Water in a piston–cylinder resting on stops: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q1

Question 1 — Water in a piston–cylinder resting on stops (Part A, equal value)

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, stop-supported piston–cylinder holds a fixed mass of wet water; the data are collected below. Find. (a) the water temperature at which the piston just lifts off the stops; (b) the heat added up to that instant.

Quantity	Value
Water mass, $m$	0.130 kg
Initial temperature, $T_1$	40 °C
Enclosed volume (piston on stops), $V_1$	0.027 m³
Piston face area, $A$	0.0347 m²
Piston mass, $m_p$	100 kg
Local gravity, $g$	9.41 m/s²
Atmospheric pressure, $P_\text{atm}$	93.7 kPa

Figure 1 — While the piston rests on the stops the volume is fixed at $V_1$; heating raises the pressure at constant volume until it reaches the value needed to support the piston.

Approach. While the piston sits on the stops the volume is constant, so heating is a constant-volume process; the piston lifts when the internal pressure reaches the piston-support pressure, and the heat added is the closed-system internal-energy change.

Pressure required to lift the piston. A force balance on the piston (atmosphere plus weight above, water pressure below) gives $$P_2 = P_\text{atm} + \frac{m_p\,g}{A} = 93.7 + \frac{(100)(9.41)}{0.0347\times10^{3}} = 93.7 + 27.1 \;\text{kPa}$$ $P_2 = 120.8$ kPa
Specific volume is fixed (constant-volume heating). The mixture keeps the same specific volume from state 1 to the lift point: $$v_2 = v_1 = \frac{V_1}{m} = \frac{0.027}{0.130} = 0.2077\ \text{m}^3/\text{kg}$$
Lift-off temperature = saturation temperature at $P_2$. At $P_2 = 120.8$ kPa the saturated-vapour specific volume ($v_g \approx 1.419\ \text{m}^3/\text{kg}$) exceeds $v_2$, so the water is still a wet mixture; its temperature is therefore the saturation temperature at $P_2$: $T_2 = T_\text{sat}(120.8\ \text{kPa}) \approx 105\ ^\circ\text{C}$ This answers part (a).
State-1 internal energy. At 40 °C, $v_f = 0.001008$, $v_g = 19.52\ \text{m}^3/\text{kg}$, so the quality is $$x_1 = \frac{v_1 - v_f}{v_g - v_f} = \frac{0.2077-0.001008}{19.52-0.001008} = 0.0106$$ $$u_1 = u_f + x_1 u_{fg} = 167.5 + (0.0106)(2261.9) = 191.5\ \text{kJ/kg}$$
State-2 internal energy. At $P_2 = 120.8$ kPa, $v_f = 0.001047$, $v_g = 1.419\ \text{m}^3/\text{kg}$: $$x_2 = \frac{0.2077-0.001047}{1.419-0.001047} = 0.146,\qquad u_2 = 440.1 + (0.146)(2071.8) = 741.9\ \text{kJ/kg}$$
Heat added (constant-volume closed system, $W=0$). The first law reduces to $Q = m(u_2-u_1)$: $$Q = 0.130\,(741.9 - 191.5) = 71.6\ \text{kJ}$$ $Q \approx 71.6$ kJ This answers part (b).

Quantity	Result
Lift-off pressure, $P_2$	120.8 kPa
(a) Lift-off temperature, $T_2$	≈ 105 °C (saturation at $P_2$)
State-1 quality / $u_1$	0.0106 / 191.5 kJ/kg
State-2 quality / $u_2$	0.146 / 741.9 kJ/kg
(b) Heat added, $Q$	≈ 71.6 kJ

Concept & Theory

Topic. Two-stage piston–cylinder process with a stop: constant-volume heating of a wet mixture followed by (not reached here) constant-pressure lift.

Key relations. Piston-support pressure $P = P_\text{atm} + m_p g/A$; quality $x=(v-v_f)/(v_g-v_f)$; wet-mixture property $u = u_f + x\,u_{fg}$; closed-system first law $Q = \Delta U + W$ with $W=0$ at constant volume.

Why this works. As long as the piston presses on the stops the gas cannot expand, so the trapped water heats at fixed specific volume and its state slides up a vertical line on the $P$–$v$ diagram. The piston only begins to rise once the water pressure climbs to the value that just balances atmospheric load plus piston weight. Because $v_2 < v_g$ at that pressure the water is still saturated, so its temperature is simply $T_\text{sat}(P_2)$ — no superheat table is needed.

Common pitfall. Assuming the pressure is constant during the first phase (it is not — volume is constant), or forgetting to check that state 2 is still inside the dome before reading $T_2=T_\text{sat}$.

Source. Çengel & Boles, Thermodynamics §3 (property tables, quality) and §4 (energy analysis of closed systems; constant-volume and spring/stop processes).

Question 1 of 8: Water in a piston–cylinder resting on stops

Question 1 — Water in a piston–cylinder resting on stops (Part A, equal value)