---
title: "Diesel cycle efficiency compared with Otto"
subject: "Internal Combustion Engines"
chapter: 1
key: "semester_6/internal_combustion_engines:diesel-cycle-versus-otto-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/diesel-cycle-versus-otto-numerical
site: BCT Engineering
---

# Diesel cycle efficiency compared with Otto

Internal Combustion Engines · Chapter 1: Overview of thermodynamics of fuel-air cycles and real cycles

- Asked in: year not given, 6 marks
- Video: [Summarize and compare Otto and Diesel Cycles](https://www.youtube.com/watch?v=jED9LbvSCUA) — Randall Manteufel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/diesel-cycle-versus-otto-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-1

## Question

A Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2.2. Air at the start of compression is at 100 kPa and 300 K. Using $c_p = 1.005$ kJ/kg K, $c_v = 0.718$ kJ/kg K and $\gamma = 1.4$, find the temperatures at the end of each process, the heat added and rejected per kg, the thermal efficiency and the mean effective pressure. Compare the efficiency with that of an Otto cycle of the same compression ratio and explain the result.

## Short answer

T2=953 K, T3=2097 K, T4=905 K; heat added 1150 kJ/kg, rejected 434 kJ/kg; efficiency 62.2%; mep 880 kPa. At the same compression ratio the Otto cycle is more efficient (68.5%), but practical CI engines use higher ratios and outperform SI engines.

## Answer

The Diesel cycle has isentropic compression (1-2), constant-pressure heat addition (2-3), isentropic expansion (3-4) and constant-volume heat rejection (4-1).

```text
 p
 |      2________3
 |     /         \
 |    /           \
 |   /             \
 |  /               4
 | /              _/
 |/____1________/____ v
```

#### Temperatures
$$
\begin{aligned}
T_2 &= T_1 r^{\gamma-1} = 300 \times 18^{0.4} = 953.3\ \text{K} \\
T_3 &= \rho\, T_2 = 2.2 \times 953.3 = 2097.3\ \text{K} \\
T_4 &= T_3 \left(\frac{\rho}{r}\right)^{\gamma-1} = 2097.3 \times \left(\frac{2.2}{18}\right)^{0.4} = 904.7\ \text{K}
\end{aligned}
$$

#### Heat and efficiency
$$
\begin{aligned}
Q_{in} &= c_p (T_3 - T_2) = 1.005(2097.3 - 953.3) = 1149.7\ \text{kJ/kg} \\
Q_{out} &= c_v (T_4 - T_1) = 0.718(904.7 - 300) = 434.2\ \text{kJ/kg} \\
W_{net} &= 1149.7 - 434.2 = 715.5\ \text{kJ/kg} \\
\eta &= \frac{715.5}{1149.7} = 0.622
\end{aligned}
$$

Check with the formula:

$$
\eta = 1 - \frac{1}{18^{0.4}} \times \frac{2.2^{1.4} - 1}{1.4 \times 1.2} = 1 - \frac{1}{3.178} \times \frac{2.0157}{1.68} = 0.622
$$

#### Mean effective pressure
$$
\begin{aligned}
v_1 &= \frac{0.287 \times 300}{100} = 0.861\ \text{m}^3/\text{kg}, \quad v_2 = \frac{0.861}{18} = 0.0478 \\
\text{mep} &= \frac{715.5}{0.861 - 0.0478} = 880\ \text{kPa}
\end{aligned}
$$

#### Comparison with Otto
For $r = 18$: $\eta_{Otto} = 1 - 18^{-0.4} = 0.685$, which is higher than the Diesel value of 0.622.

| Point | Otto | Diesel |
|---|---|---|
| Heat addition | constant volume | constant pressure |
| Efficiency at same $r$ | 68.5% | 62.2% |
| Reason | all heat added at highest temperature | part of heat added while piston moves down |

At the same compression ratio the Otto cycle is more efficient. In practice an SI engine cannot use $r = 18$ because of knock, while a CI engine compresses only air and can. So real Diesel engines (r = 16-22) are more efficient than petrol engines (r = 8-10).

**Answer:** $T_2 = 953$ K, $T_3 = 2097$ K, $T_4 = 905$ K; $Q_{in} = 1150$ kJ/kg, $Q_{out} = 434$ kJ/kg; $\eta = 62.2\%$; mep = 880 kPa.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is Q_in for the Diesel cycle calculated with c_p?
   A. Heat is added at constant pressure
   B. Heat is added at constant volume
   C. Compression is isothermal
   D. Heat is rejected at constant pressure
   Answer: A. Process 2-3 is at constant pressure, so Q_in = c_p(T3 - T2); rejection at constant volume uses c_v. (Key point: Diesel heat addition)

2. How is T3 related to T2 in the Diesel cycle?
   A. T3 = T2 x r^(gamma-1)
   B. T3 = cut-off ratio x T2
   C. T3 = T2 / cut-off ratio
   D. T3 = T2 + Q_in/c_v
   Answer: B. At constant pressure T3/T2 = v3/v2 = rho, so T3 = 2.2 x 953.3 = 2097 K. (Key point: Temperature after cut-off)

3. For the same compression ratio, how do the efficiencies compare, and why?
   A. Diesel is higher, because heat is added at constant pressure
   B. They are equal because r is the same
   C. Otto is higher, because all its heat is added at the highest temperature near TDC
   D. Diesel is higher because it has a higher peak pressure
   Answer: C. Otto gives 68.5% against 62.2% for Diesel at r = 18, since part of the Diesel heat is added while the piston moves down. (Key point: Diesel vs Otto at same r)

4. Why are real Diesel engines still more efficient than petrol engines?
   A. The Diesel cycle is ideally better than Otto at equal r
   B. Diesel fuel has a lower calorific value
   C. They have no heat loss
   D. They compress only air, so they can use a much higher compression ratio without knock
   Answer: D. SI engines are limited to r of 8-10 by knock; CI engines use r of 16-22, which outweighs the cycle disadvantage. (Key point: Why real Diesel engines win)

## Related in Internal Combustion Engines

- [Otto cycle efficiency and mean effective pressure](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Aotto-cycle-efficiency-numerical) — Internal Combustion Engines, chapter 1
- [Efficiency ratio and specific fuel consumption](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Arelative-efficiency-sfc-numerical) — Internal Combustion Engines, chapter 8
- [Heat balance sheet of a diesel engine](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Adiesel-engine-heat-balance-numerical) — Internal Combustion Engines, chapter 8
- [Volumetric efficiency from orifice air measurement](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Avolumetric-efficiency-orifice-air-consumption) — Internal Combustion Engines, chapter 8

## Related in other subjects

- [Diesel cycle: cut-off percentage, expansion temperature, efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-1500c-cr16-cutoff-percentage) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Diesel cycle with compression ratio 15:1](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adiesel-cycle-compression-15) — Power Plant Equipment, chapter 5
- [Otto cycle CR 8: state points and efficiency](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aotto-cycle-cr8-2000kj-heat) — Fundamental of Thermodynamics & Heat Transfer, chapter 6
- [Diesel cycle: cut off 8% of stroke, CR 16 to 20](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Adiesel-cr16-cutoff-8-percent) — Fundamental of Thermodynamics & Heat Transfer, chapter 6