Electrical Power Systems

264 Electrical Power Systems
Example 10.5: Two 13.8 KV, MVA three phase alternators operating in parallel supply power
to a substation through a feeder having an impedance of (1 + j1.6) ohm to positive and negative
sequence currents and (2 + j6) ohm to zero sequence currents. Each alternator has x1 = 0.8 ohm,
x2 = 0.6 ohm and x0 = 0.3 ohm and its neutral is grounded through a reactance of 0.2 ohm.
Evaluate the fault currents in each line and the potential above earth attained by the alternator
neutrals, consequent of simultaneous occurrence of earth fault on the b and c phases at the
substation.
Solution: ig. 10.14 shows the necessary circuit connection.
is:
ig. 10.14: Circuit connection of Example 10.5.
As the two identical alternators are connected in parallel, the equivalent sequence impedances
Zg1 = j0.8/2 = j0.4 ohm
Zg2 = j0.6/2 = j0.3 ohm
Zg0 = 1
( j0.3 + 3 ´ j0.2) = j0.45 ohm
2
Total sequence impedance upto the fault
Z1 = Zg1 + Zf1 = j0.4 + 1 + j1.6 = (1 + j2) ohm
Z2 = Zg2 + Zf2 = j0.3 + 1 + j1.6 = (1 + j1.9) ohm
Z0 = Zg0 + Zf0 = j0.45 + 2 + j6 = (2 + j6.45) ohm
We know
I b =
I c =
2
2 2
0 a
1 2 0 1 2
[ Z ( b –1)+( b – b)
Z ] E
ZZ + Z ( Z + Z)
[ Z ( b –1)+ ( b– b Z ] E
ZZ +Z (+ Z Z
2 )
)
2 0 a
1 2 0 1 2
...(i)
...(ii)