ¼ Dp 0:433 , (14:1) where h ¼ pumping head, ft Dp ¼ pump pressure differential, psi. As the volumetric throughput increases, the pumping head of a centrifugal pump decreases and power slightly increases. However, there exists an optimal range of flow rate where the pump efficiency is maximal. A typical ESP characteristic chart is shown in Fig. 14.4. ESPs can operate over a wide range of parameters (depths and volumes), to depths over 12,000 ft and volu- metric flow rates of up to 45,000 bbl/day. Certain operat- ing variables can severely limit ESP applications, including the following: . Free gas in oil . Temperature at depth . Viscosity of oil . Sand content fluid . Paraffin content of fluid Excessive free gas results in pump cavitation that leads to motor fluctuations that ultimately reduces run life and reliability. High temperature at depth will limit the life of the thrust bearing, the epoxy encapsula- tions (of electronics, etc.), insulation, and elastomers. Increased viscosity of the fluid to be pumped reduces the total head that the pump system can generate, which leads to an increased number of pump stages and increased horsepower requirements. Sand and paraffin content in the fluid will lead to wear and choking conditions inside the pump. 14.2.2 ESP Applications The following factors are important in designing ESP applications: . PI of the well . Casing and tubing sizes . Static liquid level ESPs are usually for high PI wells. More and more ESP applications are found in offshore wells. The outside di- ameter of the ESP down-hole equipment is determined by the inside diameter (ID) of the borehole. There must be Diffuser Impeller Figure 14.2 An internal schematic of centrifugal pump. Figure 14.3 A sketch of a multistage centrifugal pump. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 209 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/209
- 214. clearance around the outside of the pump down-hole equipment to allow the free flow of oil/water to the pump intake. The desired flow rate and tubing size will determine the total dynamic head (TDH) requirements for the ESP system. The ‘‘TDH’’ is defined as the pressure head immediately above the pump (in the tubing). This is converted to feet of head (or meters of head). This TDH is usually given in water equivalent. Thus, TDH ¼ static column of fluid (net) head + friction loss head + back- pressure head. The following procedure can be used for selecting an ESP: 1. Starting from well inflow performance relationship (IPR), determine a desirable liquid production rate qLd . Then select a pump size from the manufacturer’s specification that has a minimum delivering flow rate qLp, that is, qLp qLd . 2. From the IPR, determine the flowing bottom-hole pressure pwf at the pump-delivering flow rate qLp, not the qLd . 3. Assuming zero casing pressure and neglecting gas weight in the annulus, calculate the minimum pump depth by Dpump ¼ D À pwf À psuction 0:433gL , (14:2) where Dpump ¼ minimum pump depth, ft D ¼ depth of production interval, ft pwf ¼ flowing bottom-hole pressure, psia psuction ¼ required suctionpressureofpump,150–300 psi gL ¼ specific gravity of production fluid, 1.0 for freshwater. 4. Determine the required pump discharge pressure based on wellhead pressure, tubing size, flow rate qLp, and fluid properties. This can be carried out quickly using the computer spreadsheet HagedornBrownCorrelation.xls. 5. Calculate the required pump pressure differential Dp ¼ pdischarge À psuction and then required pumping head by Eq. (14.1). 6. From the manufacturer’s pump characteristics curve, read pump head or head per stage. Then calculate the required number of stages. 7. Determine the total power required for the pump by multiplying the power per stage by the number of stages. Example Problem 14.1 A 10,000-ft-deep well produces 32 8API oil with GOR 50 scf/stb and zero water cut through a 3-in. (2.992-in. ID) tubing in a 7-in. casing. The oil has a formation volume factor of 1.25 and average viscosity of 5 cp. Gas-specific gravity is 0.7. The surface and bottom-hole temperatures are 70 8F and 170 8F, respectively. The IPR of the well can be described by the Vogel model with a reservoir pressure 4,350 psia and AOF 15,000 stb/day. If the well is to be put in production with an ESP to produce liquid at 8,000 stb/day against a flowing wellhead pressure of 100 psia, determine the required specifications for an ESP for this application. Assume the minimum pump suction pressure is 200 psia. Solution 1. Required liquid throughput at pump is qLd ¼ (1:25)(8,000) ¼ 10,000 bbl=day: Select an ESP that delivers liquid flow rate qLp ¼ qLd ¼ 10,000 bbl=day in the neighborhood of its maximum efficiency (Fig. 14.4). 2. Well IPR gives pwfd ¼ 0:125p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 qLd =qmaxð Þ p À 1 h i ¼ 0:125(4,350)½ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 8,000=15,000ð Þ p À 1Š ¼ 2,823 psia: 3. The minimum pump depth is Dpump ¼ D À pwf À psuction 0:433gL ¼ 10,000 À 2,823 À 200 0:433(0:865) ¼ 2,997 ft: Use pump depth of 10,000 À 200 ¼ 9,800 ft. The pump suction pressure is 2,000 4,000 8,0006,000 10,000 12,000 16,00014,000 18,000 1,000 2,000 3,000 4,000 7,000 6,000 5,000 8,000 Head Feet PumpingHead(ft) Pump Rate (bbl/D) 200 400 600 800 10 20 30 40 50 60 70 80 Pump Only Eff Motor Load HP Head Efficiency Horsepower Recommended Capacity Range 0 Figure 14.4 A typical characteristic chart for a 100-stage ESP. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 210 3.1.2007 9:10pm Compositor Name: SJoearun 14/210 ARTIFICIAL LIFT METHODS
- 215. psuction ¼ 2,823 À 0:433(0:865)(10,000 À 9,800) ¼ 2,748 psia: 4. Computer spreadsheet HagedornBrownCorrelation.xls gives the required pump discharge pressure of 3,728 psia. 5. The required pump pressure differential is Dp ¼ pdischarge À psuction ¼ 3,728 À 2,748 ¼ 980 psi: The required pumping head is h ¼ Dp 0:433 ¼ 980 0:433 ¼ 2,263 feet of freshwater: 6. At throughput 10,000 bbl/day, Fig. 14.4 gives a pump- ing head of 6,000 ft for the 100-stage pump, which yields 60 ft pumping head per stage. The required num- ber of stages is (2,263)=(60) ¼ 38 stages. 7. At throughput 10,000 bbl/day, Fig. 14.4 gives the power of the 100-stage pump of 600 hp, which yields 6 hp/stage. The required power for a 38-stage pump is then (6)(38) ¼ 226 hp. The solution given by the computer spreadsheet ESP- design.xls is shown in Table 14.1. 14.3 Hydraulic Piston Pumping Hydraulic piston pumping systems can lift large volumes of liquid from great depth by pumping wells down to fairly low pressures. Crooked holes present minimal problems. Both natural gas and electricity can be used as the power source. They are also applicable to multiple completions and offshore operations. Their major disadvantages in- clude power oil systems being fire hazards and costly, power water treatment problems, and high solids produc- tion being troublesome. As shown in Fig. 14.5, a hydraulic piston pump (HPP) consists of an engine with a reciprocating piston driven by a power fluid connected by a short shaft to a piston in the pump end. HPPs are usually double-acting, that is, fluid is being displaced from the pump on both the upstroke and the downstroke. The power fluid is injected down a tubing string from the surface and is either returned to the surface through another tubing (closed power fluid) or commin- gled with the produced fluid in the production string (open power fluid). Because the pump and engine pistons are directly connected, the volumetric flow rates in the pump and engine are related through a simple equation (Cholet, 2000): qpump ¼ qeng Apump Aeng , (14:3) where qpump =flowrateoftheproducedfluidinthepump,bbl/day qeng ¼ flow rate of the power fluid, bbl/day Apump ¼ net cross-sectional area of pump piston, in:2 Aeng ¼ net cross-sectional area of engine piston, in:2 . Equation (14.3) implies that liquid production rate is proportional to the power fluid injection rate. The propor- tionality factor Apump=Aeng is called the ‘‘P/E ratio.’’ By adjusting the power fluid injection rate, the liquid produc- tion rate can be proportionally changed. Although the P/E ratio magnifies production rate, a larger P/E ratio means higher injection pressure of the power fluid. The following pressure relation can be derived from force balance in the HPP: peng,i À peng,d ¼ ppump,d À ppump,iÞ P=Eð Þ þ Fpump, À (14:4) where peng,i ¼ pressure at engine inlet, psia peng,d ¼ engine discharge pressure, psia ppump,d ¼ pump discharge pressure, psia Table 14.1 Result Given by the Computer Spreadsheet ESPdesign.xls ESPdesign.xls Description: This spreadsheet calculates parameters for ESP selection. Instruction: (1) Update parameter values in the Input data and Solution sections; and (2) view result in the Solution section. Input data Reservoir depth (D): 10,000 ft Reservoir pressure (pbar): 4,350 psia AOF in Vogel equation for IPR (qmax): 15,000 stb/day Production fluid gravity (gL): 0.865 1 for H2O Formation volume factor of production liquid (BL): 1.25 rb/stb Tubing inner diameter (dti): 2.992 in. Well head pressure (pwh): 100 psia Required pump suction pressure (psuction): 200 psia Desired production rate (qLd ): 8,000 stb/day Solution Desired bottom-hole pressure from IPR (pwfd ) ¼ 2,823 psia Desired production rate at pump (qLd ) ¼ 10,000 bbl/day Input here the minimum capacity of selected pump (qLp): 10,000 bbl/day Minimum pump setting depth (Dpump) ¼ 2,997 ft Input pump setting depth (Dpump): 9,800 ft Pump suction pressure (psuction) ¼ 2,748 psia Input pump discharge pressure (pdischarge): 3,728 psia Required pump pressure differential (Dp) ¼ 980 psia Required pumping head (h) ¼ 2,263 ft H2O Input pumping head per stage of the selected pump (hs): 60.00 ft/stage Input horse power per stage of the selected pump (hps): 6.00 hp/stage Input efficiency of the selected pump (Ep): 0.72 Required number of stages (Ns) ¼ 38 Total motor power requirement (hpmotor) ¼ 226.35 hp Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 211 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/211
- 216. ppump,i ¼ pump intake pressure, psia Fpump ¼ pump friction-induced pressure loss, psia. Equation (14.4) is also valid for open power fluid system where peng,d ¼ ppump,d . The pump friction-induced pressure loss Fpump depends on pump type, pumping speed, and power fluid viscosity. Its value can be estimated with the following empirical equation: Fpump ¼ 50gL 0:99 þ 0:01npf À Á 7:1eBqtotal À ÁN=Nmax , (14:5) where gL ¼ specific gravity of production liquid, 1.0 for H2O npf ¼ viscosity of power fluid, centistokes qtotal ¼ total liquid flow rate, bbl/day N ¼ pump speed, spm Nmax ¼ maximum pump speed, spm B ¼ 0:000514 for 23 ⁄8 -in. tubing ¼ 0:000278 for 27 ⁄8 -in. tubing ¼ 0:000167 for 31 ⁄2 -in. tubing ¼ 0:000078 for 41 ⁄2 -in. tubing. The pump intake pressure ppump,i can be determined on the basis of well IPR and desired liquid production rate qLd . If the IPR follows Vogel’s model, then for an HPP installed close to bottom hole, ppump,i can be estimated using ppump,i ¼ 0:125pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 81 À 80 qLd =qmaxð Þ p À 1 h i À Gb  D À Dp À Á , (14:6) where Gb ¼ pressure gradient below the pump, psi/ft D ¼ reservoir depth, ft Dp ¼ pump setting depth, ft. The pump discharge pressure ppump;d can be calculated based on wellhead pressure and production tubing perfor- mance. The engine discharge pressure peng;d can be calcu- lated based on the flow performance of the power fluid returning tubing. With all these parameter values known, theengineinletpressurepeng,i canbecalculatedbyEq.(14.6). Then the surface operating pressure can be estimated by ps ¼ peng,i À ph þ pf , (14:7) where ps ¼ surface operating pressure, psia ph ¼ hydrostatic pressure of the power fluid at pump depth, psia pf ¼ frictional pressure loss in the power fluid injection tubing, psi. The required input power can be estimated from the following equation: HP ¼ 1:7  10À5 qengps (14:8) Selection of HPP is based on the net lift defined by LN ¼ Dp À ppump,i Gb (14:9) and empirical value of P/E defined by P=E ¼ 10,000 LN : (14:10) The following procedure is used for selecting an HPP: 1. Starting from well IPR, determine a desirable liquid production rate qLd . Then calculate pump intake pres- sure with Eq. (14.6). 2. Calculate net lift with Eq. (14.9) and P/E ratio with Eq. (14.10). 3. Calculate flow rate at pump suction point by qLs ¼ BoqLd , where Bo is formation volume factor of oil. Then estimate pump efficiency Ep. 4. Select a pump rate ratio N=Nmax between 0.2 and 0.8. Calculate the design flow rate of pump by qpd ¼ qLs Ep N=Nmaxð Þ : 5. Based on qpd and P/E values, select a pump from the manufacturer’s literature and get rated displacement values qpump, qeng, and Nmax. If not provided, calculate flow rates per stroke by q 0 pump ¼ qpump Nmax and q 0 eng ¼ qeng Nmax : 6. Calculate pump speed by N ¼ N Nmax Nmax: 7. Calculate power fluid rate by qpf ¼ N Nmax qeng Eeng : 8. Determine the return production flow rate by qtotal ¼ qpf þ qLs for open power fluid system or qtotal ¼ qLs for closed power fluid system. 9. Calculate pump and engine discharge pressure ppump,d and peng,d based on tubing performance. 10. Calculate pump friction-induced pressure loss using Eq. (14.5). 11. Calculate required engine pressure using Eq. (14.4). 12. Calculate pressure change Dpinj from surface to engine depth in the power fluid injection tubing based on single-phase flow. It has two components: Dpinj ¼ ppotential À pfriction Pump Piston Engine Piston Down stroke Up stroke Figure 14.5 A sketch of a hydraulic piston pump. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 212 3.1.2007 9:10pm Compositor Name: SJoearun 14/212 ARTIFICIAL LIFT METHODS
- 217. 13. Calculate required surface operating pressure by pso ¼ peng,i À Dpinj: 14. Calculate required surface operating horsepower by HPso ¼ 1:7 Â 10À5 qpf pso Es , where Es is the efficiency of surface pump. Example Problem 14.2 A 10,000-ft-deep well has a potential to produce 40 8API oil with GOR 150 scf/stb and 10% water cut through a 2-in. (1.995-in. ID) tubing in a 7-in. casing with a pump installation. The oil has a formation volume factor of 1.25 and average viscosity of 5 cp. Gas- and water-specific gravities are 0.7 and 1.05, respectively. The surface and bottom-hole temperatures are 80 and 180 8F, respectively. The IPR of the well can be described by Vogel’s model with a reservoir pressure 2,000 psia and AOF 300 stb/day. If the well is to be put in production with an HPP at a depth of 9,700 ft in an open power fluid system to produce liquid at 200 stb/day against a flowing wellhead pressure of 75 psia, determine the required specifications for the HPP for this application. Assume the overall efficiencies of the engine, HHP, and surface pump to be 0.90, 0.80, and 0.85, respectively. Solution This problem is solved by computer spreadsheet HydraulicPistonPump.xls, as shown in Table 14.2. 14.4 Progressive Cavity Pumping The progressive cavity pump (PCP) is a positive displacement pump, using an eccentrically rotating sin- gle-helical rotor, turning inside a stator. The rotor is usually constructed of a high-strength steel rod, typi- cally double-chrome plated. The stator is a resilient elastomer in a double-helical configuration molded inside a steel casing. A sketch of a PCP system is shown in Fig. 14.6. Progressive cavity pumping systems can be used for lifting heavy oils at a variable flow rate. Solids and free gas production present minimal problems. They can be Table 14.2 Solution Given by HydraulicPistonPump.xls HydraulicPistonPump.xls Description: This spreadsheet calculates parameters for HPP selection. Instruction: (1) Update parameter values in the Input data and Solution sections; and (2) view result in the Solution section. Input data Reservoir depth (D): 10,000 ft Reservoir pressure (pbar): 2,000 psia AOF in Vogel equation for IPR (qmax): 300 stb/day Production fluid gravity (gL): 0.8251 1 for H2O Formation volume factor of production liquid (BL): 1.25 rb/stb Tubing inner diameter (dti): 1.995 in. B value: 0.000514 Power fluid viscosity (vpf ): 1 cs Well head pressure (pwh): 100 psia Pump setting depth (Dp): 9,700 ft Desired production rate (qLd ): 200 stb/day HPP efficiency (Ep): 0.80 Surface pump efficiency (Es): 0.85 Engine efficiency (Ee): 0.90 Pump speed ratio (N=Nmax): 0.80 Power fluid flow system (1 ¼ OPFS, 0 ¼ CPFS): 1 Solution Desired bottom-hole pressure from IPR (pwfd ) ¼ 1,065 psia Pump intake pressure (ppump) ¼ 958 psia Net lift (LN ) ¼ 7,019 ft Design pump to engine area ratio (P/E) ¼ 1.42 Flow rate at pump suction point (qLs) ¼ 250 bbl/day Design flow rate of pump (qpd ) ¼ 391 bbl/day Input from manufacturer’s literature: Pump P/E: 1.13 qp,max: 502 bbl/day qe,max: 572 bbl/day Nmax: 27 Flow rate per stroke/min in pump (q0 p) ¼ 18.59 bbl/day Flow rate per stroke/min in engine (q0 e) ¼ 21.19 bbl/day Pump speed (N) ¼ 21.60 spm Power fluid rate (qpf ) ¼ 508 bbl/day Return production flow rate (qtotal) ¼ 758 bbl/day Input pump discharge pressure by mHB correlation (ppump,d ): 2,914 psia Input engine discharge pressure by mHB correlation (peng,d ): 2,914 psia Pump friction-induced pressure loss (Fpump) ¼ 270 psi Required engine pressure (peng,i) ¼ 5,395 psia Input pressure change in the injection tubing (Dpinj): ¼ 3,450 psi Required surface operating pressure (pso) ¼ 1,945 psia Required surface horsepower (HPso) ¼ 20 hp Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 213 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/213
- 218. installed in deviated and horizontal wells. With its abil- ity to move large volumes of water, the progressing cavity pump is also used for coal bed methane, dewatering, and water source wells. The PCP reduces overall operating costs by increasing operating efficiency while reducing energy requirements. The major disadvantages of PCPs include short operating life (2–5 years) and high cost. 14.4.1 Down-Hole PCP Characteristics Proper selection of a PCP requires knowledge of PCP geometry, displacement, head, and torque requirements. Figure 14.7 (Cholet, 2000) illustrates rotor and stator geometry of PCP where D ¼ rotor diameter, in. E ¼ rotor=stator eccentricity, in. Pr ¼ pitch length of rotor, ft Ps ¼ pitch length of stator, ft. Two numbers define the geometry of the PCP: the num- ber of lobes of rotor and the number of lobes of the stator. A pump with a single helical rotor and double helical stator is described as a ‘‘1-2 pump’’ where Ps ¼ 2Pr. For a multilobe pump, Ps ¼ Lr þ 1 Lr Pr, (14:11) where Lr is the number of rotor lobes. The ratio Pr=Ps is called the ‘‘kinematics ratio.’’ Pump displacement is defined by the fluid volume pro- duced in one revolution of the rotor: V0 ¼ 0:028DEPs, (14:12) where V0 ¼ pump displacement, ft3 . Pump flow rate is expressed as Qc ¼ 7:12DEPsN À Qs, (14:13) where Qc ¼ pump flow rate, bbl/day N ¼ rotary speed, rpm Qs ¼ leak rate, bbl/day. The PCP head rating is defined by DP ¼ 2np À 1 À Á dp, (14:14) where DP ¼ pump head rating, psi np ¼ number of pitches of stator dp ¼ head rating developed into an elementary cavity, psi. PCP mechanical resistant torque is expressed as Tm ¼ 144V0DP ep , (14:15) where Tm ¼ mechanical resistant torque, lbf -ft ep ¼ efficiency. The load on thrust bearing through the drive string is expressed as Fb ¼ 4 2E þ Dð Þ2 DP, (14:16) where Fb ¼ axial load, lbf . 14.4.2 Selection of Down-Hole PCP The following procedure can be used in the selection of a PCP: 1. Starting from well IPR, select a desirable liquid flow rate qLp at pump depth and the corresponding pump intake pressure below the pump ppi. 2. Based on manufacturer’s literature, select a PCP that can deliver liquid rate QLp, where QLp qLp. Obtain the value of head rating for an elementary cavity dp. 3. Determine the required pump discharge pressure ppd based on wellhead pressure, tubing size, flow rate QLp, Drive System Coupling Drive Head Wellhead Sucker Rod Drive System Drive System Rotor Stator Centralizer Stop Bushing Figure 14.6 Sketch of a PCP system. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 214 3.1.2007 9:10pm Compositor Name: SJoearun 14/214 ARTIFICIAL LIFT METHODS
- 219. and fluid properties. This can be carried out quickly using the computer spreadsheet HagedornBrownCorre- lation.xls. 4. Calculate required pump head by DP ¼ ppd À ppi: (14:17) 5. Calculate the required number of pitches np using Eq. (14.14). 6. Calculate mechanical resistant torque with Eq. (14.15). 7. Calculate the load on thrust bearing with Eq. (14.16). 14.4.3 Selection of Drive String Sucker rod strings used in beam pumping are also used in the PCP systems as drive strings. The string diam- eter should be properly chosen so that the tensile stress in the string times the rod cross-sectional area does not exceed the maximum allowable strength of the string. The following procedure can be used in selecting a drive string: 1. Calculate the weight of the selected rod string Wr in the effluent fluid (liquid level in annulus should be consid- ered to adjust the effect of buoyancy). 2. Calculate the thrust generated by the head rating of the pump Fb with Eq. (14.16). 3. Calculate mechanical resistant torque Tm with Eq. (14.15). 4. Calculate the torque generated by the viscosity of the effluent in the tubing by Tv ¼ 2:4 Â 10À6 mf LN d3 (D À d) 1 ln ms mf ms mf À 1 ! , (14:18) where Tv ¼ viscosity-resistant torque, lbf -ft mf ¼ viscosity of the effluent at the inlet temperature, cp ms ¼ viscosity of the effluent at the surface temperature, cp L ¼ depth of tubing, ft d ¼ drive string diameter, in. 5. Calculate total axial load to the drive string by F ¼ Fb þ Wr: (14:19) 6. Calculate total torque by T ¼ Tm þ Tv: (14:20) 7. Calculate the axial stress in the string by st ¼ 4 pd3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F2d2 þ 64T2 Â 144 p , (14:21) where the tensile stress st is in pound per square inch. This stress value should be compared with the strength of the rod with a safety factor. 14.4.4 Selection of Surface Driver The prime mover for PCP can be an electrical motor, hydraulic drive, or internal-combustion engine. The min- imum required power from the driver depends on the total resistant torque requirement from the PCP, that is, Ph ¼ 1:92 Â 10À4 TN, (14:22) where the hydraulic power Ph is in hp. Driver efficiency and a safety factor should be used in driver selection from manufacturer’s literature. 14.5 Plunger Lift Plunger lift systems are applicable to high gas–liquid ratio wells. They are very inexpensive installations. Plunger automatically keeps tubing clean of paraffin and scale. But they are good for low-rate wells normally less than 200 B/D. Listiak (2006) presents a thorough discussion of this technology. E Section D D+2E D+4E Pr Ps Stator Centerline Rotor Centerline Pump Assembly Rotor Stator Figure 14.7 Rotor and stator geometry of PCP. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 215 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/215
- 220. Traditionally, plunger lift was used on oil wells. Recently, plunger lift has become more common on gas wells for de-watering purposes. As shown in Fig. 14.8, high-pressure gas wells produce gas carrying liquid water and/or condensate in the form of mist. As the gas flow velocity in the well drops as a result of the reservoir pressure depletion, the carrying capacity of the gas de- creases. When the gas velocity drops to a critical level, liquid begins to accumulate in the well and the well flow can undergo annular flow regime followed by a slug flow regime. The accumulation of liquids (liquid loading) in- creases bottom-hole pressure that reduces gas production rate. Low gas production rate will cause gas velocity to drop further. Eventually the well will undergo bubbly flow regime and cease producing. Liquid loading is not always obvious, and recognizing the liquid-loading problem is not an easy task. A thorough diagnostic analysis of well data needs to be performed. The symptoms to look for include onset of liquid slugs at the surface of well, increasing difference between the tubing and casing pressures with time, sharp changes in gradient on a flowing pressure survey, sharp drops in a production decline curve, and prediction with analytical methods. Accurate prediction of the problem is vital for taking timely measures to solve the problem. Previous investiga- tors have suggested several methods to predict the prob- lem. Results from these methods often show discrepancies. Also, some of these methods are not easy to use because of the difficulties with prediction of bottom-hole pressure in multiphase flow. Turner et al. (1969) were the pioneer investigators who analyzed and predicted the minimum gas flow rate capable of removing liquids from the gas production wells. They presented two mathematical models to describe the liquid- loading problem: the film movement model and entrained drop movement model. On the basis of analyses on field data they had, they concluded that the film movement model does not represent the controlling liquid transport mechanism. The Turner et al. entrained drop movement model was derived on the basis of the terminal-free settling velocity of liquid drops and the maximum drop diameter correspond- ing to the critical Weber number of 30. According to Turner et al. (1969), gas will continuously remove liquids from the well until its velocity drops to below the terminal velocity. The minimum gas flow rate for a particular set of conditions (pressure and conduit geometry) can be calcu- lated using a mathematical model. Turner et al. (1969) found that this entrained drop movement model gives underestimates of the minimum gas flow rates. They recommended the equation-derived values be adjusted upward by approximately 20% to ensure removal of all drops. Turner et al. (1969) believed that the discrepancy was attributed to several facts including the use of drag coefficients for solid spheres, the assumption of stagnation velocity, and the critical Weber number established for drops falling in air, not in compressed gas. The main problem that hinders the application of the Turner et al. entrained drop model to gas wells comes from the difficulties of estimating the values of gas density and pressure. Using an average value of gas-specific gravity (0.6) and gas temperature (120 8F), Turner et al. derived an expression for gas density as 0.0031 times the pressure. However, they did not present a method for calculating the gas pressure in a multiphase flow wellbore. Starting from the Turner et al. entrained drop model, Guo and Ghalambor (2005) determined the minimum kinetic energy of gas that is required to lift liquids. A four-phase (gas, oil, water, and solid particles) mist- flow model was developed. Applying the minimum kinetic energy criterion to the four-phase flow model resulted in a closed-form analytical equation for predicting the min- imum gas flow rate. Through case studies, Guo and Gha- lambor demonstrated that this new method is more conservative and accurate. Their analysis also indicates that the controlling conditions are bottom-hole conditions where gas has higher pressure and lower kinetic energy. This analysis is consistent with the observations from air- drilling operations where solid particles accumulate at bottom-hole rather than top-hole (Guo and Ghalambor, 2002). However, this analysis contradicts the results by Turner et al. (1969), that indicated that the wellhead con- ditions are, in most instances, controlling. 14.5.1 Working Principle Figure 14.9 illustrates a plunger lift system. Plunger lift uses a free piston that travels up and down in the well’s tubing string. It minimizes liquid fallback and uses the well’s energy more efficiently than in slug or bubble flow. The purpose of plunger lift is like that of other artificial lift methods: to remove liquids from the wellbore so that the well can be produced at the lowest bottom-hole pres- sures. Whether in a gas well, oil well, or gas lift well, the mechanics of a plunger lift system are the same. The plunger, a length of steel, is dropped down the tubing to the bottom of the well and allowed to travel back to the surface. It provides a piston-like interface between liquids and gas in the wellbore and prevents liquid fallback. By providing a ‘‘seal’’ between the liquid and gas, a well’s own energy can be used to efficiently lift liquids out of the wellbore. A plunger changes the rules for liquid removal. However, in a well without a plunger, gas velocity must be high to remove liquids. With a plunger, gas velocity can be very low. Unloading relies much more on the well’s ability to store enough gas pressure to lift the plunger and a liquid slug to surface, and less on critical flow rates. Plunger operation consists of shut-in and flow periods. The flow period is further divided into an unloading period and flow after plunger arrival. Lengths of these periods will vary depending on the application, producing capability of the well, and pressures. A plunger cycle starts with the shut-in period that allows the plunger to drop from the surface to the bottom of the well. At the same time, the well builds gas pressure stored either in the casing, in the fracture, or in the near wellbore region of the reservoir. The well must be shut in long enough to build reservoir pressure that will provide energy to lift both the plunger and the liquid slug to the surface against line pressure and friction. When this time and GasFlow Decreasing Gas Velocity Mist Flow Slug Flow Annular Flow Bubble Flow Figure 14.8 Four flow regimes commonly encoun- tered in gas wells. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 216 3.1.2007 9:10pm Compositor Name: SJoearun 14/216 ARTIFICIAL LIFT METHODS
- 221. pressure have been reached, the flow period is started and unloading begins. In the initial stages of the flow period, the plunger and liquid slug begin traveling to the surface. Gas above the plunger quickly flows from the tubing into the flowline, and the plunger and liquid slug follow up the hole. The plunger arrives at the surface, unloading the liquid. Initially, high rates prevail (often three to four times the average daily rate) while the stored pressure is blown down. The well can now produce free of liquids, while the plunger remains at the surface, held by the well’s pressure and flow. As rates drop, velocities eventually drop below the critical rate, and liquids begin to accumu- late in the tubing. The well is shut in and the plunger falls back to the bottom to repeat the cycle. At the end of the shut-in period, the well has built pressure. The casing pressure is at its maximum, and the tubing pressure is lower than the casing pressure. The difference is equivalent to the hydrostatic pressure of the liquid in the tubing. When the well is opened, the tubing pressure quickly drops down to line pressure, while the casing pressure slowly decreases until the plunger reaches the surface. As Lubricator/ Catcher Plunger Arrival Sensor Plunger Down Hole Bumper Spring Note: Well sketch, not to scale or correct proportion. Earth Liquid Slug Electronic Controller/ Motor Valve Figure 14.9 A sketch of a plunger lift system (courtesy Ferguson Beauregard). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 217 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/217
