Compression-Ignition (Dual) Cycle: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014, Q3

Question 3: Compression-Ignition (Dual) Cycle (equal value)

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. Dual (mixed) air-standard cycle: state 1 at $P_1=100\ \text{kPa}$, $T_1=50\ ^\circ\text{C}=323.15\ \text{K}$; compression ratio $r=V_1/V_2=13$; maximum pressure $P_3=P_4=4500\ \text{kPa}$; heat split $Q_v:Q_p=2:1$; $\gamma=1.4$, $c_v=0.718\ \text{kJ/kg}\cdot\text{K}$.

Find. The temperatures $T_1$–$T_5$ and the thermal efficiency.

Figure 2 — p–V diagram of the dual (mixed) cycle: 1→2 isentropic compression, 2→3 constant-volume heat addition to the maximum pressure, 3→4 constant-pressure heat addition, 4→5 isentropic expansion, 5→1 constant-volume heat rejection.

Approach. Fix state 2 from the isentropic compression, state 3 from the constant-volume rise to the maximum pressure, split the heat 2:1 to locate state 4, expand isentropically to state 5, then form the efficiency from the heat added and rejected.

Isentropic compression 1→2. $$T_2=T_1r^{\gamma-1}=323.15(13)^{0.4}=\boxed{901\ \text{K}},\qquad P_2=P_1r^{\gamma}=100(13)^{1.4}=3627\ \text{kPa}.$$
Constant-volume heat addition 2→3. At fixed volume the temperature scales with pressure to the maximum 4500 kPa:$$T_3=T_2\frac{P_3}{P_2}=901\frac{4500}{3627}=\boxed{1119\ \text{K}}.$$
Constant-pressure heat addition 3→4 from the 2:1 heat split. $Q_v=c_v(T_3-T_2)$ is twice $Q_p=c_p(T_4-T_3)$, so $c_v(T_3-T_2)=2c_p(T_4-T_3)$, i.e. $(T_3-T_2)=2\gamma(T_4-T_3)$:$$T_4=T_3+\frac{T_3-T_2}{2\gamma}=1119+\frac{217}{2.8}=\boxed{1196\ \text{K}}.$$ The cut-off ratio is $\rho_c=V_4/V_3=T_4/T_3=1.069$.
Isentropic expansion 4→5. Since $V_5=V_1$ and $V_3=V_2$, the expansion ratio is $V_5/V_4=r/\rho_c=13/1.069=12.16$:$$T_5=T_4\!\left(\frac{V_4}{V_5}\right)^{\gamma-1}=1196(12.16)^{-0.4}=\boxed{440\ \text{K}}.$$
Thermal efficiency. Heat added $Q_\text{in}=c_v(T_3-T_2)+c_p(T_4-T_3)=c_v(217+108.5)=325.5\,c_v$; heat rejected $Q_\text{out}=c_v(T_5-T_1)=117.2\,c_v$. Hence$$\eta=1-\frac{Q_\text{out}}{Q_\text{in}}=1-\frac{117.2}{325.5}=\boxed{0.640\ (64.0\%)}.$$

Point	Temperature	Pressure
1	323 K (50 °C)	100 kPa
2	901 K	3627 kPa
3	1119 K	4500 kPa
4	1196 K	4500 kPa
5	440 K	136 kPa
Thermal efficiency	64.0 %

Concept & Theory

Topic. Air-standard dual (mixed / Sabathé) cycle, the model for a modern compression-ignition engine.

Key relations. Isentropic $Tv^{\gamma-1}=$ const, $Pv^{\gamma}=$ const; constant-volume $T\propto P$; constant-pressure $T\propto V$; $\eta=1-Q_\text{out}/Q_\text{in}$ with $Q_v=c_v\Delta T$, $Q_p=c_p\Delta T$.

Why this works. The dual cycle adds heat partly at constant volume (rapid pressure rise near top dead centre) and partly at constant pressure (the remaining combustion as the piston begins to descend), which mirrors real diesel combustion better than the pure Diesel or Otto idealisations. Fixing the maximum pressure ties state 3 to state 2, and the prescribed 2:1 heat split then fixes the cut-off at state 4.

Common pitfall. Using $c_v$ (instead of $c_p$) for the constant-pressure heat, or forgetting that the isentropic expansion runs over $r/\rho_c$, not the full compression ratio $r$.

Source. Çengel & Boles, Thermodynamics §9 (gas power cycles — Otto, Diesel and dual cycles).

Question 3 of 8: Compression-Ignition (Dual) Cycle

Question 3: Compression-Ignition (Dual) Cycle (equal value)