Four-Stroke Engine Cycles and Indicator Diagrams Revision Notes | A Level AQA Physics

1. Why Indicator Diagrams Matter

Every time a car engine fires, it converts chemical energy into mechanical work. The question engineers ask is: how much work is actually being extracted from the gas, and how does that compare to what is theoretically possible? The answer lives in a single graph — the indicator diagram.

An indicator diagram is a pressure-volume (p-V) graph traced out as the engine completes one full cycle. The area enclosed by the loop on that graph equals the net work done by the gas per cycle. This makes it an extraordinarily powerful tool: from one diagram, you can read off power output, efficiency, and the losses introduced by real-world imperfections.

Before reading the diagrams, you need to know what is physically happening inside the cylinder during each stroke. That is where the four-stroke cycle comes in.

2. The Four-Stroke Petrol Engine Cycle

Picture a single cylinder with a piston moving up and down. The piston's position controls the volume of gas inside. Four distinct strokes — two downward, two upward — complete one full thermodynamic cycle. The crankshaft turns twice for every one complete cycle.

• Stroke 1 — Induction (intake): The piston moves down, the inlet valve opens, and a fresh mixture of petrol vapour and air is drawn into the cylinder at approximately atmospheric pressure. Volume increases, pressure stays roughly constant.
• Stroke 2 — Compression: Both valves close. The piston moves up, compressing the fuel-air mixture. Volume decreases sharply; pressure and temperature rise. No heat is exchanged with the surroundings — this is approximately adiabatic.
• Stroke 3 — Power (expansion): Near the top of the compression stroke, the spark plug fires. Combustion raises the pressure almost instantaneously at roughly constant volume. The high-pressure gas then pushes the piston down, doing work on the crankshaft. This is the only stroke that delivers useful mechanical energy.
• Stroke 4 — Exhaust: The exhaust valve opens. The piston moves back up, pushing the burned gases out of the cylinder at approximately atmospheric pressure. The cycle then repeats.

Notice that only one of the four strokes actually delivers work — the power stroke. The other three strokes consume or return energy, which is why the net work per cycle is the difference between the work done by the gas on the piston and the work done by the piston on the gas.

3. The Theoretical Petrol Cycle Indicator Diagram (Otto Cycle)

The theoretical model for the petrol engine is called the Otto cycle. On a p-V diagram, it consists of four idealised processes. Understanding this theoretical shape is essential because real diagrams are always compared against it.

Here is how the theoretical Otto cycle traces out on a p-V graph, moving around the loop in sequence:

• A to B — Adiabatic compression: Point A represents the state of the gas at the start of compression — after the induction stroke has filled the cylinder, but before compression begins (i.e. at maximum volume, approximately atmospheric pressure). The piston moves up, compressing the fuel-air mixture from $V_{max}$ to $V_{min}$; pressure rises steeply along a curve. No heat exchange occurs. Note: the induction and exhaust strokes are not part of the main thermodynamic loop — they trace along the horizontal axis at atmospheric pressure and are shown separately (see final bullet below).
• B to C — Constant-volume heat addition (combustion): Volume stays fixed at $V_{min}$; pressure jumps sharply upward as the fuel-air mixture ignites. This is a vertical line on the diagram.
• C to D — Adiabatic expansion (power stroke): Volume increases back to $V_{max}$; pressure falls along a curve. Again, no heat exchange.
• D to A — Constant-volume heat rejection (exhaust valve opens): Volume stays fixed at $V_{max}$; pressure drops sharply back to the starting value. Another vertical line.
• Induction and exhaust strokes (outside the main loop): These strokes occur at approximately atmospheric pressure and trace along the horizontal axis. Because they cancel each other out, they contribute zero net work and are not part of the A→B→C→D→A thermodynamic loop.

The enclosed area of this loop — the region bounded by the four processes — equals the net work done per cycle. A larger enclosed area means more work extracted per cycle. This is the key insight that makes indicator diagrams so useful for engineers.

4. The Diesel Engine Cycle

The diesel engine follows a different thermodynamic route. The key difference from the petrol engine is that diesel engines compress air alone — no fuel is present during compression. The compression ratio is much higher (typically 14:1 to 22:1 versus 8:1 to 12:1 for petrol), so the air temperature at the end of compression is high enough to ignite diesel fuel spontaneously when it is injected. There is no spark plug.

Because fuel injection continues as the piston begins to move down, combustion occurs over a range of volumes rather than instantaneously. This gives the diesel cycle its distinctive feature: heat addition at approximately constant pressure, rather than constant volume.

