Electrical Power Systems

Unbalanced ault Analysis 261
Example 10.3: or example 10.2, assume that the Rn = 0. ind the fault current in each phase
and voltage of the healthy phase for a double-line-to ground fault on terminals of the generator.
Solution: Using eqn. (10.36)
I a1 =
\ I a1 =
\ I a1 = – j16.56 pu
Z
1
+ (+3 )
Ea
Z2 Z0 Zf
Z + Z +3Z
2 0 f
10 .
j0.025 ´ j0.04
j0.045
+
j0.025 + j0.04
Va1 = Va2 = Va0 = Ea – Ia1 Z1 = 1 – (–j16.56) ( j0.045)
= 0.2548 pu.
Ia2 = – Va2
=
Z2
– 0.2548
= j10.192 pu
j0.025
Ia0 = – Va0
=
Z0
– 0.2548
= j6.37 pu.
j0.04
Now
Ib = b2 Ia1 + bIa2 + Ia0 \
Ib = (– 0.5 – j0.866) (– j16.56) + (– 0.5 + j0.866) ( j10.192) + j6.37
\ Ib = – 23.16 + j9.554
= 25.05 157.6 o pu.
I c = b I a1 + b 2 I a2 + I a0
\ Ic = (– 0.5 + j0.866) (– j16.56) + (– 0.5 – j0.866) ( j10.192) + j6.37
\ Ic = 23.16 + j9.554 = 25.05 22. 4°
pu.
Voltage of healthy phase,
Va = 3Va1 = 3 × 0.2548 = 0.7644 pu.
Example 10.4: A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its
positive, negative and zero sequence impedances are 0.30, 0.40 and 0.05 pu respectively. Determine
the following:
(a) What value of reactance must be placed in the generator neutral so that the fault current
for a line-to-ground fault of zero fault impedance shall not exceed the rated line current?
(b) What value of resistance in the neutral will serve the same purpose?
(c) What value of reactance must be placed in the neutral of the generator to restrict the fault
current to ground to rated line current for a double line to ground fault?
(d) What will be the magnitudes of the line currents when the ground current is restricted as
above?