Page 176 - Solutions Manual to accompany Electric Machinery Fundamentals

jX R  jX   20j   0.15   j 0.852  
Z  M 1 1   0.138  j 0.8172   0.830 80.4  
TH
R  1  1 X M  j X  0.15   j   20   0.852 

jX  20 j  


V  M V     266 0 V  255 0.41 V
TH
R  1  1 X M  j X   0.15   j   20   0.852 
The slip at pullout torque is
R
s max  2
R 2    X  X 2
TH TH 2
0.154 
s max   0.0815
 0.138  2   0.8172  1.066    2

The synchronous speed of this motor is
120 120  f 60 Hz
n  e   1800 r/min
sync
P 4
 sync   1800 r/min   2 rad      1 min    188.5 rad/s
 1 r   60 s 

This corresponds to a rotor speed of
  n  1 s   n 1  0.0815 1800 r/min  1653 r/min
max max sync
The pullout torque of the motor is

3V 2
 max  TH
 sync  R TH TH   R TH  X 2 X 2 
2




  3 255 V 2
 max  2 
  188.5 rad/s 0.138      2   0.138  0.8182    1.066 
   
 max  272.1 N m

6-17. If the motor in Problem 6-15 is to be driven from a 460-V 50-Hz power supply, what will the pullout
torque be? What will the slip be at pullout?
SOLUTION If this motor is driven from a 50 Hz source, the resistances will be unchanged and the
reactances will be increased by a ratio of 5/6. The resulting equivalent circuit is shown below.

I A
R 1 jX 1 jX 2 R 2

+ j0.890 ? 0.154 ?
0.15 ? j0.710 ?

 1  s
j16.67 ? jX M R  
V ? 2  s 

7.546 ?
-

The phase voltage must be derated by 5/6, so V  212.5 V .

170