Mass Transfer Basics Questions and Answers

Answer: C

The ratio is the Schmidt number.

The Schmidt number compares momentum diffusivity with mass diffusivity: \(Sc = \nu / D_{AB}\). It plays the same role in mass transfer that the Prandtl number plays in heat transfer, and for water vapour in air it is about 0.6.

The Prandtl number is \(\nu / \alpha\), momentum diffusivity over thermal diffusivity. The Stanton number is \(h / (\rho c_{p} V)\), a ratio of heat transferred to heat capacity of the flow. The Sherwood number is \(k_{c}L / D_{AB}\), the dimensionless mass transfer coefficient. Only the Schmidt number is formed from \(\nu\) and \(D_{AB}\) alone.

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.