Electrical Power Systems

Z 1 =
j0.691 0.23 ´
0. 691+ 0. 23
Z 2 = Z 1 = j0.172 pu
Z 0 = j3.264
I a1 =
=
0
Z
Vf
+ Z + Z
1 2 0
= j0.172 pu
0. 909
j(0.172 + 0.172 + 3.264)
=– j0.252 pu.
\ I a1 = I a2 = I a0 = – j0.252 pu.
ault current = 3 I a0 = 3 ´ (– j0.252) = – j1.341 pu.
The component of I a1 flowing towards g from the generator side
=– j0.252 ´ 023 .
= – j0.063 pu
0921 .
component of I a1 flowing towards g from the motors side is
Unbalanced ault Analysis 259
=– j0.252 ´ 0691 .
= – j0.189 pu
0921 .
Similarly, the component of Ia2 from the generator side is –j 0.063 pu and its component
from the motors side is – j0.189 pu. All of Ia0 flows towards g from motor-2.
ault currents from the generator towards g,
L
NM
a
b
c
O
Q
I
I
I P =
L
NM
1 1 1
2
b b 1
b
2
b 1
ault currents from the motors are
L
NM
c
O
Q
I a
I b
I P =
L
NM
1 1 1
2
b b 1
b
2
b 1
O L
QP
NM
O
QP
O
Q
– j0.063
P – j0.063
=
0
L
NM
O
Q
– j0.189
P – j0.189
=
– j0.252
ig. 10.13: Thevenin equivalent of positive
sequence network of ig. 10.12.
L
NM
O
Q
– j0.126
P – j0.063
pu Ans.
– j0.063
L
NM
O
Q
– j0.63
P – j0.063
pu. Ans.
– j0.063
Example 10.2: Two 11 KV, 12 MVA, 3f, star connected generators operate in parallel.
(ig. 10.13). The positive, negative and zero sequence reactances of each being j0.09, j0.05 and
j0.04 pu respectively. A single line to ground fault occurs at the terminals of one of the generators.
Estimate (i) the fault current (ii) current in grounding resistor (ii) voltage across grounding
resistor.