An AC electric motor is designed to produce 150 horsepower (hp) at the shaft at

Question

An AC electric motor is designed to produce 150 horsepower (hp) at the shaft at a speed of 3,000 rpm. The motor voltage supply is 480 VRMS. The power factor is 0.9. The motor is cooled entirely by water that flows throughout the housing. The water flows at a rate of 2 gallons per minute (gpm). It enters the motor at 70 F and exits at 90 F Determine the following: 1.) Period of the motor waveform in seconds. 2.) The electrical power input to the motor in kilowatts. 3.) The motor RMS current in amps. 4.) The motor resistance, RMotor, in ohms assuming that all losses are due to Joule heating.

Explanation / Answer

1) Time Period = 60/3000 = 0.02s

2)90F = 32.22C, 70F =21.11C

Change in temperature = 32.22C - 21.11C =11.11C

Loss due to water heating per minute = mcdT = Volume*Density*Specific Heat*Change in temperature

=2*3.785*4.1855*11.11*1000= 353.266 kJ

Power Loss due to water heating = 5.888 kW

Therefore power input = 150*745.7 + 5888 = 117.743 kW

3) Current = input power / (voltage * power factor) = 117743/(480*0.9) = 272.55 A

4) By Joule heating law, Heat produced = I^2*R = v^2/R

Resistance = 5888 / 272.5^2 = 0.079 ohm

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