Two-Phase H₂O, Polytropic ($Pv=\text{const}$) Expansion: 04-BS-10, May 2014, Q8

Question 8: Two-Phase H₂O, Polytropic ($Pv=\text{const}$) Expansion (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $P_1=500$ kPa, $x_1=0.98$, $P_2=150$ kPa, process law $Pv=\text{const}$ (polytropic, $n=1$).

Find. (a) $w$ [kJ/kg]; (b) $q$ [kJ/kg].

Fig. Q8 — $P$–$v$ path for the quasi-equilibrium $Pv=\text{const}$ expansion (state 1 two-phase at 500 kPa; state 2, at 150 kPa, works out slightly into the superheated region — see Step 2).

Approach

Fix state 1 from the given quality, then use $P_1v_1=P_2v_2$ (the stated process law) to get $v_2$ directly. Check whether $v_2$ lands inside the two-phase dome at $P_2$ or past it into superheated steam before pulling $u_2$. Work follows the closed-form $\int P\,dv$ integral for a $Pv=\text{const}$ process; heat transfer follows from the first law.

State 1. Two-phase at 500 kPa, $x_1=0.98$: $$v_1=0.3673\ \text{m}^3/\text{kg},\qquad u_1=2522.28\ \text{kJ/kg}.$$
State 2, from the process law. $P_1v_1=P_2v_2\Rightarrow v_2=P_1v_1/P_2=500\times0.3673/150=1.2244\ \text{m}^3/\text{kg}$. At 150 kPa, $v_g=1.1593\ \text{m}^3/\text{kg}$; since $v_2>v_g$, state 2 is (just) superheated, not two-phase. Matching $v_2$ on the 150 kPa isobar: $$T_2=131.13\ ^\circ\text{C},\qquad u_2=2550.75\ \text{kJ/kg}.$$
Work (part a). For $Pv=C=P_1v_1$: $w=\int_1^2 P\,dv=P_1v_1\ln(v_2/v_1)$: $$w=500\times0.3673\times\ln\!\left(\frac{1.2244}{0.3673}\right)=\boxed{221.13\ \text{kJ/kg}}.$$
Heat transfer (part b). First law, closed system: $q=\Delta u+w$: $$q=(u_2-u_1)+w=(2550.75-2522.28)+221.13=\boxed{249.60\ \text{kJ/kg}}.$$

Quantity	Result
(a) $w$	221.13 kJ/kg
(b) $q$	249.60 kJ/kg

Concept & Theory

Topic. Quasi-equilibrium polytropic ($n=1$) expansion of H₂O across the saturation boundary.

Key relations. $Pv=\text{const}$ (n=1) boundary work: $w=P_1v_1\ln(v_2/v_1)$ (the $n=1$ special case of the general $\int P\,dv$ result, which would otherwise diverge for a log-free power-law form). First law, closed system: $q=\Delta u+w$.

Why this works. The process-law constraint $Pv=\text{const}$ fixes $v_2$ from $P_2$ alone without needing to know in advance whether the endpoint lies inside or outside the two-phase dome — the phase check is a separate step performed after $v_2$ is known, not an assumption baked into the process law itself.

Common pitfall. Assuming state 2 must be two-phase because state 1 is, or using the two-phase quality formula $x_2=(v_2-v_f)/(v_g-v_f)$ without first confirming $v_2

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

Question 8 of 9: Two-Phase H₂O, Polytropic ($Pv=\text{const}$) Expansion

Question 8: Two-Phase H₂O, Polytropic ($Pv=\text{const}$) Expansion (15 marks)

Approach