# CL: Equimolar Counter-Diffusion

In a distillation column, there is a temperature gradient as you move from the bottom of the column upwards. Because of this, if we assume that the system is at equilibrium everywhere, then, as (colder) liquid flows downward and (hotter) gas flows upward they must exchange material in order to maintain equilibrium (since the equilibrium is temperature dependent). If we think about a simple system: a binary mixture, the column is at constant pressure, and the latent heats of vaporization of the two components are similar, we can argue that the flux of A exactly counterbalances the flux of B (i.e., NA = -NB).

If we are interested in the flux and concentration profile of A, the governing equation is given by (see above for assumptions):

Recall that NA is given by

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

Further recall the definition of the molar-averaged velocity V, so that we can write it 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 (equimolar counter-diffusion) we have already noted that NA = -NB, so

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

this reduces our expression for NA to

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

For a one dimensional isobaric, isothermal system we may write this as

$\displaystyle{N_A = -D\frac{dc_A}{dz}}$

Assuming that the liquid surface is at the saturation concentration and that the concentration drops to some value $\displaystyle{C_{A_\delta}}$ at some distance $\displaystyle{\delta}$, we can then solve for the molar flow by rearranging:

$\displaystyle{\dot M_A = N_A A = -DA\frac{dc_A}{dz}}$

and integrating (why did $\displaystyle{\dot M_A}$ come out of the integral?!)

$\displaystyle{\dot M_A \int_0^\delta = -DA\int_{c_{A_{sat}}}^{c_{A_\delta}}dc_A}$

to get

$\displaystyle{\dot M_A \delta = -DA(c_{A_\delta} - c_{A_{sat}})}$

which can be re-written as

$\displaystyle{\dot M_{A_{0\to\delta}} = \frac{DA}{\delta}(c_{A_{sat}} - c_{A_\delta})}$

to highlight that it is very similar to our heat transfer solution (for conduction through a rectangular slab) for a variety of reasons:

• The flux is independent of position
• The flux is linearly dependent on the driving force (concentration difference)
• We can write the flow of mass (the flux times the area) as being equal to $\displaystyle{\frac{DA}{L}\Delta c}$ much like the conduction expression $\displaystyle{\frac{kA}{L}\Delta T}$
##### OUTCOME:

Calculate the magnitude of diffusive mass flow/flux through a planar film in equimolar counter-diffusion