Page 198 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION The equivalent circuit of this induction motor is shown below:

I A
R 1 jX 1 jX 2 R 2

+ j0.386 ? 0.037 ?
0.058 ? j0.320 ?
   s
1
j9.24 ? jX M R  
 s 
V ? 2
3.663 ?
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by
the sum of the series impedances, as shown below.

I A
R 1 jX 1 jX F R F

+ 0.058 ? j0.320 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1
Z  F 1 1  1 1  2.970  j 1.512 3.333 27.0   

jX M  Z 2 j 9.24   3.70  j 0.386

The phase voltage is 460/ 3 = 266 V, so line current I L is

V 266 0 V
 I  I  
L A
R  1 jX  1 R  F jX F 0.058   j 0.32   2.97   j 1.512 

 I  I 75.2   31.2 A
L A
(b) The stator power factor is
PF  cos 31.2  0.855 lagging

(c) To find the rotor power factor, we must find the impedance angle of the rotor

  tan  1 X 2  tan  1 0.386  5.96
R
R / s 0.037 / 0.01
2
(d) The rotor frequency is

f   sf  0.01 60 Hz   0.6 Hz
r s
Therefore the rotor power factor is
PF  R cos5.96  0.995 lagging

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