Question 9.10: A 10 hp, four-pole, 50 Hz, three-phase induction motor has f...

Output power             = 10 hp = 10 × 735.5 W

= 7355 W

and slip,             S = 0.04

Friction and windage loss             = 0.03 × 7355 W

= 220.6 W

Power developed             = 7355 − 220.6 = 7134.4 W

Rotor copper loss             = S × Rotor input,       (i)

Power developed             = (1 − S) Rotor input       (ii)

From (i) and (ii),

Power developed   = (1− S) \frac{\text {Rotor copper loss}}{S}

or,            Rotor copper loss =\frac{S}{(1− S)} Power developed

Substituting values,             Rotor copper loss =\frac{0.04}{0.96} \times 7134.4 W

= 297.3 W

Rotor input − Rotor copper loss = Power developed

Therefore, Rotor input = Power developed + Rotor copper loss

= 7134.4 + 297.3

= 7431.7 W

Let input power be = X W

Out of this, 0.06 X W is wasted as stator losses. Stator output = Rotor input = 0.94 X W. Equating with actual values, 0.94 X = 7431.7 W

or,           X = 7906 W

Percentage Efficiency = \frac{\text {Output}}{\text {Input}} \times 100 =\frac{7355 \times 100}{7906}=93 per cent