Two-Phase Water Expansion, Pv = Constant: 04-BS-10, December 2018, Q7

Question 7: Two-Phase Water Expansion, 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. $m=5$ kg, $P_1=500$ kPa, $x_1=0.98$. Expands to $P_2=150$ kPa with $Pv=\text{const}$ (polytropic, $n=1$).

Find. (a) $W$ [kJ]; (b) $Q$ [kJ].

Fig. Q7 — The $Pv=\text{const}$ (hyperbolic) expansion path from a wet state ($v_1=0.3673$ m³/kg) to a state at 150 kPa where $v_2=1.2244$ m³/kg exceeds $v_g(150\ \text{kPa})=1.1593$ m³/kg, so state 2 is actually superheated, not wet.

Approach

Fix state 1 from the given quality, use the process relation to get $v_2$ directly (no iteration needed since $n=1$ makes $v_2=P_1v_1/P_2$ explicit), check whether state 2 lands inside or outside the dome at 150 kPa, and evaluate $u_2$ accordingly. The polytropic work integral for $n=1$ is the logarithmic (isothermal-form) expression, and $Q$ follows from the closed-system first law.

State 1. At 500 kPa, $x_1=0.98$: $v_f=0.001093$, $v_g=0.3748$ m³/kg $\Rightarrow v_1=v_f+x_1(v_g-v_f)=\boxed{0.3673\ \text{m}^3/\text{kg}}$; $u_f=639.54$, $u_{fg}=1921.17$ kJ/kg $\Rightarrow u_1=u_f+x_1 u_{fg}=2522.3$ kJ/kg.
State 2 from $Pv=\text{const}$. $$v_2=\frac{P_1v_1}{P_2}=\frac{500\times0.3673}{150}=\boxed{1.2244\ \text{m}^3/\text{kg}}.$$ At 150 kPa, $v_g=1.1593$ m³/kg $< v_2$, so state 2 is superheated (not wet): this specific volume at 150 kPa corresponds to $T_2\approx 131.1\ ^\circ$C, $u_2=2550.7$ kJ/kg.
Work (part a). For $Pv=\text{const}$ ($n=1$), the boundary-work integral takes the isothermal (logarithmic) form: $$w=\int_1^2 P\,dv=P_1v_1\ln\frac{v_2}{v_1}=500\times0.3673\times\ln\frac{1.2244}{0.3673} =183.65\times1.204=221.13\ \text{kJ/kg}$$ $$W=m\,w=5\times221.13=\boxed{1105.6\ \text{kJ}}.$$
Heat transfer (part b). Closed-system first law: $$q=(u_2-u_1)+w=(2550.7-2522.3)+221.13=28.4+221.13=249.60\ \text{kJ/kg}$$ $$Q=m\,q=5\times249.60=\boxed{1248.0\ \text{kJ}}.$$

Quantity	Result
(a) $W$	1105.6 kJ
(b) $Q$	1248.0 kJ

Concept & Theory

Topic. Polytropic ($n=1$, $Pv=\text{const}$) quasi-equilibrium expansion of a two-phase mixture that crosses out of the dome into the superheated region partway through.

Key relations. $Pv=\text{const}\Rightarrow v_2=P_1v_1/P_2$ explicitly (no iteration, unlike a general polytropic index). Boundary work for $n=1$: $w=P_1v_1\ln(v_2/v_1)$ (the logarithmic form, distinct from the $n\ne1$ power-law integral). Closed system first law: $q=\Delta u+w$.

Why this works. $Pv^n=\text{const}$ with $n=1$ is mathematically identical in form to an isothermal ideal-gas process, so the same $\ln(v_2/v_1)$ integral applies even though water is nowhere close to an ideal gas here — the result follows purely from the stated $P$-$v$ relationship, not from any ideal-gas assumption.

Common pitfall. Assuming state 2 is still a wet two-phase mixture just because state 1 was — always check $v_2$ against $v_g$ at the new pressure; here $v_2=1.2244$ exceeds $v_g(150\ \text{kPa})=1.1593$ m³/kg, so $u_2$ must come from the superheated table (or an equation of state), not from a quality interpolation.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 (Moving Boundary Work, Polytropic Process).

Question 7 of 9: Two-Phase Water Expansion, Pv = Constant

Question 7: Two-Phase Water Expansion, Pv = Constant (15 marks)

Approach