Indicator Diagrams and Engine Cycles

An indicator diagram is a pressure-volume (pV) graph showing the state of the working fluid in an engine cylinder through one complete cycle. The diagram is the engineer's most powerful diagnostic and design tool: the area enclosed by the loop is the net work done per cycle, the direction of the loop (clockwise or anticlockwise) tells you whether the device is a heat engine or a refrigerator, and the shape of the loop reveals the idealised cycle the engine is approximating. This lesson develops the indicator-diagram interpretation rigorously, then applies it to the two most important real-world engine cycles: the Otto cycle (the idealisation of a four-stroke petrol engine) and the Diesel cycle (the idealisation of a Diesel engine). For each cycle we derive the thermal efficiency formula in terms of the compression ratio and the adiabatic index γ, work through numerical examples, and compare petrol and Diesel performance.

Spec mapping (AQA 7408 §3.11.2 — Thermodynamics, option C): This lesson covers indicator diagrams as pV plots of an engine's working cycle; the result that net work per cycle equals the enclosed area W_net = ∮p dV; the distinction between clockwise heat-engine loops and anticlockwise refrigeration loops; the four-stage Otto cycle (two adiabatics, two isochorics) with efficiency η = 1 − 1/r^(γ−1); the four-stage Diesel cycle (two adiabatics, isobaric heat addition, isochoric heat rejection); and comparison of petrol and Diesel real-engine efficiencies. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Engine cycles are constructed from the four processes of Lesson 2 — every stage is one of isothermal, isobaric, isochoric or adiabatic. The efficiency analysis sits between the first law (Lesson 2) and the second law / Carnot bound (Lesson 4) — Carnot efficiency provides the upper limit that no Otto or Diesel cycle can exceed. The mechanical output couples to the rotational dynamics of Lesson 1 — engine torque drives a crankshaft of moment of inertia I, the flywheel smooths the pulsed torque delivery, and the power output is P = τω. Engine cycles also synoptically connect to combustion chemistry (Year 12 chemistry — calorific values of fuels) and to environmental physics (engine efficiency directly determines CO₂ output per unit useful work).

Indicator Diagrams — The Fundamental Result

For any quasi-static change of an ideal gas, the work done on the gas is W_on = −∫p dV (Lesson 2). Around a closed cycle that returns the gas to its starting state, the net change in internal energy is zero (ΔU_cycle = 0), so by the first law:

Net heat in = Net work out per cycle: Q_net = −W_on, net = ∮p dV

The integral ∮p dV is the signed area enclosed by the loop on a pV diagram. Geometrically:

A clockwise loop has positive ∮p dV, so the gas does positive net work on the surroundings. This is a heat engine: heat in at high T, work out, heat out at low T.
An anticlockwise loop has negative ∮p dV, so net work is done on the gas. This is a refrigerator / heat pump: work in, heat moved from low T to high T (covered in Lesson 5).

The magnitude of the area is the work per cycle. Multiplying by the number of cycles per second (the engine's cycle frequency) gives the mean power output.

graph LR
    A["Indicator diagram (pV loop)"] --> B{"Loop direction?"}
    B -->|Clockwise| C["Heat engine<br/>W_net out > 0<br/>η = W_net / Q_in"]
    B -->|Anticlockwise| D["Refrigerator / heat pump<br/>W_net in > 0<br/>COP = Q_out / W_in"]

    style C fill:#27ae60,color:#fff
    style D fill:#3498db,color:#fff

Worked Example — Reading Work Off a Simple Cycle

A gas is taken around the rectangular pV cycle: (V₁, p₁) → (V₂, p₁) → (V₂, p₂) → (V₁, p₂) → (V₁, p₁) with p₁ = 5.0 × 10⁵ Pa, p₂ = 1.0 × 10⁵ Pa, V₁ = 2.0 × 10⁻³ m³, V₂ = 8.0 × 10⁻³ m³, traversed clockwise. Calculate the net work output per cycle.

W_net = area of rectangle = (p₁ − p₂) × (V₂ − V₁) = (4.0 × 10⁵) × (6.0 × 10⁻³) = 2400 J.

This is a textbook idealisation but it makes the geometric point: read the area, that's the work. If the cycle runs at 30 Hz (30 cycles per second), the mean power output is 2400 × 30 = 72 kW.

