Electrical Power Systems

\ V b =
E a = 3752.8 0°
1763 . - 33. 38º + 5. 56 - 124. 55º ´ 3752. 8 0°
838 . 59°
\ Vb = 2793.5 - 165. 6°
Volts Ans.
Similarly
V c =
2 eb- bjZ2 + bb-1gZ0 E
Z + Z + Z
1 2 0
Vc = 3204.16 109. 7°
Volts. Ans.
a
Unbalanced ault Analysis 271
Example 10.9: Two alternators are operating in parallel and supplying a synchronous motor
which is receiving 60 MW power at 0.8 power factor (lagging) at 6.0 KV. Single line diagram for
this system is given in ig. 10.20. Data are given below. Compute the fault current when a
single line to ground fault occurs at the middle of the line through a fault resistance of 4.033
ohm.
ig. 10.20: Circuit diagram of Example 10.9.
Data:
G 1 & G 2 : 11 KV, 100 MVA, x g1 = 0.20 pu, x g2 = x g0 = 0.10 pu
T 1 : 180 MVA, 11.5/115 KV, x T1 = 0.10 pu
T 2 : 170 MVA, 6.6/115 KV, x T2 = 0.10 pu
M : 6.3 KV, 160 MVA, x m1 = x m2 = 0.30 pu, x m0 = 0.10 pu
Line :
x Line1 = x Line2 = 30.25 ohm, x Line0 = 60.5 ohm
Solution:
Let Base MVA = 100, Base KV = 11
\ Base voltage of transmission line would be
115I
´ 11 = 110 KV. HG 115 . KJ
\ xT1 = 01 100
2
115 .
. ´ ´I = 0.061 pu
180 HG 11 KJ
xT2 = 01 100
.
115 .
´ ´I = 0.064 pu
170 HG 11 KJ
Transmission line base impedance = 110
2
2 b g = 121 ohm.
100