Page 159 - Solutions Manual to accompany Electric Machinery Fundamentals

(d) What is the rotor’s electrical frequency at one-quarter of the rated load?

SOLUTION
(a) This machine has 10 poles, which produces a synchronous speed of

120 120  f 60 Hz
n sync  e   720 r/min
P 10
(b) The slip at rated load is
n  n 720 670


s  sync m  100%   100% 6.94%
n sync 720
(c) The motor is operating in the linear region of its torque-speed curve, so the slip at ¼ load will be

s  0.25(0.0694) 0.0171

The resulting speed is

 n   1 s n   1  0.0171 720 r/min  708 r/min
m sync
(d) The electrical frequency at ¼ load is
f  sf  0.0171 60 Hz  1.03 Hz
r e
6-4. A 50-kW, 460-V, 50-Hz, two-pole induction motor has a slip of 5 percent when operating a full-load
conditions. At full-load conditions, the friction and windage losses are 700 W, and the core losses are 600
W. Find the following values for full-load conditions:
(a) The shaft speed n m

(b) The output power in watts

(c) The load torque  load in newton-meters

(d) The induced torque  ind in newton-meters

(e) The rotor frequency in hertz
SOLUTION

(a) The synchronous speed of this machine is
120 f 120 50 Hz 
n  se   3000 r/min
sync
P 2
Therefore, the shaft speed is

 n   1 s n   1 0.05 3000 r/min   2850 r/min
m sync
(b) The output power in watts is 50 kW (stated in the problem).

(c) The load torque is
P 50 kW
  OUT   167.5 N m

load
 m  2850 r/min  2 rad    1 min 
  1 r     60 s  
(d) The induced torque can be found as follows:



P conv  P OUT  P F&W  P core  P misc  50 kW 700 W 600 W 0 W   51.3 kW
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