Motor Efficiency Calculator
Enter the motor load and measured V, I and PF to get efficiency, loss breakdown and annual running cost of the losses.
Step-by-step
- Enter valid values to begin.
Shaft output is taken as the load you enter (nameplate rating if the motor is near full load). The loss breakdown is a typical IEEE 112 allocation for a mid-size four-pole induction motor, not measured data. For accurate segregation, request the motor test report.
Motor Efficiency: Where the Missing Kilowatts Go
Core Engineering Principles
Every motor loses some of the electrical power it takes in before it reaches the shaft. That lost power turns to heat: copper loss in the stator windings (I²R), copper loss in the rotor bars, iron or core loss from the alternating field in the laminations, friction and windage, and stray load loss from leakage flux. At full load, a 50 HP premium motor wastes about 7% of its input. That doesn’t sound like much, but 7% of 40 kW running 6,000 hours a year is 16,800 kWh, and over a twenty-year life the electricity costs around 50 to 100 times the purchase price of the motor.
To find efficiency in the field, measure the electrical input with a power meter (kW, not just amps and nameplate PF) and compare it to the shaft output. The shaft output is harder; most plants estimate it from the load, from slip, or from the nameplate rating when the motor is known to be near full load. The efficiency curve is fairly flat from 50% to 100% load and falls off sharply below about 40%, which is why oversized motors are wasteful.
Losses = Pin − Pout • typical split: stator copper ~37%, rotor copper ~18%, core ~20%, friction & windage ~9%, stray load ~16%
NEC & Standard References
IEEE Std 112 is the standard test method (method B, with loss segregation) that manufacturers use to publish efficiency. NEMA MG-1 sets Premium efficiency levels, for example 93.0% for a 50 HP, 4-pole, 1800 rpm enclosed motor, and NEMA MG-1 12.58 defines nominal and minimum efficiency. IEC 60034-30-1 classes motors IE1 to IE4. In the US, the Department of Energy sets minimum efficiency for general-purpose motors under 10 CFR 431, which currently requires NEMA Premium level for most 1 to 500 HP motors. The split of losses on this page is a typical IEEE 112 allocation for a mid-size four-pole motor; your motor’s test report will differ.1. Input power = 1.732 × 460 × 58 × 0.87 / 1000 = 40.2 kW.
2. Efficiency = 37.3 / 40.2 = 92.8%.
3. Losses = 40.2 − 37.3 = 2.9 kW. Roughly 1.1 kW stator copper, 0.5 kW rotor copper, 0.6 kW core, 0.3 kW friction, 0.5 kW stray.
4. Annual loss cost = 2.9 × 6,000 × $0.10 = $1,740 a year.
If the motor were an older 88% machine the losses would be 5.1 kW, so replacement can pay back in about two years.
- Rewound motors lose efficiency. A poor rewind can cost one to two points of efficiency. Ask for a documented rewind with controlled burn-out temperature.
- Oversized motors run inefficiently. Below 40% load both efficiency and power factor fall.
- Voltage unbalance and low voltage both raise losses. A 3% unbalance can cost half a point of efficiency.
- Heat means failure. Every 10 °C rise above the insulation rating halves winding life.
- Measure at steady load. Take readings after the motor has been running 30 minutes, or the numbers will drift.