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A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively be
- a)240 V, 750 rpm
- b)240 V, 800 rpm
- c). 240 V, 780 rpm
- d)240 V, 840 rpm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor...
For same air-gap flux,
Most Upvoted Answer
A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor...
Solution:
Given data:
Power, P = 20 kW
Voltage, V = 400 V
Frequency, f = 50 Hz
Speed, N1 = 1470 rpm
Torque, T1 = 100 N-m
New frequency, f1 = 30 Hz
To keep the same air-gap flux, the voltage per phase is given by the equation:
V/f = constant
So, for the new frequency f1 = 30 Hz, the new voltage per phase V1 is given by:
V/f = V1/f1
V1 = V(f1/f) = 400(30/50) = 240 V
Now, to find the new speed N2, we can use the formula:
N1/N2 = (f1/f)
N2 = N1(f/f1) = 1470(50/30) = 2450 rpm
But the load torque is still the same, i.e., T1 = T2
So, the power developed by the motor at the new speed N2 is given by:
P2 = (T2N2)/9550
20,000 = (100N2)/9550
N2 = 1910 rpm
But this is not the same as the calculated value of N2 = 2450 rpm. This means that the load torque cannot be maintained at 100 N-m at the reduced frequency and voltage.
Therefore, we need to reduce the load torque to a value that can be maintained at the new frequency and voltage.
Now, let the new load torque be T2'. We can find the new speed N2' using the formula:
N1/N2' = (T1/T2')(f1/f)
N2' = (T2'/T1)(f/f1)N1
100 = (T2'/100)(50/30)1470
T2' = 64.29 N-m
Now, for the new load torque T2' = 64.29 N-m, the new speed N2' is given by:
N1/N2' = (T1/T2')(f1/f)
N2' = (T2'/T1)(f/f1)N1
N2' = (64.29/100)(50/30)1470 = 800 rpm
Therefore, the supply voltage and the speed for the same load torque would be 240 V and 800 rpm, respectively, which is option (b).
Given data:
Power, P = 20 kW
Voltage, V = 400 V
Frequency, f = 50 Hz
Speed, N1 = 1470 rpm
Torque, T1 = 100 N-m
New frequency, f1 = 30 Hz
To keep the same air-gap flux, the voltage per phase is given by the equation:
V/f = constant
So, for the new frequency f1 = 30 Hz, the new voltage per phase V1 is given by:
V/f = V1/f1
V1 = V(f1/f) = 400(30/50) = 240 V
Now, to find the new speed N2, we can use the formula:
N1/N2 = (f1/f)
N2 = N1(f/f1) = 1470(50/30) = 2450 rpm
But the load torque is still the same, i.e., T1 = T2
So, the power developed by the motor at the new speed N2 is given by:
P2 = (T2N2)/9550
20,000 = (100N2)/9550
N2 = 1910 rpm
But this is not the same as the calculated value of N2 = 2450 rpm. This means that the load torque cannot be maintained at 100 N-m at the reduced frequency and voltage.
Therefore, we need to reduce the load torque to a value that can be maintained at the new frequency and voltage.
Now, let the new load torque be T2'. We can find the new speed N2' using the formula:
N1/N2' = (T1/T2')(f1/f)
N2' = (T2'/T1)(f/f1)N1
100 = (T2'/100)(50/30)1470
T2' = 64.29 N-m
Now, for the new load torque T2' = 64.29 N-m, the new speed N2' is given by:
N1/N2' = (T1/T2')(f1/f)
N2' = (T2'/T1)(f/f1)N1
N2' = (64.29/100)(50/30)1470 = 800 rpm
Therefore, the supply voltage and the speed for the same load torque would be 240 V and 800 rpm, respectively, which is option (b).
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A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer?.
A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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ample number of questions to practice A 3-phase, 20 kW, 400 V, 1470 rpm, 50 Hz squirrel cage induction motor develops a torque of 100 N-m at a speed of 1400 rpm. if the motor is connected to a 30 Hz supply; for keeping the same air-gap flux, the supply voltage and the speed for the same load torque would respectively bea)240 V, 750 rpm b)240 V, 800 rpmc). 240 V, 780 rpm d)240 V, 840 rpmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
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