Page 195 - Solutions Manual to accompany Electric Machinery Fundamentals

voltage V  (and hence its starting current) by simply re-connecting the stator windings in Y during
starting, and then restoring the connections to  when the motor comes up to speed. Answer the
following questions about this type of starter.
(a) How would the phase voltage at starting compare with the phase voltage under normal running
conditions?
(b) How would the starting current of the Y-connected motor compare to the starting current if the
motor remained in a -connection during starting?
SOLUTION

(a) The phase voltage at starting would be 1 / 3 = 57.7% of the phase voltage under normal running
conditions.

(b) Since the phase voltage decreases to 1 / 3 = 57.7% of the normal voltage, the starting phase
current will also decrease to 57.7% of the normal starting current. However, since the line current for the
original delta connection was 3 times the phase current, while the line current for the Y starter
connection is equal to its phase current, the line current is reduced by a factor of 3 in a Y- starter.
For the -connection: I L ,   3 I ,  

For the Y-connection: I L ,Y  I  ,Y
But I ,   3I ,   Y , so I L ,  3I , L Y

6-26. A 460-V 50-hp six-pole -connected 60-Hz three-phase induction motor has a full-load slip of 4 percent,
an efficiency of 91 percent, and a power factor of 0.87 lagging. At start-up, the motor develops 1.75
times the full-load torque but draws 7 times the rated current at the rated voltage. This motor is to be
started with an autotransformer reduced voltage starter.
(a) What should the output voltage of the starter circuit be to reduce the starting torque until it equals
the rated torque of the motor?

(b) What will the motor starting current and the current drawn from the supply be at this voltage?
SOLUTION

(a) The starting torque of an induction motor is proportional to the square of V TH ,

  V  2  V  2
start2  TH2  T 2
 start1   V TH1     V  T 1 

If a torque of 1.75  rated is produced by a voltage of 460 V, then a torque of 1.00  rated would be produced
by a voltage of

1.00  rated   V T 2  2
1.75  rated   460 V  

 460 V 2
V   348 V
2
T
1.75
(b) The motor starting current is directly proportional to the starting voltage, so
 348 V 

I L 2    L 1  I  0.757  L 1  I  0.757 7I rated   5.30 I rated
 460 V 
The input power to this motor is
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