# Factors determining torque

When a 3-phase induction motor is operating on a
certain load with slip s, then

Induced e.m.f. per phase in rotor = sF2 volts

(where E2 = standstill rotor induced e.m.f.)

Rotor impedance per phase, Z2= R2 + jsX2

(where R2 and X2 are rotor resistance and standstill reactance per phase)

ROTOR OUTPUT, LOSSES AND EFFICIENCY

Example 12. The power input to a 3-phase induction motor is 50 k W. The stator-losses total 1.2 kW. If the motor is running with a slip of 3 per cent, find the total mechanical power developed,

Solution. Stator power input = 50 kW

Stator losses = 1.2 kW

Slip                                  , = 3% (= 0.03)

Mechanical power developed:

Rotor input = stator output = stator input – stator losses

= 50 – 1.2 = 48.8 kW

Now mechanical power developed

= (1 – s) rotor input

= (1- 0.03) × 48.8

= 47.34 kW. (Ans.)

Example 13.A 6-pole, 50-Hz, 3-phase induction motor running on full-load with 3% slip develops a torque of 160N-m at its pulley rim. The friction and windage losses are 210 Wand the stator copper and iron losses equal 1640 W. Calculate:

(i)                           Rotor output,

(ii)                        Rotor copper loss, and

(iii)                      Efficiency at full-load.

Solution. Number of poles, p = 6

Supply frequency, f = 50 Hz

Slip, s = 3% or 0.03

Torque developed at full-load, Tf = 160 N-m

Friction and windage losses = 210 W

Stator copper and iron losses = 1640 W

Example 14. The power input to the rotor of a 440 V, 50-Hz, 3-phase, 6-pole induction motor is 50 kW. It is observed that the rotor e.m.f makes 120 complete cycles per minute. Calculate:

(i)                           Slip,

(ii)                        Rotor speed,

(iii)                      Rotor copper loss/phase,

(iv)                      Mechanical power developed, and

(v)                         Rotor resistance/phase if rotor current is 50 A,

Solution. Supply voltage = 440 V

Supply frequency, f = 50-Hz

Number of poles, p = 6

Rotor input = 50 kW or 50000 W

Example 15.A 50 H.P., 6-pole, 50 Hz, slip-ring induction motor runs at 960 r.p.m. on full-load with a rotor current of 40 A. Allowing 300 W for copper loss in the short-circuiting gear and 1200 W for mechanical losses, find the resistance R2 per phase of the 3-phase rotor winding.

(GATE, 1998)

Solution. Pout = 50 H.P. = 50 × 735.5 = 36775 W ;

p = 6 ;

f = 50 Hz ;

N = 960 r.p.m. ;

I2 = 40 A ;

copper loss in the short-circuiting gear = 300 W ; mechanical loss = 1200 W.

Resistance, R2:

Rotor output = motor output + short-circuiting gear loss + mechanical losses

= 36775 + 300 + 1200 = 38275 W