MHE: Analyzing Mass Transfer Equipment (tower-type, not mixed)

In order to analyze mass transfer equipment, we need to recognize that there are two ways that we may denote the amount of material being transferred:

Furthermore, we must recognize that the transport rate equations that we have encountered thus far for interphase transport, such as:

$\displaystyle{N_A = k_G(p_{A,G}-p_{A,i}) = K_G(p_{A,G}-p_A^*)}$

yield an answer with units of mass transferred per unit time per unit perpendicular area. In other words, this expression gives us an answer based on the area of contact between the two phases (a quantity that in many cases is immeasurable)

If we multiply these equations by a, the interfacial area per unit volume, and dz, a differential height section of a tower, we get an expression:

$\displaystyle{N_A a dz= k_Ga(p_{A,G}-p_{A,i}) dz= K_Ga(p_{A,G}-p_A^*)dz}$

which now has units of mass transferred per unit time per unit cross-sectional area. (Note: This step is essentiall multiplying by the interfacial area, Ai = aV, and then dividing by the cross-sectional area, Acs.)


The capacity coefficient is defined as the product of the convective mass transfer coefficient (either single-phase or interphase/overall) and the interfacial area per unit volume.

Writing a mass balance for species A over a differential segment of a tower (denoting the two phase flows as L and G), and then dividing by cross-sectional area, we get:

$\displaystyle{\frac{G}{A_{cs}}(y_{A,z+dz} - y_{A,z}) = \frac{L}{A_{cs}}(x_{A,z+dz} - x_{A,z})}$


Here we have implicitly assumed that we have a dilute solution so that the total flows L and G do not change as species A is transferred.

For a sufficiently small differential section, we can write this as:

$\displaystyle{\frac{G}{A_{cs}} dy_A = \frac{L}{A_{cs}} dx_A}$

Noting that either of these could be written as being equal to our modified flux (above; i.e., per unit cross-sectional area), we can write:

$\displaystyle{N_A a dz= K_Ga(p_{A,G}-p_A^*) dz = \frac{G}{A_{cs}}dy}$

Using our dilute assumption again, we can write the partial pressures in such a way as to get:

$\displaystyle{N_A a dz= K_GaP_{tot}(y_{A,G}-y_A^*) dz = \frac{G}{A_{cs}} dy}$

Re-arranging and preparing to integrate gives us:

$\displaystyle{ \int_o^L dz= \frac{G}{A_{cs}K_GaP_{tot}}\int_0^L\frac{dy}{(y_{A,G}-y_A^*)}}$

The problem with this expression is that yA* varies along the length (as the xA,G varies). Because of this, the easiest way to handle this type of problem is to note the following: For a dilute solution, we have already seen that the mass balance leads to a straight line (above), but the equilibrium expression will also lead to a straight line: yA* PTOT = HxA,G, with a constant value of H.

With this linear dependence of both of the equations, we can re-write the above equation in terms of the difference between yA,G and yA*, so that with:

$\displaystyle{\Delta = (y_{A,G}-y_A^*)}$

In order to make out change in variables, we can show that:

$\displaystyle{\frac{d\Delta}{dy_a}=\frac{\Delta_L -\Delta_0}{y_{A,L}-y_{A,0}}}$

therefore, our integral becomes:

$\displaystyle{\int_0^L dz = \frac{G}{A_{cs}K_GaP_{tot}} \frac{y_{A,G,L}-y_{A,G,0}}{\Delta_L-\Delta_o} \int_0^L \frac{d\Delta}{\Delta}}$

So that integrating gives:

$\displaystyle{L = \frac{G}{A_{cs}K_GaP_{tot}} \frac{y_{A,G,L}-y_{A,G,0}}{\Delta_L-\Delta_o} \ln \frac{\Delta_L}{\Delta_0}}$

and rearranging gives:

$\displaystyle{M_A = N_AaA_{CS}L =G (y_{A,G,L}-y_{A,G,0})}$

$\displaystyle{= K_GaP_{tot}LA_{CS} (y_{A,G}-y_A^*)_{LM}= K_GaVP_{tot} (y_{A,G}-y_A^*)_{LM}}$


Calculate the "average" driving force in a continuous contactor


Calculate the mass exchange in a continuous contactor and/or the size of the column