Page 283 - Solutions Manual to accompany Electric Machinery Fundamentals

The slip s  0.05 , so R 2 / s   2.80  / 0.05 56 

56  j 2.56  60.5j 
Z   28.82  j 28.05 
56  j 2.56  j 60.5
F
   R   /2 s  jX  2   jX 
Z  2 M
B  R   /2 s  jX  jX
2 2 M
The slip s  0.05 , so R 2   / 2 s     2.80  / 2 0.05  1.436 
 j 1.436   j 60.5  2.56
Z   1.321 j 2.486 
B
1.436  j 2.56  j 60.5
(a) The input current is

V
I 
1
R  1 jX  1 0.5Z  F 0.5Z B

120 0 V
I  1  j 2.00  0.5  2.56  j  28.82  0.5  28.05  j 2.486  1.321  4.862   46.2 A



P  IN VI cos  120 V 4.862 A cos 46.2  403.8 W
(b) The air-gap power is
P AG,F I 1 2   F 0.5R    4.862 A 2   14.41   340.6 W

P AG,B I 1 2   B 0.5R    4.862 A 2   0.661   15.6 W

P AG  P AG,F  P AG,B  340.6 W 15.6 W  325 W
(c) The power converted from electrical to mechanical form is
P  1 s P   1 0.05 340.6 W  323 W
conv,F AG,F
P    1 s P   1 0.05 15.6 W  14.8 W
conv,B AG,B

P  P  P  323 W 14.8 W  308 W
conv conv,F conv,B
(d) The output power is

P OUT  P conv  P  rot 308 W 51 W  257 W
(e) The induced torque is
P 325 W
 ind   AG   2 rad    1 min   1.72 N m


sync
 1800 r/min  1 r     60 s  
(f) The load torque is

P 257 W
 load  OUT   1.44 N m

 m  0.95 1800 r/min   2 rad    1 min 

  1 r     60 s  
(g) The overall efficiency is

P 257 W
  OUT  100%   100% 63.6%

P IN 403.8 W

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