Electrical Power Systems

266 Electrical Power Systems
14. 39 174. 09°
Ic = 796 . 0
134. 23°
´ ° KA
34.33
\ Ic = 3336 . 39. 86°
KA = 3336 39. 86°
KA
The total fault currents to ground, equal to 3 Ia0, flows from ground to neutral through a
combined neutral grounding reactance j0.2 × 0.2 / (0.2 + 0.2) = j0.1W. Therefore, there is a
voltage drop from ground to neutral equal to (3 Ia0 × j0.1), or a potential rise from ground to
neutral given by (– 3Ia0 × j0.1).
3 EaZ2 Potential rise =
´ j 01
ZZ 1 2 + Z0bZ1 + Z2g
b . g
= 3 7.96 0 2.147 62.24
´ °´ °
34.33 134.23°
´ bj01 . g
= 0.1493 18° KV = 149.3 18° Volt Ans.
Example 10.6: Athree phase synchronous generator with solidly grounded neutral is subjected
to a line-to-line fault on phases b and c accompanied by a ground fault on phase a. Assume that
synchronous generator was running on no load. Develop and draw the sequence networks
simulating the above fault condition.
Solution:
ig. 10.15: Circuit diagram of Example 10.6.
There is a line to ground fault on phase a, therefore V a = 0. urther, the phases b and c are
short circuited.