Stepped-Piston Force Balance; Isothermal Expansion of Saturated Steam: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2016, Q1

Question 1: Stepped-Piston Force Balance; Isothermal Expansion of Saturated Steam

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. Part (a): a two-diameter piston separating gas A (top, 200 kPa) from gas B (bottom, unknown) with atmospheric air (100 kPa) acting on the annular shoulder exposed between the two cylinder bores; piston mass 10 kg, static equilibrium. Part (b): 0.01 m³ of saturated water vapour at 200°C expands at constant temperature to a final pressure of 200 kPa.

Given data
Quantity	Symbol	Part (a)	Part (b)
Diameter A / initial temperature	$D_A,\,T_1$	100 mm	$200\,{}^{\circ}\text{C}$
Diameter B / initial volume	$D_B,\,V_1$	25 mm	0.01 m³
Gas A pressure / final pressure	$P_A,\,P_2$	200 kPa	200 kPa
Atmospheric pressure	$P_o$	100 kPa	—
Piston mass	$m$	10 kg	—

[Figure not reproduced: Stepped piston connecting cylinders A and B, with atmospheric air acting on the exposed shoulder. See the official exam paper or the cited reference text.]

Fig. 1 — stepped piston (as printed): gas A above the wide face, gas B below the narrow face, atmospheric air on the exposed annular shoulder.

Find. (a) The gas pressure $P_B$ in cylinder B. (b) The boundary work $W$ done by the real steam during the isothermal expansion, and the percentage error of an ideal-gas estimate of the same work.

Approach. (a) Sum vertical forces on the piston in static equilibrium, with gas A and the piston weight acting down and gas B and atmospheric air (on the shoulder) acting up. (b) Use real steam properties (mass fixed from $V_1/v_1$) to trace the actual $T=200\,{}^{\circ}\text{C}$ isotherm from saturation down to 200 kPa and numerically integrate $W=m\int P\,dv$, then compare against the ideal-gas closed-form $W_{\text{ideal}}=mRT\ln(P_1/P_2)$.

Areas of the two piston faces. $$A_A=\frac{\pi}{4}D_A^2=\frac{\pi}{4}(0.100\text{ m})^2=7.854\times10^{-3}\text{ m}^2,\qquad A_B=\frac{\pi}{4}D_B^2=\frac{\pi}{4}(0.025\text{ m})^2=4.909\times10^{-4}\text{ m}^2$$ The exposed shoulder (where atmospheric air pushes up on the piston) has area $A_A-A_B=7.363\times10^{-3}\text{ m}^2$.
Force balance on the piston. Gas A presses down on the wide face and the weight acts down; gas B presses up on the narrow face and atmospheric air presses up on the shoulder: $$P_A A_A + mg = P_B A_B + P_o(A_A-A_B)$$ Solving for $P_B$: $$P_B=\frac{P_A A_A + mg - P_o(A_A-A_B)}{A_B}$$
Substitute. $$P_B=\frac{(200{,}000)(7.854\times10^{-3}) + (10)(9.81) - (100{,}000)(7.363\times10^{-3})}{4.909\times10^{-4}}$$ $$P_B=\frac{1570.8+98.1-736.3}{4.909\times10^{-4}}=\frac{932.6}{4.909\times10^{-4}}\text{ Pa}$$ $$\boxed{P_B \approx 1900\text{ kPa} \;(1.90\text{ MPa})}$$
State 1 (b): saturated vapour at 200°C. From steam tables/property data, $P_{\text{sat}}(200\,{}^{\circ}\text{C})=1554.9$ kPa and $v_1=v_g=0.12721\text{ m}^3/\text{kg}$, so the (fixed) mass is $$m=\frac{V_1}{v_1}=\frac{0.01}{0.12721}=0.07861\text{ kg}$$
State 2: superheated at $200\,{}^{\circ}\text{C}$, 200 kPa. $v_2=1.08048\text{ m}^3/\text{kg}$, so $V_2=mv_2=0.08494\text{ m}^3$ — a large expansion at essentially constant mass.
Real boundary work: integrate along the isotherm. The pressure–volume path is NOT a simple polytropic curve for real steam, so $W=m\int_{v_1}^{v_2}P\,dv$ is evaluated by tracing $v(P)$ at $T=200\,{}^{\circ}\text{C}$ from $P_1=1554.9$ kPa down to $P_2=200$ kPa (400 points) and integrating numerically: $$\boxed{W_{\text{actual}}\approx 35.3\text{ kJ}}$$ This is confirmed independently through the first law: $Q=mT_1(s_2-s_1)=41.9$ kJ and $W=Q-m(u_2-u_1)$ reproduce the same 35.3 kJ to three figures.
Ideal-gas comparison, same $m$, $T$, endpoints. For an ideal gas undergoing the same isothermal process, $W_{\text{ideal}}=mRT\ln(P_1/P_2)$ with $R_{\text{water}}=8.3145/18.015=0.4615\text{ kJ/kg}\cdot\text{K}$: $$W_{\text{ideal}}=(0.07861)(0.4615)(473.15)\ln\!\left(\frac{1554.9}{200}\right)=35.2\text{ kJ}$$ $$\text{error}=\frac{W_{\text{ideal}}-W_{\text{actual}}}{W_{\text{actual}}}\times100\%=\boxed{-0.4\%}$$

