# Diffusion through a Porous Material

In a steady-state situation when gas flows through a porous material at a low velocity (laminar flow), the following empirical formula, Darcy’s model, is valid:

P»x = (4.301)

Where k represents the permeability of the matter (m2). The kinematic viscosity of the gas is denoted by v(m2/s), and p is the density of the gas for the total volume—or if the real density of the gas is d, p = 4>d, where is the volume percentage of the gas in the porous material. It is also seen that $ gives the percentage of the free cross-sectional area of the gas in the material:

P _ m(g)/V __ V(g) _ A(g)L _ A(g)* *

V d m(g)/V(g) V AL A ‘* *

In Eq. (4.301) velocity ux is the real velocity of the gas in the pores: ux = qv/A(g) = qv/(<p)A, where qv is the volume flow.

From Darcy’s equation we can determine a formula for the counterforce produced by the porous material to the flowing or diffusing component A. If this counterforce is found, it can be added to the diffusion resistance force caused by component B to component A; hence the sum of these two forces represents the total diffusion resistance.

For a porous material the linear momentum equation can be written as

PLh=-4,% + f’”*> (4302) * *

Where fmx represents the resistance force between the gas and the material, the flow friction. The term is important in Eq. (4.302). It comes from the fact that while p appears on the left side of Eq. (4.302), the balance is constructed

For the mixture of the material and the gas, and therefore the pressure must be calculated for the surface area of the total material tf>p. When <f) is held constant, independent of x, Eq. (4,302) is obtained.

In a steady-state case, Eq. (4.302) is simplified to

F ~ №* *

Tmx 00*’

And with Eq. (4.301),

With v = r/d = rj4>/p. If the flow velocity is zero, Eq. (4.303) can be interpreted as saying that the resistance force is linearly proportional to the velocity difference between the gas and the material and also linearly proportional to the dynamic viscosity of the gas.

Equation (4.303) is valid but it is lacking something. The resistance force f^x that applies to the component A has to be found, and not that for the whole mixture. The force applying to the whole mixture fmx is the sum of the partial forces and f^x:

F = fW + (4 304)

Fmx I mx / mx ‘ ».o-v rj* *

Assuming that the force is divided along the ratio of the mass flows, Eq. (4.303) gives

= — J^’Ax (4.305a)

Fmx = • (4.305b)

Summing (4.305a) and (4.305b), fmx is obtained for Eq. (4.303). This is due to the fact that pAvAx + pBvBx = pux.

We now consider the resistance force f^x caused by the diffusion. This force resists the diffusion flow in a porous material together with f^x. Writing the linear momentum equation for component A in accordance with Eq. (4.302),

P^ = ~4>^A + F! iAx, + f(/x) • (4.306i * *

This gives a model for f^x, Eq. (4.305b), but not a model for force fjx While

Force fmx gives the flow force caused by the material, it is normal to represent this

Fact so that fjx^ gives the pure diffusion resistance force that is not caused by the material. This requires treating fj^ independently from the material or porosity. For <f> = 1 or k = <=°, where f^J = 0, Eq. (4.306) gives

Pa^ = ~^ + ^ • (4307) In a steady-state case the results are

FdV = (4.308)

In a steady-state case at constant pressure (p = pA+ pB = constant), Pick’s law (Eq. (4.273)) is valid:

Pa(vax~vx) = jit’’* *

With cA = pA/{RT).* *

Instead of the mean velocity x weighted with the molar fractions, a velocity weighted with the mass fractions—the mass center velocity ux—can be used. Using Eq. (4.271), the above equation becomes

(4.309)

Equations (4.308) and (4.309) give a formula for the diffusion resistance force f(fx}:

Ft* = T5~Pa(v*x-“x) • (4.310)* *

Substituting the formulas for forces and (Eqs. (4.310) and (4.305a)) in Eq. (4.306) gives

DyAx _ ,dpA M RT, , VA$~ ,. — .* *

Pa dt *dx MaMbDabPa(Vax Ux k VAx’ 4-311)* *

In the steady-state case Eq. (4.311) is simplified to

4R, Јf‘,,‘—Jt7Tv — = -4b? • <4-312>

The aim is to solve this equation for the term vAx, or actually p^v^, which is the diffusion flow density of component A.

An important case is

"Bx |

= 0. (4.313)

This case applies to drying processes (B = dry air), condensation, and absorption, such as the diffusion of sulfur dioxide gas through a calcium oxide.

When Eq. (4.313) is valid, pux = Pa^ax + Pbvbx ~ PAyAxi and ux — (Pa^p)vax (p = Pa + Pb) • Therefore, in this case

V,4x = JVAX ■ (4.314)

Substituting Eq. (4.314) in Eq. (4.312) and solving for the mass flow density gives

PAVAx = ….. Kt’o—— TF* ■ (4*315)* *

M R7 Pb V<p* *

MAMBDAB P pk* *

This can be written as

Pb _ Pb^b _ P Pa__ ^b__ p pM p M

And

U = JL = JL. p 4>d <t> * *

Substituting these into Eq. (4.315) after grouping terms,

P^‘= p M, n <■ — ^ ■ <«i«> + ~—TrЈDARvl |

_________ P Mad $Pa* *

MAn <p P-Pa rT ab dx P-P’aKTDa^* *

The term

E =—————————— T (4.317)

1 + ^ A D v—* *

P-PaRTDab ‘* *

Is the resistance factor caused by the material. Using Eq. (4.317), the diffusion flow is

“ ~ep^rMD^ ■ <4’318) * *

The term p/{p-pA) derives from assumption (4.313) representing the effects of Stefan flow. If e = 1, Eq. (4.318) gives the diffusion flow in a free space.

Equation (4.318) indicates that if k—> the diffusion resistance re

Mains under 1, namely, ej = 4>- This is easy to understand, as 4> represents that part of the material cross-sectional surface through which the vapor diffuses.

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