Diesel cycle efficiency compared with Otto

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).

 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.

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.