Page 196 - Solutions Manual to accompany Electric Machinery Fundamentals

P 50 hp 746 W/hp 
P  OUT   41.0 kW
IN
 0.91
The rated current is equal to

P  41.0 kW
I rated  IN   59.15 A

3 V T PF   3 460 V 0.87 
Therefore, the motor starting current is


I L 2  5.445 rated   I 5.30 59.15 A   313.5 A
The turns ratio of the autotransformer that produces this starting voltage is
N SE  N C  460 V  1.322
N C 348 V

so the current drawn from the supply will be
I 313.5 A
I  start   237 A
line
1.377 1.322
6-27. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and
with a load of about 25 percent of the rated value for the machine. If the rotor resistance of this machine
is doubled by inserting external resistors into the rotor circuit, explain what happens to the following:
(a) Slip s
(b) Motor speed n
m
(c) The induced voltage in the rotor
(d) The rotor current

(e)  ind

(f) P out
(g) P RCL

(h) Overall efficiency 

SOLUTION
(a) The slip s will increase.

(b) The motor speed n m will decrease.
(c) The induced voltage in the rotor will increase.

(d) The rotor current will increase.
(e) The induced torque will adjust to supply the load’s torque requirements at the new speed. This will
depend on the shape of the load’s torque-speed characteristic. For most loads, the induced torque will
decrease.

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