Class lecture on Hydrology by Rabindra Ranjan saha Lecture 13

Class lecture on Hydrology by Rabindra Ranjan saha Lecture 13

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  1 Lecture 13 Flood routing Floodrouting is the technique of determining the flood hydrograph sections. The hydrologic analysis of problems such as flood forecasting, flood protection, reservoir design and spillway design invariably include flood routing. Types of Flood Routing Broadly: 1. Reservoir routing Two types 2.Channel routing
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  2 Lecture 13 (contd.) 1)Reservoir routing The effect of a flood wave entering a reservoir is studied in reservoir routing to predict the variations of reservoir elevation and outflow discharge with time. This form of reservoir routing is essential  in the design of the capacity of spillways and other reservoir outlet structures  and also in the location and sizing of the capacity of reservoir to meet specific requirements. Reservoir Routing is done, knowing the volume – elevation characteristic of the reservoir and the outflow-elevation relationship for the spillways and other outlet structures in the reservoir.
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  3 Lecture 13 (contd.) 2)Channel routing The change in the shape of a hydrograph as it travels down a channel is studied by channel routing, by considering a channel reach and an output hydrograph at the upstream end. This form of routing aims to predict the flood hydrograph at various sections of the reach. Information on the flood-peak and duration of high- water levels obtained by channel routing is of utmost importance in flood-forecasting operations and flood- protection works.
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  4 Broadly classification ofchannel routing methods 1. Hydrologic routing : hydrologic –routing methods employ essentially the equation of continuity 2. Hydraulic routing method: On the other hand hydraulic routing method employ the continuity equation together with the equation of motion of unsteady flow. The basic equations used in the hydraulic routing, known as St. Vansant equations. Two categories1.Hydrolgic routing 2.Hydraulic routing Lecture 13 (contd.)
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  5 Basic equations The passageof a flood hydrograph through a reservoir or a channel reach is an unsteady –flow. The equation of continuity used in all hydrologic routing as the primary equation states that the difference between the inflow and outflow rate is equal to the rate of change of storage, i.e. I – Q = dS/dt (9-25) where, I = Inflow rate, Q = Out flow rate and S = Storage. For a time interval t the difference between the total inflow volume and total out flow volume in a reach is equal to the change in storage in that reach (From Eq-9-25) I t – Q t = S (9-26) Lecture 13 (contd.)
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  6 Where, I =average inflow in time t , Q = average outflow in time t and S = change in storage. By taking I = (I1 + I2 ) /2 and Q = (Q1 + Q2) /2 and S = S2 – S1 with suffixes 1 and 2 denote the beginning and end of time interval t then the Eq(9-26) is written as (I1 + I2 ) t /2 - (Q1 + Q2) t /2 = S2 – S1 (9-27) Lecture 13 (contd.)
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  7 1.Hydrologic Channel Routing Thetotal storage in the channel reach can be expressed as S = K [ x Im + ( I - x) Qm ] (9-28) where, K and x are the coefficients and m = a constant exponent. The values m varies from 0.6 for rectangular channels to a value of about 1.0 for natural channels. Lecture 13 (contd.)
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  8 Muskingum Equation Using m= 1.0 in Equation (9-28) for a linear relationship for S in terms of I and Q as S = K [ x I + (1- x) Q] (9-29) this relationship is known as the Muskingum equation . Lecture 13 (contd.)
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  AccumulatedstorageS Discharge t Mean Storage Attenuation Lag S Time Accumulation Storage Release fromStorage S Time 0 Figure 9-2 : Hydrograph and storage in Channel routing In Equation(9-29) i.e. S = K [ x I + (I- x) Q] the parameter x is known as weighing factor and takes a value between 0 and 0.5. When x = 0, obviously the storage S is a function of discharge only. The Eq (9 -29) reduces to S = KQ (9-30) Lecture 13 (contd.)
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  10 Such a storageis known as linear storage or linear reservoir. When x = 0, both inflow and outflow are important in determining the storage. The coefficient K is known as storage-time constant and has the dimensions of time. It is approximately equal to the time of travel of a flood wave through the channel reach. Lecture-13
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  11 Estimation K andx The Figure 9-2 shows a typical inflow and outflow hydrograph through a reach. Note that the outflow peak does not occur at the point of intersection of the inflow and outflow hydrographs . Now from the continuity Eq 9-27 we get the increment in storage (S) at any time t and time element t can be calculated: (I1 + I2) t /2 - (Q1+ Q2) t /2 = S Summation of the various incremental storage values can also be calculated from the S vs time relationship (Figure 9-2). Lecture 13 (contd.)
