Page 161 - Solutions Manual to accompany Electric Machinery Fundamentals

I A
R 1 jX 1 jX F R F

+
0.10 ? j0.21 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1
Z  F 1 1  1 1  1.318  j 0.386 1.374 16.3   

jX M  Z 2 j 10   1.40  j 0.21

The phase voltage is 208/ 3 = 120 V, so line current I L is
V 120 0 V

I  L I  A  
R  1 jX  1 R  F jX F 0.10   j 0.21   1.318   j 0.386 

I  L I  A 78.0  22.8 A
(b) The stator copper losses are

P SCL  3I A 2 1  R   3 78.0 A 2  0.10   1825 W

R
2
(c) The air gap power is P  3I 2 2  3I R
2
AG
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 78.0 A   1.318   24.0 kW
AG
2
s A F
(d) The power converted from electrical to mechanical form is

  P  1 s P   1 0.05 24.0 kW   22.8 kW
conv AG
(e) The induced torque in the motor is
P 24.0 kW
  AG   127.4 N m

 sync  1800 r/min  2 rad    1 min 
ind
  1 r     60 s  
(f) The output power of this motor is



P OUT  P conv  P mech  P core  P misc  22.8 kW 500 W 400 W 0 W  21.9 kW
The output speed is

 n   1 s n   1 0.05 1800 r/min   1710 r/min
m sync
Therefore the load torque is
155