Uncontrolled Expansion into an Evacuated Compartment: 04-BS-10, May 2016, Q9

Question 9: Uncontrolled Expansion into an Evacuated Compartment (15 marks)

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Given. Insulated (adiabatic), rigid, two-compartment vessel. Initial: steam at $P_1=1.0$ MPa, $T_1=500\ ^\circ$C fills compartment A; compartment B is evacuated. Valve opens, steam expands to fill the whole vessel at $P_2=0.1$ MPa.

Find. (a) $T_2$ [°C]; (b) $V_A/V_{total}$ [%]; (c) $s_{gen}$ [kJ/kg·K].

Approach

The whole two-compartment vessel is rigid and insulated, so across the entire (fixed-mass) system there is no heat transfer and no boundary work — the total internal energy is conserved, $u_2=u_1$ (this is not a throttling process with $h_2=h_1$; that relation applies to steady flow through a valve from an external line, not to an uncontrolled expansion inside a single closed, insulated vessel). The final temperature is then whatever satisfies $u(P_2,T_2)=u_1$ at the given final pressure. The fraction of the vessel initially occupied by the steam follows from the specific volumes ($V\propto v$ at fixed mass), and the entropy produced is simply $s_2-s_1$ since this isolated system exchanges no entropy with its surroundings.

Initial state. At 1.0 MPa, 500°C: $u_1=3124.99$ kJ/kg, $v_1=0.35411$ m$^3$/kg, $s_1=7.7641$ kJ/kg·K.
Final temperature (part a). Energy conservation for the isolated, insulated, rigid vessel (no work crosses its outer boundary, no heat transfer): $$u_2=u_1=3124.99\ \text{kJ/kg}.$$ Solving for $T_2$ at $P_2=0.1$ MPa such that $u(P_2,T_2)=3124.99$ kJ/kg: $$T_2=\boxed{495.69\ ^\circ\text{C}}\quad(v_2=3.5456\ \text{m}^3/\text{kg},\ s_2=8.8242\ \text{kJ/kg}\cdot\text{K}).$$
Vessel-fraction initially occupied by steam (part b). Since the total mass is fixed and the vessel is rigid, $V\propto v$ at each state, so the initial fraction is simply the ratio of specific volumes: $$\frac{V_A}{V_{total}}=\frac{v_1}{v_2}=\frac{0.35411}{3.5456}=\boxed{0.0999\ (10.0\%)}.$$
Entropy generation (part c). The combined vessel is isolated (no mass, heat, or work crosses its outer boundary), so all of the entropy increase is internally generated: $$s_{gen}=s_2-s_1=8.8242-7.7641=\boxed{1.060\ \text{kJ/kg}\cdot\text{K}}.$$ This is strictly positive, as required for the highly irreversible uncontrolled (unresisted) expansion into a vacuum.

Quantity	Result
(a) $T_2$	495.69°C
(b) $V_A/V_{total}$	9.99% (≈10%)
(c) $s_{gen}$	1.060 kJ/kg·K

Concept & Theory

Topic. Uncontrolled (free) expansion of steam into an evacuated compartment within a single rigid, insulated vessel.

Key relations. Isolated-system energy conservation: $u_2=u_1$ (not $h_2=h_1$). Fixed-mass, rigid-vessel volume fraction: $V_A/V_{total}=v_1/v_2$. Isolated-system entropy generation: $s_{gen}=s_2-s_1$ (no entropy transfer terms, since $Q=0$ for the whole vessel).

Why this works. Defining the system as the entire two-compartment vessel (both the initially-filled and initially-evacuated sides together) makes it a closed, rigid, insulated system: no mass, heat, or work crosses its outer boundary at any point during the process, so the first law reduces to $\Delta U=0$ regardless of how violently irreversible the internal process is.

Common pitfall. Confusing this uncontrolled expansion with throttling (Joule-Thomson) through a valve fed from an external line, which conserves enthalpy ($h_2=h_1$) because flow work is done on the fluid crossing the valve boundary from the supply line. Here there is no external line — both compartments are part of the same closed system, so no flow-work term arises and internal energy (not enthalpy) is what is conserved.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 (Energy Analysis of Closed Systems § Uncontrolled Expansion Examples).

Question 9 of 9: Uncontrolled Expansion into an Evacuated Compartment

Question 9: Uncontrolled Expansion into an Evacuated Compartment (15 marks)

Approach