Page 50 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION
(a) The transformer supplies a load of 80 MVA at 0.8 PF lagging. Therefore, the secondary line current
of the transformer is

I LS  S  80,000,000 VA  402 A
LS 3V V  3 115,000
The base apparent power is S base  100 MVA , and the base line voltage on the secondary side is
V LS ,base  115 kV , so the base value of the secondary line current is

I LS ,base  S base  100,000,000 VA  502 A
3V LS ,base  V  3 115,000

so the per-unit secondary current is
I 402 A
 I LS    1  cos 0.8  0.8  36.87
LS ,pu
I LS ,pu 502 A
The per-unit phasor diagram is shown below:
V P V 1.039 1.7 P  


V = 1.00°
S


 I 0.8  I = 0.8-31.8°
36.87

(b) The per-unit primary voltage of this transformer is

 V  V  I 1.0 0     Z  36.87  0.8 0.015  j 0.06  1.039 1.7  
P S EQ
and the voltage regulation is

1.039 1.0
VR   100% 3.9%

1.0
(c) The secondary side of this transfer is Y-connected, so the base phase voltage of the low voltage
(secondary) side of this transformer is:
V 115 kV
V  S ,base  LS ,base   66.4 kV
3 3
The base impedance of the transformer referred to the low-voltage side is:

3 V 2  3 66.4 kV 2
Z   S   ,base  133 
base
S 100 MVA
base
Each per-unit impedance is converted to actual ohms referred to the low-voltage side by multiplying it by
this base impedance. The resulting equivalent circuit is shown below:

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