Page 299 - Solutions Manual to accompany Electric Machinery Fundamentals



V 208 30 V


 I ab   20.8 10 A

ab

Z  10 20 

V 208 90 V

 I bc   20.8   110 A
bc Z  10 20 




V 208 150 V


 I ca   20.8 130 A
ca


Z  10 20 







 I  I  I 20.8 10 A 20.8 130 A 36   20 A
a ab ca






 I  I I 20.8   110 A 20.8 10 A 36   140 A
b bc ab







 I  I I 20.8 130 A 20.8   110 A 36 100 A
c ca bc
A-6. Figure PA-4 shows a small 480-V distribution system. Assume that the lines in the system have zero
impedance.

(a) If the switch shown is open, find the real, reactive, and apparent powers in the system. Find the total
current supplied to the distribution system by the utility.
(b) Repeat part (a) with the switch closed. What happened to the total current supplied? Why?
SOLUTION
(a) With the switch open, the power supplied to each load is
V 2  4 V  80 2

P 1  3  cos  3 cos 30  5 . 9 86 kW
Z 10 
V 2 480 V  2
Q  3  sin  3 sin 30  34.56 kvar
1
Z 10 
V 2  V 277 2


P 2  3  cos  3 cos 36.87  46 . 04 kW
Z 4 
V 2 277 V  2
Q  3  sin  3 sin 36.87  34.53 kvar
2
Z 4 
P TOT  P 1  P 2  59 . 86 kW  46.04 kW  105.9 kW

Q TOT  Q  1 Q  2 34.56 kvar 34.53 kvar  69.09 kvar
The apparent power supplied by the utility is
S TOT  P TOT 2  Q TOT 2  126.4 kVA
The power factor supplied by the utility is
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