Electrical Power Systems

Power System Components and Per Unit System 107
Solution: Assuming,
(MVA) B = 100 and (KV) B = 33 in the generator circuit.
\ x g = 0.15 pu
In the motor circuit
(KV) B, line = 33 110
´ = 113.43 kV
32
32
(KV) B, motor = 113. 43 ´ = 33 kV.
110
( )
Now ZB = 33
= 10.89 W
100
( )
ZB, T1 = ZB, T2 = 32
W = 9.309 W
110
\ xT1(W) = 0.08 × 9.309 W = 0.744 W
0. 744
\ xT1, new (pu) = = 0.0683 pu
10. 89
\ xT2, new (pu) = 0.0683 pu
Z B, line =
\ x line(pu) =
2
2
( 113. 43)
100
2
= 128.66 W
60
= 0.466 pu
128. 66
x motor 1 (pu) = 020 100
40
x motor-2 (pu) = 020 100
30
33
. ´ ´ HG I KJ = 0.413 pu
30
x motor-3 (pu) = 020 100
30
33
. ´ ´ HG I KJ = 0.551 pu
20
30
33
. ´ ´ HG I KJ = 0.826 pu.
igure 5.15 shows the per-unit reactance diagram.
ig. 5.15: Per-unit reactance diagram.
2
2
2