22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2019
Question 1 of 8: Irreversible adiabatic air compression & Freon-12 vessel charging
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas processes and entropy generation, polytropic compression, rigid-vessel charging, the reciprocating air compressor, throttling/flash separation, the steam turbine, and vapour-compression refrigeration/heat-pump cycles; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady radial conduction through a cylindrical wall with convection, conduction with uniform internal generation, transient (lumped) cooling by combined convection and radiation, and the effectiveness–NTU method for shell-and-tube exchangers. Steam and Freon-12 (R-12) properties are evaluated, which reproduces the IAPWS steam tables and the standard R-12 property tables to graphing accuracy; enthalpy differences (the only quantities used) are datum-independent. Air and combustion gases are treated as ideal with constant specific heats ($\gamma=1.4$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005$, $c_v=0.718\ \text{kJ/kg}\cdot\text{K}$).
Question 1 — Irreversible adiabatic air compression & Freon-12 vessel charging (Part A, equal value)
Find. Compression work $W$, entropy change $\Delta S$, whether a reversible process between the same end states would add or reject heat, and the polytropic exponent $n$; then sketch both paths on $p$–$v$ and $T$–$s$ axes.
The reversible adiabat is vertical on $T$–$s$ ($\Delta s=0$) and reaches only $T_{2s}=293\ ^\circ\text{C}$; the real process generates entropy, ending hotter (320 °C) and to the right.
Approach. With $Q=0$ the first law gives the work directly from the internal-energy change; the entropy change of an ideal gas follows the $T$–$P$ form; the sign of the entropy change decides the reversible-path heat direction; the polytropic index comes from the $T$–$P$ polytropic relation.
Work done on the gas (first law, adiabatic). For a closed system with $Q=0$, $W_\text{on}=\Delta U=m\,c_v\,(T_2-T_1)$:
$$W_\text{on}=(0.050)(0.718)(593.15-293.15)=\boxed{10.77\ \text{kJ}}$$
Work is done on the air, so this raises its internal energy.
Entropy change (ideal-gas $T$–$P$ relation).
$$\Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}=1.005\ln\frac{593.15}{293.15}-0.287\ln 10=0.0475\ \text{kJ/kg}\cdot\text{K}$$
$$\Delta S=m\,\Delta s=(0.050)(0.0475)=\boxed{2.37\ \text{J/K}}$$
The rise is the entropy generated by the irreversibility (an adiabatic process can only have $\Delta S\ge0$).
Reversible path between the same end states — heat direction. For any reversible process, $\displaystyle \Delta S=\int \frac{\delta Q}{T}$. Because the two fixed end states have $\Delta S=+2.37\ \text{J/K}>0$, the reversible connection must have $\int\delta Q/T>0$:
$$\text{heat would have to be } \boxed{\textbf{added}}\ \big(Q_\text{rev}\approx T\,\Delta S>0\big).$$
(The real process rejects no heat; the reversible one with identical entropy rise must absorb it.)
Polytropic exponent (T–P polytropic relation). For $pv^{\,n}=\text{const}$, $\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{\frac{n-1}{n}}$, so
$$\frac{n-1}{n}=\frac{\ln(T_2/T_1)}{\ln(P_2/P_1)}=\frac{\ln 2.0234}{\ln 10}=0.3061\;\Rightarrow\; n=\frac{1}{1-0.3061}=\boxed{1.441}$$
Since $n>\gamma=1.4$, the real process ends hotter than the isentropic $T_{2s}=T_1(P_2/P_1)^{(\gamma-1)/\gamma}=566\ \text{K}=293\ ^\circ\text{C}$ — consistent with entropy having been generated.
Quantity
Result
Work done on the air
10.77 kJ
Entropy change $\Delta S$
+2.37 J/K (generated)
Reversible-path heat
added ($Q_\text{rev}\approx+0.70$ kJ)
Polytropic exponent $n$
1.441 ($>\gamma$)
Part (b) — Charged mass of Freon-12
Given. Rigid unit $V=0.050\ \text{m}^3$, evacuated, charged with R-12 to $P=250$ kPa at $T=20\ ^\circ\text{C}$. At 250 kPa the R-12 saturation temperature is $\approx-6\ ^\circ\text{C}$, so at 20 °C the charge is superheated vapour.
Find. Mass of Freon-12 held under this condition.
Approach. Read the superheated-R-12 specific volume at 250 kPa and 20 °C, then $m=V/v$.
Specific volume of the superheated charge. From the superheated R-12 table at 0.25 MPa, 20 °C (appended to the exam):
$$v=0.0762\ \text{m}^3/\text{kg}$$
Mass from the rigid volume.
$$m=\frac{V}{v}=\frac{0.050}{0.0762}=\boxed{0.656\ \text{kg}}$$
Check — the remaining 1(b) sub-parts are incomplete as given.
The source retains only disconnected fragments of the other sub-questions (a charged mass as saturated vapour, a liquid+vapour split for a stated total mass, and a "heat removed" during charging). The standard method for each is: for a saturated-vapour charge, $m=V/v_g(20\ ^\circ\text{C})$; for a two-phase charge of known total mass $m_t$, the specific volume $v=V/m_t$ fixes the quality $x=(v-v_f)/(v_g-v_f)$ and hence the liquid and vapour masses $m_f=(1-x)m_t$, $m_g=x\,m_t$; the heat removed during a slow (isothermal) charge is $Q=\sum m_i u_i - m_t u_t$ over the incoming stream and stored inventory. These are presented as method only, because the numerical data are not reproduced here.
