Rigid Tank Charged from a Steam Supply Line: 04-BS-10, May 2015, Q9

Question 9: Rigid Tank Charged from a Steam Supply Line (15 marks)

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Given. Rigid tank, $m_1=0.5$ kg at $700$ kPa, $320\ ^\circ$C (initial). Supply line: constant $1.5$ MPa, $320\ ^\circ$C. Final tank state: $1.0$ MPa, $320\ ^\circ$C.

Find. (a) $m_2$; (b) $Q$.

Approach

The tank volume is fixed, so the initial specific volume (from the given initial state) fixes the tank volume, and the final specific volume (from the given final state) then gives the final mass directly. The unsteady-flow energy balance for a charging process ($m_2u_2-m_1u_1=Q+\dot m_{in}h_{line}$, with no work crossing the rigid boundary) then isolates the heat transfer, using the supply line's own enthalpy (not internal energy) for the entering mass since flow work at the valve is real.

Tank volume and initial/final specific volumes. At $700$ kPa, $320\ ^\circ$C: $v_1=0.38523$ m$^3$/kg, $u_1=2831.68$ kJ/kg. Tank volume: $$V=m_1v_1=0.5\times0.38523=0.19261\ \text{m}^3.$$ At $1.0$ MPa, $320\ ^\circ$C: $v_2=0.26786$ m$^3$/kg, $u_2=2826.50$ kJ/kg.
Final mass (part a). $$m_2=\frac{V}{v_2}=\frac{0.19261}{0.26786}=\boxed{0.7191\ \text{kg}}.$$
Supply-line enthalpy and entering mass. At the line condition ($1.5$ MPa, $320\ ^\circ$C): $h_{line}=3082.43$ kJ/kg. Entering mass $m_{in}=m_2-m_1=0.7191-0.5=0.2191$ kg.
Heat transfer (part b). Unsteady energy balance, rigid tank (no boundary work): $$Q=(m_2u_2-m_1u_1)-m_{in}h_{line}$$ $$=(0.7191\times2826.50-0.5\times2831.68)-(0.2191\times3082.43)$$ $$=(2032.32-1415.84)-675.15=\boxed{-58.66\ \text{kJ}}.$$ Negative sign: heat is transferred from the steam in the tank to the surroundings.

Quantity	Result
(a) $m_2$	0.7191 kg
(b) $Q$	−58.66 kJ (heat OUT of the tank)

Concept & Theory

Topic. Unsteady (uniform-flow) charging of a rigid tank from a constant-state supply line.

Key relations. Uniform-flow energy balance for charging a rigid, insulated- valve tank: $m_2u_2-m_1u_1=Q+m_{in}h_{line}$ (no boundary work since the tank volume is fixed).

Why this works. The entering steam carries flow work ($Pv$) across the valve in addition to its internal energy, so its energy content on entry must be counted as enthalpy $h_{line}$, not $u_{line}$ — using $u_{line}$ instead would understate the energy added and give the wrong (and wrong-signed) heat transfer.

Common pitfall. Assuming a charging process with an insulated valve must have $Q=0$ for the whole tank — "insulated valve" only means the throttling itself is adiabatic; the tank walls are a separate boundary, and here $Q\neq0$ is required to keep the final temperature equal to both the initial and supply-line temperature (320°C) despite the pressure rising from 700 kPa. All three temperatures being numerically equal is a deliberate feature of this problem, not a simplification that forces $Q=0$.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis of Control Volumes, Uniform-Flow Processes).

Question 9 of 9: Rigid Tank Charged from a Steam Supply Line

Question 9: Rigid Tank Charged from a Steam Supply Line (15 marks)

Approach