Spring-Loaded Piston Heating and Adiabatic Water Pumping: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q1

Question 1: Spring-Loaded Piston Heating and Adiabatic Water Pumping (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.

Part (a) — heat added against a linear spring

Quantity	Value
Initial pressure / temperature	$P_1=1$ atm $=101.3$ kPa, $T_1=20$ °C $=293.15$ K
Initial volume	$V_1=20$ cm³ $=20\times10^{-6}$ m³
Piston area (adopted)	$A=4$ cm² $=4\times10^{-4}$ m²
Spring constant	$k=100$ N/cm $=1\times10^{4}$ N/m
Final pressure	$P_2=3$ atm $=303.98$ kPa
Air properties (cold-air standard)	$R=287$ J/kg·K, $c_v=718$ J/kg·K

Figure 1a — As heat is added the air expands, the piston rises a distance $x=\Delta V/A$, and the spring force $kx$ makes the pressure rise linearly with volume from 1 atm to 3 atm.

Approach. The initially undeformed spring means $P_1=P_o$; as the gas expands the spring force makes $P$ increase linearly with $V$, so the boundary work is the area under a straight $P$–$V$ line and the heat follows from the closed-system energy balance $Q=\Delta U+W$.

Relate pressure rise to volume change. A displacement $x$ compresses the spring by $kx$ and lifts the pressure by $kx/A$, so with $x=\Delta V/A$, $\;P-P_1=\dfrac{k\,\Delta V}{A^{2}}$. Solving for the volume change to reach $P_2$: $$\Delta V=\frac{(P_2-P_1)\,A^{2}}{k}=\frac{(202650)(4\times10^{-4})^{2}}{1\times10^{4}}=3.24\times10^{-6}\ \text{m}^3=3.24\ \text{cm}^3$$ so $V_2=V_1+\Delta V=23.24$ cm³.
Find the final temperature (ideal gas). With $P_1V_1/T_1=P_2V_2/T_2$, $$T_2=T_1\frac{P_2V_2}{P_1V_1}=293.15\times\frac{(3)(23.24)}{(1)(20)}=1022\ \text{K}$$
Mass of air. $m=\dfrac{P_1V_1}{RT_1}=\dfrac{(101325)(20\times10^{-6})}{(287)(293.15)}=2.41\times10^{-5}$ kg.
Boundary work (straight $P$–$V$ line). The work is the trapezoidal area under the process line: $$W=\frac{P_1+P_2}{2}\,\Delta V=\frac{(101325+303975)}{2}(3.24\times10^{-6})=0.657\ \text{J}$$
Energy balance. $\Delta U=mc_v(T_2-T_1)=(2.41\times10^{-5})(718)(1022-293.15)=12.6$ J, hence $$Q=\Delta U+W=12.6+0.66$$ ==**$Q\approx13.3$ J of heat must be added.**==

Check — piston area reading.

The paper prints the area as "0.04 cm²", but with $V_1=20$ cm³ that implies a 5 m long cylinder and renders the spring negligible (the spring effect scales as $A^{2}$). The self-consistent reading is $A=4$ cm² (a 5 cm long cylinder), which makes the spring the point of the problem; that value is adopted. With the literal 0.04 cm² the process is essentially constant-volume and $Q\approx10.1$ J.

Part (b) — adiabatic pumping power

Quantity	Value
Feed temperature	$T_1=40$ °C
Pressures	$p_1=6.8$ atm, $p_2=68.0$ atm ($\Delta p=61.2$ atm $=6.201$ MPa)
Mass flow	$\dot m=35$ kg/min $=0.5833$ kg/s
Specific volume	$v=v_f(40\text{ °C})=0.001008$ m³/kg

Approach. For an adiabatic, reversible, steady-flow pump the shaft work equals $\int v\,dp$; because the liquid is essentially incompressible $v$ is constant and the integral collapses to $v\,\Delta p$.

Steady-flow work of a reversible pump. Neglecting kinetic and potential changes, the shaft work per unit mass is $w=\displaystyle\int_{1}^{2} v\,dp$. Treating the water as incompressible ($v=$ const) gives $$w=v\,(p_2-p_1)\qquad\Rightarrow\qquad \boxed{\dot W=\dot m\,v\,(p_2-p_1)}$$ which is the requested relationship.
Substitute the data. With $\Delta p=61.2\times101325=6.201\times10^{6}$ Pa, $$w=(0.001008)(6.201\times10^{6})=6250\ \text{J/kg}$$ $$\dot W=\dot m\,w=(0.5833)(6250)$$ ==**$\dot W\approx3646$ W ($\approx3.65$ kW)** using the given data.==

Check — reconciling the 3772 W target.

Using $v_f(40\text{ °C})=0.001008$ m³/kg and 1 atm $=101.325$ kPa, the printed pressures give $\dot W=3646$ W. The exam's quoted 3772 W is recovered if the feed pressures are 7.0 → 70.0 atm ($\Delta p=63$ atm, close to 6.8/68.0), which yields $\dot W\approx3753$ W, or with a marginally larger $v$. Either way the pumping power is ≈ 3.6–3.8 kW; the derived relation $\dot W=\dot m v\,\Delta p$ is the graded content.

Quantity	Result
(a) Volume change / final $T$	$\Delta V=3.24$ cm³, $T_2=1022$ K
(a) Boundary work	0.66 J
(a) Heat added	≈ 13.3 J
(b) Pump specific work	≈ 6250 J/kg
(b) Pumping power	≈ 3.65 kW (exam: 3772 W)

Concept & Theory

Topic. Closed-system heating against a linear spring (part a) and reversible steady-flow pump work on an incompressible liquid (part b).

Key relations. Linear-spring pressure $P=P_1+k\,\Delta V/A^{2}$; boundary work $W=\int P\,dV=\tfrac12(P_1+P_2)\Delta V$; closed-system $Q=\Delta U+W$ with $\Delta U=mc_v\Delta T$; pump work $w=\int v\,dp\to v\,\Delta p$ for $v=$ const.

Why this works. A spring attached to the piston makes the resisting pressure a linear function of displacement, so the $P$–$V$ path is a straight line and its area (the work) is a simple trapezoid; the ideal-gas law then fixes $T_2$ and the energy balance the heat. For the pump, the reversible steady-flow work is $\int v\,dp$ (not $\int P\,dV$), and because a liquid's specific volume barely changes with pressure the integral is just $v\,\Delta p$.

Common pitfall. Using $\int P\,dV$ instead of $\int v\,dp$ for the pump (they differ for open systems), or forgetting that a small piston area makes the spring term $k\Delta V/A^{2}$ enormous — the area must be read consistently with the stated volume.

Source. Çengel & Boles, Thermodynamics §4 (closed-system boundary work, spring-loaded pistons) and §7 (reversible steady-flow work, pumps).