Non-equimolar diffusion in a distillation vapour film: 23-Chem-A3 Heat and Mass Transfer, May 2016, Q3

Question 3 — Non-equimolar diffusion in a distillation vapour film (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. Binary vapour film, $T=350\text{ K}$, $P=500\text{ mm Hg}=66.7\text{ kPa}$, $C_T=P/RT=22.9\text{ mol/m}^3$; $y_A=0.7$ at the interface, $0.5$ in the bulk; film $L=0.5\text{ mm}$; $D_{AB}=2\times10^{-5}\text{ m}^2/\text{s}$; latent heats $\lambda_A=1.5\,\lambda_B$.

Find. (a) $N_A,N_B$; (b) $y_A,y_B$ and $dC_A/dz,\,dC_B/dz$ at $z=L/2$.

Coupled profiles across the film. A evaporates (interface → bulk); B condenses in the opposite sense at $1.5\times$ the molar rate, so the mixture drifts toward the interface.

Approach. An adiabatic interface ties the two fluxes through the latent-heat balance, giving a fixed flux ratio; substitute that ratio into the general one-dimensional flux equation and integrate across the film for the rates, then integrate to the mid-plane for the local state.

Flux ratio from the energy balance. Latent heat released by condensing B equals that absorbed vaporising A: $\lambda_A N_A=-\lambda_B N_B$, so with $\lambda_A=1.5\lambda_B$, $$N_B=-1.5\,N_A,\qquad N=N_A+N_B=-0.5\,N_A .$$
General flux with drift. $N_A=-C_TD_{AB}\dfrac{dy_A}{dz}+y_A N$. Inserting $N=-0.5N_A$, $$N_A\,(1+0.5\,y_A)=-C_TD_{AB}\frac{dy_A}{dz}.$$
Integrate across the film (part a). Separating and integrating $y_A:0.7\!\to\!0.5$ over $z:0\!\to\!L$, $$N_A=\frac{2\,C_TD_{AB}}{L}\ln\frac{1+0.5(0.7)}{1+0.5(0.5)}=\frac{2(22.9)(2\times10^{-5})}{5\times10^{-4}}\ln\frac{1.35}{1.25}.$$ $$\boxed{N_A=1.41\times10^{-4}\ \text{kmol m}^{-2}\text{s}^{-1},\qquad N_B=-2.12\times10^{-4}\ \text{kmol m}^{-2}\text{s}^{-1}.}$$ A migrates from interface to bulk (enriching the vapour); B moves the other way at $1.5\times$ the rate.
Mid-plane mole fractions (part b). Integrating only to $z=L/2$ gives $\ln(1+0.5y_A)=\ln1.35-\tfrac12(0.0770)$, so $$y_A(\tfrac{L}{2})=0.598,\qquad y_B(\tfrac{L}{2})=0.402 .$$
Mid-plane concentration gradients. From step 2, $\dfrac{dy_A}{dz}=-\dfrac{N_A(1+0.5y_A)}{C_TD_{AB}}=-400\ \text{m}^{-1}$, hence (with $C_T$ constant) $$\frac{dC_A}{dz}=C_T\frac{dy_A}{dz}=-9.2\times10^{3}\ \text{mol m}^{-4},\qquad \frac{dC_B}{dz}=+9.2\times10^{3}\ \text{mol m}^{-4}.$$ The gradients are equal and opposite because the total molar concentration is fixed.

Quantity	Value
$N_A$ (MVC)	+1.41 × 10⁻⁴ kmol m⁻² s⁻¹
$N_B$ (LVC)	−2.12 × 10⁻⁴ kmol m⁻² s⁻¹
$y_A,\,y_B$ at mid-plane	0.598, 0.402
$dC_A/dz,\,dC_B/dz$ at mid-plane	−9.2 × 10³, +9.2 × 10³ mol m⁻⁴

Concept & Theory

Topic: Non-equimolar counter-diffusion in distillation, fixed by an adiabatic latent-heat balance.

Key relations: Energy tie $\lambda_AN_A=-\lambda_BN_B$; drift-flux $N_A=-C_TD_{AB}\,dy_A/dz+y_AN$; integrated across the film, $N_A=\dfrac{C_TD_{AB}}{\phi L}\ln\dfrac{1-\phi\,y_{A,L}}{1-\phi\,y_{A,0}}$ with $\phi=1+N_B/N_A$ (here $\phi=-0.5$, which supplies the factor 2).

Why this works: Unlike equimolar distillation ($N_A=-N_B$), unequal latent heats make $|N_B|>|N_A|$, giving a net molar flux toward the interface. That drift term $y_AN$ curves the profiles slightly and must be kept; integrating the exact flux law across the film yields the rates without assuming a linear profile.

Common pitfall: Assuming equimolar counter-diffusion (ignoring the latent-heat ratio) drops the drift term and gives the wrong $N_B$; also, the driving-force logarithm uses $(1+0.5y_A)$, not the inert-gas $p_{BM}$ of a stagnant-B problem.

Source: Coulson & Richardson Vol. 1, §10.2 (diffusion in distillation, unequal molar latent heats); BSL §18 (diffusion with a moving mixture).

Question 3 of 7: Non-equimolar diffusion in a distillation vapour film

Question 3 — Non-equimolar diffusion in a distillation vapour film (25 marks)