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9.1.8

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Publicada em: Negócios, Tecnologia
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9.1.8

  1. 1. Power in = √3×VL×IL×cosø
  2. 2. Where: Power in (Pin) = Watts or kilowatts V = Supply voltage volts I = Stator Current in amperes ø = Power factor
  3. 3. Efficiency is a comparison between input and output power expressed as a percentage.
  4. 4. Example Calculation for Efficiency: A 40 kW 415 volt three phase induction motor requires a full load current of 70 amps per terminal at a power factor of 0.88. What is the motorʼs full load efficiency?
  5. 5. Answer: The motor has a power output of 40 kW the motor has an input of power determined by Pin = √3 x VL x IL x Cos ø
  6. 6. Pin = √3×415×70×.88 = 44.278 kW
  7. 7. Efficiency is Power out x 100 % Power in
  8. 8. Therefore efficiency is 40, 000 x 100 = 90.3% 44,278

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