The Otto Cycle — Idealised Petrol Engine

The Otto cycle is the idealisation of a four-stroke spark-ignition engine (the standard petrol/gasoline engine). It consists of four reversible stages:

1 → 2: Adiabatic (isentropic) compression. The piston rises, compressing the air-fuel mixture without heat exchange. V drops from V₁ to V₂ = V₁/r where r = V₁/V₂ is the compression ratio. p and T rise according to pV^γ = const.
2 → 3: Isochoric heat addition. At top dead centre, the spark fires and the mixture combusts almost instantaneously — too fast for the piston to move appreciably. Volume is approximately constant; pressure and temperature jump as heat Q_in is released by the fuel.
3 → 4: Adiabatic (isentropic) expansion. The hot high-pressure gas drives the piston down. Adiabatic (no heat exchange during the rapid power stroke). V rises back from V₂ to V₁.
4 → 1: Isochoric heat rejection. At bottom dead centre, the exhaust valve opens. Volume is approximately constant; pressure and temperature drop as heat Q_out is dumped to the atmosphere (in real engines this corresponds to the exhaust stroke and a fresh intake stroke, but the idealised cycle treats them together as a single heat-rejection process).

graph LR
    S1["State 1<br/>BDC, low p<br/>after intake"] -->|"1→2 adiabatic<br/>compression"| S2["State 2<br/>TDC, high p, hot"]
    S2 -->|"2→3 isochoric<br/>spark + combustion<br/>Q_in"| S3["State 3<br/>TDC, very high p<br/>very hot"]
    S3 -->|"3→4 adiabatic<br/>power stroke<br/>expansion"| S4["State 4<br/>BDC, moderate p<br/>warm"]
    S4 -->|"4→1 isochoric<br/>exhaust<br/>Q_out"| S1

    style S3 fill:#e74c3c,color:#fff
    style S1 fill:#3498db,color:#fff

Efficiency of the Otto Cycle

The thermal efficiency of any heat engine is η = W_net / Q_in. For the Otto cycle, on a monatomic-or-diatomic ideal gas, heat is only exchanged on the two isochoric stages:

Q_in = n C_v (T₃ − T₂) (isochoric heating, 2→3)
Q_out = n C_v (T₄ − T₁) (isochoric rejection, 4→1, taken as positive)

By the first law W_net = Q_in − Q_out = n C_v [(T₃ − T₂) − (T₄ − T₁)].

η = 1 − Q_out/Q_in = 1 − (T₄ − T₁) / (T₃ − T₂)

The two adiabatic stages give:

1→2: T₁ V₁^(γ−1) = T₂ V₂^(γ−1), so T₂/T₁ = (V₁/V₂)^(γ−1) = r^(γ−1).
3→4: T₃ V₂^(γ−1) = T₄ V₁^(γ−1), so T₃/T₄ = (V₁/V₂)^(γ−1) = r^(γ−1).

So T₂/T₁ = T₃/T₄, which means T₄/T₁ = T₃/T₂, which means (T₄−T₁)/(T₃−T₂) = T₁/T₂ = 1/r^(γ−1).

Otto-cycle efficiency: η_Otto = 1 − 1/r^(γ−1)

The efficiency depends only on the compression ratio r and the adiabatic index γ. It does not depend on the temperatures, the gas quantities, or the fuel. Higher r → higher efficiency.

Worked Example — Otto-Cycle Efficiency

Question: A petrol engine has a compression ratio of 9.0. Calculate the theoretical Otto-cycle efficiency (use γ = 1.4 for air).

Solution:

η = 1 − 1/9^0.4 = 1 − 1/2.408 = 1 − 0.415 = 0.585 or 58.5%.

The real efficiency of a road-going petrol engine is approximately 25–30%, much lower than the Otto-cycle ideal. The shortfall is caused by (i) finite-time combustion (so 2→3 is not strictly isochoric), (ii) heat leakage through cylinder walls (so 1→2 and 3→4 are not strictly adiabatic), (iii) friction within the engine, (iv) pumping losses on the intake and exhaust strokes, and (v) gas-mixture properties that deviate from a pure ideal-gas treatment. Real engines also cannot push r above about 12 in petrol applications without triggering autoignition (knock) before the spark fires.

Exam Tip: When asked to "calculate the efficiency", state whether you are giving the Otto/Diesel ideal efficiency or the real-engine efficiency. Mark schemes typically want the formula-based ideal value unless a specific real-engine scenario is given.

The Diesel Cycle — Idealised Diesel Engine

The Diesel cycle differs from the Otto cycle in how heat is added:

1 → 2: Adiabatic compression of pure air (no fuel yet) to very high T and p.
2 → 3: Isobaric heat addition. Fuel is injected into the very hot compressed air and ignites spontaneously (no spark plug). Combustion is slower than petrol's, so the piston moves outward during combustion, keeping pressure approximately constant. Volume rises from V₂ to V₃; this is the cut-off ratio r_c = V₃/V₂.
3 → 4: Adiabatic expansion. Power stroke.
4 → 1: Isochoric heat rejection. Exhaust.

