Question 5 of 6: Two-Film Interphase Transfer in SO₂ Absorption
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) — variable-conductivity conduction, resistance networks, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — double-pipe exchanger sizing, evaporator heat load, diffusion through stagnant films; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations and two-film interphase transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — three in Part A (Heat Transfer) and three in Part B (Mass Transfer, printed as Problems 1–3 of Part B and numbered Questions 4–6 here). At least two problems from each part must be attempted and only the first two from each section are marked. Open-book, three-hour paper. All six problems are solved here for completeness.
Paper note
This is the 16-CHEM-A3 Heat and Mass Transfer paper of May 2019. The printed paper is five pages (cover notes, Part A Problems 1–3, Part B Problems 1–3), followed by two sheets of graph paper and a periodic table.
Part A — Heat Transfer
Question 5: Two-Film Interphase Transfer in SO₂ Absorption (25 points)
Given. Equilibrium $P_A=25X_A$ (atm); total pressure $P=10$ atm; bulk point $Y_A=0.04$, $X_A=0.01$; film coefficients $k_y'=10$, $k_x'=8$ kmol/m²·hr (per mole fraction). SO₂ diffuses through stagnant air on the gas side and stagnant water on the liquid side. The primed (equimolar-counterdiffusion) coefficients therefore need the stagnant-film correction $k_y=k_y'/(1-y_A)_{iM}$, $k_x=k_x'/(1-x_A)_{iM}$.
Find. The interface mole fractions $X_{Ai}$, $Y_{Ai}$ and the SO₂ molar flux $N_A$.
Figure 5 — Two-film picture. SO₂ falls from $Y_A=0.040$ in the bulk gas to $Y_{Ai}=0.0364$ at the interface, then from $X_{Ai}=0.0146$ to $X_A=0.010$ across the liquid film; the interface point ($X_{Ai},Y_{Ai}$) lies on the equilibrium line and the flux is continuous across it.
Approach. Recast the equilibrium into mole-fraction form $Y^{*}=(25/P)X$. Equate the gas-film and liquid-film flux expressions; their ratio fixes the tie-line slope $-k_x/k_y$. Intersect that tie line with the equilibrium line to locate the interface. Because the log-mean inert fractions depend on the unknown interface, start with $k'$ (inert factors $=1$), then correct and repeat until the interface stops moving (Geankoplis Example 10.4-1 method). The flux then follows from either film.
Equilibrium in mole fractions. Dividing $P_A=25X_A$ by the total pressure, $Y^{*}=P_A/P=(25/10)X_A=2.5\,X_A$, so the equilibrium line has slope $m=2.5$ and passes through the origin.
Flux continuity fixes the tie line. For $A$ diffusing through stagnant $B$ in each phase, the same flux crosses both films:$$N_A=\frac{k_y'}{(1-y_A)_{iM}}(Y_A-Y_{Ai})=\frac{k_x'}{(1-x_A)_{iM}}(X_{Ai}-X_A)\;\Rightarrow\;\text{slope}=-\frac{k_x}{k_y}.$$
First trial (inert factors $=1$). With slope $-k_x'/k_y'=-0.8$ and $Y_{Ai}=2.5X_{Ai}$:$$0.04-2.5X_{Ai}=0.8(X_{Ai}-0.01)\;\Rightarrow\;0.048=3.3\,X_{Ai}\;\Rightarrow\;X_{Ai}=0.01455,\quad Y_{Ai}=0.03636.$$
Stagnant-film correction and iteration. The log-mean inert fractions are$$(1-y_A)_{iM}=\frac{(1-0.03636)-(1-0.04)}{\ln[(1-0.03636)/(1-0.04)]}=0.9618,\qquad (1-x_A)_{iM}=\frac{(1-0.01)-(1-0.01455)}{\ln[(1-0.01)/(1-0.01455)]}=0.9877,$$so $k_y=10/0.9618=10.40$ and $k_x=8/0.9877=8.10$, and the slope becomes $-8.10/10.40=-0.779$. Re-solving $0.04-2.5X_{Ai}=0.779(X_{Ai}-0.01)$ gives $X_{Ai}=0.01458$. A further pass changes nothing in the fourth significant figure, so$$\boxed{X_{Ai}=0.0146},\qquad \boxed{Y_{Ai}=2.5(0.01458)=0.0364}.$$
Molar flux. From the gas film,$$N_A=k_y(Y_A-Y_{Ai})=10.40(0.0400-0.03644)=\boxed{0.0371\ \text{kmol/m}^2\text{hr}},$$and the liquid film checks it: $k_x(X_{Ai}-X_A)=8.10(0.01458-0.0100)=0.0371$ kmol/m²·hr. The dilute first trial ($k'$ uncorrected) would give $0.0364$, about 2% low. SO₂ transfers from gas to liquid (absorption), consistent with the bulk gas lying above equilibrium ($Y^*=2.5(0.01)=0.025<0.04$). Comparing resistances, $1/k_y=0.096$ against $m/k_x=2.5/8.10=0.309$, so about 76% of the resistance lies in the liquid film.
