Page 166 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION To get the maximum torque at starting, the s max must be 1.00. Therefore,

R
s  2
max 2
R TH 2   TH  X 2  X
R
1.00  2
 0.0959  2   0.2066  0.210    2

R  0.428 
2
Since the existing resistance is 0.070 , an additional 0.358  must be added to the rotor circuit. The
resulting torque-speed characteristic is:

6-9. If the motor in Problem 6-5 is to be operated on a 50-Hz power system, what must be done to its supply
voltage? Why? What will the equivalent circuit component values be at 50 Hz? Answer the questions in
Problem 6-5 for operation at 50 Hz with a slip of 0.05 and the proper voltage for this machine.
SOLUTION If the input frequency is decreased to 50 Hz, then the applied voltage must be decreased by 5/6
also. If this were not done, the flux in the motor would go into saturation, since

1
   v dt
N T

and the period T would be increased. At 50 Hz, the resistances will be unchanged, but the reactances will
be reduced to 5/6 of their previous values. The equivalent circuit of the induction motor at 50 Hz is
shown below:

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