Page 173 - Solutions Manual to accompany Electric Machinery Fundamentals

I A
R 1 jX 1 jX 2 R 2

+ j1.066 ? 0.154 ?
0.15 ? j0.852 ?
 1  s
j20 ? jX M R  
V ? 2  s 

7.546 ?
-

(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

+
0.15 ? j0.852 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1
Z  F 1 1  1 1  6.123 j 3.25 6.932 28.0   

jX M  Z 2 j 20   7.70  j 1.066

The phase voltage is 460/ 3 = 266 V, so line current I L is
V 266 V
0
I  L I  A  
R  1 jX  1 R  F jX F 0.15   j 0.852   6.123   j 3.25 

I  L I  A 35.5  33.2 A
(b) The stator power factor is
PF  cos 33.2  0.83 7 lagging

u
(c) To find the rotor power factor, we m st find the impedance angle of the rotor
X 1.066
  tan  1 2  tan  1  7.88
R
R 2 / s 7.70
(d) The rotor frequency is

f  r s  sf  0.02 60 H  z  1.2 Hz
Therefore the rotor power factor is

PF  R cos7.88  0.991 lagging

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