Discussion
Mass transfer coefficient for equimolar counter-diffusion of water vapour in air is 0.4 m s-1 (based on concentration difference). Mass diffusivity of water vapour in air is 3 ×10-4 m2 s-1. For 100 µm diameter droplet, the Sherwood Number (NSh) is equal to
Answer: C
The Sherwood number equals one-third of the mass transfer coefficient (numerically).
The Sherwood number is \(N_{Sh} = k_{c}d/D_{AB}\). With \(k_{c} = 0.4\) m/s, \(d = 100\ \mu\text{m} = 10^{-4}\) m and \(D_{AB} = 3 \times 10^{-4}\) m2/s:
\[ \begin{aligned} N_{Sh} &= \frac{0.4 \times 10^{-4}}{3 \times 10^{-4}} \\ &= \frac{0.4}{3} \approx 0.133 \end{aligned} \]
Because \(d/D_{AB} = 1/3\) in these units, the number is one-third of 0.4. It is not the coefficient itself (0.4). Diameter divided by diffusivity is 1/3 s/m, a dimensional quantity, and three times the diffusivity is \(9 \times 10^{-4}\); neither equals 0.133.
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