Page 199 - Solutions Manual to accompany Electric Machinery Fundamentals

(e) The stator copper losses are

P SCL  3I A 2 1  R   3 75.2 A 2   0.058   984 W

R
2
(f) The air gap power is P  3I 2 2  3I R
AG
2
s A F
R
2
/
(Note that 3IR is equal to 3I 2 2 , since the only resistance in the original rotor circuit was R s ,
2
A
F
s 2
and the resistance in the Thevenin equivalent circuit is R . The power consumed by the Thevenin
F
equivalent circuit must be the same as the power consumed by the original circuit.)
R 2
P  3I 2 2  3I 2  R   3 75.2 A  2.97   50.4 kW
AG
2
s A F
(g) The power converted from electrical to mechanical form is
  P  1 s P   1 0.01 50.4 kW  49.9 kW
conv AG
(h) The synchronous speed of this motor is
120 120  f 60 Hz
n  se   1800 r/min
sync
P 4
 sync   1800 r/min   2 rad      1 min    188.5 rad/s
 1 r   60 s 
Therefore the induced torque in the motor is
P 50.4 kW
  AG   267.4 N m

ind
 sync  1800 r/min  2 rad    1 min 
  1 r     60 s  
(i) The output power of this motor is



P OUT  P conv  P mech  P core  P misc  49.9 kW 650 W 600 W 150 W  48.5 kW
The output speed is
 n   1 s n   1 0.01 1800 r/min  1782 r/min
m sync
Therefore the load torque is

P 48.5 kW
  OUT   260 N m

load
 m  1782 r/min  2 rad    1 min 
  1 r     60 s  
(j) The overall efficiency is
P P
  OUT  100%  OUT  100%
P IN 3V I A cos

48.5 kW

   100% 94.5%

  3 266 V 75.2 A cos 31.2 


(k) The motor speed in revolutions per minute is 1782 r/min. The motor speed in radians per second is
  m  1782 r/min   2 rad      1 min    186.6 rad/s
 1 r   60 s 

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