Question 9 of 9: Rigid Tank Filled from a Supply Line, Isothermal
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 9: Rigid Tank Filled from a Supply Line, Isothermal (15 marks)
Given. $V=0.5$ m$^3$, initially evacuated. Supply line: $P_{line}=100$ kPa,
$T_{line}=21\ ^\circ$C. Final tank pressure $P_2=100$ kPa. Tank temperature held constant at
$T=21\ ^\circ$C ($=294.15$ K) throughout by heat transfer.
Find. $Q$ [kJ].
Approach
This is an unsteady-flow (charging) problem, but with the unusual twist that heat transfer
actively holds the tank temperature equal to the supply-line temperature throughout, so the final
internal energy equals the line's internal energy exactly. Apply the uniform-flow energy balance
with the mass balance ($m_{in}=m_2$, since the tank started evacuated) to solve directly for $Q$.
Final mass in the tank. Ideal gas at $P_2=100$ kPa, $T=294.15$ K:
$$m_2=\frac{P_2V}{RT}=\frac{100\times0.5}{0.287\times294.15}=\boxed{0.5923\ \text{kg}}.$$
Energy balance. Uniform-flow first law for the tank (no initial mass, no work,
mass entering carries the line's flow enthalpy $h_{line}$):
$$Q=m_2u_2-m_{in}h_{line}=m_2u_2-m_2h_{line}=m_2(u_2-h_{line}).$$
Since the tank temperature is held at $T_{line}$ throughout, $u_2=u_{line}$ (ideal gas, $u=u(T)$
only), so:
$$u_2-h_{line}=u_{line}-h_{line}=-RT_{line}.$$
Heat transfer.
$$Q=-m_2RT_{line}=-0.5923\times0.287\times294.15\ \text{kJ}.$$
Recognizing $m_2RT_{line}=P_2V$ exactly (the same ideal-gas law used to find $m_2$), this collapses
to the closed form:
$$Q=-P_2V=-100\times0.5=\boxed{-50.00\ \text{kJ}}.$$
The negative sign shows heat must be REJECTED from the tank to hold its temperature constant while
being charged — the flow work carried in by the entering air would otherwise raise the tank's
temperature above the line's.
