Question 9.5: 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

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 \times 50}{2}=3000 rpm

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

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

for       P = 8, N_{s}=\frac{120 \times 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 \times f=0.04 \times 50=2  Hz