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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?
    Most Upvoted Answer
    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.
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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 kW
    Stator o/p = 40 - 1 = 39 kW
    Slip = = 0.025
    Rotor 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?.
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