25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016

Question 1 of 8: Stepped Piston & Isothermal Steam Expansion

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations, May 2016 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 1: Stepped Piston & Isothermal Steam Expansion

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.

(a) Stepped-piston force balance

Given. A vertical stepped piston connects cylinder A (top, wide) and cylinder B (bottom, narrow), with the annular step exposed to atmospheric air in the gap between the cylinders.

Given data
Quantity	Symbol	Value
Diameter, cylinder A	$D_A$	100 mm
Diameter, cylinder B	$D_B$	25 mm
Piston mass	$m$	10 kg
Gas A pressure	$P_A$	200 kPa
Atmospheric air	$P_0$	100 kPa

Find. The gas pressure $P_B$ inside cylinder B.

Stepped piston: gas A pushes down on the 100 mm face, atmosphere pushes up on the annular step, gas B pushes up on the 25 mm face.

Approach. Write a vertical force balance on the piston, with atmospheric air acting on the exposed annular shoulder ($A_A-A_B$) between the two cylinder bores.

Piston cross-sectional areas. $A_A=\dfrac{\pi}{4}D_A^2=\dfrac{\pi}{4}(0.100)^2=7.854\times10^{-3}\text{ m}^2$, $A_B=\dfrac{\pi}{4}D_B^2=\dfrac{\pi}{4}(0.025)^2=4.909\times10^{-4}\text{ m}^2$.
Force balance (downward positive). Gas A and the piston weight push down; atmosphere on the shoulder and gas B push up: $$P_A A_A + mg = P_0(A_A-A_B) + P_B A_B$$ Substituting $mg=10(9.81)=98.1\text{ N}$: $200{,}000(7.854\times10^{-3})+98.1 = 100{,}000(7.363\times10^{-3})+P_B(4.909\times10^{-4})$.
Solve for $P_B$. $1668.9\text{ N} = 736.3\text{ N} + P_B(4.909\times10^{-4})$, so $P_B=\dfrac{932.6}{4.909\times10^{-4}}$. $$\boxed{P_B \approx 1899.8\text{ kPa} \approx 1.90\text{ MPa}}$$

(b) Isothermal expansion of saturated steam

Given. Saturated water vapour at 200°C expands isothermally and quasi-statically from $V_1=0.01\text{ m}^3$ down to $P_2=200\text{ kPa}$.

Steam properties along the $T=200^\circ\text{C}$ isotherm (Sonntag & Borgnakke Table B.1.1/B.1.3)
$P$ (kPa)	$v$ (m³/kg)	State
1554.9 (sat.)	0.12736	1 — initial, saturated vapour
800	0.26088	intermediate
400	0.53434	intermediate
200	1.08034	2 — final, superheated

Find. The actual work done, and the percent error from treating the steam as an ideal gas.

Approach. Read $v_1$ (saturated vapour) and the $P$–$v$ points along the $200^\circ\text{C}$ isotherm from the tables, get the mass from $m=V_1/v_1$, then numerically integrate $w=\int P\,dv$ (real fluid, not ideal gas) and compare with the ideal-gas closed form.

Mass in the cylinder. $v_1=0.12736\text{ m}^3/\text{kg}$ (sat. vapour at 200°C), so $$m=\frac{V_1}{v_1}=\frac{0.01}{0.12736}=0.07852\text{ kg}$$
Actual work (real steam). Table pairs along the isotherm are integrated segment-by-segment, treating $Pv$ as linear in $v$ within each tabulated interval ($P=\frac{a+bv}{v}\Rightarrow\int P\,dv=a\ln\frac{v_j}{v_i}+b(v_j-v_i)$), summed from the saturated-vapour state (1554.9 kPa) down to 200 kPa. Ten tabulated points give a specific work of $449.1\text{ kJ/kg}$, so $$\boxed{W_{\text{actual}}=m\!\int P\,dv \approx 35.26\text{ kJ}}$$
Ideal-gas comparison. For an isothermal ideal-gas process, $w_{\text{ideal}}=RT\ln(v_2/v_1)$ with $R=0.4615\text{ kJ/kg}\cdot\text{K}$ (steam gas constant) and $T=473.15\text{ K}$: $$w_{\text{ideal}}=0.4615(473.15)\ln\!\left(\frac{1.08034}{0.12736}\right)=167.9\text{ kJ/kg}\quad\Rightarrow\quad W_{\text{ideal}}=m\,w_{\text{ideal}}\approx 36.66\text{ kJ}$$
Percent error. $$\text{error}=\frac{W_{\text{ideal}}-W_{\text{actual}}}{W_{\text{actual}}}\times100\%=\frac{36.66-35.26}{35.26}\times100\%$$ $$\boxed{\approx 3.95\%\text{ (ideal-gas model overpredicts the work)}}$$

Question 1 — final results
Quantity	Value
Gas pressure in cylinder B	$P_B \approx 1899.8\text{ kPa}$ (1.90 MPa)
Mass of steam	$m = 0.0785\text{ kg}$
Actual work of expansion	$W_{\text{actual}} \approx 35.26\text{ kJ}$
Ideal-gas work estimate	$W_{\text{ideal}} \approx 36.66\text{ kJ}$
Error from ideal-gas assumption	$\approx 3.95\%$

Topic: Piston force balances and real-fluid vs. ideal-gas $P$-$v$-$T$ behaviour.

Key relations: vertical force balance $\sum F=0$ on a stepped piston; $w=\int P\,dv$ for a quasi-equilibrium process; ideal-gas isothermal work $w=RT\ln(v_2/v_1)$.

Why this works: A piston in mechanical equilibrium transmits pressure forces through every distinct area it presents to a fluid or the atmosphere — a stepped geometry simply adds an extra term for the exposed shoulder. For the steam expansion, because $w=\int P\,dv$ depends only on the process path in $P$-$v$ space, tabulated real-fluid data must be integrated numerically; steam near 1.5–0.2 MPa and 200°C is far from the ideal-gas limit (small compressibility factor deviation), so the ideal-gas formula measurably overpredicts the work.

Common pitfall: forgetting the atmospheric force on the exposed shoulder area (treating the piston as if $P_0$ acts over the full $A_A$, or not at all); assuming steam behaves ideally at these moderate pressures without checking against the real $P$-$v$-$T$ data.

Source: Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics, Ch.2 (energy balance and work) & Table B.1; Cengel & Boles, Thermodynamics: An Engineering Approach, Ch.3.

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