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A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction motor has a rotor impedance of 0.02 + 0.08 Ω/phase and negligible stator impedance at standstill. Consider the following statements.
1. The breakdown slip at which maximum torque occurs is 0.25
2. The starting torque is 47% of the maximum torque
3. The maximum torque developed by the motor is 1364.4 Nm
Q. Which of the above statements is/are correct?
  • a)
    1 and 2
  • b)
    1,2 and 3
  • c)
    2 and 3
  • d)
    1 only
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction...
The breakdown slip is,

The ratio of the torque developed at any slip s to the maximum torque is

For starting torque s = 1, so

Starting torque is 47% of maximum torque.
Synchronous speed, 

Maximum power developed by the motor
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Most Upvoted Answer
A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction...
J ohms per phase. The stator winding resistance and leakage reactance are negligible. The motor is operating with a slip of 0.05 and a rotor current of 20 A per phase.

To determine the following:

a) The rotor copper loss per phase

b) The rotor core loss per phase

c) The mechanical power developed

d) The output power

e) The input power

f) The efficiency

g) The power factor

h) The torque developed

Solution:

a) The rotor copper loss per phase can be calculated using the formula:

Pcopper = 3I2R2

where I is the current per phase and R2 is the rotor resistance per phase.

Pcopper = 3 x 202 x 0.02 = 2.4 kW

b) The rotor core loss per phase can be assumed to be negligible.

c) The mechanical power developed can be calculated using the formula:

Pmech = (1 - s) x Pout

where s is the slip and Pout is the output power.

Pout = 3VphIphcos(θ)

where Vph is the phase voltage, Iph is the phase current, and θ is the phase angle between voltage and current.

Vph = 208/√3 = 120 V

Iph = I = 20 A

θ = 0 (since the rotor impedance is purely resistive)

Pout = 3 x 120 x 20 x 1 = 7.2 kW

Pmech = (1 - 0.05) x 7.2 = 6.84 kW

d) The output power is already calculated in part c) as 7.2 kW.

e) The input power can be calculated as:

Pin = Pout + Ploss

where Ploss is the total losses in the motor.

Ploss = Pcopper + Protor + Pcore

where Protor is the rotor iron loss per phase.

From the given data, Protor can be assumed to be negligible.

Ploss = Pcopper + Pcore = 2.4 kW

Pin = 7.2 + 2.4 = 9.6 kW

f) The efficiency can be calculated as:

η = Pout / Pin

η = 7.2 / 9.6 = 0.75 or 75%

g) The power factor can be calculated as:

PF = Pout / (3VphIph)

PF = 7.2 / (3 x 120 x 20) = 0.02 or 0.02 lagging

h) The torque developed can be calculated using the formula:

T = (1 - s) x (3VphIph / ω)

where ω is the angular frequency in radians per second.

ω = 2πf = 2π x 60 = 377 rad/s

T = (1 - 0.05) x (3 x 120 x 20 / 377) = 94.5 Nm.
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A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction motor has a rotor impedance of 0.02 + 0.08 Ω/phase and negligible stator impedance at standstill. Consider the following statements.1. The breakdown slip at which maximum torque occurs is 0.252. The starting torque is 47% of the maximum torque3. The maximum torque developed by the motor is 1364.4 NmQ.Which of the above statements is/are correct?a)1 and 2b)1,2 and 3c)2 and 3d)1 onlyCorrect answer is option 'A'. Can you explain this answer?
Question Description
A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction motor has a rotor impedance of 0.02 + 0.08 Ω/phase and negligible stator impedance at standstill. Consider the following statements.1. The breakdown slip at which maximum torque occurs is 0.252. The starting torque is 47% of the maximum torque3. The maximum torque developed by the motor is 1364.4 NmQ.Which of the above statements is/are correct?a)1 and 2b)1,2 and 3c)2 and 3d)1 onlyCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction motor has a rotor impedance of 0.02 + 0.08 Ω/phase and negligible stator impedance at standstill. Consider the following statements.1. The breakdown slip at which maximum torque occurs is 0.252. The starting torque is 47% of the maximum torque3. The maximum torque developed by the motor is 1364.4 NmQ.Which of the above statements is/are correct?a)1 and 2b)1,2 and 3c)2 and 3d)1 onlyCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 208 V, 8 pole 60 Hz three phase star connected wound rotor induction motor has a rotor impedance of 0.02 + 0.08 Ω/phase and negligible stator impedance at standstill. Consider the following statements.1. The breakdown slip at which maximum torque occurs is 0.252. The starting torque is 47% of the maximum torque3. The maximum torque developed by the motor is 1364.4 NmQ.Which of the above statements is/are correct?a)1 and 2b)1,2 and 3c)2 and 3d)1 onlyCorrect answer is option 'A'. Can you explain this answer?.
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