22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2017
Question 2 of 8: Air-Standard Dual (Mixed) Cycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, filling of evacuated vessels, air-standard dual and gas-turbine (turbojet) cycles, and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radial composite-cylinder conduction with convection, external cross-flow over a cylinder, internal-flow temperature decay, radiation between concentric spheres with a shield, and the ε–NTU cross-flow heat-exchanger method. Ammonia and ideal-gas air properties are read from the tables appended to the examination paper; the turbojet uses cold-air-standard constant specific heats.
Paper format: National Examination 16-Mec-A1, May 2017, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three questions from one part and two from the other). All eight questions are solved in full below.
Ideal-gas air table ($u$, $h$, $v_r$) — variable specific heats
Given. The tabulated cycle data above. Find. the cut-off ratio $r_c$, the constant-volume pressure ratio $r_p$, the heat added and rejected per unit mass, and the thermal efficiency.
Figure 2 — Dual cycle: 1→2 isentropic compression, 2→3 constant-volume heat addition, 3→4 constant-pressure heat addition, 4→5 isentropic expansion, 5→1 constant-volume heat rejection. The $T$–$S$ panel is drawn to scale from the solved states (both isentropic legs are vertical; $T_5=822$ K sits just below $T_2=867$ K); the $P$–$V$ panel is schematic, with the clearance volume exaggerated for legibility.
Approach. Use the air table's relative volume $v_r$ to carry the two isentropic legs, read $u$ and $h$ at each state for the heat terms, and get $r_p$ and $r_c$ from the constant-volume and constant-pressure temperature ratios; the efficiency is $1-q_{out}/q_{in}$.
State 1 and isentropic compression 1→2. At $T_1=298.15$ K the table gives $u_1=212.7$ kJ/kg and $v_{r1}=631.1$. For the isentropic compression $v_{r2}=v_{r1}/r_v=631.1/16.5=38.25$, which the table inverts to $T_2=867.0$ K with $u_2=647.2$ kJ/kg.
Constant-volume pressure ratio (2→3). At constant volume $r_p=\dfrac{p_3}{p_2}=\dfrac{T_3}{T_2}=\dfrac{1443.15}{867.0}$.
$r_p = 1.665$
Cut-off ratio (3→4). At constant pressure $r_c=\dfrac{V_4}{V_3}=\dfrac{T_4}{T_3}=\dfrac{1868.15}{1443.15}$.
$r_c = 1.294$
Isentropic expansion 4→5 to $V_5=V_1$. Since $V_5/V_4=V_1/V_4=r_v/r_c=16.5/1.294=12.75$, $\;v_{r5}=v_{r4}\,(r_v/r_c)=3.49(12.75)=44.5$, giving $T_5=822.4$ K and $u_5=610.5$ kJ/kg.
Heat rejected (5→1, constant volume) and efficiency. $q_{out}=u_5-u_1=610.5-212.7=397.8$ kJ/kg, so
$$\eta_{th}=1-\frac{q_{out}}{q_{in}}=1-\frac{397.8}{1026.4}$$
$\eta_{th}\approx0.612$ (61.2 %), with $w_{net}=q_{in}-q_{out}=628.6$ kJ/kg
Check — table interpolation.
All state properties are linearly interpolated from the appended ideal-gas air table (e.g. $T_2$ from $v_r$ between 860 K and 880 K, $u_3$ between 1440 K and 1460 K, $h_4$ between 1850 K and 1900 K). A neighbouring-row read can move the boxed numbers by a few tenths of a per cent; the efficiency is stable at ≈ 61 %.
