The power drawn from the supply is the shaft power divided by efficiency; multiplying it by operating time gives consumption: E (kWh) = P2 × load factor / η × hours.

Worked example

Assumptions: 7.5 kW 4-pole IE3 motor, full load (load factor 1), catalogue efficiency 90.4%, 8 hours a day, 22 working days a month.

  • Input power: 7.5 / 0.904 ≈ 8.30 kW
  • Monthly operating time: 8 × 22 = 176 hours
  • Monthly consumption: 8.30 × 176 ≈ 1,461 kWh

The result can also be checked with nameplate data: P1 = √3 × U × I × cos φ = 1.732 × 400 V × 14.3 A × 0.84 ≈ 8.32 kW.

Points to watch:

  • A motor rarely runs at full load all the time; at part load the input power falls and efficiency also changes. The catalogue gives efficiencies at 3/4 and 1/2 load separately.
  • The most reliable figure is a measurement with a clamp meter or energy analyser while the machine is in normal production.
  • To get the cost, multiply the kWh by the tariff unit price.

Where to start reducing consumption is explained in our electric motor energy savings article.