Air-Standard Diesel Cycle, Variable Specific Heats: 04-BS-10, December 2016, 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. $r=16$, $r_c=2$, $P_1=95$ kPa, $T_1=300$ K. Variable specific heats.

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

Approach

Solve the isentropic compression $1\to2$ via the variable-cp relation, apply the cutoff ratio to get $T_3=r_cT_2$ directly (constant-pressure heat addition), then solve the isentropic expansion $3\to4$ the same way. Heat added, heat rejected, and net work follow from $u$ and $h$ differences; MEP is net work divided by the swept specific volume $v_1-v_2$.

Isentropic compression (1→2). Solving $s^\circ(T_2)-s^\circ(T_1)= R\ln(v_1/v_2)=R\ln(r)$ for $T_2$: $$T_2=\boxed{861.35\ \text{K}},\qquad h_2=1016.24\ \text{kJ/kg}.$$
Constant-pressure heat addition (2→3). With cutoff ratio $r_c=v_3/v_2=2$ and constant pressure, $T_3=r_cT_2$: $$T_3=2\times861.35=\boxed{1722.70\ \text{K}}\quad\textbf{(a)},\qquad h_3=2034.37\ \text{kJ/kg}.$$ $$q_{in}=h_3-h_2=2034.37-1016.24=1018.13\ \text{kJ/kg}.$$
Isentropic expansion (3→4). Expansion ratio $v_4/v_3=r/r_c=8$; solving $s^\circ(T_4)-s^\circ(T_3)=R\ln(8)$ gives $$T_4=\boxed{881.20\ \text{K}},\qquad h_4=1038.37\ \text{kJ/kg}.$$
Heat rejected and net work. Constant-volume rejection $4\to1$: $$q_{out}=u_4-u_1=(h_4-RT_4)-(h_1-RT_1)=1038.37-252.85-(426.30-86.10)=445.27\ \text{kJ/kg}.$$ $$w_{net}=q_{in}-q_{out}=1018.13-445.27=572.87\ \text{kJ/kg}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{572.87}{1018.13}=\boxed{0.5627\ (56.3\%)}\quad\textbf{(b)}.$$
Mean effective pressure (part c). $v_1=RT_1/P_1=0.287\times300/95=0.9063$ m$^3$/kg, $v_2=v_1/r=0.05665$ m$^3$/kg. $$\text{MEP}=\frac{w_{net}}{v_1-v_2}=\frac{572.87}{0.9063-0.05665}=\frac{572.87}{0.8497} =\boxed{674.2\ \text{kPa}}\quad\textbf{(c)}.$$

Quantity	Result
(a) $T_3$	1722.70 K
(b) $\eta_{th}$	0.5627 (56.3%)
(c) MEP	674.2 kPa

Concept & Theory

Topic. Air-standard Diesel cycle with variable specific heats, evaluated via the ideal-gas $s^\circ(T)$ table function instead of constant $k$.

Key relations. Isentropic legs: $s^\circ(T_2)-s^\circ(T_1)=R\ln(v_1/v_2)$; cutoff ratio sets $T_3=r_cT_2$ directly since heat addition is at constant pressure; MEP $= w_{net}/(v_1-v_2)$.

Why this works. At Diesel-cycle peak temperatures (here above 1700 K), air's specific heats rise noticeably above their room-temperature values, so the variable-cp $s^\circ(T)$ approach is materially more accurate than a constant-$k=1.4$ shortcut, especially for $T_3$ and $T_4$.

Common pitfall. Using the compression-ratio isentropic relations for the expansion leg too — the expansion ratio is $r/r_c=8$, not $r=16$, since the cutoff ratio already consumed part of the stroke during constant-pressure heat addition.

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

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

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

Approach