Enter the motor horsepower (rated), the load (as a percent of rated load), and the input power into the calculator to determine the motor efficiency of an electric motor.
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Motor Efficiency Formula
\eta = \left(\frac{0.7457 \cdot HP \cdot L}{P_{in}}\right)\times 100- η = motor efficiency (%)
- HP = rated motor horsepower (hp)
- L = load as a decimal fraction (e.g., 0.75 for 75% load)
- Pin = electrical input power (kW)
- 0.7457 = unit conversion factor (1 hp = 0.7457 kW)
When output power is measured directly: η (%) = (Pout / Pin) x 100, with both values in the same units (watts or kW).
NEMA Premium Efficiency by Motor Size
Full-load nominal efficiencies per NEMA MG-1 Table 12-12 for 4-pole TEFC induction motors. NEMA Premium (IEC IE3 equivalent) has been the U.S. federal minimum for most motors 1 hp and above since EISA 2007 took full effect. The efficiency gap between standard and premium is largest at small motor sizes and narrows above 50 hp.
| Rated HP | NEMA Premium (%) | Energy Efficient (%) | Standard Efficiency (%) | Annual kWh Savings vs. Standard (at 4,000 hrs/yr, 75% load) |
|---|---|---|---|---|
| 1 | 85.5 | 82.5 | 76.8 | ~130 kWh |
| 5 | 89.5 | 87.5 | 84.0 | ~280 kWh |
| 10 | 91.7 | 89.5 | 86.5 | ~430 kWh |
| 25 | 93.6 | 92.4 | 90.2 | ~680 kWh |
| 50 | 94.5 | 93.0 | 91.7 | ~980 kWh |
| 100 | 95.4 | 94.1 | 92.4 | ~1,740 kWh |
| 200 | 96.2 | 95.0 | 93.6 | ~2,780 kWh |
Efficiency at Partial Load
Motor efficiency peaks near 75% of rated load. Fixed losses (iron core, friction, windage) are constant regardless of output; at light loads they consume a larger share of input power, dragging efficiency down. Copper (I²R) losses grow with load, causing a modest dip above peak efficiency.
| Load (%) | Typical Efficiency Range | Practical Note |
|---|---|---|
| 25% | 70 to 82% | Fixed losses dominate; strong candidate for motor downsizing |
| 50% | 85 to 91% | Acceptable; efficiency rising toward peak |
| 75% | 88 to 93% | Peak efficiency zone for most induction motors |
| 100% | 87 to 92% | Slightly below peak due to increased copper losses |
A motor consistently running below 50% load is a candidate for downsizing. Replacing a 20 hp motor running at 25% load with a properly-sized 5 hp motor typically recovers 8 to 12 efficiency points and reduces annual energy costs proportionally.
IEC and NEMA Efficiency Classes
| IEC Class | NEMA Equivalent | Typical Efficiency Range | Regulatory Status |
|---|---|---|---|
| IE1 | Standard | 76 to 91% | Not legal for new motors in EU (since 2021) or US (since 2016) |
| IE2 | Energy Efficient (EPAct) | 82 to 94% | Still permitted for some exempted motor types |
| IE3 | NEMA Premium | 85 to 96% | Current federal minimum for most motors 0.75 kW and above (US and EU) |
| IE4 | Super Premium | 87 to 97% | Voluntary; common in permanent magnet and synchronous reluctance motors |
Motor Loss Breakdown
Understanding which losses are fixed versus load-dependent is key to diagnosing motor efficiency issues and making correct sizing decisions.
| Loss Type | Cause | Load Dependent? | Share of Total Losses |
|---|---|---|---|
| Copper (I²R) | Resistance heating in stator and rotor windings | Yes (proportional to I²) | 35 to 45% |
| Core (Iron) | Hysteresis and eddy currents in laminations | No (fixed with voltage) | 20 to 25% |
| Friction and Windage | Bearing friction, cooling fan drag | Mostly no | 10 to 15% |
| Stray Load | Harmonic fields, leakage flux | Yes | 10 to 15% |
Motor Efficiency Definition
Motor efficiency is the ratio of mechanical output power (shaft power) to electrical input power, expressed as a percentage. A motor rated at 92% efficiency converts 92 watts of every 100 watts drawn from the supply into useful shaft work; the remaining 8 watts become heat, noise, and friction losses inside the motor.
Example Problem
How to calculate motor efficiency?
- First, determine the input power.
For this example, we are analyzing an electric motor in a car that receives its power from a battery pack. The input power is 85 kW.
- Next, determine the maximum horsepower.
The maximum rated horsepower of the engine/motor is 150 hp.
- Next, determine the load.
The percentage of the max horsepower that the engine is running at is .65 or 65%.
- Finally, calculate the motor efficiency.
Using the formula above, the motor efficiency is found to be (0.7457 * 150 * 0.65 / 85) * 100 = 85.54 %.