- 222. the plunger nears the surface, the liquid on top of the plunger may surge through the system, causing spikes in line pressure and flow rate. This continues until the plun- ger reaches the surface. After the plunger surfaces, a large increase in flow rate will produce higher tubing pressures and an increase in flowline pressure. Tubing pressure will then drop very close to line pressure. Casing pressure will reach its minimum either on plunger arrival or after, as the casing blows down and the well produces with minimal liquids in the tubing. If the well stays above the critical unloading rate, the casing pressure will remain fairly con- stant or may decrease further. As the gas rate drops, liquids become held up in the tubing and casing pressure will increase. Upon shut in, the casing pressure builds more rapidly. How fast depends on the inflow performance and reservoir pressure of the well. The tubing pressure will increase quickly from line pressure, as the flowing gas friction ceases. It will eventually track casing pressure (less the liquid slug). Casing pressure will continue to increase to maximum pressure until the well is opened again. As with most wells, maximum plunger lift production occurs when the well produces against the lowest possible bottom-hole pressure. On plunger lift, the lowest average bottom-hole pressures are almost always obtained by shut- ting the well in the minimum amount of time. Practical experience and plunger lift models demonstrate that lifting large liquid slugs requires higher average bottom-hole pressure. Lengthy shut-in periods also increase average bottom-hole pressure. So the goal of plunger lift should be to shut the well in the minimum amount of time and produce only enough liquids that can be lifted at this minimum buildup pressure. What is the minimum shut-in time? The absolute min- imum amount of time for shut-in is the time it takes the plunger to reach the bottom. The well must be shut-in in this length of time regardless of what other operating conditions exist. Plungers typically fall between 200 and 1,000 ft/min in dry gas and 20 and 250 ft/min in liquids. Total fall time varies and is affected by plunger type, amount of liquids in the tubing, the condition of the tubing (crimped, corkscrewed, corroded, etc.), and the deviation of the tubing or wellbore. The flow period during and after plunger arrival is used to control liquid loads. In general, a short flow period brings in a small liquid load, and a long flow period brings in a larger liquid load. By controlling this flow time, the liquid load is controlled. So the well can be flowed until the desired liquid load has entered the tubing. A well with a high GLR may be capable of long flow periods without requiring more than minimum shut-in times. In this case, the plunger could operate as few as 1 or 2 cycles/ day.Conversely,a wellwith alow GLRmaynever beable to flow after plunger arrival and may require 25 cycles/day or more. In practice, if the well is shutting in for only the minimum amount of time, it can be flowed as long as possible to maintain target plunger rise velocities. If the well is shutting in longer than the minimum shut-in time, there should be little or no flow after the plunger arrives at the surface. 14.5.2 Design Guideline Plunger lift systems can be evaluated using rules of thumb in conjunction with historic well production or with a mathematical plunger model. Because plunger lift installa- tions are typically inexpensive, easy to install, and easy to test, most evaluations are performed by rules of thumb. 14.5.2.1 Estimate of Production Rates with Plunger Lift The simplest and sometimes most accurate method of determining production increases from plunger lift is from decline curve analysis. Gas and oil reservoirs typically have predictable declines, either exponential, harmonic, or hyperbolic. Initial production rates are usually high enough to produce the well above critical rates (unloaded) and establish a decline curve. When liquid loading occurs, a marked decrease and deviation from normal decline can be seen. By unloading the well with plunger lift, a normal decline can be reestablished. Production increases from plunger lift will be somewhere between the rates of the well when it started loading and the rate of an extended decline curve to the present time. Ideally, decline curves would be used in concert with critical velocity curves to predetermine when plunger lift should be installed. In this manner, plunger lift will maintain production on a steady decline and never allow the well to begin loading. Another method to estimate production is to build an inflow performance curve based on the backpressure equa- tion. This is especially helpful if the well has an open annu- lus and casing pressure is known. The casing pressure gives a good approximation of bottom-hole pressure. The IPR curve can be built based on the estimated reservoir pressure, casing pressure, and current flow rate. Because the job of plunger lift is to lower the bottom-hole pressure by remov- ing liquids, the bottom-hole pressure can be estimated with no liquids. This new pressure can be used to estimate a production rate with lower bottom-hole pressures. 14.5.2.2 GLR and Buildup Pressure Requirements There are two minimum requirements for plunger lift operation: minimum GLR and buildup pressure. For the plunger lift to operate, there must be available gas to provide the lifting force, in sufficient quantity per barrel of liquid for a given well depth. 14.5.2.2.1 Rules of Thumb As a rule of thumb, the minimum GLR requirement is considered to be about 400 scf/bbl/1,000 ft of well depth, that is, GLRmin ¼ 400 D 1,000 , (14:23) where GLRmin ¼ minimum required GLRforplunger lift, scf/bbl D ¼ depth to plunger, ft. Equation (14.23) is based on the energy stored in a com- pressed volume of 400 scf of gas expanding under the hydrostatic head of a barrel of liquid. The drawback is that no consideration is given to line pressures. Excessively high line pressures, relative to buildup pressure may in- crease the requirement. The rule of thumb also assumes that the gas expansion can be applied from a large open annulus without restriction. Slim-hole wells and wells with packers that require gas to travel through the reservoir or through small perforations in the tubing will cause a greater restriction and energy loss. This increases the minimum requirements to as much as 800–1,200 scf/bbl/1,000 ft. Well buildup pressure is the second requirement for plunger operation. This buildup pressure is the bot- tom-hole pressure just before the plunger begins its ascent (equivalent to surface casing pressure in a well with an open annulus). In practice, the minimum shut-in pres- sure requirement for plunger lift is equivalent to 1½ times maximum sales line pressure. The actual requirement may be higher. The rule works well in intermediate-depth wells (2,000–8,000 ft) with slug sizes of 0.1–0.5 barrels/cycle. It breaks down for higher liquid volumes, deeper wells (due to increasing friction), and excessive pressure restrictions at the surface or in the wellbore. An improved rule for minimum pressure is that a well can lift a slug of liquid equal to about 50–60% of the difference between shut-in casing pressure and maximum sales line pressure. This rule gives Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 218 3.1.2007 9:10pm Compositor Name: SJoearun 14/218 ARTIFICIAL LIFT METHODS
- 223. pc ¼ pL max þ psh fsl , (14:24) where pc ¼ required casing pressure, psia pLmax ¼ maximum line pressure, psia psh ¼ slug hydrostatic pressure, psia fsl ¼ slug factor, 0.5–0.6. This rule takes liquid production into account and can be used for wells with higher liquid production that require more than 1–2 barrels/cycle. It is considered as a conser- vative estimate of minimum pressure requirements. To use Eq. (14.24), first the total liquid production on plunger lift and number of cycles possible per day should be estimated. Then the amount of liquid that can be lifted per cycle should be determined. That volume of liquid per cycle is converted into the slug hydrostatic pressure using the well tubing size. Finally, the equation is used to estimate required casing pressure to operate the system. It should be noted that a well that does not meet minimum GLR and pressure requirements could still be plunger lifted with the addition of an external gas source. Design at this point becomes more a matter of the econo- mics of providing the added gas to the well at desired pressures. 14.5.2.2.2 Analytical Method Analytical plunger lift design methods have been developed on the basis of force balance. Several studies in the literature address the addition of makeup gas to a plunger installation through either exist- ing gas lift operations, the installation of a field gas supply system, or the use of wellhead compression. Some of the studies were presented by Beeson et al. (1955), Lebeaux and Sudduth (1955), Foss and Gaul (1965), Abercrombie (1980), Rosina (1983), Mower et al. (1985), and Lea (1981, 1999). The forces acting on the plunger at any given point in the tubing include the following: 1. Stored casing pressure acting on the cross-section of the plunger 2. Stored reservoir pressure acting on the cross-section of the plunger 3. Weight of the fluid 4. Weight of the plunger 5. Friction of the fluid with the tubing 6. Friction of the plunger with the tubing 7. Gas friction in the tubing 8. Gas slippage upward past the plunger 9. Liquid slippage downward past the plunger 10. Surface pressure (line pressure and restrictions) acting against the plunger travel Several publications have been written dealing with this approach. Beeson et al. (1955) first presented equations for high GLR wells based on an empirically derived analysis. Foss and Gaul (1965) derived a force balance equation for use on oil wells in the Ventura Avenue field. Mower et al. (1985) presented a dynamic analysis of plunger lift that added gas slippage and reservoir inflow and mathemati- cally described the entire cycle (not just plunger ascent) for tight-gas/very high GLR wells. The methodology used by Foss and Gaul (1965) was to calculate a casing pressure required to move the plunger and liquid slug just before it reached the surface, called Pcmin. Since Pcmin is at the end of the plunger cycle, the energy of the expanding gas from the casing to the tubing is at its minimum. Adjusting Pcmin for gas expansion from the casing to the tubing during the full plunger cycle results in the pressure required to start the plunger at the begin- ning of the plunger cycle, or Pcmax. The equations below are essentially the same equations presented by Foss and Gaul (1956) but are summarized here as presented by Mower et al. (1985). The Foss and Gaul model is not rigorous, because it assumes constant friction associated with plunger rise velocities of 1,000 ft/ min, does not calculate reservoir inflow, assumes a value for gas slippage past the plunger, assumes an open unre- stricted annulus, and assumes the user can determine unloaded gas and liquid rates independently of the model. Also, this model was originally designed for oil well operation that assumed the well would be shut-in on plunger arrival, so the average casing pressure, Pcavg, is only an average during plunger travel. The net result of these assumptions is an overprediction of required casing pressure. If a well meets the Foss and Gaul (1956) criteria, it is almost certainly a candidate for plunger lift. 14.5.2.3 Plunger Lift Models 14.5.2.3.1 Basic Foss and Gaul Equations (modified by Mower et al) The required minimum casing pressure is expressed as Pcmin ¼ Pp þ 14:7 þ Pt þ Plh þ Plf À Á  Vslug  à  1 þ D K , (14:25) where Pc min ¼ required minimum casing pressure, psia Pp ¼ Wp=At, psia Wp ¼ plunger weight, lbf At ¼ tubing inner cross-sectional area, in:2 Plh ¼ hydrostatic liquid gradient, psi/bbl slug Plf ¼ flowing liquid gradient, psi/bbl slug Pt ¼ tubing head pressure, psia Vslug ¼ slug volume, bbl D ¼ depth to plunger, ft K ¼ characteristic length for gas flow in tubing, ft. Foss and Gaul suggested an approximation where K and Plh þ Plf are constant for a given tubing size and a plunger velocity of 1,000 ft/min: To successfully operate the plunger, casing pressure must build to Pcmax given by Pcmax ¼ Pcmin Aa þ At Aa : (14:26) The average casing pressure can then be expressed as Pcavg ¼ Pcmin 1 þ At 2Aa , (14:27) where Aa is annulus cross-sectional area in squared inch. The gas required per cycle is formulated as Vg ¼ 37:14FgsPcavgVt Z Tavg þ 460 À Á , (14:28) where Vg ¼ required gas per cycle, Mscf Fgs ¼ 1 þ 0:02 (D=1,000), modified Foss and Gaul slippage factor Vt ¼ At(D À VslugL), gas volume in tubing, Mcf L ¼ tubing inner capacity, ft/bbl Z ¼ gascompressibilityfactorinaveragetubingcondition Tavg ¼ average temperature in tubing, 8F. The maximum number of cycles can be expressed as Tubing size (in.) K (ft) Plh þ Plf (psi/bbl) 23 ⁄8 33,500 165 27 ⁄8 45,000 102 31 ⁄2 57,600 63 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 219 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/219
- 224. NC max ¼ 1440 D Vr þ DÀVslugL Vfg þ VslugL Vfl , (14:29) where NC max ¼ the maximum number of cycles per day Vfg ¼ plunger falling velocity in gas, ft/min Vfl ¼ plunger falling velocity in liquid, ft/min Vr ¼ plunger rising velocity, ft/min. The maximum liquid production rate can be expressed as qL max ¼ NC maxVslug: (14:30) The required GLR can be expressed as GLRmin ¼ Vg Vslug : (14:31) Example Problem 14.3: Plunger Lift Calculations Calculate required GLR, casing pressure, and plunger lift operating range for the following given well data: Gas rate: 200 Mcfd expected when unloaded Liquid rate: 10 bbl/day expected when unloaded Liquid gradient: 0.45 psi/ft Tubing, ID: 1.995 in. Tubing, OD: 2.375 in. Casing, ID: 4.56 in. Depth to plunger: 7,000 ft Line pressure: 100 psi Available casing pressure: 800 psi Reservoir pressure: 1200 psi Average Z factor: 0.99 Average temperature: 140 8F Plunger weight: 10 lb Plunger fall in gas: 750 fpm Plunger fall in liquid: 150 fpm Plunger rise velocity: 1,000 fpm Solution The minimum required GLR bya rule of thumb is GLRmin ¼ 400 D 1,000 ¼ 400 7,000 1,000 ¼ 2,800 scf=bbl: The well’s GLR of 2,857 scf/bbl is above 2,800 scf/bbl and is, therefore, considered adequate for plunger lift. The minimum required casing pressure can be estimated using two rules of thumb. The simple rule of thumb gives pc ¼ 1:5pL max ¼ (1:5)(100) ¼ 150 psi: To calculate the minimum required casing pressure with the improved rule of thumb, the slug hydrostatic pressure needs to be known. For this case, assuming 10 cycles/day, equivalent to a plunger trip every 2.4 hours, and 10 bbls of liquid, the plunger will lift 1 bbl/cycle. The hydrostatic pressure of 1 bbl of liquid in 23 ⁄8 -in. tubing with a 0.45-psi/ft liquid gradient is about 120 psi. Then pc ¼ pL max þ psh fsl ¼ 100 þ 120 0:5 to 0:6 ¼ 300 to 340 psi: Since the well has 800 psi of available casing pressure, it meets the pressure requirements for plunger lift. The Foss and Gaul–type method can be used to deter- mine plunger lift operating range. Basic parameters are given in Table 14.3. Since the Foss and Gaul–type calculations involve de- termination of Z-factor values in Eq. (14.28) at different pressures, a spreadsheet program PlungerLift.xls was developed to speed up the calculation procedure. The solution is given in Table 14.4. It was given that the estimated production when unloaded is 200 Mcfd with 10 bbl/day of liquid (GLR ¼ 200=10 ¼ 20 Mscf=bbl), and the maximum casing pressure buildup is 800 psi. From the Table 14.4, find casing pressure of about 800 psi, GLR of 20 Mscf/bbl, and production rates of 10 bbl/day. This occurs at slug sizes between about 0.25 and 3 bbl. The well will operate on plunger lift. 14.6 Hydraulic Jet Pumping Figure 14.10 shows a hydraulic jet pump installation. The pump converts the energy from the injected power fluid (water or oil) to pressure that lifts production fluids. Because there is no moving parts involved, dirty and gassy fluids present no problem to the pump. The jet pumps can be set at any depth as long as the suction pressure is sufficient to prevent pump cavitation problem. The disadvantage of hydraulic jet pumps is their low efficiency (20–30%). 14.6.1 Working Principle Figure 14.11 illustrates the working principle of a hydraulic jet pump. It is a dynamic-displacement pump that differs from a hydraulic piston pump in the manner in which it increases the pressure of the pumped fluid with a jet nozzle. The power fluid enters the top of the pump from an injection tubing. The power fluid is then accelerated through the nozzle and mixed with the produced fluid in the throat of the pump. As the fluids mix, the momentum of the power fluid is partially transferred to the produced fluid and increases its kinetic energy (velocity head). Table 14.3 Summary of Calculated Parameters Tubing inner cross-sectional area (At) ¼ 3:12 in:2 Annulus cross-sectional area (Aa) ¼ 11:90 in:2 Plunger-weight pressure (Pp) ¼ 3.20 psi Slippage factor (Fgs) ¼ 1.14 Tubing inner capacity (L) ¼ 258.80 ft/bbl The average temperature (Tavg) ¼ 600 8R Table 14.4 Solution Given by Spreadsheet Program PlungerLift.xls Vslug (bbl) PCmin (psia) PCmax (psia) PCavg (psia) Z Vt (Mcf) Vg (Mscf) NCmax (cyc/day) qLmax (bbl/day) GLRmin (Mscf/bbl) 0.05 153 193 173 0.9602 0.1516 1.92 88 4.4 38.44 0.1 162 205 184 0.9624 0.1513 2.04 87 8.7 20.39 0.25 192 243 218 0.9744 0.1505 2.37 86 21.6 9.49 0.5 242 306 274 0.9689 0.1491 2.98 85 42.3 5.95 1 342 432 387 0.9499 0.1463 4.20 81 81.3 4.20 2 541 684 613 0.9194 0.1406 6.61 75 150.8 3.31 3 741 936 838 0.8929 0.1350 8.95 70 211.0 2.98 4 940 1,187 1,064 0.8666 0.1294 11.21 66 263.6 2.80 5 1,140 1,439 1,290 0.8457 0.1238 13.32 62 309.9 2.66 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 220 3.1.2007 9:10pm Compositor Name: SJoearun 14/220 ARTIFICIAL LIFT METHODS
- 225. Some of the kinetic energy of the mixed stream is con- verted to static pressure head in a carefully shaped diffuser section of expanding area. If the static pressure head is greater than the static column head in the annulus, the fluid mixture in the annulus is lifted to the surface. 14.6.2 Technical Parameters The nomenclatures in Fig. 14.11 are defined as p1 ¼ power fluid pressure, psia q1 ¼ power fluid rate, bbl/day p2 ¼ discharge pressure, psia q2 ¼ q1 þ q3, total fluid rate in return column, bbl/day p3 ¼ intake pressure, psia q3 ¼ intake (produced) fluid rate, bbl/day Aj ¼ jet nozzle area, in.2 As ¼ net throat area, in.2 At ¼ total throat area, in.2 . The following dimensionless variables are also used in jet pump literature (Cholet, 2000): R ¼ Aj At (14:32) M ¼ q3 q1 (14:33) H ¼ p2 À p3 p1 À p2 (14:34) h ¼ MH, (14:35) where R ¼ dimensionless nozzle area M ¼ dimensionless flow rate H ¼ dimensionless head h ¼ pump efficiency. 14.6.3 Selection of Jet Pumps Selection of jet pumps is made on the basis of manufacturer’s literatures where pump performance charts are usually avail- able. Figure 14.12 presents an example chart. It shows the effect of M on H and h. For a given jet pump specified by R value, there exists a peak efficiency hp. It is good field practice to attempt to operate the pump at its peak efficiency. If Mp and Hp are used to denote M and H at the peak efficiency, respectively, pump parameters should be designed using Mp ¼ q3 q1 (14:36) and Hp ¼ p2 À p3 p1 À p2 , (14:37) Production inlet chamber Pump tubing Casing Nozzle Power fluid Throat Diffuser Combined fluid return Well production Figure 14.10 Sketch of a hydraulic jet pump installation. P1 q1 Nozzle P3 q3 P2 q2 Diffuser A B Aj As At Throat Figure 14.11 Working principle of a hydraulic jet pump. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 221 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/221
- 226. where Mp and Hp values can be determined from the given performance chart. If the H scale is not provided in the chart, Hp can be determined by Hp ¼ hp Mp : (14:38) The power fluid flow rate and pump pressure differential are related through jet nozzle size by q1 ¼ 1214:5Aj ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p1 À p3 g1 r , (14:39) where g1 is the specific gravity of the power fluid, q1 is in bbl/day, and p1 and p3 are both in psi. The following procedure can be taken to select a jet pump: 1. Select a desired production rate of reservoir fluid q3 based on well IPR. Determine the required bottom- hole pressure pwf . 2. Design a pump setting depth D and estimate required pump intake pressure p3 based on pwf and flow gradi- ent below the pump. 3. From manufacturer’s literature, choose a pump with R value and determine Mp and Hp values for the pump based on pump performance curves. 4. Calculate power fluid rate q1 by q1 ¼ q3 Mp : 5. Based on tubing flow performance, calculate the re- quired discharge pressure p2,r using production rate q2 ¼ q1 þ q3. This step can be performed with the spreadsheet program HagedornBrownCorrelation.xls. 6. Determine the power fluid pressure p1 required to provide power fluid rate q1 with Eq. (14.39), that is, p1 ¼ p3 þ g1 q1 1214:5Aj 2 : 7. Determine the available discharge pressure p2 from the pump with Eq. (14.37), that is, p2 ¼ p3 þ Hpp1 1 þ Hp : 8. If the p2 value is greater than p2,r value with a reason- able safety factor, the chosen pump is okay to use, and go to Step 9. Otherwise, go to Step 3 to choose a different pump. If no pump meets the requirements for the desired production rate q3 and/or lifting pres- sure p2,r, go to Step 2 to change pump setting depth or reduce the value of the desired fluid production rate q3. 9. Calculate the required surface operating pressure ps based on the values of p1 and q1 and single-phase flow in tubing. 10. Calculate input power requirement by HP ¼ 1:7 Â 10À5 q1ps, where HP ¼ required input power, hp ps ¼ required surface operating pressure, psia. Summary This chapter provides a brief introduction to the principles of electrical submersible pumping, hydraulic piston pump- ing, hydraulic jet pumping, progressive cavity pumping, and plunger lift systems. Design guidelines are also pre- sented. Example calculations are illustrated with spread- sheet programs. References abercrombie, b. Plunger lift. In: The Technology of Artifi- cial Lift Methods (Brown, K.E., ed.), Vol. 2b. Tulsa: PennWell Publishing Co., 1980, pp. 483–518. beeson, c.m., knox, d.g., and stoddard, j.h. Plunger lift correlation equations and nomographs. Presented at AIME Petroleum Branch Annual meeting, 2–5 Octo- ber 1955, New Orleans, Louisiana. Paper 501-G. brown, k.e. The Technology of Artificial Lift Methods, Vol. 2b. Tulsa: PennWell Publishing Co., 1980. Figure 14.12 Example jet pump performance chart. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 222 3.1.2007 9:10pm Compositor Name: SJoearun 14/222 ARTIFICIAL LIFT METHODS
- 227. Centrilift-Hughes, Inc. Oilfield Centrilift-Hughes Submers- ible Pump Handbook. Claremore, Oaklahoma, 1998. cholet, h. Well Production Practical Handbook. Paris: Editions TECHNIP, 2000. foss, d.l. and gaul, r.b. Plunger lift performance criteria with operating experience: Ventura Avenue field. Dril- ling Production Practices API 1965:124–140. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston, TX: Gulf Publishing Co., 2005. guo, b. and ghalambor, a. Gas Volume Requirements for Underbalanced Drilling Deviated Holes. PennWell Books Tulsa, Oaklahoma, 2002. lea, j.f. Plunger lift vs velocity strings. Energy Sources Technology Conference Exhibition (ETCE ’99), 1–2 February 1999, Houston Sheraton Astrodome Hotel in Houston, Texas. lea, j.f. Dynamic analysis of plunger lift operations. Pre- sented at the 56th Annual Fall Technical Conference and Exhibition, 5–7 October 1981, San Antonio, Texas. Paper SPE 10253. lebeaux, j.m. and sudduth, l.f. Theoretical and practical aspects of free piston operation. JPT September 1955:33–35. listiak, s.d. Plunger lift. In: Petroleum Engineering Hand- book (Lake, L., ed.). Dallas: Society of Petroleum Engineers, 2006. mower, l.n., lea, j.f., beauregard, e., and ferguson, p.l. Defining the characteristics and performance of gas-lift plungers. Presented at the SPE Annual Technical Con- ference and Exhibition held in 22–26 September 1985, Las Vegas, Nevada. SPE Paper 14344. rosina, l. A study of plunger lift dynamics [Masters Thesis], University of Tulsa, 1983. turner, r.g., hubbard, m.g., and dukler, a.e. Analysis and prediction of minimum flow rate for the continu- ous removal of liquids from gas wells. J. Petroleum Technol. November 1969. Problems 14.1 A 9,000-ft-deep well produces 26 8API oil with GOR 50 scf/stb and zero water cut through a 3-in. (2.992- in. ID) tubing in a 7-in. casing. The oil has a forma- tion volume factor of 1.20 and average viscosity of 8 cp. Gas-specific gravity is 0.75. The surface and bottom-hole temperatures are 70 and 160 8F, re- spectively. The IPR of the well can be described by Vogel’s model with a reservoir pressure 4,050 psia and AOF 12,000 stb/day. If the well is put in produc- tion with an ESP to produce liquid at 7,000 stb/day against a flowing well head pressure of 150 psia, determine the required specifications for an ESP for this application. Assume the minimum pump suction pressure is 220 psia. 14.2 A 9,000-ft-deep well has a potential to produce 35 8API oil with GOR 120 scf/stb and 10% water cut through a 2-in. (1.995-in. ID) tubing in a 7-in. casing with a pump installation. The oil has a forma- tion volume factor of 1.25 and average viscosity of 5 cp. Gas- and water-specific gravities are 0.75 and 1.05, respectively. The surface and bottom-hole tem- peratures are 70 and 170 8F, respectively. The IPR of the well can be described by Vogel’s model with a reservoir pressure 2,000 psia and AOF 400 stb/day. If the well is to put in production with a HPP at depth of 8,500 ft in an open power fluid system to produce liquid at 210 stb/day against a flowing well head pressure of 65 psia, determine the required specifications for the HPP for this applica- tion. Assume the overall efficiencies of the engine, HHP, and surface pump to be 0.90, 0.80, and 0.85, respectively. 14.3 Calculate required GLR, casing pressure, and plun- ger lift operating range for the following given well data: Gas rate: 250 Mcfd expected when unloaded Liquid rate: 12 bbl/day expected when unloaded Liquid gradient: 0.40 psi/ft Tubing, ID: 1.995 in. Tubing, OD: 2.375 in. Casing, ID: 4.56 in. Depth to plunger: 7,000 ft Line pressure: 120 psi Available casing pressure: 850 psi Reservoir pressure: 1250 psi Average Z-factor: 0.99 Average temperature: 150 8F Plunger weight: 10 lb Plunger fall in gas: 750 fpm Plunger fall in liquid: 150 fpm Plunger rise velocity: 1,000 fpm Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap14 Final Proof page 223 3.1.2007 9:10pm Compositor Name: SJoearun OTHER ARTIFICIAL LIFT METHODS 14/223
- 228. Part IV Production Enhancement Good production engineers never stop looking for opportunities to improve the performance of their production systems. Performance enhancement ideas are from careful examinations and thorough analyses of production data to find the controlling factors affecting the performance. Part IV of this book presents procedures taken in the petroleum industry for identifying well problems and means of solving the problems. Materials are presented in the following four chapters. Chapter 15: Well Problem Identification Chapter 16: Matrix Acidizing Chapter 17: Hydraulic Fracturing Chapter 18: Production Optimization Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 225 22.12.2006 6:14pm
- 229. 15 Well Problem Identification Contents 15.1 Introduction 15/228 15.2 Low Productivity 15/228 15.3 Excessive Gas Production 15/231 15.4 Excessive Water Production 15/231 15.5 Liquid Loading of Gas Wells 15/231 Summary 15/241 References 15/241 Problems 15/242 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 227 22.12.2006 6:14pm
- 230. 15.1 Introduction The engineering work for sustaining and enhancing oil and gas production rates starts from identifying problems that cause low production rates of wells, quick decline of the desirable production fluid, or rapid increase in the undesir- able fluids. For oil wells, these problems include . Low productivity . Excessive gas production . Excessive water production . Sand production For gas wells, the problems include . Low productivity . Excessive water production . Liquid loading . Sand production Although sand production is easy to identify, well testing and production logging are frequently needed to identify the causes of other well problems. 15.2 Low Productivity The lower than expected productivity of oil or gas well is found on the basis of comparison of the well’s actual production rate and the production rate that is predicted by Nodal analysis. If the reservoir inflow model used in the Nodal analysis is correct (which is often questionable), the lower than expected well productivity can be attributed to one or more of the following reasons: . Overestimate of reservoir pressure . Overestimate of reservoir permeability (absolute and relative permeabilities) . Formation damage (mechanical and pseudo skins) . Reservoir heterogeneity (faults, stratification, etc.) . Completion ineffectiveness (limited entry, shallow per- forations, low perforation density, etc.) . Restrictions in wellbore (paraffin, asphaltane, scale, gas hydrates, sand, etc.) The first five factors affect reservoir inflow performance, that is, deliverability of reservoir. They can be evaluated on the basis of pressure transient data analyses. The true production profile from different zones can be obtained based on production logging such as temperature and spinner flow meter logs. An example is presented in Fig. 15.1, which shows that Zone A is producing less than 10% of the total flow, Zone B is producing almost 70% of the total rate, and Zone C is contributing about 25% of the total production. The last factor controls well deliverability. It can be evaluated using data from production logging such as flowing gradient survey (FGS). The depth interval with high-pressure gradient is usually the interval where the depositions of paraffins, asphaltanes, scales, or gas hydrates are suspected. 15.2.1 Pressure Transient Data Analysis Pressure transient testing plays a key role in evaluating exploration and development prospects. Properly designed well tests can provide reservoir engineers with reservoir pressure, reserves (minimum economic or total), and flow capacity, all of which are essential in the reservoir evalu- ation process. Some of the results one can obtain from pressure transient testing include the following: . Initial reservoir pressure . Average reservoir pressure . Directional permeability . Radial effective permeability changes from the wellbore . Gas condensate fallout effect on flow Depth Temperature % flow from spinner flowmeter 150 0 25 50 75 100 160 sp g - ray 7100 7150 7200 7250 7300 Fill 7350 Zone A Zone B Zone C Figure 15.1 Temperature and spinner flowmeter-derived production profile (Economides et al., 1994). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 228 22.12.2006 6:14pm 15/228 PRODUCTION ENHANCEMENT