• Stroke 1 — Induction: Air only is drawn in as the piston descends. Pressure is approximately atmospheric.
• Stroke 2 — Compression: Air is compressed adiabatically to a much smaller volume than in a petrol engine. Temperature rises to around 700–900 °C.
• Stroke 3 — Power: Diesel fuel is injected and ignites. Combustion occurs at roughly constant pressure as the piston moves down. This is the defining feature of the diesel cycle. Expansion then continues adiabatically.
• Stroke 4 — Exhaust: Burned gases are expelled as the piston rises.

On the theoretical diesel cycle indicator diagram, the heat-addition process appears as a horizontal line (constant pressure) rather than the vertical line seen in the Otto cycle. The adiabatic compression and expansion curves are steeper than in the Otto cycle because of the higher compression ratio.

5. Real Indicator Diagrams — Deviations from Theory

The theoretical diagrams assume idealised conditions that never exist in a real engine. When you measure the actual pressure inside a cylinder throughout a real cycle, the resulting indicator diagram looks noticeably different — and understanding why is exactly what examiners test.

The most important deviations to know are listed below. Each one reduces the enclosed area of the diagram, which means less net work per cycle and lower efficiency.

• Rounded corners: The theoretical diagram has sharp corners at the transitions between processes (for example, the instant switch from compression to combustion). In reality, valves take time to open and close, and combustion is not instantaneous, so the corners are smoothed into curves.
• Combustion is not truly instantaneous or constant-volume: In a petrol engine, the flame front takes a finite time to propagate across the cylinder. Pressure rises over a small but non-zero crank angle, so the B-to-C line tilts rather than being perfectly vertical.
• Heat losses to cylinder walls: The theoretical cycle assumes adiabatic compression and expansion, but real cylinders lose heat to the metal walls throughout the cycle. This reduces the peak pressure reached and lowers the pressure during expansion, shrinking the enclosed area.
• Friction and pumping losses: Real gases have viscosity, and real pistons experience friction. The exhaust and induction strokes are not perfectly at atmospheric pressure — the exhaust stroke requires slightly above atmospheric pressure to push gases out, and the induction stroke draws in at slightly below atmospheric. This creates a small negative loop at the bottom of the diagram that subtracts from the net work.
• Blow-by past piston rings: Some gas leaks past the piston rings, reducing the effective pressure during the power stroke.

The net effect is that the real indicator diagram has a smaller enclosed area than the theoretical diagram. This directly means less work per cycle and lower indicated power than the theoretical maximum. Comparing the two diagrams allows engineers to identify and quantify these losses.

6. Using Indicator Diagrams to Calculate Power and Efficiency

The indicator diagram is not merely a qualitative tool — it gives you quantitative data. The two key quantities you can extract are indicated power and engine efficiency.

The factor of 2 in the denominator for $n$ is a common source of error. In a four-stroke engine, the crankshaft completes two full revolutions for every one power stroke per cylinder. So if the engine runs at 3000 rpm (50 rev/s), there are only 25 power strokes per second per cylinder.

The difference between indicated power and brake power is the power lost to friction within the engine — friction between piston rings and cylinder walls, in bearings, and in driving ancillaries. This is called friction power: $P_{friction} = P_{indicated} - P_{brake}$.

Worked example: Calculating Indicated Power from an Indicator Diagram

A four-stroke petrol engine has a swept volume per cylinder of $4.0 \times 10^{-4}$ m³. The area enclosed by the indicator diagram is measured as $1.5 \times 10^{-3}$ cm² on a diagram where the pressure axis scale is $2.0 \times 10^{5}$ Pa per cm and the volume axis scale is $1.0 \times 10^{-4}$ m³ per cm. The engine runs at 3000 rpm. Calculate the indicated power per cylinder.

