Diesel Cycle Calculator
Compute the theoretical thermal efficiency of a Diesel engine cycle from its compression ratio and cut-off ratio. Use this to compare engine designs or understand how operating parameters limit maximum efficiency.
About this calculator
The Diesel cycle models the ideal thermodynamic operation of a compression-ignition engine. Air is compressed isentropically (process 1→2), fuel burns at constant pressure (2→3), the gas expands isentropically (3→4), and heat is rejected at constant volume (4→1). The thermal efficiency is: η = [1 − (1/r_c^(γ−1)) × (r_co^γ − 1)/(γ × (r_co − 1))] × 100%, where r_c is the compression ratio, r_co is the cut-off ratio (V₃/V₂, the ratio of volume after to before combustion), and γ = 1.4 for air. Higher compression ratios always improve efficiency, while a larger cut-off ratio (more fuel burned) reduces efficiency relative to the Otto cycle. Real diesel engines achieve compression ratios of 14–22 and cut-off ratios of 1.5–4, yielding indicated thermal efficiencies of 40–55%.
How to use
Suppose a diesel engine has a compression ratio r_c = 18 and a cut-off ratio r_co = 2. Step 1 — Compute r_c^(γ−1): 18^0.4 ≈ 3.107. Step 2 — Compute r_co^γ: 2^1.4 ≈ 2.639. Step 3 — Evaluate the cut-off bracket: (2.639 − 1)/(1.4 × (2 − 1)) = 1.639/1.4 = 1.171. Step 4 — Apply formula: η = [1 − (1/3.107) × 1.171] × 100 = [1 − 0.3219 × 1.171] × 100 = [1 − 0.3769] × 100 ≈ 62.3%. This is the ideal cycle efficiency; real engine brake thermal efficiency will be lower due to friction and heat losses.
Frequently asked questions
Why is the Diesel cycle more efficient than the Otto cycle at the same compression ratio?
At the same compression ratio, the Otto cycle (spark-ignition, constant-volume heat addition) actually has higher theoretical efficiency than the Diesel cycle, because the Diesel cycle adds heat at constant pressure over a finite volume expansion, lowering the average temperature at which heat is added. However, diesel engines operate at much higher compression ratios (14–22) than gasoline engines (8–12) because there is no fuel-air mixture present during compression to auto-ignite prematurely. This higher compression ratio more than compensates, giving diesel engines a real-world efficiency advantage over comparable gasoline engines.
What does the cut-off ratio represent in a Diesel cycle?
The cut-off ratio r_co = V₃/V₂ is the ratio of the cylinder volume at the end of combustion (constant-pressure heat addition) to the volume at the start of combustion (top dead center). It is directly related to how much fuel is injected: a larger cut-off ratio means more fuel burned per cycle and greater power output, but it also lowers thermal efficiency. Engineers select the cut-off ratio to balance power demand against efficiency targets. At light load (small fuel injection), the cut-off ratio approaches 1, and Diesel cycle efficiency approaches that of the Otto cycle for the same compression ratio.
How does compression ratio affect Diesel cycle efficiency in practice?
Increasing compression ratio raises the temperature and pressure at the end of compression, allowing more complete combustion and a larger thermodynamic temperature ratio between the hot and cold reservoirs, which directly improves Carnot-like efficiency. In the Diesel efficiency formula, r_c appears as r_c^(γ−1) in the denominator of the loss term — a larger r_c shrinks this term and pushes efficiency toward 100% theoretically. Practical limits include mechanical stress on pistons and bearings, increased NOₓ emissions due to higher peak temperatures, and the need for stronger (heavier) engine structures. Modern truck diesel engines balance these factors at compression ratios around 16–18.