- 231. . Near wellbore damage/stimulation . Rate-dependent skin . Boundary identification . Partial penetration effect on flow . Effective fracture length . Effective fracture conductivity . Dual-porosity characteristics (storativity and transmis- sivity ratios) The theoretical basis of pressure transient data analysis is beyond the scope of this book. It can be found elsewhere (Chaudhry, 2004; Horne, 1995; Lee et al., 2003). Modern computer software packages are available for data anal- yses. These packages include PanSystem (EPS, 2004) and F.A.S.T. WellTest (Fekete, 2003). The following subsec- tions briefly present some principles of data analyses that lead to deriving reservoir properties directly affecting well productivity. Reservoir Pressure. Reservoir pressure is a key param- eter controlling well deliverability. A simple way to deter- mine the magnitude of initial reservoir pressure may be the Horner plot of data from pressure buildup test if the reservoir boundary was not reached during the test. If the boundary effects are seen, the average reservoir pressure can be estimated on the basis of the extrapolated initial reservoir pressure from Horner plot and the MBH plot (Dake, 2002). Effective Permeability. The effective reservoir perme- ability that controls the well’s deliverability should be derived from the flow regime that prevails in the reservoir for long-term production. To better understand the flow regimes, the commonly used equationsdescribingflow in oil reservoirs are summarized first in this subsection. Similar equations for gas reservoirs can be found in Lee et al. (2003). Horizontal Radial Flow. For vertical wells fully pene- trating nonfractured reservoirs, the horizontal radial flow can be mathematically described in consistent units as pwf ¼ pi À qBm 4pkhh ln kht fmctr2 w þ 2S þ 0:80907 ! , (15:1) where pwf ¼ flowing bottom-hole pressure pi ¼ initial reservoir pressure q ¼ volumetric liquid production rate B ¼ formation volume factor m ¼ fluid viscosity kh ¼ the average horizontal permeability h ¼ pay zone thickness t ¼ flow time f ¼ initial reservoir pressure ct ¼ total reservoir compressibility rw ¼ wellbore radius S ¼ total skin factor. Horizontal Linear Flow. For hydraulically fractured wells, the horizontal linear flow can be mathematically described in consistent units as pwf ¼ pi À qBm 2pkyh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pkyt fmctx2 f s þ S # , (15:2) where xf is fracture half-length and ky is the permeability in the direction perpendicular to the fracture face. Vertical Radial Flow. For horizontal wells as depicted in Fig. 15.2, the early-time vertical radial flow can be mathematically described in consistent units as pwf ¼ pi À qBm 4pkyzL ln kyzt fmctr2 w þ 2S þ 0:80907 ! , (15:3) where L is the horizontal wellbore length and kyz is the geometric mean of horizontal and vertical permeabilities, that is, kyz ¼ ffiffiffiffiffiffiffiffiffi kykz p : (15:4) Horizontal Pseudo-Linear Flow. The pseudo-linear flow toward a horizontal wellbore can be mathematically described in consistent units as pwf ¼ pi À qBm 2pky h À Zwð Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4pkyt fmctL2 s þ S # : (15:5) Horizontal Pseudo-Radial Flow. The pseudo-radial flow toward a horizontal wellbore can be mathematically described in consistent units as pwf ¼ pi À qBm 4pkhh ln kht fmctr2 w þ 2S þ 0:80907 ! : (15:6) For vertical wells fully penetrating nonfractured reser- voirs, it is usually the average (geometric mean) of hori- zontal permeabilities, kh, that dominates long-term production performance. This average horizontal perme- ability can be derived from the horizontal radial flow regime. For wells draining relatively small portions of hydraulically fractured reservoir segments, it is usually the permeability in the direction perpendicular to the frac- ture face that controls long-term production performance. This permeability can be derived from the horizontal lin- ear flow regime. For horizontal wells draining relatively large portions of nonfractured reservoir segments, it is usu- ally again the geometric mean of horizontal permeabilities x L y z h Zw Zw h z x y Figure 15.2 Notations for a horizontal wellbore. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 229 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/229
- 232. that dominates long-term production performance. This average horizontal permeability can be derived from the pseudo-radial flow regime. For vertical wells partially pene- trating nonfractured reservoirs, both horizontal and vertical permeabilities influence long-term production performance. These permeabilities can usually be derived from the hemi- spherical flow regime. Flow regimes are usually identified using the diagnostic pressure derivative p0 defined as p0 ¼ dDp d ln (t) ¼ t dDp dt , (15:7) where t is time and Dp is defined as Dp ¼ pi À pwf (15:8) for drawdown tests, where pi and pwf are initial reservoir pressure and flowing bottom-hole pressure, respectively. For pressure buildup tests, the Dp is defined as Dp ¼ psw À pwfe, (15:9) where pws and pwfe are ship-in bottom-hole pressure and the flowing bottom-hole pressure at the end of flow (before shut-in), respectively. For any type of radial flow (e.g., horizontal radial flow, vertical radial flow, horizontal pseudo-radial flow), the diagnostic derivative is derived from Eqs. (15.1), (15.3), and (15.6) as p0 ¼ dDp d ln (t) ¼ qBm 4pkHR , (15:10) where k is the average permeability in the flow plane (kh or kyz) and kh ¼ ffiffiffiffiffiffiffiffiffiffiffi kx ky p HR is the thickness of the radial flow (h or L). Apparently, the diagnostic derivative is constant over the radial flow time regime. The plot of p0 versus t data should show a trend of straight line parallel to the t-axis. For linear flow (e.g., flow toward a hydraulic fracture), the diagnostic derivative is derived from Eq. (15.2) as p0 ¼ dDp d ln (t) ¼ qB 4hxf ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mt pfctky s : (15:12) For pseudo-linear flow (e.g., flow toward a horizontal well), the diagnostic derivative is derived from Eq. (15.5) as p0 ¼ dDp d ln (t) ¼ qB 2L(h À zw) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi mt pfctky s : (15:13) Taking logarithm of Eqs. (15.12) and (15.13) gives log p0 ð Þ ¼ 1 2 log tð Þ þ log qB 4hxf ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m pfctky r (15:14) and log p0 ð Þ ¼ 1 2 log tð Þ þ log qB 2L(h À zw) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m pfctky r : (15:15) Equations (15.13) and (15.14) indicate that the signature of the linear flow regime is the 1 ⁄2 slope on the log-log plot of diagnostic derivative versus time. Once the flow regimes are identified, permeabilities associated with the flow regime can be determined based on slope analyses. For any types of radial flow, Eqs. (15.1), (15.3), and (15.6) indicate that plotting of bottom-hole pressure versus time data on a semilog scale will show a trend with a constant slope mR, where mR ¼ À qBm 4pkHR : (15:16) Then the average permeability in the flow plane (kh or kyz) can be estimated by k ¼ À qBm 4pHRmR : (15:17) For any types of linear flow, Eqs. (15.2) and (15.5) indicate that plotting of the bottom-hole pressure versus the square-root of time data will show a trend with a constant slope mL, where mL ¼ À qB HLXL ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m pfctky r , (15:18) where HL ¼ h and XL ¼ 2xf for linear flow, and HL ¼ h À Zw and XL ¼ L for pseudo-linear flow, respec- tively. Then the permeability in the flow plane can be estimated by ky ¼ m pfct qB mLHLXL 2 : (15:19) If a horizontal well is tested for a time long enough to detect the pseudo-radial flow, then it is possible to estimate other directional permeabilities by kx ¼ k2 h ky (15:20) and kz ¼ k2 yz ky : (15:21) Although kx and kz are not used in well productivity analysis, they provide some insight about reservoir anisot- ropy. Skin Factor. Skin factor is a constant that is used to adjust the flow equation derived from the ideal condition (homogeneous and isotropic porous media) to suit the applications in nonideal conditions. It is an empirical fac- tor employed to consider the lumped effects of several aspects that are not considered in the theoretical basis when the flow equations were derived. The value of the skin factor can be derived from pressure transient test analysis with Eqs. (15.1), (15.2), (15.3), (15.5), and (15.6). But its value has different meanings depending on flow regime. A general expression of the skin factor is S ¼ SD þ SCþu þ SP þ X SPS, (15:22) where SD is damage skin during drilling, cementing, well completion, fluid injection, and even oil and gas produc- tion. Physically, it is due to plugging of pore space by external or internal solid particles and fluids. This com- ponent of skin factor can be removed or averted with well stimulation operations. The SCþu is a skin component due to partial completion and deviation angle, which make the flow pattern near the wellbore deviate from ideal radial flow pattern. This skin component is not removable in water coning and gas coning systems. The SP is a skin component due to the nonideal flow condition around the perforations associated with cased-hole completion. It depends on a number of parameters including perforation density, phase angle, perforation depth, diameter, com- pacted zone, and others. This component can be mini- mized with optimized perorating technologies. The SSPS represents pseudo-skin components due to non–Darcy flow effect, multiphase effect, and flow convergence near the wellbore. These components cannot be eliminated. It is essential to know the magnitude of components of the skin factor S derived from the pressure transient test data analysis. Commercial software packages are available for decomposition of the skin factor on the basis of well completion method. One of the packages is WellFlo (EPS, 2005). Example Problem 15.1 A horizontal wellbore was placed in a 100-ft thick oil reservoir of 0.23 porosity. Oil formation Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 230 22.12.2006 6:14pm 15/230 PRODUCTION ENHANCEMENT
- 233. volume factor and viscosity are 1.25 rb/stb and 1 cp, respectively. The total reservoir compressibility factor is 10À5 psiÀ1 . The well was tested following the schedule shown in Fig. 15.3. The measured flowing bottom-hole pressures are also presented in Fig. 15.3. Estimate directional permeabilities and skin factors from the test data. Solution Figure 15.4 presents a log-log diagnostic plot of test data. It clearly indicates a vertical radial flow at early time, a pseudo-linear flow at mid-time, and the beginning of a pseudo-radial flow at late time. The semi-log analysis for the vertical radial flow is shown in Fig. 15.5, which gives kyz ¼ 0:9997 md and near-wellbore skin factor S ¼ À0:0164. The square-root time plot analysis for the pseudo-linear flow is shown in Fig. 15.6, which gives the effective well- bore length of L ¼ 1,082:75 ft and a skin factor due to convergence of S ¼ 3:41. The semi-log analysis for the horizontal pseudo-radial flow is shown in Fig. 15.7, which gives kh ¼ 1:43 md and pseudo-skin factor S ¼ À6:17. Figure 15.8 shows a match between the measured and model-calculated pressure responses given by an optimiza- tiontechnique.This matchwas obtainedusing the following parameter values: kh ¼ 1:29 md kz ¼ 0:80 md S ¼ 0:06 L ¼ 1,243 ft: To estimate the long-term productivity of this horizontal well, the kh ¼ 1:29 md and S ¼ 0:06 should be used in the well inflow equation presented in Chapter 3. 15.3 Excessive Gas Production Excessive gas production is usually due to channeling be- hind the casing (Fig. 15.9), preferential flow through high- permeability zones (Fig. 15.10), gas coning (Fig. 15.11), and casing leaks (Clark and Schultz, 1956). The channeling behind the casing and gas coning prob- lems can be identified based on production logging such as temperature and noise logs. An example is depicted in Fig. 15.12, where both logs indicate that gas is being produced from an upper gas sand and channeling down to some perforations in the oil zone. Excessive gas production of an oil well could also be due to gas production from unexpected gas zones. This can be identified using production logging such as temperature and density logs. An example is presented in Fig. 15.13, where both logs indicate gas production from the thief zone B. 15.4 Excessive Water Production Excessive water production is usually from water zones, not from the connate water in the pay zone. Water enters the wellbore due to channeling behind the casing (Fig. 15.14), preferential flow through high-permeability zones (Fig. 15.15), water coning (Fig. 15.16), hydraulic fracturing into water zones, and casing leaks. Figure 15.17 shows how to identify fracture height using prefracture and postfracture temperature logs to tell whether the hydraulic fracture has extended into a water zone. In addition to those production logging tools that are mentioned in the previous section, other production log- ging tools can be used for identifying water-producing zones. Fluid density logs are especially useful for identify- ing water entries. Comparison between water-cut data and spinner flowmeter log can sometimes give an idea of where the water is coming from. Figure 15.18 shows a spinner flowmeter log identifying a watered zone at the bottom of a well with a water-cut of nearly 50%. 15.5 Liquid Loading of Gas Wells Gas wells usually produce natural gas-carrying liquid water and/or condensate in the form of mist. As the gas flow velocity in the well drops because of reservoir pres- sure depletion, the carrying capacity of the gas decreases. Figure 15.3 Measured bottom-hole pressures and oil production rates during a pressure drawdown test. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 231 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/231
- 234. When the gas velocity drops to a critical level, liquids begin to accumulate in the well and the well flow can undergo an annular flow regime followed by a slug flow regime. The accumulation of liquids (liquid loading) increases the bottom-hole pressure, which reduces gas production rate. A low gas production rate will cause gas velocity to drop further. Eventually, the well will undergo a bubbly flow regime and cease producing. Several measures can be taken to solve the liquid-loading problem. Foaming the liquid water can enable the gas to lift water from the well. Using smaller tubing or creating a lower wellhead pressure sometimes can keep mist flowing. The well can be unloaded by gas-lifting or pumping the liquids out of the well. Heating the wellbore can prevent oil condensation. Down-hole injection of water into an underlying disposal zone is another option. However, liquid-loading is not always obvious and recognizing the liquid-loading problem is not an easy task. A thorough diagnostic analysis of well data needs to be performed. The symptoms to look for include onset of liquid slugs at the surface of well, increasing difference between the tubing and casing pressures with time, sharp changes in gradient on a flowing pressure survey, and sharp drops in production decline curve. 15.5.1 The Turner et al. Method Turner et al. (1969) were the pioneer investigators who analyzed and predicted the minimum gas flow rate to prevent liquid-loading. They presented two mathematical 1,000 100 DeltaPandP-Derivative(psi) 10 0.001 0.01 0.1 1 Elapsed Time (hours) Vertical Radial Flow Pseudo Linear Flow Begging of Pseudo Radial Flow 10 100 1,000 Figure 15.4 Log-log diagnostic plot of test data. Model results Two no-flow boundaries-homogeneous Infinitely acting Kbar = 0.9997 md Radial Flow Plot 6,000 5,500 5,000 4,500 4,000 3,500 0.001 0.01 0.1 1 Elapsed Time (hours) Pressure(psia) 10 100 1,000 S = −0.0164 Figure 15.5 Semi-log plot for vertical radial flow analysis. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 232 22.12.2006 6:14pm 15/232 PRODUCTION ENHANCEMENT
- 235. models to describe the liquid-loading problem: the film- movement model and the entrained drop movement model. On the basis of analyses on field data, they con- cluded that the film-movement model does not represent the controlling liquid transport mechanism. Turner et al.’s entrained drop movement model was de- rived on the basis of the terminal free settling velocity of liquid drops and the maximum drop diameter corresponding to the critical Weber number of 30. Turner et al.’s terminal slip velocity equation is expressed in U.S. field units as vsl ¼ 1:3s1=4 rL À rg 1=4 C 1=4 d r 1=2 g : (15:23) According to Turner et al., gas will continuously remove liquids from the well until its velocity drops to below the Figure 15.6 Square-root time plot for pseudo-linear flow analysis. Figure 15.7 Semi-log plot for horizontal pseudo-radial flow analysis. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 233 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/233
- 236. terminal slip velocity. The minimum gas flow rate (in MMcf/D) for a particular set of conditions (pressure and conduit geometry) can be calculated using Eqs. (15.23) and (15.24): QgslMM ¼ 3:06pvslA Tz (15:24) Figure 15.19 shows a comparison between the results of Turner et al.’s entrained drop movement model. The map shows many loaded points in the unloaded region. Turner et al. recommended the equation-derived values be adjusted upward by approximately 20% to ensure removal of all drops. Turner et al. believed that the discrepancy was attributed to several facts including the use of drag coeffi- cients for solid spheres, the assumption of stagnation velo- city, and the critical Weber number established for drops falling in air, not in compressed gas. The main problem that hinders the application of Turner et al.’s entrained drop model to gas wells comes from the difficulties of estimating the values of fluid den- sity and pressure. Using an average value of gas-specific gravity (0.6) and gas temperature (120 8F), Turner et al. derived an expression for gas density as 0.0031 times the pressure. However, they did not present a method for calculating the gas pressure in a multiphase flow wellbore. Figure 15.8 Match between measured and model calculated pressure data. Well bore Casing leak Bad cement job High pressure Gas zone Oil zone Figure 15.9 Gas production due to channeling behind the casing (Clark and Schultz, 1956). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 234 22.12.2006 6:14pm 15/234 PRODUCTION ENHANCEMENT
- 237. The spreadsheet program TurnerLoading.xls has been developed for quick calculation associated with this book. Turner et al.’s entrained drop movement model was later modified by a number of authors. Coleman et al. (1991) suggested to use Eq. (15.23) with a lower constant value. Nosseir et al. (2000) expanded Turner et al.’s entrained drop model to more than one flow regime in a well. Lea and Nickens (2004) made some corrections to Turner et al.’s simplified equations. However, the original drawbacks (neglected transport velocity and multiphase flow pressure) with Turner et al.’s approach still remain unsolved. 15.5.2 The Guo et al. Method Starting from Turner et al.’s entrained drop model, Guo et al. (2006) determined the minimum kinetic energy of gas that is required to lift liquids. A four-phase (gas, oil, water, and solid particles) mist-flow model was developed. Applying the minimum kinetic energy criterion to the four-phase flow model resulted in a closed-form analytical equation for predicting the minimum gas flow rate. 15.5.2.1 Minimum Kinetic Energy Kinetic energy per unit volume of gas can be expressed as Ek ¼ rgv2 g 2gc : (15:25) Substituting Eq. (15.23) into Eq. (15.25) gives an expres- sion for the minimum kinetic energy required to keep liquid droplets from falling: Eksl ¼ 0:026 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s rL À rg Cd v u u t (15:26) If the value of drag coefficient Cd ¼ 0:44 (recommended by Turner et al.) is used and the effect of gas density is neglected (a conservative assumption), Eq. (15.26) becomes Well Bore Gas-Oil Contact Intermediate permeability Intermediate permeability Low permeability Low permeability High permeability Figure 15.10 Gas production due to preferential flow through high-permeability zones (Clark and Schultz, 1956). Well Bore Gas Cap Oil Zone Figure 15.11 Gas production due to gas coning (Clark and Schultz, 1956). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 235 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/235
- 238. Eksl ¼ 0:04 ffiffiffiffiffiffiffiffiffi srL p : (15:27) In gas wells producing water, typical values for water–gas interfacial tension and water density are 60 dynes/cm and 65 lbm=ft3 , respectively. This yields the minimum kinetic energy value of 2:5 lbf -ft=ft3 . In gas wells producing conden- sate, typical values for condensate–gas interfacial tension and condensate density are 20 dynes/cm and 45 lbm=ft3 , respectively. This yields the minimum kinetic energy value of 1:2 lbf -ft=ft3 . The minimum gas velocity required for transporting the liquid droplets upward is equal to the minimum gas ve- locity required for floating the liquid droplets (keeping the droplets from falling) plus the transport velocity of the droplets, that is, vgm ¼ vsl þ vtr: (15:28) The transport velocity vtr may be calculated on the basis of liquid production rate, geometry of the conduit, and liquid volume fraction, which is difficult to quantify. Instead of trying to formulate an expression for the transport velocity vtr, Guo et al. used vtr as an empir- ical constant to lump the effects of nonstagnation ve- locity, drag coefficients for solid spheres, and the critical Weber number established for drops falling in air. On the Temperature Noise Amplitude Gas zone Oil production zone Gas Gas Oil Oil 600 Hz Figure 15.12 Temperature and noise logs identifying gas channeling behind casing (Economides et al., 1994). Depth A B C D Temperature (ЊF) Fluid density (g/cc) Figure 15.13 Temperature and fluid density logs identifying a gas entry zone (Economides et al., 1994). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 236 22.12.2006 6:14pm 15/236 PRODUCTION ENHANCEMENT
- 239. basis of the work by Turner et al., the value of vtr was taken as 20% of vsl in this study. Use of this value results in vgm % 1:2vsl: (15:29) Substituting Eqs. (15.23) and (15.29) into Eq. (15.25) results in the expression for the minimum kinetic energy required for transporting the liquid droplets as Ekm ¼ 0:0576 ffiffiffiffiffiffiffiffiffi srL p : (15:30) For typical gas wells producing water, this equation yields the minimum kinetic energy value of 3:6 lbf -ft=ft3 . For typical gas wells producing condensate, this equation gives the minimum kinetic energy value of 1:73 lbf -ft=ft3 . These numbers imply that the required minimum gas pro- duction rate in water-producing gas wells is approximately twice that in condensate-producing gas wells. To evaluate the gas kinetic energy Ek in Eq. (15.25) at a given gas flow rate and compare it with the minimum required kinetic energy Ekm in Eq. (15.30), the values of gas density rg and gas velocity vg need to be determined. Expressions for rg and vg can be obtained from ideal gas law: rg ¼ 2:7Sgp T (15:31) vg ¼ 4:71 Â 10À2 TQG Aip (15:32) Substituting Eqs. (15.31) and (15.32) into Eq. (15.25) yields Ek ¼ 9:3 Â 10À5 SgTQ2 G A2 i p : (15:33) Equation (15.33) indicates that the gas kinetic energy decreases with increased pressure, which means that the controlling conditions are bottom-hole conditions where gas has higher pressure and lower kinetic energy. This analysis is consistent with the observations from air- drilling operations where solid particles accumulate at the bottom-hole rather than at the top-hole. However, this analysis is in contradiction with the results by Turner et al., which indicated that the wellhead conditions are in most instances, controlling. Low pressure Oil reservoir High pressure water sand Casing leak Water channel along Bad cement job Figure 15.14 Water production due to channeling behind the casing. Well Bore Low permeability Low permeability Intermediate permeability High permeability Figure 15.15 Preferential water flow through high-permeability zones. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 237 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/237
- 240. 15.5.2.2 Four-Phase Flow Model To accurately predict the bottom-hole pressure p in Eq. (15.33), a gas-oil-water-solid four-phase mist-flow model was developed by Guo et al. (2006). According to the four-phaseflowmodel,theflowingpressurepatdepthL can be solved numerically from the following equation: 144b p À phf À Á þ 1 À 2bm 2 ln 144p þ mð Þ2 þ n 144phf þ m À Á2 þ n b P À Phf À Á þ 1 À 2bm 2 ln P þ mð Þ2 þ n Phf þ m À Á2 þ n À m þ b c n À bm2 ffiffiffi n p tanÀ1 144p þ m ffiffiffi n p À tanÀ1 144phf þ m ffiffiffi n p ! ¼ a 1 þ d2 e À Á L, (15:34) where a ¼ 15:33SsQs þ 86:07SwQw þ 86:07SoQo þ 18:79SgQG 103TavQG Â cos (u), (15:35) b ¼ 0:2456Qs þ 1:379Qw þ 1:379Qo 103TavQG , (15:36) c ¼ 6:785 Â 10À6 TavQG Ai , (15:37) d ¼ Qs þ 5:615 Qw þ Qoð Þ 600Ai , (15:38) e ¼ 6f gDh cos (u) , (15:39) fM ¼ 1 1:74 À 2 log 20 Dh 2 4 3 5 2 , (15:40) Well Bore Oil Zone Water Cone Figure 15.16 Water production due to water coning. 8,800 9,000 9,200 9,400 9,600 9,800 10,000 10,200 10,400 175 80ЊC 93ЊC 108ЊC 121ЊC 135ЊC 200 225 Temperature (ЊF) 250 275 3,170 3,110 3,050 2,990 2,930 HoleDepth(ft) Holedepth(m) 2,870 2,810 Static Log Profiles Separate Fracture Top Post Frac Profile Thermal Conductivity Effects Pre Frac Profile 2,750 2,690 Figure 15.17 Prefracture and postfracture temperature logs identifying fracture height (Dobkins, 1981). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 238 22.12.2006 6:14pm 15/238 PRODUCTION ENHANCEMENT
- 241. Spinner Speed RPS Production Profile 0 0 10 B/D 4,860 0.0% 4.1% 21.0% 14.4% 12.1% 48.4 % 13,400 13,300 13,200 13,100 13,000 12,900 12,800 12,700 Figure 15.18 Spinner flowmeter log identifying a watered zone at bottom. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 239 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/239
- 242. m ¼ cde 1 þ d2e , (15:41) and n ¼ c2 e 1 þ d2eð Þ 2 , (15:42) where A ¼ cross-sectional area of conduit, ft2 Dh ¼ hydraulic diameter, in. fM ¼ Moody friction factor g ¼ gravitational acceleration, 32:17 ft=s2 L ¼ conduit length, ft p ¼ pressure, psia phf ¼ wellhead flowing pressure, psia QG ¼ gas production rate, Mscf/day Qo ¼ oil production rate, bbl/day Qs ¼ solid production rate, ft3 =day Qw ¼ water production rate, bbl/day Sg ¼ specific gravity of gas, air ¼ 1 So ¼ specific gravity of produced oil, freshwater ¼ 1 Sw ¼ specific gravity of produced water, freshwater ¼ 1 Ss ¼ specific gravity of produced solid, freshwater ¼ 1 Tav ¼ the average temperature in the butting, 8R «0 ¼ pipe wall roughness, in. u ¼ inclination angle, degrees. 15.5.2.3 Minimum Required Gas Production Rate A logical procedure for predicting the minimum required gas flow rate Qgm involves calculating gas density rg, gas velocity vg, and gas kinetic energy Ek at bottom-hole con- dition using an assumed gas flow rate QG, and compare the Ek with Ekm. If the Ek is greater than Ekm, the QG is higher than the Qgm. The value of QG should be reduced and the calculation should be repeated until the Ek is very close to Ekm. Because this procedure is tedious, a simple equation was derived by Guo et al. for predicting the minimum required gas flow rate in this section. Under the minimum unloaded condition (the last point of the mist flow regime), Eq. (15.33) becomes Ekm ¼ 9:3  10À5 SgTbhQgm 2 A2 i p , (15:43) which gives p ¼ 9:3  10À5 SgTbhQgm 2 A2 i Ekm : (15:44) Substituting Eq. (15.44) into Eq. (15.34) results in 144ba1 þ 1 À 2bm 2 ln a2 À m þ b c n À bm2 ffiffiffi n p  tanÀ1 b1 À tanÀ1 b2  à ¼ g, (15:45) where a1 ¼ 9:3  10À5 SgTbhQ2 gm A2 i Ekm À phf , (15:46) a2 ¼ 1:34  10À2 SgTbhQ2 gm A2 i Ekm þ m 2 þ n 144phf þ m À Á2 þ n , (15:47) b1 ¼ 1:34  10À2 SgTbhQ2 gm A2 i Ekm þ m ffiffiffi n p , (15:48) b2 ¼ 144phf þ m ffiffiffi n p , (15:49) and g ¼ a 1 þ d2 e À Á L: (15:50) All the parameter values should be evaluated at Qgm. The minimum required gas flow rate Qgm can be solved from Eq. (15.45) with a trial-and-error or numerical method such as the Bisection method. It can be shown that Eq. (15.45) is a one-to-one function of Qgm for Qgm values greater than zero. Therefore, the Newton– Raphson iteration technique can also be used for solving Qgm. Commercial software packages such as MS Excel can be used as solvers. In fact, the Goal Seek function built into MS Excel was used for generating solutions presented in this chapter. The spreadsheet program is named GasWellLoading.xls. Example Problem 15.2 To demonstrate how to use Eq. (15.45) for predicting the minimum unloading gas flow rate, consider a vertical gas well producing 0.70 specific gravity gas and 50 bbl/day condensate through a 2.441-in. inside diameter (ID) tubing against a wellhead pressure of 900 psia. Suppose the tubing string is set at a depth of 10,000 ft, and other data are given in Table 15.1. Solution The solution given by the spreadsheet program GasWellLoading.xls is shown in Table 15.2. 0 2,000 4,000 6,000 8,000 10,000 12,000 12,00010,0008,0006,0004,0002,0000 Calculated Minimum Flow Rate (Mcf/D) TestFlowRate(Mcf/D) Unloaded Nearly loaded up Loaded up Questionable ? Figure 15.19 Calculated minimum flow rates with the Turner et al. model and test flow rates. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 240 22.12.2006 6:14pm 15/240 PRODUCTION ENHANCEMENT
- 243. 15.5.3 Comparison of the Turner et al. and the Guo et al. Methods Figure 15.20 illustrates Eq. (15.45)–calculated minimum flow rates mapped against the test flow rates for the same wells used in Fig. 15.19. This map shows six loaded points in the unloaded region, but they are very close to the boundary. This means the Guo et al. method is more accurate than the Turner et al. method in estimating the minimum flow rates. Summary This chapter presents a guideline to identifying problems commonly encountered in oil and gas wells. Well test analysis provides a means of estimating properties of indi- vidual pay zones. Production logging analysis identifies fluid entries to the wellbore from different zones. The Guo et al. method is more accurate than the Turner et al. method for predicting liquid-loading problems in gas pro- duction wells. References chaudhry, a.c. Oil Well Testing Handbook. Burlington: Gulf Professional Publishing, 2004. clark, n.j. and schultz, w.p. The analysis of problem wells. Petroleum Engineer September 1956;28:B30– B38. coleman, s.b., clay, h.b., mccurdy, d.g., and norris, l.h., iii. A new look at predicting gas well loading-up. JPT (March 1991), Trans. AIME 1991;291:329. dake, l.p. Fundamentals of Reservoir Engineering. Amsterdam: Elsevier, 2002. dobkins, t.a. Improved method to determine hydraulic fracture height. JPT April 1981:719–726. economides, m.j., hill, a.d., and ehlig-economides, c. Petroleum Production Systems. New Jersey: Prentice Hall PTR, 1994. E-Production Services, Inc. FloSystem User Manual. Edinburgh: E-Production Services, Inc., 2005. E-Production Services, Inc. PanSystem User Manual. Edinburgh: E-Production Services, Inc., 2004. fekete., f.a.s.t. WellTest User Manual. Calgary: Fekete Associates, Inc., 2003. guo, b., ghalambor, a., and xu, c. A systematic approach to predicting liquid loading in gas well. SPE Produc- tion Operations J. February 2006. horne, r.n. Modern Well Test Analysis: A Computer-Aided Approach. New York: Petroway Publishing, 1995. lea, j.f. and nickens, h.v. Solving gas-well liquid-loading problems. SPE Prod. Facilities April 2004:30. lee, j.w., rollins, j.b., and spivey, j.p. Pressure Transient Testing. Richardson: Society of Petroleum Engineers, 2003. Table 15.1 Basic Parameter Values for Example Problem 15.1 Gas-specific gravity 0:7 (air ¼ 1) Hole inclination 0 degrees Wellhead temperature 608 Geothermal gradient 0.01 8F/ft Condensate gravity 60 8API Water-specific gravity 1:05 (water ¼ 1) Solid-specific gravity 2:65 (water ¼ 1) Interfacial tension 20 dyne/cm Tubing wall roughness 0.000015 in. Table 15.2 Result Given by the Spreadsheet Program GasWellLoading.xls Calculated Parameters Hydraulic diameter 0.2034 ft Conduit cross-sectional area 0.0325 ft2 Average temperature 570 8R Minimum kinetic energy 1.6019 lb-ft/ft3 a ¼ 2.77547E-05 b ¼ 1.20965E-07 c ¼ 875999.8117 d ¼ 0.10598146 e ¼ 0.000571676 fM ¼ 0.007481992 m ¼ 53.07387106 n ¼ 438684299.6 Solution Critical gas production rate 1,059 Mscf/day Pressure ( p) ¼ 1,189 psia Objective function f(Qgm) ¼ À1:78615E-05 0 2,000 4,000 6,000 8,000 10,000 12,000 12,00010,0008,0006,0004,0002,0000 Calculated Minimum Flow Rate (Mcf/D) TestFlowRate(Mcf/D) Unloaded Nearly loaded up Loaded up Questionable ? Figure 15.20 The minimum flow rates given by the Guo et al. model and the test flow rates. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 241 22.12.2006 6:14pm WELL PROBLEM IDENTIFICATION 15/241
