Insulated Vessel with Evacuated Compartment — Free Expansion of Steam: 04-BS-10, May 2014, Q9

Question 9: Insulated Vessel with Evacuated Compartment — Free Expansion of Steam (15 marks)

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Given. Rigid, insulated, two-compartment vessel. Initial (filled) compartment: $P_1=1.0$ MPa, $T_1=500\ ^\circ$C. Other compartment initially evacuated. Final state (after the valve opens and steam fills the whole vessel): $P_2=0.1$ MPa.

Find. (a) $T_2$ [$^\circ$C]; (b) $v_1/v_2$ (initial volume fraction, %); (c) $s_{gen}$ [kJ/kg·K].

Approach

This is an unrestrained (free) expansion of a fixed mass of steam into a rigid, insulated container: no work crosses the overall boundary ($W=0$, rigid) and no heat crosses it ($Q=0$, insulated), so the first law for this closed system collapses to $u_2=u_1$ exactly. Fix state 1, match $u_2=u_1$ at the given final pressure to get $T_2$, then compute the volume ratio and entropy generation from the two fully-defined states.

State 1. $P_1=1.0$ MPa, $T_1=500\ ^\circ$C (superheated): $$u_1=3124.99\ \text{kJ/kg},\qquad v_1=0.3541\ \text{m}^3/\text{kg},\qquad s_1=7.7641\ \text{kJ/kg}\cdot\text{K}.$$
Energy balance ⇒ final temperature (part a). Rigid ($W=0$), insulated ($Q=0$), closed system (fixed total mass, no flow terms): $$u_2=u_1=3124.99\ \text{kJ/kg}.$$ Matching this internal energy on the $P_2=0.1$ MPa isobar (superheated, since $u_2$ is far above the saturated-vapor value at 0.1 MPa): $$T_2=\boxed{495.69\ ^\circ\text{C}}.$$ (Real steam is not a perfect ideal gas, so $T_2$ comes out slightly below $T_1$ rather than exactly equal to it, unlike the ideal-gas limit of this classic "Joule free expansion" problem.)
Volume fraction initially occupied by steam (part b). The same fixed mass occupies $V_1$ (one compartment) initially and the full vessel volume $V_2$ finally, so $V_1/V_2=v_1/v_2$. Reading $v_2$ at the now-known final state ($P_2=0.1$ MPa, $T_2=495.69\ ^\circ$C): $$v_2=3.5456\ \text{m}^3/\text{kg}.$$ $$\frac{V_1}{V_2}=\frac{v_1}{v_2}=\frac{0.3541}{3.5456}=\boxed{0.0999\ (9.99\%)}.$$
Entropy generation (part c). Since the process is adiabatic ($Q=0$), the entire entropy increase of the system is generated internally — there is no entropy-transfer term to subtract: $$s_2=8.8242\ \text{kJ/kg}\cdot\text{K}\ \ (\text{at }P_2=0.1\ \text{MPa},\,T_2=495.69\ ^\circ\text{C}).$$ $$s_{gen}=s_2-s_1=8.8242-7.7641=\boxed{1.0601\ \text{kJ/kg}\cdot\text{K}}.$$

Quantity	Result
(a) $T_2$	495.69°C
(b) Initial steam volume fraction	9.99%
(c) $s_{gen}$	1.0601 kJ/kg·K

Concept & Theory

Topic. Unrestrained (free) expansion of a fixed mass into an evacuated compartment — a rigid, insulated closed system.

Key relations. First law with $Q=W=0$: $u_2=u_1$. Volume conservation with fixed mass: $V_1/V_2=v_1/v_2$. For an adiabatic process, $s_{gen}=\Delta s_{system}$ directly (no heat-transfer entropy term).

Why this works. Free expansion is the canonical example of an internally irreversible adiabatic process: because the steam does no work on anything (it expands into a vacuum, pushing against zero back-pressure) and exchanges no heat, energy conservation forces $u_2=u_1$ regardless of how irreversible the expansion is — all of that irreversibility instead shows up as a jump in entropy, computed independently from $u_2=u_1$ once $T_2$ and hence $s_2$ are known.

Common pitfall. Assuming $T_2=T_1$ by analogy with the ideal-gas free-expansion result ($T$ constant because $u=u(T)$ only for an ideal gas). Real steam's internal energy also depends on pressure/volume, so matching $u_2=u_1$ at the much lower final pressure yields a $T_2$ that is close to, but measurably below, $T_1$.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy).

Question 9 of 9: Insulated Vessel with Evacuated Compartment — Free Expansion of Steam

Question 9: Insulated Vessel with Evacuated Compartment — Free Expansion of Steam (15 marks)

Approach