Question 4.14: A slip-ring-type three-phase induction motor rotates at a sp...

We know the synchronous speed of the rotating magnetic field produced is expressed as

N_{s} = \frac{120  f}{P}

Here, f = 50 Hz but number of poles of the stator winding has not been mentioned. The number of poles can be 2, 4, 6, 8, etc.

for                      P=2,     N_{s} = \frac{120  ×  50}{2} = 3000 rpm

for                      P=4,     N_{s} = \frac{120  ×  50}{4} = 1500 rpm

for                      P=6,     N_{s} = \frac{120  ×  50}{6} = 1000 rpm

for                      P=8,     N_{s} = \frac{120  ×  50}{8} = 750 rpm

We know that an induction motor runs at a speed slightly less than the synchronous speed. Here, N_{r} = 1440 rpm (given). Synchronous speed corresponding to this rotor speed must, therefore, be 1500 rpm Thus, N_{r} = 1440 rpm and N_{s} = 1500 rpm.

Slip,                    S = \frac{N_{s}  –  N_{r}}{N_{s}} = \frac{1500  –  1440}{1500} = 0.04

Rotor frequency,    f_{r} = S × f = 0.04  × 50 = 2 Hz