Question 1 of 7: Evaporation into a Closed Chamber (Stefan Diffusion)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.
Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 1: Evaporation into a Closed Chamber (Stefan Diffusion) (Part A — equal value)
Given. A sealed 1 m³ chamber of air at $T=293$ K, $P_T=101.3$ kPa, initial water partial pressure $p_0=0.8$ kPa. A pool at 303 K (surface partial pressure = its saturation value) evaporates through a stagnant gas film $L=0.25$ mm over area $A=0.01$ m².
Find. (i) the mass of water that must evaporate to raise the chamber to 90% saturation at 293 K, and (ii) the time required.
Figure 1 — The pool at 303 K sets the water partial pressure at the liquid face ($p_s=4.3$ kPa); water diffuses through the 0.25 mm stagnant film into the bulk, whose partial pressure climbs from 0.8 to 2.07 kPa.
Approach. Water diffuses through a stagnant air film (Stefan diffusion); the instantaneous flux drives an unsteady mass balance on the well-mixed chamber, which integrates to the time.
Water that must evaporate. The chamber vapour rises from $p_0$ to $p_f$; by the ideal-gas law the added moles are $$n=\frac{(p_f-p_0)V}{RT}=\frac{(2.07-0.8)\times10^{3}\,(1)}{8.314(293)}=0.521\ \text{mol}\;\Rightarrow\; m=0.521\times18.015=\boxed{9.39\ \text{g}}.$$
Stefan flux through the stagnant film. With air stationary, the molar flux of water from the surface (1) to the bulk (2) is $$N_A=\frac{D\,P_T}{RTL}\ln\!\frac{P_T-p}{P_T-p_s},$$ where $p$ is the (rising) bulk partial pressure and $p_s=4.3$ kPa at the liquid face.
Unsteady balance on the chamber. Accumulation equals inflow, $\dfrac{V}{RT}\dfrac{dp}{dt}=A\,N_A$. The $RT$ cancels, giving $$V\,\frac{dp}{dt}=\frac{A\,D\,P_T}{L}\ln\!\frac{P_T-p}{P_T-p_s}\;\Rightarrow\; t=\frac{V L}{A D P_T}\int_{p_0}^{p_f}\frac{dp}{\ln[(P_T-p)/(P_T-p_s)]}.$$
Evaluate. The pre-factor $\dfrac{VL}{ADP_T}=\dfrac{(1)(0.25\times10^{-3})}{(0.01)(0.24\times10^{-4})(101.3\times10^{3})}=1.028\times10^{-2}$ s/Pa, and the integral (numerically, with $p$ in Pa) is $4.43\times10^{4}$ Pa, so $$t=1.028\times10^{-2}\times4.43\times10^{4}=\boxed{456\ \text{s}\approx 7.6\ \text{min}}.$$
Check (film temperature): the gas film spans 303 K (surface) to 293 K (bulk). The diffusivity is quoted once, so both the flux and the chamber accumulation are evaluated at 293 K; using the film-mean 298 K would shorten $t$ by only ~2%. Bulk flow of air is neglected as instructed.
