Air-Standard Diesel Cycle, Variable Specific Heats: 04-BS-10, May 2018, Q6

Question 6: Air-Standard Diesel Cycle, Variable Specific Heats (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. $r=v_1/v_2=16$, $r_c=v_3/v_2=2$, $P_1=95$ kPa, $T_1=300$ K. Variable specific heats (air tables).

Find. (a) $T_3$ [K]; (b) $\eta_{th}$; (c) MEP [kPa].

Approach

Follow the four-process air-standard Diesel cycle using the variable-specific-heat air table functions throughout: isentropic compression (1→2) via the standard-state entropy function, constant-pressure heat addition (2→3) sized by the cutoff ratio, isentropic expansion (3→4) back to the compression-ratio volume, then close the cycle with constant-volume heat rejection.

Isentropic compression (1→2). Solving $s_2^\circ(T_2)-s_1^\circ(T_1)=R\ln(v_2/v_1)^{-1}=-R\ln r$ for $T_2$ (variable-$c_p$ air table) gives $$T_2=\boxed{861.4\ \text{K}},\qquad h_2=1016.24\ \text{kJ/kg}.$$
Constant-pressure heat addition (2→3), part (a). The cutoff ratio fixes the volume ratio at constant pressure, and for an ideal gas $T_3/T_2=v_3/v_2=r_c$: $$T_3=r_c\,T_2=2\times861.4=\boxed{1722.7\ \text{K}},\qquad h_3=2034.37\ \text{kJ/kg}.$$ $$q_{in}=h_3-h_2=2034.37-1016.24=1018.13\ \text{kJ/kg}.$$
Isentropic expansion (3→4). The expansion ratio is $v_4/v_3=r/r_c=8$; solving $s_4^\circ(T_4)-s_3^\circ(T_3)=R\ln(r/r_c)$ gives $$T_4=881.2\ \text{K},\qquad h_4=1038.37\ \text{kJ/kg}.$$
Heat rejected and thermal efficiency (part b). Constant-volume rejection (4→1) uses internal energies, $u=h-RT$: $$q_{out}=u_4-u_1=(h_4-RT_4)-(h_1-RT_1)=445.27\ \text{kJ/kg}.$$ $$w_{net}=q_{in}-q_{out}=1018.13-445.27=572.87\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{572.87}{1018.13}=\boxed{0.5627\ (56.3\%)}.$$
Mean effective pressure (part c). Specific volumes at states 1 and 2 from the ideal gas law: $$v_1=\frac{RT_1}{P_1}=\frac{0.287\times300}{95}=0.9063\ \text{m}^3/\text{kg},\qquad v_2=v_1/r=0.05664\ \text{m}^3/\text{kg}.$$ $$\text{MEP}=\frac{w_{net}}{v_1-v_2}=\frac{572.87}{0.9063-0.05664}=\frac{572.87}{0.8496} =\boxed{674.2\ \text{kPa}}.$$

Quantity	Result
(a) $T_3$	1722.7 K
(b) $\eta_{th}$	0.5627 (56.3%)
(c) MEP	674.2 kPa

Concept & Theory

Topic. Air-standard Diesel cycle analyzed with variable (temperature-dependent) specific heats via the air property tables, rather than the constant-$k$ cold-air-standard shortcuts.

Key relations. Isentropic process via standard-state entropy: $s_2^\circ-s_1^\circ=R\ln(P_2/P_1)$ (or the equivalent volume-ratio form). Constant-pressure heat addition: $T_3=r_c T_2$. MEP $=w_{net}/(v_1-v_2)$.

Why this works. At Diesel-cycle peak temperatures (roughly 900–1750 K here), air's true $c_p$ rises meaningfully above its room-temperature value, so the variable-$c_p$ (table) method gives a materially different (and more accurate) result than the constant-$k=1.4$ cold-air-standard shortcut — this is why the question explicitly asks to "account for the variation of specific heats with temperature."

Common pitfall. Using $T_4/T_3=(v_3/v_4)^{k-1}$ (the constant-$k$ isentropic relation) instead of solving the standard-state entropy equation for $T_4$ — the constant-$k$ shortcut silently reintroduces the cold-air-standard assumption the question asks you to avoid.

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

Question 6 of 9: Air-Standard Diesel Cycle, Variable Specific Heats

Question 6: Air-Standard Diesel Cycle, Variable Specific Heats (15 marks)

Approach