Power and Efficiency - Physics for Pilot Assessments

Two machines can do exactly the same job and still differ in the two ways that matter commercially: how quickly they do it, and how much energy they waste doing it. The first is power and the second is efficiency. By the end of this lesson you should be able to calculate a power from an amount of work and a time, run that calculation backwards for either of them, work out an efficiency from an input and a useful output, and say why an efficiency above 100 per cent is not a difficult claim but an impossible one.

Power is a rate, not an amount

Lift a hundred bricks onto a scaffold and the work done is the same whether you take an hour over it or ten minutes: the same weights have been raised the same height, so the joules are identical. What differs is the rate. Power is the amount of work done, or equivalently the amount of energy transferred, in each second — so it is work divided by time. That makes power and energy two genuinely different quantities, which is worth saying explicitly because they are the most frequently confused pair in this module: energy is a quantity of stuff, measured in joules, while power is how fast that stuff is being moved, measured in joules per second.

Power

P = \frac{W}{t}

$P$: power, in watts (W)
$W$: the work done, or energy transferred, in joules (J)
$t$: the time taken, in seconds (s)

The unit is the watt, written with a capital W because it commemorates James Watt, and one watt is one joule per second. That definition is all the unit is: a 60-watt lamp converts 60 joules of electrical energy every second it is switched on, and a 2000-watt kettle takes 2000 joules from the mains every second. Where the numbers get big, the usual prefixes apply — a kilowatt is 1000 watts and a megawatt is a million watts, since the prefix mega- means a million. Note the unfortunate collision of symbols in the formula above: the W on the left of the unit sentence is watts and the W in the numerator is work in joules. Physics has more quantities than letters, and this is one of the places where you have to read from the context.

The two rearrangements follow the usual pattern and both are examined. Multiply both sides of $\text{power} = \frac{\text{work}}{\text{time}}$ by the time: on the right the times cancel, leaving work = power × time. That is the form that tells you how many joules a device has consumed — multiply its rating in watts by the number of seconds it ran. Divide that version by the power instead and the powers cancel, giving $\text{time} = \frac{\text{work}}{\text{power}}$ , which answers how long a job will take at a given rate.

Reading the units backwards

It is worth being able to reconstruct a watt from what you already know, because a certain kind of item asks which combination of quantities gives a power and offers four plausible-looking products. Work back from the definition. A watt is a joule per second. A joule is a newton multiplied by a metre. So a watt is a newton-metre per second — a force, times a distance, divided by a time. Any expression that reduces to that is a power, and anything that does not, is not: a newton times a metre is an energy, not a power, and a newton divided by a square metre is a pressure from the earlier module. The habit of checking the units of an answer catches more mistakes than any other single technique in physics.

That reconstruction also hands you a second and very useful way of writing power. If a force F pushes something a distance d in a time t, the work is Fd and the power is Fd divided by t. But d divided by t is the speed, so the power is the force multiplied by the speed. No new physics has appeared — it is work over time with the division done in a different order — but it is the form you want whenever something is being pushed steadily along at a constant speed against a resisting force.

Power when a force pushes at a steady speed

P = F v

$P$: power, in watts (W)
$F$: the force applied along the direction of movement, in newtons (N)
$v$: the speed, in metres per second (m/s)

So a locomotive pulling with 20 000 newtons at 30 metres per second is delivering 600 000 watts, or 600 kilowatts, and the same locomotive at half the speed against the same resistance delivers half the power. Notice that this form contains no time at all, because the speed has swallowed it.

Efficiency: how much of it was useful

No machine turns all of the energy you feed it into the thing you wanted. A motor lifting a load also warms its own windings and bearings and makes a noise; a filament lamp asked for light delivers far more heat than light; brakes exist specifically to turn kinetic energy into thermal energy. The energy that ends up somewhere other than where you wanted it is called wasted energy, and — from the previous lesson — it has not vanished. Conservation of energy holds exactly, so the total energy going in equals the useful energy out plus the wasted energy out, always. Efficiency is the fraction of the input that came out as the useful part.

Efficiency as a fraction

\text{efficiency} = \frac{E_\text{useful}}{E_\text{total in}}

$E_\text{useful}$: the energy transferred usefully — the output you actually wanted, in joules (J)
$E_\text{total in}$: the total energy supplied, in joules (J)
$\text{efficiency}$: a bare ratio between 0 and 1, with no units, because joules have been divided by joules

Because it is a ratio of two energies, the units cancel and efficiency has none; it is a plain number between zero and one. It is usually quoted as a percentage instead, which means nothing more than multiplying that fraction by a hundred, since per cent means per hundred.

The same thing as a percentage

\text{efficiency (\%)} = \frac{E_\text{useful}}{E_\text{total in}} \times 100

$E_\text{useful}$: the useful energy out, in joules (J)
$E_\text{total in}$: the total energy in, in joules (J)
$100$: the factor that turns a fraction into a percentage

One consequence is absolute rather than merely usual: the useful output can never exceed the total input, so the fraction can never be greater than one and the percentage can never exceed a hundred. A device claiming otherwise would be creating energy, which the previous lesson ruled out. In practice every real device falls short of a hundred per cent, because there is always some friction, some warming, some sound — so an efficiency strictly below one is the normal state of the world, and an efficiency of exactly one is an idealisation used to simplify a question.

