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23-Chem-A3 Heat and Mass Transfer · December 2018

Question 4 of 6: Diffusion of HCl through a Stagnant Water Film

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — conduction resistance networks, critical radius, convection correlations, radiation exchange between surfaces; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — heat-exchanger sizing and diffusion through stagnant films; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential (Rayleigh) distillation and penetration-theory absorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations. Property data from Perry's Chemical Engineers' Handbook (9th ed.).

Exam format: six problems worth 25 points each — Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6). The rubric asks for at least two problems from each part; every problem is solved here for completeness. Open-book, three-hour paper.

Part A — Heat Transfer

Part B — Mass Transfer

Question 4: Diffusion of HCl through a Stagnant Water Film (25 points)

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. Film thickness $z=4$ mm $=0.004$ m; boundary 1 is 12 wt% HCl ($\rho_1=1060.7$ kg/m³), boundary 2 is 4 wt% HCl ($\rho_2=1020.15$ kg/m³); $D_{AB}=2.5\times10^{-9}$ m²/s; water ($B$) is stagnant (non-diffusing). $M_{\text{HCl}}=36.46$, $M_{\text{H}_2\text{O}}=18.02$ g/mol.

QuantityValue
Film thickness $z$0.004 m
Boundary 112 wt% HCl, $\rho=1060.7$ kg/m³
Boundary 24 wt% HCl, $\rho=1020.15$ kg/m³
Diffusivity $D_{AB}$$2.5\times10^{-9}$ m²/s

Find. The molar diffusion flux $N_A$ of HCl through the stagnant film.

HCl diffusion N᳐12 wt%4 wt%boundary 1boundary 2stagnant water film, z = 4 mmwater is essentially non-diffusing (Stefan drift correction x_BM)
Figure 4 — HCl diffusing through the stagnant liquid water film. The mole fraction falls from $x_{A1}=0.0631$ at boundary 1 to $x_{A2}=0.0202$ at boundary 2 over 4 mm; because water does not diffuse, a small counter-drift correction $x_{BM}$ applies (diffusion of $A$ through stagnant $B$).

Approach. Convert each wall composition from weight fraction to mole fraction and to molar concentration (using the solution density and mean molar mass), then apply the diffusion-through-stagnant-$B$ flux relation with the log-mean solvent fraction.

  1. Wall mole fractions. Per 100 g of solution, at boundary 1: $n_A=12/36.46=0.329$, $n_B=88/18.02=4.884$, so $x_{A1}=0.329/5.213=\boxed{0.0631}$. At boundary 2: $n_A=4/36.46=0.110$, $n_B=96/18.02=5.328$, so $x_{A2}=0.110/5.438=0.0202$.
  2. Molar concentration of the solution. The mean molar mass is $M=100/(n_A+n_B)$, so $c=\rho/M$. At the two faces $c_1=1060.7/19.18=55.29$ and $c_2=1020.15/18.39=55.47$ kmol/m³; the mean is $c=55.4\ \text{kmol/m}^3$ (note $c$ comes from the liquid density, not $P/RT$).
  3. Log-mean stagnant-solvent fraction. With $x_{B1}=0.9369$, $x_{B2}=0.9798$,$$x_{BM}=\frac{x_{B2}-x_{B1}}{\ln(x_{B2}/x_{B1})}=0.958.$$Because the solution is dilute, this correction is only about 4% above unity.
  4. Flux (diffusion of $A$ through stagnant $B$).$$N_A=\frac{D_{AB}\,c\,(x_{A1}-x_{A2})}{z\,x_{BM}}=\frac{(2.5\times10^{-9})(55.4)(0.0631-0.0202)}{(0.004)(0.958)}=\boxed{1.55\times10^{-6}\ \text{kmol/m}^2\text{s}}.$$
QuantityResult
Mole fractions $x_{A1}$ / $x_{A2}$0.0631 / 0.0202
Mean concentration $c$55.4 kmol/m³
Log-mean solvent $x_{BM}$0.958
Diffusion flux $N_A$$1.55\times10^{-6}$ kmol/m²·s