Electrical Power Systems

196 Electrical Power Systems
If = Vo
²
Ig1 =
²
Ig2 =
Base current
x T = j0.06 pu
=
j012
.
1
= – j8.33 pu
j 012 .
j015
.
´ – j833
.
jb01
. + 015 . g b g
=– j5.0 pu
j010
.
´ – j833
.
jb01
. + 015 . g b g = – j3.33 pu
15 ´ 1000
IB = = 787.3 Amp.
3 ´ 11
²
\ Ig1 = – j5 × 787.3 = – j3.936 KA.
²
Ig2 = – j3.33 × 787.3 = – j2.621 KA.
I f =– j8.33 × 787.3 = – j6.557 KA.
8.3 CURRENT LIMITING REACTORS
The short circuit current is large enough to do considerable damage mechanically and thermally.
The interrupting capacities of circuit breakers to handle such current would be very large. To
reduce this high fault current, artificial reactances are sometimes connected between bus
sections. These current limiting reactors are usually consist of insulated copper strip embeded
in concrete formers. This is necessary to withstand the high mechanical forces produced by the
current in the neighbouring conductors.
Example 8.6: The estimated short circuit MVA at the bus bars of a generating station-1 is 900
MVA and at another generating station-2 of 600 MVA. Generator voltage at each station is 11.2
KV. The two stations are interconnected by a reactor of reactance 1 ohm per phase. Compute the
fault MVA at each station.
Solution:
SC MVA of generating station-1 = 900 MVA
SC MVA of generating station-2 = 600 MVA
Assume base MVA = 100
\ x 1 =
Base MVA
SC MVA
\ x2 = 100
= 0.166 pu
600
100
= = 0.111 pu
900
ig. 8.7(a)
ig. 8.7(b)
ig. 8.7(c)