Line to Line & Double Line to Ground Fault On Power System

Line to Line & Double Line to Ground Fault On Power System

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  Name :- SmitShah -140410109096 T.Y Electrical 2 Sem 6 Subject:-Electrical power system 2 Topic :-Line to Line & Double Line to Ground Fault On Power System 1
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  Line-‐to-‐Line Faults To representa line-‐to-‐line fault through impedance Zf the hypothetical stubs on the three lines at the fault are connected as shown. Bus k is again the fault point P, and without any loss of generality, the line-‐to-‐line fault is regarded as being on phases b and c. Clearly: Ifb Ifc a Zf k k c I fa I  0 fa k b I fb  I fc Vkb Vkc  Z If fb 2
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  Line‐To‐Line Fault Three‐phase generatorwith a fault through an impedance Zf between phase b and c. Ia=0 Zs N ZsZs Ea Eb Ec Ib Va ZfIc Vb Vc Assuming the generator is initially on no‐load, the boundary conditions at the fault point are: Vb  Vc  Z fIb Ia  0 Ib  I c 0 3
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  Line‐To‐Line Fault 4 Substituting forIa = 0, and Ic = ‐Ib, the symmetrical components of the currents from are: From the above equation, we find that:  0a I0 (10.68) ba 3 11 2 I  (a  a )I a 12 2 I  (a  a)Ib (10.69) (10.70) 3
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  Line‐To‐Line Fault Also, from(10.69) and (10.70), we note that: 1 2 aa I  I From (10.16), we have: (10.71) 1 20 b a aaa V  V 0  a2 V1  aV 2 a a a V  V V V (10.16) aa  Zf Ib (a  a)(V V ) V V  (V 0  a2 V1  aV 2 )  (V 0  aV1  a2 V 2 ) b c a a a a a a 2 1 2 (10.72) 0 1 2 2 aaacV  V  aV  a V Substituti ng for V1 and V 2 from (10.54) and noting I 2  I1 , we get : a a a a a f b(a  a)[Ea  (Z  Z )I ]  Z I2 1 2 1 (10.73) (10.54) 2 2 2 aa V 0  0  Z 0 I 0 a a V  0  Z I V1  E  Z1 I1 a a a Substituti ng for I b from (10.69), we get : 1 3Ia a a f (a  a2 )(a2  a) E  (Z1  Z 2 )I 1  Z (10.74) ba 11 2I  (a  a )I 3 (10.69) 5
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  Line‐To‐Line Fault aSince (a a2 )(a2  a)  3,solving for I1 results in : f a Ea (Z1  Z 2  Z ) I1  (10.75) The fault current is b I  I  (a2  a)I 1 c a or abI   j 3I1(10.77) (10.78) 6
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  Line‐To‐Line Fault Eq. (10.71)and (10.75) can be represented by connecting the positive and negative – sequence networks as shown in the following figure. 1 2 aa I  I 1 aE I  f a (Z1  Z 2  Z ) 7
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  Line-‐to-‐Line Faults Thus, allthe fault conditions are satisfied by connecting the positive-‐ and negative-‐sequence networks in parallel through impedance Zf as was shown. The zero-‐sequence network is inactive and does not enter into the line-‐to-‐line calculations. The equation for the positive-‐sequence current in the fault can be determined directly from the circuit as: 8
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  Double Line-‐to-‐Ground Faults Againit is clear, with fault taken on phases b and c, that the relations at fault bus k are: Ifa Ifb Ifc a Zf I fa 0 Vkb  Vkc  Zf Ifb  I fc k k b k c 9
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  Double Line‐To‐Ground Fault Figure10.14 shows a three‐phase generator with a fault on phases b and c through an impedance Zf to ground. Assuming the generator is initially on no‐ load, the boundary conditions at the fault point are Vb Vc  Z f (Ib  Ic ) (10.79) (10.80) I  I0  I1  I2  0 a a a a From (10.16), the phase voltages Vb and Vc are 10
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  Double Line‐To‐Ground Fault 02 1 2 aaa V V 0  aV1  a2 V 2 Vb V  a V aV aac a (10.81) (10.82) SinceVb  Vc , from above we note that 1 2 aaV V (10.83) Substituting for the symmetrical components of current in (10.79), we get V  Z (I 0  a2 I1  aI 2  I 0  aI 1  a2 I 2 ) (b) f a a a a a a  Z (2I 0  I1  I 2 ) f a a a f aI 0  3Z (10.84) 11
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  Substituti ng forV from (10.84) and for V 2 from (10.83) into (10.81), we have :b a a a f a a a  V 0 V1 3Z I 0  V 0  (a2  a)V1 (10.85) E  Z1 I1 Substituti ng for the symmetrical components of voltage from (10.54) into (10.85) aand solving for I 0 ,we get : 0 0 a a a (Z  3Z ) I   f Also, substituting for the symmetrical components of voltage in (10.83), we obtain E  Z1 I1 (10.86) Z2a I 2   a a Substituti ng for I 0 and I2 into (10.80) and solving for I 1 , we get: a a a (10.87) Ea a 2 0 1 Z 2  Z 0  3Z f Z (Z  3Zf ) Z  I1  (10.88) 12
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  Equation (10.86) -(10.88) can be represented by connecting the positive - sequence impedance in series with the paralel combination of the negative - sequence and zero - sequence networks as shown in the equivalent circuit of figure 10.15. The value of I1 found from (10.86) is substituted in (10.86) and (10.87),a and I0 and I2 are found. The phase current are then found from (10.8). a a Finally, the fault current is obtained from (10.89)I  I  I  3I0 c abf Figure 10.15 Sequence network connection for double line‐to‐ground fault 13
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  Double Line-‐to-‐Ground Faults Fora bolted fault Zf is set equal to 0. When Zf = ∞, the zero-‐ sequence circuit becomes an open circuit, no zero-‐sequence current an flow, and the equations revert back to those for the line-‐to-‐line fault. Again we observe that the sequence currents, once calculated, can be treated as negative injections into the sequence networks at the fault bus k and the sequence voltage changes at all buses of the system can then be calculated from the bus impedance matrices, as we have done in all along. 14
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