Exercise 1 Solution

Part a)

Arbitrarily set the origin of the coordinate, l, at the left end of the diffusion chamber. The value of l on the right-hand end of the chamber is L, equal to 0.05 m. Equation 23 is applicable because diffusion occurs in the molecular regime with no pressure gradient. At steady state, the total diffusive flux of each component is constant with respect to time and position along the diffusion path. Thus, Equation 23 is integrated to obtain Equation Exercise Solution 1-1.

\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )L}\textup{ln}\left\{\frac{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(L)}{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(0)} \right\} (Exercise Solution 1-1)


xA(L) = mole fraction of argon at l = L
xA(0) = mole fraction of argon at l = 0

The molar concentration is computed from the ideal gas law, C = p/RT,

with: p = 1 × 105 Pa, R = 8.205 m3 Pa/deg-mole and T = 298 K.

\displaystyle C=\frac{p}{RT}=\frac{10^{5}\ \textup{Pa}}{8.205\frac{\textup{m}^{3}\ \textup{Pa}}{\textup{deg\ mole}}298\ \textup{K}}=40.9\frac{\textup{moles}}{\textup{m}^{3}}

Other parameter values are:

xA(L) = 0, xA(0) = 1, \displaystyle M_{AB}^{0.5} = 3.158, L = 0.05 m and D = 2.37 × 10-5 m2s-1.

The total diffusive flux of argon is:

\displaystyle N_{A}^{D}=\frac{(2.37\times 10^{-5})(40.9){\textup{ln}(3.158^{-1})}}{(1-3.158)(0.05)}=0.0103\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

The flux of argon is positive because it is in the direction of increasing l.

The flux of helium can be calculated from Equation Exercise Solution 1-1 after interchanging the subscripts. However, the flux of helium is computed more easily from Graham’s law:

\displaystyle N_{B}^{D}=-M_{AB}^{0.5}N_{A}^{D}=-3.158(0.0103)=-0.0325\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

The corresponding values of flux computed from Fick’s law (Equation 11) are 0.019 moles/(m2s) for species A and -0.019 moles/(m2s) for species B. The Fick’s law calculation results in a rather large error in this case as there is significant disparity between the molecular weights of the gas components.

Part b)

The non-equimolar flux or net flux is given by Equation 5.

\displaystyle N^{D}=N_{A}^{D}+N_{B}^{D}=0.0103-0.0325=-0.0222\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

This is the flux of the gas as a whole. It is engendered entirely by diffusion, occurs without loss of momentum by viscous shear, and is in the direction of diffusion of the lighter component (helium).

Part c)

Integrate Equation 23 subject to xA = xA(l) at position l and xA = 1 at l = 0 or use Equation Exercise Solution 1-1 to obtain Exercise Solution 1-2.

\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )l}\textup{ln}\left\{\frac{1-\left ( 1-M_{AB}^{0.5} \right )x_{A}(l)}{M_{AB}^{0.5}} \right\} (Exercise Solution 1-2)

However, xA(L) = 0, consequently, Equation Exercise Solution 1-2 becomes Equation Exercise Solution 1-3.

\displaystyle N_{A}^{D}=\frac{DC}{\left ( 1-M_{AB}^{0.5} \right )L}\textup{ln}\left\{\frac{1}{M_{AB}^{0.5}} \right\} (Exercise Solution 1-3)

The desired result is obtained by combining these two equations and solving for xA(l) to obtain Exercise Solution 1-4.

\displaystyle x_{A}(l)=\frac{1-\textup{exp}\left\{(1-l/L)\textup{ln}(M_{AB}^{0.5}) \right\}}{1-M_{AB}^{0.5}} (Exercise Solution 1-4)

Part d)

Equation 10 is combined with Graham’s law to obtain Exercise Solution 1-5.

\displaystyle J_{A}=N_{A}^{D}\left\{1-x_{A}(1-M_{AB}^{0.5}) \right\} (Exercise Solution 1-5)

Then Equation Exercise Solution 1-4 is substituted for xA to arrive at Equation Exercise Solution 1-6.

\displaystyle J_{A}(l)=N_{A}^{D}\textup{exp}\left\{(1-l/L)\textup{ln}(M_{AB}^{0.5}) \right\} (Exercise Solution 1-6)

Equation Exercise Solution 1-6 shows how the equimolar flux varies over the length of the diffusion chamber.

At a point midway between the supply headers (l/L = 0.5), we have

JA = (0.0103) exp{0.5 ln(3.158)} = 0.0183 \frac{\textup{moles}}{\textup{m}^{2}\ s}

The corresponding helium flux follows from Equation Exercise Solution 1-6 by interchanging the subscripts to obtain:

JB = (−0.0325) exp{0.5 ln(0.317)} = −0.0183 \frac{\textup{moles}}{\textup{m}^{2}\ s}

These molar fluxes are equal in magnitude and opposite in direction so they satisfy Equation 7. This numerical computation applies to a particular point, and we leave it to the reader to demonstrate that Equation Exercise Solution 1-6 and its counterpart for component B sum to zero at all points. Of course, this must be the case because Equation 7 is at the heart of the analysis leading to the starting point for this example, Equation 23.

Return to Exercise 1


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