Page 296 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION Since the transmission lines are lossless in this power system, the full voltage generated by G
1
will be present at each of the loads.
(a) Since this load is Y-connected, the phase voltage is

480 V
V   277 V
1 
3
The phase current can be derived from the equation P  3V I cos as follows:

P 100 kW
I   1 3 cos    3 277 V 0.9   133.7 A

V

(b) Since this load is -connected, the phase voltage is
V  480 V

2
The phase current can be derived from the equation S  3V I as follows:
 
S 80 kVA
I  2    55.56 A
  3V  3 480 V
(c) The real and reactive power supplied by the generator when the switch is open is just the sum of the
real and reactive powers of Loads 1 and 2.
P  100 kW
1

Q  1 P tan P tan    1  cos PF  100 kW tan25.84   48.4 kvar
P  2 S cos  80 kVA 0.8    64 kW

Q  2 S sin  80 kVA 0.6   48 kvar


P  G P  1 P  2 100 kW 64 kW 164 kW
Q  G Q  1 Q  2 48.4 kvar 48 kvar  96.4 kvar

P Q
(d) The line current when the switch is open is given by I  , where   tan  1 G .
L
3 V L cos P G
  tan  1 Q G  tan  1 96.4 kvar  30.45
P G 164 kW

P 164 kW
I    228.8 A
L
3 V L cos   3 480 V cos  30.45

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