Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — December 2015. 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 (water/steam, air, N₂, CO₂, moist air)
were computed from high-accuracy equations of state
in place of printed property-table interpolation; every boxed numeric result.
Question 8: Carnot Power Cycle with Air (15 marks)
Given. $m=0.5$ kg air. $\eta_{th}=0.50$. Process 1→2 (isothermal
expansion): $Q_{12}=40$ kJ. State 1: $P_1=700$ kPa, $V_1=0.12$ m$^3$. Constant properties
($k=1.4$, $R=0.287$ kJ/kg·K, $c_v=0.7175$ kJ/kg·K).
Find. (a) $T_H$, $T_L$ [K]; (b) $V_2$ [m$^3$]; (c) $W$ and $Q$ for each of the
four processes [kJ].
Fig. Q8 — P–V diagram of the Carnot power
cycle (1→2 isothermal expansion at $T_H$, 2→3 adiabatic expansion, 3→4 isothermal
compression at $T_L$, 4→1 adiabatic compression).
Approach
The Carnot cycle's maximum temperature is fixed directly by the given state-1 conditions via the
ideal-gas law; the minimum temperature then follows from the given thermal efficiency
($\eta_{th}=1-T_L/T_H$ for any Carnot engine). The isothermal-expansion heat transfer (equal to its
work, since $\Delta U=0$) fixes $V_2$. The two adiabatic legs share the same volume ratio
$(T_H/T_L)^{1/(k-1)}$, which fixes $V_3$ and $V_4$; each process's work and heat transfer then follow
from the standard ideal-gas isothermal/adiabatic relations.
Maximum temperature (part a). From the ideal-gas law at state 1:
$$T_H=\frac{P_1V_1}{mR}=\frac{700\times0.12}{0.5\times0.287}=\boxed{585.37\ \text{K}}.$$
Minimum temperature (part a). Carnot efficiency $\eta_{th}=1-T_L/T_H$:
$$T_L=T_H(1-\eta_{th})=585.37\times0.50=\boxed{292.68\ \text{K}}.$$
Volume at end of isothermal expansion (part b). For an ideal gas, isothermal
$Q_{12}=W_{12}=mRT_H\ln(V_2/V_1)$:
$$V_2=V_1\exp\!\left(\frac{Q_{12}}{mRT_H}\right)=0.12\times\exp\!\left(\frac{40}{0.5\times0.287\times585.37}\right)=\boxed{0.1932\ \text{m}^3}.$$
Adiabatic legs (states 3, 4). Both adiabatic legs share the same volume ratio
$(T_H/T_L)^{1/(k-1)}=5.657$ (a general property of the ideal-gas Carnot cycle):
$$V_3=V_2\times5.657=1.0929\ \text{m}^3,\qquad V_4=V_1\times5.657=0.6788\ \text{m}^3.$$
Process 1→2: isothermal expansion at $T_H$.
$$W_{12}=Q_{12}=\boxed{40.00\ \text{kJ}}.$$
Process 3→4: isothermal compression at $T_L$. Heat rejected
$Q_L=Q_{12}\,T_L/T_H$ (Carnot ratio):
$$Q_{34}=W_{34}=-Q_{12}\frac{T_L}{T_H}=-40\times\frac{292.68}{585.37}=\boxed{-20.00\ \text{kJ}}.$$
Process 4→1: adiabatic compression, $T_L\rightarrow T_H$. $Q_{41}=0$;
by symmetry with process 2→3:
$$W_{41}=-mc_v(T_H-T_L)=\boxed{-105.00\ \text{kJ}},\qquad Q_{41}=0.$$
Cross-check. $W_{net}=W_{12}+W_{23}+W_{34}+W_{41}=40+105-20-105=20.00$ kJ,
and $\eta_{th}=W_{net}/Q_{12}=20.00/40=0.50$ — matches the given 50% exactly.
