Page 169 - Solutions Manual to accompany Electric Machinery Fundamentals


n  n 3580 3440
SR  nl fl  100%   100% 4.07%

n fl 3440
6-11. The input power to the rotor circuit of a six-pole, 60 Hz, induction motor running at 1100 r/min is 5 kW.
What is the rotor copper loss in this motor?

SOLUTION This synchronous speed of this motor is
120 f 120 60 Hz 
n  se   1200 r/min
sync
P 6
The slip of the rotor is
n  n 1200 1100


s  sync nl  100%   100% 8.33%
nl
n sync 1200
The air gap power is the input power to the rotor, so
P AG  5 kW

The power converted from electrical to mechanical form is
  P  1 s P   1   0.0833 5 kW  4584 W
conv AG
The rotor copper losses are the difference between the air gap power and the power converted to
mechanical form, so

P  P  P  5000 W 4584 W  416 W
RCL AG conv
6-12. The power crossing the air gap of a 60 Hz, four-pole induction motor is 25 kW, and the power converted
from electrical to mechanical form in the motor is 23.2 kW.
(a) What is the slip of the motor at this time?
(b) What is the induced torque in this motor?
(c) Assuming that the mechanical losses are 300 W at this slip, what is the load torque of this motor?

SOLUTION
(a) The synchronous speed of this motor is

120 f 120 60 Hz 
n  se   1800 r/min
sync
P 4
The power converted from electrical to mechanical form is

P  1 s  P
conv AG
so
P 23.4 kW
s  conv   0.064
1
1
P AG 25 kW
or 6.4%.
(b) The speed of the motor is

 n   1 s n   1   0.064 1800 r/min   1685 r/min
m sync
The induced torque of the motor is

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