Page 285 - Solutions Manual to accompany Electric Machinery Fundamentals

P 120.9 W

  OUT   0.658 N m
 m  0.975 1800 r/min   2 rad    1 min 
load

  1 r     60 s  
(g) The overall efficiency is
P 120.9 W
  OUT  100%   100% 53.5%

P IN 226.1 W
(h) The stator power factor is

PF cos 60.9   0.486 lagging
9-3. Suppose that the motor in Problem 9-1 is started and the auxiliary winding fails open while the rotor is
accelerating through 400 r/min. How much induced torque will the motor be able to produce on its main
winding alone? Assuming that the rotational losses are still 51 W, will this motor continue accelerating or
will it slow down again? Prove your answer.

SOLUTION At a speed of 400 r/min, the slip is

1800 r/min 400 r/min
s   0.778
1800 r/min
The impedances Z and Z are:
B
F
R s  jX  jX 
/
Z  2 2 M
F
R 2 / s  jX  2 jX M
The slip s  0.778 , so R 2 / s   2.80  / 0.778 3.60  
 j 3.60   j 60.5  2.56
Z  F  3.303 j 2.645 
3.60  j 2.56  j 60.5

   R   /2 s  2  jX   jX
Z  2 M
B
2  R   /2 s  jX  2 jX M
The slip s  0.778 , so R 2    / 2 s  2.80   2 0.778 /   2.291 
2.291 j 2.56  60.5j 
Z  B  2.106  j 2.533

2.291 j 2.56  j 60.5
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  3.303 0.5  2.645  j 2.533  2.106   17.21   47.6 A
The air-gap power is

P AG,F I 1 2   F 0.5R    17.21 A 2   3.303   978.3 W

P AG,B I 1 2   B 0.5R    17.21 A 2   2.106   623.8 W

P AG  P AG,F  P AG,B  978.3 W 623.8 W  354.5 W
The power converted from electrical to mechanical form is

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