Diffusion of water vapour through stagnant air

Model

Water vapour (A) diffuses upward through stagnant air (B). Air does not move, so $N_B = 0$ and there is a net bulk flow of the mixture (Stefan flow). For steady state, constant temperature and pressure:

$$N_A = \frac{D_{AB}P}{\bar R T\,z}\,\ln\frac{P-p_{A2}}{P-p_{A1}}$$

where $z$ is the diffusion length, $p_{A1}$ the vapour pressure at the water surface and $p_{A2}$ at the top.

Data

• $P = 101\,325$ Pa, $T = 298$ K, $z = 0.15$ m
• $p_{A1} = 3170$ Pa (saturated at the water surface), $p_{A2} = 0$ (dry air at top)
• $D_{AB} = 0.256\times10^{-4}$ m$^2$/s, $\bar R = 8.314$ J/mol K
• $A_c = \pi(0.01)^2/4 = 7.854\times10^{-5}$ m$^2$

Molar flux

$$\begin{aligned} \frac{D_{AB}P}{\bar RTz} &= \frac{0.256\times10^{-4}\times101\,325}{8.314\times298\times0.15} = 6.98\times10^{-3}\ \text{mol/m}^2\text{s}\\ \ln\frac{101\,325-0}{101\,325-3170} &= \ln(1.0323) = 0.03178\\ N_A &= 6.98\times10^{-3}\times0.03178 = 2.22\times10^{-4}\ \text{mol/m}^2\text{s} \end{aligned}$$

That is $2.22\times10^{-7}$ kmol/m$^2$s.

Evaporation rate

$$\begin{aligned} \dot n &= N_AA_c = 2.2185\times10^{-4}\times7.854\times10^{-5} = 1.742\times10^{-8}\ \text{mol/s}\\ \dot m &= \dot n\times18\ \text{g/mol} = 3.14\times10^{-7}\ \text{g/s}\\ &= 3.14\times10^{-7}\times3600 = 1.13\times10^{-3}\ \text{g/h} = 1.13\ \text{mg/h} \end{aligned}$$

Answer: $N_A \approx 2.22\times10^{-4}\ \text{mol/m}^2$s; evaporation rate about $1.13$ mg per hour.