Exercise 4 Solution

Part a)

The water table constitutes a zero-flux boundary for both air and carbon dioxide. The air flux must be zero everywhere because the flux is zero at the boundary, there are no sources or sinks for air, and the system is at steady state. On the other hand, carbon dioxide is generated by an aerobic source present near the base of the vadose zone. The carbon dioxide migrates upward toward the ground surface at a steady rate.

A qualitative understanding of this problem is described by the following reasoning. Because the source is at the base, the concentration of carbon dioxide must decrease in the upward direction. It follows that the concentration of air must decrease in the downward direction and, therefore, air must diffuse downward from the surface toward the base of the vadose zone in response to the gradient of mole fraction. The rate of downward diffusion of air must be canceled by upward advection in the bulk gas; otherwise, the air would not be stagnant. The concentration gradient drives upward migration of carbon dioxide, enhanced by upward advection. The motion of the gas as a whole (phase motion) is the sum of the non-equimolar flux and the viscous flux. The non-equimolar flux is directed downward because it always occurs in the direction of diffusion of the constituent with the smaller molecular weight (air in the present case).

Writing Equation 4 with air (species A) and with CO2 (species B) results in the following expression.

NA + NB = ND + Nv

The air is assumed to be stagnant so, NA = 0, which leads to Equation Exercise Solution 4-1.

NB = ND + Nv (Exercise Solution 4-1)

The CO2 flux accounts for the motion of the fluid as a whole (i.e., the phase motion). It is comprised of a non-equimolar flux and a viscous flux.

Writing Equation 26 for the stagnant air component (NA = 0) and combining it with Equation 2 results in Equation Exercise Solution 4-2 for the viscous flux.

\displaystyle N^{v}=\frac{DCdx_{A}/dl}{x_{A}\left\{1-(1-M_{AB}^{0.5})x_{A} \right\}} (Exercise Solution 4-2)

This expression is used for the viscous flux in Equation 26, written for the carbon dioxide, to obtain Equation Exercise Solution 4-3.

\displaystyle N_{B}=-\frac{DCdx_{B}/dl}{1-(1-M_{BA}^{0.5})x_{B}} \displaystyle +\frac{x_{B}DCdx_{A}/dl}{x_{A}\left\{1-(1-M_{AB}^{0.5})x_{A} \right\}} (Exercise Solution 4-3)

This result is simplified with xA + xB = 1 and M_{BA}^{0.5}=M_{AB}^{-0.5} to obtain the differential equation shown as Equation Exercise Solution 4-4.

\displaystyle N_{B}=\frac{DCdx_{A}/dl}{x_{A}}=-\frac{DCdx_{B}/dl}{1-x_{B}} (Exercise Solution 4-4)

The flux of carbon dioxide is constant (because the system is at steady state), so integration between the points z1 and z2, where the mole fractions of CO2 are xB(z1) and xB(z2), respectively, provides the equation we seek, Equation Exercise Solution 4-5.

\displaystyle N_{B}=\frac{DC}{z_{2}-z_{1}}\textup{ln}\left\{\frac{1-x_{B}(z_{2})}{1-x_{B}(z_{1})} \right\} (Exercise Solution 4-5)

Equation Exercise Solution 4-4 reduces to Fick’s law when one species is present in trace concentrations (e.g., xB << 1). Indeed, it is common practice to employ Fick’s law for the estimation of flux by the so-called gradient method (e.g., Tracy, 2015; Johnson et al., 2006; Maier and Schaak-Kirchner, 2014). Such practice is satisfactory if CO2 is present in only trace quantity, but masks the fact that non-equimolar and viscous fluxes significantly affect the transport, as demonstrated in the following computations.

Part b)

 Exercise Solution 4-5 produces a value for NB.

C = p/RT = \displaystyle \frac{8.3\times 10^{4}}{(8.205)(294.6)}=34.34\ \frac{\textup{moles}}{\textup{m}^{3}}

\displaystyle N_{B}=\frac{(4.7\times 10^{-6})(34.34)}{1.29}\textup{ln}\frac{(1-0.0013)}{(1-0.0583)} \displaystyle =7.5\times 10^{-6}\ \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

With the flux of carbon dioxide known, the phase (bulk gas) flux is as follows.

N = NA + NB = NB = 7.5 × 10−6 \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

For the conditions set out in this example (i.e., negligible pressure diffusion), the non-equimolar flux is given by (Exercise 2).

\displaystyle N^{D}=-\frac{(1-M_{BA}^{0.5})DCdx_{B}/dl}{1-(1-M_{BA}^{0.5})x_{B}}

This is integrated to obtain the following expression.

\displaystyle N^{D}=\frac{DC}{z_{2}-z_{1}}\textup{ln}\left [ \frac{1-(1-M_{BA}^{0.5})x_{B}(z_{2})}{1-(1-M_{BA}^{0.5})x_{B}(z_{1})} \right ]

This is evaluated with the data provided to obtain a value for ND.

ND = −1.7 × 10−6 \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

The viscous flux is then calculated.

Nv = NBND = 7.5 × 10−6 − (−1.7 × 10−6) = 9.2 × 10−6 \frac{\textup{moles}}{\textup{m}^{2}\ \textup{s}}

We have the remarkable result that the viscous flux exceeds the magnitude of the total mole flux of CO2. Even though the viscous flux is the dominant flux, the associated pressure gradient can be expected to be quite small provided that the intrinsic permeability is sufficiently large to assure the molecular diffusion regime prevails.

Return to Exercise 4


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