Two-Phase Piston-Cylinder — Lift-off Temperature and Heat Added: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2017, Q1

Question 1: Two-Phase Piston-Cylinder — Lift-off Temperature and Heat Added

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. A rigid-volume, two-phase (liquid+vapour) mass of water sits below a piston that is initially resting on stops; heat is added slowly at constant volume until the in-cylinder pressure is large enough to lift the piston, after which the process would continue at constant pressure — but only the lift-off point is asked for here.

Given data
Quantity	Symbol	Value
Mass of water	$m$	0.130 kg
Initial temperature	$T_1$	$40\,{}^{\circ}\text{C}$
Enclosed volume (piston on stops)	$V_1$	$0.027\text{ m}^3$
Piston face area	$A$	$0.0347\text{ m}^2$
Piston mass	$m_p$	100 kg
Local gravitational acceleration	$g$	$9.41\text{ m/s}^2$
Atmospheric pressure	$P_{atm}$	93.7 kPa

Find. (a) The temperature $T_2$ at which the piston starts to lift off the stops; (b) the heat $Q$ added up to that point.

Piston-cylinder device: water (liquid+vapour) below a piston resting on stops, heated from below; the piston is held down only by its own weight plus atmospheric pressure acting from above.

Approach. Fix state 1 from $T_1$ and the specific volume $v_1=V_1/m$ (a two-phase state, since $v_1$ lies between $v_f$ and $v_g$ at $40\,{}^{\circ}\text{C}$). While the piston sits on the stops the volume cannot change, so heating from state 1 to the lift-off state 2 is a constant-volume process; state 2's pressure is fixed by a force balance on the piston, and its temperature follows because $v_2=v_1$ still lies inside the dome at that pressure. The heat added is then just $Q=m(u_2-u_1)$ since $W=0$ throughout (piston stationary the whole time).

State 1 — specific volume and quality. $$v_1=\frac{V_1}{m}=\frac{0.027}{0.130}=0.2077\text{ m}^3/\text{kg}$$ At $40\,{}^{\circ}\text{C}$: $v_f=0.001008\text{ m}^3/\text{kg}$, $v_g=19.515\text{ m}^3/\text{kg}$, and since $v_f\lt v_1\lt v_g$ the water is a saturated liquid–vapour mixture with $$x_1=\frac{v_1-v_f}{v_g-v_f}=\frac{0.2077-0.001008}{19.515-0.001008}=0.01059$$ so $u_1=u_f+x_1u_{fg}=191.48\text{ kJ/kg}$. Because state 1 is two-phase, its pressure is simply $P_{sat}(40\,{}^{\circ}\text{C})=7.385\text{ kPa}$ — far below what is needed to lift the piston, confirming the piston really is resting on the stops at the start.
Lift-off pressure — force balance on the piston. The piston lifts once the pressure below it can support both its own weight and the atmosphere pressing down from above: $$\begin{aligned} P_2A&=P_{atm}A+m_pg \\ P_2&=P_{atm}+\frac{m_pg}{A}=93.7+\frac{100\times9.41}{0.0347\times1000}=93.7+27.12 \end{aligned}$$ $$\boxed{P_2=120.82\text{ kPa}}$$
State 2 — temperature at lift-off, part (a). Volume is unchanged while the piston is stationary, so $v_2=v_1=0.2077\text{ m}^3/\text{kg}$. At $P_2=120.82\text{ kPa}$, $v_g=1.4193\text{ m}^3/\text{kg}$, which is still well above $v_2$, so state 2 is also two-phase and its temperature is simply the saturation temperature at $P_2$: $$\boxed{T_2=T_{sat}(120.82\text{ kPa})=105.0\,{}^{\circ}\text{C}}$$ At $P_2$: $v_f=0.001047\text{ m}^3/\text{kg}$, giving $x_2=(v_1-v_f)/(v_g-v_f)=(0.2077-0.001047)/(1.4193-0.001047)=0.1457$ and $u_2=u_f+x_2u_{fg}=741.93\text{ kJ/kg}$.
Heat added — part (b). The process 1→2 is at constant volume (piston stationary throughout), so the boundary work is zero and the closed-system first law reduces to $Q=\Delta U$: $$Q=m(u_2-u_1)=0.130\times(741.93-191.48)=0.130\times550.45$$ $$\boxed{Q=71.56\text{ kJ}}$$

Question 1 — results
Quantity	Value
Lift-off pressure $P_2$	120.82 kPa
(a) Lift-off temperature $T_2$	$105.0\,{}^{\circ}\text{C}$
Quality at lift-off $x_2$	0.1457
(b) Heat added $Q$	71.56 kJ

Topic. Two-phase closed-system heating against piston stops (constant-volume then constant-pressure staging).

Key relations. $v=V/m$; two-phase quality $x=(v-v_f)/(v_g-v_f)$; piston force balance $P_{lift}=P_{atm}+m_pg/A$; closed-system first law $Q=\Delta U+W$, with $W=0$ at constant volume.

Why this works. As long as the in-cylinder pressure is below the lift-off value, the piston's weight plus the atmosphere above it holds it firmly on the stops, so the enclosed volume literally cannot change — the process is constant-volume by construction, not by assumption. Once the pressure reaches $P_{lift}$ the piston becomes a free body in equilibrium and any further heat addition raises quality at constant pressure instead; the question only asks up to that transition, so the simple $Q=\Delta U$ form applies without ever needing the constant-pressure stage.

Common pitfall. Assuming state 1's pressure already equals the lift-off pressure (and trying to iterate on it) instead of recognising state 1 is an independent two-phase state fixed by $T_1$ and $v_1$ alone; also failing to check that $v_2$ still lies below $v_g$ at $P_2$, which is what confirms state 2 is two-phase (not superheated) and lets $T_2$ be read directly as a saturation temperature.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 (energy analysis of closed systems, moving-boundary work).

Question 1 of 8: Two-Phase Piston-Cylinder — Lift-off Temperature and Heat Added

Question 1: Two-Phase Piston-Cylinder — Lift-off Temperature and Heat Added