Page 717 - Basic Electrical Engineering

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. 8.3)

(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

Thus,

Example 8.10 A 10 hp, four-pole, 50 Hz, three-phase induction motor has
friction and windage loss of 3 per cent of output. Calculate at full load the

rotor copper loss, rotor input for a full-load slip of 4 per cent. If at this load
stator loss is 6 per cent of the input power, calculate the efficiency.

Solution:

Output power = 10 hp = 10 × 735.5 W