- 244. nosseir, m.a., darwich, t.a., sayyouh, m.h., and sallaly, m.e. A new approach for accurate prediction of loading in gas wells under different flowing conditions. SPE Prod. Facilities November 2000;15(4):245. turner, r.g., hubbard, m.g., and dukler, a.e. Analysis and prediction of minimum flow rate for the continu- ous removal of liquids from gas wells. JPT November 1969, Trans. AIME 1969;246:1475. Problems 15.1 Consider a gas well producing 50 bbl/d of condensate and 0.1 cubic foot of sand through a 2.441-in. I.D. tubing against a wellhead pressure of 500 psia. Sup- pose the tubing string is set at a depth of 8,000 ft, use the following data and estimate the minimum gas production rate before the gas well gets loaded. 15.2 Consider a gas well producing 50 bbl/day of water and 0:2 ft3 of sand through a 2.441-in. ID tubing against a wellhead pressure of 600 psia and tempera- ture of 80 8F. Suppose the tubing string is set at a depth of 9,000 ft and geothermal gradient is 0.01 8F/ft, estimate the minimum gas production rate before the gas well gets loaded. 15.3 Consider a gas well producing 80 bbl/day of water and 0:1 ft3 of sand through a 1.995-in. ID tubing against a wellhead pressure of 400 psia and tempera- ture of 70 8F. Suppose the tubing string is set at a depth of 7,000 ft and geothermal gradient is 0.01 8F/ft, estimate the minimum gas production rate before the gas well gets loaded. 15.4 Consider a gas well producing 70 bbl/day of oil and 0:1 ft3 of sand through a 1.995-in. ID tubing against a wellhead pressure of 600 psia and temperature of 80 8F. Suppose the tubing string is set at a depth of 6,000 ft and geothermal gradient is 0.01 8F/ft, esti- mate the minimum gas production rate before the gas well gets loaded.Gas-specific gravity: 0:75 (air ¼ 1) Hole inclination: 0 degrees Wellhead temperature: 60 8F Geothermal gradient: 0.01 8F/ft Condensate gravity: 60 8API Water-specific gravity: 1:07 (water ¼ 1) Solid-specific gravity: 2:65 (water ¼ 1) Oil–gas interface tension: 20 dyne/cm Tubing wall roughness: 0.000015 in. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap15 Final Proof page 242 22.12.2006 6:14pm 15/242 PRODUCTION ENHANCEMENT
- 245. 16 Matrix Acidizing Contents 16.1 Introduction 16/244 16.2 Acid–Rock Interaction 16/244 16.3 Sandstone Acidizing Design 16/244 16.4 Carbonate Acidizing Design 16/247 Summary 16/248 References 16/248 Problems 16/249 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 243 21.12.2006 2:30pm
- 246. 16.1 Introduction Matrix acidizing is also called acid matrix treatment. It is a technique to stimulate wells for improving well inflow per- formance. In the treatment, acid solution is injected into the formation to dissolve some of the minerals to recover per- meability of sandstones (removing skin) or increase per- meability of carbonates near the wellbore. After a brief introduction to acid–rock interaction, this chapter focuses on important issues on sandstone acidizing design and carbonate acidizing design. More in-depth information can be found from Economides and Nolte (2000). 16.2 Acid–Rock Interaction Minerals that are present in sandstone pores include mont- morillonite (bentonite), kaolinite, calcite, dolomite, sider- ite, quartz, albite (sodium feldspar), orthoclase, and others. These minerals can be either from invasion of external fluid during drilling, cementing, and well comple- tion or from host materials that exist in the naturally occurring rock formations. The most commonly used acids for dissolving these minerals are hydrochloric acid (HCl) and hydrofluoric acid (HF). 16.2.1 Primary Chemical Reactions Silicate minerals such as clays and feldspars in sandstone pores are normally removed using mixtures of HF and HCl, whereas carbonate minerals are usually attacked with HCl. The chemical reactions are summarized in Table 16.1. The amount of acid required to dissolve a given amount of mineral is determined by the stoichiom- etry of the chemical reaction. For example, the simple reaction between HCl and CaCO3 requires that 2 mol of HCl is needed to dissolve 1 mol of CaCO3. 16.2.2 Dissolving Power of Acids A more convenient way to express reaction stoichiometry is the dissolving power. The dissolving power on a mass basis is called gravimetric dissolving power and is defined as b ¼ Ca nmMWm naMWa , (16:1) where b ¼ gravimetric dissolving power of acid solution, lbmmineral=lbm solution Ca ¼ weight fraction of acid in the acid solution nm ¼ stoichiometry number of mineral na ¼ stoichiometry number of acid MWm= molecular weight of mineral MWa ¼ molecular weight of acid. For the reaction between 15 wt% HCl solution and CaCO3, Ca ¼ 0:15, nm ¼ 1, na ¼ 2, MWm ¼ 100:1, and MWa ¼ 36:5. Thus, b15 ¼ (0:15) (1)(100:1) (2)(36:5) ¼ 0:21 lbm CaCO3=lbm 15 wt% HCl solution: The dissolving power on a volume basis is called volumet- ric dissolving power and is related to the gravimetric dis- solving power through material densities: X ¼ b ra rm , (16:2) where X ¼ volumetric dissolving power of acid solution, ft3 mineral=ft3 solution ra ¼ density of acid, lbm=ft3 rm ¼ density of mineral, lbm=ft3 16.2.3 Reaction Kinetics The acid–mineral reaction takes place slowly in the rock matrix being acidized. The reaction rate can be evaluated experimentally and described by kinetics models. Research work in this area has been presented by many investigators including Fogler et al. (1976), Lund et al. (1973, 1975), Hill et al. (1981), Kline and Fogler (1981), and Schechter (1992). Generally, the reaction rate is affected by the characteristics of mineral, properties of acid, reservoir temperature, and rates of acid transport to the mineral surface and removal of product from the surface. Detailed discussion of reaction kinetics is beyond the scope of this book. 16.3 Sandstone Acidizing Design The purpose of sandstone acidizing is to remove the dam- age to the sandstone near the wellbore that occurred dur- ing drilling and well completion processes. The acid treatment is only necessary when it is sure that formation damage is significant to affect well productivity. A major formation damage is usually indicated by a large positive skin factor derived from pressure transit test analysis in a flow regime of early time (see Chapter 15). 16.3.1 Selection of Acid The acid type and acid concentration in acid solution used in acidizing is selected on the basis of minerals in the formation and field experience. For sandstones, the typical treatments usually consist of a mixture of 3 wt% HF and 12 wt% HCl, preceded by a 15 wt% HCl preflush. McLeod (1984) presented a guideline to the selection of acid on the basis of extensive field experience. His recommendations for sandstone treatments are shown in Table 16.2. McLeod’s recommendation should serve only as a starting point. When many wells are treated in a particular forma- tion, it is worthwhile to conduct laboratory tests of the responses of cores to different acid strengths. Figure 16.1 shows typical acid–response curves. Table 16.1 Primary Chemical Reactions in Acid Treatments Montmorillonite (Bentonite)-HF/HCl: Al4Si8O20(OH)4 þ 40HF þ 4Hþ $ 4AlFþ 2 þ 8SiF4 þ 24H2O Kaolinite-HF/HCl: Al4Si8O10(OH)8 þ 40HF þ 4Hþ $ 4AlFþ 2 þ 8SiF4 þ 18H2O Albite-HF/HCl: NaAlSi3O8 þ 14HF þ 2Hþ $ Naþ þ AlFþ 2 þ 3SiF4 þ 8H2O Orthoclase-HF/HCl: KAlSi3O8 þ 14HF þ 2Hþ $ Kþ þ AlFþ 2 þ 3SiF4 þ 8H2O Quartz-HF/HCl: SiO2 þ 4HF $ SiF4 þ 2H2O SiF4 þ 2HF $ H2SiF6 Calcite-HCl: CaCO3 þ 2HCl ! CaCl2 þ CO2 þ H2O Dolomite-HCl: CaMg(CO3)2 þ 4HCl ! CaCl2 þ MgCl2 þ 2CO2 þ 2H2O Siderite-HCl: FeCO3 þ 2HCl ! FeCl2 þ CO2 þ H2O Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 244 21.12.2006 2:30pm 16/244 PRODUCTION ENHANCEMENT
- 247. 16.3.2 Acid Volume Requirement The acid volume should be high enough to remove near- wellbore formation damage and low enough to reduce cost of treatment. Selection of an optimum acid volume is complicated by the competing effects. The volume of acid needed depends strongly on the depth of the damaged zone, which is seldom known. Also, the acid will never be distributed equally to all parts of the damaged forma- tion. The efficiency of acid treatment and, therefore, acid volume also depends on acid injection rate. To ensure that an adequate amount of acid contacts most of the damaged formation, a larger amount of acid is necessary. The acid preflush volume is usually determined on the basis of void volume calculations. The required minimum acid volume is expressed as Va ¼ Vm X þ VP þ Vm, (16:3) where Va ¼ the required minimum acid volume, ft3 Vm ¼ volume of minerals to be removed, ft3 VP ¼ initial pore volume, ft3 and Vm ¼ p r2 a À r2 w À Á 1 À fð ÞCm, (16:4) VP ¼ p r2 a À r2 w À Á f, (16:5) where ra ¼ radius of acid treatment, ft rw ¼ radius of wellbore, ft f ¼ porosity, fraction Cm ¼ mineral content, volume fraction. Example Problem 16.1 A sandstone with a porosity of 0.2 containing 10 v% calcite (CaCO3) is to be acidized with HF/HCl mixture solution. A preflush of 15 wt% HCl solution is to be injected ahead of the mixture to dissolve the carbonate minerals and establish a low pH environment. If the HCl preflush is to remove all carbonates in a region within 1 ft beyond a 0.328-ft radius wellbore before the HF/HCl stage enters the formation, what minimum preflush volume is required in terms of gallon per foot of pay zone? Table 16.2 Recommended Acid Type and Strength for Sandstone Acidizing HCl Solubility 20% Use HCl Only High-perm sand ( k 100 md) High quartz (80%), low clay ( 5%) 10% HCl-3% HFa High feldspar ( 20%) 13.5% HCl-1.5% HFa High clay ( 10%) 6.5% HCl-1% HFb High iron chlorite clay 3% HCl-0.5% HFb Low-perm sand ( k 10 md) Low clay ( 5%) 6% HCl-1.5% HFc High chlorite 3% HCl-0.5% HFd a Preflush with 15% HCl. b Preflush with sequestered 5% HCl. c Preflush with 7.5% HCl or 10% acetic acid. d Preflush with 5% acetic acid. 300 250 200 150 100 PercentofOriginalPermeability 50 Pore Volumes of Acid 0.1 1 10 8 wt% HF 4 wt% HF 2 wt% HF 100 1,000 Berea sandstone 80ЊF-100 psi Figure 16.1 Typical acid response curves (Smith and Hendrickson, 1965). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 245 21.12.2006 2:30pm MATRIX ACIDIZING 16/245
- 248. Solution Volume of CaCO3 to be removed: Vm ¼ p r2 a À r2 w À Á 1 À fð ÞCm ¼ p 1:3282 À 0:3282 À Á 1 À 0:2ð Þ(0:1) ¼ 0:42 ft3 CaCO3=ft pay zone Initial pore volume: VP ¼ p r2 a À r2 w À Á f ¼ p 1:3282 À 0:3282 À Á (0:2) ¼ 1:05 ft3 =ft pay zone Gravimetric dissolving power of the 15 wt% HCl solution: b ¼ Ca ymMWm yaMWa ¼ (0:15) (1)(100:1) (2)(36:5) ¼ 0:21 lbm CaCO3=lbm 15 wt% HCl solution Volumetric dissolving power of the 15 wt% HCl solution: X ¼ b ra rm ¼ (0:21) (1:07)(62:4) (169) ¼ 0:082 ft3 CaCO3=ft3 15 wt% HCl solution The required minimum HCl volume Va ¼ Vm X þ VP þ Vm ¼ 0:42 0:082 þ 1:05 þ 0:42 ¼ 6:48 ft3 15 wt% HCl solution=ft pay zone ¼ (6:48)(7:48) ¼ 48 gal 15 wt% HCl solution=ft pay zone The acid volume requirement for the main stage in a mud acid treatment depends on mineralogy and acid type and strength. Economides and Nolte (2000) provide a listing of typical stage sequences and volumes for sandstone acidizing treatments. For HCl acid, the volume requirement increases from 50 to 200 gal/ft pay zone with HCl solubility of HF changing from less than 5% to 20%. For HF acid, the volume requirement is in the range of 75–100 gal/ft pay zone with 3.0–13.5% HCl and 0.5–3.0% HF depending on mineralogy. Numerous efforts have been made to develop a rigorous method for calculating the minimum required acid volume in the past 2 decades. The most commonly used method is the two-mineral model (Hekim et al., 1982; Hill et al., 1981; Taha et al., 1989). This model requires a numerical tech- nique to obtain a general solution. Schechter (1992) pre- sented an approximate solution that is valid for Damkohler number being greater than 10. This solution approximates the HF fast-reacting mineral front as a sharp front. Readers are referred to Schechter (1992) for more information. Because mud acid treatments do not dissolve much of the formation minerals but dissolve the materials clogging the pore throats, Economides and Nolte (2000) suggest taking the initial pour volume (Eq. [16.5]) within the radius of treat- mentastheminimumrequiredacidvolumeforthemainstage of acidizing treatment. Additional acid volume should be considered for the losses in the injection tubing string. 16.3.3 Acid Injection Rate Acid injection rate should be selected on the basis of mineral dissolution and removal and depth of damaged zone. Selecting an optimum injection rate is a difficult process because the damaged zone is seldom known with any accuracy and the competing effects of mineral disso- lution and reaction product precipitation. Fortunately, research results have shown that acidizing efficiency is relatively insensitive to acid injection rate and that the highest rate possible yields the best results. McLeod (1984) recommends relatively low injection rates based on the observation that acid contact time with the forma- tion of 2–4 hours appears to give good results. da Motta (1993) shows that with shallow damage, acid injection rate has little effect on the residual skin after 100 gal/ft of injection rate; and with deeper damage, the higher the injection rate, the lower the residual skin. Paccaloni et al. (1988) and Paccaloni and Tambini (1990) also report high success rates in numerous field treatments using the high- est injection rates possible. There is always an upper limit on the acid injection rate that is imposed by formation breakdown (fracture) pres- sure pbd . Assuming pseudo–steady-state flow, the max- imum injection rate limited by the breakdown pressure is expressed as qi,max ¼ 4:917 Â 10À6 kh pbd À p À Dpsf À Á ma ln 0:472re rw þ S , (16:6) where qi ¼ maximum injection rate, bbl/min k ¼ permeability of undamaged formation, md h ¼ thickness of pay zone to be treated, ft pbd ¼ formation breakdown pressure, psia p ¼ reservoir pressure, psia Dpsf ¼ safety margin, 200 to 500 psi ma ¼ viscosity of acid solution, cp re ¼ drainage radius, ft rw ¼ wellbore radius, ft S ¼ skin factor, ft. The acid injection rate can also be limited by surface injection pressure at the pump available to the treatment. This effect is described in the next section. 16.3.4 Acid Injection Pressure In most acid treatment operations, only the surface tubing pressure is monitored. It is necessary to predict the surface injection pressure at the design stage for pump selection. The surface tubing pressure is related to the bottom-hole flowing pressure by psi ¼ pwf À Dph þ Dpf , (16:7) where psi ¼ surface injection pressure, psia pwf ¼ flowing bottom-hole pressure, psia Dph ¼ hydrostatic pressure drop, psia Dpf ¼ frictional pressure drop, psia. The second and the third term in the right-hand side of Eq. (16.7) can be calculated using Eq. (11.93). However, to avert the procedure of friction factor determination, the following approximation may be used for the frictional pressure drop calculation (Economides and Nolte, 2000): Dpf ¼ 518r0:79 q1:79 m0:207 1,000D4:79 L, (16:8) where r ¼ density of fluid, g=cm3 q ¼ injection rate, bbl/min m ¼ fluid viscosity, cp D ¼ tubing diameter, in. L ¼ tubing length, ft. Equation (16.8) is relatively accurate for estimating fric- tional pressures for newtonian fluids at flow rates less than 9 bbl/min. 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- 249. Example Problem 16.2 A 60-ft thick, 50-md sandstone pay zone at a depth of 9,500 ft is to be acidized with an acid solution having a specific gravity of 1.07 and a viscosity of 1.5 cp down a 2-in. inside diameter (ID) coil tubing. The formation fracture gradient is 0.7 psi/ft. The wellbore radius is 0.328 ft. Assuming a reservoir pressure of 4,000 psia, drainage area radius of 1,000 ft, and a skin factor of 15, calculate (a) the maximum acid injection rate using safety margin 300 psi. (b) the maximum expected surface injection pressure at the maximum injection rate. Solution (a) The maximum acid injection rate: qi,max ¼ 4:917 Â 10À6 kh pbd À p À Dpsf À Á ma ln 0:472re rw þ S ¼ 4:917 Â 10À6 (50)(60) (0:7)(9,500) À 4,000 À 300ð Þ (1:5) ln 0:472(1,000) (0:328) þ 15 ¼ 1:04 bbl=min (b) The maximum expected surface injection pressure: pwf ¼ pbd À Dpsf ¼ (0:7)(9,500) À 300 ¼ 6,350 psia Dph ¼ (0:433)(1:07)(9,500) ¼ 4,401 psi Dpf ¼ 518r0:79 q1:79 m0:207 1,000D4:79 L ¼ 518(1:07)0:79 (1:04)1:79 (1:5)0:207 1,000(2)4:79 (9,500) ¼ 218 psi psi ¼ pwf À Dph þ Dpf ¼ 6,350 À 4,401 þ 218 ¼ 2,167 psia 16.4 Carbonate Acidizing Design The purpose of carbonate acidizing is not to remove the damage to the formation near the wellbore, but to create wormholes through which oil or gas will flow after stimulation. Figure 16.2 shows wormholes created by acid dissolution of limestone in a laboratory (Hoefner and Fogler, 1988). Carbonate acidizing is a more difficult process to pre- dict than sandstone acidizing because the physics is much more complex. Because the surface reaction rates are very high and mass transfer often plays the role of limiting step locally, highly nonuniform dissolution patterns are usually created. The structure of the wormholes depends on many factors including flow geometry, injection rate, reaction kinetics, and mass transfer rates. Acidizing de- sign relies on mathematical models calibrated by labora- tory data. 16.4.1 Selection of Acid HCl is the most widely used acid for carbonate matrix acidizing. Weak acids are suggested for perforating fluid and perforation cleanup, and strong acids are recom- mended for other treatments. Table 16.3 lists recom- mended acid type and strength for carbonate acidizing (McLeod, 1984). All theoretical models of wormhole propagation predict deeper penetration for higher acid strengths, so a high concentration of acid is always preferable. 16.4.2 Acidizing Parameters Acidizing parameters include acid volume, injection rate, and injection pressure. The acid volume can be calculated with two methods: (1) Daccord’s wormhole propagation model and (2) the volumetric model, on the basis of desired penetration of wormholes. The former is optimistic, whereas the latter is more realistic (Economides et al., 1994). Based on the wormhole propagation model presented by Daccord et al. (1989), the required acid volume per unit thickness of formation can be estimated using the follow- ing equation: Vh ¼ pfD2=3 q 1=3 h r df wh bNAc (16:9) where Vh ¼ required acid volume per unit thickness of formation, m3 =m f ¼ porosity, fraction D ¼ molecular diffusion coefficient, m2 =s qh ¼ injection rate per unit thickness of formation, m3 =sec-m rwh ¼ desired radius of wormhole penetration, m df ¼ 1:6, fractal dimension b ¼ 105 Â 10À5 in SI units NAc ¼ acid capillary number, dimensionless, where the acid capillary number is defined as NAc ¼ fbga (1 À f)gm , (16:10) Figure 16.2 Wormholes created by acid dissolution of limestone (Hoefner and Fogler, 1988; courtesy AIChE). Table 16.3 Recommended Acid Type and Strength for Carbonate Acidizing Perforating fluid: 5% acetic acid Damaged perforations: 9% formic acid 10% acetic acid 15% HCl Deep wellbore damage: 15% HCl 28% HCl Emulsified HCl Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 247 21.12.2006 2:30pm MATRIX ACIDIZING 16/247
- 250. where ga ¼ acid specific gravity, water ¼ 1:0 gm ¼ mineral specific gravity, water ¼ 1:0. Based on the volumetric model, the required acid volume per unit thickness of formation can be estimated using the following equation: Vh ¼ pf r2 wh À r2 w À Á PVð Þbt, (16:11) where (PV)bt is the number of pore volumes of acid injected at the time of wormhole breakthrough at the end of the core. Apparently, the volumetric model requires data from laboratory tests. Example Problem 16.3 A 28 wt% HCl is needed to propagate wormholes 3 ft from a 0.328-ft radius wellbore in a limestone formation (specific gravity 2.71) with a porosity of 0.15. The designed injection rate is 0.1 bbl/ min-ft, the diffusion coefficient is 10À9 m2 =sec, and the density of the 28% HCl is 1:14 g=cm3 . In linear core floods, 1.5 pore volume is needed for wormhole breakthrough at the end of the core. Calculate the acid volume requirement using (a) Daccord’s model and (b) the volumetric model. Solution (a) Daccord’s model: b ¼ Ca ymMWm yaMWa ¼ (0:28) (1)(100:1) (2)(36:5) ¼ 0:3836 lbm CaCO3=lbm 28 wt% HCl solution: NAc ¼ fbga (1 À f)gm ¼ (0:15)(0:3836)(1:14) (1 À 0:15)(2:71) ¼ 0:0285 qh ¼ 0:1 bbl= min -ft ¼ 8:69 Â 10À4 m3 =sec-m rwh ¼ 0:328 þ 3 ¼ 3:328 ft ¼ 1:01 m Vh ¼ pfD2=3 q 1=3 h r df wh bNAc ¼ p(0:15)(10À9 )2=3 (8:69 Â 10À4 )1=3 (1:01)1:6 (1:5 Â 10À5)(0:0285) ¼ 0:107 m3 =m ¼ 8:6 gal=ft (b) Volumetric model: Vh ¼ pf r2 wh À r2 w À Á PVð Þbt ¼ p(0:15)(3:3282 À 0:3282 )(1:5) ¼ 7:75 ft3 =ft ¼ 58 gal=ft: This example shows that the Daccord model gives opti- mistic results and the volumetric model gives more realistic results. The maximum injection rate and pressure for carbon- ate acidizing can be calculated the same way as that for sandstone acidizing. Models of wormhole propagation predict that wormhole velocity increases with injection rate to the power of 1 ⁄2 to 1. Therefore, the maximum injection rate is preferable. However, this approach may require more acid volume. If the acid volume is con- strained, a slower injection rate may be preferable. If a sufficient acid volume is available, the maximum injection rate is recommended for limestone formations. However, a lower injection rate may be preferable for dolomites. This allows the temperature of the acid entering the for- mation to increase, and thus, the reaction rate increases. The designed acid volume and injection rate should be adjusted based on the real-time monitoring of pressure during the treatment. Summary This chapter briefly presents chemistry of matrix acidizing and a guideline to acidizing design for both sandstone and carbonate formations. More in-depth materials can be found in McLeod (1984), Economides et al. (1994), and Economides and Nolte (2000). References daccord, g., touboul, e., and lenormand, r. Carbonate acidizing: toward a quantitative model of the worm- holing phenomenon. SPEPE Feb. 1989:63–68. da motta, e.p. Matrix Acidizing of Horizontal Wells, Ph.d. Dissertation. Austin: University of Texas at Austin, 1993. economides, m.j., hill, a.d., and ehlig-economides, c. Petroleum Production Systems. Englewood Cliffs, NJ: Prentice Hall, 1994. economides, m.j. and nolte, k.g. Reservoir Stimulation, 3rd edition. New York: John Wiley Sons, 2000. fogler, h.s., lund, k., and mccune, c.c. Predicting the flow and reaction of HCl/HF mixtures in porous sandstone cores. SPEJ Oct. 1976, Trans. AIME, 1976;234:248–260. hekim, y., fogler, h.s., and mccune, c.c. The radial movement of permeability fronts and multiple reaction zones in porous media. SPEJ Feb. 1982:99–107. hill, a.d. and galloway, p.j. Laboratory and theoretical modeling of diverting agent behavior. JPT June 1984:1157–1163. hill, a.d., lindsay, d.m., silberberg, i.h., and schechter, r.s. Theoretical and experimental studies of sandstone acidizing. SPEJ Feb. 1981;21:30–42. hoefner, m.l. and fogler, h.s. Pore evolution and channel formation during flow and reaction in porous media. AIChE J. Jan. 1988;34:45–54. lund, k., fogler, h.s., and mccune, c.c. Acidization I: the dissolution of dolomite in hydrochloric acid. Chem. Eng. Sci. 1973;28:691. lund, k., fogler, h.s., mccune, c.c., and ault, j.w. Acidization II: the dissolution of calcite in hydro- chloric acid. Chem. Eng. Sci. 1975;30:825. mcleod, h.o., jr. Matrix acidizing. JPT 1984;36:2055– 2069. paccaloni, g. and tambini, m. Advances in matrix stimu- lation technology. JPT 1993;45:256–263. paccaloni, g., tambini, m., and galoppini, m. Key factors for enhanced results of matrix stimulation treatment. Presented at the SPE Formation Damage Control Symposium held in Bakersfield, California on Febru- ary 8–9, 1988. SPE Paper 17154. schechter, r.s. Oil Well Stimulation. Englewood Cliffs, NJ: Prentice Hall, 1992. smith, c.f., and hendrickson, a.r. Hydrofluoric acid stimulation of sandstone reservoirs. JPT Feb. 1965, Trans. AIME 1965;234:215–222. taha, r., hill, a.d., and sepehrnoori, k. Sandstone acid- izing design with a generalized model. SPEPE Feb. 1989:49–55. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 248 21.12.2006 2:30pm 16/248 PRODUCTION ENHANCEMENT
- 251. Problems 16.1 For the reaction between 20 wt% HCl solution and calcite, calculate the gravimetric and volumetric dis- solving power of the acid solution. 16.2 For the reaction between 20 wt% HCl solution and dolomite, calculate the gravimetric and volumetric dissolving power of the acid solution. 16.3 A sandstone with a porosity of 0.18 containing 8 v% calcite is to be acidized with HF/HCl mixture solu- tion. A preflush of 15 wt% HCl solution is to be injected ahead of the mixture to dissolve the carbon- ate minerals and establish a low-pH environment. If the HCl preflush is to remove all carbonates in a region within 1.5 ft beyond a 0.328-ft-radius wellbore before the HF/HCl stage enters the formation, what minimum preflush volume is required in terms of gallon per foot of pay zone? 16.4 A sandstone with a porosity of 0.15 containing 12 v% dolomite is to be acidized with HF/HCl mixture so- lution. A preflush of 15 wt% HCl solution is to be injected ahead of the mixture to dissolve the carbon- ate minerals and establish a low-pH environment. If the HCl preflush is to remove all carbonates in a region within 1.2 feet beyond a 0.328-ft-radius well- bore before the HF/HCl stage enters the formation, what minimum preflush volume is required in terms of gallon per foot of pay zone? 16.5 A 30-ft thick, 40-md sandstone pay zone at a depth of 9,000 ft is to be acidized with an acid solution having a specific gravity of 1.07 and a viscosity of 1.2 cp down a 2-in. ID coil tubing. The formation fracture gradient is 0.7 psi/ft. The wellbore radius is 0.328 ft. Assuming a reservoir pressure of 4,000 psia, drainage area radius of 1,500 ft and skin factor of 10, calculate (a) the maximum acid injection rate using safety margin 200 psi. (b) the maximum expected surface injection pressure at the maximum injection rate. 16.6 A 40-ft thick, 20-md sandstone pay zone at a depth of 8,000 ft is to be acidized with an acid solution having a specific gravity of 1.07 and a viscosity of 1.5 cp down a 2-in. ID coil tubing. The formation fracture gradient is 0.65 psi/ft. The wellbore radius is 0.328 ft. Assuming a reservoir pressure of 3,500 psia, drainage area radius of 1,200 ft, and skin factor of 15, calculate (a) the maximum acid injection rate using a safety margin of 400 psi. (b) the maximum expected surface injection pressure at the maximum injection rate. 16.7 A 20 wt% HCl is needed to propagate wormholes 2 ft from a 0.328-ft radius wellbore in a limestone formation (specific gravity 2.71) with a porosity of 0.12. The designed injection rate is 0.12 bbl/min-ft, the diffusion coefficient is 10À9 m2 =sec, and the den- sity of the 20% HCl is 1:11 g=cm3 . In linear core floods, 1.2 pore volume is needed for wormhole breakthrough at the end of the core. Calculate the acid volume requirement using (a) Daccord’s model and (b) the volumetric model. 16.8 A 25 wt% HCl is needed to propagate wormholes 3 ft from a 0.328-ft radius wellbore in a dolomite formation (specific gravity 2.87) with a porosity of 0.16. The designed injection rate is 0.15 bbl/min-ft, the diffusion coefficient is 10À9 m2 =sec, and the den- sity of the 25% HCl is 1:15 g=cm3 . In linear core floods, 4 pore volumes is needed for wormhole breakthrough at the end of the core. Calculate the acid volume requirement using (a) Daccord’s model and (b) the volumetric model. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap16 Final Proof page 249 21.12.2006 2:30pm MATRIX ACIDIZING 16/249
- 252. 17 Hydraulic Fracturing Contents 17.1 Introduction 17/252 17.2 Formation Fracturing Pressure 17/252 17.3 Fracture Geometry 17/254 17.4 Productivity of Fractured Wells 17/256 17.5 Hydraulic Fracturing Design 17/258 17.6 Post-Frac Evaluation 17/262 Summary 17/264 References 17/264 Problems 17/265 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 251 3.1.2007 9:19pm Compositor Name: SJoearun
- 253. 17.1 Introduction Hydraulic fracturing is a well-stimulation technique that is most suitable to wells in low- and moderate-permeability reservoirs that do not provide commercial production rates even though formation damages are removed by acidizing treatments. Hydraulic fracturing jobs are carried out at well sites using heavy equipment including truck-mounted pumps, blenders, fluid tanks, and proppant tanks. Figure 17.1 illustrates a simplified equipment layout in hydraulic frac- turing treatments of oil and gas wells. A hydraulic fractur- ing job is divided into two stages: the pad stage and the slurry stage (Fig. 17.2). In the pad stage, fracturing fluid only is injected into the well to break down the formation and create a pad. The pad is created because the fracturing fluid injection rate is higher than the flow rate at which the fluid can escape into the formation. After the pad grows to a desirable size, the slurry stage is started. During the slurry stage, the fracturing fluid is mixed with sand/proppant in a blender and the mixture is injected into the pad/fracture. After filling the fracture with sand/proppant, the fracturing job is over and the pump is shut down. Apparently, to reduce the injection rate requirement, a low leaf-off frac- turing fluid is essential. Also, to prop the fracture, the sand/ proppant should have a compressive strength that is high enough to resist the stress from the formation. This chapter concisely describes hydraulic fracturing treatments. For detailed information on this subject, see Economides and Nolte (2000). This chapter focuses on the following topics: . Formation fracturing pressure . Fracture geometry . Productivity of fractured wells . Hydraulic fracturing design . Post-frac evaluation 17.2 Formation Fracturing Pressure Formation fracturing pressure is also called breakdown pressure. It is one of the key parameters used in hydraulic fracturing design. The magnitude of the parameter de- pends on formation depth and properties. Estimation of the parameter value begins with in situ stress analysis. Fracturing fluid Proppant Blender Pumper Figure 17.1 Schematic to show the equipment layout in hydraulic fracturing treatments of oil and gas wells. Open Open Open Open Fill fracture with sand/proppant Inject frac fluid/proppant mixture OpenClosed Closed ClosedClosed Create fracture Inject fluid only Pad stage Slurry stage Figure 17.2 A schematic to show the procedure of hydraulic fracturing treatments of oil and gas wells. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 252 3.1.2007 9:19pm Compositor Name: SJoearun 17/252 PRODUCTION ENHANCEMENT