All the energy that goes in comes out somewhere. The useful branch and the wasted branch add up to the input exactly, and the efficiency is the useful branch as a fraction of the whole.

The figure is also the fastest route to the wasted energy, which items ask for as often as they ask for the efficiency itself. Since input equals useful plus wasted, the wasted energy is the input minus the useful part — no separate formula required, just subtraction. And because dividing a useful power by a total power cancels the seconds in exactly the way dividing energies cancels the joules, the same efficiency calculation works with powers in watts instead of energies in joules, provided both figures are powers.

Example 4 — efficiency and waste from an input and an output

A motor draws 500 joules of electrical energy and delivers 200 joules of useful lifting. What is its efficiency, as a fraction and as a percentage, and how much energy is wasted?

Efficiency is the useful output divided by the total input: $\text{efficiency} = \frac{200}{500}$
Both numbers divide by 100, giving $\frac{2}{5} = 0.4$ . So the efficiency is 0.4.
As a percentage, 0.4 × 100 = 40 per cent.
The wasted energy is the input minus the useful output: 500 − 200 = 300 joules, which ends up as thermal energy in the windings and bearings, plus a little sound.
Sanity-check the direction: the wasted 300 joules is more than the useful 200, so the efficiency must be below half — and 40 per cent is.

Example 5 — working backwards to the input needed

A pump is 25 per cent efficient and a job needs 400 joules of useful output. How much energy must be supplied, and how much will be wasted?

Turn the percentage into a fraction first: 25 per cent is $\frac{25}{100} = 0.25$ .
Start from $\text{efficiency} = \frac{\text{useful}}{\text{total in}}$ and multiply both sides by the total in: efficiency × total in = useful. Now divide both sides by the efficiency, and the efficiencies on the left cancel: $\text{total in} = \frac{\text{useful}}{\text{efficiency}}$ .
So the input needed is $\frac{400}{0.25}$ . Dividing by 0.25 is the same as multiplying by 4, so the input is 1600 joules.
The wasted energy is 1600 − 400 = 1200 joules.
Check the efficiency: $\frac{400}{1600} = 0.25$ , which is 25 per cent. Correct. Notice the direction of the trade — a low efficiency means the input must be much larger than the output, never smaller.

Example 6 — efficiency from powers, and a power from a force

A machine takes 1000 watts of electrical power and lifts a load steadily at 0.5 metres per second against a weight of 1200 newtons. What is its useful output power, and what is its efficiency?

For the useful output, use power = force × speed: $P = 1200 \times 0.5 = 600\ \text{W}$
Both figures are now powers, so the efficiency is the useful power divided by the input power: $\frac{600}{1000} = 0.6$ , which is 60 per cent.
The wasted power is 1000 − 600 = 400 watts, appearing as heat and sound at a rate of 400 joules every second.
Cross-check with energies over one second: in one second 1000 joules go in and 600 joules of lifting come out, and $\frac{600}{1000}$ is the same 0.6. Dividing both figures by the same time never changes the ratio.

Why it matters in aviation

Work says how much energy a climb needs; power says how quickly the engines can supply it, and that is what sets the rate of climb. The energy delivered each second has to cover the drag first, and only what is left over goes into gaining height — so climb performance is never a property of the engine alone. The same power on a cleaner airframe climbs better, and in the cruise, where no height is being gained, the aircraft settles at the speed where the power available and the power the air demands are equal.

Efficiency accounts for the rest of the fuel. Only a fraction of the chemical energy released ends up as useful work on the airflow, the remainder leaving as heat and noise, and that fraction changes with speed and with altitude. It is also why there is no single best speed to fly. One speed minimises the energy spent per unit of distance, and it is the one to choose when the aim is to get as far as possible on the fuel aboard; another minimises the energy spent per unit of time, and it is the one to choose when the aim is to stay airborne as long as possible. The two are different because distance and time are different denominators.

What the test asks

Two distractors, and both are about a category rather than a calculation. The first treats power and energy as interchangeable: an item quotes a device in watts and asks for the energy it uses, and one option gives the rating straight back with the unit swapped to joules. A rating in watts is a rate, so it has to be multiplied by a number of seconds before it becomes an energy — and the corresponding reverse trap gives you joules and asks for a power, expecting you to hand back the joules unchanged. Read the units on the answer options first. The second is the impossible efficiency: an option above 100 per cent, or an option that quietly divides the input by the output instead of the output by the input, which gives 2.5 rather than 0.4 for the numbers in Example 4. Anything above one is wrong, so if your answer exceeds a hundred per cent you have divided the wrong way round. Expect also a wasted-energy item where the trap is to give the useful output instead of the difference, and expect a question in which the useful and wasted figures are given and the input is what is missing — in which case you add them.