What motor efficiency is, and what it is not
Efficiency is mechanical power out divided by electrical power in, at a stated operating point. The difference between the two is losses, and in an induction motor those losses fall into five groups: stator I²R in the windings, rotor I²R in the bars and end rings, core loss from hysteresis and eddy currents in the laminations, friction and windage in bearings and fan, and stray load loss that does not fit the other four. Each behaves differently with load, which is why efficiency is a curve and not a number.
Two of those loss groups scale roughly with the square of load — the I²R terms — and three are roughly constant. That combination produces the characteristic shape: efficiency climbs steeply from zero load, peaks somewhere between about 60% and 90% of rated load depending on the design, and falls off slightly at full load and beyond. It is the reason a lightly loaded motor is inefficient: the constant losses are being paid for out of a small output.
Power factor behaves differently and worse. The magnetising current an induction motor draws is nearly independent of load, so at light load that reactive current dominates and the power factor collapses. A motor at 25% load can show a good-looking current reading and a poor power factor, which means the apparent power — and therefore the conductor and transformer loading — is far higher than the useful work justifies. The power factor correction calculator sizes the capacitance that fixes the electrical symptom, though not the underlying oversizing.
The nameplate efficiency is a full-load value, determined by a standard test procedure — IEEE 112 Method B in North America, IEC 60034-2-1 elsewhere — under controlled conditions with a balanced sinusoidal supply. A field measurement will not reproduce it exactly, and it is not supposed to.
Measuring input power without fooling yourself
Input power is √3 × V × I × PF for a three-phase motor and V × I × PF for single-phase. The √3 comes from the relation between line and phase quantities in a balanced three-phase system, not from there being three phases — that factor is already in the line-to-line voltage.
The power factor term is where field measurements go wrong. Volts times amps is apparent power in volt-amperes; multiplying by power factor is what turns it into real power in watts. Assume unity power factor and you understate the input power, which inflates the calculated efficiency — sometimes above 100%, which is the calculator's cue to tell you the measurement is wrong. Use a true-power meter or a power-quality analyser rather than a clamp ammeter and an assumption.
Voltage unbalance is the second trap. A few percent of unbalance produces a much larger current unbalance and a real increase in rotor heating, and it makes the average of three phase currents a poor stand-in for a proper three-wattmeter measurement. If your phase currents differ by more than a few percent, measure power properly rather than computing it from an average.
The third trap is the drive. On the output of a variable-frequency drive the voltage is a switched waveform and the current is non-sinusoidal, so an averaging meter reads neither correctly and the concept of displacement power factor is no longer the whole story. Measure on the input side of the drive if you want the energy the system consumes, and treat drive and motor as one unit.
Shaft output is the other half of the equation and the harder half. Direct torque measurement is best. Failing that, the duty point of the driven machine — a pump curve at a measured flow and head, a fan at a measured airflow and static pressure — gives a defensible figure. The input-power load estimate in this calculator is the fallback: it compares measured input power with the input power the motor would draw at full load, and it inherits whatever error the nameplate efficiency carries.
Worked example: a 10 hp motor at three-quarter load
A 10 hp, 460 V three-phase motor with a nameplate efficiency of 91.7% drives a pump. The analyser reads 460 V, 9.1 A and a power factor of 0.85. The pump curve at the measured flow and head says the shaft is taking 7.5 hp.
- Input power. √3 × 460 × 9.1 × 0.85. First √3 × 460 = 796.743. Times 9.1 = 7,250.36. Times 0.85 = 6,162.81 W, or 6.1628 kW.
- Output power. 7.5 hp × 745.6999 = 5,592.75 W, or 5.5927 kW.
- Efficiency. 5,592.75 / 6,162.81 = 0.90751, so 90.75%.
- Losses. 6.1628 − 5.5927 = 0.5701 kW, all of it becoming heat in the machine.
- Load from shaft output. 7.5 / 10 = 75.00%.
- Load from input power. Full-load input would be 10 × 745.6999 / 0.917 = 7,456.999 / 0.917 = 8,131.95 W = 8.1319 kW. So 6.1628 / 8.1319 = 75.79%.
- Against nameplate. 90.75 − 91.70 = −0.95 percentage points, which is what you expect at 75% load: efficiency at part load is normally a little below the full-load nameplate figure for a motor of this size.
The two load estimates agree to within 0.8 percentage points, which is the reassuring outcome — they are computed from different measurements and would diverge if either the shaft estimate or the electrical reading were wrong.
Energy: at 4,000 hours a year the motor consumes 6.1628 × 4,000 = 24,651 kWh, costing 24,651 × $0.12 = $2,958. Delivering the same 5.5927 kW of shaft power at the nameplate 91.7% would need 5.5927/0.917 = 6.0989 kW, or 24,396 kWh and $2,927. The 0.95-point efficiency shortfall is worth about $31 a year here — small, which is the honest answer, and it is why efficiency projects on small motors rarely pay unless the motor is replaced anyway.