- 254. Consider a reservoir rock at depth H as shown in Fig. 17.3. The in situ stress caused by the weight of the overburden formation in the vertical direction is expressed as sv ¼ rH 144 , (17:1) where sv ¼ overburden stress, psi r ¼ the average density of overburden formation, lb=ft3 H ¼ depth, ft. The overburden stress is carried by both the rock grains and the fluid within the pore space between the grains. The contact stress between grains is called effective stress (Fig. 17.4): s0 v ¼ sv À app, (17:2) where s0 v ¼ effective vertical stress, psi a ¼ Biot’s poro-elastic constant, approximately 0.7 pp ¼ pore pressure, psi. The effective horizontal stress is expressed as s0 h ¼ n 1 À n s0 v, (17:3) where n is Poison’s ratio. The total horizontal stress is expressed as sh ¼ s0 h þ app: (17:4) Because of the tectonic effect, the magnitude of the hori- zontal stress may vary with direction. The maximum hori- zontal stress may be sh,max ¼ sh,min þ stect, where stect is called tectonic stress. Based on a failure criterion, Terzaghi presented the following expression for the breakdown pressure: pbd ¼ 3sh,min À sh,max þ T0 À pp, (17:5) where T0 is the tensile strength of the rock. Example Problem 17.1 A sandstone at a depth of 10,000 ft has a Poison’s ratio of 0.25 and a poro-elastic constant of 0.72. The average density of the overburden formation is 165 lb=ft3 . The pore pressure gradient in the sandstone is 0.38 psi/ft. Assuming a tectonic stress of 2,000 psi and a tensile strength of the sandstone of 1,000 psi, predict the breakdown pressure for the sandstone. Gas Oil Water ptf H pwt p pe Figure 17.3 Overburden formation of a hydrocarbon reservoir. Figure 17.4 Concept of effective stress between grains. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 253 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/253
- 255. Solution Overburden stress: sv ¼ rH 144 ¼ (165)(10,000) 144 ¼ 11,500 psi Pore pressure: pp ¼ (0:38)(10,000) ¼ 3,800 psi The effective vertical stress: s0 v ¼ sv À app ¼ 11,500 À (0:72)(3,800) ¼ 8,800 psi The effective horizontal stress: s0 h ¼ n 1 À n s0 v ¼ 0:25 1 À 0:25 8,800ð Þ ¼ 2,900 psi The minimum horizontal stress: sh,min ¼ s0 h þ app ¼ 2,900 þ (0:72)(3,800) ¼ 5,700 psi The maximum horizontal stress: sh,max ¼ sh,min þ stect ¼ 5,700 þ 2,000 ¼ 7,700 psi Breakdown pressure: pbd ¼ 3sh,min À sh,max þ T0 À pp ¼ 3(5,700) À 7,700 þ 1,000 À 3,800 ¼ 6,600 psi 17.3 Fracture Geometry It is still controversial about whether a single fracture or multiple fractures are created in a hydraulic fracturing job. Whereas both cases have been evidenced based on the information collected from tiltmeters and microseismic data, it is commonly accepted that each individual fracture is sheet-like. However, the shape of the fracture varies as predicted by different models. 17.3.1 Radial Fracture Model A simple radial (penny-shaped) crack/fracture was first presented by Sneddon and Elliot (1946). This occurs when there are no barriers constraining height growth or when a horizontal fracture is created. Geertsma and de Klerk (1969) presented a radial fracture model showing that the fracture width at wellbore is given by ww ¼ 2:56 mqi 1 À nð ÞR E !1 4 , (17:6) where ww ¼ fracture width at wellbore, in. m ¼ fluid viscosity, cp qi ¼ pumping rate, bpm R ¼ the radius of the fracture, ft E ¼ Young’s modulus, psi. Assuming the fracture width drops linearly in the radial direction, the average fracture width may be expressed as w ¼ 0:85 mqi 1 À nð ÞR E !1 4 : (17:7) 17.3.2 The KGD Model Assuming that a fixed-height vertical fracture is propagated in a well-confined pay zone (i.e., the stresses in the layers above and below the pay zone are large enough to prevent fracture growth out of the pay zone), Khristianovich and Zheltov (1955) presented a fracture model as shown in Fig. 17.5. The model assumes that the width of the crack at any distance from the well is independent of vertical position, which is a reasonable approximation for a frac- ture with height much greater than its length. Their solution included the fracture mechanics aspects of the fracture tip. They assumed that the flow rate in the fracture was con- stant, and that the pressure in the fracture could be approxi- mated by a constant pressure in the majority of the fracture body, except for a small region near the tip with no fluid penetration, and hence, no fluid pressure. This concept of fluid lag has remained an element of the mechanics of the fracture tip. Geertsma and de Klerk (1969) gave a much simpler solution to the same problem. The solution is now referred to as the KGD model. The average width of the KGD fracture is expressed as w ¼ 0:29 qim 1 À nð Þx2 f Ghf #1=4 p 4 , (17:8) where w ¼ average width, in. qi ¼ pumping rate, bpm Area of highest flow resistance w(x,t) w(o,t) Approximate elliptical shape of fracture xf x ux rw hf Figure 17.5 The KGD fracture geometry. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 254 3.1.2007 9:19pm Compositor Name: SJoearun 17/254 PRODUCTION ENHANCEMENT
- 256. m ¼ fluid viscosity, cp G ¼ E=2(1 þ n), shear modulus, psia hf ¼ fracture height, ft 17.3.3 The PKN model Perkins and Kern (1961) also derived a solution for a fixed- height vertical fracture as illustrated in Fig. 17.6. Nordgren (1972) added leakoff and storage within the fracture (due to increasing width) to the Perkins and Kern model, deriving what is now known as the PKN model. The average width of the PKN fracture is expressed as w ¼ 0:3 qim 1 À nð Þxf G !1=4 p 4 g , (17:9) where g % 0:75. It is important to emphasize that even for contained fractures, the PKN solution is only valid when the fracture length is at least three times the height. The three models discussed in this section all assume that the fracture is planar, that is, fracture propagates in a particular direction (perpendicular to the minimum stress), fluid flow is one-dimensional along the length (or radius) of the fracture, and leakoff behavior is governed by a simple expression derived from filtration theory. The rock in which the fracture propagates is assumed to be a continuous, homogeneous, isotropic linear elastic solid, and the fracture is considered to be of fixed height (PKN and KGD) or completely confined in a given layer (radial). The KGD and PKN models assume respectively that the fracture height is large or small relative to length, while the radial model assumes a circular shape. Since these models were developed, numerous extensions have been made, which have relaxed these assumptions. 17.3.4 Three-Dimensional and Pseudo-3D Models The planar 2D models discussed in the previous section are deviated with significant simplifying assumptions. Although their accuracies are limited, they are useful for understanding the growth of hydraulic fractures. The power of modern computer allows routine treatment designs to be made with more complex models, which are solved numerically. The biggest limitation of the simple models is the requirement to specify the fracture height or to assume that a radial fracture will develop. It is not always obvious from data such as logs where, or whether, the fracture will be contained. In addition, the fracture height will usually vary from the well to the tip of the fracture, as the pressure varies. There are two major types of pseudo–three-dimensional (P3D) models: lumped and cell based. In the lumped (or elliptical) models, the fracture shape is assumed to consist of two half-ellipses joined at the center. The horizontal length and wellbore vertical tip extensions are calculated at each time-step, and the assumed shape is made to match these positions. Fluid flow is assumed to occur along streamlines from the perforations to the edge of the ellipse, with the shape of the streamlines derived from simple analytical solutions. In cell-based models, the fracture shape is not prescribed. The fracture is treated as a series of connected cells, which are linked only via the fluid flow from cell to cell. The height at any cross-section is calcu- lated from the pressure in that cell, and fluid flow in the vertical direction is generally approximated. Lumped models were first introduced by Cleary (1980), and numerous papers have since been presented on their use (e.g., Cleary et al., 1994). As stated in the 1980 paper, ‘‘The heart of the formulae can be extracted very simply by a non-dimensionalization of the governing equations; the remainder just involves a good physics-mathematical choice of the undetermined coefficients.’’ The lumped models implicitly require the assumption of a self-similar fracture shape (i.e., one that is the same as time evolves, except for length scale). The shape is generally assumed to consist of two half-ellipses of equal lateral extent, but with different vertical extent. In cell-based P3D models, the fracture length is discre- tized into cells along the length of the fracture. Because only one direction is discretized and fluid flow is assumed to be essentially horizontal along the length of the fracture, the model can be solved much more easily than planar 3D models. Although these models allow the calculation of fracture height growth, the assumptions make them pri- marily suitable for reasonably contained fractures, with length much greater than height. Figure 17.6 The PKN fracture geometry. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 255 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/255
- 257. Planar 3D models: The geometry of a hydraulic fracture is defined by its width and the shape of its periphery (i.e., height at any distance from the well and length). The width distri- bution and the overall shape change as the treatment is pumped, and during closure. They depend on the pressure distribution, which itself is determined by the pressure gra- dients caused by the fluid flow within the fracture. The relation between pressure gradient and flow rate is very sensitive to fracture width, resulting in a tightly coupled calculation. Although the mechanics of these processes can be described separately, this close coupling complicates the solution of any fracture model. The nonlinear relation be- tween width and pressure and the complexity of a moving- boundary problem further complicate numerical solutions. Clifton and Abou-Sayed (1979) reported the first numerical implementation of a planar model. The solution starts with a small fracture, initiated at the perforations, divided into a number of equal elements (typically 16 squares). The ele- ments then distort to fit the evolving shape. The elements can develop large aspect ratios and very small angles, which are not well handled by the numerical schemes typically used to solve the model. Barree (1983) developed a model that does not show grid distortion. The layered reservoir is divided into a grid of equal-size rectangular elements, over the entire region that the fracture may cover. Simulators based on such models are much more com- putationally demanding than P3D-based simulators, be- cause they solve the fully 2D fluid-flow equations and couple this solution rigorously to the elastic-deformation equations. The elasticity equations are also solved more rigorously, using a 3D solution rather than 2D slices. Computational power and numerical methods have im- proved to the point that these models are starting to be used for routine designs. They should be used whenever a significant portion of the fracture volume is outside the zone where the fracture initiates or where there is signifi- cant vertical fluid flow. Such cases typically arise when the stress in the layers around the pay zone is similar to or lower than that within the pay. Regardless of which type of model is used to calculate the fracture geometry, limited data are available on typical treatments to validate the model used. On commercial treatments, the pressure history during the treatment is usually the only data available to validate the model. Even in these cases, the quality of the data is questionable if the bottom-hole pressure must be inferred from the surface pressure. The bottom-hole pressure is also not sufficient to uniquely determine the fracture geometry in the absence of other information, such as that derived from tiltmeters and microseismic data. If a simulator incorporates the correct model, it should match both treating pressure and fracture geometry. Table 17.1 summarizes main features of fracture models in different categories. Commercial packages are listed in Table 17.2. 17.4 Productivity of Fractured Wells Hydraulically created fractures gather fluids from reser- voir matrix and provide channels for the fluid to flow into wellbores. Apparently, the productivity of fractured wells depends on two steps: (1) receiving fluids from formation and (2) transporting the received fluid to the wellbore. Usually one of the steps is a limiting step that controls the well-production rate. The efficiency of the first step depends on fracture dimension (length and height), and the efficiency of the second step depends on fracture per- meability. The relative importance of each of the steps can be analyzed using the concept of fracture conductivity defined as (Argawal et al., 1979; Cinco-Ley and Sama- niego, 1981): FCD ¼ kf w kxf , (17:10) where FCD ¼ fracture conductivity, dimensionless kf ¼ fracture permeability, md w ¼ fracture width, ft xf ¼ fracture half-length, ft. Table 17.1 Features of Fracture Geometry Models A. 2D models Constant height Plain strain/stress Homogeneous stress/elastic properties Engineering oriented: quick look Limited computing requirements B. Pseudo-3D (2D Â 2D) models Limited height growth Planar frac properties of layers/adjacent zones State of stress Specialized field application Moderate computer requirements C. Fully 3D models Three-dimensional propagation Nonideal geometry/growth regimes Research orientated Large database and computer requirements Calibration of similar smaller models in conjunction with laboratory experiments Table 17.2 Summary of Some Commercial Fracturing Models Software name Model type Company Owner PROP Classic 2D Halliburton Chevron 2D Classic 2D ChevronTexaco CONOCO 2D Classic 2D CONOCO Shell 2D Classic 2D Shell TerraFrac Planar 3D Terra Tek ARCO HYRAC 3D Planar 3D Lehigh U. S.H. Advani GOHFER Planar 3D Marathon R. Barree STIMPLAN Pseudo–3D ‘‘cell’’ NSI Technologies M. Smith ENERFRAC Pseudo–3D ‘‘cell’’ Shell TRIFRAC Pseudo–3D ‘‘cell’’ S.A. Holditch Association FracCADE Pseudo–3D ‘‘cell’’ Schlumberger EAD sugar-land PRACPRO Pseudo–3D ‘‘parametric’’ RES, Inc. GTI PRACPROPT Pseudo–3D ‘‘parametric’’ Pinnacle Technologies GTI MFRAC-III Pseudo–3D ‘‘parametric’’ Meyer Associates Bruce Meyer Fracanal Pseudo–3D ‘‘parametric’’ Simtech A. Settari Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 256 3.1.2007 9:19pm Compositor Name: SJoearun 17/256 PRODUCTION ENHANCEMENT
- 258. In the situations in which the fracture dimension is much less than the drainage area of the well, the long-term productivity of the fractured well can be estimated assum- ing pseudo-radial flow in the reservoir. Then the inflow equation can be written as q ¼ kh pe À pwf À Á 141:2Bm ln re rw þ Sf , (17:11) where Sf is the equivalent skin factor. The fold of increase can be expressed as J Jo ¼ ln re rw ln re rw þ Sf , (17:12) where J ¼ productivity of fractured well, stb/day-psi Jo ¼ productivity of nonfractured well, stb/day-psi. The effective skin factor Sf can be determined based on fracture conductivity and Fig. 17.7. It is seen from Fig. 17.7 that the parameter Sf þ ln xf =rw À Á approaches a constant value in the range of FCD 100, that is, which gives Sf % 0:7 À ln xf =rw À Á , (17:13) meaning that the equivalent skin factor of fractured wells depends only on fracture length for high-conductivity frac- tures, not fracture permeability and width. This is the situation in which the first step is the limiting step. On the other hand, Fig. 17.7 indicates that the parameter Sf þ ln xf =rw À Á declines linearly with log (FCD) in the range of FCD 1, that is, Sf % 1:52 þ 2:31 log rwð Þ À 1:545 log kf w k À 0:765 log xf À Á : (17:14) Comparing the coefficients of the last two terms in this relation indicates that the equivalent skin factor of frac- tured well is more sensitive to the fracture permeability and width than to fracture length for low-conductivity fractures. This is the situation in which the second step is the limiting step. The previous analyses reveal that low-permeability res- ervoirs, leading to high-conductivity fractures, would benefit greatly from fracture length, whereas high-perme- ability reservoirs, naturally leading to low-conductivity fractures, require good fracture permeability and width. Valko et al. (1997) converted the data in Fig. 17.7 into the following correlation: sf þ ln xf rw ¼ 1:65 À 0:328u þ 0:116u2 1 þ 0:180u þ 0:064u2 þ 0:05u3 (17:15) where u ¼ ln (FCD) (17:16) Example Problem 17.2 A gas reservoir has a permeability of 1 md. A vertical well of 0.328-ft radius draws the reservoir from the center of an area of 160 acres. If the well is hydraulically fractured to create a 2,000-ft long, 0.12-in. wide fracture of 200,000 md permeability around the center of the drainage area, what would be the fold of increase in well productivity? Solution Radius of the drainage area: re ¼ ffiffiffiffi A p r ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (43,560)(160) p r ¼ 1,490 ft Fracture conductivity: FCD ¼ kf w kxf ¼ (200,000)(0:12=12) (1)(2,000=2) ¼ 2 Figure 17.7 reads Sf þ ln xf =rw À Á % 1:2, which gives Sf % 1:2 À ln xf =rw À Á ¼ 1:2 À ln 1,000=0:328ð Þ ¼ À6:82: The fold of increase is J Jo ¼ ln re rw ln re rw þ Sf ¼ ln 1,490 0:328 ln 1,490 0:328 À 6:82 ¼ 5:27: In the situations in which the fracture dimension is com- parable to the drainage area of the well, significant error may result from using Eq. (17.12), which was derived based Figure 17.7 Relationship between fracture conductivity and equivalent skin factor (Cinco-Ley and Samaniego, 1981). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 257 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/257
- 259. on radial flow. In these cases, the long-term productivity of the well may be estimated assuming bilinear flow in the reservoir. Pressure distribution in a linear flow reservoir and a linear flow in a finite conductivity fracture is illus- trated in Fig. 17.8. An analytical solution for estimating fold of increase in well productivity was presented by Guo and Schechter (1999) as follows: J Jo ¼ 0:72 ln re rw À 3 4 þ So ze ffiffiffi c p þ Sð Þ 1 1Àe À ffiffic p xf À 1 2xf ffiffi c p , (17:17) where c ¼ 2k zewkf and ze are distance between the fracture and the boundary of the drainage area. 17.5 Hydraulic Fracturing Design Hydraulic fracturing designs are performed on the basis of parametric studies to maximize net present values (NPVs) of the fractured wells. A hydraulic fracturing design should follow the following procedure: 1. Select a fracturing fluid 2. Select a proppant 3. Determine the maximum allowable treatment pressure 4. Select a fracture propagation model 5. Select treatment size (fracture length and proppant concentration) 6. Perform production forecast analyses 7. Perform NPV analysis A complete design must include the following components to direct field operations: . Specifications of fracturing fluid and proppant . Fluid volume and proppant weight requirements . Fluid injection schedule and proppant mixing schedule . Predicted injection pressure profile 17.5.1 Selection of Fracturing Fluid Fracturing fluid plays a vital role in hydraulic fracture treat- ment because it controls the efficiencies of carrying proppant and filling in the fracture pad. Fluid loss is a major fracture design variable characterized by a fluid-loss coefficient CL and a spurt-loss coefficient Sp. Spurt loss occurs only for wall-building fluids and only until the filter cake is estab- lished. Fluid loss into the formation is a more steady process than spurt loss. It occurs after the filter cake is developed. Excessive fluid loss prevents fracture propagation because of insufficient fluid volume accumulation in the fracture. Therefore, a fracture fluid with the lowest possible value of fluid-loss (leak-off) coefficient CL should be selected. The second major variable is fluid viscosity. It affects transporting, suspending, and deposition of proppants, as well as back-flowing after treatment. The viscosity should be controlled in a range suitable for the treatment. A fluid viscosity being too high can result in excessive injection pressure during the treatment. However, other considerations may also be major for particular cases. They are compatibility with reservoir fluids and rock, compatibility with other materials (e.g., resin-coated proppant), compatibility with operating pressure and temperature, and safety and environmental concerns. 20 180 340 500 660 820 980 1,140 1,300 0 60 140 220 300 380 460 540 620 700 780 860 940 1,020 1,100 1,180 1,260 1,340 0 2,000 Pressure(psi) Distance inthedirection perpendicularto thefracture(ft.) Distance in fracture direction(ft.) Figure 17.8 Relationship between fracture conductivity and equivalent skin factor. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 258 3.1.2007 9:19pm Compositor Name: SJoearun 17/258 PRODUCTION ENHANCEMENT
- 260. 17.5.2 Selection of Proppant Proppant must be selected on the basis of in situ stress conditions. Major concerns are compressive strength and the effect of stress on proppant permeability. For a vertical fracture, the compressive strength of the proppant should be greater than the effective horizontal stress. In general, bigger proppant yields better permeability, but proppant size must be checked against proppant admittance criteria through the perforations and inside the fracture. Figure 17.9 shows permeabilities of various types of proppants under fracture closure stress. Example Problem 17.3 For the following situation, esti- mate the minimum required compressive strength of 20/ 40 proppant. If intermediate-strength proppant is used, estimate the permeability of the proppant pack: Formation depth: 10,000 ft Overburden density: 165 lbm=ft3 Poison’s ratio: 0.25 Biot constant: 0.7 Reservoir pressure: 6,500 psi Production drawdown: 2,000 and 4,000 psi Solution The initial effective horizontal stress: s0 h ¼ n 1 À n rH 144 À app ¼ 0:25 1 À 0:25 (165)(10,000) 144 À (0:7)(6500) ! ¼ 2,303 psi The effective horizontal stress under 2,000-psi pressure drawdown: s0 h ¼ n 1 À n rH 144 À app ¼ 0:25 1 À 0:25 (165)(10,000) 144 À (0:7)(4500) ! ¼ 2,770 psi The effective horizontal stress under 4,000-psi pressure drawdown: s0 h ¼ n 1 À n rH 144 À app ¼ 0:25 1 À 0:25 (165)(10,000) 144 À (0:7)(2500) ! ¼ 3,236 psi Therefore, the minimum required proppant compressive strength is 3,236 psi. Figure 17.9 indicates that the pack of the intermediate-strength proppants will have a perme- ability of about kf ¼ 500 darcies. 17.5.3 The maximum Treatment Pressure The maximum treatment pressure is expected to occur when the formation is broken down. The bottom-hole pressure is equal to the formation breakdown pressure pbd and the expected surface pressure can be calculated by psi ¼ pbd À Dph þ Dpf , (17:18) where psi ¼ surface injection pressure, psia pbd ¼ formation breakdown pressure, psia Dph ¼ hydrostatic pressure drop, psia Dpf ¼ frictional pressure drop, psia. The second and the third term in the right-hand side of Eq. (17.18) can be calculated using Eq. (11.93) (see Chapter 11). However, to avert the procedure of friction factor determination, the following approximation may be used for the frictional pressure drop calculation (Economides and Nolte, 2000): Dpf ¼ 518r0:79 q1:79 m0:207 1,000D4:79 L, (17:19) where r ¼ density of fluid, g=cm3 q ¼ injection rate, bbl/min m ¼ fluid viscosity, cp D ¼ tubing diameter, in. L ¼ tubing length, ft. Equation (17.19) is relatively accurate for estimating fric- tional pressures for newtonian fluids at low flow rates. Figure 17.9 Effect of fracture closure stress on proppant pack permeability (Economides and Nolte, 2000). Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 259 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/259
- 261. Example Problem 17.4 For Example Problem 17.1, predict the maximum expected surface injection pressure using the following additional data: Specific gravity of fracturing fluid: 1.2 Viscosity of fracturing fluid: 20 cp Tubing inner diameter: 3.0 in. Fluid injection rate: 10 bpm Solution Hydrostatic pressure drop: Dph ¼ (0:433)(1:2)(10,000) ¼ 5,196 psi Frictional pressure drop: Dpf ¼ 518r0:79 q1:79 m0:207 1,000D4:79 L ¼ 518(1:2)0:79 (10)1:79 (20)0:207 1,000(3)4:79 (10,000) ¼ 3,555 psi Expected surface pressure: psi ¼ pbd À Dph þ Dpf ¼ 6,600 À 5,196 þ 3,555 ¼ 4,959 psia 17.5.4 Selection of Fracture Model An appropriate fracture propagation model is selected for the formationcharacteristicsandpressurebehavioronthebasisof in situ stresses and laboratory tests. Generally, the model should be selected to match the level of complexity required for the specific application, quality and quantity of data, allo- cated time to perform a design, and desired level of output. Modeling with a planar 3D model can be time consuming, whereas the results from a 2D model can be simplistic. Pseudo-3D models provide a compromise and are most often used in the industry. However, 2D models are still attractive in situations in which the reservoir conditions are simple and wellunderstood.Forinstance,tosimulateashortfracturetobe createdinathicksandstone,theKGDmodelmaybebeneficial. To simulate a long fracture to be created in a sandstone tightly bondedbystrongoverlayingandunderlayingshales,thePKN modelismoreappropriate.Tosimulatefrac-packinginathick sandstone, the radial fracture model may be adequate. It is always important to consider the availability and quality of inputdatainmodelselection:garbage-ingarbage-out(GIGO). 17.5.5 Selection of Treatment Size Treatment size is primarily defined by the fracture length. Fluid and proppant volumes are controlled by fracture length, injectionrate,andleak-off properties.Ageneralstatementcan be made that the greater the propped fracture length and greater the proppant volume, the greater the production rate of the fractured well. Limiting effects are imposed by technical and economical factors such as available pumping rate and costs of fluid and proppant. Within these constraints, the optimum scale of treatment should be ideally determined based on the maximum NPV. This section demonstrates how to design treatment size using the KGD fracture model for simplicity. Calculation procedure is summarized as follows: 1. Assume a fracture half-length xf and injection rate qi, calculate the average fracture width ww using a selected fracture model. 2. Based on material balance, solve injection fluid volume Vinj from the following equation: Vinj ¼ Vfrac þ VLeakoff , (17:20) where Vinj ¼ qiti (17:21) Vfrac ¼ Af ww (17:22) VLeakoff ¼ 2KLCLAf rp ffiffiffi ti p (17:23) KL ¼ 1 2 8 3 h þ p(1 À h) ! (17:24) rp ¼ h hf (17:25) Af ¼ 2xf hf (17:26) h ¼ Vfrac Vinj (17:27) Vpad ¼ Vinj 1 À h 1 þ h (17:28) Since KL depends on fluid efficiency h, which is not known in the beginning, a numerical iteration procedure is required. The procedure is illustrated in Fig. 17.10. 3. Generate proppant concentration schedule using: cp(t) ¼ cf t À tpad tinj À tpad « , (17:29) where cf is the final concentration in ppg. The proppant concentration in pound per gallon of added fluid (ppga) is expressed as c0 p ¼ cp 1 À cp=rp (17:30) and « ¼ 1 À h 1 þ h : (17:31) 4. Predict propped fracture width using w ¼ Cp 1 À fp À Á rp , (17:32) where Cp ¼ Mp 2xf hf (17:33) Mp ¼ ccp(Vinj À Vpad ) (17:34) ccp ¼ cf 1 þ « (17:35) Example Problem 17.5 The following data are given for a hydraulic fracturing treatment design: Pay zone thickness: 70 ft Young’s modulus of rock: 3 Â 106 psi Poison’s ratio: 0.25 Fluid viscosity: 1.5 cp Leak-off coefficient: 0:002 ft= min1=2 Proppant density: 165 lb=ft3 Proppant porosity: 0.4 Fracture half-length: 1,000 ft Assume a KL value ti V inj = qiti V frac = Afw qiti = Afw + 2KLCLAfrp ti V pad = Vinj 1−h 1+h h = V frac V inj KL = 1 2 8 3 h + p(1-h) Figure 17.10 Iteration procedure for injection time calculation. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 260 3.1.2007 9:19pm Compositor Name: SJoearun 17/260 PRODUCTION ENHANCEMENT
- 262. Fracture height: 100 ft Fluid injection rate: 40 bpm Final proppant concentration: 3 ppg Assuming KGD fracture, estimate a. Fluid volume requirement b. Proppant mixing schedule c. Proppant weight requirement d. Propped fracture width Solution a. Fluid volume requirements: The average fracture width: ww ¼ 0:29 qim(1 À n)x2 f Ghf #1=4 p 4 ¼ 0:29 (40)(1:5)(1 À 0:25)(1,000)2 (3Â106) 2(1þ0:25) (70) 2 4 3 5 1=4 4 ¼ 0:195 in: Fracture area: Af ¼ 2xf hf ¼ 2(1,000)(100) ¼ 2 Â 105 ft2 Fluid volume based on volume balance: qiti ¼ Af ww þ 2KLCLAf rp ffiffiffi ti p : Assuming KL ¼ 1:5, (40)(5:615)ti ¼ (2 Â 105 ) 0:195 12 þ 2(1:5)(2 Â 10À3 ) Â (2 Â 105 ) 70 100 ffiffiffi ti p gives ti ¼ 37 min. Check KL value: Vinj ¼ qit ¼ (40)(42)(37) ¼ 6:26 Â 104 gal Vfrac ¼ Af ww ¼ (2 Â 105 ) 0:195 12 (7:48) ¼ 2:43 Â 104 gal h ¼ Vfrac Vinj ¼ 2:43 Â 104 6:26 Â 104 ¼ 0:3875 KL ¼ 1 2 3 8 h þ p(1 À h) ! ¼ 1 2 3 8 (0:3875) þ p(1 À 0:3875) ! ¼ 1:48 OK Pad volume: « ¼ 1 À h 1 þ h ¼ 1 À 0:3875 1 þ 0:3875 ¼ 0:44 Vpad ¼ Vinj« ¼ (6:26 Â 104 ) 0:44ð Þ ¼ 2:76 Â 104 gal It will take 17 min to pump the pad volume at an injection rate of 40 bpm. b. Proppant mixing schedule: cp(t) ¼ (3) t À 17 37 À 17 0:44 gives proppant concentration schedule shown in Table 17.3. Slurry concentration schedule is plotted in Fig. 17.11. c. Proppant weight requirement: ccp ¼ cf 1 þ « ¼ 3 1 þ 0:44 ¼ 2:08 ppg Mp ¼ ccp(Vinj À Vpad ) ¼ (2:08)(6:26 Â 104 À 2:76 Â 104 ) ¼ 72,910 lb d. Propped fracture width: Cp ¼ Mp 2xf hf ¼ 72,910 2(1,000)(100) ¼ 0:3645 lb=ft3 w ¼ Cp (1 À fp)rp ¼ 0:3645 (1 À 0:4)(165) ¼ 0:00368 ft ¼ 0:04 in: 17.5.6 Production forecast and NPV Analyses The hydraulic fracturing design is finalized on the basis of production forecast and NPV analyses. The information Table 17.3 Calculated Slurry Concentration t (min) cp (ppg) 0 0 17 0.00 20 1.30 23 1.77 26 2.11 29 2.40 32 2.64 35 2.86 37 3.00 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Injection Time (min) SlurryConcentration(ppg) 0 10 15 20 25 30 35 405 Figure 17.11 Calculated slurry concentration. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 261 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/261
