Induction Motor MCQs Part 12

Contents

- 1 The value of full-load slip of a 3-phase induction motor ranges from
- 2 In 3 phase induction motor, for higher efficient, the slip should be
- 3 A 3-phase has 4 pole star-connected stator winding and runs on 230 V, 60 Hz supply. The rotor resistance is 0.2 Ω/phase. The standstill reactance is 0.8 Ω/phase. The full load slip and ratio of stator to rotor turns are 3% and 1.8 respectively. The rotor input is
- 4 The efficiency of induction motor can be greater than 1-s
- 5 A 415 V, 20 kW, 4-pole, 50 Hz, Y-connected induction motor has full-load slip of 3%. The output torque of machine at full-load is:

## The value of full-load slip of a 3-phase induction motor ranges from

The value of full-load slip of a 3-phase induction motor ranges from

- 2% to 5%
- 10% to 20%
- 20% to 30%
- 90% to 100%

Correct answer: 1. 2% to 5%

## In 3 phase induction motor, for higher efficient, the slip should be

In 3 phase induction motor, for higher efficient, the slip should be:

- As small as possible
- Large
- Very large
- 1

Correct answer: 1. As small as possible

## A 3-phase has 4 pole star-connected stator winding and runs on 230 V, 60 Hz supply. The rotor resistance is 0.2 Ω/phase. The standstill reactance is 0.8 Ω/phase. The full load slip and ratio of stator to rotor turns are 3% and 1.8 respectively. The rotor input is

A 3-phase has 4 pole star-connected stator winding and runs on 230 V, 60 Hz supply. The rotor resistance is 0.2 Ω/phase. The standstill reactance is 0.8 Ω/phase. The full load slip and ratio of stator to rotor turns are 3% and 1.8 respectively. The rotor input is:

- 1600 W
- 2400 W
- 3800 W
- 4600 W

Correct answer: 2. 2400 W

Solution:

Rotor impedance per phase at slip (s) = 0.03 is:

Z2 = (R2/s) + jX2

Z2 = 0.2/0.03 + j0.8 = 6.66666666666667 + 0.8j = 6.71 Ω

Stator voltage/phase (E1) = 230/√3 = 133 V

Rotor voltage/phase (E2) = 1/1.8 = 133 = 73.77 V

Rotor current/phase (I2) = E2/Z2 = 73.77/6.71 = 10.99 A

——————————-

Rotor Cu loss at 3% slip = 3. I2²R2 = 3 * 10.99² * 0.2 = 72 W

——————————

Rotor input = Rotor Cu loss/s = 72/0.03 = 2400 W

## The efficiency of induction motor can be greater than 1-s

The efficiency of induction motor can be greater than 1-s:

- True
- False
- Sometimes true, sometimes false
- None of above

Correct answer: 2. False

## A 415 V, 20 kW, 4-pole, 50 Hz, Y-connected induction motor has full-load slip of 3%. The output torque of machine at full-load is:

A 415 V, 20 kW, 4-pole, 50 Hz, Y-connected induction motor has full-load slip of 3%. The output torque of machine at full-load is:

- 89.95 Nm
- 131.3 Nm
- 162.4 Nm
- 211.3 Nm

Correct answer: 2. 131.3 Nm

Solution:

Synchronous speed (Ns) = 120f/P = 120*50/4 = 1500 RPM

————-

Motor speed (N) = Ns ( 1 – s) = 1500 * ( 1 – 0.03) = 1455 RPM

————-

Angular velocity of motor (ωm) is:

ωm = 1455 RPM = 1455/60 RPS = 1455/60 * 2π rad/s = 152.37 rad/s

————-

Output torque Tsh = Output power/ωm in rad/s = 20 * 10^3/152.37 = 131.3 Nm