Induction Motor MCQs Part 18

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Induction Motor MCQs Part 18

A wound rotor motor is used in practical applications that require

A wound rotor motor is used in practical applications that require

  1. Less costly motor is not required
  2. High rotor resistance is required during
  3. High starting torque is required
  4. Speed control is required

Correct answer: 3. High starting torque is required

A high starting torque can be obtained in 3 phase induction motor by doing which of following

A high starting torque can be obtained in 3 phase induction motor by doing which of following:

  1. Decreasing rotor resistance
  2. Increasing rotor reactance
  3. Increasing rotor resistance
  4. None of these

Correct answer: 3. Increasing rotor resistance

The number of rotor slots in a squirrel cage motor are __________ rotor slots

The number of rotor slots in a squirrel cage motor are __________ rotor slots

  1. More than
  2. Less than
  3. Either of above
  4. None of above

Correct answer: 3. Either of above

A 3-phase induction motor is connected to a normal supply voltage. EMF induced between slip rings

A 3-phase induction motor is connected to a normal supply voltage. EMF induced between slip rings at standstill is 50. The resistance and standstill reactance per phase are 0.7 Ω and 3.5 Ω respectively. The rotor (star-connected) is joined to a star connected resistance of 4 Ω/phase. The rotor phase current at starting will be:

  1. 2.25 A
  2. 3.56 A
  3. 4.93 A
  4. 9.92 A

Correct answer: 3. 4.93 A

Solution:

At standstill, rotor EMF per phase (E2) = 50/√3 = 28.87 V

Rotor impedance/phase at standstill Z2 = √{4 + 0.7}² + 3.5² = 5.86 Ω

Rotor current/phase at starting (I2) = E2/Z2 = 28.87/5.86 = 4.93 A

A 3-phase, 50 Hz, 8 pole induction motor running with a slip of 4% is taking 20 kW

A 3-phase, 50 Hz, 8 pole induction motor running with a slip of 4% is taking 20 kW. Stator losses amount 0.5 kW. If mechanical torque lost in friction is 16 Nm, what is BHP:

  1. 23.45
  2. 48.54
  3. 69.64
  4. 99.56

Correct answer: 1. 23.45

Solution:

Stator output = Stator input – Stator losses = 20 – 0.5 = 19.5 kW

Stator output = Rotor input = 19.5 kW

———————-

Rotor Cu loss = s * Rotor input = 0.04 * 19.5 = 0.78 kW

Rotor output = 19.5 – 0.78 = 18.72 kW

———————-

Ns = 120 * f/P = 120 * 50/8 = 750 RPM

N = (1 – s) * Ns = (1 – 0.04) * 750 = 720 RPM

———————-

Total torque (T) = 9.55 * Rotor output/N

T = 9.55 * 18.72 * 1000 / 720 = 248 Nm

———————-

Tsh = T – T (lost) = 248 – 16 = 232 Nm

———————-

BHP = 2π N Tsh / 60*746 = 2π * 720 * 232 / 60 * 746 = 23.45

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