Otto cycle efficiency and mean effective pressure

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.

 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%).