2.3 Film model
Steadystate ordinary molecular diffusion
In separation processes, it is of greatest interest to consider the mass transfer across an interface between a gas and a liquid (absorption, stripping, flash evaporation, distillation) or between two liquids (liquid liquid extraction).
In this section, the theoretical models to quantitatively describe the mass transfer across such an interface will be presented.
This theory is well known as film model. The key assumption is that the whole resistance to the diffusional mass transfer is entirely residing in the fluids themselves.
Thus meaning that the molecules pass across the interface without encountering resistance, but to diffuse into the bulk of the fluid, they have still to cross a thin, stagnant region, called film, where the whole resistance to mass transfer is present.
To better clarify the "role" of the film layer within this theory, we can compare it to a kind of membrane which a liquid must pass across, encountering, of course, a certain resistance. In both sides of the membrane, the liquids are homogeneous and the concentration is uniform. However, the concentrations are different on each side of the membrane and this difference is the driving force for the mass transfer across the membrane.
Please, be careful that this membrane represents the liquid film and not the interface, which, in accordance with the film theory, doesn't make any resistance to mass transfer!
Onefilm model
In the onefilm model, only one film (on the liquid side) is considered for the mass transfer resistance. We assume to have a pure gas of components A diffusing (absorbing) into a nonvolatile liquid B. Therefore no resistance is present in the gas side, and at the interface the equilibrium condition is given by the Henry's law:
c_{Ai} = H_{A} p_{A}
In the liquid phase, the assumption is made that all particles of component A, diffusing into the liquid film layer, will further diffuse into the liquid bulk. This is true since the film is assumed to be very thin. Bulk flow of component A is neglected so that the mass transfer of component A into the liquid B is totally due to the molecular diffusion.
From Fick's law is:
N_{A} =  D_{AB}[ dc_{A}/dz ]
In steadystate, we assume N_{A} = const.:
dN_{A}/dz = d [D_{AB} (dc_{A}/dz)]/dz = 0
d^{2}c_{A}/dz^{2} = 0
or working with the more convenient mole fraction: x_{A} = c_{A}/c being c the total, constant concentration.
N_{A} =  D_{AB}c [ dx_{A}/dz ]
and applying the material balance equation:
dN_{A}/dz = 0 with
x_{z=0} = x_{Ai} and x_{z=d} = x_{A, bulk}
N_{A} =  D_{AB}c [ (x_{Ai}  x_{A,bulk}) / d ]
Finally the ratio [D_{AB} / d] is replaced with the mass transfer coefficient k_{c}. Now the rate of mass transfer can be written as:
N_{A} = k_{c}c [x_{Ai}  x_{A,bulk}]
N_{A} = k_{c} [ c_{Ai}  c_{A,bulk}]
Therefore the onefilm model postulates that the change in concentration, entirely due to the molecular diffusion, follows a linear behaviour within a film layer of given thickness d.
The rate of mass transfer N_{A} is directly proportional to the change of concentration, which is the driving force for mass transfer.
Twofilms model
Let us now investigate the situation in which the diffusional resistances are considered in both the fluids in contact. Therefore now two films are present, gasside and liquidside, as shown in the picture.
In accordance with the film model theory, also in this case the entire resistance to molecular diffusion is generated by the two films so that it is possible to assume that the liquid and the gas phases are in equilibrium at the interface:
y_{i} = f_{eq.}(x_{i})
and if Henry's law can be applied, the equilibrium can be assumed linear:
y_{i} = m x_{i}
In steady state, the rate at which the molecules reach the interface from the gasside must be equal to the rate of diffusion in the liquidside, since no accumulation or depletion of molecules occurs at the interface. N is then equal to:
(1) gasside: N = k_{G}c_{G} [y  y_{i}] = k_{y} [y  y_{i}]
(2) liquidside: N = k_{L}c_{L} [x_{i}  x] = k_{x} [x_{i}  x]
where k_{y}and k_{x} are the locally applicable mass transfer coefficients.
Since (1) and (2) must be equal, it is possible to write:
k_{y} [y  y_{i}] = k_{x} [x_{i}  x]
The ratio  [k_{x} / k_{y}] = (y  y_{i}) / (x_{i}  x) is then the slope of the line connecting the equilibrium composition with the bulk composition, as shown in the picture here below.
Therefore if the locally mass transfer coefficients are known, as well as the bulk and equilibrium compositions, the rate of mass transfer can be calculated.
However, in experimental determination in real life, it is impossible with any physical instrumentation to get so close (molecularly close!) to the interface to sample and analyze the gas and liquid compositions at equilibrium. Therefore the only composition which can be measured is the bulk composition for gas and liquid.
To calculate the rate of mass transfer from experimental available data, a more appropriate expression for N must be found.
The overall mass transfer coefficient (MTC)
Let us look the here above graph (fig.3) and let us consider the overall driving force.
Dy = y  y* = y  mx
N = K_{Y} [y  y*]
where K_{Y} is the overall mass transfer coefficient.
To distinguish from the local coefficient, it is normally indicated with capital letter.
To relate the overall mass transfer_coefficient with the local one the following steps can be followed:
(a) N = K_{Y} [(y  y_{i}) + (y_{i}  y*)]
(b) N = K_{Y} [(y  y_{i}) + m (x_{i}  x)]
from (1) and (2) it is:
y  y_{i} = N / k_{y} ;
x_{i}  x = N / k_{x}
finally rearranging:
(3)
1 / K_{Y} = 1 / k_{y} + m / k_{x};
(4) 1 / K_{X} = 1 / (m k_{y}) + 1 / k_{x};
If the gas diffusing into the liquid is very soluble, the absorption is favoured, the slope of the equilibrium line, m, is very small and the second term of the second member in equation (3) is very small. In this case the following assumption is valid:
(3 bis)
1 / K_{Y} = 1 / k_{y} ;
On the other way round, if the gas diffusing into the liquid is not very soluble, the absorption is not favoured and the slope of the equilibrium line, m, is very large. The second term of the second member in equation (4) can be neglected and the following assumption is valid:
(4 bis)
1 / K_{X} = 1 / k_{x} ;
