Engine Power and Efficiency Calculations Revision Notes | A Level AQA Physics

1. Why Engine Power Calculations Matter

Every internal combustion engine takes in chemical energy from fuel and tries to convert it into useful mechanical work at the output shaft. The problem is that real engines are never perfectly efficient — energy leaks away as heat through the cylinder walls, and mechanical friction between moving parts steals power before it ever reaches the wheels.

To analyse this properly, AQA A Level physics breaks engine power into four distinct quantities: input power, indicated power, brake power, and friction power. Each one tells you something different about where energy is going — and together they let you calculate three separate efficiency ratios. Mastering these is the key to answering any engine question on the paper.

2. Input Power — Energy Entering the Engine

Before the engine can do anything useful, it must burn fuel. The rate at which chemical energy enters the engine is called the input power. Think of it as the power of the fuel supply — how much energy per second is being released by combustion.

The calorific value is a property of the fuel itself — petrol has a calorific value of roughly $44 \times 10^6$ J kg$^{-1}$. The fuel flow rate is how quickly the engine is consuming that fuel. Multiply the two together and you get the power flowing into the engine every second.

3. Indicated Power — What the Gas Delivers to the Piston

Not all of the input power becomes useful work — some is lost as heat through the cylinder walls and exhaust gases. The power that the hot, expanding gas actually delivers to the piston is called the indicated power. You calculate it directly from the engine's pressure-volume (p-V) cycle diagram.

On a p-V diagram, the engine cycle traces out a closed loop. The area enclosed by that loop equals the net work done per cycle by the gas on the piston. To find the power, you scale that work by how many times the cycle repeats each second and by how many cylinders are doing it simultaneously.

A four-stroke engine completes one power stroke every two crankshaft revolutions, so its cycle frequency is half the rotational frequency in revolutions per second. This is a detail examiners test — make sure you use the cycle frequency, not the engine speed in rpm, without converting correctly.

Worked example: Calculating Indicated Power from a p-V Loop

A four-cylinder, four-stroke engine runs at 3000 rpm. The area of the p-V loop for one cylinder is 500 J. Calculate the indicated power.

1. $n = \frac{3000}{60} = 50 \text{ rev s}^{-1}$ — Divide rpm by 60 to get rev s$^{-1}$. Always do this conversion before anything else.
2. $f_{\text{cycle}} = \frac{50}{2} = 25 \text{ cycles s}^{-1}$ — A four-stroke engine completes one full thermodynamic cycle every two crankshaft revolutions, so the cycle frequency is half the rotational frequency. This is the most common error students make here.
3. $\text{indicated power} = 500 \times 25 \times 4$ — Area of loop = 500 J, cycle frequency = 25 Hz, number of cylinders = 4. [M1] for correct formula and substitution.
4. $\text{indicated power} = 500 \times 25 \times 4 = 50\,000 \text{ W} = 50 \text{ kW}$ — $500 \times 25 = 12\,500$; $12\,500 \times 4 = 50\,000$ W. [A1] for correct answer with units.

Answer: Indicated power = 50 kW

4. Brake Power and Friction Power — What Reaches the Shaft

Even after the gas has done its work on the piston, the power must travel through connecting rods, crankshaft, bearings, and gearbox before it reaches the output shaft. All of those moving surfaces rub against each other, and that friction converts some of the indicated power into heat. What survives at the output shaft is the brake power.

The name 'brake power' comes from the historical method of measuring it: a friction brake (a Prony brake or dynamometer) was clamped to the output shaft to measure the torque it could sustain at a given speed. Today, a dynamometer does the same job electronically, but the name has stuck.

This relationship is the engine's power budget. Every watt of indicated power either reaches the shaft as brake power or is dissipated as friction power — there is no third option. If you know any two of the three quantities, you can always find the third.

Worked example: Finding Brake Power and Friction Power

An engine's output shaft produces a torque of 200 N m at an angular velocity of 150 rad s$^{-1}$. The indicated power is 35 kW. Calculate the brake power and the friction power.

1. $P = T\omega = 200 \times 150$ — Substitute torque = 200 N m and angular velocity = 150 rad s$^{-1}$ directly. [M1]
2. $P = 200 \times 150 = 30\,000 \text{ W} = 30 \text{ kW}$ — $200 \times 150 = 30\,000$ W. [A1] for correct value with units.
3. $\text{friction power} = \text{indicated power} - \text{brake power} = 35\,000 - 30\,000$ — Convert indicated power to watts: 35 kW = 35 000 W. Keep units consistent throughout. [M1]
4. $\text{friction power} = 35\,000 - 30\,000 = 5\,000 \text{ W} = 5 \text{ kW}$ — $35\,000 - 30\,000 = 5\,000$ W. [A1] This 5 kW is entirely wasted as heat inside the engine.

