Electrical Power Systems

274 Electrical Power Systems
we know fault current
\ If =
3Vf
Z1 + Z2 + Z0 + 3 Rf
\ If =
3 ´ 1153 . 1190 .
j0. 162 + j0162 . + j0.
197 + 0. 10
\ |If| = 6.514 pu = 6514 . ´
100
KA
3 ´ 110
= 3.418 KA Ans.
Exercise 10 :
10.1 ig. 10.23 show a power network. The far end of the line is connected to an infinite bus.
Data for this system are given below. (a) Draw the positive, negative and zero-sequence
networks. (b) when both the generators and infinite bus operating at 1.0 pu voltage on no
load, a L-G fault occurs at one of the terminals of transformer T1 . Calculate the fault
current.
ig. 10.23: Circuit connection of P 10.1.
Data Positive Negative Zero
G 1 0.15 0.15 0.08
G 2 0.25 0.25 –
T 1 & T 2 0.15 0.15 0.15
Line 0.20 0.20 0.40
Infinite bus 0.15 0.15 0.05
Ans: I f = – j6.732 pu.
10.2 ig. 10.24 shows a sample power network. A double line to ground fault occurs on phases
b and c at point in the system. ind the fault current in phase c of G1. Both the
machines are rated 1.2 MVA, 0.6 KV with x 1 = x 2 = 0.10 pu. and x 0 = 0.05 pu. Transformers
are 1.2 MVA each with leakage reactance of 0.05 pu. Transmission line reactances are
x L2 = x L2 = 0.20 pu. and x L0 = 0.40 pu.–on a base of 1.2 MVA, 3.3 KV. The reactances of
the neutral grounding reactors are 0.50 pu. on the MVA base of the machines.
Ans: 5.266 KA
ig. 10.24: Circuit diagram of P 10.2.
10.3 A 11 KV, 30 MVA generator has x1 = x2 = 0.2 pu and x0 = 0.05 pu. (i) a line to ground fault
occurs on the generator terminal. ind the fault current and line to line voltages during
fault conditions. Assume that the generator neutral is solidly grounded.
Ans: 10. 5 – 90°
KA.