Non-Ideal Air-Standard Diesel Cycle: 04-BS-10, December 2019, Q7

Question 7: Non-Ideal Air-Standard Diesel Cycle (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.

Check — reconstructed missing values. The compression ratio (15), $p_1=100$ kPa, $T_1=300$ K, and the expansion efficiency (90%) are all legible/confirmed. Representative round-number values of $T_{max}=2200$ K and $\eta_c=95\%$ are adopted (paired with the given $\eta_e=90\%$) so the question can still be solved and checked in full; the method below applies unchanged to any other $T_{max}$/$\eta_c$ pair.

Given. $CR=v_1/v_2=15$, $P_1=100$ kPa, $T_1=300$ K, $T_{max}=T_3=2200$ K (assumed), $\eta_c=0.95$ (assumed), $\eta_e=0.90$ (given). Cold-air-standard properties: $c_p=1.005$, $c_v=0.718$ kJ/kg·K, $k=1.4$.

Find. (a) $w_{net}$ [kJ/kg]; (b) $\eta_{th}$.

Fig. Q7 — $p$–$v$ diagram: 1→2 actual (inefficient) adiabatic compression, 2→3 constant-$P$ heat addition to $T_{max}$, 3→4 actual (inefficient) adiabatic expansion back to $v_1$, 4→1 constant-volume heat rejection.

Approach

Get the isentropic compression temperature from the compression ratio, then apply the compression efficiency to find the ACTUAL state 2 (same volume, more work, higher $T$). Heat addition at constant pressure to $T_{max}$ fixes the cutoff ratio. The isentropic expansion temperature follows from the volume ratio $v_4/v_3=CR/r_c$; applying the expansion efficiency gives the actual state 4. Net work MUST be taken as $q_{in}-q_{out}$ (full-cycle energy balance) rather than the difference of the two adiabatic-stroke works alone, since that difference omits the boundary work done during the constant-pressure heat-addition stroke.

Actual compression (1→2). $$T_{2s}=T_1\,CR^{k-1}=300\times15^{0.4}=886.3\ \text{K},\qquad w_{c,a}=\frac{c_v(T_{2s}-T_1)}{\eta_c}=\frac{0.718(886.3-300)}{0.95}=443.08\ \text{kJ/kg}$$ $$T_2=T_1+\frac{w_{c,a}}{c_v}=300+\frac{443.08}{0.718}=\boxed{917.1\ \text{K}}.$$
Heat addition and cutoff ratio (2→3). $$r_c=\frac{T_3}{T_2}=\frac{2200}{917.1}=\boxed{2.399},\qquad q_{in}=c_p(T_3-T_2)=1.005(2200-917.1)=\boxed{1289.31\ \text{kJ/kg}}.$$
Actual expansion (3→4). $v_4/v_3=CR/r_c=15/2.399=6.253$: $$T_{4s}=\frac{T_3}{(v_4/v_3)^{k-1}}=\frac{2200}{6.253^{0.4}}=1056.7\ \text{K},\qquad w_{e,a}=\eta_e\,c_v(T_3-T_{4s})=0.90\times0.718(2200-1056.7)=738.75\ \text{kJ/kg}$$ $$T_4=T_3-\frac{w_{e,a}}{c_v}=2200-\frac{738.75}{0.718}=\boxed{1171.1\ \text{K}}.$$
Heat rejected (4→1, constant volume). $$q_{out}=c_v(T_4-T_1)=0.718(1171.1-300)=\boxed{625.45\ \text{kJ/kg}}.$$
Net work (part a) and thermal efficiency (part b). Full-cycle energy balance ($\Delta U=0$ over one complete cycle) requires $w_{net}=q_{in}-q_{out}$: $$w_{net}=1289.31-625.45=\boxed{663.85\ \text{kJ/kg}}$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{663.85}{1289.31}=\boxed{0.5149\ (51.5\%)}.$$

Quantity	Result
(a) $w_{net}$	663.85 kJ/kg
(b) $\eta_{th}$	0.5149 (51.5%)

Concept & Theory

Topic. Air-standard Diesel cycle with non-ideal (inefficient) compression and expansion strokes.

Key relations. $T_{2s}=T_1CR^{k-1}$ (constant-volume-ratio isentropic compression); actual work from the stroke's own isentropic-efficiency definition; cutoff ratio $r_c=T_3/T_2$ from the constant-pressure heat addition; net work from the FULL-CYCLE energy balance $w_{net}=q_{in}-q_{out}$.

Why this works. Because the working fluid returns to state 1 at the end of every cycle, $\Delta U_{cycle}=0$ identically, so net work must equal net heat transferred — regardless of how irreversible the individual adiabatic strokes are. Computing net work instead as $w_{e,a}-w_{c,a}$ silently drops the $p\,dv$ boundary work performed during the constant-pressure heat-addition stroke (2→3), understating $w_{net}$ by roughly half in this problem.

Common pitfall. Treating this like an open-system (turbine/compressor) cycle where $w=\Delta h$ captures everything — a closed piston-cylinder cycle's net work needs the whole-cycle $q_{in}-q_{out}$ check, since boundary work during constant-pressure/constant-volume strokes isn't automatically embedded in the adiabatic strokes' $\Delta u$ terms.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Diesel Cycle).

Question 7 of 9: Non-Ideal Air-Standard Diesel Cycle

Question 7: Non-Ideal Air-Standard Diesel Cycle (15 marks)

Approach