CL: Two-resistence Model in Fluid-Fluid Mass Transfer

We have already shown how two heat flows in parallel can be combined using a resistance analogy. Here, we will examine mass transfer across a fluid-fluid boundary using the same logic.

Here we have a convective mass transfer in the liquid phase (from the bulk to the "surface" or interface) given by:

$\displaystyle{N_{A,L} = k_L(C_{A,L}-C_{A,i})}$

Similarly we have convection in the gas phase (or the second liquid phase) which is given by:

$\displaystyle{N_{A,G} = k_G(p_{A,i} - p_{A,G})}$

At steady state, these two fluxes need to be equal (why?!( so that we can write the ratio of mass transfer coefficients as:

$\displaystyle{\frac{k_L}{k_G} = \frac{(p_{A,i}-p_{A,G})}{(C_{A,L}-C_{A,i})}}$

Clearly if the transfer in one phase is considerably faster than in the other phase, the driving force in the "fast" phase goes to zero; however, in general, the convection coefficents might be of similar magnitude so that both phases must be considered.

In contrast to the case of combining radiation and convection (or any modes of heat transport, as we will see), combining these two mass transfers using resistances is not as simple (unless it is two liquids, how do we handle that case?). The main difficulty arising from problems combining the driving forces (concentrations versus pressures).

In order to handle this, we define two new variables: pA* as the pressure of A in equilibrium with the bulk concentration CA,L; and CA* as the concentration of A in equilibrium with the gas pressure pA,G.

NOTE:

While we know that, at the interphase, an equilibrium expression of the type pA,i = mCA,i can be written (Henry's Law), the "equilibrium" concentrations defined here are fictitious because we are using bulk concentrations!

Using these definitions, we can then write the mass flux across the interface as:

$\displaystyle{N_A = K_G(p_A^* - p_{A,G}) = K_L(C_{A,L}-C_A^*)}$

where KG and KL are overall transfer coefficients that combine both phase convection constants, kG and kL.

The origin of the relationship between "big" and "little" k's can be shown by rearranging the flux relation:

$\displaystyle{\frac{1}{K_G} = \frac{(p_A^*-p_{A,G})}{N_A} = \frac{(p_A^*-p_{A,i})+(p_{A,i}-p_{A,G})}{N_A}=}$

$\displaystyle{ \frac{m(C_{A,L}-C_{A,i})+(p_{A,i}-p_{A,G})}{N_A}= \frac{m}{k_L}+\frac{1}{k_G}}$

OUTCOME:

Use the two-resistence model to perform fluid-fluid mass transfer calculations