Question 3 of 7: Dissolution of a polymer film into a flowing solvent
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (vapour pressures, diffusivities, Henry constants) is stated in the question, so each answer is self-contained.
Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer and gas absorption; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer mass transfer and diffusion with reaction; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (species diffusion, sphere in a stagnant medium); Treybal, Mass-Transfer Operations (3rd ed.) — packed-tower and wetted-wall design; supporting data from Perry's Chemical Engineers' Handbook (9th ed.).
Question 3: Dissolution of a polymer film into a flowing solvent (Part B — equal value)
Given. Liquid MEK flows over a flat, soluble polymer strip; polymer (A) dissolves from the wall into the moving solvent (B). The concentration boundary layer grows along the plate, so this is a classic laminar flat-plate convective mass-transfer problem (all data in cgs).
Quantity
Value
Diffusivity $D_{AB}$
$3.0\times10^{-6}$ cm²/s
Kinematic viscosity $\nu_B$
$6.0\times10^{-3}$ cm²/s
Surface solubility $c_{A,s}$
0.04 g/cm³
Volumetric flow $Q$
30 cm³/s
Channel width $w$ / depth $h$
10 cm / 2.0 cm
Film length $L$
20 cm
Find. (a) $Sc$ and $\overline{Sh}_L$; (b) the average dissolution flux $\overline{N_A}$ (g of polymer/cm²·s).
Approach. Get the mean velocity from the channel cross-section, form $Re_L$ and $Sc$, apply the laminar flat-plate correlation $\overline{Sh}_L=0.664\,Re_L^{1/2}Sc^{1/3}$, then convert to a coefficient and multiply by the surface solubility.
Schmidt number. $$Sc=\frac{\nu_B}{D_{AB}}=\frac{6.0\times10^{-3}}{3.0\times10^{-6}}=\boxed{2000}.$$ A large $Sc$ (thin concentration layer inside a thicker momentum layer) is typical of liquids.
Mean solvent velocity. From the channel cross-section, $$v=\frac{Q}{w\,h}=\frac{30}{(10)(2.0)}=1.5\ \text{cm/s}.$$
Plate Reynolds number. On the 20-cm film length, $$Re_L=\frac{vL}{\nu_B}=\frac{(1.5)(20)}{6.0\times10^{-3}}=5000\ (\ll 5\times10^5,\ \text{laminar}).$$
Average Sherwood number. $$\overline{Sh}_L=0.664\,Re_L^{1/2}Sc^{1/3}=0.664\,(5000)^{1/2}(2000)^{1/3}=\boxed{592}.$$
Average mass-transfer coefficient. $$\overline{k_c}=\frac{\overline{Sh}_L\,D_{AB}}{L}=\frac{(592)(3.0\times10^{-6})}{20}=8.87\times10^{-5}\ \text{cm/s}.$$
Average dissolution flux. With $c_{A\infty}\approx0$, $$\overline{N_A}=\overline{k_c}\,(c_{A,s}-c_{A\infty})=(8.87\times10^{-5})(0.04)=\boxed{3.55\times10^{-6}\ \text{g/cm}^2\!\cdot\!\text{s}}.$$
Quantity
Result
Schmidt number $Sc$
2000
Reynolds number $Re_L$
5000 (laminar)
Average Sherwood number $\overline{Sh}_L$
592
Average coefficient $\overline{k_c}$
$8.87\times10^{-5}$ cm/s
Average flux $\overline{N_A}$
$3.55\times10^{-6}$ g/cm²·s
Check
The velocity comes from the full channel cross-section ($w\times h=10\times2.0$ cm), not the pan length; $Re_L=5000$ confirms laminar flow, validating the $0.664\,Re^{1/2}Sc^{1/3}$ correlation. The solid density $\rho_{A,\text{solid}}$ and solvent density $\rho_B$ are not needed for the flux — they would only enter a film-recession (thickness-vs-time) calculation, which the question does not ask.
