Page 76 - Solutions Manual to accompany Electric Machinery Fundamentals

2
V  S 
Z
( , , ) pu on base 2  ( , , ) pu on base 1  base 1 V 2  base 2 base 2 S (2-60)
R
X
RX
Z
base 1
2
8314 V  1000 kVA 
R 2,pu  0.020  8314 V 2   500 kVA  0.040
2
8314 V  1000 kVA 
X 2,pu  0.085  8314 V 2   500 kVA  0.170
The per-unit impedance of the transmission line is
Z 1.5 j 10 
Z  line   0.00723 j 0.0482
line,pu
Z 207.4 
base2
The per-unit impedance of Load 1 is


Z 0.45 36.87 
Z  load1   1.513 j 1.134
load1,pu
Z base3 0.238 
The per-unit impedance of Load 2 is
Z  j 0.8 
Z load2,pu  Z load2  0.238    j 3.36
base3
The resulting per-unit, per-phase equivalent circuit is shown below:
0.010 j0.040 0.00723 j0.0482 0.040 j0.170

T Line T
1 2
1.513
10° + - L L
j1.134 1 2
-j3.36

(b) With the switch opened, the equivalent impedance of this circuit is


Z  0.010  j 0.040 0.00723 j 0.0482 0.040  j 0.170 1.513  j 1.134
EQ

Z EQ  1.5702  j 1.3922  2.099 41.6 
The resulting current is
10

I   0.4765   41.6

2.099 41.6 
The load voltage under these conditions would be

 V Z  I  0.4765   41.6  1.513 j 1.134   0.901   4.7
Load,pu Load
V Load  V Load,pu base3  0.901 480 V   432 V
V
The power supplied to the load is

2
P Load,pu  I R Load  0.4765  1.513  2  0.344
P Load  P Load,pu S base  0.344 1000 kVA   344 kW

The power supplied by the generator is
P  VI cos   0.4765 cos41.6   0.356
1
G ,pu
70