Page 200 - Solutions Manual to accompany Electric Machinery Fundamentals

(l) The power flow diagram for this motor is

50.4 kW 49.9 kW

48.5 kW
51.3 kW

650 W 150 W

600 W 500 W
984 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

+ j0.386 ? 0.037 ?
0.058 ? j0.320 ?

j9.24 ? 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

+
0.058 ? j0.320 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1

Z  F 1 1  1 1  0.0341 j 0.3707 0.372 84.7   
jX M  Z 2 j 9.24   0.037  j 0.386

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 0.058   j 0.32   0.0341   j 0.3707 
L A
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