Question 4.6: A six-pole, three-phase synchronous generator driven at 1000...

The synchronous generator is rotated at a synchronous speed of 1000 rpm by a prime mover. The synchronous speed N_{s} is expressed as

N_{s}=\frac{120 f}{p}     where f is the frequency of the generated EMF

or,               f=\frac{N_{s} ×p}{120}

=\frac{1000×6}{120}=50  Hz

This synchronous generator now supplies power to the induction motor at a frequency of 50 Hz. The synchronous speed of the rotating magnetic field produced in the motor is

N_{s}=\frac{120 f}{p}=\frac{120× 50 }{2}= 3000  rpm (for  p=2)

and               N_{s}=\frac{120× 50 }{4}= 1500  rpm (for  p=4)

The motor speed is 1440 rpm, which should be slightly less than the synchronous speed. Logically, the number of poles of the motor must be 4.

Percentageslip,          S=\frac{N_{s} – N_{r} }{N_{s}}× 100=\frac{(1500-1440)× 100}{1500} = 4 per cent