Electrical Power Systems

206 Electrical Power Systems
\ I f =
Base current for 220 KV side
1
= – j5.78 pu
j 0. 173
100 ´ 1000
IB = = 262.43 Amp.
3 ´ 220
\ I f = 5.78 × 262.43 = 1.516 KA.
Base current on 11 KV side
ault level = 5.78 pu = 5.78 × 100 = 578 MVA.
220
= IB ´
HG I
11 KJ
= 5248.6 Amp.
= 262 43
ault current supplied by the two generators
\ I fg1 =
220
11
. ´ HG I KJ
= 5248.6 × (– j5.78) = 30. 34 – 90ºKA
0. 272
´ 30. 34 – 90º
KA
0522 .
\ Ifg1 = 15. 8 – 90ºKA
I fg2 = 025
.
´ 30. 34 – 90º
KA
0522 .
\ Ifg2 = 14. 53 – 90ºKA
Example 8.14: ig. 8.15 shows a system having four synchronous generators each rated 11.2
KV, 60 MVA and each having a subtransient reactance of 16%. ind (a) fault level for a fault on
one of the feeders (near the bus with x = 0). (b) the reactance of the current limiting reactor
x R to limit the fault level to 860 MVA for a fault on one of the feeders near the bus.
ig. 8.15: Sample power system of Example 8.14.
Solution:
Set Base MVA = 60, Base voltage = 11.2 KV.
x² = x² = x² = x²
= 16% = 0. 16
g 1 g 2 g 3 g 4 pu