DC Motor Efficiency Calculator

What a motor's efficiency actually tells you

Efficiency is the fraction of the electricity going in that comes back out as turning at the shaft. Everything else becomes heat. Feed this calculator the electrical side, meaning voltage, current and power factor, and the mechanical side, meaning either shaft power or torque with speed, and it returns the efficiency at that working point, the losses in watts, and what those losses cost you over a run of hours you specify. A 400 V three-phase motor drawing 15 A at a power factor of 0.85 while delivering 7.5 kW to its load is running at 84.9%, throwing away 1.333 kW as heat.

Quick start. A clamp meter on one phase and the nameplate are usually enough. Read the current with the machine doing its normal work, take the voltage at the terminals, and get the power factor from a power meter if you have one. Enter the shaft power the load actually demands, not the rated figure on the housing.

What each field wants, in the order it appears

Supply - direct current, single-phase or three-phase. This decides the formula: UI for direct current, UI x PF for single phase, 1.732 x UI x PF for three phase. Nothing is assumed if you skip it.
Voltage at the motor terminals, in volts, measured under load rather than copied from the plate.
Current in amps, read while the motor carries the load you care about.
Power factor, for either alternating-current supply. It is the one input that moves the answer most, and there is no default value here on purpose.
How you know the output - a shaft power in kilowatts or horsepower, or a torque in newton meters or pound-feet together with a speed in revolutions per minute.
Running hours, optional - how long the motor works over the period you want costed, for example a year of shifts.
Energy price per kilowatt-hour, optional and in whatever currency you use. Leave both of these empty and the calculator simply does not talk about money.
Reference efficiency, optional - the figure from a datasheet or from a motor you are considering instead. The result then shows what that machine would draw for the same shaft work, and the difference in energy and money.

Five motors measured on the bench

The reference case, three-phase. 400 V, 15 A, power factor 0.85, 7.5 kW at the shaft. Input 8.833 kW, efficiency 84.9%, losses 1.333 kW, which is 15.1% of everything drawn. The supply also has to carry 10.39 kVA of apparent power, a number that matters for cable and breaker sizing even though it does no work.

A small single-phase motor. 230 V, 6 A, power factor 0.8, 800 W at the shaft. Input 1.104 kW, efficiency 72.46%, losses 304 W. Small single-phase machines are simply worse, and 27.54% of the input turning into heat is ordinary for this size.

A direct-current motor. 48 V, 20 A, 800 W at the shaft. Input is simply 960 W, efficiency 83.33%, losses 160 W. There is no power factor on a direct-current supply, so the calculator does not ask for one and does not show apparent power.

Output known as torque instead of power. 48 Nm at 1450 rpm on the same three-phase supply. That works out to 7.288 kW at the shaft and an efficiency of 82.51%, with losses of 1.545 kW. Horsepower works too: 10 hp is 7.457 kW and gives 84.42% against the same input.

A badly oversized motor. The same supply, but the load only needs 3 kW. Efficiency collapses to 33.96% and 5.833 kW goes straight to heat. The machine is not broken. The same supply reading covered 7.5 kW of useful work in the first case and is now carrying a 3 kW load, which is the signature of a motor far larger than the job.

Losses in watts, and what they cost to run

The result builds this table for whatever you enter. Here it is for the reference case: 7.5 kW of shaft work over 4,000 hours at 0.18 per kilowatt-hour. It answers the question a purchase decision really asks, which is not what efficiency you have but what the difference between two efficiencies is worth.

If efficiency were	Input power	Lost as heat	Energy lost	Cost of losses
75%	10 kW	2.5 kW	10,000 kWh	1,800
80%	9.375 kW	1.875 kW	7,500 kWh	1,350
84.9% (the reference case)	8.833 kW	1.333 kW	5,334 kWh	960.09
88%	8.523 kW	1.023 kW	4,091 kWh	736.36
92%	8.152 kW	652.2 W	2,609 kWh	469.57
96%	7.813 kW	312.5 W	1,250 kWh	225

The gap between the first and last rows is 8,750 kWh and 1,575 over the period, for one motor. That is the arithmetic behind every argument for replacing an old machine rather than rewinding it, and it gets stronger the more hours the motor runs.

