Question 9.9: A 15 hp, three-phase, four-pole, 50 Hz induction motor has f...
A 15 hp, three-phase, four-pole, 50 Hz induction motor has full-load speed of 1455 rpm. The friction and windage loss of the motor at this speed is 600 W. Calculate the rotor copper loss.
N_{s}=\frac{120 f}{P}=\frac{120 \times 50}{4}=1500 rpm
Slip, S=\frac{N_{s}-N_{r}}{N_{s}}=\frac{1500-1455}{1500}=0.03
Motor output available at the shaft = 15 hp
= 15 × 735.5 W
= 11032 W
Now, let us look at the power flow diagram
Power developed by rotor = Shaft power output + Friction and Windage losses
= 11032 + 600
= 11632 W
We know the relation
Rotor copper loss = S × Rotor input (See eq. 9.3)
f_{r}=Sfi.e., rotor frequency = Slip × Stator frequency (9.3)
(Slip × Rotor input) is lost as rotor copper loss
The remaining power, i.e., (1 − S) Rotor input, is developed as the rotor power.
Therefore,
Power developed by rotor = (1 − S) Rotor input
=\frac{(1 − S) \text {Rotor copper loss}}{S}Rotor copper loss =\frac{S}{(1 − S)} × Rotor power developed
Thus,
=\frac{0.3 \times 11632}{(1 – 0.03)} W
= 360 W
