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Motor Efficiency

η=100 PoutPin\eta = \frac{100 \, P_{out}}{P_{in}}

Worked example: 18.5 kW out of 20 kW in → 92.5% — press Try an example to run it live, then adjust anything.

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Motor Efficiency explained

MηPinPout

The difference between input and output is heat: stator and rotor I²R losses, core losses, windage and friction. A 10 hp motor at 88% efficiency swallows 7457/0.88 ≈ 8474 W to deliver 7457 W of shaft power, dumping about a kilowatt into the room — which is why motor rooms need ventilation and why efficiency shows up twice in an energy audit, once as electricity and once as cooling load.

Efficiency is not constant: it peaks near 75–100% of rated load and falls off a cliff below about 40%, which is the strongest argument against habitually oversizing motors. Since the 1990s, minimum efficiencies have been legislated — IE3 "premium" class in the IEC world, EPAct and NEMA Premium in the US — and the gain from an IE1 to an IE3 machine, a few points, usually repays the price difference within a year of continuous running.

Motor Efficiency formula

η=100 PoutPin\eta = \frac{100 \, P_{out}}{P_{in}}
Where
  • η\eta= Efficiency (%)
  • PoutP_{out}= Output power (W)
  • PinP_{in}= Input power (W)

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Part of this formula set

Motor calculations

5 formulasFull-load current, synchronous speed, slip, efficiency and input power — reading a motor nameplate and checking what it actually draws.