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  12 If an inflowand outflow hydrograph set is available for a given reach, values of S at various time intervals can be determined by the above technique. By choosing a trial value of x, values of S at any time t are plotted against the corresponding [ x l + (1-x) Q] values. If the value of x is chosen correctly, a straight-line relationship as given by Equation(2-29): (S = K [ x I + (1- x) Q] ) will result. However, if an incorrect value of x is used, the plotted points will trace a looping curve. By trial and error, a value of x is to be so chosen that the data very nearly describe a straight line Figure 9-3. The inverse slope of this straight line will give the value of K Normally , for natural channels, the value of x lies between 0 to 0.30.For a given reach, the values of x and K are assumed to be constant. Lecture-13(contd.)
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  13 Lecture-13(contd.) Example 9-3 The followingtable shows the inflow outflow hydrographs were observed in a river reach. Estimate the values of K and x applicable to this reach for use in the Muskingum equation. Table Time (h) o 6 12 18 24 30 36 42 48 54 60 66 Inflow (m3/s) 5 20 50 50 32 22 15 10 7 5 5 5 Outflow (m3/s) 5 6 12 29 38 35 29 23 17 13 9 7
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  14 Solution: Using a timeincrement t = 6 h, the calculations are performed in a tabular manner as in following table:- The incremental storage S and S are calculated in by subtracting inflow from outflow respectively. It is advantageous to use the units (m3/s .h) for storage terms. The equation: I t – Q t = S; t (I – Q) Average = S Lecture-13(contd.)
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  15 Figure9-3:DeterminationofKand xforachannelreach Lecture-13(contd.)
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  16 Calculation of x: Fort = 0 h First trial: x=0.35 x =xl + (1-x)Q 0.35 × 5 +(1-0.35) × 5 =5 Second Trial x = 0.30 0.30 × 5 +(1-0.30) × 5 =5 Third Trial x = 0.25 0.25 × 5 +(1-0.25) × 5 = 5 Lecture-13(contd.)
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  17 Similarly, For t =(6 -0) h = 6 h x= [ x I + (1- x) Q] First trial x = 0.35 0.35 × 20 +(1-0.35) × 6 = 10.9 Second Trial x = 0.30 0.30 × 20 +(1-0.30) × 6 =10.2 Third trial x = 0.25 0.25 × 20 +(1-0.25) × 6 = 9.5 t = (12 -6) h = 6 h x= xl + (1-x)Q x = 0.35, 0.30, 0.25 0.35 × 50 +(1-0.35) × 12 = 25.3 0.30 × 50 +(1-0.30) × 12 = 23.4 0.25 × 50 +(1-0.25) × 12 = 21.5 and so on Lecture-13(contd.)
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  18 Calculation of slopeK For slope, K: From above graph: Considering, S =400 m3/s .h, then the value of Q i.e. vertical from the graph for x=0.25 is equal to 30, then S = KQ From above Figure-3, K = 400/30 = 13.3 h Similarly for any value of S, K can be calculated by getting Q from graph. Fill up the Following the Table and compute the values. Lecture-13(contd.)
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  19 0 5 50 - - 0 5.0 5.0 5.0 7.0 42 6 20 6 14 - - 42 10.9 10.2 9.5 26.0 156 12 50 12 38 - - 198 25.3 23.4 21.5 29.5 177 18 50 29 21 375 36.4 35.3 34.3 7.5 45 24 32 38 -6 420 35.9 36.2 36.5 -9.5 -57 30 22 35 -13 420+ (-57)=363 30.5 31.1 31.8 -13.5 -81 36 15 29 -14 282 24.1 24.8 25.5 -13.5 -81 42 10 23 -13 201 18.5 19.1 19.8 -11.5 -69 48 7 17 -10 132 13.5 14.0 14.5 -9 -54 54 5 13 -8 78 10.2 10.6 11.0 -6 -36 60 5 9 -4 42 7.6 7.8 8.0 -3 -18 66 5 7 -2 24 6.3 6.4 6.5 Time (h) I m3/s Q (m3/s) (I-Q) Average (I-Q)/2 S = t (I – Q)/2 (m3/s.h S=∑S (m3/s ..h) [ x I + (1- x) Q] ( m3/s) x = 0.35 x = 0.30 x = 0.25 1 2 3 4 5 6 7 8 9 10 Lecture-13(contd.) Calculation Table