Page 168 - Solutions Manual to accompany Electric Machinery Fundamentals

(d) The power converted from electrical to mechanical form is
P conv  1 s P  AG  1 0.05 17.1 kW  16.25 kW

(e) The induced torque in the motor is

P 17.1 kW

 ind  AG   108.9 N m

sync  1500 r/min   2 rad      1 min  
 1 r   60 s 
(f) In the absence of better information, we will treat the mechanical and core losses as constant despite
the change in speed. This is not true, but we don’t have reason for a better guess. Therefore, the output
power of this motor is




P OUT  P conv  P mech  P core  P misc  16.25 kW 500 W 400 W 0 W  15.35 kW
The output speed is

 n   1 s n   1 0.05 1500 r/min   1425 r/min
m sync
Therefore the load torque is
P 15.35 kW

  OUT   102.9 N m
 m  1425 r/min  2 rad    1 min 
load
  1 r     60 s  
(g) The overall efficiency is
P P
  OUT  100%  OUT  100%
P IN 3V I A cos

15.35 kW

   100% 83.4%


  3 100 V 66.0 A cos21.7
(h) The motor speed in revolutions per minute is 1425 r/min. The motor speed in radians per second is
  m  1425 r/min   2 rad      1 min    149.2 rad/s
 1 r   60 s 
6-10. A three-phase 60-Hz two-pole induction motor runs at a no-load speed of 3580 r/min and a full-load
speed of 3440 r/min. Calculate the slip and the electrical frequency of the rotor at no-load and full-load
conditions. What is the speed regulation of this motor [Equation (3-68)]?

SOLUTION The synchronous speed of this machine is 3600 r/min. The slip and electrical frequency at no-
load conditions is

n  n 3600 3580

s  sync nl  100%   100% 0.56%

nl
n 3600
sync

f r ,nl  e  sf  0.0056 60 Hz   0.33 Hz
The slip and electrical frequency at full load conditions is
n  n 3600 3440


s  sync nl  100%   100% 4.44%
fl
n sync 3600

f r ,fl  e  sf  0.0444 60 Hz   2.67 Hz
The speed regulation is
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