CL: Diffusion Through a Stagnant Gas

Consider a tank of some volatile liquid, A, which is vented to the atmosphere through a chimney that has a stiff breeze blowing by it. Once we have reached steady state, there will be a mixture of A and air (which we will call B) within the chimney. The A is evaporating and flowing (diffusing) up through the column until it is ultimately blown away by the wind. The B, on the other hand, is completely stagnant (not moving up the chimney nor down the chimney).

This is clear if you consider that, at steady state with no accumulation, any B that moves down the chimney must exit the bottom and that any that moves up the chimney must be replaced at the bottom (if either of these is not met, there will be accumulation). Also note that this is not the case for the A since there is a steady stream of A being supplied by the liquid in the tank. In this case there is no net flux of B (NB = 0), but there is a non-zero flux of A ($N_A \neq 0$).

By definition, NA is given by

$\displaystyle{N_A = -c_{total}D\nabla y_A + c_AV}$

where the molar-averaged velocity V, we can written for a binary system as

$\displaystyle{V = \frac{c_Av_A + c_Bv_B}{c_A+c_B} = \frac{N_A +N_B}{c_{total}}}$

In this particular problem (diffusion through a stagnant gas) we have already noted that NB = 0, so

$\displaystyle{V = \frac{N_A + 0}{c_{total}}=\frac{N_A}{c_{total}}}$


We can have a non-zero bulk flow even in a "diffusion" problem (as long as the total fluxes do not cancel out). We must determine if there is a flow or not from examining the total mass flux term itself! (This is the most important difference between heat and mass transfer.)

Plugging in our expression for V, we reduces our expression for NA to

$\displaystyle{N_A = -c_{total}D\nabla y_A + y_AN_A}$

we can solve this for NA to give

$\displaystyle{N_A = -\frac{c_{total}D\nabla y_A}{1-y_A}}$

which in one dimension becomes

$\displaystyle{N_A = -\frac{c_{total}D\frac{dy_A}{dz}}{1-y_A}}$

We can then solve for the flux by integrating between z=0,L and note that the corresponding concentrations are yA = yAsat, 0

$\displaystyle{N_A\int_0^L dz = -c_{total}D\int_{y_{A_{sat}}}^0 \frac{dy_A}{1-y_A}}$

(Here we note that $N_A \neq F(z)$ by analogy to our previous example as well as heat transfer...if it did change with z, we would have accumulation!)

Integrating gives

$N_A L = c_{total}D(\ln{[1-0]} - \ln{[1-y_{A_{sat}}]}) = -c_{total}D\ln{[1-y_{A_{sat}}]}$

so that NA is

$\displaystyle{N_A = -\frac{c_{total}D}{L}\ln{[1- y_{A_{sat}}]}}$

which we note is not a function of z, as we stipulated.


Calculate the magnitude of diffusive mass flow/flux through a planar stagnant film