Question 4 of 9: Air-Standard Dual Cycle with Non-Ideal Compression/Expansion

Question 4: Air-Standard Dual Cycle with Non-Ideal Compression/Expansion (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 dual (mixed Otto–Diesel) cycle, cold-air-standard properties ($c_v=0.718$, $c_p=1.005$ kJ/kg·K, $R=0.287$ kJ/kg·K, $k=1.4$). Compression ratio $r=9$; $P_1=100$ kPa, $T_1=300$ K; total heat addition $q_{in}=1400$ kJ/kg, split $\tfrac23$ at constant volume and $\tfrac13$ at constant pressure; compression isentropic efficiency $\eta_{comp}=85\%$; expansion isentropic efficiency $\eta_{exp}=90\%$.

Find. (a) net work per unit mass; (b) thermal efficiency; (c) mean effective pressure (MEP).

Approach

Because the compression and expansion strokes are each non-ideal (given isentropic efficiencies), fix the real state after each stroke from its ideal (isentropic) counterpart divided/multiplied by the stated efficiency, then apply the const-$V$/const-$P$ heat-addition split directly to those real states; net work follows from an overall energy balance ($w_{net}=q_{in}-q_{out}$) and is cross-checked by summing the actual expansion, compression and constant-pressure boundary work terms directly.

1. Compression 1→2, actual. Ideal (isentropic) exit temperature $T_{2s}=T_1 r^{k-1}=300(9)^{0.4}=722.5$ K, ideal work $w_s=c_v(T_{2s}-T_1)=303.33$ kJ/kg. With $\eta_{comp}=0.85$ the actual (larger) work drives the actual, hotter state 2: $$w_{comp}=\frac{w_s}{\eta_{comp}}=\frac{303.33}{0.85}=\boxed{356.86\ \text{kJ/kg}}\quad\Rightarrow\quad T_2=T_1+\frac{w_{comp}}{c_v}=797.0\ \text{K}\ (P_2=2391.1\ \text{kPa}).$$
2. Constant-volume heat addition 2→3. $q_v=\tfrac23(1400)=933.33$ kJ/kg: $$T_3=T_2+\frac{q_v}{c_v}=797.0+\frac{933.33}{0.718}=2096.9\ \text{K}\quad(P_3=P_2\,T_3/T_2=6290.8\ \text{kPa}).$$
3. Constant-pressure heat addition 3→4. $q_p=\tfrac13(1400)=466.67$ kJ/kg: $$T_4=T_3+\frac{q_p}{c_p}=2096.9+\frac{466.67}{1.005}=2561.3\ \text{K},\qquad \frac{V_4}{V_3}=\frac{T_4}{T_3}=1.2214.$$
4. Expansion 4→5, actual, $V_5=V_1$. Ideal isentropic exit temperature with expansion ratio $V_4/V_5=(V_4/V_3)/r=0.1357$: $$T_{5s}=T_4\left(\frac{V_4}{V_5}\right)^{k-1}=1152.1\ \text{K},\qquad w_s=c_v(T_4-T_{5s})=1011.75\ \text{kJ/kg}.$$ With $\eta_{exp}=0.90$: $$w_{exp}=\eta_{exp}\,w_s=\boxed{910.58\ \text{kJ/kg}}\quad\Rightarrow\quad T_5=T_4-\frac{w_{exp}}{c_v}=1293.1\ \text{K}.$$
5. Heat rejected, net work and thermal efficiency (parts a, b). Heat is rejected at constant volume, 5→1: $$q_{out}=c_v(T_5-T_1)=713.02\ \text{kJ/kg}\quad\Rightarrow\quad w_{net}=q_{in}-q_{out}=1400-713.02=\boxed{686.98\ \text{kJ/kg}}.$$ As a check, summing the actual boundary-work terms directly ($w_{exp}$ out, $w_{comp}$ in, plus the constant-pressure work $R(T_4-T_3)$) gives the identical 686.98 kJ/kg. $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{686.98}{1400}=\boxed{49.07\%}.$$
6. Mean effective pressure (part c). $v_1=RT_1/P_1=0.86100$ m³/kg, $v_2=v_1/r=0.09567$ m³/kg: $$\text{MEP}=\frac{w_{net}}{v_1-v_2}=\frac{686.98}{0.86100-0.09567}=\boxed{897.6\ \text{kPa}}.$$