Electrical Power Systems

Power System Components and Per Unit System 109
Example 5.6: Draw the per-unit impedance diagram of the system shown in ig. 5.17. Assumed
base values are 100 MVA and 100 kV.
G1 : 50 MVA, 12.2 kV, x g1 = 0.10 pu
G2 : 20 MVA, 13.8 kV, x g2 = 0.10 pu
T 1 : 80 MVA, 12.2/132 kV, x T1 = 0.10 pu
T 2 : 40 MVA, 13.8/132 kV, x T2 = 0.10 pu
ig. 5.17: Sample power system.
Load : 50 MVA, 0.80 pf lagging operating at 124 kV.
Solution: Base kV in the transmission line = 100 kV
12. 2
Base kV in the generator circuit G1 = 100 × = 9.24 kV.
132
13. 8
Base kV in the generator circuits G2 = 100 ´ = 10.45 kV
132
Now, or G1,
xg1, new = xg1, old × (MVA)
(MVA)
(MVA) B, new = (MVA) B = 100
B, new
B, old
(KV)
´
(KV)
(MVA) B, old = Rated MVA of G1 = 50 MVA
(KV) B, old = 12.2 kV
(KV) B, new = 9.24 kV
x g1, old = x g1 = 0.10 pu
\ xg1, new = 010 100
.
50
Similarly for G2,
x g2, new = 010 100
.
12. 2
924 .
2
2
B, old
2
B, new
´ ´ HG I KJ pu = 0.3486 pu.
20
or T1, xT1, new = 01 100
.
13. 8
10. 45
2
´ ´ HG I KJ pu = 0.8719 pu
80
or T2, x T2, new = 01 100
.
12. 2
924 .
2
´ ´ HG I KJ pu = 0.2179 pu.
40
13. 8
10. 45
´ ´ HG I KJ pu = 0.33 pu.
2