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04-BS-10 · December 2013

Question 4 of 9: Air-Standard Diesel Cycle — Variable Specific Heats

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-10, Thermodynamics — December 2013. 3 hours, Closed-Book Exam (approved calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and charts supplied in an appendix). Part A: answer 2 of Questions 1-3 (20 marks each). Part B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the first two Part-A and first four Part-B questions as they appear in the answer book are marked. All nine questions (Part A complete, Part B complete) are solved below for completeness.

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 8th ed. All state properties (R-134a, water/steam, moist air, N₂, CO₂, air) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric result.

Question 4: 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. Air-standard Diesel cycle, variable specific heats. Compression ratio $r=v_1/v_2=16$; cutoff ratio $r_c=v_3/v_2=2$; $P_1=95$ kPa, $T_1=27\ ^\circ\text{C}=300$ K.

QuantityValue
$r=v_1/v_2$16
$r_c=v_3/v_2$2
$P_1,\ T_1$95 kPa, 300 K

Find. (a) $T_3$ (after heat addition); (b) $\eta_{th}$; (c) MEP.

Specific volume v (m³/kg)P (kPa)Q4 — Air-standard Diesel cycle (P–v)1234
Fig. Q4 — P–v diagram of the air-standard Diesel cycle (1→2 isentropic compression · 2→3 constant-pressure heat addition · 3→4 isentropic expansion · 4→1 constant-volume heat rejection).

Approach

Find $T_2$ from the isentropic compression relation written with variable specific heats ($s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$, with $P_2/P_1$ eliminated via the ideal-gas law in terms of the known volume ratio $r$) using root-finding; get $T_3$ directly from the constant-pressure condition $T_3=r_cT_2$; repeat the entropy root-find for the isentropic expansion 3→4 using the volume ratio $v_4/v_3=r/r_c$; then take $q_{in}$, $q_{out}$, $w_{net}$, $\eta_{th}$ and MEP from the resulting enthalpies/internal energies.

  1. Isentropic compression (1→2). With $P_2/P_1=(T_2/T_1)\,r$ (ideal gas, $v_1/v_2=r$), the isentropic condition $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$ is solved for $T_2$: $$T_2=861.35\ \text{K}\quad(h_2=890.9\ \text{kJ/kg}).$$
  2. Constant-pressure heat addition (2→3, part a). $P_2=P_3$, so ideal-gas $v_3/v_2=T_3/T_2=r_c$ gives $T_3$ directly: $$T_3=r_c\,T_2=2\times861.35=\boxed{1722.70\ \text{K}}\quad(h_3=1909.0\ \text{kJ/kg}).$$
  3. Isentropic expansion (3→4). Since $v_4=v_1$ and $v_3=r_cv_2$, $v_4/v_3=r/r_c=16/2=8$. Solving $s^\circ(T_4)-s^\circ(T_3)=R\ln[(T_4/T_3)/8]$ for $T_4$: $$T_4=881.20\ \text{K}\quad(h_4=910.9\ \text{kJ/kg}).$$
  4. Heat added and heat rejected. Constant-$P$ heat addition uses $\Delta h$; constant-$v$ heat rejection uses $\Delta u=\Delta h-R\Delta T$: $$q_{in}=h_3-h_2=1909.0-890.9=1018.13\ \text{kJ/kg}.$$ $$q_{out}=u_4-u_1=(h_4-RT_4)-(h_1-RT_1)=445.27\ \text{kJ/kg}.$$
  5. Thermal efficiency (part b). $$w_{net}=q_{in}-q_{out}=1018.13-445.27=572.87\ \text{kJ/kg},\qquad \eta_{th}=\frac{w_{net}}{q_{in}}=\frac{572.87}{1018.13}=\boxed{56.27\%}.$$
  6. Mean effective pressure (part c). $v_1=RT_1/P_1=0.287\times300/95=0.9063$ m³/kg; $v_2=v_1/r=0.0566$ m³/kg: $$\text{MEP}=\frac{w_{net}}{v_1-v_2}=\frac{572.87}{0.9063-0.0566}=\boxed{674.2\ \text{kPa}}.$$
QuantityResult
(a) $T_3$1722.70 K
(b) $\eta_{th}$56.27%
(c) MEP674.2 kPa