Page 175 - Solutions Manual to accompany Electric Machinery Fundamentals

(l) The equivalent circuit of this induction motor at starting conditions is shown below:

I A
R 1 jX 1 jX 2 R 2

+ 0.15 ? j0.852 ? j1.066 ? 0.154 ?

j20 ? 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

+ 0.15 ? j0.852 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1

Z  F 1 1  1 1  0.139  j 1.013 1.023 82.2   
jX M  Z 2 j 20   0.154  j 1.066

The phase voltage is 460/ 3 = 266 V, so line current I L is

V 266 0 V
I  I   
A
L
R  1 jX  1 R  F jX F 0.15   j 0.852   0.139   j 1.023 

I  L I  A 140.2   81.2 A
The starting kVA of the motor is

S start  3V  A  I   3 266 V 140 A   111.7 kVA
The locked rotor kVA/hp is
111.7 kVA
kVA/hp   4.47
25 hp

Therefore this motor is Starting Code Letter D.

6-16. For the motor in Problem 6-15, what is the pullout torque? What is the slip at the pullout torque? What is
the rotor speed at the pullout torque?

SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit
from the rotor back to the power supply, and then using that with the rotor circuit model.
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