Distinguish shaft horsepower, electrical real power, and apparent power. Work a 10 hp example with 90% efficiency and 0.8 power factor, and see why the idealized shortcut is lower.
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The calculation
Running kVA = shaft hp × 0.7457 ÷ motor efficiency ÷ power factor. An idealized hp × 0.746 ÷ PF shortcut assumes 100% motor efficiency.
Worked example
10 mechanical hp is about 7.46 kW at the shaft. At 90% motor efficiency it needs about 8.29 kW real electrical input. With 0.8 power factor this is about 10.36 kVA apparent input. Ignoring efficiency gives an idealized 9.325 kVA.
The idealized 9.325-kVA shortcut assumes 100% efficiency when horsepower names motor shaft output. Use motor efficiency and power factor for a realistic running estimate; starting current and equipment ratings require separate consideration.
Video chapters
- 0:00 — Why horsepower does not equal kVA
- 0:18 — Convert shaft power to kilowatts
- 0:32 — Account for motor efficiency
- 0:49 — Account for power factor
- 1:05 — Combine the complete estimate
- 1:24 — What the simple shortcut assumes
- 1:44 — Check the estimate and its limits
- 2:04 — State both assumptions
Read the full transcript
A motor marked ten horsepower describes mechanical power delivered at its shaft. Kilovolt-amperes describe apparent electrical power supplied to the motor. Between them are two different effects: electrical losses inside the motor, and power factor on the supply. We need both to estimate a real motor's running kVA.
One mechanical horsepower is about zero point seven four five seven kilowatts. Ten horsepower is therefore about seven point four six kilowatts of shaft power. That conversion changes units; it does not yet include motor losses or reactive electrical demand.
Suppose the motor is ninety percent efficient at this operating point. The shaft receives ninety percent of the real electrical power going in. Divide seven point four five seven kilowatts by zero point nine. The input is about eight point two nine kilowatts; the difference becomes losses, mostly heat.
Power factor is real kilowatts divided by apparent kilovolt-amperes. At a power factor of zero point eight, divide the eight point two nine input kilowatts by zero point eight. The motor's estimated running apparent power is about ten point three six kilovolt-amperes.
In one line, multiply shaft horsepower by zero point seven four five seven, then divide by motor efficiency and by power factor. Here that is ten times zero point seven four five seven, divided by zero point nine, then divided by zero point eight. Keep the two divisors separate because they describe different physical effects.
A common shortcut omits efficiency: horsepower times zero point seven four six, divided by power factor. For ten horsepower and zero point eight, it returns nine point three two five kilovolt-amperes. That is an idealized result if the horsepower is shaft output, because it assumes the motor loses no power.
Reverse the full estimate: ten point three six kVA times zero point eight power factor gives about eight point two nine electrical kilowatts. Multiply by zero point nine efficiency and you recover roughly seven point four six shaft kilowatts, or ten horsepower. This is a running estimate; motor starting current can be much higher.
Horsepower to kVA is not a fixed unit conversion for a real motor. Convert shaft horsepower to kilowatts, account for motor efficiency, then divide the electrical kilowatts by power factor. For our stated ninety-percent and zero-point-eight assumptions, ten horsepower needs about ten point three six running kVA, not the lossless shortcut's nine point three two five.
Sources
- US Department of Energy — Motor-efficiency guide (PDF)
- US Department of Energy — Motor power-factor guide (PDF)
- NIST — SI conversion factors, Appendix B.8