Question 5 of 9: Rigid Tank Charged from a Steam Supply Line
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.
Question 5: Rigid Tank Charged from a Steam Supply Line (15 marks)
Since $T_1$, $T_{line}$, and $T_2$ are all at 320°C but different pressures, all three states
must first be checked against $T_{sat}$ at their respective pressures to confirm they are
superheated (not saturated) before pulling properties. The tank's fixed volume links $m_1v_1=V=
m_2v_2$, giving $m_2$; the unsteady-flow energy balance $Q+m_{in}h_{line}=m_2u_2-m_1u_1$ (using the
supply line's enthalpy, since flow work crosses the boundary with the entering mass) then gives
$Q$.
Confirm all three states are superheated. $T_{sat}(700\ \text{kPa})=164.95\
^\circ$C, $T_{sat}(1.5\ \text{MPa})=198.29\ ^\circ$C, $T_{sat}(1.0\ \text{MPa})=179.88\ ^\circ$C
— all well below $320\ ^\circ$C, so all three states are superheated vapour.
Initial state and tank volume. At 700 kPa, 320°C: $v_1=0.385227$ m$^3$/kg,
$u_1=2831.68$ kJ/kg.
$$V=m_1v_1=0.5\times0.385227=0.192613\ \text{m}^3\ \text{(fixed thereafter)}.$$
Final state and mass (part a). At 1.0 MPa, 320°C: $v_2=0.267859$ m$^3$/kg,
$u_2=2826.50$ kJ/kg.
$$m_2=\frac{V}{v_2}=\frac{0.192613}{0.267859}=\boxed{0.7191\ \text{kg}}\quad\textbf{(a)}.$$
Mass entering: $m_{in}=m_2-m_1=0.7191-0.5=0.2191$ kg.
Supply-line enthalpy. At 1.5 MPa, 320°C: $h_{line}=3082.43$ kJ/kg (the
enthalpy, not internal energy, is used here since flow work accompanies the entering mass across
the boundary).
Heat transfer (part b). Unsteady-flow energy balance,
$Q=(m_2u_2-m_1u_1)-m_{in}h_{line}$:
$$Q=(0.7191\times2826.50-0.5\times2831.68)-0.2191\times3082.43$$
$$=(2032.57-1415.84)-675.36=616.73-675.36=\boxed{-58.66\ \text{kJ}}\quad\textbf{(b)}.$$
The negative sign shows heat is rejected from the tank: the entering supply steam carries
more energy (via its higher pressure and flow work) than is needed to reach the stated final state
at the same temperature as the supply, so the tank must lose heat to the surroundings to keep
$T_2=320\ ^\circ$C rather than ending up hotter.