- 263. of the selected fracture half-length xf and the calculated fracture width w, together with formation permeability (k) and fracture permeability (kf ), can be used to predict the dimensionless fracture conductivity FCD with Eq. (17.10). The equivalent skin factor Sf can be estimated based on Fig. 17.7. Then the productivity index of the fractured well can be calculated using Eq. (17.11). Produc- tion forecast can be performed using the method presented in Chapter 7. Comparison of the production forecast for the fractured well and the predicted production decline for the unstimu- lated well allows for calculations of the annual incremental cumulative production for year n for an oil well: DNp,n ¼ Nf p,n À Nnf p,n, (17:36) where DNp,n ¼ predicted annual incremental cumulative production for year n Nf p,n ¼ forecasted annual cumulative production of fractured well for year n Nnf p,n ¼ predicted annual cumulative production of nonfractured well for year n. If Eq. (17.36) is used for a gas well, the notations DNp,n, Nf p,n, and Nnf p,n should be replaced by DNp,n, Nf p,n, and Nnf p,n, respectively. The annual incremental revenue above the one that the unstimulated well would deliver is expressed as DRn ¼ $ð ÞDNp,n, (17:37) where ($) is oil price. The present value of the future revenue is then NPVR ¼ Xm n¼1 DRn 1 þ ið Þn , (17:38) where m is the remaining life of the well in years and i is the discount rate. The NPV of the hydraulic fracture project is NPV ¼ NPVR À cost: (17:39) The cost should include the expenses for fracturing fluid, proppant, pumping, and the fixed cost for the treatment job. To predict the pumping cost, the required hydraulic horsepower needs to be calculated by HHP ¼ qipsi 40:8 : (17:40) 17.6 Post-Frac Evaluation Post-frac evaluation can be performed by pressure match- ing, pressure transient data analysis, and other techniques including pumping radioactive materials stages and run- ning tracer logs, running production logging tools, and conducting back-pressure and performing Nodal analysis. 17.6.1 Pressure Matching Pressure matching with a computer software is the first step to evaluate the fracturing job. It is understood that the more refined the design model is, the more optional parameters we have available for pressure matching and the more possible solutions we will get. The importance of capturing the main trend with the simplest model possible can only be beneficial. Attention should be paid to those critical issues in pressure matching such as fracture con- finement. Therefore, all the lumped pseudo-3D models developed for processing speed of pressure-matching ap- plications are widely used. The final result of the net pressure-matching process should ideally be an exact superposition of the simulation on the pumping record. A perfect match is obtainable by adjusting controlling parameter of a fracture simulator, but this operation is quite time consuming and is not the goal of the exercise. Perfect matches are sometimes proposed by manually changing the number of fractures during the propagation. Unfortunately, there is no inde- pendent source that can be used to correlate a variation of the number of fractures. The option of multiple fractures is not available to all simulators. Nevertheless, much pres- sure adjustment can be obtained by changing parameters controlling the near-wellbore effect. Example parameters are the number of perforations, the relative erosion rate of perforation with proppant, and the characteristics of fracture tortuosity. These parameters have a major impact on the bottom-hole response but have nothing to do with the net pressure to be matched for fracture geometry estimate. Matching the Net Pressure during Calibration Treat- ment and the Pad. The calibration treatment match is part of the set of analysis performed on-site for redesign of the injection schedule. This match should be reviewed before proceeding with the analysis of the main treatment itself. Consistency between the parameters obtained from both matches should be maintained and deviation recognized. The first part of the treatment-match process focusing on the pad is identical to a match performed on the cali- bration treatment. The shut-in net pressure obtained from a minifrac (calibration treatment decline) gives the magni- tude of the net pressure. The pad net pressure history (and low prop concentration in the first few stages) is adjusted by changing either the compliance or the tip pressure. The Nolte–Smith Plot (Nolte and Smith, 1981) provides indi- cation of the degree of confinement of the fracture. A positive slope is an indication of confinement, a negative slope an indication of height growth, and a zero slope an indication of toughness-dominated short fracture or mod- erate height growth. Using 2D Models. In general, when the fracture is confined (PKN model) and viscous dominated, we either decrease the height of the zone or increase the Young’s modulus to obtain higher net pressure (compliance is $h=E). For a radial fracture (KGD model), we adjust the tip pressure effect to achieve net pressure match. If the fractured formation is a clean sand section and the fracture is confined or with moderate height growth, the fracture height should be fixed to the pay zone. In a layered forma- tion/dirty sandstone, the fracture height could be adjusted because any of the intercalated layers may or may not have been broken down. The fracture could still be confined, but the height cannot a priori be set as easily as in the case of a clean sand zone section. Unconsolidated sands show low Young’s modulus ( $5 Â 105 psi), this should not be changed to match the pressure. A low Young modulus value often gives insufficient order of magnitude of net pressure because the viscous force is not the dominating factor. The best way to adjust a fracture elastic model to match the behavior of a loosely consolidated sand is to increase the ‘‘apparent toughness’’ that controls the tip effect propagating pressure. Using Pseudo-3D Models. Height constraint is adjusted by increasing the stress difference between the pay-zone and the bounding layer. Stiffness can be increased with an increase of the Young modulus of all the layers that are fractured or to some extent by adding a small shale layer with high stress in the middle of the zone (pinch-point effect). Very few commercial fracturing simulators actually use a layer description of the modulus. All of the lumped 3D models use an average value. Tip effect can also be adjusted by changing toughness (Meyer et al., 1990). For some simulators, the users have no direct control of this effect, as an apparent toughness is recalculated from the rock toughness and fluid-lag effect. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 262 3.1.2007 9:19pm Compositor Name: SJoearun 17/262 PRODUCTION ENHANCEMENT
- 264. Simulating controlled height growth with a pseudo-3D model can be tricky. Height growth is characterized by a slower rate of pressure increase than in the case of a confined fracture. To capture the big picture, a simplifica- tion to a three-layer model can help by reducing the num- ber of possible inputs. Pressure-matching slow height growth of a fracture is tedious and lengthy. In the first phase, we should adjust the magnitude of the simulated net pressure. The match can be considered excellent if the difference between the recorded pressure and the simulated pressure is less than 15% over the length of the pad. The pressure matching can be performed using data from real-time measurements (Wright et al., 1996; Burton et al., 2002). Computer simulation of fracturing operations with recorded job parameters can yield the following frac- ture dimensions: . Fracture height . Fracture half-length . Fracture width A typical pressure matching with a pseudo-3D fracturing model is shown in Fig. 17.12 (Burton et al., 2002). Efficiency and Leakoff. The first estimate of effi- ciency and leakoff is obtained from the calibration treat- ment decline analysis. The calibration treatment provides a direct measurement of the efficiency using the graphical G-plot analysis and the 3 ⁄4 rules or by using time to closure with a fracturing simulator. Then calibration with a model that estimates the geometry of the fracture provides the corresponding leakoff coefficient (Meyer and Jacot, 2000). This leakoff coefficient determination is model dependent. Propped Fracture Geometry. Once we have obtained both a reasonable net pressure match, we have an estimate of length and height. We can then directly calculate the average width expressed in mass/area of the propped frac- ture from mass balance. The propped geometry given by any simulator after closure should not be any different. Post-propped Frac Decline. The simulator-generated pressure decline is affected by the model of extension recession that is implemented and by the amount of sur- face area that still have leakoff when the simulator cells are packed with proppant. It is very unlikely that the simula- tor matches any of those extreme cases. The lumped solu- tion used in FracProPT does a good job of matching pressure decline. The analysis methodology was indeed developed around pressure matching the time to closure. The time to closure always relates to the efficiency of the fluid regardless of models (Nolte and Smith, 1981). 17.6.2 Pressure Buildup Test Analysis Fracture and reservoir parameters can be estimated using data from pressure transient well tests (Cinco-Ley and Samaniego, 1981; Lee and Holditch, 1981). In the pressure transient well-test analysis, the log-log plot of pressure derivative versus time is called a diagnostic plot. Special slope values of the derivative curve usually are used for identification of reservoir and boundary models. The transient behavior of a well with a finite-conductivity fracture includes several flow periods. Initially, there is a fracture linear flow characterized by a half-slope straight line; after a transition flow period, the system may or may not exhibit a bilinear flow period, indicated by a one-fourth–slope straight line. As time increases, a for- mation linear flow period might develop. Eventually, the system reaches a pseudo-radial flow period if the drain- age area is significantly larger than the fracture dimension (Fig. 17.13). During the fracture linear flow period, most of the fluid entering the wellbore comes from the expansion of the system within the fracture. The behavior in the period occurs at very small amounts of time, normally a few seconds for the fractures created during frac-packing operations. Thus, the data in this period, even if not distorted by wellbore storage effect, are still not of prac- tical use. 1,500 2,000 2,500 3,000 3,500 4,000 Treatment Time(min) BHP(psi) −2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 SlurryRate(bbl/min) PropConc(PPA) BHP (Job data) BHP (PropFRAC) Slurry Rate (bbl/min) Prop Conc (PPA) 0 10 15 20 25 30 35 405 Figure 17.12 Bottom-hole pressure match with three-dimensional fracturing model PropFRAC. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 263 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/263
- 265. The bilinear flow regime means two linear flows occur simultaneously. One flow is a linear flow within the frac- ture and the other is a linear flow in the formation toward the fracture. Bilinear flow analysis gives an estimate of fracture length and conductivity. A calculated pressure distribution during a bilinear flow is illustrated in Fig. 17.2 (Guo and Schechter, 1999). The formation linear flow toward the fracture occurs after the bilinear flow. Linear flow analysis yields an esti- mate of formation permeability in the direction perpen- dicular to the fracture face. If the test time is long enough and there is no boundary effect, a system pseudo-radial flow will eventually occur. Pseudo-radial flow analysis provides an estimate of formation permeability in the radial direction. The reader is referred to Chapter 15 for analysis and interpretation of pressure transient data. It is important to note that by nomeansdoes the pressure- match procedure and the pressure transient data analysis give details of the fracture geometry such as fracture width near the wellbore, which frequently dominates the post- treatment well performance. The fracture width near the wellbore can be significantly lower than that in the region away from the wellbore.This can occur becauseof a number of mishaps. Overdisplacement of proppant leads to the frac- ture unsupported near the wellbore, resulting in fracture closure. Fluid backflow reduces the amount of proppant near the wellbore, which results in less fracture width sup- ported. If the proppant grains do not have compressive strength to withstand the stress concentration in the near- wellboreregion,theywillbecrushedduringfractureclosure, resulting in tight fracture near the wellbore. The reduced fracture width near the wellbore affects well productivity because of the fracture choking effect. Post-treatment flow tests should be run to verify well performance. The effect of near-wellbore fracture geometry on post- treatment well production is of special significance in deviated and horizontal wells (Chen and Economides, 1999). This is because a fracture from an arbitrarily oriented well ‘‘cuts’’ the wellbore at an angle, thereby limiting the communication between the wellbore and the reservoir. This feature of fluid entry to the wellbore itself causes the fracture-choking effect, even though the near- wellbore fracture is perfectly propped. Certainly, a hori- zontal well in the longitudinal to the fracture direction and using 180-degree perforation phasing that can be oriented will eliminate the problem. However, to align the horizon- tal wellbore in the longitudinal to the fracture direction, the horizontal wellbore has to be drilled in the direction parallel to the maximum horizontal stress direction. The orientation of the stress can be obtained by running tests in a vertical pilot hole of the horizontal well. Special log imaging (e.g., FMI and FMS) can be run in combination with an injection test at small-rate MDT or large-scale minifrac to fracture the formation and read directly the image in the wellbore after the fracture has been created. 17.6.3 Other evaluation techniques In addition to the pressure-matching and pressure buildup data analyses, other techniques can be used to verify the fracture profile created during a fracpack operation. These techniques include (1) pumping radioactive materials in the proppant stages and running tracer logs to verify the fracture heights, (2) running production logging tools to determine the production profiles, and (3) conducting back-pressure and performing Nodal analysis to verify the well deliverability. Summary This chapter presents a brief description of hydraulic frac- turing treatments covering formation fracturing pressure, fracture geometry, productivity of fractured wells, hydraulic fracturing design, and post-frac evaluation. More in-depth discussions can be found from Economides et al. (1994) and Economides and Nolte (2000). References argawal, r.g., carter, r.d., and pollock, c.b. Evaluation and prediction of performance of low- permeability gas wells stimulated by massive hydraulic fracturing. JPT March 1979, Trans. AIME 1979;267:362–372. barree, r.d. A practical numerical simulator for three dimensional fracture propagation in heterogeneous media. Proceedings of the Reservoir Simulation Sym- posium, San Francisco, CA, 403-411 Nov. 1983. SPE 12273. burton, r.c., davis, e.r., hodge, r.m., stomp, r.j., palthe, p.w., and saldungaray, p. Innovative completion design and well performance evaluation for effective Frac-packing of long intervals: a case study from the West Natuna Sea, Indonesia. Presented at the SPE International Petroleum Conference and Exhibition held 10–12 February 2002, in Villahermosa, Mexico. Paper SPE 74351. chen, z. and economides, m.j. Effect of near-wellbore fracture geometry on fracture execution and post-treat- ment well production of deviated and horizontal wells. SPE Prod. Facilities August 1999. cinco-ley, h. and samaniego, f. Transient pressure analysis for fractured wells. J. Petroleum Technol. September 1981. cleary, m.p. Comprehensive design formulae for hydraulic fracturing. Presented at the SPE Annual Technology Conference held in Dallas, Texas, Septem- ber 1980. SPE 9259. Fracture linear flow: Pseudo-radial flow Bilinear flow: Reservoir linear flow Figure 17.13 Four flow regimes that can occur in hydraulically fractured reservoirs. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 264 3.1.2007 9:19pm Compositor Name: SJoearun 17/264 PRODUCTION ENHANCEMENT
- 266. cleary, m.p., coyle, r.s., teng, e.y., cipolla, c.l., meehan, d.n., massaras, l.v., and wright, t.b. Major new developments in hydraulic fracturing, with documented reductions in job costs and increases in normalized production. Presented at the 69th Annual Technical Conference and Exhibition of the Society of Petroleum Engineers, held in New Orleans, Louisiana, 25–28 September 1994. SPE 28565. clifton, r.j. and abou-sayed, a.s. On the computation of the three-dimensional geometry of hydraulic fractures. Presented at the SPE/DOE Low Perm. Gas Res. Sympo- sium, held in Denver, Colorado, May 1979. SPE 7943. economides, m.j., hill, a.d., and ehlig-economides, c. Petroleum Production Systems, Upper Saddle River, New Jersey, Prentice Hall PTR, 1994. economides, m.j. and nolte, k.g. Reservoir Stimulation, 3rd edition. New York: John Wiley Sons, 2000. geertsma, j. and de klerk, f. A rapid method of predict- ing width and extent of hydraulic induced fractures. J. Petroleum Technol. Dec. 1969;21:1571–1581. guo, b. and schechter, d.s. A simple and rigorous IPR equation for vertical and horizontal wells intersecting long fractures. J. Can. Petroleum Technol. July 1999. khristianovich, s.a. and zheltov, y.p. Formation of vertical fractures by means of highly viscous liquid. In: Proceedings of the SPE Fourth World Petroleum Congress held in Rome, Section II. 1955, pp. 579–586. lee, w.j. and holditch, s.a. Fracture evaluation with pressure transient testing in low-permeability gas res- ervoirs. J. Petroleum Technol. September 1981. meyer, b.r., cooper, g.d., and nelson, s.g. Real-time 3-D hydraulic fracturing simulation: theory and field case studies. Presented at the 65th Annual Technical Con- ference and Exhibition of the Society of Petroleum Engineers, held in New Orleans, Louisiana, 23–26 Sep- tember 1990. Paper SPE 20658. meyer, b.r. and jacot, r.h. Implementation of fracture calibration equations for pressure dependent leakoff. Presented at the 2000 SPE/AAPG Western Regional Meeting, held in Long Beach, California, 19–23 June 2000. Paper SPE 62545. nolte, k.g. and smith, m.b. Interpretation of fracturing pressures. J. Petroleum Technol. September 1981. nordgren, r.p. Propagation of vertical hydraulic fracture. SPEJ Aug. 1972:306–314. perkins, t.k. and kern, l.r. Width of Hydraulic Fracture. J. Petroleum Technol. Sept. 1961:937–949. sneddon, i.n. and elliott, a.a. The opening of a griffith crack under internal pressure. Quart. Appl. Math. 1946;IV:262. valko, p., oligney, r.e., economides, m.j. High permeability fracturing of gas wells. Gas TIPS. October 1997;3:31–40. wright, c.a., weijers, l., germani, g.a., maclvor, k.h., wilson, m.k., and whitman, b.a. Fracture treatment design and evaluation in the Pakenham field: a real- data approach. Presented at the SPE Annual Technical Conference and Exhibition, held in Denver, Colorado, 6–9 October 1996. Paper SPE 36471. Problems 17.1 A sandstone at a depth of 8,000 ft has a Poison’s ratio of 0.275 and a poro-elastic constant of 0.70. The average density of the overburden formation is 162 lb=ft3 . The pore–pressure gradient in the sand- stone is 0.36 psi/ft. Assuming a tectonic stress of 1,000 psi and a tensile strength of the sandstone of 800 psi, predict the breakdown pressure for the sand- stone. 17.2 A carbonate at a depth of 12,000 ft has a Poison’s ratio of 0.3 and a poro-elastic constant of 0.75. The average density of the overburden formation is 178 lb=ft3 . The pore–pressure gradient in the sand- stone is 0.35 psi/ft. Assuming a tectonic stress of 2,000 psi and a tensile strength of the sandstone of 1,500 psi, predict the breakdown pressure for the sandstone. 17.3 A gas reservoir has a permeability of 5 md. A vertical well of 0.328-ft radius draws the reservoir from the center of an area of 320 acres. If the well is hydraul- ically fractured to create a 2,000-ft long, 0.15-in. wide fracture of 200,000-md permeability around the cen- ter of the drainage area, what would be the fold of increase in well productivity? 17.4 A reservoir has a permeability of 100 md. A vertical well of 0.328-ft radius draws the reservoir from the center of an area of 160 acres. If the well is hydraul- ically fractured to create a 2,800-ft long, 0.12-in. wide fracture of 250,000-md permeability around the cen- ter of the drainage area, what would be the fold of increase in well productivity? 17.5 For the following situation, estimate the minimum required compressive strength of 20/40 proppant. If high-strength proppant is used, estimate the perme- ability of the proppant pack: Formation depth: 12,000 ft Overburden density: 165 lbm=ft3 Poison’s ratio: 0.25 Biot constant: 0.72 Reservoir pressure: 6,800 psi Production drawdown: 3,000 psi 17.6 For the Problem 17.5, predict the maximum expected surface injection pressure using the following addi- tional data: Specific gravity of fracturing fluid: 1.1 Viscosity of fracturing fluid: 10 cp Tubing inner diameter: 3.0 in. Fluid injection rate: 20 bpm 17.7 The following data are given for a hydraulic fractur- ing treatment design: Pay zone thickness: 50 ft Young’s modulus of rock: 4 Â 106 psi Poison’s ratio: 0.25 Fluid viscosity: 1.25 cp Leakoff coefficient: 0:003 ft= min1=2 Proppant density: 185 lb=ft3 Proppant porosity: 0.4 Fracture half length: 1,200 ft Fracture height: 70 ft Fluid injection rate: 35 bpm Final proppant concentration: 5 ppg Assuming KGD fracture, estimate a. Fluid volume requirement b. Proppant mixing schedule c. Proppant weight requirement d. Propped fracture width 17.8 Predict the productivity index of the fractured well described in Problem 17.7. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap17 Final Proof page 265 3.1.2007 9:19pm Compositor Name: SJoearun HYDRAULIC FRACTURING 17/265
- 267. 18 Production Optimization Contents 18.1 Introduction 18/268 18.2 Naturally Flowing Well 18/268 18.3 Gas-Lifted Well 18/268 18.4 Sucker Rod–Pumped Well 18/269 18.5 Separator 18/270 18.6 Pipeline Network 18/272 18.7 Gas-Lift Facility 18/275 18.8 Oil and Gas Production Fields 18/276 18.9 Discounted Revenue 18/279 Summary 18/279 References 18/279 Problems 18/280 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 267 4.1.2007 10:04pm Compositor Name: SJoearun
- 268. 18.1 Introduction The term ‘‘production optimization’’ has been used to describe different processes in the oil and gas industry. A rigorous definition of the term has not been found from the literature. The book by Beggs (2003) ‘‘Production Optimization Using NODAL Analysis’’ presents a systems analysis approach (called NODAL analysis, or Nodal analysis) to analyze performance of production systems. Although the entire production system is analyzed as a total unit, interacting components, electrical circuits, complex pipeline networks, pumps, and compressors are evaluated individually using this method. Locations of excessive flow resistance or pressure drop in any part of the network are identified. To the best of our understanding, production optimiza- tion means determination and implementation of the optimum values of parameters in the production system to maximize hydrocarbon production rate (or discounted revenue) or to minimize operating cost under various tech- nical and economical constraints. Because a system can be defined differently, the production optimization can be performed at different levels such as well level, platform/ facility level, and field level. This chapter describes production optimization of systems defined as . Naturally flowing well . Gas-lifted well . Sucker rod–pumped well . Separator . Pipeline network . Gas lift facility . Oil and gas production fields In the upstream oil and gas production, various appro- aches and technologies are used to address different as- pects of hydrocarbon production optimization. They serve to address various business objectives. For example, on- line facility optimizer addresses the problem of maximizing the value of feedstock throughput in real time. This chap- ter presents principals of production optimization with the aids of computer programs when necessary. 18.2 Naturally Flowing Well A naturally flowing well may be the simplest system in production optimization. The production rate from a sin- gle flowing well is dominated by inflow performance, tub- ing size, and wellhead pressure controlled by choke size. Because the wellhead pressure is usually constrained by surface facility requirements, there is normally not much room to play with the choke size. Well inflow performance is usually improved with well-stimulation techniques including matrix acidizing and hydraulic fracturing. While matrix-acidizing treat- ment is effective for high-permeability reservoirs with significant well skins, hydraulic-fracturing treatment is more beneficial for low-permeability reservoirs. Inflow equations derived from radial flow can be used for pre- dicting inflow performance of acidized wells, and equa- tions derived from both linear flow and radial flow may be employed for forecasting deliverability of hydraulically fractured wells. These equations are found in Chapter 15. Figure 18.1 illustrates inflow performance relationship (IPR) curves for a well before and after stimulation. It shows that the benefit of the stimulation reduces as bottom-hole pressure increases. Therefore, after predicting inflow performance of the stimulated well, single-well Nodal analysis needs to be carried out. The operating points of stimulated well and nonstimulated wells are compared. This comparison provides an indication of whether the well inflow is the limiting step that controls well deliverability. If yes, treatment design may proceed (Chapters 16 and 17) and economic evaluation should be performed (see Section 18.9). If no, optimization of tubing size should be investigated. It is not true that the larger the tubing size is, the higher the well deliverability is. This is because large tubing reduces the gas-lift effect in oil wells. Large tubing also results in liquid loading of gas wells due to the inadequate kinetic energy of gas flow required to lift liquid. The optimal tubing size yields the lowest frictional pressure drop and the maximum production rate. Nodal analysis can be used to generate tubing performance curve (plot of operating rate vs tubing size) from which the optimum tubing size can be identified. Figure 18.2 shows a typical tubing performance curve. It indicates that a 3.5-in. inner diameter (ID) tubing will give a maximum oil production rate of 600 stb/day. However, this tubing size may not be considered optimal because a 3.0-in. ID tubing will also deliver a similar oil production rate and this tubing may be cheaper to run. An economics evaluation should be performed (see Section 18.9). 18.3 Gas-Lifted Well The optimization of individual gas-lift wells mainly focuses on determining and using the optimal gas-lift gas injection rate. Overinjection of gas-lift gas is costly and results in lower oil production rate. The optimal gas Before Stimulation After Stimulation BottomHolePressure(pwf) Production Rate(q) Figure 18.1 Comparison of oil well inflow performance relationship (IPR) curves before and after stimulation. OperatingRate(stb/day) 625 500 375 250 125 0 Inside Diameter of Tubing (in.) 0 1.25 2.5 3.75 5 Figure 18.2 A typical tubing performance curve. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 268 4.1.2007 10:04pm Compositor Name: SJoearun 18/268 PRODUCTION ENHANCEMENT
- 269. injection rate can be identified from a gas-lift perfor- mance curve, which can be generated using Nodal analy- sis software such as WellFlo (1997). Figure 18.3 presents a typical gas-lift performance curve. It shows that a 5.0-MMscf/day gas injection rate will give a maximum oil production rate of 260 stb/day. However, this gas injection rate may not be the optimum rate because slightly lower gas injection rates will also deliver a similar oil production rate with lower high-pressure gas con- sumption. An economics evaluation should be performed on a scale of a batch of similar wells (see Section 18.9). 18.4 Sucker Rod–Pumped Well The potential of increasing oil production rate of a normal sucker rod–pumped well is usually low. Optimization of this type of well mainly focuses on two areas: . Improving the volumetric efficiency of the plunger pump . Improving the energy efficiency of the pumping unit Estimating the volumetric efficiency of plunger pump and improving the energy efficiency of the pumping unit require the use of the information from a dynamometer card that records polished rod load. Figure 18.4 demon- strates a theoretical load cycle for elastic sucker rods. However, because of the effects of acceleration and fric- tion, the actual load cycles are very different from the theoretical cycles. Figure 18.5 demonstrates an actual load cycle of a sucker rod under normal working condi- tions. It illustrates that the peak polished rod load can be significantly higher than the theoretical maximum pol- ished rod load. Much information can be obtained from the dynamom- eter card. The procedure is illustrated with the parameters shown in Fig. 18.6. The nomenclature is as follows: C ¼ calibration constant of the dynamometer, lb/in. D1 ¼ maximum deflection, in. D2 ¼ minimum deflection, in. D3 ¼ load at the counterbalance line (CB) drawn on the dynamometer card by stopping the pumping unit at the position of maximum counterbalance effect (crank arm is horizontal on the upstroke), in. A1 ¼ lower area of card, in:2 A2 ¼ upper area of card, in:2 . The following information can be obtained from the card parameter values: Peak polished rod load: PPRL ¼ CD1 Minimum polished rod load: MPRL ¼ CD2 Range of load: ROL ¼ C(D1 À D2) Average upstroke load: AUL ¼ CðA1 þ A2Þ L Average downstroke load: ADL ¼ CA1 L Work for rod elevation: WRE ¼ A1 converted to ft-lb Work for fluid elevation and friction: WFEF ¼ A2 converted to ft-lb Approximate ‘‘ideal’’ counterbalance: AICB ¼ PPRL þ MPRL 2 Actual counterbalance effect: ACBE ¼ CD3 Correct counterbalance: CCB ¼ (AUL þ ADL)=2 ¼ C A1 þ A2 2 À ) L Polished rod horsepower: PRHP ¼ CSNA2 33,000(12)L 300 240 180 120 OperatingRate(stb/day) 80 0 0 1.5 3 Lift Gas Injection Rate (MMscf/day) 4.5 6 Figure 18.3 A typical gas lift performance curve of a low-productivity well. Polished Rod Position PolishedRodLoad S PRLmin PRLmax Work Figure 18.4 Theoretical load cycle for elastic sucker rods. s F1 F0 =gross plunger load Wrf = weight of rods in fluid Minimum polished rods in fluid F2 PolishedRodLoad Bottom of stroke Top of stroke Polished Rod Position Peak polished rod load. pprl Polished rod card for pumping Speed greater than zero. n0 Polished rod card for pumping spee. n=0 Figure 18.5 Actual load cycle of a normal sucker rod. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 269 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/269
