Exercise 3 Solution

From Equation 8c, the condition NA + NB = 0 requires that the viscous flux cancel the non-equimolar flux: Nv = ND.

The value of the non-equimolar flux (ND = −0.0222 \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}) computed in Exercise 1 applies here because the diffusive fluxes are not appreciably affected by the pressure gradient. So Nv = −0.0222 \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}} and, from Equation 2, N^{v}=\frac{k_{g}}{\mu RT}pdp/dl = 0.0222 \frac{\textup{moles}}{\textup{m}^{2}\ s}.

This is integrated to obtain the following expression

\displaystyle \frac{k_{g}}{2\mu RTL}\left ( p_{0}^{2}-p_{L}^{2} \right )=0.0222

which can be used to directly calculate the desired pressure difference. However, computations are facilitated by rewriting the left side using:

\displaystyle \frac{1}{2}\left ( p_{0}^{2}-p_{L}^{2} \right )=\bar{p}\left ( p_{0}-p_{L} \right )

where, \bar{p} is the average pressure in the chamber.

\displaystyle p_{0}-p_{L}=0.0222\frac{\mu L}{Ck_{g}}=\frac{(0.0222)(2.3\times 10^{-5})(0.05)}{(1\times 10^{-12})(40.9)}=624\ \textup{Pa}

The pressure drop is quite small (about 6 cm H2O) relative to the 1 × 105 Pa pressure maintained in the right-hand header, justifying the assumption that the mean pressure is closely approximated by the controlled pressure in the right-hand header.

Return to Exercise 3


Flux Equations for Gas Diffusion in Porous Media Copyright © 2021 by David B. McWhorter. All Rights Reserved.