Page 174 - Solutions Manual to accompany Electric Machinery Fundamentals

(e) The stator copper losses are

P SCL  3I A 2 1  R   3 35.5 A 2  0.15   567 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 ,
/
A
2
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 35.5 A   6.123   23.15 kW
AG
2
s A F
(g) The power converted from electrical to mechanical form is

  P  1 s P   1 0.02 23.15 kW  22.69 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 23.15 kW
  AG   122.8 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  22.69 kW 400 W 400 W 150 W  21.74 kW
The output speed is
 n   1 s n   1 0.02 1800 r/min  1764 r/min
m sync
Therefore the load torque is

P 21.74 kW

  OUT   117.7 N m
load
 m  1764 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

21.74 kW

   100% 91.7%


  3 266 V 35.5 A cos 33.2 

(k) The motor speed in revolutions per minute is 1764 r/min. The motor speed in radians per second is
  m  1764 r/min   2 rad      1 min    184.7 rad/s
 1 r   60 s 

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