Motor power calculation: HP and kW the simple way

Production / Utilities June 27, 2026 11 min read 2,000 words

Whether you are sizing a conveyor drive, checking a pump motor, or converting a nameplate from kW to HP, motor power comes down to two formulas: power from torque and speed, and electrical input from voltage and current. Here is each one, a full worked example, and the efficiency and power-factor details that separate a right-sized motor from a wasteful one.

The two formulas you actually need

Almost every motor question is one of these:

  • Mechanical (shaft) power from torque and speed: P(kW) = T × N / 9550, where T is torque in Nm and N is speed in rpm.
  • Electrical input power for a three-phase motor: P(kW) = √3 × V × I × PF / 1000, where V is line voltage, I is line current and PF is power factor.

The constant 9550 bundles the unit conversions (it is 60,000 / 2π). Memorise it — it turns Nm and rpm straight into kW.

Power from torque and speed

When you know the load torque and the running speed, mechanical power is direct. Say a mixer needs 120 Nm at 1450 rpm:

P = T × N / 9550 P = 120 × 1450 / 9550 = 18.2 kW at the shaft. That is the power the load demands; the motor must deliver at least this on its output shaft.

If your speed is in rad/s rather than rpm, use the cleaner form P(W) = T(Nm) × ω(rad/s). The 9550 version simply saves you converting rpm every time.

Electrical input power and current

The nameplate kW is shaft output. The motor draws more than that from the supply because of losses, so:

Input power = output power / efficiency. A 18.5 kW motor at 91% efficiency draws 18.5 / 0.91 = 20.3 kW electrical input.

To find the running current, rearrange the three-phase formula:

I = P(input, W) / (√3 × V × PF) For 20,300 W at 415 V and PF 0.86: I = 20300 / (1.732 × 415 × 0.86) = 32.8 A. This is the full-load current you size cables and overloads around.

Worked example: a conveyor drive

A belt conveyor needs to move 8000 kg/h up a 4 m rise at 0.5 m/s. Work it through:

StepCalculationResult
Lifting powerm·g·h per second = (8000/3600) × 9.81 × 487 W vertical
Friction + belt load (allow ~3×)realistic conveyor demand~2.6 kW
Drive efficiency (gearbox 95%)2.6 / 0.952.74 kW
Service factor (1.25)2.74 × 1.253.42 kW
Next standard frameround up3.7 kW (5 HP)

The takeaway: the raw lifting power was tiny, but friction, efficiency and a sensible margin set the real motor at 3.7 kW. Skipping those steps is how plants end up with motors that trip on start-up.

kW to HP conversion table

To convert, multiply kW by 1.341 for imperial HP (or 1.3596 for metric HP). Common ratings:

kWHP (approx)Typical use
0.751.0Small pump, fan
2.23.0Conveyor, blower
3.75.0Compressor, hydraulic pack
7.510Machine tool main drive
1520Large pump, mixer
3750Process fan, big compressor

Sizing a motor without over-spending

In 14 years on the plant floor, the most common waste I saw was oversized motors. A motor running at 40% load sits at a poor power factor and lower efficiency, so it quietly burns energy for years. To size well:

  1. Calculate the true load power (torque × speed, or pump/fan duty).
  2. Divide by drive efficiency and motor efficiency.
  3. Add a service factor of 15–25% — not 100%.
  4. Pick the next standard frame up, then check it runs near 75–100% load.

A motor loaded to 75–95% of rating is in its efficiency and power-factor sweet spot. For pump and fan drives, remember the affinity laws — power rises with the cube of speed, so a small speed increase is a big power increase. If energy cost is the target, our OEE guide and cycle time guide help connect motor loading to real output.

Watch the part-load power factor Power factor on an induction motor falls from ~0.86 at full load to ~0.5 at quarter load. An oversized motor not only wastes real power, it drags your plant power factor down and can attract a utility penalty. Size to the load, not to "just in case".

Common motor calculation mistakes

  • Confusing input and output. Nameplate kW is shaft output; supply draw is higher by the efficiency loss.
  • Forgetting √3. The 1.732 factor is mandatory for three-phase — leaving it out understates power by 42%.
  • Ignoring drive losses. Gearboxes, belts and couplings each take a few percent.
  • Sizing on running power only. Starting torque for high-inertia or loaded starts can need a larger frame.
Run the numbers in one click MetricMech's free motor HP and power calculator does torque-to-power, kW-to-HP, and three-phase current in one place — enter torque and rpm, or voltage, current and power factor, and read the result.

If your work also involves preparing inspection drawings for machined motor or pump components, CadNexa's auto-ballooning tool turns a PDF drawing into a numbered inspection sheet in minutes — handy when the same parts need an FAI. Browse more shop-floor calculators on the MetricMech templates page.

RR
Rajadurai R
Founder, MetricMech & CadNexa · 14 years plant-head experience