Efficiency from energy outputs:
formula and worked example

What efficiency tells you about a machine

Efficiency compares what you got with what you paid. Every real machine loses some energy as heat, sound or friction, so the useful output is always less than the input. The ratio tells you how much of the input does what you actually wanted.

It is how you compare a motor with another, a boiler with an older one, or a charging system with its alternatives. A higher figure means less energy wasted for the same result, and usually less to pay for.

Formula and how to define the inputs

• Useful energy: the part of the output that does the intended job, in joules or kWh.
• Input energy: the total energy supplied to the device over the same period, in the same unit.

Both values must be energies over the same time span and in the same unit. If you measure power instead, take both as power at the same moment, which gives the same ratio.

The definition of useful is a choice. For a light bulb, useful means light, not heat. For a heater, it is heat delivered into the room. Change the definition and the efficiency changes with it.

740 J of useful work from 1,000 J

1. 1Take the ratio: 740 ÷ 1,000 = 0.74.
2. 2Convert to a percentage: 0.74 × 100 = 74%.
3. 3Find the loss: 1,000 - 740 = 260 J, which is 26% of the input.
4. 4Check the losses add up: 74% + 26% = 100%.

Energy is conserved, so the 260 J has not vanished. It has left as heat in the surroundings, as sound, or as other energy forms that you are not counting as useful.

What a given efficiency means

An efficiency of 74% says that for each 1,000 J paid for, 740 J does the job. Energy costs scale accordingly: to get 740 J of useful output you must buy 1,000 J, which is about 1.35 times the useful amount.

• Below 100%: normal for every real device, since some energy is always dissipated.
• Above 100%: a sign of a mistake or of mixing definitions, since energy cannot be created.
• Heat pumps can quote coefficients above 1, but that compares heat moved with electricity used, a different quantity.

What counts as good depends on the technology. Electric motors, engines, lamps and heaters occupy very different ranges, so compare within a category.

Gains and chains

With the same 1,000 J input, raising useful output by 10% from 740 J to 814 J lifts efficiency from 74% to 81.4%. That is a gain of 7.4 percentage points, not 10: percentage points and percentage change are different things.

Stages in series multiply. A motor at 90%, a gearbox at 90% and a pump at 90% give 0.9 × 0.9 × 0.9 ≈ 0.729, about 73% overall. Improving the weakest stage usually pays off most.

Ways efficiency figures mislead

• Comparing power in watts against energy in joules without matching them over the same time.
• Counting waste heat as useful when the heat is not the product, or vice versa.
• Averaging stage efficiencies instead of multiplying them.
• Quoting a peak efficiency at the best operating point as if it held across all loads.

Using efficiency to cost a job

If a task needs 7.4 MJ of useful work and the machine is 74% efficient, the energy to buy is 7.4 ÷ 0.74 = 10 MJ. Compare machines on that purchased energy and the running cost difference becomes visible.

Check part-load behaviour as well. Many machines run far below their rated efficiency at light loads, which matters if yours is rarely working at full output.

Read labels carefully when comparing machines. Electrical energy, fuel energy and heat can all be given in joules or kilowatt-hours, but fuel heating values follow different conventions. Make sure the input energy is stated the same way for both machines, or the efficiency comparison will be skewed from the start.