Page 182 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION From the DC test,
13.5 V
2R   R  0.105 
1
64 A 1
I DC

+

R
1

V
DC

R 1 R 1

-

In the no-load test, the line voltage is 208 V, so the phase voltage is 120 V. Therefore,
V 120 V
X  1 X M     5.00 @ 60 Hz

I A ,nl 24.0 A
In the locked-rotor test, the line voltage is 24.6 V, so the phase voltage is 14.2 V. From the locked-rotor
test at 15 Hz,
V 14.2 V
Z R  jX       0.220 
LR
LR
LR
I A ,LR 64.5 A
P  2200 W 
  cos  1 LR  cos   1   36.82
LR
S   
LR   3 24.6 V 64.5 A   
Therefore,
R LR  Z LR LR   cos 0.220  cos 36.82  0.176  
 R  1 R  2 0.176 
 R  2 0.071 


X  LR  Z LR s LR   in 0.2202  sin 36.82   0.132
At a frequency of 60 Hz,
 60 Hz
X LR    15 Hz   X  LR  0.528 

For a Design Class B motor, the split is X  0.211  and X  0.317  . Therefore,
2
1
X  5.000   0.211   4.789 
M
The resulting equivalent circuit is shown below:

176