When a 3-phase induction motor is operating on a

certain load with slip s, then

Induced e.m.f. per phase in rotor = sF_{2} volts

(where E_{2} = standstill rotor induced e.m.f.)

Rotor impedance per phase, Z_{2}= R_{2} + jsX_{2}

(where R_{2} and X_{2} 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, T_{f} = 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 R _{2} per phase of the 3-phase rotor winding. *

**(GATE, 1998) **

**Solution.** P_{out} = 50 H.P. = 50 × 735.5 = 36775 W ;

p = 6 ;

f = 50 Hz ;

N = 960 r.p.m. ;

I_{2} = 40 A ;

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

**Resistance, ****R _{2}: **

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

= 36775 + 300 + 1200 = 38275 W

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