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

In this slide given the information about symmetrical fault , its derivations , examples etc..

1. Name :- Smit Shah -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

2. Line-­‐to-­‐Line Faults To represent a line-­‐to-­‐line fault through impedance Z f 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: I fb I fc a Z f k k c I fa I  fa k b I fb   I fc V kb  V kc  Z I f f b 2

3. Line‐To‐Line Fault Three‐phase generator with a fault through an impedance Z f between phase b and c . I a =0 Z s N Z s Z s E a E b E c I b V a Z f I c V b V c Assuming the generator is initially on no‐load, the boundary conditions at the fault point are: V b  V c  Z f I b I a  I b  I c  3

4. Line‐To‐Line Fault 4 Substituting for I a = 0, and I c = ‐I b , the symmetrical components of the currents from are: From the above equation, we find that:  a I (10.68) b a 3 1 1 2 I  ( a  a ) I a 1 2 2 I  ( a  a ) I b (10.69) (10.70) 3

5. Line‐To‐Line Fault Also, from (10.69) and (10.70), we note that: 1 2 a a I   I From (10.16), we have: (10.71) 1 2 b a a a a V  V  a 2 V 1  aV 2 a a a V  V  V  V (10.16) a a  Z f I b  ( a  a ) ( V  V ) V  V  ( V  a 2 V 1  aV 2 )  ( V  aV 1  a 2 V 2 ) b c a a a a a a 2 1 2 (10.72) 0 1 2 2 a a a c V  V  aV  a V Substituti ng for V 1 and V 2 from (10.54) and noting I 2   I 1 , we get : a a a a a f b ( a  a )[ E a  ( Z  Z ) I ]  Z I 2 1 2 1 (10.73) (10.54) 2 2 2 a a V   Z I a a V   Z I V 1  E  Z 1 I 1 a a a Substituti ng for I b from (10.69), we get : 1 3 I a a a f ( a  a 2 )( a 2  a ) E  ( Z 1  Z 2 ) I 1  Z (10.74) b a 1 1 2 I  ( a  a ) I 3 (10.69) 5

6. Line‐To‐Line Fault a Since (a  a 2 )(a 2  a)  3, solving for I 1 results in : f a E a ( Z 1  Z 2  Z ) I 1  (10.75) The fault current is b I   I  ( a 2  a ) I 1 c a or a b I   j 3 I 1 (10.77) (10.78) 6

7. 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 a a I   I 1 a E I  f a ( Z 1  Z 2  Z ) 7

8. Line- ­‐to-­‐Line Faults Thus , all the fault conditions are satisfied by connecting the positive-­‐ and negative-­‐sequence networks in parallel through impedance Z f 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

9. Double Line-­‐to-­‐Ground Faults Again it is clear, with fault taken on phases b and c, that the relations at fault bus k are: I fa I fb I fc a Z f I fa  V kb  V kc  Z f  I fb  I fc  k k b k c 9

10. Double Line‐To‐Ground Fault Figure 10.14 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 V b  V c  Z f ( I b  I c ) (10.79) (10.80) I  I  I 1  I 2  a a a a From (10.16), the phase voltages V b and V c are 10

11. Double Line‐To‐Ground Fault 0 2 1 2 a a a V  V  aV 1  a 2 V 2 V b  V  a V  aV a a c a (10.81) (10.82) SinceV b  V c , from above we note that 1 2 a a V  V (10.83) Substituting for the symmetrical components of current in (10.79), we get V  Z ( I  a 2 I 1  aI 2  I  aI 1  a 2 I 2 ) ( b ) f a a a a a a  Z (2 I  I 1  I 2 ) f a a a f a I  3 Z (10.84) 11

12. Substituti ng for V from (10.84) and for V 2 from (10.83) into (10.81), we have : b a a a f a a a  V  V 1 3 Z I  V  ( a 2  a ) V 1 (10.85) E  Z 1 I 1 Substituti ng for the symmetrical components of voltage from (10.54) into (10.85) a and solving for I , we get : a a a ( Z  3 Z ) I   f Also, substituting for the symmetrical components of voltage in (10.83), we obtain E  Z 1 I 1 (10.86) Z 2 a I 2   a a Substituti ng for I and I 2 into (10.80) and solving for I 1 , we get : a a a (10.87) E a a 2 0 1 Z 2  Z  3 Z f Z ( Z  3 Z f ) Z  I 1  (10.88) 12

13. 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 I 1 found from (10.86) is substituted in (10.86) and (10.87), a and I and I 2 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  3 I c a b f Figure 10.15 Sequence network connection for double line‐to‐ground fault 13

14. Double Line-­‐to-­‐Ground Faults For a bolted fault Z f 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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