C2 — Helium diffusion through a glass tube wall: 23-Chem-B1 Transport Phenomena, May 2014, Q6

Question 6: C2 — Helium diffusion through a glass tube wall

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. Steady radial diffusion of dissolved He through a cylindrical glass wall, no reaction.

Symbol	Value
Inner / outer radius $R_i,R_o$	$0.050$ m, $0.055$ m
Surface concentrations $c_i,c_o$	$1200$, $0\ \text{mol}\,\text{m}^{-3}$
Diffusivity $D$	$2.3\times10^{-10}\ \text{m}^2\text{s}^{-1}$
Mid-wall radius	$0.0525$ m

Find. (a) molar rate of He per unit tube length; (b) He concentration at $r=52.5$ mm.

Fig. C2: Radial diffusion of dissolved helium through the annular glass wall (inner radius $R_i$, outer $R_o$). With fixed surface concentrations and no reaction, the molar flow per unit length is constant and the concentration falls logarithmically across the wall.

Approach. This is the mass-transfer analogue of steady conduction through a cylindrical wall: Fick’s law with a constant molar flow per length integrates to a logarithmic profile.

Constant molar flow per length. At steady state the molar rate of He crossing any radius is the same: $W_A=N_A(2\pi r)=\text{const}$ (per unit length). Fick’s law $N_A=-D\,dc/dr$ then gives $W_A=-D(2\pi r)\dfrac{dc}{dr}.$
Separate and integrate. $\dfrac{W_A}{2\pi D}\dfrac{dr}{r}=-dc$; integrating from the inner to the outer surface, $$\frac{W_A}{2\pi D}\ln\frac{R_o}{R_i}=c_i-c_o.$$
(a) Diffusion rate per unit length. Solving for $W_A$ with $\ln(0.055/0.050)=0.09531$: $$W_A=\frac{2\pi D\,(c_i-c_o)}{\ln(R_o/R_i)}=\frac{2\pi(2.3\times10^{-10})(1200)}{0.09531}=\boxed{1.82\times10^{-5}\ \text{mol}\,\text{m}^{-1}\text{s}^{-1}}.$$
(b) Concentration profile. Integrating from $R_i$ to a general $r$ gives $c(r)=c_i+(c_o-c_i)\dfrac{\ln(r/R_i)}{\ln(R_o/R_i)}$. At $r=0.0525$ m, $\ln(0.0525/0.050)=0.04879$, so $$c(0.0525)=1200+(0-1200)\frac{0.04879}{0.09531}=1200-614.3=\boxed{585.7\ \text{mol}\,\text{m}^{-3}}.$$ Note this is below the arithmetic mean (600): because the diffusion area $2\pi r$ is smallest at the inner surface, the gradient is steepest there, so the logarithmic profile lies below a straight line joining the two surface values.

Check — modelling assumptions

The $25^\circ\text{C}$ / $4$ bar state fixes the solubility-controlled inner-surface concentration ($c_i=1200\ \text{mol}\,\text{m}^{-3}$) and does not otherwise enter the transport calculation. Diffusivity and surface concentrations are constant, and the outer surface is a perfect sink ($c_o=0$).

Quantity	Value
$\ln(R_o/R_i)$	$0.09531$
(a) He rate per unit length $W_A$	$1.82\times10^{-5}\ \text{mol}\,\text{m}^{-1}\text{s}^{-1}$
(b) Concentration at $r=52.5$ mm	$585.7\ \text{mol}\,\text{m}^{-3}$

Concept & Theory

Topic. Steady radial mass diffusion through a cylindrical wall (heat–mass transfer analogy).

Key relations. Constant flow $W_A=-2\pi rD\,dc/dr$; cylindrical result $W_A=2\pi D(c_i-c_o)/\ln(R_o/R_i)$; logarithmic profile $c(r)=c_i+(c_o-c_i)\ln(r/R_i)/\ln(R_o/R_i)$.

Why this works. With no reaction, conservation forces the same molar flow through every cylindrical surface; dividing that constant flow by the growing area $2\pi r$ makes the gradient fall as $1/r$, which integrates to a logarithm — exactly the mass-transfer twin of conduction through a pipe wall (diffusivity $D$ playing the role of $k$).

Common pitfall. Using diameters instead of radii in the area/geometry (the ratio $R_o/R_i=D_o/D_i$ is safe, but the $2\pi r$ area is not), or applying a planar-slab formula to what is a curved (logarithmic) geometry.

Source. Welty et al., Ch. 26 (steady molecular diffusion, cylindrical); Bird, Stewart & Lightfoot, Ch. 18; Incropera & DeWitt, Ch. 14 (species diffusion, analogy to conduction).

Question 6 of 6: C2 — Helium diffusion through a glass tube wall

Question 6: C2 — Helium diffusion through a glass tube wall