Page 186 - Solutions Manual to accompany Electric Machinery Fundamentals


f   sf  0.02 60 H  z  1.2 Hz
r s
Therefore the rotor power factor is
PF  R cos7.50  0.991 lagging

st tor copper losses are
(e) The a
P SCL  3I A 2 1  R   3 11.7 A  2 0.54   222 W

R
2
(f) The air gap power is P AG  3I 2 2 s 2  3I R F
A
R
2
(Note that 3IR F is equal to 3I 2 2 s 2 , since the only resistance in the original rotor circuit was R s ,
/
A
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 AG  3I 2 2 2  3I A 2 F  R   3 11.7 A   17.98   7.38 kW
s
(g) The power converted from electrical to mechanical form is
  P  1 s P   1 0.02 7.38 kW  7.23 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 7.38 kW
 ind  AG   39.2 N m


sync  1800 r/min   2 rad      1 min  
 1 r   60 s 
(i) The output power of this motor is



P  P  P  P  P  7.23 kW 150 W 150 W 50 W  6.88 kW
OUT conv mech core misc
The outp ut speed is
 n   1 s  n  1 0.02 1800 r/min  1764 r/min
m sync
Therefore the load torque is

P 6.88 kW

 load     2 rad    1 min   37.2 N m
OUT

m
 1764 r/min  1 r     60 s  
(j) The overall efficiency is
P P
  OUT  100%  OUT  100%
P IN 3V I A cos

6.88 kW
   100% 90.5% 

  3 266 V 11.7 A cos 35.5 


180