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Torque and RPM to kW and HP Calculator with Annual Energy Cost

Convert between torque, speed and power without hunting for the constant. Enter shaft torque and RPM to get kW, hp and lb-ft, then see the electrical input and a yearly energy cost for the duty cycle.

Your numbers

Nm

Torque at the motor shaft or load shaft. 1 Nm = 0.7376 lb-ft.

1,450 RPM

Rated speed of the motor or the driven shaft.

92 %

Nameplate efficiency at load. IE3 motors above 15 kW are typically 91 to 95 percent.

h
days
$/kWh

Your results

Shaft power
18.2212
P(kW) = T(Nm) x n(RPM) / 9549.3.
Annual energy cost
$9,507
Shaft power (mechanical hp)
24.435
Torque (lb-ft)
88.5074
Electrical input power
19.8057
Annual energy
79,223

Steady-state estimate at constant load. Part-load efficiency, power factor, drive losses and variable duty cycles change real consumption.

Save results:

Shaft power (kW) vs speed (RPM)

Sweeps Shaft speed from 500 to 3600, holding your other inputs fixed.

Machine-level costing from your ERP

Want energy and machine data feeding product costing? Ask for an optional expert review.

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How to use it

Enter the torque at the shaft and the speed, the motor efficiency from the nameplate or datasheet, how many hours per day and days per year it runs, and your electricity tariff. The result shows shaft power in kW and hp, the electrical input after losses, the yearly energy in kWh and the cost. The chart sweeps speed at constant torque, which shows how power grows linearly with RPM, a useful check when a variable frequency drive changes speed.

  • Use the load torque, not the motor rated torque, for actual consumption
  • Get efficiency at the actual load point
  • Include idle hours at no load only if the motor still draws power
  • Convert kW to hp by multiplying by 1.341

The formulas

Power in kW equals torque in Nm times speed in RPM divided by 9,549.3, which is 60,000 divided by 2 pi. Mechanical horsepower is 1.34102 times kW, and in US units hp equals torque in lb-ft times RPM divided by 5,252. Electrical input equals shaft power divided by efficiency. Annual energy is input power times hours per day times days per year. Metric horsepower (PS) differs slightly from mechanical hp, so check which one a nameplate uses.

Worked example

A conveyor drive needs 120 Nm at 1,450 RPM. Shaft power is about 18.2 kW or 24.4 hp. With a motor at 92 percent efficiency, electrical input is about 19.8 kW. Running 16 hours a day for 250 days gives about 79,000 kWh a year, which at 0.12 per kWh costs roughly 9,500. Raising efficiency by two points saves around 200 dollars a year on this drive, so efficiency upgrades pay back mainly on large, long-running motors.

Common mistakes and ERP notes

Common errors are using nameplate kW as the real load, forgetting drive and gearbox losses, and mixing metric and mechanical horsepower. Oversized motors run at poor part-load efficiency. In ERP asset records, store the motor rating, speed, efficiency class and duty on the equipment so maintenance can plan replacements and so energy costing for products can use real machine data instead of flat overhead rates.

  • Measure real current when possible
  • Account for gearbox and VFD losses separately
  • Record nameplate data on the equipment master

Frequently Asked Questions

What is the formula for kW from torque and RPM?

Power in kW equals torque in Nm multiplied by speed in RPM, divided by 9,549.3. The constant comes from 60,000 divided by 2 pi. For US units, hp equals torque in lb-ft times RPM divided by 5,252.

How do I convert kW to hp?

Multiply kW by 1.34102 for mechanical horsepower. Metric horsepower, often written PS or CV, is about 1.3596 times kW, so always confirm which unit the equipment documentation uses before comparing ratings.

Why does energy cost use electrical input rather than shaft power?

You pay for electricity drawn from the supply, which is shaft power divided by efficiency. Motor, drive and gearbox losses all show up in the bill, so using shaft power alone would understate cost.

Does this model a variable load?

No, it assumes a steady load for the hours entered. For variable duty, calculate each load band separately or use average power from a logger, then sum the energy across bands.

If machine energy costs reach products only through flat overhead rates, we can help allocate real machine-level data in your costing setup.