Two-Phase Expansion with Pv = Constant: 04-BS-10, May 2016, Q8

Question 8: Two-Phase Expansion with Pv = Constant (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. Initial: $P_1=500$ kPa, $x_1=0.98$ (two-phase). Final: $P_2=150$ kPa. Process: $Pv=\text{constant}$ (polytropic, $n=1$).

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

Approach

Fix the initial specific volume and internal energy from the given quality, then use the $Pv=C$ process relation to locate the final specific volume at the given final pressure. For an $n=1$ polytropic process the boundary work integrates to $w=C\ln(v_2/v_1)=P_1v_1\ln(v_2/v_1)$. Checking the final specific volume against the saturated-vapor value at 150 kPa determines whether the final state is still two-phase or has crossed into superheated vapor, which is needed to evaluate $u_2$ for the energy balance.

Initial state. At 500 kPa, $x_1=0.98$: $v_f=0.001093$, $v_g=0.3749$ m$^3$/kg, so $$v_1=v_f+x_1(v_g-v_f)=0.001093+0.98\times0.37381=\boxed{0.36733\ \text{m}^3/\text{kg}}.$$ $u_1=u_f+x_1u_{fg}=2522.28$ kJ/kg.
Final specific volume from $Pv=C$. $$C=P_1v_1=500\times0.36733=183.67\ \text{kPa}\cdot\text{m}^3/\text{kg}.$$ $$v_2=\frac{C}{P_2}=\frac{183.67}{150}=\boxed{1.2244\ \text{m}^3/\text{kg}}.$$ Since $v_g(150\ \text{kPa})=1.1593$ m$^3$/kg $
Boundary work (part a). For $Pv=C$ ($n=1$ polytropic), $w=\int P\,dv=C\ln(v_2/v_1)$: $$w=183.67\times\ln\!\left(\frac{1.2244}{0.36733}\right)=183.67\times1.2040 =\boxed{221.13\ \text{kJ/kg}}.$$
Heat transfer (part b). First-law energy balance, closed system: $$q=(u_2-u_1)+w=(2550.75-2522.28)+221.13=28.47+221.13=\boxed{249.60\ \text{kJ/kg}}.$$

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

Concept & Theory

Topic. Polytropic ($n=1$, $Pv=$const) boundary-work expansion starting from a two-phase state, closed piston-cylinder system.

Key relations. Boundary work for $Pv=C$: $w=C\ln(v_2/v_1)$ (the $n=1$ special case of the general polytropic work integral, which would otherwise be $w=(P_1v_1-P_2v_2)/(n-1)$). Closed-system energy balance: $q=\Delta u+w$.

Why this works. $Pv=$const is mathematically identical in form to an isothermal ideal-gas process, but here the working fluid is a real two-phase/superheated substance, so $v_2$ must be found from the process relation first and then the corresponding property state (two-phase or superheated) looked up independently — there is no shortcut formula linking $T$ to $v$ the way there is for an ideal gas.

Common pitfall. Assuming the final state stays two-phase just because the initial state was — here $v_2=1.2244$ m$^3$/kg exceeds $v_g$ at 150 kPa, so the substance has fully vaporized and become superheated partway through the expansion; using a two-phase quality formula at state 2 would be invalid.

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

Question 8 of 9: Two-Phase Expansion with Pv = Constant

Question 8: Two-Phase Expansion with Pv = Constant (15 marks)

Approach