Ideal Air-Standard Dual Cycle: 04-BS-10, Undated paper, Q4

Question 4: Ideal Air-Standard Dual Cycle (Part B – 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, $k=1.4$; ideal, no losses). 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 ($2\to3$) and $\tfrac13$ at constant pressure ($3\to4$).

Find. (a) $T_3$, $T_4$; (b) $w_{net}$; (c) $\eta_{th}$; (d) MEP.

Approach

Walk the five states of the dual cycle in order — isentropic compression $1\to2$, constant-volume heat addition $2\to3$, constant-pressure heat addition $3\to4$, isentropic expansion $4\to5$ back to $V_1$, constant-volume heat rejection $5\to1$ — using cold-air-standard relations throughout.

Isentropic compression, state 2. $$T_2=T_1\,r^{\,k-1}=300\times9^{0.4}=722.47\ \text{K}.$$
Constant-volume heat addition — part (a), $T_3$. $q_{2\to3}=\tfrac23\times1400=933.33$ kJ/kg. $$T_3=T_2+\frac{q_{2\to3}}{c_v}=722.47+\frac{933.33}{0.718}=722.47+1299.90=\boxed{2022.4\ \text{K}}.$$
Constant-pressure heat addition — part (a), $T_4$. $q_{3\to4}=\tfrac13\times1400=466.67$ kJ/kg. $$T_4=T_3+\frac{q_{3\to4}}{c_p}=2022.4+\frac{466.67}{1.005}=2022.4+464.35=\boxed{2486.7\ \text{K}}.$$ Cutoff ratio $r_c=T_4/T_3=1.2296$.
Isentropic expansion, state 5. Expansion volume ratio $V_5/V_4=r/r_c=7.318$: $$T_5=T_4\left(\frac{r_c}{r}\right)^{k-1}=2486.7\times\left(\frac{1.2296}{9}\right)^{0.4}=1121.6\ \text{K}.$$
Heat rejected, net work, thermal efficiency — parts (b), (c). $$q_{out}=c_v(T_5-T_1)=0.718\times(1121.6-300)=589.9\ \text{kJ/kg}.$$ $$w_{net}=q_{in}-q_{out}=1400-589.9=\boxed{810.1\ \text{kJ/kg}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{810.1}{1400}=\boxed{0.5786\ (57.9\%)}.$$
Mean effective pressure — part (d). $$v_1=\frac{RT_1}{P_1}=\frac{0.287\times300}{100}=0.8610\ \text{m}^3/\text{kg},\quad v_2=v_1/r=0.09567\ \text{m}^3/\text{kg}.$$ $$MEP=\frac{w_{net}}{v_1-v_2}=\frac{810.1}{0.8610-0.09567}=\boxed{1058.5\ \text{kPa}}.$$

Quantity	Value
(a) $T_3$ (end of const.-vol. heat addition)	2022.4 K
(a) $T_4$ (end of const.-press. heat addition)	2486.7 K
(b) $w_{net}$	810.1 kJ/kg
(c) $\eta_{th}$	57.86%
(d) MEP	1058.5 kPa

Topic. Air-standard dual (mixed Otto–Diesel) cycle.

Key relations. Isentropic: $T_2/T_1=r^{k-1}$; constant volume: $T_3=T_2+q_v/c_v$; constant pressure: $T_4=T_3+q_p/c_p$, $r_c=T_4/T_3$; isentropic expansion: $T_5=T_4(r_c/r)^{k-1}$; $MEP=w_{net}/(v_1-v_2)$.

Why this works. The dual cycle splits heat addition across a constant-volume leg (models the very fast initial burn) and a constant-pressure leg (models the slower burn as the piston starts to move) — it sits between the idealized Otto cycle (100% constant-volume) and Diesel cycle (100% constant-pressure), and collapses to either as the split fraction goes to 0 or 1.

Common pitfall. Applying the isentropic-expansion volume ratio as $r$ (the overall compression ratio) instead of the correct, smaller $r/r_c$ — the piston has already moved from $V_2$ to $V_4=r_c V_2$ during the constant-pressure leg, so only $V_5/V_4=r/r_c$ remains for the power stroke.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §9-6 (dual cycle).

Question 4 of 9: Ideal Air-Standard Dual Cycle

Question 4: Ideal Air-Standard Dual Cycle (Part B – 15 marks)

Approach