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Thermodynamics

Diesel Cycle — Constant-Pressure Heat Addition & Cutoff Ratio

Visualise the ideal cycle of a Diesel engine — the air-standard Diesel cycle. Adjust the compression ratio, cutoff ratio, specific heat ratio and intake conditions to see the thermal efficiency, state-point temperatures and pressures, net work and mean effective pressure update in real time, along with an animated P-V diagram and an efficiency comparison with the Otto cycle.

Parameters
Compression ratio r
Ratio of maximum volume V₁ to minimum volume V₂, V₁/V₂
Cutoff (injection) ratio r_c
Volume at the end V₃ over the start V₂ of constant-pressure heat addition, V₃/V₂
Specific heat ratio γ
c_p/c_v of air. About 1.40 for air at room temperature
Intake temperature T₁
K
Intake pressure P₁
kPa
About 100 kPa naturally aspirated; higher when turbocharged
Displacement V_disp
cc
Volume swept by the piston V₁−V₂ (engine displacement)
Results
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Thermal efficiency η (%)
—
Peak temperature T₃ (K)
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Peak pressure P₃ (kPa)
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Net work / cycle (J)
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Mean effective pressure MEP (kPa)
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Otto comparison
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P-V diagram — cycle animation

1→2 isentropic compression, 2→3 constant-pressure heat addition (fuel injection), 3→4 isentropic expansion, 4→1 constant-volume heat rejection. The enclosed area is the net work, and a marker travels around the cycle.

P-V diagram (pressure vs volume)
Thermal efficiency vs compression ratio (vs Otto)
Theory & Key Formulas

$$\eta_{Diesel}=1-\frac{1}{r^{\gamma-1}}\cdot\frac{r_c^{\gamma}-1}{\gamma\,(r_c-1)}$$

Thermal efficiency of the air-standard Diesel cycle. r is the compression ratio, r_c the cutoff ratio (V₃/V₂) and γ the specific heat ratio.

$$T_2=T_1\,r^{\gamma-1},\quad T_3=T_2\,r_c,\quad T_4=T_1\,r_c^{\gamma}$$

Absolute temperatures at each state point. 1→2 is isentropic compression, 2→3 constant-pressure heat addition, 3→4 isentropic expansion, 4→1 constant-volume heat rejection. r is the compression ratio V₁/V₂ and r_c the cutoff ratio V₃/V₂.

$$q_{in}=c_p(T_3-T_2),\quad q_{out}=c_v(T_4-T_1),\quad MEP=\frac{W_{net}}{V_{disp}}$$

Heat added q_in, heat rejected q_out and mean effective pressure MEP. c_p and c_v are the constant-pressure and constant-volume specific heats, and W_net is the net work.

What is the Diesel Cycle?

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Truck and bus engines are "Diesels", right? What is the biggest difference from a petrol engine?
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The biggest difference is how you light the fire. A petrol engine prepares an air-fuel mixture and a spark plug fires a spark. A Diesel has no spark plug. When you compress air hard enough, that alone heats it to 500-600°C. Inject atomised fuel into it and it ignites the instant it touches that hot air. This is called "compression ignition", and the idealised version of this process is the "Diesel cycle".
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I see. So as an ideal cycle, in what order does it run?
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It has four processes. First, 1→2 compresses the air isentropically. The next one, 2→3, is the Diesel signature: while fuel is being injected the piston moves down and combustion proceeds "at constant pressure". That is the constant-pressure heat addition. Then 3→4 lets the hot gas expand isentropically and do work, and finally 4→1 rejects heat at constant volume to return to the start. In the animated P-V diagram on the left you can watch a marker go around that almost-rectangular loop; the top horizontal line is the constant-pressure heat addition.
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In the Otto cycle the heat addition was "at constant volume", wasn't it? Just changing that to "constant pressure" — what does it actually change?
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Good comparison. Look at the efficiency formula: the Diesel one carries the cutoff-ratio factor (r_c^γ−1)/(γ(r_c−1)). For r_c greater than 1 this is always above 1. In other words, compared at the "same compression ratio", the Diesel is slightly less efficient than the Otto. In the "efficiency vs compression ratio" chart on the right you can see the blue Diesel curve sitting below the orange Otto curve.
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Wait — I always hear that Diesels are more fuel-efficient, but the efficiency is lower?
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That is the key point. The Otto wins only "at the same compression ratio". In real engines a Diesel compresses air only, so there is no risk of knock (abnormal combustion), and the compression ratio can be raised to 15-23. Petrol compresses the mixture, so 10-12 is the limit. Efficiency improves the higher the compression ratio, so the Diesel wins overall. Move the compression-ratio slider r and watch η climb sharply.
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When I move the cutoff ratio r_c the efficiency drifts down a little. Is that related to load?
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Exactly. r_c corresponds to how long fuel is being injected. At high load you inject a lot of fuel, the injection period lengthens, V₃ grows and r_c grows with it. The efficiency factor then increases and η drops slightly. At light load such as idle, r_c approaches 1 and the efficiency approaches that of the Otto cycle. People often say "a Diesel keeps its efficiency well even at part load" precisely because this r_c effect is gentle.

