Motor kW Calculator
Calculate the electrical input kW of an induction motor from shaft HP, efficiency, and power factor.
Common conversions
| Input | Result |
|---|---|
| 1 HP @ η 0.85 | 0.88 kW |
| 5 HP @ η 0.86 | 4.34 kW |
| 7.5 HP @ η 0.88 | 6.36 kW |
| 10 HP @ η 0.90 | 8.29 kW |
| 15 HP @ η 0.91 | 12.29 kW |
| 20 HP @ η 0.92 | 16.21 kW |
| 25 HP @ η 0.93 | 20.05 kW |
| 50 HP @ η 0.94 | 39.67 kW |
The math behind it
- kW = (25 × 0.7457) / 0.91
- kW = 18.6425 / 0.91
- kW ≈ 20.49 kW
Everything you need to know
A motor's nameplate horsepower or kW rating describes mechanical output at the shaft, not the electrical power it draws from the panel. The difference is lost to copper, iron, friction, and windage losses inside the machine. NEMA Premium and IE3/IE4 motors typically run 88% to 96% efficient at full load, with larger frames toward the top of that range and small fractional-HP motors toward the bottom.
Reading the nameplate
A typical induction motor nameplate lists rated HP or kW (shaft output), voltage, full-load amps (FLA), power factor, efficiency, service factor, and NEMA or IEC frame size. The HP/kW figure is always output, never input, on both NEMA and IEC nameplates. FLA is measured at rated voltage, rated load, and rated frequency, and it already accounts for the motor's efficiency and power factor, so there's no need to calculate FLA from kW if it's already printed on the plate.
Where efficiency losses go
Four loss categories separate input kW from output kW: stator and rotor copper losses (I²R heating in the windings, usually the largest single category), iron losses (hysteresis and eddy currents in the laminated core), friction and windage (bearings and the cooling fan), and stray load losses (leakage flux that grows with load). A motor with 90% efficiency wastes 10% of its input as heat across these four paths, which is why continuously run motors benefit from upgrading to a higher IE class.
From input kW to full-load amps
Once input kW is known, convert to current with the standard kW-to-amps formula for the supply type (single-phase or three-phase) using the motor's nameplate power factor. For code compliance in North America, compare the calculated FLA against NEC Table 430.250, which lists standardized full-load current values by HP and voltage; use the table value for conductor and breaker sizing whenever it differs from a calculation.
Common applications
NEC requires conductors rated at 125% of motor FLA. Starting from input kW gives you the FLA needed for proper conductor sizing.
Drive ratings list output kW or HP. Match the drive's continuous output to the motor's input kW with at least 10% headroom for harmonics.
Replacing an 88% motor with a 95% IE3 unit cuts input kW by about 7% for the same shaft output, meaningful over 8,000 run-hours a year.
Backup power systems must supply the motor's electrical input kW, not its shaft HP. A 50 HP motor at 94% efficiency needs about 39.7 kW of generator capacity for that load alone.
Common mistakes
Using HP × 0.7457 alone gives shaft kW. Always divide by η for the input kW used in service sizing.
Motors lose 3-8% efficiency below 50% load. Don't oversize for 'safety'; the extra capacity gets paid for every hour of operation.
A 1.15 SF motor rated 10 HP can handle 11.5 HP briefly, but input-kW calculations for normal duty should use the 10 HP rating, not 11.5.
A 1 HP motor might run at 82% efficiency while a 100 HP motor from the same product line runs at 95%. Use the nameplate value, not a single rule of thumb, whenever it's available.
Frequently asked questions
What input power does a 10 HP motor draw at 90% efficiency?+
About 8.29 kW. Divide the shaft output (10 HP x 0.7457 = 7.457 kW) by the efficiency (0.90) to get electrical input power.
What is a typical power factor for an induction motor?+
About 0.85 at full load for a standard 4-pole design. Power factor drops toward 0.5-0.6 at light load, which is why lightly loaded motors draw more current per kW than the nameplate suggests.
What efficiency should I assume if the nameplate doesn't list one?+
Use 0.88 for older or budget motors and 0.93 to 0.95 for NEMA Premium/IE3 motors. Frame size matters too; a 1 HP motor rarely exceeds 82-85% while a 100 HP motor can reach 95-96%.
Why is input kW always higher than output kW?+
Because of internal losses. Copper losses in the windings, iron losses in the core, friction and windage, and stray load losses all convert some input electricity to heat instead of shaft work.
Does a higher service factor mean higher input kW?+
Not at rated load. Service factor (commonly 1.15) is headroom for brief overload, not the motor's normal operating point. Size input kW from the rated HP, not the service-factor HP, for normal duty calculations.
Do NEMA Premium motors really save meaningful energy?+
Yes. Moving from an 88% efficient motor to a 95% IE3 motor cuts input kW by about 7% for the same shaft output, which adds up over thousands of run-hours a year.
How much does efficiency drop at partial load?+
By roughly 3% to 8% below 50% load compared to full-load efficiency. Oversizing a motor 'for safety' usually means running it in this less efficient range.
Is nameplate kW on a European motor the input or the output?+
Output, the same convention as NEMA nameplates. Divide by efficiency to get input kW regardless of whether the motor is rated to IEC or NEMA standards.
How do I get full-load amps from motor input kW?+
Divide input kW (times 1000) by voltage and power factor for single-phase, or by √3 x voltage x power factor for three-phase. Check NEC Table 430.250 first since it takes precedence over a calculated FLA for code compliance.
Does power factor change how much input kW a motor uses?+
No. Power factor affects current and kVA, not real power. A motor's input kW depends only on shaft HP and efficiency; PF matters when converting that kW to amps.