Electrical Power Systems

Circuit model under fault condition is shown in ig. 8.15(a)
xeq = j 016 .
4
= j0.04
(a) fault level= 1
= 25.0 pu = 25 × 60 MVA
004 .
= 1500 MVA. Ans.
(b) The generators G1 and G2 will supply 1
2 1500 ´ =
750 MVA, directly to the fault. Therefore, the
fault MVA from G3 and G4 must be limited to
(860 – 750) = 110 MVA. The reactance of G3 and
G4 together is 016 .
= 0.08 pu.
2
Thus,
1
=
xR + 008 .
110
60
\ x R =0.465 pu
2 b g = 2.09 ohm
Base impedance = 112 .
60
\ xR = 0.465 × 2.09 = 0.97 ohm.
Symmetrical ault 207
Example 8.15: ig. 8.16 shows a power system network. Each of the alternators G1 and G2 is
rated at 125 MVA, 11 KV and has a subtransient reactance of 0.21 pu. Each of the transformers
is rated at 125 MVA, 11/132 KV and has a leakage reactance of 0.06 pu. ind (a) fault MVA and
(b) fault current for a fault at bus 5.
ig. 8.16: Power system network of Example 8.15.
Solution:
Set Base MVA =125, Base Voltage = 11 KV
Base voltage for transmission line = 132 KV
2 b g ohm.
Base impedance for the transmission line = 132
125
= 139.392 ohm.
40
\ x34 = j = j0.286 pu,
139. 392
ig. 8.15(a)