- 270. Example Problem 18.1 Analyze the dynamometer card shown in Fig. 18.6 assuming the following parameter values: S ¼ 45 in: N ¼ 18:5 spm C ¼ 12,800 lb=in: D1 ¼ 1:2 in: D2 ¼ 0:63 in: L ¼ 2:97 in: A1 ¼ 2:1 in:2 A2 ¼ 1:14 in:2 Solution Peak polished rod load: PPRL ¼ (12,800)(1:20) ¼ 15,400 lb Minimum polished rod load: MPRL ¼ (12,800)(0:63) ¼ 8,100 lb Average upstroke load: AUL ¼ (12,800)(1:14 þ 2:10) 2:97 ¼ 14,000 lb Average downstroke load: ADL ¼ (12,800)(2:10) 2:97 ¼ 9,100 lb Correct counterbalance: CCB ¼ (12,800)(2:10 þ 1:14 2 ) 2:97 ¼ 11,500 lb Polished rod horsepower: PRHP ¼ (12,800)(45)(18:5)(1:14) 33,000(12)(2:97) ¼ 10:3 hp The information of the CCB can be used for adjusting the positions of counterweights to save energy. In addition to the dimensional parameter values taken from the dynamometer card, the shape of the card can be used for identifying the working condition of the plunger pump. The shapeofthedynamometercardsare influencedby . Speed and pumping depth . Pumping unit geometry . Pump condition . Fluid condition . Friction factor Brown (1980) listed 13 abnormal conditions that can be identified from the shape of the dynamometer cards. For example, the dynamometer card shown in Fig. 18.7 indicates synchronous pumping speeds, and the dynamometer card depicted in Fig. 18.8 reveals a gas-lock problem. 18.5 Separator Optimization of the separation process mainly focuses on recovering more oil by adjusting separator temperature and pressure. Field experience proves that lowering the operating temperature of a separator increases the liquid recovery. It is also an efficient means of handling high- pressure gas and condensate at the wellhead. A low-tem- perature separation unit consists of a high-pressure separa- tor, pressure-reducing chokes, and various pieces of heat exchange equipment. When the pressure is reduced by the use of a choke, the fluid temperature decreases because of the Joule–Thomson or the throttling effect. This is an irreversible adiabatic process whereby the heat content of the gas remains the same across the choke but the pressure and temperature of the gas stream are reduced. Generally at least 2,500–3,000 psi pressure drop is required from wellhead flowing pressure to pipeline pres- sure for a low-temperature separation unit to pay out in increased liquid recovery. The lower the operating tem- perature of the separator, the lighter the liquid recovered will be. The lowest operating temperature recommended for low-temperature units is usually around À20 F. This is constrained by carbon steel embitterment, and high-alloy steels for lower temperatures are usually not economical for field installations. Low-temperature separation units are normally operated in the range of 0–20 8F. The actual temperature drop per unit pressure drop is affected by several factors including composition of gas stream, gas and liquid flow rates, bath temperature, and ambient tem- perature. Temperature reduction in the process can be estimated using the equations presented in Chapter 5. L Zero line D3 C.B. A2 A1 D1 D2 Figure 18.6 Dimensional parameters of a dynamometer card. A B C D F E G Figure 18.7 A dynamometer card indicating synchron- ous pumping speeds. Figure 18.8 A dynamometer card indicating gas lock. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 270 4.1.2007 10:04pm Compositor Name: SJoearun 18/270 PRODUCTION ENHANCEMENT
- 271. Gas expansion pressures for hydrate formation can be found from the chart prepared by Katz (1945) or Guo and Ghalambor (2005). Liquid and vapor phase densities can be predicted by flash calculation. Following the special requirement for construction of low-temperature separation units, the pressure-reducing choke is usually mounted directly on the inlet of the high-pressure separator. Hydrates form in the downstream of the choke because of the low gas temperature and fall to the bottom settling section of the separator. They are heated and melted by liquid heating coils located in the bottom of the separator. Optimization of separation pressure is performed with flash calculations. Based on the composition of well- stream fluid, the quality of products from each stage of separation can be predicted, assuming phase equilibriums are reached in the separators. This requires the knowledge of the equilibrium ratio defined as ki ¼ yi xi , (18:1) where ki ¼ liquid/vapor equilibrium ratio of compound i yi ¼ mole fraction of compound i in the vapor phase xi ¼ mole fraction of compound i in the liquid phase. Accurate determination of ki values requires computer simulators solving the Equation of State (EoS) for hydro- carbon systems. Ahmed (1989) presented a detailed procedure for solving the EoS. For pressures lower than 1,000 psia, a set of equations presented by Standing (1979) provide an easy and accurate means of determining ki values. According to Standing, ki can be calculated by ki ¼ 1 p 10aþcFi , (18:2) where a ¼ 1:2 þ 4:5 Â 10À4 p þ 1:5 Â 10À9 p2 (18:3) c ¼ 0:89 À 1:7 Â 10À4 p À 3:5 Â 10À8 p2 (18:4) Fi ¼ bi 1 Tbi À 1 T (18:5) bi ¼ log pci 14:7 1 Tbi À 1 Tci , (18:6) where pc ¼ critical pressure, psia Tb ¼ boiling point, 8R Tc ¼ critical temperature, 8R. Consider 1 mol of fed-in fluid and the following equation holds true on the basis of mass balance: nL þ nV ¼ 1, (18:7) where nL ¼ number of mole of fluid in the liquid phase nV ¼ number of mole of fluid in the vapor phase. For compound i, zi ¼ xinL þ yinV , (18:8) where zi is the mole fraction of compound i in the fed-in fluid. Combining Eqs. (18.1) and (18.8) gives zi ¼ xinL þ kixinV , (18:9) which yields xi ¼ zi nL þ kinV : (18:10) Mass balance applied to Eq. (18.10) requires XNc i¼1 xi ¼ XNc i¼1 zi nL þ kinV ¼ 1, (18:11) where Nc is the number of compounds in the fluid. Com- bining Eqs. (18.1) and (18.8) also gives zi ¼ yi ki nL þ yinV , (18:12) which yields yi ¼ ziki nL þ kinV : (18:13) Mass balance applied to Eq. (18.13) requires XNc i¼1 yi ¼ XNc i¼1 ziki nL þ kinV ¼ 1: (18:14) Subtracting Eq. (18.14) from Eq. (18.11) gives XNc i¼1 zi nL þ kinV À XNc i¼1 ziki nL þ kinV ¼ 0, (18:15) which can be rearranged to obtain XNc i¼1 zi(1 À ki) nL þ kinV ¼ 0: (18:16) Combining Eqs. (18.16) and (18.7) results in XNc i¼1 zi(1 À ki) nV (ki À 1) þ 1 ¼ 0: (18:17) This equation can be used to solve for the number of mole of fluid in the vapor phase nv. Then, xi and yi can be calculated with Eqs. (18.10) and (18.13), respectively. The apparent molecular weights of liquid phase (MW) and vapor phase (MW) can be calculated by MWL a ¼ XNc i¼1 xiMWi (18:18) MWV a ¼ XNc i¼1 yiMWi, (18:19) where MWi is the molecular weight of compound i. With the apparent molecular weight of the vapor phase known, the specific gravity of the vapor phase can be determined, and the density of the vapor phase in lbm=ft3 can be calculated by rV ¼ MWV a p zRT : (18:20) The liquid phase density in lbm=ft3 can be estimated by the Standing method (1981), that is, rL ¼ 62:4goST þ 0:0136Rsgg 0:972 þ 0:000147 Rs ffiffiffiffi gg go q þ 1:25 T À 460ð Þ h i1:175 , (18:21) where goST ¼ specific gravity of stock-tank oil, water gg ¼ specific gravity of solution gas, air ¼ 1 Rs ¼ gas solubility of the oil, scf/stb. Then the specific volumes of vapor and liquid phases can be calculated by VVsc ¼ znV RTsc psc (18:22) VL ¼ nLMWL a L , (18:23) Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 271 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/271
- 272. where VVsc ¼ specific volume of vapor phase under standard condition, scf/mol-lb R ¼ gas constant, 10:73 ft3 -psia/lb mol-R Tsc ¼ standard temperature, 520 8R psc ¼ standard pressure, 14.7 psia VL ¼ specific volume of liquid phase, ft3 /mol-lb. Finally, the gas–oil ratio (GOR) in the separator can be calculated by GOR ¼ VVsc VL : (18:24) Specific gravity and the American Petroleum Institute (API) gravity of oil at the separation pressure can be calculated based on liquid density from Eq. (18.21). The lower the GOR, the higher the API gravity, and the higher the liquid production rate. For gas condensates, there exists an optimum separation pressure that yields the lower GOR at a given temperature. Example Problem 18.2 Perform flash calculation under the following separator conditions: Pressure: 600 psia Temperature: 200 8F Specific gravity of stock-tank oil: 0.90 water ¼ 1 Specific gravity of solution gas: 0.70 air ¼ 1 Gas solubility (Rs): 500 scf/stb Solution The flash calculation can be carried out using the spreadsheet program LP-Flash.xls. The results are shown in Table 18.1. 18.6 Pipeline Network Optimization of pipelines mainly focuses on de-bottle- necking of the pipeline network, that is, finding the most restrictive pipeline segments and replacing/adding larger segments to remove the restriction effect. This requires the knowledge of flow of fluids in the pipe. This section pre- sents mathematical models for gas pipelines. The same principle applies to oil flow. Equations for oil flow are presented in Chapter 11. 18.6.1 Pipelines in Series Consider a three-segment gas pipeline in a series of total length L depicted in Fig. 18.9a. Applying the Weymouth equation to each of the three segments gives p2 1 À p2 2 ¼ ggT zL1 D 16=3 1 qhpb 18:062Tb 2 (18:25) p2 2 À p2 3 ¼ ggT zL2 D 16=3 2 qhpb 18:062Tb 2 (18:26) p2 3 À p2 4 ¼ ggT zL3 D 16=3 3 qhpb 18:062Tb 2 : (18:27) Adding these three equations gives p2 1 À p2 4 ¼ ggT z L1 D 16=3 1 þ L2 D 16=3 2 þ L3 D 16=3 3 !  qhpb 18:062Tb 2 (18:28) or qh ¼ 18:062Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 1 À p2 4 ggTz L1 D 16=3 1 þ L2 D 16=3 2 þ L3 D 16=3 3 ! v u u u u t : (18:29) Capacity of a single-diameter (D1) pipeline is expressed as q1 ¼ 18:062Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 1 À p2 4 ggTz L D 16=3 1 ! v u u u u t : (18:30) Dividing Eq. (18.29) by Eq. (18.30) yields qt q1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L D 16=3 1 ! L1 D 16=3 1 þ L2 D 16=3 2 þ L3 D 16=3 3 ! v u u u u u u u t : (18:31) 18.6.2 Pipelines in Parallel Consider a three-segment gas pipeline in parallel as depicted in Fig. 18.9b. Applying the Weymouth equation to each of the three segments gives q1 ¼ 18:062 Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)D 16=3 1 ggT zL v u u t (18:32) q2 ¼ 18:062 Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)D 16=3 2 ggT zL v u u t (18:33) q3 ¼ 18:062 Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2)D 16=3 3 ggT zL v u u t : (18:34) Adding these three equations gives qt ¼ q1 þ q2 þ q3 ¼ 18:062 Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2) ggTzL s  ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 3 q : (18:35) Dividing Eq. (18.35) by Eq. (18.32) yields qt q1 ¼ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 3 q ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q : (18:36) 18.6.3 Looped Pipelines Consider a three-segment looped gas pipeline depicted in Fig. 18.10. Applying Eq. (18.35) to the first two (parallel) segments gives Gas composition Compound Mole fraction C1 0.6599 C2 0.0869 C3 0.0591 i-C4 0.0239 n-C4 0.0278 i-C5 0.0157 n-C5 0.0112 C6 0.0181 C7þ 0.0601 N2 0.0194 CO2 0.0121 H2S 0.0058 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 272 4.1.2007 10:04pm Compositor Name: SJoearun 18/272 PRODUCTION ENHANCEMENT
- 273. qt ¼ q1 þ q2 ¼ 18:062 Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( p2 1 À p2 2) ggTzL s ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q (18:37) or p2 1 À p2 3 ¼ ggTzL1 ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q 2 qtpb 18:062Tb 2 : (18:38) Applying the Weymouth equation to the third segment (with diameter D3) yields p2 3 À p2 2 ¼ ggTzL3 D 16=3 3 qtpb 18:062Tb 2 : (18:39) Adding Eqs. (18.38) and (18.39) results in p2 1 À p2 2 ¼ ggTz qtpb 18:062Tb 2  L1 ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q 2 þ L3 D 16=3 3 0 B B B @ 1 C C C A (18:40) or qt ¼ 18:062Tb pb  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 1 À p2 2 À Á ggTz L1 ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q 2 þ L3 D 16=3 3 0 B B B @ 1 C C C A v u u u u u u u u t : (18:41) Capacity of a single-diameter (D3) pipeline is expressed as q3 ¼ 18:062Tb pb ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 1 À p2 2 ggTz L D 16=3 3 ! v u u u u t : (18:42) Dividing Eq. (18.41) by Eq. (18.42) yields qt q3 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L D 16=3 3 ! L1 ffiffiffiffiffiffiffiffiffiffiffi D 16=3 1 q þ ffiffiffiffiffiffiffiffiffiffiffi D 16=3 2 q 2 þ L3 D 16=3 3 0 B B B @ 1 C C C A v u u u u u u u u u u u t : (18:43) Let Y be the fraction of looped pipeline and X be the increase in gas capacity, that is, (b) (a) L1 L L3 D1 D3p1 p2 D2 L2 p3 p4 L D1 D3 p1 p2 D2 p2qt q1 q2 q3 qt q Figure 18.9 Sketch of (a) a series pipeline and (b) a parallel pipeline. L1 L L3 D1 D2 p3 D3p1 p2qt q1 q2 qt Figure 18.10 Sketch of a looped pipeline. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 273 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/273
- 274. Y ¼ L1 L , X ¼ qt À q3 q3 : (18:44) If, D1 ¼ D3, Eq. (18.43) can be rearranged as Y ¼ 1 À 1 1 þ Xð Þ2 1 À 1 1 þ R2:31 D À Á2 , (18:45) where RD is the ratio of the looping pipe diameter to the original pipe diameter, that is, RD ¼ D2=D3. Equation (18.45) can be rearranged to solve for X explicitly X ¼ 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À Y 1 À 1 1 þ R2:31 D À Á2 !v u u t À 1: (18:46) The effects of looped line on the increase of gas flow rate for various pipe diameter ratios are shown in Fig. 18.11. This figure indicates an interesting behavior of looping: The increase in gas capacity is not directly proportional to the fraction of looped pipeline. For example, looping of 40% of pipe with a new pipe of the same diameter will increase only 20% of the gas flow capacity. It also shows that the benefit of looping increases with the fraction of looping. For example, looping of 80% of the pipe with a new pipe of the same diameter will increase 60%, not 40%, of gas flow capacity. Example Problem 18.3 Consider a 4-in. pipeline that is 10 miles long. Assuming that the compression and delivery pressures will maintain unchanged, calculate gas capacity increases by using the following measures of improvement: (a) replace 3 miles of the 4-in. pipeline by a 6-in. pipeline segment; (b) place a 6-in. parallel pipeline to share gas Table 18.1 Flash Calculation with Standing’s Method for ki Values Flash calculation nv ¼ 0:8791 Compound zi ki zi(ki À 1)=[nv(ki À 1) þ 1] C1 0.6599 6.5255 0.6225 C2 0.0869 1.8938 0.0435 C3 0.0591 0.8552 À0:0098 i-C4 0.0239 0.4495 À0:0255 n-C4 0.0278 0.3656 À0:0399 i-C5 0.0157 0.1986 À0:0426 n-C5 0.0112 0.1703 À0:0343 C6 0.0181 0.0904 À0:0822 C7þ 0.0601 0.0089 À0:4626 N2 0.0194 30.4563 0.0212 CO2 0.0121 3.4070 0.0093 H2S 0.0058 1.0446 0.0002 Sum: 0.0000 nL ¼ 0.1209 Compound xi yi xiMWi yiMWi C1 0.1127 0.7352 1.8071 11.7920 C2 0.0487 0.0922 1.4633 2.7712 C3 0.0677 0.0579 2.9865 2.5540 i-C4 0.0463 0.0208 2.6918 1.2099 n-C4 0.0629 0.0230 3.6530 1.3356 i-C5 0.0531 0.0106 3.8330 0.7614 n-C5 0.0414 0.0070 2.9863 0.5085 C6 0.0903 0.0082 7.7857 0.7036 C7þ 0.4668 0.0042 53.3193 0.4766 N2 0.0007 0.0220 0.0202 0.6156 CO2 0.0039 0.0132 0.1709 0.5823 H2S 0.0056 0.0058 0.1902 0.1987 Apparent molecular weight of liquid phase: 23.51 80.91 Apparent molecular weight of vapor phase: 0.76 Specific gravity of liquid phase: water ¼ 1 Specific gravity of vapor phase: 0.81 air ¼ 1 Input vapor phase z factor: 0.958 Density of liquid phase: 47.19 lbm=ft3 Density of vapor phase: 2.08 lbm=ft3 Volume of liquid phase: 0.04 bbl Volume of vapor phase: 319.66 scf GOR: 8,659 scf/bbl API gravity of liquid phase: 56 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 274 4.1.2007 10:04pm Compositor Name: SJoearun 18/274 PRODUCTION ENHANCEMENT
- 275. transmission; and (c) loop 3 miles of the 4-in. pipeline with a 6-in. pipeline segment. Solution (a) Replace a portion of pipeline: L ¼ 10 mi L1 ¼ 7 mi L2 ¼ 3 mi D1 ¼ 4 in: D2 ¼ 6 in: qt q1 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 416=3 7 416=3 þ 3 616=3 v u u u u u t ¼ 1:1668, or 16:68% increase in flow capacity: (b) Place a parallel pipeline: D1 ¼ 4 in: D2 ¼ 6 in: qt q1 ¼ ffiffiffiffiffiffiffiffiffiffi 416=3 p þ ffiffiffiffiffiffiffiffiffiffi 616=3 p ffiffiffiffiffiffiffiffiffiffi 416=3 p ¼ 3:9483, or 294:83% increase in flow capacity: (c) Loop a portion of the pipeline: L ¼ 10 mi L1 ¼ 7 mi L2 ¼ 3 mi D1 ¼ 4 in: D2 ¼ 6 in: qt q3 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 416=3 L1 ffiffiffiffiffiffiffiffiffiffi 416=3 p þ ffiffiffiffiffiffiffiffiffiffi 616=3 p 2 þ L3 416=3 0 B @ 1 C A v u u u u u u u u t ¼ 1:1791, or 17:91% increase in flow capacity: Similar problems can also be solved using the spreadsheet program LoopedLines.xls. Table 18.2 shows the solution to Example Problem 18.3 given by the spreadsheet. 18.7 Gas-Lift Facility Optimization of gas lift at the facility level mainly focuses on determination of the optimum lift-gas distribution among the gas-lifted wells. If lift-gas volume is not limited by the capacity of the compression station, every well should get the lift-gas injection rate being equal to its optimal gas injection rate (see Section 18.3). If limited lift-gas volume is available from the compression station, the lift gas should be assigned first to those wells that will produce more incrementals of oil production for a given incremental of lift-gas injection rate. This can be done by calculating and comparing the slopes of the gas-lift performance curves of individual wells at the points of adding more lift-gas injection rate. This principle can be illustrated by the following example problem. Example Problem 18.4 The gas-lift performance curves of two oil wells are known based on Nodal analyses at well level. The performance curve of Well A is presented in Fig. 18.3 and that of Well B is in Fig. 18.12. If a total lift-gas injection rate of 1.2 to 6.0 MMscf/day is available to the two wells, what lift-gas flow rates should be assigned to each well? Solution Data used for plotting the two gas-lift performance curves are shown in Table 18.3. Numerical derivatives (slope of the curves) are also included. At each level of given total gas injection rate, the incre- mental gas injection rate (0.6 MMscf/day) is assigned to one of the wells on the basis of their performance curve slope at the present gas injection rate of the well. The procedure and results are summarized in Table 18.4. The results indicate that the share of total gas injection rate by wells depends on the total gas rate availability and per- formance of individual wells. If only 2.4 MMscf/day of gas is available, no gas should be assigned to Well A. If only 3.6 MMscf/day of gas is available, Well A should share one-third of the total gas rate. If only 6.0 MMscf/day of gas is available, each well should share 50% of the total gas rate. 0 20 40 60 80 100 120 140 160 180 200 Looped Line (%) IncreaseinFlowRate(%) RD = 0.4 RD = 0.6 RD = 0.8 RD = 1.0 RD = 1.2 RD = 1.4 RD = 1.6 RD = 1.8 RD = 2.0 1000 10 20 30 40 50 60 70 80 90 Figure 18.11 Effects of looped line and pipe diameter ratio on the increase of gas flow rate. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 275 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/275
- 276. 18.8 Oil and Gas Production Fields An oil or gas field integrates wells, flowlines, separation facilities, pump stations, compressor stations, and trans- portation pipelines as a whole system. Single-phase and multiphase flow may exist in different portions in the system. Depending on system complexity and the objective of optimization task, field level production optimization can be performed using different approaches. 18.8.1 Types of Flow Networks Field-level production optimization deals with complex flow systems of two types: (1) hierarchical networks and (2) nonhierarchical networks. A hierarchical network is defined as a treelike converging system with multiple in- flow points (sources) and one outlet (sink). Figure 18.13 illustrates two hierarchical networks. Flow directions in this type of network are known. Fluid flow in this type of network can be simulated using sequential solving approach. Commercial software to perform this type of computation are those system analysis (Nodal analysis) programs such as FieldFlo and PipeSim, among others. A nonhierarchical network is defined as a general system with multiple inflow points (sources) and multiple outlets (sinks). Loops may exist, so the flow directions in some portions of the network are not certain. Figure 18.14 presents a nonhierarchical network. Arrows in this figure represent flow directions determined by a computer pro- gram. Fluid flow in this type of network can be simulated using simultaneous solving approaches. Commercial soft- ware to perform this type of computations include ReO, GAP, HYSYS, FAST Piper, and others. 18.8.2 Optimization Approaches Field-level production optimizations are carried out with two distinct approaches: (a) the simulation approach and (b) the optimization approach. 18.8.2.1 Simulation Approach The simulation approach is a kind of trial-and-error approach. A computer program simulates flow conditions (pressures and flow rates) with fixed values of variables in each run. All parameter values are input manually before each run. Different scenarios are investigated with differ- ent sets of input data. Optimal solution to a given problem is selected on the basis of results of many simulation runs with various parameter values. Thus, this approach is more time consuming. 18.8.2.2 Optimization Approach The optimization approach is a kind of intelligence-based approach. It allows some values of parameters to be determined by the computer program in one run. The parameter values are optimized to ensure the objective function is either maximized (production rate as the objective function) or minimized (cost as the objective function) under given technical or economical constraints. Apparently, the optimization approach is more efficient than the simulation approach. 18.8.3 Procedure for Production Optimization The following procedure may be followed in production optimization: 1. Define the main objective of the optimization study. The objectives can be maximizing the total oil/gas production rate or minimizing the total cost of operation. 2. Define the scope (boundary) of the flow network. 3. Based on the characteristics of the network and fluid type, select a computer program. 4. Gather the values of component/equipment param- eters in the network such as well-inflow performance, tubing sizes, choke sizes, flowline sizes, pump capacity, compressor horsepower, and others. 5. Gather fluid information including fluid compositions and properties at various points in the network. 6. Gather the fluid-flow information that reflects the cur- rent operating point, including pressures, flow rates, and temperatures at all the points with measurements. OperatingRate(stb/day) 2,000 1,500 1,200 800 400 0 Life Gas Injection Rate (MMscf/day) 0 1.5 3 4.5 6 Figure 18.12 A typical gas lift performance curve of a high-productivity well. Figure 18.13 Schematics of two hierarchical networks. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 276 4.1.2007 10:04pm Compositor Name: SJoearun 18/276 PRODUCTION ENHANCEMENT
- 277. Table 18.2 Solution to Example Problem 18.3 Given by the Spreadsheet LoopedLines.xls LoopedLines.xls This spreadsheet computes capacities of series, parallel, and looped pipelines. Input data Original pipe ID: 4 in. Total pipeline length: 10 mi Series pipe ID: 4 6 4 in. Segment lengths: 7 3 0 mi Parallel pipe ID: 4 6 0 in. Looped pipe ID: 4 6 4 in. Segment lengths: 3 7 mi Solution Capacity improvement by series pipelines: ¼ 1.1668 qh ¼ 3:23Tb pb ffiffiffi 1 f s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p2 1 À p2 2 À Á D5 g TTzzL s Capacity improvement by parallel pipelines: ¼ 3.9483 Capacity improvement by looped pipelines: ¼ 1.1791 Blue square = Flow station BLOQUE VIII Total gas from PDE_Cl_1 TO TI TO BA Black square = Low Pt entre manifold Green square = Compressor plant Red square = High pressure manifold Black line = Low pressure gas Purple square = Gas lift manifold Green line = High pressure wet gas Red line = High pressure dry gas Figure 18.14 An example of a nonhierarchical network. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 277 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/277
- 278. 7. Construct a computer model for the flow network. 8. Validate equipment models for each well/equipment in the network by simulating and matching the current operating point of the well/equipment. 9. Validate the computer model at facility level by simu- lating and matching the current operating point of the facility. 10. Validate the computer model at field level by simulat- ing and matching the current operating point of the field. 11. Run simulations for scenario investigations with the computer model if a simulation-type program is used. 12. Run optimizations with the computer model if an optimization-type program is used. 13. Implement the result of optimization with an open- loop or closed-loop method. 18.8.4 Production Optimization Software Commercial software packages are available for petroleum production optimization at all levels. Field-level optimiza- tion can be performed with ReO, GAP, HYSYS, FAST Piper, among others. This section makes a brief introduc- tion to these packages. 18.8.4.1 ReO The software ReO (EPS, 2004) is a compositional produc- tion simulator that can simulate and optimize highly non- hierarchical networks of multiphase flow. Its optimizer technology is based on sequential linear programming techniques. Because the network is solved simultaneously rather than sequentially, as is the case for nodal analysis techniques, the system can optimize and simulate account- ing for targets, objectives, and constraints anywhere in the network. A key feature of ReO is that it is both a production simulation and an optimization tool. Simulation deter- mines the pressures, temperatures, and fluid flow rates within the production system, whereas optimization deter- mines the most economical production strategy subject to engineering or economic constraints. The economic mod- eling capability inherent within ReO takes account of the revenues from hydrocarbon sales in conjunction with the production costs, to optimize the net revenue from the field. The ReO Simulation option generates distributions of pressure, temperature, and flow rates of water, oil, and gas in a well-defined network. The ReO Optimization option determines optimum parameter values that will lead to the maximum hydrocarbon production rate or the minimum operating cost under given technical and economical constraints. ReO addresses the need to opti- mize production operations, that is, between reservoir and facilities, in three main areas: . To aid in the design of new production capacity, both conceptual and in detail . To optimize production systems either off-line or in real time . To forecast performance and create production profiles for alternative development scenarios ReO integrates complex engineering calculations, practical constraints, and economic parameters to determine the optimal configuration of production network. It can be employed in all phases of field life, from planning through development and operations, and to enable petroleum, production, facility, and other engineers to share the same integrated model of the field and perform critical analysis and design activities such as the following: . Conceptual design in new developments Table 18.3 Gas Lift Performance Data for Well A and Well B Oil production rate (stb/day) Slope of performance curve (stb/MMscf) Lift gas injection rate (MMscf/day) Well A Well B Well A Well B 0.6 0 740 242 850 1.2 145 1,250 150 775 1.8 180 1,670 54 483 2.4 210 1,830 46 142 3 235 1,840 33 13 3.6 250 1,845 17 6 4.2 255 1,847 8 0 4.8 259 1,845 4 À56 5.4 260 1,780 À3 À146 6 255 1,670 Table 18.4 Assignments of Different Available Lift Gas Injection Rates to Well A and Well B Total lift gas Gas injection rate before assignment (stb/day) Slope of performance curve (stb/MMscf) Lift gas assignment (MMscf/day) Gas injection rate after assignment (stb/day) (MMscf/day) Well A Well B Well A Well B Well A Well B Well A Well B 1.2 0 0 242 850 0 1.2 0 1.2 1.8 0 1.2 242 775 0 0.6 0 1.8 2.4 0 1.8 242 483 0 0.6 0 2.4 3 0 2.4 242 142 0.6 0 0.6 2.4 3.6 0.6 2.4 242 142 0.6 0 1.2 2.4 4.2 1.2 2.4 150 142 0.6 0 1.8 2.4 4.8 1.8 2.4 54 142 0 0.6 1.8 3 5.4 1.8 3 54 13 0.6 0 2.4 3 6 2.4 3 46 13 0.6 0 3 3 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 278 4.1.2007 10:04pm Compositor Name: SJoearun 18/278 PRODUCTION ENHANCEMENT
