Page 77 - Solutions Manual to accompany Electric Machinery Fundamentals

Q G ,pu  VI sin   0.4765 sin 41.6   0.316
1
S G ,pu  VI   0.4765   0.4765
1
P  G P G ,pu S base  0.356 1000 kVA   356 kW

Q  G Q G ,pu S base  0.316 1000 kVA   316 kVAR
S  S S  0.4765 1000 kVA   476.5 kVA
G G ,pu base
The power factor of the generator is
PF cos 41.6   0.748 lagging

(c) With the switch closed, the equivalent impedance of this circuit is
1.513 j 1.134  j 3.36 


Z EQ  0.010  j 0.040 0.00723 j 0.0482 0.040  j 0.170   j 1.513   j 3.36  1.134 



Z EQ  0.010  j 0.040 0.00788  j 0.0525 0.040  j 0.170 (2.358  j 0.109)

Z  2.415 j 0.367  2.443 8.65 
EQ
The resulting current is

10

I   0.409  8.65

2.443 8.65
The load voltage under these conditions would be

 V Z  I  0.409   8.65  2.358  j 0.109   0.966   6.0
Load,pu Load
V Load  V Load,pu base3  0.966 480 V   464 V
V
The power supplied to the two loads is the power supplied to the resistive component of the parallel
combination of the two loads: 2.358 pu.

2
2
P Load,pu  I R Load  0.409  2.358   0.394
P  P S  0.394 1000 kVA   394 kW
Load Load,pu base
The power supplied by the generator is
P G ,pu  VI cos   0.409 cos6.0   0.407
1
1
Q G ,pu  VI sin   0.409 sin6.0   0.0428
S  VI   0.409   0.409
1
G ,pu
P  G P G ,pu S base  0.407 1000 kVA   407 kW
Q  G Q G ,pu S base  0.0428 1000 kVA   42.8 kVAR
S  G S G ,pu S base  0.409 1000 kVA   409 kVA

The power factor of the generator is

PF cos 6.0   0.995 lagging
(d) The transmission losses with the switch open are:
2
2
P line,pu  I R line  0.4765  0.00723   0.00164
P line  P line ,pu S base  0.00164 1000 kVA  1.64 kW

The transmission losses with the switch closed are:
2
2
P line,pu  I R line  0.409  0.00723   0.00121
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