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04-BS-10 · May 2014

Question 4 of 9: Air-Standard Dual 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 — May 2014. 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, O₂, air) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric result.

Question 4: Air-Standard Dual 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. Compression ratio $r=17$, cutoff ratio $r_c=V_3/V_x=1.2$, $P_1=95$ kPa, $T_1=310$ K. Pressure doubles during the constant-volume leg ($P_x=2P_2$). Variable specific heats (temperature-dependent air properties, via the ideal-gas standard-entropy function $s^\circ(T)$).

Find. (a) $q_v$ and $q_p$ [kJ/kg]; (b) $w_{net}$ [kJ/kg]; (c) $q_{out}$ [kJ/kg]; (d) $\eta_{th}$.

Entropy s (kJ/kg·K)T (°C)Q4 — Air-standard dual cycle, variable specific heats (T-s, no dome)12x34
Fig. Q4 — T–s state points for the air-standard dual cycle (1→2 isentropic compression, 2→x constant-volume heat addition, x→3 constant-pressure heat addition, 3→4 isentropic expansion; the closing 4→1 constant-volume heat-rejection leg is a vertical drop in $s$ at the left edge, omitted from the drawn path for clarity).

Approach

Solve the isentropic compression with the temperature-dependent $s^\circ(T)$ relation $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)\,r]$ by root-finding. The constant-volume leg fixes $T_x=2T_2$ directly from the stated pressure doubling (ideal gas, $v$ fixed $\Rightarrow T\propto P$). The constant-pressure leg fixes $T_3=r_c\,T_x$ directly (ideal gas, $P$ fixed $\Rightarrow T\propto V$). The isentropic expansion uses the mirrored root-finding relation $s^\circ(T_4)-s^\circ(T_3)=R\ln[(T_4/T_3)(r_c/r)]$. All enthalpies/entropies come from real-air property calls at a fixed low reference pressure, standing in for the printed variable-$c_p$ ideal-gas air table.

  1. Isentropic compression (1→2). Solving $s^\circ(T_2)-s^\circ(T_1)=R\ln[(T_2/T_1)\times17]$ for $T_2$: $$T_2=904.95\ \text{K}\ (631.80\ ^\circ\text{C}).$$
  2. Constant-volume heat addition (2→x), pressure doubles. With $v$ fixed, $T_x/T_2=P_x/P_2=2$: $$T_x=2\times904.95=1809.89\ \text{K}\ (1536.74\ ^\circ\text{C}).$$ $$q_v=u_x-u_2=(h_x-RT_x)-(h_2-RT_2)=\boxed{817.32\ \text{kJ/kg}}.$$
  3. Constant-pressure heat addition (x→3), cutoff ratio. With $P$ fixed, $T_3=r_c\,T_x=1.2\times1809.89=2171.87$ K ($1898.72\ ^\circ$C): $$q_p=h_3-h_x=\boxed{452.22\ \text{kJ/kg}}.$$
  4. Isentropic expansion (3→4). Solving $s^\circ(T_4)-s^\circ(T_3)=R\ln[(T_4/T_3)(r_c/r)]$ for $T_4$: $$T_4=944.64\ \text{K}\ (671.49\ ^\circ\text{C}).$$
  5. Heat rejection (part c) and net work (part b). $$q_{out}=u_4-u_1=(h_4-RT_4)-(h_1-RT_1)=\boxed{491.14\ \text{kJ/kg}}.$$ $$q_{in}=q_v+q_p=817.32+452.22=1269.54\ \text{kJ/kg}.$$ $$w_{net}=q_{in}-q_{out}=1269.54-491.14=\boxed{778.40\ \text{kJ/kg}}.$$
  6. Thermal efficiency (part d). $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{778.40}{1269.54}=\boxed{0.6131\ (61.3\%)}.$$
QuantityResult
(a) $q_v$817.32 kJ/kg
(a) $q_p$452.22 kJ/kg
(b) $w_{net}$778.40 kJ/kg
(c) $q_{out}$491.14 kJ/kg
(d) $\eta_{th}$0.6131 (61.3%)