- 279. . Equipment sizing, evaluation and selection . Daily production optimization, on-line or off-line . Problem and bottleneck detection/diagnosis . Production forecasting . Reservoir management . Data management Target and penalty functions are used in ReO within a valid region. This type of ‘‘target’’ is required to find the best compromise among conflicting objectives in a system. An example might be ensuring maximum production by driving down wellhead pressure in a gas field while main- taining optimum intake pressures to a compressor train. One of the most important aspects of modeling produc- tion systems is the correct calculation of fluid PVT prop- erties. Variable detail and quality often characterizes the PVT data available to the engineer, and ReO is designed to accommodate this. If complete compositional analysis has been performed, this can be used directly. If only Black Oil data are available, ReO will use a splitting technique to define a set of components to use in the compositional description. This approach means that dif- ferent fluids, with different levels of detailed description can be combined into the same base set of components. Where wells are producing fluids of different composition, the mixing of these fluids is accurately modeled in the system. The composition is reported at all the nodes in the network. This is highly valuable in fields with differing wells compositions. The facility models available in ReO for gas networks include pipeline, chokes (both variable and fixed diam- eter), block valves, standard compressors (polytropic model), heat exchangers (intercoolers), gas and gas con- densate wells, sinks (separators, gas export and delivery points, flares, or vents), manifolds, links (no pressure loss pipelines), and flanges (no flow constraint). Production constraints may be defined at any point within the production system in terms of pressure and/or flow rate along with objective functions for maximizing and minimizing flow rate or pressure in terms of sales revenues and costs. ReO is seamlessly integrated with the program WellFlo application. WellFlo may be run from within ReO and new well models may be defined or existing well models used to simulate inflow and tubing performance. The most complex application of ReO has been in Latin America where a network system including several hundred wells is optimized on a daily basis through a SCADA system. This system includes a low-pressure gas-gathering network integrated with a number of compressor trains and a high-pressure gas injection and distribution network. 18.8.4.2 HYSYS HYSYS is an integrated steady-state and dynamic process simulator (AspenTech, 2005). HYSYS creates simulation models for the following: . Plant design . Performance monitoring . Troubleshooting . Operational improvement . Business planning . Asset management HYSYS offers an integrated set of intuitive and interactive simulation and analysis tools and real-time applications. It provides rapid evaluations of safe and reliable designs through quick creation of interactive models for ‘‘what if’’ studies and sensitivity analysis. HYSYS Upstream is for handling petroleum fluids and RefSYS is for handling multiunit modeling and simulation of refinery systems. HYSYS interfaces with applications such as Microsoft Excel and Visual Basic and features ActiveX compliance. 18.8.4.3 FAST Piper FAST Piper (Fekete, 2001) is a gas pipeline, wellbore, and reservoir deliverability model that enables the user to op- timize both existing and proposed gas-gathering systems. FAST Piper is designed to be a ‘‘quick and simple looking tool’’ that can solve very complicated gathering system designs and operating scenarios. Developed and supported under Microsoft Windows 2000 and Windows XP, FAST Piper deals with critical issues such as multiphase flow, compressors, contracts, rate limitations, multiple wells, multiple pools, gas com- position tracking, among others. The Key Features FAST Piper include the following: . Allows matching of current production conditions . Analyzes ‘‘what-if’’ scenarios (additional wells, com- pression, contracts, etc.) . Integrated the coal bed methane (CBM) reservoir model allowing the user to predict the total gas and water production of an interconnected network of CBM wells, while incorporating compressor capacity curves, facility losses, and pipeline friction losses. 18.9 Discounted Revenue The economics of production optimization projects is evaluated on the basis of discounted revenue to be gener- ated by the projects. The most widely used method for calculating the discounted revenue is to predict the net present value (NPV) defined as NPV ¼ NPVR À cost, (18:47) where NPVR ¼ Xm n¼1 DRn 1 þ ið Þn , (18:48) where m is the remaining life of the system in years, and i is the discount rate. The annual incremental revenue after optimization is expressed as DRn ¼ $ð ÞDNp, n, (18:49) where ($) is oil or gas price and the DNp, n is the predicted annual incremental cumulative production for year n, which is expressed as DNp, n ¼ Nop p, n À Nno p, n, (18:50) where Nop p, n ¼ forcasted annual cumulative production of optimized system for year n Nno p, n ¼ predicted annual cumulative production of non-optimized well for year n. Summary This chapter presents principles of production optimization of well, facility, and field levels. While well- and facility- level optimization computations can be carried out using Nodal analysis approach, field-level computations fre- quently require simulators with simultaneous solvers. Pro- duction optimization is driven by production economics. References ahmed, t. Hydrocarbon Phase Behavior. Houston: Gulf Publishing Company, 1989. AspenTech. Aspen HYSYS. Aspen Technology, Inc., 2005. Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 279 4.1.2007 10:04pm Compositor Name: SJoearun PRODUCTION OPTIMIZATION 18/279
- 280. beggs, h.d. Production Optimization Using NODAL Ana- lysis, 2nd edition. Tulsa: OGCI, Inc., Petroskils, LLC., and H. Dale Beggs, 2003. brown, k.e. The Technology of Artificial Lift Methods, Vol. 2a. Tulsa, OK: Petroleum Publishing Co., 1980. edinburgh Petroleum Services. FloSystem User Documen- tation. Edinburgh: Edinburgh Petroleum Services, Ltd., 1997. E-Production Solutions. ReO User Documentation. Edinburgh: E-Production Solutions, 2004. Fekete Associates. Fekete Production Optimization. Fekete Associates, Inc., Calgary, Canada, 2001. guo, b. and ghalambor, a. Natural Gas Engineering Handbook. Houston, TX: Gulf Publishing Company, 2005. standing, m.b. A set of equations for computing equilib- rium ratios of a crude oil/natural gas system at pres- sures below 1,000 psia. J. Petroleum Technol. Trans. AIME 1979;31(Sept):1193. standing, m.b. Volume and Phase Behavior of Oil Field Hydrocarbon Systems, 9th edition. Dallas: Society of Petroleum Engineers, 1981. Problems 18.1 Analyze the dynamometer card shown in Figure 18.7 (scale ¼ 1: 1:5) assuming the following parameter values: S ¼ 40 in. N ¼ 20 spm C ¼ 12,500 lb=in. 18.2 Perform flash calculation under the following separ- ator conditions: Pressure: 500 psia Temperature: 150 8F Specific gravity of stock-tank oil: 0:85 (water ¼ 1) Specific gravity of solution gas: 0:65 (air ¼ 1) Gas solubility (Rs): 800 scf/stb 18.3 Consider a 6-in. pipeline that is 20 miles long. Assuming that the compression and delivery pres- sures will remain unchanged, calculate gas-capacity increases using the following measures of improve- ment: (a) replace 10 miles of the pipeline by a 8-in. pipeline segment; (b) place an 8-in. parallel pipeline to share gas transmission; and (c) loop 10 miles of the pipeline with an 8-in. pipeline segment. 18.4 The gas lift performance data of four oil wells are as follows: If a total lift gas injection rate of 12 MMscf/ day is available to the four wells, what lift gas flow rates should be assigned to each well? Gas Composition Compound Mole fraction C1 0.6899 C2 0.0969 C3 0.0591 i-C4 0.0439 n-C4 0.0378 i-C5 0.0157 n-C5 0.0112 C6 0.0081 C7þ 0.0101 N2 0.0094 CO2 0.0021 H2S 0.0058 Lift gas injection rate (MMscf/day) Oil production rate (stb/day) Well A Well B Well C Well D 0.6 80 740 870 600 1.2 145 1,250 1,450 1,145 1.8 180 1,670 1,800 1,180 2.4 210 1,830 2,100 1,210 3 235 1,840 2,350 1,235 3.6 250 1,845 2,500 1,250 4.2 255 1,847 2,550 1,255 4.8 259 1,845 2,590 1,259 5.4 260 1,780 2,600 1,260 6 255 1,670 2,550 1,255 Guo, Boyun / Computer Assited Petroleum Production Engg 0750682701_chap18 Final Proof page 280 4.1.2007 10:04pm Compositor Name: SJoearun 18/280 PRODUCTION ENHANCEMENT
- 281. Appendix A: Unit Conversion Factors Quantity U.S. Field unit To SI unit To U.S. Field unit SI unit Length (L) feet (ft) 0.3084 3.2808 meter (m) mile (mi) 1.609 0.6214 kilometer (km) inch (in.) 25.4 0.03937 millimeter (mm) Mass (M) ounce (oz) 28.3495 0.03527 gram (g) pound (lb) 0.4536 2.205 kilogram (kg) lbm 0.0311 32.17 slug Volume (V) gallon (gal) 0.003785 264.172 meter3 (m3 ) cu. ft. (ft3 ) 0.028317 35.3147 meter3 (m3 ) barrel (bbl) 0.15899 6.2898 meter3 (m3 ) Mcf (1,000 ft3 , 60 8F, 14:7 psia) 28.317 0.0353 Nm3 (15 8C, 101:325 kPa) sq. ft (ft2 ) 9:29 Â 10À2 10.764 meter2 (m2 ) Area (A) acre 4:0469 Â 103 2:471 Â 10À4 meter2 (m2 ) sq. mile 2.59 0.386 (km)2 Pressure (P) lb=in:2 (psi) 6.8948 0.145 kPa (1000 Pa) psi 0.0680 14.696 atm psi/ft 22.62 0.0442 kPa/m inch Hg 3:3864 Â 103 0:2953 Â 10À3 Pa Temperature (t) F 0.5556(F-32) 1.8Cþ32 C Rankine (8R) 0.5556 1.8 Kelvin (K) Energy/work (w) Btu 252.16 3:966 Â 10À3 cal Btu 1.0551 0.9478 kilojoule (kJ) ft-lbf 1.3558 0.73766 joule (J) hp-hr 0.7457 1.341 kW-hr Viscosity (m) cp 0.001 1,000 PaÁs lb/ftÁsec 1.4882 0.672 kg/(m-sec) or (PaÁs) lbf-s=ft2 479 0.0021 dyne-s=cm2 (poise) Thermal conductivity (k) Btu-ft=hr-ft2 -F 1.7307 0.578 W/(mÁK) Specific heat (Cp) Btu/(lbmÁ8F) 1 1 cal/(gÁ8C) Btu/(lbmÁ8F) 4:184 Â 103 2:39 Â 10À4 J.(kgÁK) Density (P) lbm=ft3 16.02 0.0624 kg=m3 Permeability (k) md 0.9862 1.0133 mD ( ¼ 10À15 m2 ) md ( ¼ 10À3 darcy) 9:8692 Â 10À16 1:0133 Â 1015 m2 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach 0750682701_appendix Final Proof page 282 5.1.2007 9:59pm Compositor Name: PDjeapradaban 282 APPENDICES
- 282. Appendix B: The Minimum Performance Properties of API Tubing Nom. (in.) O.D. (in.) Grade Wt per ft with couplings (lb) Inside diameter (in.) Drift diameter (in.) O.D. of upset (in.) O.D. of Cplg. (in.) Collapse resistance (psi) Internal yield pressure (psi) Joint yield strength (lb) Non-Upset Upset Non-Upset Upset Non-Upset Upset 3 ⁄4 1.050 F-25 1.20 0.824 0.730 1.315 1.660 5,960 4,710 8,320 H-40 1.20 0.824 0.730 1.315 1.660 7,680 7,530 13,300 J-55 1.20 0.824 0.730 1.315 1.660 10,560 10,360 18,290 C-75 1.14 1.20 0.824 0.730 1.315 1.313 1.660 14,410 14,120 11,920 24,950 N-80 1.20 0.824 0.730 1.315 1.660 15,370 15,070 26,610 1 1.315 F-25 1.80 1.049 0.955 1.469 1.900 5,540 4,430 12,350 H-40 1.80 1.049 0.955 1.469 1.900 7,270 7,080 19,760 J-55 1.80 1.049 0.955 1.469 1.900 10,000 9,730 27,160 C-75 1.70 1.80 1.049 0.955 1.469 1.660 1.900 13,640 13,270 20,540 37,040 N-80 1.80 1.049 0.955 1.469 1.900 14,650 14,160 39,510 11 ⁄4 1.660 F-25 2.40 1.380 1.286 1.812 2.200 4,400 3,690 16,710 H-40 2.40 1.380 1.286 1.812 2.200 6,180 5,910 26,740 J-55 2.40 1.380 1.286 1.812 2.200 8,490 8,120 36,770 C-75 2.30 2.40 1.380 1.286 1.812 2.054 2.200 11,580 11,070 29,120 50,140 N-80 2.40 1.380 1.286 1.812 2.200 12,360 11,800 53,480 11 ⁄2 1.900 F-25 2.75 2.90 1.610 1.516 2.094 2.200 2.500 3,920 3,340 11,930 19,900 H-40 2.75 2.90 1.610 1.516 2.094 2.200 2.500 5,640 5,350 19,090 31,980 J-55 2.75 2.90 1.610 1.516 2.094 2.200 2.500 7,750 7,350 26,250 43,970 C-75 2.75 2.90 1.610 1.516 2.094 2.200 2.500 10,570 10,020 35,800 59,960 N-80 2.75 2.90 1.610 1.516 2.094 2.200 2.500 11,280 10,680 38,180 63,960 2 2.375 F-25 4.00 2.041 1.947 2.875 3,530 3,080 18,830 F-25 4.60 4.70 1.995 1.901 2.594 2.875 3.063 4,160 3,500 22,480 32,600 H-40 4.00 2.041 1.947 2.875 5,230 4,930 30,130 H-40 4.60 4.70 1.995 1.901 2.594 2.875 3.063 5,890 5,600 35,960 52,170 J-55 4.00 2.041 1.947 2.875 7,190 6,770 41,430 J-55 4.60 4.70 1.995 1.901 2.594 2.875 3.063 8,100 7,700 49,440 71,730 C-75 4.00 2.041 1.947 2.875 9,520 9,230 56,500 C-75 4.60 4.70 1.995 1.901 2.594 2.875 3.063 11,040 10,500 67,430 97,820 C-75 5.80 5.95 1.867 1.773 2.594 2.875 3.063 14,330 14,040 96,560 126,940 N-80 4.00 2.041 1.947 2.875 9,980 9,840 60,260 N-80 4.60 4.70 1.995 1.901 2.594 2.875 3.063 11,780 11,200 71,920 104,340 N-80 5.80 5.95 1.867 1.773 2.594 2.875 3.063 15,280 14,970 102,980 135,400 P-105 4.60 4.70 1.995 1.901 2.594 2.875 3.063 15,460 14,700 94,400 136,940 P-105 5.80 5.95 1.867 1.773 2.594 2.875 3.063 20,060 19,650 135,170 177,710 21 ⁄2 2.875 F-25 6.40 6.50 2.441 2.347 3.094 3.500 3.668 3,870 3,300 32,990 45,300 H-40 6.40 6.50 2.441 2.347 3.094 3.500 3.668 5,580 5,280 52,780 72,480 J-55 6.40 6.50 2.441 2.347 3.094 3.500 3.668 7,680 7,260 72,570 99,660 C-75 6.40 6.50 2.441 2.347 3.094 3.500 3.668 10,470 9,910 98,970 135,900 C-75 8.60 8.70 2.259 2.165 3.094 3.500 3.668 14,350 14,060 149,360 186,290 (Continued ) Guo,Boyun/PetroleumProductionEngineering,AComputer-AssistedApproach0750682701_appendixFinalProofpage2835.1.20079:59pmCompositorName:PDjeapradaban APPENDICES283
- 283. Appendix B: (Continued ) Nom. (in.) O.D. (in.) Grade Wt per ft with couplings (lb) Inside diameter (in.) Drift diameter (in.) O.D. of upset (in.) O.D. of Cplg. (in.) Collapse resistance (psi) Internal yield pressure (psi) Joint yield strength (lb) Non-Upset Upset Non-Upset Upset Non-Upset Upset N-80 6.40 6.50 2.441 2.347 3.094 3.500 3.668 11,160 10,570 105,560 144,960 N-80 8.60 8.70 2.259 2.165 3.094 3.500 3.668 15,300 15,000 159,310 198,710 P-105 6.40 6.50 2.441 2.347 3.094 3.500 3.668 14,010 13,870 138,550 190,260 P-105 8.60 8.70 2.259 2.165 3.094 3.500 3.668 20,090 19,690 209,100 260,810 3 3.500 F-25 7.70 3.068 2.943 4.250 2,970 2,700 40,670 F-25 9.20 9.3 2.992 2.867 3.750 4.250 4.500 3,680 3,180 49,710 64,760 F-25 10.20 2.922 2.797 4.250 4,330 3,610 57,840 H-40 7.70 3.068 2.943 4.250 4,630 4,320 65,070 H-40 9.20 9.3 2.992 2.867 3.750 4.250 4.500 5,380 5,080 79,540 103,610 H-40 10.20 2.922 2.797 4.250 6,060 5,780 92,550 J-55 7.70 3.068 2.943 4.250 5,970 5,940 89,470 J-55 9.20 9.3 2.992 2.867 3.750 4.250 4.500 7,400 6,980 109,370 142,460 J-55 10.20 2.922 2.797 4.250 8,330 7,940 127,250 C-75 7.70 3.068 2.943 4.250 7,540 8,100 122,010 C-75 9.20 9.3 2.992 2.867 3.750 4.250 4.500 10,040 9,520 149,140 194,260 C-75 10.20 2.922 2.797 4.250 11,360 10,840 173,530 C-75 12.70 12.95 2.750 2.625 3.750 4.250 4.500 14,350 14,060 230,990 276,120 N-80 7.70 3.068 2.943 4.250 7,870 8,640 130,140 N-80 9.20 9.3 2.992 2.867 3.750 4.250 4.500 10,530 10,160 159,080 207,220 N-80 10.20 2.922 2.797 4.250 12,120 11,560 185,100 N-80 12.70 12.95 2.750 2.625 3.750 4.250 4.500 15,310 15,000 246,390 294,530 P-105 9.20 9.3 2.992 2.867 3.750 4.250 4.500 13,050 13,340 208,790 271,970 P-105 12.70 12.95 2.750 2.625 3.750 4.250 4.500 20,090 19,690 323,390 386,570 31 ⁄2 4.000 F-25 9.50 3.548 3.423 4.750 2,630 2,470 15,000 F-25 11.00 3.476 3.351 4.250 5.000 3,220 2,870 76,920 H-40 9.50 3.548 3.423 4.750 4,060 3,960 72,000 H-40 11.00 3.476 3.351 4.250 5.000 4,900 4,580 123,070 J-55 9.50 3.548 3.423 4.750 5,110 5,440 99,010 J-55 11.00 3.476 3.351 4.250 5.000 6,590 6,300 169,220 C-75 9.50 3.548 3.423 4.750 6,350 7,420 135,010 C-75 11.00 3.476 3.351 4.250 5.000 8,410 8,600 230,760 N-80 9.50 3.548 3.423 4.750 6,590 7,910 144,010 N-80 11.00 3.476 3.351 4.250 5.000 8,800 9,170 246,140 4 4.500 F-25 12.60 12.75 3.958 3.833 4.750 5.200 5.563 2,870 2,630 65,230 90,010 H-40 12.60 12.75 3.958 3.833 4.750 5.200 5.563 4,500 4,220 104,360 144,020 J-55 12.60 12.75 3.958 3.833 4.750 5.200 5.563 5,720 5,790 143,500 198,030 C-75 12.60 12.75 3.958 3.833 4.750 5.200 5.563 7,200 7,900 195,680 270,030 N-80 12.60 12.75 3.958 3.833 4.750 5.200 5.563 7,500 8,440 208,730 288,040 Guo,Boyun/PetroleumProductionEngineering,AComputer-AssistedApproach0750682701_appendixFinalProofpage2845.1.20079:59pmCompositorName:PDjeapradaban 284APPENDICES
- 284. Index A Acid, 10/129, 16/244–249 volume, 16/245–249 Acidizing, 16/243 design,16/243–244, 16/247–248 models, 16/248 Acidmineral reaction, 16/244 kinetics, 16/244, 16/247 stoichiometry, 16/244 American Gas Association, 11/157 American Petroleum Institute, 1/6, 1/17, 2/26 9/110, 12/163, 18/272 Annular flow, 4/46, 4/48, 14/216, 15/232 API gravity, 2/20, 2/26, 5/65, 11/145, 11/157, 12/170, 12/179–180, 18/272, 18/274 Artificial lift, 3/30, 4/57, 5/66, 12/159, 12/162, 12/164, 12/179, 13/182, 13/184, 13/205, 14/207, 14/208, 14/209, 14/216, 14/222, 18/279 method, 3/30, 4/57, 5/66, 12/159, 13/182–206, 14/207, 14/208–223, 18/279 B Boiling point, 18/271 Buckling, 9/110, 9/112–115, 11/151 Buoyancy, 9/111–112, 12/179, 14/215 C Capacity, 1/11, 10/121–122, 10/124, 10/126–131, 11/137, 11/142–143, 11/148, 11/153, 12/162, 13/188–189, 13/191–192, 13/203, 14/210–211, 14/216, 14/219–220, 15/228, 15/231, 18/272–280 Carbon dioxide, 2/23, 2/25, 10/118, 11/151 Carbonate acidizing, 16/243–244, 16/247–248 design, 16/243–244, 16/247 Casings, 1/5 Cavitation, 14/209, 14/220 Centrifugal, 1/10–11, 10/118–119, 11/137, 11/142–143, 11/157, 13/188, 13/190–193, 13/206, 14/208–209 efficiency, 10/120, 10/125, 10/127, 10/129, 11/135, 11/137, 11/139–140, 11/142–143, 11/148, 11/150, 11/152–153, 11/157, 12/162, 12/169–170, 12/172–174, 12/177, 12/180, 13/183, 13/188–192, 13/206 volumetric, 1/5, 1/10, 2/20, 4/48, 4/52, 5/65, 7/88, 7/92, 8/98, 11/135–137, 11/139–141, 11/148, 12/159, 12/170, 12/172–174, 12/177, 12/180, 13/188–189, 14/209, 14/211, 15/229, 16/244, 16/246–249, 18/269 horsepower, 11/135–137, 11/139–140, 11/142–143, 11/157, 12/173, 12/173, 12/177, 13/189–192, 13/205–206, 14/208–210, 14/213, 17/262, 18/270, 18/276 actual, 3/42, 5/65, 11/140, 11/142, 11/145–148, 11/152, 12/168, 13/190–191, 13/196, 13/201, 14/218, 15/228, 18/269–270 brake, 11/136, 11/157, 12/173, 13/190, 13/206 isentropic, 5/60, 5/62, 5/64, 11/138, 11/142, 13/187, 13/189–191 Channeling, 15/231, 15/234, 15/236–237 Chokes, 1/5, 1/7, 1/17, 5/60–62, 5/64–66, 13/182, 13/187, 18/270, 18/279 Coating, 4/48, 11/148–149, 11/153 Collapse, 4/48, 9/110 –112, 9/114, 11/150–151, 11/157 Completion, 3/30, 9/111–112, 9/115, 12/162, 12/177, 14/208, 14/211, 15/228, 15/230, 16/244, 17/264 Compressibility, 2/20–23, 2/25–27, 3/30, 3/33–35, 3/43, 4/50, 4/53–56, 4/58, 5/66, 6/82–84, 6/86, 7/88–89, 7/93, 7/95–96, 8/98, 10/121, 11/142–143, 11/146, 13/186, 13/190–192, 13/200, 14/219, 15/229, 15/231 Compressor, 1/3–4, 1/10–11, 10/118, 10/126, 11/133–134, 11/136–140, 11/142–143, 11/146, 11/156–157, 13/182–183, 13/185, 13/187–193, 13/197, 13/205–206, 18/268, 18/276–277, 18/279 Conductivity, 11/152–153, 15/229, 15/238, 17/256–258, 17/262–264 Corrosion, 1/12, 10/126, 10/129, 11/148–149, 11/151–152, 11/157 Critical point, 1/5 Cylinders, 12/162, 13/189 D Damage characterization Decline curve analysis, 14/218 constant fractional decline, 8/98 harmonic decline, 8/98, 8/100–103 hyperbolic decline, 8/98, 8/100–101, 8/103 Dehydration, 10/117, 10/118, 10/121, 10/125–129, 10/132 cooling, 10/125–126, 10/128 glycol, 10/126–132, stripping still, 10/127–128, 10/131–132 Density of gas, 2/24, 10/121 Dewatering, 14/214 Downhole, 12/162, 12/179 Drilling, 1/6, 4/57, 5/66, 10/127, 11/157, 14/216, 14/223, 15/230, 16/244 mud, 10/127 Drums, 11/139 Drying, 10/126 Dynamometer cards, 12/174, 12/177, 18/270 E Economics, 1/4, 7/88, 14/219, 18/268–269, 18/279 Enthalpy, 11/157, 13/189 Entropy, 11/157, 13/189 Equation of state, 2/26, 18/217 Exploration well, 3/39 F Fittings, 1/7, 5/66 Flow metering, 5/66 Flow efficiency, 10/125, 11/153 Flow regime, 3/30–31, 3/42, 4/48–49, 4/51, 4/53, 5/60, 5/63–65, 7/88, 11/144, 14/216, 15/229–230, 15/232, 15/235, 15/240, 16/244, 17/264 Flowline, 1/4, 1/7, 1/11, 1/13, 1/15, 5/63–64, 5/67, 6/75–76, 6/85, 10/124, 10/132, 11/143, 11/150–151, 11/153, 13/183, 14/217, 14/218, 18/276 Fluid, 1/1, 1/4–5, 1/7–8, 1/11–12, 2/20, 2/22, 2/26, 3/30, 3/33–35, 3/37, 3/40, 3/43, 4/46–48, 4/51, 4/57, 5/60, 5/63–64, 5/66, 6/70, 6/72, 6/79, 6/82, 6/84, 7/88–92, 7/94, 9/110–115, 10/118, 10/120, 10/132, 11/134–136, 11/138, 11/143–146, 11/150–154, 11/156–159, 12/162, 12/164–165, 12–169–173, 12/177–179, 13/182–184, 13/186, 13/192–193, 13/196–206, 14/208–215, 14/219–223, 15/228–231, 15/234, 15/236, 15/241, 16/244, 16/247, 17/252–256, 17/258–266, 18/269–272, 18/276–279 loss, 17/258 volume, 7/89, 14/214, 17/258, 17/260–261, 17/265 Formation damage, 15/228, 16/244–245, 17/252 Formation volume factor, 2/20, 2/22, 2/25–26, 3/30, 3/33–35, 3/40, 3/43, 4/50–51, 6/72, 6/76, 6/78, 6/85–86, 7/88, 7/93, 7/95, 12/170, 12/173, 12/179, 14/210, 14/211–213, 14/223, 25/339 Forming, 1/7, 6/74, 6/84, 17/262, 17/264 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Index Final Proof page 285 28.12.2006 10:57am
- 285. Fracture direction, 17/258, 17/264 Friction, 1/6, 1/11, 4/46–48, 4/50–51, 4/53, 4/55, 4/57, 5/60, 6/72, 6/81, 11/137, 11/142–148, 11/157, 12/162, 12/168–169, 12/172–173, 13/183–188, 13/191, 14/210, 14/212–213, 14/216, 14/218–219, 15/240, 16/246, 17/259–160, 18/268–270, 18/279 factor, 4/46–57, 6/72, 6/81, 11/144–148, 11/157, 13/184, 15/240, 16/246, 17/259, 18/270 pressure drop, 4/57, 16/246, 17/259–260, 18/268 G Gas, 2/22–27, 4/50, 6/82–86 compressibility, 2/22–27, 4/50, 6/82–86, 10/121, 11/142–143, 13/186, 13/191–192, 13/200, 14/219, 13/200, 14/219 compressors, 10/126, 11/156 condensate, 1/4, 4/56, 5/64–66, 10/120, 10/124, 15/228, 18/272, 18/279 flow, 1/4, 1/9–10, 3/42, 4/48, 4/53, 5/60–67, 6/70, 6/72–75, 6/82, 10/118, 10/121, 10/126–132, 11/136–137, 11/139–140, 11/142–148, 11/157, 13/182, 13/185, 13/187–188, 13/190–196, 13/205–206, 14/216, 14/219, 15/231–232, 15/234–235, 15/237, 15/240, 18/268, 18/274–275, 18/280 formation volume factor, 2/22, 2/25–26 gravity, 11/146, 11/148, 13/201, 13/206 injection rate, 13/183–187, 18/268–269, 18/275–276, 18/278, 18/280 lift, 1/10, 4/46, 5/66, 12/159, 13/181–185, 13/187, 13/189, 13/191–195, 13/197–201, 13/203–206, 14/208, 14/216, 14/219, 14/223, 15/232, 18/268–269, 18/275–278, 18/280 lines, 11/149–151 pipelines, 10/125, 11/143, 11/149, 11/152, 18/272 transmission, 11/147, 11/157, 18/280 viscosity, 2/21, 2/23–24, 2/26, 4/52, 5/61, 6/75–76, 6/82–83, 6/85–86, 7/93, 11/144, 13/187, 13/200, 13/206 well deliverability, 6/71, 6/82–83 well performance, 6/85 Gathering lines, 11/150–151 GOR, 6/82–84, 6/86, 10/118, 10/125, 11/143, 11/148, 11/153, 13/182, 13/185, 13/188–189, 13/203, 14/210, 14/213, 14/219, 14/223, 16/246, 17/256, 17/265, 18/268, 18/272, 18/274, 18/280 Gravel pack, 9/112 H Harmonic decline, 8/97, 8/98, 8/100–103 HCl preflush, 16/244, 16/245, 16/249 Head, 1/4–1/10, 1/13, 1/15, 1/17, 3/30, 3/37, 4/46, 4/50–55, 4/57–58, 5/60, 5/63–64, 5/66–67, 6/70–82, 6/84–86, 7/88, 7/93, 7/95–96, 10/120, 10/127, 10/129, 11/136, 11/142–143, 11/148, 11/150–151, 11/157–158, 2/162, 12/168, 12/172–173, 12/177, 12/180, 13/184–185, 13/187–188, 13/191–192, 13/198–199, 3/201–202, 13/205–206, 14/208–216, 14/218–221, 14/223, 15/232, 15/237, 15/240–242, 16/245, 16/249 rating, 14/214–215 Heavy oil, 14/213 Hoop stress, 11/150–152 Horizontal well, 3/31–33, 3/42, 9/111, 14/214, 15/229–230, 16/248, 17/264–265 Hydrates, 1/7, 5/60, 5/62, 10/125–126, 11/152, 15/228 Hydraulically fractured well, 15/229 Hydraulic fracturing, 9/112, 15/225, 15/231, 17/251–255, 17/257–261, 17/263–265 Hydraulic piston pumping, 12/159, 14/207–208, 14/211, 14/222 Hyperbolic decline, 8/97–98, 8/100–101, 8/103 Hydrogen sulfide, 2/23, 10/118, 10/126, 10/129, 11/151 Hydrostatic pressure, 11/149–151, 13/199, 14/212, 14/217, 14/219–220, 16/246, 17/259–260 I Inflow performance relationship (IPR), 3/29–30, 3/32, 3/42–43, 14/210 Injected gas, 13/183 Insulation, 11/148, 11/152–154, 11/156, 11/158, 14/209 Interest, 2/20, 11/145 IPR curve, 3/29, 3/32–43, 7/88, 7/92,13/183, 14/218 J Jet pumping, 12/159, 14/207–208, 14/220, 14/222 Jet pumps, 14/220–221 K KGD model, 17/254, 17/262 L Laminar flow, 4/46–47, 9/112, 11/144 Leak, 1/7, 1/12, 10/129, 12/177, 12/179, 13/185, 13/189, 14/214, 15/231, 15/234, 15/237, 17/255, 17/258, 17/260, 17/263, 17/265 Lifting, 12/172, 13/182, 13/184, 14/208, 14/213, 14/218, 14/222, 15/232 Line pipe, 1/11, 11/145 Line size, 1/11 Liquid holdup, 4/48, 4/51–53 Liquid phase, 4/48, 4/50, 4/52, 10/120, 10/124 M Maintenance, 1/5, 1/11, 7/95, 10/127, 11/137, 13/183, 13/188 Manifolds, 1/11, 1/11, 11/143, 13/187, 13/205, 13/206 Matrix acidizing, 16/224, 16/247, 16/248 Measurement, 1/7, 2/20, 2/23, 4/48, 4/50, 4/57, 5/66, 7/94, 17/263 Meters, 14/210 Methane, 14/214 Mist flow, 4/56, 14/216, 15/232, 15/235, 15/238, 15/239 Molecular weight, 2/22, 2/24, 2/26–27, 11/143, 13/192, 16/244 Mole fraction, 2/22, 2/24–27, 4/51, 6/72, 6/85 Mollier diagram, 13/189, 13/190 Motor, 10/128, 11/137, 11/140, 11/142, 12/162, 12/174, 13/189, 4/208–211, 14/215, 14/217 Multilateral, 3/37, 6/79, 6/80– 6/85 Multiphase flow 3/41, 4/45–46, 4/48, 4/53, 4/57, 5/59, 5/63, 5/66, 13/184, 14/216, 15/234–235 3/41, 4/46, 4/48, 4/53, 4/57, 5/63, 13/184, 14/216, 15/234, 15/235 Multiphase fluid, 4/46, 5/64, 11/145 N Natural gas, 1/10–11, 2/20–26, 4/53, 4/57, 5/60, 5/62, 5/67, 10/125–128, 10/132, 11/134, 11/136–137, 11/140–142, 11/144, 11/146, 11/148, 11/152, 11/157, 13/186 –187, 13/189–200, 13/205–206, 15/231 composition of, 10/118, 15/230 water content, 10/125–126, 10/129–130, 10/132 Nodal analysis, 6/70–71, 6/74–75, 6/77, 6/80, 6/84, 7/88–90, 7/92–94, 15/228, 17/262, 17/264 O Offshore, 1/9, 1/11, 9/114, 10/118, 10/125, 11/148, 11/151–152, 12/162, 13/182, 14/208–14/209, 14/211 Operations, 1/11, 12/162, 13/182, 14/208, 14/211 Oil properties, 2/20, 5/66 Operating costs, 10/127, 14/214 Operating pressure, 6/71, 6/76, 10/120, 10/122, 10/124, 10/128–132, 11/147, 11/150, 13/190, 13/192, 13/197–199, 14/212–213, 14/222, 17/258 Operators, 10/120, 10/132, 11/140, 11/150 Orifice, 5/61–62, 5/64, 5/67, 13/182, 13/184–185, 13/187, 13/192, 13/194, 13/198, 13/200 Charts, 2/21–23, 4/52, 4/54, 5/64, 6/74, 6/78, 6/80, 10/121, 11/145, 13/187, 14/221 expansion factor, 2/26 Outflow performance curve, 6/70, 6/74–76, 13/183, 6/185 P Panhandle equation 11/146 Packer, 1/5–7, 9/112–115, 13/183, 13/184, 13/203–205, 14/218 Paraffin, 1/7, 14/209, 14/215, 15/228 Parametric study, 17/256, 17/258 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Index Final Proof page 286 28.12.2006 10:57am 286 INDEX
- 286. Pay thickness, 6/82, 6/82, 6/86 Perforating, 16/247 Performance curve, 5/60, 6/70, 6/74–76, 13/183–185, 14/218, 14/222, 18/268, 18/269, 18/275, 18/276, 18/278 Permeability, 3/30, 3/32–35, 3/37, 3/39–41, 3/43, 6/82, 6/83, 6/86, 7/88, 7/93, 7/95, 15/228–231, 15/235, 15/237, 16/244–246, 16/248, 17/252, 17/256, 17/257, 17/259, 17/262, 17/264, 17/265, 18/268 Petroleum hydrocarbons, 1/4, 1/11, 10/126, 10/127, 10/129, 10/131, 11/157 Phase behavior, 10/120 Pipeline, 1/4, 1/9–1/11, 1/16–1/17, 2/20, 9/114, 10/120, 10/125, 10/126, 11/133, 11/134, 11/136, 11/143–158, 18/267, 18/268, 18/270, 18/272–277, 18/278, 18/280 PKN model, 17/255, 17/260, 17/261 Platform, 1/9, 1/11, 1/15, 1/17, 10/188, 10/132, 11/150, 11/151, 18/268 Polymer, 11/152 Pore space, 3/34, 15/230, 17/253 Porosity, 3/30, 3/33–3/35, 3/43, 7/88, 7/93, 7/95, 7/96, 15/229, 15/230, 16/245–249, 17/260, 17/265 Positive-displacement pump, 14/209 Precipitation, 16/246 Preflush/postflush, 16/244, 16/245, 16/249 Pressure, 1/5, 1/7 drop, 4/46, 4/48, 4/53, 4/57, 4/60, 5/66, 7/88–90, 7/95, 10/128, 11/114, 12/172, 13/183, 13/197, 16/246, 17/259, 17/260, 18/268, 18/270 gauge, 1/6–1/8, 5/60, 10/118 traverse, 4/46, 4/48, 4/53, 4/55, 4/58, 13/198 Processing plant, 1/1 Produced water, 4/50, 15/240 Production, 1/11 facility, 1/11, 11/143, 18/278 injection well, 3/31 logging, 15/228, 15/231, 15/241, 17/262, 17/264 Progressing cavity pump (PCP), 14/213–215, 14/241 Proppant, 17/252, 17/258–265 Pseudo–steady–state flow, 3/31–3/34, 3/37/, 3/43, 7/87, 7/88, 7/90, 7/92, 7/94, 7/95, 8/98, 16/246 Pump, 1/4, 1/9–10, 1/16–17, 10/128, 10/129, 10/131, 11/133–136, 11/156, 12/162, 12/164, 12/165, 12/172, 14/208, 14/209, 14/220, 14/221, 17/252, 18/268 intake pressure, 14/212–214, 14/222 Jet, 12/159, 14/207–208, 14/220–222 PVT, 2/20–2/21, 2/23, 2/26–2/27, 7/94, 18/279 R Range, 1/10–11, 2/20, 2/26–27, 3/36, 6/70, 6/73, 6/75, 6/78, 10/125, 11/136, 11/142, 11/144, 11/145, 11/147, 12/126, 12/172, 13/188–191, 13/196, 14/209, 14/210, 14/220, 14/223, 16/246, 17/257, 17/258, 18/269, 18/270 Real gas, 2/23–24, 2/26, 3/30–31, 7/94, 10/121, 11/142, 11/143, 11/147, 13/189 Regulation, 5/60 Relative permeability, 3/34, 3/39–41, 3/43, Relative roughness, 4/46, 4/47, 4/55, 4/57, 4/58, 6/70, 6/71, 6/75, 6/76, 6/85, 11/144–147 Reserves, 2/26, 15/228 Reservoir, 1/4–7, 1/17, 2/20–21, 2/25, 3/30–43, 4/46, 4/55, 4/58, 5/60, 6/70–86, 7/88–95, 8/98, 9/110, 11/153, 11/157, 13/182–184, 13/186, 13/192, 13/199, 13/201, 14/208, 14/210–213, 14/216, 14/218–220, 14/222, 14/223, 15/228–231, 15/237, 15/241, 15/242, 16/244, 16/246–249, 17/252, 17/253, 17/256–260, 17/263–265, 18/268, 18/278, 18/279 engineering, 2/20, 3/42, 7/94, 15/241 hydrocarbons, 1/4, thickness, 3/30 Reynolds number, 4/46, 4/47, 4/50, 4/53, 4/55, 5/60–62, 6/81, 11/144–148, 11/157, 13/187 Riser, 1/11, 11/151, 11/152, 11/157 S Safety, 1/3, 1/7, 1/11–13, 1/15–16, 9/111, 10/118, 11/150–151, 12/170, 12/172–174, 13/185, 13/187–188, 14/215, 14/222, 16/246–247, 17/258 Sandstone, 16/243–248, 17/253, 17/260, 17/262 Acidizing, 16/243–248 Saturated oil, 1/4, 3/34, 3/36, 3/38, 7/88 Saturations, 3/37, 7/89 Scales, 11/148, 15/228 Separator, 1/3–4, 1/8–9, 10/118–129, 10/131–132, 13/183, 14/208, 18/267–268, 18/270–272, 18/279 SI units, 2/21, 2/24–25, 4/51–52, 4/54, 6/72–74, 6/78–80, 13/188, 16/247 Single-phase flow, 3/31, 3/34, 3/37, 7/88–89, 14/212, 14/222 Slug, 1/9, 4/48–49, 10/118, 10/125, 10/127, 13/192, 13/201–203, 14/216–220, 15/232 Specific gravity, 2/20–26, 4/47, 4/49–55, 5/61–65, 6/70–80, 6/82–85, 7/88–89, 7/93, 10/122, 10/128–131, 11/143–146, 11/148, 12/168, 12/172–173, 13/187–188, 13/190–191, 13/193, 13/200, 14/210, 14/212, 14/216, 14/222, 15/234, 15/240–241, 16/247–248, 17/260, 18/271–272, 18/274 Spread, 13/193–195 Stability, 11/148–149, 11/153 Stabilization, 11/149 Steady-state, 7/90, 7/92, 7/94, 8/98, 11/146, 11/152, 13/182, 13/192, 16/246, 18/279 Storage, 1/16, 10/120, 10/124, 11/136, 11/151, 13/204, 17/255, 17/263 Stress, 9/110–112, 11/149–152, 12/165, 12/170, 12/173–174, 12/177–179, 14/215, 17/252–256, 17/259–260, 17/262, 17/264 Subsea, 1/11, 11/152 Sucker rod pumping, 12/159, 12/161–163, 12/165, 12/167, 12/169, 12/171, 12/173–175, 12/177, 12/179 Surface equipment, 1/6–7, 5/60, 12/174 System analysis, 6/70, 6/84, 13/184–186, 18/276 T Temperature, 1/4–5, 1/12, 2/20–25, 3/30, 4/50–56, 5/60–65, 6/70–80, 6/82–84, 7/88, 7/92–93, 9/112–114, 10/120–122, 10/124–127, 10/129–130, 10/132, 11/138–140, 11/142–144, 11/146, 11/148, 11/150, 11/152–156, 12/162, 13/186–193, 13/200–201, 13/203, 14/208–210, 14/213, 14/215–216, 14/219–220, 15/228, 15/231, 15/234, 15/236, 15/238, 15/240–241, 16/244, 16/248, 17/258, 18/270–272, 18/276, 18/278 Thermal conductivity, 11/152–153, 15/238 Thermodynamic, 4/46, 4/53 Torque, 12/162–163, 12/166–171, 12/173, 12/177, 12/179, 14/214–215 Transient flow, 3/30–31, 3/33, 3/39, 7/87–88, 7/90, 7/92–93, 7/95, 11/153 Transportation, 1/4, 1/9, 1/11, 9/107, 10/118, 11/133–137, 11/139, 11/141, 11/143, 11/145, 11/147, 11/149, 11/151, 11/153, 11/155–156, 18/276 Transmission lines, 11/136, 11/147 Tubing movement, 9/114 Turbulent flow, 4/46–48, 4/55, 6/81, 9/112, 11/144–146 Two-phase flow, 1/5, 3/32, 3/34, 3/38, 3/41, 3/48, 5/63–64, 5/66, 7/88, 7/90–91, 7/94, 11/149, 11/151, 13/198 Two-phase reservoirs, 3/34–35, 3/39 U Unsaturated oil, 3/35 Undersaturated oil, 1/4, 3/33, 3/35–37, 7/88–89 Reservoirs, 3/33, 7/88 Units, 1/17, 2/21, 2/24–25, 4/46, 4/51–54, 4/56, 5/60, 6/72–74, 6/78–81, 8/99, 10/118, 10/121, 10/128, 10/132, 11/140, 11/144–146, 11/150–151, 12/162–163, 12/165, 12/168–169, 12/171, 12/174–175, 13/188, 14/208–209, 15/229, 15/233, 16/247, 18/270–271 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Index Final Proof page 287 28.12.2006 10:57am INDEX 287
- 287. V Valves, 1/6–8, 1/10, 10/119–120, 10/126, 11/136, 13/181–183, 13/188–201, 13/203–205, 18/279 Velocity, 1/8, 1/11, 4/46–47, 4/49, 4/52–53, 5/60–63, 10/121, 10/126, 10/137–138, 11/144–145, 11/154, 12/165, 13/188, 13/203, 14/216, 14/218–220, 15/231–237, 15/240, 16/248 Vertical lift performance (VLP), 4/46, 13/183 Viscosity, 2/20–21, 2/23–24, 2/26, 3/30, 3/33–35, 3/39–40, 4/46, 4/48, 4/52–54, 4/58, 5/61, 6/74–76, 6/78, 6/80, 6/82–83, 7/88, 7/93, 7/95, 10/127, 11/144, 11/147, 12/172, 13/187, 13/200, 14/209–210, 14/212–213, 14/215, 15/229, 15/231, 16/246–247, 17/254–255, 17/258–260 W Wall thickness, 1/11, 9/110, 11/148–152, 11/154 Water, 1/4 coning, 15/230–231, 15/238 flow, 13/183, 15/237 production, 4/50–52, 4/57, 9/112, 13/202, 15/227–228, 15/231, 15/237–238, 15/240, 18/279 Well, 1/5–9, 2/20–21, 3/30–43, 4/46, 4/55, 5/60, 6/69–71, 7/88–90, 8/98–99, 8/101, 9/107, 10/118–122, 11/153, 12/159, 13/182–185, 14/208–214, 15/225–235, 16/244, 17/252, 18/267–269 deliverability, 1/1, 3/30, 6/69–71, 6/73, 6/76–77, 6/79, 6/81–84, 13/205, 15/228–229, 17/264, 18/268 operation, 14/219 productivity, 1/4, 3/42, 12/172, 15/228–230, 16/244, 17/257–258, 17/264, test, 15/241, 17/263 Wellbore flow, 4/46, 6/70 Wellhead, 1/4–7, 1/9, 1/13, 1/15, 3/30, 3/37, 4/46, 4/50–51, 5/60, 5/63–64, 6/70–72, 6/74–82, 6/84, 7/93, 10/127, 11/150–151, 12/173, 13/184–185, 13/198–199, 13/201, 13/205, 14/210, 14/212–214, 14/216, 14/219, 15/232, 15/237, 15/240–241, 18/268, 18/270, 18/279 Weymouth equation, 11/146–148, 13/187, 18/272–273 Wormhole, 16/247–248 Y Yield stress, 9/110–112 Z Z-factor, 2/23–25, 5/65, 7/92–93, 11/146, 13/188, 14/220 Guo, Boyun / Petroleum Production Engineering, A Computer-Assisted Approach Index Final Proof page 288 28.12.2006 10:57am 288 INDEX