IE2 / IE3 / IE4 running-cost estimate

What the numbers mean

Rated kW

Shaft output. Electrical input (kW) = output ÷ (η/100). A 7.5 kW motor at 90% efficiency draws about 8.33 kW from the grid at full load.

Efficiency η

Nameplate or test-report percentage at the stated load (usually 100% or 75% of rated). IEC class (IE2/IE3/IE4) is a band, not one number for every kW and pole count. This tool never invents η — type the figure from the two plates you are comparing.

IE2 / IE3 / IE4

Efficiency classes in IEC 60034-30-1. Higher class → higher η at the same kW/poles, so less kWh for the same shaft work. Minimum legal class depends on the country and the rating.

Hours / year

Running hours. 6000 h is a typical process pump (about 16 h × 365). A fan that runs 2000 h saves less than the same kW on continuous duty.

Load factor

Average load ÷ rated load. 0.8 means you treat the motor as 80% loaded all year. It is a filter, not a measured duty cycle.

Tariff

Price per kWh in your currency. The table is a number; keep the same currency on every row.

Annual energy

kWh/year ≈ (kW_output / η) × hours × load factor. Cost = kWh × tariff. Ignore this as a payback certificate — use it to see if an IE3 upgrade is even in the right order of magnitude.

Worked numbers (you still type η)

Example only, using 7.5 kW, 6000 h, load 0.8, tariff 0.12. Efficiencies below are sample nameplate-style figures so you can see the formula — replace them with your plates.