Electric motor efficiency calculation

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The discussion focuses on calculating the efficiency of three-phase electric motors, emphasizing the correct understanding of power factor in the formula. The efficiency is defined as the ratio of work output to work input, with work input calculated as voltage times current times power factor times the square root of three. It is clarified that the formula provided pertains to power, not work, and highlights the need for accurate measurement of mechanical output, often requiring a calibrated dynamometer. Additionally, the operating conditions, such as motor temperature, can significantly affect efficiency calculations. Accurate efficiency assessment requires careful consideration of these factors.
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Hello everyone,

I am trying to calculate the electric efficiency of some 3 phase electric motors but I am a little confused about the power factor.

efficiency=work out/work in

work out=rated W of my motor (hp*745.7)

work in= V*A*PF*√3

Is the work in correct?

Thanks
 
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Your formula is for POWER. Work is Power * time. Balanced 3 phase P = V*A*PF*√3.

Work = Energy = P*T ( You pay for Kilowatt Hours - not Kilowatts)

However the trick in this type of project is typically accurately measuring the mechanical output - needing a calibrated dyno - or other device.

Also - for efficiency it is important (or more meaningful) to define what the operating point is for the efficiency calculation / measurements. --e.g. A Cold motor is more efficient then a warm motor - etc.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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