Water Vapour Diffusing Through a Conical Stack: 23-Chem-A3 Heat and Mass Transfer, December 2013, Q2

Question 2: Water Vapour Diffusing Through a Conical Stack (equal value)

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. Stagnant-air diffusion in a frustum whose cross-section varies with height, so the molar flux changes with position but the molar flow rate $W_A$ is constant.

Quantity	Value
$T$, $P$	327.15 K, $1.0\times10^{5}$ Pa
Height $H$	10 m
Diameters (bottom / top)	0.80 m / 0.40 m ⇒ radii 0.40 / 0.20 m
$p_{A}$ bottom / top	2500 Pa / 500 Pa ($y=0.025 / 0.005$)
$D_{AB}$	$3\times10^{-5}$ m²/s

Find. (a) mass rate of water vapour (kg/h); (b) molar flux at top and bottom.

Figure 2 — Conical stack (frustum): radius shrinks linearly from 0.40 m at the base to 0.20 m at the top over a 10 m height; water vapour diffuses upward through stagnant air.

Approach. Write the constant molar flow $W_A=N_A A(z)$, insert the Stefan-diffusion flux, separate variables and integrate the geometric factor $\int dz/A(z)$ analytically, then back out the fluxes from $N_A=W_A/A$.

Radius profile and area. The radius varies linearly, $r(z)=0.40-0.02z$ (m), so $A(z)=\pi r(z)^2$ with $z$ measured upward from the base.
Constant molar flow with Stefan drift. Because air is stagnant, $W_A = N_A A(z) = -\dfrac{D_{AB}\,c}{1-y_A}\,A(z)\dfrac{dy_A}{dz}$, with $c=P/RT=1\times10^{5}/(8.314\times327.15)=36.8\ \text{mol/m}^3.$ Separating variables, $W_A\displaystyle\int_0^{H}\frac{dz}{A(z)} = D_{AB}\,c\,\ln\frac{1-y_{A,top}}{1-y_{A,bot}}.$
Geometric integral. With $r=0.40-0.02z$, $$\int_0^{10}\frac{dz}{\pi(0.40-0.02z)^2} = \frac{1}{0.02\pi}\Big[\frac{1}{0.20}-\frac{1}{0.40}\Big] = 39.8\ \text{m}^{-1}.$$
Solve for the molar flow. $$W_A = \frac{D_{AB}\,c\,\ln\frac{0.995}{0.975}}{39.8} = \frac{(3\times10^{-5})(36.8)(0.02030)}{39.8} = \boxed{5.63\times10^{-7}\ \text{mol/s}}.$$
Mass rate (part a). Multiplying by $M=18.0$ g/mol and 3600 s/h, $\dot m = (5.63\times10^{-7})(0.01802)(3600) = \boxed{3.65\times10^{-5}\ \text{kg/h}}.$
Fluxes at the two ends (part b). Since $W_A$ is constant, the flux is largest where the area is smallest. With $A_{bot}=\pi(0.40)^2=0.503\ \text{m}^2$ and $A_{top}=\pi(0.20)^2=0.126\ \text{m}^2$, $$N_{A,bot}=\frac{W_A}{A_{bot}} = 1.12\times10^{-6},\qquad N_{A,top}=\frac{W_A}{A_{top}} = 4.48\times10^{-6}\ \text{mol/m}^2\text{s}.$$ The top flux is four times the bottom flux — exactly the area ratio $(0.40/0.20)^2$.

Quantity	Result
Molar flow $W_A$	$5.63\times10^{-7}$ mol/s
(a) Mass rate	3.65×10⁻⁵ kg/h
(b) Flux at bottom	$1.12\times10^{-6}$ mol/m²·s
(b) Flux at top	$4.48\times10^{-6}$ mol/m²·s

Concept & Theory

Topic. Steady Stefan diffusion through a variable cross-section (the molar flow, not the flux, is conserved).

Key relations. $W_A=N_A A(z)=\text{const}$; $W_A\int_0^H dz/A(z)=D_{AB}c\,\ln[(1-y_{top})/(1-y_{bot})]$; $A(z)=\pi r(z)^2$ with $r$ linear in $z$.

Why this works. A steady-state mass balance on any slice requires the same number of moles per second to pass every cross-section; only the flux adjusts as the area changes. Putting $A(z)$ inside the integral handles the taper exactly, and the dilute mole fractions make the Stefan log term a small (2%) correction.

Common pitfall. Treating the flux as constant and using an average area — that violates the mole balance. Integrate $dz/A(z)$, not $A(z)\,dz$.

Source. Geankoplis §6.2D (diffusion through varying cross-sectional area); Bird, Stewart & Lightfoot §18.2.

Question 2 of 6: Water Vapour Diffusing Through a Conical Stack

Question 2: Water Vapour Diffusing Through a Conical Stack (equal value)