1. $W_{net} = \text{diagram area (cm}^2\text{)} \times \text{pressure scale} \times \text{volume scale}$\n$W_{net} = 1.5 \times 10^{-3} \text{ cm}^2 \times 2.0 \times 10^{5} \text{ Pa cm}^{-1} \times 1.0 \times 10^{-4} \text{ m}^3 \text{ cm}^{-1}$\n$W_{net} = 1.5 \times 10^{-3} \times 2.0 \times 10^{5} \times 1.0 \times 10^{-4}$ J — The area on the paper (in cm²) must be multiplied by both axis scales to convert to actual work in joules. Each cm on the pressure axis represents a certain pressure (Pa/cm), and each cm on the volume axis represents a certain volume (m³/cm). Multiplying cm² × Pa/cm × m³/cm gives Pa·m³ = J.
2. $W_{net} = 1.5 \times 2.0 \times 1.0 \times 10^{-3+5-4}$ J\n$= 3.0 \times 10^{-2}$ J\n$= 0.030$ J — Combine the powers of ten: $-3 + 5 + (-4) = -2$. So $W_{net} = 3.0 \times 10^{-2}$ J per cycle.
3. $\text{Engine speed} = 3000 \text{ rpm} = \frac{3000}{60} \text{ rev/s} = 50 \text{ rev/s}$\n$n = \frac{50}{2} = 25 \text{ power strokes per second}$ — For a four-stroke engine, divide revolutions per second by 2 because one power stroke occurs every two revolutions of the crankshaft.
4. $P_{indicated} = W_{net} \times n$\n$P_{indicated} = 0.030 \text{ J} \times 25 \text{ s}^{-1}$\n$P_{indicated} = 0.75 \text{ W}$ — Power is energy per unit time. Multiplying net work per cycle by the number of cycles per second gives power in watts. [M1] for correct formula, [M1] for correct value of n, [A1] for final answer.
5. 0.75 W per cylinder. — Note: the axis scales and diagram area used in this example are chosen to keep the arithmetic straightforward for practice purposes and do not represent realistic engine operating conditions. A real 400 cc cylinder at 3000 rpm would typically produce hundreds of watts per cylinder. When working through exam questions, always use the scales and data as given — the method is identical regardless of the magnitude of the numbers.

Answer: $P_{indicated} = 0.75$ W per cylinder

Worked example: Calculating Engine Efficiency from Power Data

A four-cylinder four-stroke engine has an indicated power of 45 kW. A brake test measures the useful output power at the crankshaft as 38 kW. Calculate (a) the mechanical efficiency and (b) the friction power.

1. $\eta = \frac{P_{brake}}{P_{indicated}}$ — Mechanical efficiency is the ratio of useful output power to the power delivered by the working gas. [M1]
2. $\eta = \frac{38 \text{ kW}}{45 \text{ kW}} = \frac{38}{45}$ — Both powers are in kW, so the units cancel. No conversion needed when the units are the same.
3. $\eta = 0.844... \approx 0.84$ (to 2 significant figures) As a percentage: $\eta \approx 84\%$ — 38 ÷ 45 = 0.8444... Rounding to 2 significant figures gives 0.84. [A1]
4. $P_{friction} = P_{indicated} - P_{brake}$ $P_{friction} = 45 \text{ kW} - 38 \text{ kW} = 7 \text{ kW}$ — The power not delivered to the output shaft is lost to internal friction. [B1]

Answer: (a) Mechanical efficiency $\eta \approx 0.84$ (84%). (b) Friction power $= 7$ kW.

7. Interpreting Unfamiliar Cycle Diagrams

The AQA specification explicitly states that questions may be set on cycles other than the petrol and diesel cycles. When this happens, all essential information will be provided — you will not be expected to recall the details of, say, a Stirling cycle from memory. What you are expected to do is apply the core principles you already know.

Here is the interpretive toolkit you need for any unfamiliar p-V diagram:

• Enclosed area = net work per cycle. A larger loop means more work. If the loop is traversed clockwise, the cycle is a heat engine (net work done by the gas). If anticlockwise, it is a refrigerator or heat pump (net work done on the gas).
• Horizontal lines = constant-pressure (isobaric) processes. The gas expands or contracts at fixed pressure.
• Vertical lines = constant-volume (isochoric) processes. Pressure changes with no change in volume — no work is done during these sections ($W = p \Delta V = 0$ when $\Delta V = 0$).
• Curves = adiabatic or isothermal processes. A curve that is steeper than an isothermal through the same point indicates an adiabatic process; a curve that is less steep (shallower) than an isothermal through the same point indicates an isothermal process. This distinction arises because the adiabatic index $\gamma > 1$, making adiabatic curves steeper on a p-V diagram.
• The mean indicated pressure can always be found by dividing the enclosed area (in joules) by the swept volume (in m³), regardless of the cycle shape.

When a question gives you an unfamiliar diagram, start by identifying the enclosed area, then check whether any sides are horizontal (constant pressure) or vertical (constant volume). From there, the same formulae for indicated power and efficiency apply exactly as they do for petrol and diesel cycles.

8. Exam Technique and Mark-Winning Strategies

Indicator diagram questions appear in the engineering physics section of AQA A Level Physics Paper 3. They reward students who are precise with language and careful with the factor-of-two trap in power calculations. Here is how to secure every available mark.