The key difference: heat is added at constant pressure (isobaric, 2→3) rather than constant volume.

Efficiency of the Diesel Cycle

η_Diesel = 1 − (1/γ) × (r_c^γ − 1) / (r^(γ−1)(r_c − 1))

where r is the compression ratio (V₁/V₂) and r_c is the cut-off ratio (V₃/V₂). The derivation tracks Q_in = nC_p(T₃ − T₂) (isobaric) and Q_out = nC_v(T₄ − T₁) (isochoric) through the adiabatic relations.

For r_c → 1 (very fast combustion, isobaric phase shrinks to nothing), the Diesel formula reduces to the Otto formula. For r_c > 1, the Diesel efficiency is lower than the Otto efficiency at the same compression ratio. So why are real Diesel engines more efficient than real petrol engines? Because real Diesels use much higher compression ratios (typically 14–22, vs 8–12 for petrol) — and the η-gain from a higher r outweighs the η-loss from the isobaric combustion. At the same r, Otto wins; at the actually-used r values, Diesel wins.

Worked Example — Diesel-Cycle Efficiency

Question: A Diesel engine has a compression ratio of 18 and a cut-off ratio of 2.5. Calculate the theoretical Diesel-cycle efficiency (γ = 1.4).

Solution:

η_Diesel = 1 − (1/γ) × (r_c^γ − 1) / [r^(γ−1)(r_c − 1)] = 1 − (1/1.4) × (2.5^1.4 − 1) / [18^0.4 × (2.5 − 1)] = 1 − 0.714 × (3.40 − 1) / [3.18 × 1.5] = 1 − 0.714 × 2.40 / 4.77 = 1 − 0.714 × 0.503 = 1 − 0.359 = 0.641 or 64.1%.

Real Diesel engines achieve typical efficiencies of 35–40% — again, the ideal cycle overstates real-world performance for the same reasons listed above.

Comparing Petrol and Diesel

Feature	Petrol (Otto)	Diesel
Heat addition	Isochoric (constant V)	Isobaric (constant p)
Compression ratio (typical)	8 – 12	14 – 22
Ignition	Spark plug	Compression (autoignition)
Fuel	Petrol (volatile, low autoignition T)	Diesel (less volatile, higher autoignition T)
Ideal efficiency at typical r	~55 – 60%	~60 – 65%
Real efficiency (typical)	25 – 30%	35 – 40%
Power : weight	Higher	Lower
CO₂ per unit useful work	Higher	Lower (because efficiency higher)

Worked Example — Cycle Work from pV Data

Question: An idealised Otto cycle uses 1.0 × 10⁻³ mol of gas (γ = 1.4) with V₁ = 5.0 × 10⁻⁴ m³ and a compression ratio of 8. State 1 is at p₁ = 1.0 × 10⁵ Pa, T₁ = 300 K. The heat added per cycle is Q_in = 6.0 J. Calculate (a) the efficiency, (b) the net work output, (c) Q_out, (d) the mean power if the engine runs at 25 cycles per second.

Solution:

(a) η = 1 − 1/r^(γ−1) = 1 − 1/8^0.4 = 1 − 1/2.297 = 1 − 0.4353 = 0.565 or 56.5%.

(b) W_net = η × Q_in = 0.565 × 6.0 = 3.39 J per cycle.

(d) Mean power P̄ = W_net × frequency = 3.39 × 25 = 84.7 W.

(For context, a typical lawnmower engine produces ~3 kW, a road-car engine ~50–150 kW peak — so this micro-cycle example would correspond to a very small model engine.)

Indicator Diagrams and Engine Cycles

Indicator Diagrams and Engine Cycles

Indicator Diagrams — The Fundamental Result

Worked Example — Reading Work Off a Simple Cycle

The Otto Cycle — Idealised Petrol Engine

Efficiency of the Otto Cycle

Worked Example — Otto-Cycle Efficiency

The Diesel Cycle — Idealised Diesel Engine

Efficiency of the Diesel Cycle

Worked Example — Diesel-Cycle Efficiency

Comparing Petrol and Diesel

Worked Example — Cycle Work from pV Data

Worked Example: Petrol Otto Cycle with Heat Addition — Theoretical Efficiency and Peak Conditions

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