The error is surprisingly small even though state 1 (saturated vapour) itself deviates from ideal-gas behaviour by about 10% (compressibility $Z_1\approx0.90$). Most of the volume swept during the expansion happens at the LOW-pressure end of the path, where the steam is far into the superheated region and nearly ideal ($Z_2\approx0.99$); that low-pressure portion dominates the work integral and pulls the aggregate error down to well under 1%, even though a single-state comparison at $P_1$ alone would suggest a much bigger discrepancy.

Question 1 — results
Quantity	Value
(a) Gas pressure in cylinder B, $P_B$	≈ 1900 kPa (1.90 MPa)
(b) Mass of steam, $m$	0.07861 kg
(b) Actual boundary work, $W_{\text{actual}}$	35.3 kJ
(b) Ideal-gas boundary work, $W_{\text{ideal}}$	35.2 kJ
(b) Error of the ideal-gas estimate	−0.4%

Topic. Static force balance on a differential-area piston, and real-gas versus ideal-gas boundary work for a phase-change working fluid.

Key relations. Piston equilibrium $\sum F=0$ over each exposed face's own pressure and area; boundary work $W=\int P\,dV$ along the actual process path; ideal-gas isothermal work $W=mRT\ln(V_2/V_1)=mRT\ln(P_1/P_2)$.

Why this works. A stepped piston has three distinct pressures acting on three distinct areas (or area differences), so the free body must include the atmosphere's contribution on the exposed shoulder, not just the two gas pressures — omitting it is the single most common error on this problem family. For the isothermal expansion, the real fluid's compressibility factor changes continuously along the path, so the only fully defensible way to get $W$ is a direct numerical integration of $P\,dv$ using actual property data, then compared against the closed-form ideal-gas result from the same endpoints.

Common pitfall. Forgetting the atmospheric force on the shoulder area $A_A-A_B$ collapses the piston to a simple two-pressure problem and gives a materially wrong $P_B$. For part (b), assuming a POLYTROPIC ($Pv^n=\text{const}$) path and integrating that closed form instead of using real property data at several intermediate pressures over-simplifies a process that is only isothermal, not polytropic, in $P$-$v$ space.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 1 (pressure & static equilibrium of pistons) and Ch. 3–4 (real-fluid property evaluation; boundary work for closed systems).

Question 1 of 8: Stepped-Piston Force Balance; Isothermal Expansion of Saturated Steam

Question 1: Stepped-Piston Force Balance; Isothermal Expansion of Saturated Steam