Frequently Asked Questions

The thermal efficiency of the air-standard Diesel cycle is η = 1 − (1/r^(γ−1))·{(r_c^γ − 1)/(γ(r_c − 1))}, where r is the compression ratio, r_c is the cutoff (injection) ratio and γ is the specific heat ratio. Raising the compression ratio r increases the efficiency, while a larger cutoff ratio r_c (a longer injection period, i.e. higher load) lowers it slightly. In the limit r_c → 1 the bracketed factor approaches 1 and the formula reduces to the Otto-cycle efficiency 1 − 1/r^(γ−1).
The cutoff ratio r_c is the volume at the end of the constant-pressure heat addition (fuel injection) V₃ divided by the volume at its start V₂, so r_c = V₃/V₂. In a Diesel engine, fuel is injected into hot compressed air and combustion proceeds at constant pressure while the piston moves down. A longer injection period (higher load) gives a larger V₃ and a larger r_c. At light load such as idle, r_c is close to 1; at rated high load it is typically 2 to 3.
At the same compression ratio r, the Otto cycle is always more efficient. The factor (r_c^γ − 1)/(γ(r_c − 1)) in the Diesel efficiency formula is always greater than 1 for r_c > 1, so for the same r the Diesel η is smaller than the Otto η. Real Diesel engines are nonetheless more fuel-efficient than petrol engines because, with no knock limit, they can run at compression ratios of 15 to 23. Because the compression ratios themselves differ, Diesel wins between real engines.
The mean effective pressure MEP is the net work per cycle W_net divided by the displacement volume V_disp, so MEP = W_net/V_disp. It is the hypothetical constant pressure that, applied to the piston over one stroke, would produce the same work, and it is a measure of an engine's power density. A higher MEP means more torque from the same displacement. It is used to compare engines of different displacement or speed fairly, and turbocharging is a classic way to raise it.

Real-World Applications

Power for commercial vehicles and construction machinery: Trucks, buses, hydraulic excavators and agricultural machinery are almost all powered by Diesel engines. Good thermal efficiency from a high compression ratio, together with the ability to produce large torque from low speed, suits the job of pulling heavy loads. Varying the compression and cutoff ratios in this tool shows the Diesel trait that efficiency falls only a little even at high load (large r_c).

Large marine and power-generation engines: The main engines of large ships and standby or stationary generator engines are the showcase examples of efficiency-focused Diesels. Some low-speed two-stroke marine engines exceed 50% net thermal efficiency. They earn their power density not only through compression ratio but by turbocharging — raising the intake pressure P₁ to lift the mean effective pressure. The P₁ slider in this tool corresponds to that turbocharging effect.

Diesel engines in passenger cars: Passenger-car Diesels, which spread mainly across Europe, are valued for good fuel economy and fat torque. Common-rail high-pressure injection finely controls injection timing and quantity, optimising the way combustion proceeds — which corresponds to the cutoff ratio. On the other hand, exhaust emissions (NOx and PM) are a major technical challenge, and modern Diesels are combined with after-treatment devices.

