Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — December 2016. 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, interpolation not required). 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 (water/steam, R-134a, air, N₂, CO₂)
were computed from high-accuracy equations of state
in place of printed property-table interpolation; every boxed numeric result.
Given. $m=1.5$ kg air, $\eta_{th}=0.50$, $Q_{in}=40$ kJ (isothermal expansion
$1\to2$). State 1: $P_1=700$ kPa, $V_1=0.12$ m$^3$. Ideal gas, constant specific heats
($c_v=0.718$ kJ/kg·K, $k=1.4$).
Find. (a) $T_H$, $T_L$ [K]; (b) $V_2$ [m$^3$]; (c) $W$, $Q$ for each of the 4
processes.
Fig. Q9 — P–v diagram of the Carnot power cycle (1→2
isothermal expansion at $T_H$ · 2→3 isentropic expansion to $T_L$ · 3→4
isothermal compression at $T_L$ · 4→1 isentropic compression back to $T_H$), vertices
connected by straight segments for clarity rather than the true curved isotherms/isentropes.
Approach
State 1 fixes $T_H$ directly via the ideal-gas law, since $P_1$, $V_1$, and $m$ are all given.
The stated Carnot efficiency then gives $T_L$, and the given heat transfer during the isothermal
leg gives $V_2$ from $Q_{in}=mRT_H\ln(V_2/V_1)$. The two isentropic legs' work follows from
$\Delta U=mc_v\Delta T$ (since $Q=0$ on them), and the isothermal-compression heat/work is fixed by
the Carnot ratio $Q_{out}/Q_{in}=T_L/T_H$.
Maximum and minimum temperatures (part a). Ideal gas at state 1:
$$T_H=\frac{P_1V_1}{mR}=\frac{700\times0.12}{1.5\times0.287}=\boxed{195.12\ \text{K}}.$$
$$T_L=T_H(1-\eta_{th})=195.12\times0.50=\boxed{97.56\ \text{K}}\quad\textbf{(a)}.$$
Volume at end of isothermal expansion (part b). Since $mRT_H=P_1V_1=84.0$ kJ
exactly,
$$\ln\frac{V_2}{V_1}=\frac{Q_{in}}{mRT_H}=\frac{40}{84.0}=0.4762,\qquad
V_2=V_1\,e^{0.4762}=0.12\times1.6099=\boxed{0.1932\ \text{m}^3}\quad\textbf{(b)}.$$
Process 1→2, isothermal expansion at $T_H$. $\Delta U=0$, so
$$W_{12}=Q_{12}=Q_{in}=\boxed{40.00\ \text{kJ}}.$$
Process 2→3, isentropic expansion $T_H\to T_L$. $Q_{23}=0$;
$$W_{23}=-\Delta U=mc_v(T_H-T_L)=1.5\times0.718\times(195.12-97.56)=\boxed{105.08\ \text{kJ}}.$$
Process 3→4, isothermal compression at $T_L$. By the Carnot ratio
$Q_{out}/Q_{in}=T_L/T_H$:
$$Q_{out}=Q_{in}\frac{T_L}{T_H}=40\times0.50=20.00\ \text{kJ (heat rejected)},$$
so with the sign convention that heat/work leaving the gas is negative,
$$Q_{34}=W_{34}=\boxed{-20.00\ \text{kJ}}.$$
Process 4→1, isentropic compression $T_L\to T_H$. $Q_{41}=0$; by symmetry
with Step 4 (equal and opposite $\Delta U$),
$$W_{41}=-mc_v(T_H-T_L)=\boxed{-105.08\ \text{kJ}}.$$
Consistency check. Net work over the cycle:
$$W_{net}=W_{12}+W_{23}+W_{34}+W_{41}=40.00+105.08-20.00-105.08=20.00\ \text{kJ}.$$
$$\eta_{th,check}=\frac{W_{net}}{Q_{in}}=\frac{20.00}{40}=0.50,$$
exactly matching the given 50% efficiency and confirming the four process values above.