Answer: Brake power = 30 kW; Friction power = 5 kW

5. The Three Efficiency Ratios

Now that you have all four power quantities, you can calculate three distinct efficiency ratios. Each one answers a different question about the engine's performance, and AQA expects you to know all three definitions precisely.

Think of the engine as a two-stage energy conversion. Stage one: combustion converts chemical energy (input power) into mechanical work on the piston (indicated power) — this is captured by thermal efficiency. Stage two: the piston's work is transmitted through the mechanical drivetrain to the shaft (brake power) — this is captured by mechanical efficiency. The product of both stages gives the overall efficiency.

There is an elegant relationship between the three: overall efficiency equals thermal efficiency multiplied by mechanical efficiency. You can verify this algebraically — the indicated power cancels in the product $\frac{\text{indicated}}{\text{input}} \times \frac{\text{brake}}{\text{indicated}} = \frac{\text{brake}}{\text{input}}$. This is a useful internal consistency check in multi-part exam questions.

Worked example: Calculating All Three Efficiency Ratios

An engine has an input power of 90 kW, an indicated power of 54 kW, and a brake power of 45 kW. Calculate the overall, thermal, and mechanical efficiencies. Express each as a percentage.

1. $\eta_{\text{overall}} = \frac{\text{brake power}}{\text{input power}} = \frac{45\,000}{90\,000}$ — Use brake power over input power. Keep units consistent — both in W (or both in kW, the ratio is the same). [M1]
2. $\eta_{\text{overall}} = \frac{45}{90} = 0.50 = 50\%$ — $45 \div 90 = 0.50$. Multiply by 100 to express as a percentage. [A1]
3. $\eta_{\text{thermal}} = \frac{\text{indicated power}}{\text{input power}} = \frac{54\,000}{90\,000}$ — Indicated power over input power. [M1]
4. $\eta_{\text{thermal}} = \frac{54}{90} = 0.60 = 60\%$ — $54 \div 90 = 0.60$. So 60% of the fuel's chemical energy becomes mechanical work on the piston. [A1]
5. $\eta_{\text{mechanical}} = \frac{\text{brake power}}{\text{indicated power}} = \frac{45\,000}{54\,000}$ — Brake power over indicated power. [M1]
6. $\eta_{\text{mechanical}} = \frac{45}{54} = 0.833\ldots \approx 83.3\%$ — $45 \div 54 = 0.8\overline{3}$. This means about 83% of the power delivered to the piston actually reaches the shaft. [A1]
7. $0.60 \times 0.833 = 0.50$ ✓ — $0.60 \times 0.8\overline{3} = 0.50$. The product of thermal and mechanical efficiencies equals the overall efficiency, confirming the answers are internally consistent.

Answer: Overall efficiency = 50%; Thermal efficiency = 60%; Mechanical efficiency ≈ 83.3%

6. Linking the Power Terms — The Full Engine Picture

It helps to visualise all four power quantities as a cascade. Chemical energy enters at the top as input power. Thermal losses (heat through walls, exhaust) reduce this to indicated power. Friction losses then reduce indicated power further to brake power. Friction power is the gap between indicated and brake.

Notice that friction power (9 kW in this example) is the difference between indicated power (54 kW) and brake power (45 kW). The thermal losses account for the drop from input (90 kW) to indicated (54 kW) — a loss of 36 kW. Together, these two loss mechanisms explain why real engines have overall efficiencies well below 100%.

7. Exam Technique and Common Mistakes

Engine power questions on AQA papers tend to follow a predictable pattern: you are given some data, asked to calculate two or three power quantities, and then asked to calculate one or more efficiency ratios. The marks are almost always split between method marks (correct formula and substitution) and accuracy marks (correct numerical answer with units). Showing your formula before substituting is the single most reliable way to secure method marks even if you make an arithmetic slip.

The four-stroke cycle frequency trap catches a large number of students every year. When an engine speed is given in rpm, you must divide by 60 to get rev s$^{-1}$, and then divide by 2 again to get the cycle frequency for a four-stroke engine. Skipping the second division doubles your indicated power and makes every subsequent efficiency calculation wrong.