How to read the result
Start with the load percentage, not the efficiency. A motor between roughly 60% and 100% of rated load is in the region where its efficiency and power factor are close to nameplate, and there is little to gain. Below about 40% both fall away, power factor faster than efficiency, and the fix is a smaller motor or a variable-speed drive rather than anything done to the existing machine. The calculator warns when you are in that region.
Then compare the two load estimates. They come from independent measurements — one mechanical, one electrical — so agreement between them is real evidence that both are right. A large gap means one of them is wrong, and the input-power estimate is the more suspect of the two because it assumes the motor is meeting its nameplate efficiency, which is the very thing you are testing.
Read a low measured efficiency as a symptom and go looking for the cause. Voltage unbalance, a partially failed capacitor on a single-phase machine, worn or over-greased bearings, a blocked cooling fan, a shaft misalignment loading the bearings, and a stator that has been rewound without care all reduce efficiency by measurable amounts. A rewind performed with excessive burn-out temperature damages the interlaminar insulation and costs efficiency permanently.
Be careful about what the annual cost comparison means. It is the cost of the efficiency shortfall at this one operating point, not the cost of the motor. The far larger number is usually the energy the load itself demands: a pump throttled by a valve or a fan controlled by a damper wastes more than any motor inefficiency, and reducing speed on a centrifugal load cuts power roughly as the cube of speed. Fix the system before you fix the motor.
How efficiency and power factor move with load
| Load | Efficiency relative to full-load value | Power factor relative to full-load value | What dominates |
|---|---|---|---|
| 25% | Substantially lower | Much lower | Constant core, friction and windage losses against a small output; magnetising current dominates the total |
| 50% | Slightly lower | Noticeably lower | I²R losses have fallen fourfold; constant losses still significant |
| 75% | At or near peak | Close to full-load value | Best balance of constant and load-dependent losses |
| 100% | Nameplate value | Nameplate value | The point at which both are defined and tested |
| 115% (service factor) | Slightly lower | Similar or slightly lower | I²R losses rise faster than output; temperature rise increases |
This table gives directions, not numbers, deliberately: published part-load curves differ enough between ratings and designs that quoting a single figure would be misleading. Manufacturers publish part-load efficiency and power factor at 50%, 75% and 100% load for their own machines — use those.
What makes a field efficiency measurement wrong
- Assuming the power factor. Volts times amps is apparent power. Using it as if it were real power understates the input and can produce efficiencies above 100%, which is the calculator's signal that the measurement, not the motor, is at fault.
- Averaging unbalanced phase currents. A few percent of voltage unbalance produces several times that in current unbalance. Where the phases differ noticeably, measure power with a proper three-phase instrument rather than computing it from an average current.
- Measuring on the output of a variable-frequency drive. The waveform is switched and non-sinusoidal, so ordinary meters misread both voltage and current. Measure on the drive input if you want the energy the system uses.
- Guessing the shaft power. The output term is as important as the input one. A pump or fan duty point from a measured flow and head is defensible; a nameplate horsepower assumed to be delivered is not, because the motor is almost never at exactly full load.
- Comparing a part-load measurement with a full-load nameplate. Nameplate efficiency is defined at rated load. A motor at 50% load will normally measure below it and be perfectly healthy.
- Ignoring temperature. Winding resistance rises with temperature, so a motor measured cold shows lower I²R loss than the same motor at operating temperature. Let it reach thermal equilibrium first.
- Reading efficiency gains as system savings. A one-point efficiency improvement on a motor driving a throttled pump is a rounding error next to what the throttle valve is dissipating.
Standards, ratings and when replacement pays
Motor efficiency is tested to a standard, and which standard matters. IEEE 112 Method B is the North American procedure, with an indirect measurement of stray load loss; IEC 60034-2-1 is the international equivalent. Efficiency values determined by different methods are not directly comparable, and a nameplate quoting one standard cannot be checked against a figure derived under another. NEMA MG 1 defines the nominal and minimum efficiency values that go on the nameplate, and Premium Efficiency levels are specified there for the common ratings.
The energy case for a high-efficiency motor is driven by hours, not by the efficiency gap. A two-point efficiency improvement on a motor running 8,000 hours a year at 50 kW saves roughly a megawatt-hour a month; the same two points on a motor running 500 hours a year at 3 kW saves almost nothing. Always compute the saving from the actual duty rather than from the rating, which is precisely what the annual table on this page does.
Right-sizing usually beats re-specifying. A motor at 30% load is throwing away power factor and efficiency, and no amount of premium efficiency in a machine of the same rating fixes it. The uncomfortable part is that oversizing is often deliberate — someone allowed margin on margin — so establishing the real load with a measurement is the first step. The motor torque calculator converts the shaft power at a duty point into the torque the machine really has to produce.
Finally, keep the electrical consequences in view. A motor's current sets conductor size and protection under NEC Article 430, and improving efficiency reduces current for the same output — see the motor full-load amps calculator and the motor overload and breaker sizing calculator for that side of the design. Correcting power factor reduces current too, without changing efficiency at all, because it removes reactive current rather than losses.