---
title: "Otto cycle efficiency and mean effective pressure"
subject: "Internal Combustion Engines"
chapter: 1
key: "semester_6/internal_combustion_engines:otto-cycle-efficiency-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/otto-cycle-efficiency-numerical
site: BCT Engineering
---

# Otto cycle efficiency and mean effective pressure

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

- Asked in: year not given, 4+4 marks
- Video: [Otto Cycle Efficiency Derivation, Thermodynamics, Thermal Engineering](https://www.youtube.com/watch?v=bRrRPDzHLnI) — Reference Book
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/otto-cycle-efficiency-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-1

## Question

Derive an expression for the air-standard efficiency of the Otto cycle in terms of the compression ratio. An Otto cycle works with a compression ratio of 8. At the start of compression the air is at 100 kPa and 300 K, and 1500 kJ/kg of heat is added at constant volume. Taking $c_v = 0.718$ kJ/kg K, $\gamma = 1.4$ and $R = 0.287$ kJ/kg K, find the maximum temperature and pressure, the net work per kg and the mean effective pressure.

## Short answer

Isentropic compression gives T2=689.2 K, p2=1838 kPa. Constant-volume heat addition yields Tmax=2778 K, pmax=7410 kPa. Efficiency is 56.5%, net work 847 kJ/kg, and mep is 1124 kPa (11.24 bar).

## Answer

#### Derivation of Otto cycle efficiency
The Otto cycle is the ideal air-standard cycle of a spark-ignition engine. It has four processes: 1-2 isentropic compression, 2-3 constant-volume heat addition, 3-4 isentropic expansion and 4-1 constant-volume heat rejection.

```text
 p
 |        3
 |       /|\
 |      / | \         1-2 compression
 |     /  |  \        2-3 heat in (v = c)
 |    2   |   \       3-4 expansion
 |   /    |    \      4-1 heat out (v = c)
 |  /      \    4
 | /        \__/
 |/_______1__/______ v
     v2        v1
```

For 1 kg of air:

$$
\begin{aligned}
Q_{in} &= c_v (T_3 - T_2) \\
Q_{out} &= c_v (T_4 - T_1) \\
\eta &= 1 - \frac{Q_{out}}{Q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}
\end{aligned}
$$

For the isentropic processes, with $r = v_1/v_2 = v_4/v_3$:

$$
\frac{T_2}{T_1} = r^{\gamma-1}, \qquad \frac{T_3}{T_4} = r^{\gamma-1}
$$

So $T_3/T_2 = T_4/T_1$, which gives $(T_4 - T_1)/(T_3 - T_2) = T_1/T_2$. Therefore

$$
\eta_{Otto} = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{r^{\gamma-1}}
$$

The efficiency depends only on $r$ and $\gamma$. It rises as $r$ increases.

#### Numerical
$$
\begin{aligned}
T_2 &= 300 \times 8^{0.4} = 689.2\ \text{K} \\
p_2 &= 100 \times 8^{1.4} = 1837.9\ \text{kPa} \\
T_3 &= T_2 + \frac{Q_{in}}{c_v} = 689.2 + \frac{1500}{0.718} = 2778.4\ \text{K} \\
p_3 &= p_2 \frac{T_3}{T_2} = 1837.9 \times \frac{2778.4}{689.2} = 7409\ \text{kPa} \\
\eta &= 1 - 8^{-0.4} = 0.5647 \\
W_{net} &= \eta\, Q_{in} = 0.5647 \times 1500 = 847.1\ \text{kJ/kg}
\end{aligned}
$$

Check: $T_4 = T_3/8^{0.4} = 1209.3$ K, so $Q_{out} = 0.718(1209.3 - 300) = 652.9$ kJ/kg and $1500 - 652.9 = 847.1$ kJ/kg.

Specific volumes:

$$
\begin{aligned}
v_1 &= \frac{RT_1}{p_1} = \frac{0.287 \times 300}{100} = 0.861\ \text{m}^3/\text{kg} \\
v_2 &= v_1/8 = 0.1076\ \text{m}^3/\text{kg} \\
\text{mep} &= \frac{W_{net}}{v_1 - v_2} = \frac{847.1}{0.861 - 0.1076} = 1124\ \text{kPa}
\end{aligned}
$$

**Answer:** $T_{max} = 2778$ K, $p_{max} = 7.41$ MPa, $W_{net} = 847$ kJ/kg, mep = 11.24 bar (efficiency 56.5%).

## Check yourself

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

1. The air-standard Otto cycle efficiency depends on which quantities?
   A. Compression ratio and gamma only
   B. Compression ratio and maximum temperature
   C. Heat added and compression ratio
   D. Cut-off ratio and compression ratio
   Answer: A. eta = 1 - 1/r^(gamma-1), so only r and gamma matter; the heat added does not change it. (Key point: Otto efficiency formula)

2. Which relation lets (T4 - T1)/(T3 - T2) be replaced by T1/T2 in the derivation?
   A. p2 = p3 because heat is added at constant pressure
   B. T3/T2 = T4/T1, from both isentropic processes having the same volume ratio
   C. T2 = T4 because both are at the same volume
   D. Q_in = Q_out for the cycle
   Answer: B. Since T2/T1 = T3/T4 = r^(gamma-1), it follows that T3/T2 = T4/T1, which gives the ratio T1/T2. (Key point: Key step in derivation)

3. How is the maximum temperature T3 found in the numerical?
   A. T3 = T2 + Q_in/c_p, because heat is added at constant pressure
   B. T3 = T1 x r^(gamma-1)
   C. T3 = T2 + Q_in/c_v, because heat is added at constant volume
   D. T3 = T4 x r
   Answer: C. Constant-volume heat addition gives Q_in = c_v(T3 - T2), so T3 = 689.2 + 1500/0.718 = 2778 K. (Key point: Maximum temperature)

4. How is the mean effective pressure of the cycle obtained?
   A. Net work divided by v1
   B. Heat added divided by v1 - v2
   C. Maximum pressure divided by compression ratio
   D. Net work divided by swept volume (v1 - v2)
   Answer: D. mep = W_net/(v1 - v2) = 847.1/(0.861 - 0.1076) = 1124 kPa. (Key point: Mean effective pressure)

## Related in Internal Combustion Engines

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- [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
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