Page 187 - Solutions Manual to accompany Electric Machinery Fundamentals

(k) The m otor speed in revolutions per minute is 1764 r/min. The motor speed in radians per second is

  m  1764 r/min   2 rad      1 min    184.7 rad/s
 1 r   60 s 

(l) The power flow diagram for this motor is

7.38 kW 7.23 kW

6.88 kW
7.60 kW

150 W 50 W

150 W 150 W
222 W

(m) The equivalent circuit of this induction motor at starting conditions is shown below:
I A
R 1 jX 1 jX 2 R 2

+ j3.209 ? 0.488 ?
0.54 ? j2.093 ?

j51.12 ? jX M
V ?

-

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  0.435  2.864 2.90 81.3   j  

jX M  Z 2 j 51.12   0.488  j 3.209

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