Electrical Power Systems

Z L (pu) = Z
L (ohm) (. 08+ j 03 .)
=
Z 464 .
B3
= (0.1724 + j 0.0646) pu.
Per-unit circuit is shown in ig. 5.11.
IL (pu) = VS
ZT
( pu)
( pu)
Z T(pu) = j 0.10 + j0.048 + j0.1548
+ 0.1724 + j0.0646
= 0. 4058 64. 86°
0. 956 0°
\ IL(pu) =
2. 355 64. 86
0. 4058 64. 86°
= - °pu
IL(Amp) = IL(pu) × IB3 = 2. 355 - 64. 86°´ 25. 4
= 59. 83 - 64. 86°
Amp.
Power System Components and Per Unit System 105
Example 5.3: igure 5.12 shows single-line diagram of a power system. The ratings of the
generators and transformers are given below:
G1 : 25 MVA, 6.6 kV, xg1 = 0.20 pu
G2 : 15 MVA, 6.6 kV, xg2 = 0.15 pu
G3 : 30 MVA, 13.2 kV, xg3 = 0.15 pu
T1 : 30 MVA, 6.6 D – 115 Y kV, xT1 = 0.10 pu
T2 : 15 MVA, 6.6 D – 115 Y kV, xT2 = 0.10 pu
T3 : Single-phase unit each rated 10 MVA, 6.9/69 kV, xT3 = 0.10 pu.
Draw per-unit circuit diagram using base values of 30 MVA and 6.6 kV in the circuit of
generator-1.
ig. 5.12: Single-line diagram.
ig. 5.11: Per-unit circuit.
Solution: The chosen base values are 30 MVA and 6.6 kV in the generator 1 circuit.
Consequently, the transmission line base voltage of Line-1 is 115 kV. or generator-2 base
voltage is also 6.6 kV.