Page 162 - Solutions Manual to accompany Electric Machinery Fundamentals

P 21.9 kW

  OUT   122.3 N m
load
 m  1710 r/min  2 rad    1 min 
  1 r     60 s  
(g) The overall efficiency is

P P
  OUT  100%  OUT  100%
P IN 3V I A cos

21.9 kW

   100% 84.6%
  3 120 V 78.0 A cos22.8


(h) The motor speed in revolutions per minute is 1710 r/min. The motor speed in radians per second is
  m  1710 r/min   2 rad      1 min    179 rad/s
 1 r   60 s 

6-6. For the motor in Problem 6-5, what is the slip at the pullout torque? What is the pullout torque of this
motor?

SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model.

jX R  jX   10j   0.10   j 0.21  
Z  M 1 1   0.0959  j 0.2066   0.2278 65.1   
R  1  1 X M  j X  0.10   j   10   0.21 
TH

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