How the saving is calculated
Both motors deliver the same mechanical power to the shaft. The more efficient motor draws less electrical power for it:
Pshaft = Prated · load
Saving (kW) = Pshaft ÷ ηold − Pshaft ÷ ηnew
Saving per year = Saving (kW) · hours · electricity price
Worked example
A 30 kW pump motor runs 6,000 hours a year at 75 % load. Replacing an IE2 motor (92.3 %) with an IE4 motor (94.9 %) lowers the power draw by 0.67 kW and saves about 4,000 kWh per year. At 0.15 per kWh that is around 600 a year, so an investment of 1,500 pays back in 2.5 years.
Typical efficiencies, 4-pole motors at 50 Hz
| Rated power | IE1 | IE2 | IE3 | IE4 |
|---|---|---|---|---|
| 1.1 kW | 75.0 % | 81.4 % | 84.1 % | 87.2 % |
| 5.5 kW | 84.7 % | 87.7 % | 89.6 % | 91.9 % |
| 15 kW | 88.7 % | 90.6 % | 92.1 % | 93.9 % |
| 30 kW | 90.7 % | 92.3 % | 93.6 % | 94.9 % |
| 75 kW | 92.7 % | 93.9 % | 95.0 % | 96.0 % |
Minimum values from IEC 60034-30-1. Actual motors are often slightly better; use the data sheet values when you have them.
When replacing pays off
The higher the running hours and the electricity price, the faster a premium motor pays back. Motors that run more than 4,000 hours a year almost always justify IE3 or IE4. If a motor runs at varying load, a variable speed drive usually saves far more than the motor class alone, especially on pumps and fans.