Chapter 10 Symmetrical Faults

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G1 X G1 =J0.2 X 12 =J0.8 G2 X G2 = J0.4 Bus 1 Bus 2 X 13 = J0.4 X 23 = J0.4 Bus 3 Figure 1 Solution The fault is simulated by switching on an impedance Z f at bus 3 as shown in the figure below: G1 X G1 =J0.2 X 12 =J0.8 G2 X G2 = J0.4 Bus 1 Bus 2 X 13 = J0.4 X 23 = J0.4 Bus 3 Z f =J0.16 By Thevenin’s theorem, the changes in the network voltage due to the fault impedance Z f can be represented by a voltage source V th with all other voltage sources short circuited as shown below: 2
Bus 1 X 12 =J0.8 X G1 =J0.2 X G2 = J0.4 Bus 2 X 13 = J0.4 X 23 = J0.4 Bus 3 V th Z f =J0.16 From the above figure, the fault current at bus 3 is I 3 (F) = V th is the Thevenin voltage or the pre-fault voltage V 3 (0). Z 33 is the Thevenin impedance viewed from the faulted bus 3. Z f is the fault impedance. Since the generators are running at their rated values, all pre-fault bus voltages are equal to 1.0 pu i.e V th = V 3 (0) = 1.0 pu To find the Thevenin impedance, we note that the impedances between buses 1, 2 and 3 form a delta network. We transform this delta network to a star network with the following equivalent impedances: Z 1s = Z 2s = = j0.2 pu Z 3s = = j0.1 pu The star connected network is shown below: 3
Page 1: Chapter 10 Symmetrical Faults Intro
Page 5 and 6: Z 33 = Z eq + Z 3s = j0.34 pu From
Page 7 and 8: nodal analysis where the elements o
Page 9 and 10: V i (F) = V i (0) - Z ik I k (F) (9
Page 11 and 12: Y bus = The inverse of the above ma
Page 13 and 14: V 1 (F) = 0.76 and V 3 (F) = 0.32 a
Page 15: = 0.0825 + j0.3163 = 0.327ㄥ75.36

Chapter 10 Symmetrical Faults

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