Guide

Electric Motor Efficiency & IE Class Guide

Motor efficiency classes (IE1–IE5) explained, with a calculator to estimate energy savings from upgrading to higher efficiency motors. Includes payback period estimation.

10 min readUpdated July 2026

IE Efficiency Classes Explained

Electric motors consume approximately 45% of all electricity used globally. Even small efficiency improvements can save significant energy and money over a motor's 15–20 year operating life. The International Electrotechnical Commission (IEC) defines efficiency classes under IEC 60034-30:

ClassLevelStatusTypical Efficiency (22 kW, 4-pole)
IE1StandardBanned in most markets (EU, AU, US)87.6%
IE2HighBanned in EU/AU for most ratings89.8%
IE3PremiumMinimum legal standard in EU, AU, US92.7%
IE4Super PremiumAvailable, not mandated94.5%
IE5Ultra PremiumEmerging technology96.0%+

Efficiencies shown are for a 22 kW, 4-pole motor at full load. Actual values vary by manufacturer and rating.

Energy Savings Calculator

Inputs

Continuous = 8,760

AU average ≈ $0.25–$0.35

Results

Annual Cost Savings

$2,824.7

Annual Energy Savings

11,298.6 kWh

Power Savings

1.9 kW

Reduction in input power

CO₂ Reduction

7,683.1 kg/year

Based on AU grid factor 0.68 kg/kWh

How Savings Are Calculated

The input power (electrical power consumed) differs from the rated output power (shaft power) based on the motor's efficiency:

Input Power = Output Power / Efficiency

For a 22 kW motor at 87% efficiency:

Old: 22 / 0.87 = 25.3 kW input
New: 22 / 0.94 = 23.4 kW input
Savings = 1.9 kW

Over 6,000 hours/year at $0.25/kWh:

Annual Savings = 1.9 kW × 6,000 hrs × $0.25 = $2,850/year

When to Upgrade

  • When a motor fails: Replacing with IE4 or IE5 is almost always the right choice, as the efficiency premium pays back within 1–3 years.
  • When a motor is rewound: Each rewind reduces efficiency by 0.5–1%. After 2–3 rewinds, replacing with a new high-efficiency motor is more economical.
  • When a motor is oversized: A motor running at less than 40% load operates at reduced efficiency. Downsizing to a correctly-rated motor can improve efficiency by 3–5%.
  • When variable loads exist: Adding a Variable Speed Drive (VSD) can save 30–50% on energy for variable-load applications like pumps and fans (affinity laws: power ∝ speed³).
  • When running continuously: Motors running 4,000+ hours per year benefit most from efficiency upgrades. The savings accumulate fast enough for short paybacks.

Payback Period Estimation

The payback period for upgrading a motor depends on the price difference between the efficiency classes, the annual savings, and the operating hours:

Payback (years) = Price Premium / Annual Savings

Example: Upgrading from IE3 ($2,500) to IE4 ($3,200) for a 22 kW motor running 6,000 hours/year:

Price premium: $3,200 − $2,500 = $700
Annual savings: $2,850 (from calculator above)
Payback: $700 / $2,850 = 0.25 years (3 months)

Key insight: For motors running more than 4,000 hours/year, the energy cost over the motor's lifetime (15+ years) is typically 50–100× the purchase price. Always select the highest efficiency motor available — the purchase price is a tiny fraction of the total cost of ownership.

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