Motor Efficiency Savings
Calculate energy savings from upgrading motor efficiency class (IE2→IE3→IE4)
Example price — enter actual rate
For payback calculation
Energy savings
Example price — enter your actual electricity rate.
How does it work?
Electric motors account for approximately 40–45% of global electricity consumption. Upgrading to a higher efficiency class is one of the most cost-effective energy-saving measures.
IE efficiency classes (IEC 60034-30-1):
- IE1 — Standard efficiency (older motors)
- IE2 — High efficiency (minimum in EU for ≥ 0.75 kW)
- IE3 — Premium efficiency (required in EU for ≥ 0.75 kW since 2021)
- IE4 — Super premium efficiency (voluntary)
Formula: Savings (kWh) = (P_output/η_old − P_output/η_new) × operating hours
Example
Example: 11 kW motor, 4000 h/year, electricity 0.18 €/kWh (example price).
IE2 efficiency 90.5% → input: 11 / 0.905 = 12.15 kW
IE3 efficiency 91.7% → input: 11 / 0.917 = 11.99 kW
Savings: (12.15 − 11.99) × 4000 = 640 kWh/year
Monetary: 640 × 0.18 = €115/year (example price)
Assumptions
- •Efficiency is constant across all operating hours — it varies with load in practice
- •Motor operates at rated output power (full load)
- •Electricity price is an example — use your actual contracted rate
- •IE efficiency values are typical for an 11 kW 4-pole motor