The power factor decides more than you would like

On an alternating-current supply the power factor sits inside the input power, so an error there lands directly on the efficiency. Same motor, same 400 V and 15 A, same 7.5 kW at the shaft, only the power factor changing:

Power factor entered	Input power	Efficiency it reports
0.70	7.275 kW	refused, that is above 100%
0.75	7.794 kW	96.23%
0.80	8.314 kW	90.21%
0.85	8.833 kW	84.9%
0.90	9.353 kW	80.19%
0.95	9.873 kW	75.97%

Ten hundredths of power factor move the answer by 11.32 percentage points, more than the whole gap between a basic motor and a premium one. A guessed power factor makes the exercise pointless. If you cannot measure it, measure the input power directly with a meter and work backwards.

Why this tool will not stamp an IE class on your motor

Plenty of calculators take an efficiency figure and print a class next to it. That is not how the classes work. IEC 60034-30-1 sets a separate limit for every combination of rated power, pole count and supply frequency, and those limits climb steeply with size, so the same percentage can sit above the highest class on a small motor and below the lowest one on a large one. A single ladder of thresholds is therefore wrong for most machines, quietly and in both directions.

This calculator reports what your two measurements establish: the efficiency at this working point, the losses, and what they cost. The class belongs to the nameplate, where it was determined at rated load under a defined test method. To compare against it, type that rated efficiency into the reference field.

Questions from people holding a clamp meter

My reading came out above 97%. What went wrong?

Almost certainly an input, not the motor. Induction motors do not reach that, and the calculator says so rather than congratulating you. The three usual causes are a power factor taken from the nameplate while the machine runs lightly loaded, a clamp meter on the wrong range or around more than one conductor, and a shaft power copied from the nameplate instead of the load applied. The table above shows how far one of these alone moves the number.

Why is my efficiency so low when the motor is healthy?

Because efficiency is a property of the working point, not of the machine. A motor reaches its best figure somewhere around three quarters of rated load and falls away on either side, steeply once it drops below a quarter. The 33.96% example above is a perfectly good motor turning a load that needs 3 kW from a machine built for far more. The fix is a smaller motor or a variable-speed drive, not a repair.

What is the difference between the kilowatts and the kVA it shows?

Apparent power in kVA is voltage times current with no power factor applied, so it is what the cables and the breaker carry. Real power in kW is what the meter bills you for. In the reference case, 10.39 kVA against 8.833 kW, and the gap is exactly the power factor. Direct-current supplies have no such gap, so the tile disappears there.

Can I work out the efficiency from torque and speed instead?

Yes, and it is often the more honest route, because a torque transducer measures what the load really takes. Shaft power is torque x 2πn / 60, with torque in newton meters and speed in revolutions per minute: 48 Nm at 1450 rpm is 7.288 kW. Pound-feet work too.

How are the internal losses split up?

Between the windings, the core, friction and windage, and stray load losses. The proportions depend on the design and on how hard the machine works, and they cannot be recovered from one input and one output measurement. So the total is reported and the split is left alone, rather than applying fixed percentages that would look precise and mean nothing.

Is replacing an old motor with a premium one worth it?

The reference field answers that in money rather than in percentage points. In the case above, a motor rated 91.4% would draw 8.206 kW for the same shaft work instead of 8.833 kW, so it saves 627.8 W continuously, which over 4,000 hours is 2,511 kWh and 451.99. Compare that against the price of the motor and the installation. The more hours it runs, the shorter the answer gets.

Related tools

RLC Circuit Calculator

Resonant frequency, Q and bandwidth for the coil and capacitor side of a drive or a filter - See the calculator

Magnetic Force Calculator

Where the shaft force comes from: current in a field, worked out in newtons - See the calculator

Electrical Power Calculator

Watts, volt-amps and kilowatt-hours from any two electrical readings, for pinning down the input side before you come here - See the calculator

Amps Calculator

Watts to amps with breaker and wire sizing, which is where the apparent power figure above actually gets used - See the calculator

Mechanical Power Calculator

Power from work and time on the mechanical side, the other way of arriving at the number in the shaft-power box - See the calculator

Torque Calculator

Newton meters and pound-feet from a force and a lever arm, for when the torque itself is what you are chasing - See the calculator

Thermal Efficiency Calculator

The same ratio for furnaces, boilers and heat engines, where the input is fuel rather than electricity - See the calculator

Ohm's Law Calculator

Volts, amps and ohms, for checking a supply reading that looks wrong before you trust the efficiency it produced - See the calculator

Parameters