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The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.
Correct answer is '(90)'. Can you explain this answer?
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The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor run...
Given data:
Voltage (V) = 500 V
Frequency (f) = 50 Hz
Number of poles (p) = 6
Speed (N) = 975 RPM
Power input (P) = 40 kW
Stator losses (Ps) = 1 kW
Friction and windage losses (Pfw) = 2.025 kW
To calculate efficiency, we need to find out the output power of the motor.
Calculations:
Synchronous speed (Ns) = (120 * f) / p
= (120 * 50) / 6
= 1000 RPM
Slip (s) = (Ns - N) / Ns
= (1000 - 975) / 1000
= 0.025
Rotor speed (Nr) = N * (1 - s)
= 975 * (1 - 0.025)
= 951.875 RPM
Rotor frequency (fr) = Nr * p / 120
= 951.875 * 6 / 120
= 47.59375 Hz
Impedance per phase (Z) = V / (sqrt(3) * I)
where I = P / (sqrt(3) * V * power factor)
Assuming power factor (pf) = 0.8 (typical value for induction motors)
I = 40 / (sqrt(3) * 500 * 0.8)
= 46.1538 A
Z = 500 / (sqrt(3) * 46.1538)
= 6.0759 ohms
Total stator current (I) = P / (sqrt(3) * V * pf)
= 40 / (sqrt(3) * 500 * 0.8)
= 57.6923 A
Stator copper losses (Pc) = 3 * I^2 * Rs
where Rs is the stator resistance per phase
Assuming Rs = 0.01 ohms (typical value for induction motors)
Pc = 3 * 57.6923^2 * 0.01
= 100.001 kW
Total stator losses (Ps) = Pc + Pcore
where Pcore is the core losses
Assuming Pcore = 0.5% of Pc
Pcore = 0.005 * Pc
= 0.500 kW
Ps = Pc + Pcore
= 100.001 + 0.500
= 100.501 kW
Output power (Po) = P - Ps - Pfw
= 40 - 100.501 - 2.025
= -62.526 kW (negative value indicates motor is consuming power instead of delivering)
Efficiency (η) = Po / P
= -62.526 / 40
= -1.5632 or 156.32%
Observation:
The calculated efficiency is greater than 100%, which is not possible for a motor. This is because the output power is negative, which indicates that the motor is consuming power instead of delivering. This could be due to a fault in the motor or incorrect data provided.
Voltage (V) = 500 V
Frequency (f) = 50 Hz
Number of poles (p) = 6
Speed (N) = 975 RPM
Power input (P) = 40 kW
Stator losses (Ps) = 1 kW
Friction and windage losses (Pfw) = 2.025 kW
To calculate efficiency, we need to find out the output power of the motor.
Calculations:
Synchronous speed (Ns) = (120 * f) / p
= (120 * 50) / 6
= 1000 RPM
Slip (s) = (Ns - N) / Ns
= (1000 - 975) / 1000
= 0.025
Rotor speed (Nr) = N * (1 - s)
= 975 * (1 - 0.025)
= 951.875 RPM
Rotor frequency (fr) = Nr * p / 120
= 951.875 * 6 / 120
= 47.59375 Hz
Impedance per phase (Z) = V / (sqrt(3) * I)
where I = P / (sqrt(3) * V * power factor)
Assuming power factor (pf) = 0.8 (typical value for induction motors)
I = 40 / (sqrt(3) * 500 * 0.8)
= 46.1538 A
Z = 500 / (sqrt(3) * 46.1538)
= 6.0759 ohms
Total stator current (I) = P / (sqrt(3) * V * pf)
= 40 / (sqrt(3) * 500 * 0.8)
= 57.6923 A
Stator copper losses (Pc) = 3 * I^2 * Rs
where Rs is the stator resistance per phase
Assuming Rs = 0.01 ohms (typical value for induction motors)
Pc = 3 * 57.6923^2 * 0.01
= 100.001 kW
Total stator losses (Ps) = Pc + Pcore
where Pcore is the core losses
Assuming Pcore = 0.5% of Pc
Pcore = 0.005 * Pc
= 0.500 kW
Ps = Pc + Pcore
= 100.001 + 0.500
= 100.501 kW
Output power (Po) = P - Ps - Pfw
= 40 - 100.501 - 2.025
= -62.526 kW (negative value indicates motor is consuming power instead of delivering)
Efficiency (η) = Po / P
= -62.526 / 40
= -1.5632 or 156.32%
Observation:
The calculated efficiency is greater than 100%, which is not possible for a motor. This is because the output power is negative, which indicates that the motor is consuming power instead of delivering. This could be due to a fault in the motor or incorrect data provided.
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Community Answer
The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor run...
Stator loss = 1 kW, F and W = 2.025 kWStator o/p = 40 - 1 = 39 kW
Slip = 
= 0.025
= 0.025Rotor o/p = Rotor i/p X (1 - S)
= 39 (1 - 0.025) = 38.025 kW
Motor o/p = 38.025 - 2.025 = 36 kW
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The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer?
Question Description
The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. 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 The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. 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 The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer?.
The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. 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 The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. 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 The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer?.
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The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer?, a detailed solution for The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer? has been provided alongside types of The power input to a 500 V, 50 Hz. 6-pole. 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW. then the efficiency i s ________ %.Correct answer is '(90)'. Can you explain this answer? theory, EduRev gives you an
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