Question 6 of 9: Rigid Vessel of Saturated Water Vapor, Cooled at Constant Volume
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 6: Rigid Vessel of Saturated Water Vapor, Cooled at Constant Volume (15 marks)
Given. $V=0.05$ m$^3$. State 1: saturated vapor at $P_1=100$ kPa. Final state:
$T_2=75\ ^\circ$C. Rigid vessel (constant $v$, no boundary work).
State
Description
P
T
v (m³/kg)
u (kJ/kg)
1
Sat. vapor
100 kPa
99.63°C
1.6939
2505.55
2
Two-phase (final)
38.60 kPa
75°C
1.6939
1200.34
Find. (b) $P_2$; (c) $Q$ [kJ].
Fig. Q6 — P–v sketch (part a): the
constant-volume cooling process is a vertical line from saturated vapor at 100 kPa straight down
into the two-phase dome to 75°C ($x_2=0.410$). Dome windowed to the 60–160°C range
around the process for legibility.
Approach
Fix the mass and specific volume from the known initial saturated-vapor state; since the vessel
is rigid, that same specific volume is held throughout. At the final temperature, check that this
$v$ lies between $v_f$ and $v_g$ (confirming a two-phase final state), compute the quality from the
lever rule, and read off $P_2$ and $u_2$ at that quality. The first law for a closed, rigid system
(no work) then gives the heat transfer directly from $\Delta U$.
Initial state and mass. Saturated vapor at 100 kPa: $v_g=1.6939$ m$^3$/kg,
$u_1=2505.55$ kJ/kg.
$$m=\frac{V}{v_g}=\frac{0.05}{1.6939}=\boxed{0.02952\ \text{kg}}.$$
Final state (part b). Rigid vessel $\Rightarrow v_2=v_1=1.6939$ m$^3$/kg. At
$75\ ^\circ$C: $v_f=0.001026$ m$^3$/kg, $v_g=4.1289$ m$^3$/kg, and since $v_f
Heat transfer (part c). $u_2=u_f+x_2\,u_{fg}=1200.34$ kJ/kg at this quality.
Rigid vessel $\Rightarrow W=0$, so the first law reduces to $Q=\Delta U$:
$$Q=m(u_2-u_1)=0.02952\times(1200.34-2505.55)=\boxed{-38.53\ \text{kJ}}.$$
The negative sign confirms heat is rejected FROM the water, consistent with cooling.
