C1 — Length-average mass-transfer coefficient over a membrane: 23-Chem-B1 Transport Phenomena, May 2017, Q5

Question 5: C1 — Length-average mass-transfer coefficient over a membrane (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
Membrane length	$L$	$10\ \text{cm}$
Average velocity	$U$	$5\ \text{cm/s}$
Kinematic viscosity	$\nu$	$0.01\ \text{cm}^2/\text{s}$
Diffusivity	$D_{AB}$	$5\times10^{-6}\ \text{cm}^2/\text{s}$

Find. The length-average mass-transfer coefficient $k_c$ over the membrane.

Fig. C1: A thin concentration boundary layer grows over the membrane. Because $Sc\gg1$ it stays much thinner than the momentum layer, and the flat-plate Sherwood correlation gives the length-average coefficient.

Approach. Confirm the flow is laminar, evaluate the Schmidt number, apply the laminar flat-plate length-average Sherwood correlation (the mass-transfer analogue of the Blasius heat-transfer result), and convert to $k_c$.

Reynolds number over the plate. $$Re_L=\frac{UL}{\nu}=\frac{(5)(10)}{0.01}=5000.$$ This is well below the flat-plate transition ($\sim5\times10^{5}$), so the boundary layer is laminar over the whole membrane.
Schmidt number. $$Sc=\frac{\nu}{D_{AB}}=\frac{0.01}{5\times10^{-6}}=2000,$$ a large value typical of a solute diffusing in a liquid.
Length-average Sherwood number. For laminar flow over a flat plate, $$\overline{Sh}_L=0.664\,Re_L^{1/2}Sc^{1/3}=0.664(5000)^{1/2}(2000)^{1/3}=0.664(70.7)(12.60)\ \Longrightarrow\ \boxed{\overline{Sh}_L\approx592.}$$
Mass-transfer coefficient. $$k_c=\frac{\overline{Sh}_L\,D_{AB}}{L}=\frac{(592)(5\times10^{-6})}{10}\ \Longrightarrow\ \boxed{k_c\approx2.96\times10^{-4}\ \text{cm/s}.}$$

Quantity	Result
Reynolds number	$Re_L=5000$ (laminar)
Schmidt number	$Sc=2000$
Length-average Sherwood number	$\overline{Sh}_L\approx592$
Mass-transfer coefficient	$k_c\approx2.96\times10^{-4}\ \text{cm/s}$

Concept & Theory

Topic. Laminar external mass transfer; the heat–mass transfer analogy on a flat plate.

Key relations. $Re_L=UL/\nu$; $Sc=\nu/D_{AB}$; length-average $\overline{Sh}_L=0.664\,Re_L^{1/2}Sc^{1/3}$; $k_c=\overline{Sh}_L\,D_{AB}/L$.

Why this works. The concentration boundary layer obeys the same convection–diffusion equation as the thermal layer, so the Blasius heat-transfer correlation carries over with $Pr\to Sc$ and $Nu\to Sh$. The coefficient $0.664$ is the average over $0$–$L$ (twice the trailing-edge local value, since the local coefficient $\propto x^{-1/2}$). The large $Sc$ makes the concentration layer thin, which is why a boundary-layer treatment is accurate.

Common pitfall. Using the local ($0.332$) instead of the length-average ($0.664$) constant when the question asks for the average coefficient; and failing to check that $Re_L$ is below transition before choosing the laminar correlation.

Source. Welty et al., Ch. 30 (convective mass transfer, flat-plate correlations); Bird–Stewart–Lightfoot §22 (concentration boundary layers).

Question 5 of 6: C1 — Length-average mass-transfer coefficient over a membrane

Question 5: C1 — Length-average mass-transfer coefficient over a membrane (25 marks)