Question 5 of 9: Air-Standard Diesel Cycle from Given End-State Pressure/Temperature

Question 5: Air-Standard Diesel Cycle from Given End-State Pressure/Temperature (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Air-standard Diesel cycle, variable specific heats. State 1 (start of compression): $P_1=95$ kPa, $T_1=300$ K. State 3 (end of heat addition, constant-pressure leg): $P_3=7.2$ MPa, $T_3=2150$ K. Diesel topology: $1\to2$ isentropic compression, $2\to3$ constant-pressure heat addition ($P_2=P_3$), $3\to4$ isentropic expansion to $v_4=v_1$, $4\to1$ constant-volume heat rejection.

Find. (a) $r=v_1/v_2$; (b) $r_c=v_3/v_2$; (c) $\eta_{th}$; (d) MEP [kPa].

Approach

Since state 2 shares state 3's pressure (constant-pressure heat addition), $P_2=7.2$ MPa is known immediately; find $T_2$ from the isentropic relation $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$ by root-finding. The ideal-gas law then gives $v_1$, $v_2$, $v_3$ directly, and hence $r$ and $r_c$. Close the cycle with $v_4=v_1$ and solve the expansion leg's isentropic condition (now expressed via the known volume ratio $v_3/v_4$) for $T_4$, then form the energy balance.

1. Isentropic compression $1\to2$ (part a, en route). $P_2=P_3=7.2$ MPa (constant-$P$ heat addition fixes this). Solve $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)=0.287\ln(7200/95)=1.2421$ kJ/kg·K for $T_2$: $$T_2=\boxed{978.73\ \text{K}}\ (705.58\ ^\circ\text{C}),\qquad h_2=1148.29\ \text{kJ/kg}.$$ $$v_1=\frac{RT_1}{P_1}=\frac{0.287\times300}{95}=0.90632\ \text{m}^3/\text{kg},\qquad v_2=\frac{RT_2}{P_2}=\frac{0.287\times978.73}{7200}=0.03901\ \text{m}^3/\text{kg}.$$ $$r=\frac{v_1}{v_2}=\frac{0.90632}{0.03901}=\boxed{23.23}.$$
2. Cutoff ratio (part b). At constant $P$, $v_3/v_2=T_3/T_2$: $$v_3=\frac{RT_3}{P_3}=\frac{0.287\times2150}{7200}=0.08570\ \text{m}^3/\text{kg},\qquad r_c=\frac{v_3}{v_2}=\frac{0.08570}{0.03901}=\boxed{2.197}.$$
3. Isentropic expansion $3\to4$, closing the cycle at $v_4=v_1$. Solve $s^\circ(T_4)-s^\circ(T_3)=R\ln\!\big[(T_4/T_3)(v_3/v_4)\big]$ for $T_4$ (root-find, since $T_4$ appears on both sides): $$T_4=\boxed{1030.49\ \text{K}}\ (757.34\ ^\circ\text{C}),\qquad h_4=1207.39\ \text{kJ/kg}\quad (P_4=RT_4/v_4=326.3\ \text{kPa}).$$
4. Thermal efficiency (part c). $q_{in}=h_3-h_2=2566.67-1148.29=1418.38$ kJ/kg (constant-$P$ leg); $u_1=h_1-RT_1=426.30-86.10=340.20$ kJ/kg, $u_4=h_4-RT_4=1207.39-295.75=911.64$ kJ/kg; $q_{out}=u_4-u_1=571.45$ kJ/kg (constant-$v$ leg). $$w_{net}=q_{in}-q_{out}=1418.38-571.45=846.93\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{846.93}{1418.38}=\boxed{0.5971\ (59.7\%)}.$$
5. Mean effective pressure (part d). $$\text{MEP}=\frac{w_{net}}{v_1-v_2}=\frac{846.93}{0.90632-0.03901}=\boxed{976.5\ \text{kPa}}.$$