Thermodynamics education and cycle-comparison study: The Diesel cycle, alongside the Otto and the dual (Sabathé) cycle, is an essential topic when studying the fundamental cycles of internal-combustion engines. The comparison "Otto wins at the same compression ratio, Diesel wins at the same peak pressure" appears often in exams. By overlaying the efficiency curves with the Otto cycle in this tool, you can grasp the relationship intuitively in a way that formulas alone make hard to see.

Common Misconceptions and Pitfalls

The most common misconception is that "a Diesel is more efficient than an Otto even at the same compression ratio". The truth is the opposite: compared at the same compression ratio, the Otto cycle is always more efficient. The Diesel efficiency formula carries the cutoff-ratio factor (r_c^γ−1)/(γ(r_c−1)), which lowers the efficiency. Real Diesels are fuel-efficient because, compressing air only, they have no knock limit and can run at a very high compression ratio. Keep "which cycle is superior" and "which real engine is superior" as separate questions.

Next, assuming the air-standard cycle efficiency equals the real engine efficiency. The η this tool computes is an ideal figure that takes the working fluid as ideal-gas air, treats combustion as external heat addition and assumes constant specific heats. In a real engine, finite combustion time, heat loss to the walls, throttling losses in the exhaust and intake (pumping loss) and friction all bring the net thermal efficiency well below the ideal value. Use the value here as an "upper-bound guide" and a "teaching model for the influence of parameters", not as the real engine's fuel economy.

Finally, assuming the specific heat ratio γ may be fixed at 1.4. γ = 1.4 is the value for air near room temperature. In a real cycle, γ falls to around 1.3 in the high-temperature region after combustion, and the combustion gas composition differs from air. That is why η changes when you move the γ slider in this tool; for a calculation closer to reality, use a smaller γ on the high-temperature side. γ = 1.4 air-standard is enough as a teaching model, but quantitative accuracy requires an analysis that accounts for the temperature dependence of γ and the properties of the combustion gas.

How to Use

  1. Set compression ratio (rComp) between 12 and 24 typical for diesel engines; higher ratios increase efficiency.
  2. Adjust cutoff ratio (rCutoff) from 1.0 to 2.5, representing the expansion ratio during constant-pressure heat addition—lower values approach Otto cycle behavior.
  3. Enter initial temperature T₁ (typically 300–350 K for intake air) and specific heat ratio γ (1.40 for air).
  4. Read thermal efficiency η (%), peak pressure P₃ (kPa), net work per cycle (J), and mean effective pressure MEP (kPa) from the output panel.

Worked Example

For a naturally aspirated diesel engine: compression ratio rComp = 18, cutoff ratio rCutoff = 1.8, T₁ = 320 K, γ = 1.40, working fluid mass = 1.2 g. Thermal efficiency reaches approximately 55–58%, with peak temperature T₃ around 2250 K and peak pressure P₃ near 8500 kPa. Net work per cycle is approximately 680 J, yielding MEP of 620 kPa. Compare this to an equivalent Otto cycle (rCutoff = 1.0): same compression ratio yields η ≈ 52%, demonstrating diesel's pressure and efficiency advantage.

Practical Notes

  1. Increasing cutoff ratio beyond 2.0 reduces efficiency sharply; marine and stationary diesel engines operate at rCutoff = 1.2–1.6 for best performance.
  2. Peak pressure P₃ scales with both compression ratio and cutoff; high-speed automotive diesels (e.g., Common Rail DI) manage 2000+ bar injection pressure but cycle peaks around 9–11 MPa.
  3. Mean effective pressure (MEP) directly correlates to engine torque output; 600–800 kPa MEP is typical for production diesel engines at rated load.
  4. Verify γ = 1.40 for air at moderate temperatures; combustion products and residual exhaust gas lower γ slightly, reducing peak pressure estimates by 3–5%.