Page 185 - Solutions Manual to accompany Electric Machinery Fundamentals

(m) What is the starting code letter for this motor?
(n) What is the maximum acceptable temperature r ise in this motor, given its insulation class?

(o) What does the service factor of this motor mean?
S OLUTION The equivalent circuit of this induction motor is shown below:

I A
R 1 jX 1 jX 2 R 2
+ j3.209 ? 0.488 ?
0.54 ? j2.093 ?

 1  s
j51.12 ? jX M R  
V ? 2  s 

23.912 ?
-

(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

+ 0.54 ? j2.093 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1

Z  F 1 1  1 1  17.98  1 j 1.10 21.13 31.7   
jX M  Z 2 j 51.12   24.41 j 3.209

The phase voltage is 460/ 3 = 266 V, so line current I L is

V 266 0 V
I  I   
R  1 jX  1 R  F jX F 0.54   j 2.093   17.98   j 11.10 
L A
I  L I  A 11.70  35.5 A

(b) The stator power factor is
PF  cos 35.5  0.81 4 lagging

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

X 3.209
  tan  1 2  tan  1  7.50
R
R 2 / s 24.4
(d) The rotor frequency is

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