C2 — Evaporation of a water film from a blackboard: 23-Chem-B1 Transport Phenomena, May 2016, Q6

Question 6: C2 — Evaporation of a water film from a blackboard (25 marks)

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.

Quantity	Symbol	Value
Film thickness	$\delta_0$	1 mm $=0.1$ cm
Diffusional gap	$L$	1 cm
Vapor pressure (25 °C)	$p_{A}^{sat}$	0.0313 atm
Ambient relative humidity	—	10 % $\Rightarrow p_{A,\infty}=0.1\,p_A^{sat}$
Diffusion coefficient	$D_{AB}$	0.282 cm$^2$/s
Total pressure / temperature	$P,\ T$	1 atm, 298 K

Find. (b) the steady evaporative flux $N_A$; (c) the flux–recession relation; (d) the drying time.

Fig. C2: Water vapour (A) diffuses through a 1 cm stagnant-air (B) gap from the saturated film surface to 10 % humidity ambient. Stefan diffusion of A through stagnant B fixes the steady flux; a liquid mass balance sets the drying time.

Approach. Model steady diffusion of water vapour (A) through the stagnant air (B) gap — Stefan diffusion — to get the flux, relate that flux to the shrinking liquid depth by a mass balance, and integrate for the time to remove the 1 mm film.

(a) Problem definition & assumptions. Take $z$ from the water surface ($z=0$, saturated: $p_A=p_A^{sat}$) to the ambient plane ($z=L=1$ cm, $p_A=0.1\,p_A^{sat}$). Assumptions: steady one-dimensional diffusion; isothermal 25 °C; ideal gas; air (B) stagnant and insoluble in water so $N_B=0$ (Stefan condition); the film recedes quasi-steadily (its motion is slow compared with vapour diffusion).
(b) Stefan flux of A through stagnant B. With $c=P/RT$ constant and $N_B=0$, $$N_A=\frac{D_{AB}\,P}{R\,T\,L}\ln\!\left(\frac{P-p_{A,\infty}}{P-p_A^{sat}}\right).$$ Numerically, $p_A^{sat}=0.0313$ atm, $p_{A,\infty}=0.00313$ atm, $R=82.06\ \text{cm}^3\text{atm/mol K}$: $$N_A=\frac{0.282(1)}{82.06(298)(1)}\ln\!\frac{1-0.00313}{1-0.0313}=3.30\times10^{-7}\ \text{mol/cm}^2\text{s}.$$ (The mixture is dilute, so this is within 2 % of the simple Fick estimate.)
(c) Flux–recession relation. A mass balance on the liquid film of depth $\delta$ (density $\rho_L$, molar mass $M_A$) equates the molar loss to the vapour flux: $$\rho_L\frac{d\delta}{dt}=-N_A\,M_A\quad\Longrightarrow\quad \frac{d\delta}{dt}=-\frac{N_A M_A}{\rho_L}.$$
(d) Drying time. With $N_A$ essentially constant (the 1 cm gap dwarfs the $\le1$ mm film), integrate from $\delta_0$ to 0: $$t=\frac{\rho_L\,\delta_0}{N_A\,M_A}=\frac{(1\ \text{g/cm}^3)(0.1\ \text{cm})}{(3.30\times10^{-7})(18\ \text{g/mol})}\;\Longrightarrow\;\boxed{t\approx1.68\times10^{4}\ \text{s}\approx4.7\ \text{h}}.$$

Quantity	Result
Steady evaporative flux	$N_A\approx3.30\times10^{-7}\ \text{mol/cm}^2\text{s}$
Recession relation	$d\delta/dt=-N_A M_A/\rho_L$
Time to dry 1 mm film	$t\approx1.68\times10^{4}\ \text{s}\approx4.7\ \text{h}$

Concept & Theory

Topic. Stefan diffusion (A through stagnant B) with a slowly receding liquid interface.

Key relations. Stefan flux $N_A=\tfrac{D_{AB}P}{RTL}\ln\tfrac{P-p_{A,\infty}}{P-p_A^{sat}}$; liquid balance $\rho_L\,d\delta/dt=-N_A M_A$; drying time $t=\rho_L\delta_0/(N_A M_A)$.

Why this works. Because air cannot cross the water surface, its diffusive flux must be cancelled by a convective drift — the "Stefan wind" — which enhances the transport of vapour by the log-mean factor. The interface recedes quasi-steadily because vapour diffusion across the 1 cm gap is far faster than the liquid depletes, so a single constant flux governs the whole drying process.

Common pitfall. Using the saturation pressure at the wrong temperature, or dropping the stagnant-B log factor when the driving pressures are small (here it matters only at the 2 % level, but it is essential when $p_A$ approaches $P$).

Source. Bird, Stewart & Lightfoot, Transport Phenomena, §18.2 (diffusion through a stagnant gas film / Stefan tube); Welty et al., Ch. 26.

Question 6 of 6: C2 — Evaporation of a water film from a blackboard

Question 6: C2 — Evaporation of a water film from a blackboard (25 marks)