22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2016
Question 1 of 8: Stepped-Piston Pressure and Isothermal Steam Expansion
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power cycles, and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — cylindrical composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU heat-exchanger method. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and engine-oil data from standard tables.
Paper format: National Examination 07-Mec-A1, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three from one part, two from the other). All eight questions are solved in full below.
Question 1: Stepped-Piston Pressure and Isothermal Steam Expansion (equal value)
Given. A stepped piston (mass 10 kg) sealing gas A at 200 kPa on a 100 mm face and gas B on a 25 mm face, with the annular step open to atmosphere at 100 kPa; and, separately, 0.01 m³ of saturated steam at 200 °C expanded isothermally to 200 kPa. Find. The gas pressure in cylinder B, and the isothermal work done together with the error incurred by an ideal-gas assumption.
Part (a) — pressure in cylinder B
Quantity
Value
Cylinder A diameter / area
$D_A=0.100$ m, $A_A=7.854\times10^{-3}$ m²
Cylinder B diameter / area
$D_B=0.025$ m, $A_B=4.909\times10^{-4}$ m²
Gas pressure in A
200 kPa
Atmospheric pressure on step
100 kPa
Piston mass
10 kg
Figure 1 — Free body of the stepped piston: gas A presses down on the large face $A_A$; gas B presses up on the small face $A_B$; atmosphere presses up on the annular step $A_A-A_B$; the 10 kg weight acts down.
Approach. A single vertical force balance on the stepped piston relates the two gas pressures, the atmospheric pressure on the exposed annular step, and the piston weight.
Resolve the piston areas. $A_A=\tfrac{\pi}{4}D_A^2=\tfrac{\pi}{4}(0.100)^2=7.854\times10^{-3}\ \text{m}^2$, $A_B=\tfrac{\pi}{4}(0.025)^2=4.909\times10^{-4}\ \text{m}^2$, and the annular step area exposed to the atmosphere is $A_A-A_B=7.363\times10^{-3}\ \text{m}^2$.
Write the vertical equilibrium (up positive). Gas A acts down on the large face, gas B up on the small face, atmosphere up on the annulus, and gravity down on the mass: $$P_B A_B + P_o(A_A-A_B) - P_A A_A - mg = 0.$$
Solve for the gas pressure in B. $$P_B=\frac{P_A A_A + mg - P_o(A_A-A_B)}{A_B}.$$ Substituting $P_A=200\,000$ Pa, $P_o=100\,000$ Pa, $mg=10(9.81)=98.1$ N: $$P_B=\frac{1570.8+98.1-736.3}{4.909\times10^{-4}}=\boxed{1.90\times10^{6}\ \text{Pa}\approx1900\ \text{kPa}}.$$
Physical check. Because cylinder B presents only a small area, a high pressure there is needed to balance the large downward force from the 200 kPa gas acting on the ten-fold larger face A (the weight and the atmospheric step term are secondary), so $P_B\gg P_A$ is expected.
Quantity
Result
Large / small piston area
$7.854\times10^{-3}$ / $4.909\times10^{-4}$ m²
Gas pressure in cylinder B
≈ 1900 kPa (1.90 MPa)
Part (b) — isothermal expansion work and the ideal-gas error
State
Condition
$v$ (m³/kg)
$u$ (kJ/kg)
$s$ (kJ/kg·K)
1 — initial
sat. vapour, 200 °C (1554.9 kPa)
0.12721
2594.2
6.4302
2 — final
200 °C, 200 kPa (superheated)
1.0805
2654.6
7.5081
Approach. The mass follows from the initial saturated-vapour volume; for a reversible isothermal process the heat is $Q=mT\Delta s$, and the boundary work then follows from the closed-system energy balance.
Fix the mass from the initial state. $m=\dfrac{V_1}{v_g(200\,{}^\circ\text{C})}=\dfrac{0.01}{0.12721}=0.0786\ \text{kg}.$
Heat for the reversible isothermal process. With $T=473.15$ K constant, $$Q=mT(s_2-s_1)=0.0786(473.15)(7.5081-6.4302)=40.1\ \text{kJ}.$$
Boundary work from the energy balance. $\Delta U=m(u_2-u_1)=0.0786(60.4)=4.75$ kJ, so $$W=Q-\Delta U=40.1-4.75=\boxed{35.3\ \text{kJ}}.$$
Ideal-gas estimate. Treating the steam as ideal, the isothermal work is $W_\text{id}=P_1V_1\ln(P_1/P_2)=1554.9(0.01)\ln\!\frac{1554.9}{200}=31.9\ \text{kJ}.$
Error of the ideal-gas assumption. $$\text{error}=\frac{W-W_\text{id}}{W}=\frac{35.3-31.9}{35.3}=\boxed{9.8\%}.$$ The shortfall traces directly to the compressibility of saturated steam: $Z_1=P_1v_g/RT=0.906$, so the ideal model overstates the specific volume (and understates the mass and work) by about 10 %.
Check
The work is obtained without a graphical $\int P\,dV$ because the reversible isothermal path lets $Q=T\Delta S$ be evaluated exactly from the tabulated entropies. The ideal-gas comparison here assumes the fluid is ideal throughout, so the inferred mass ($P_1V_1/RT$) — not the true tabulated mass — is used, isolating the equation-of-state error. Using the true mass in the ideal formula instead would mask the effect (<0.5 %).
