Diesel cycle with compression ratio 15:1

Assumptions: the engine works on the air-standard Diesel cycle (isentropic compression 1–2, constant-pressure heat addition 2–3, isentropic expansion 3–4, constant-volume heat rejection 4–1). The heat added is taken as 1675 kJ/kg (1675 MJ/kg is clearly a misprint), $R = 0.287$ kJ/kg K, $c_p = 1.005$, $c_v = 0.718$ kJ/kg K, $\gamma = 1.4$.

Data: $r = 15$, $p_1 = 0.1$ MPa, $T_1 = 40 + 273 = 313$ K.

(i) Maximum temperature and pressure

Process 1–2 (isentropic compression):

$$\begin{aligned} T_2 &= T_1 r^{\gamma-1} = 313 \times 15^{0.4} = 924.66\ \text{K} \\ p_2 &= p_1 r^{\gamma} = 0.1 \times 15^{1.4} = 4.431\ \text{MPa} \end{aligned}$$

Process 2–3 (heat added at constant pressure):

$$T_3 = T_2 + \frac{q_s}{c_p} = 924.66 + \frac{1675}{1.005} = 2591.32\ \text{K}$$

Maximum temperature $T_3 = 2591.3$ K; maximum pressure $p_3 = p_2 = 4.431$ MPa.

(ii) Cut-off ratio

$$\rho = \frac{v_3}{v_2} = \frac{T_3}{T_2} = \frac{2591.32}{924.66} = 2.802$$

(iii) Net work per kg of air

$$\begin{aligned} T_4 &= T_3\left(\frac{\rho}{r}\right)^{\gamma-1} = 2591.32\left(\frac{2.802}{15}\right)^{0.4} = 1324.65\ \text{K} \\ q_r &= c_v (T_4 - T_1) = 0.718(1324.65 - 313) = 726.37\ \text{kJ/kg} \\ W_{net} &= q_s - q_r = 1675 - 726.37 = 948.63\ \text{kJ/kg} \end{aligned}$$

(iv) Thermal efficiency

$$\eta = \frac{W_{net}}{q_s} = \frac{948.63}{1675} = 0.5664 = 56.64\%$$

Check with the formula $\eta = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{\rho^{\gamma}-1}{\gamma(\rho-1)} = 56.64\%$.

(v) Mean effective pressure

$$\begin{aligned} v_1 &= \frac{R T_1}{p_1} = \frac{0.287 \times 313}{100} = 0.8983\ \text{m}^3/\text{kg} \\ v_2 &= \frac{v_1}{r} = 0.05989\ \text{m}^3/\text{kg} \\ p_m &= \frac{W_{net}}{v_1 - v_2} = \frac{948.63}{0.8983 - 0.05989} = 1131.5\ \text{kPa} \end{aligned}$$

Answer: $T_{max}$ = 2591.3 K, $p_{max}$ = 4.431 MPa, cut-off ratio = 2.802, $W_{net}$ = 948.6 kJ/kg, $\eta$ = 56.64 %, MEP = 1131.5 kPa ≈ 11.31 bar.