Equation (9.23) shows that full extraction can only be obtained for an infinitely large organic solvent current. Improved extraction can be obtained

with a cascade arrangement, such as that shown in figure 9.4. For the two — stage system shown in figure 9.4 the value of the fractional recovery becomes, for y2 = 0,

(9.24)

which is larger than the value given by equation (9.23). Generalizing to N stages, assuming that yN = 0 and defining = DIe/If we obtain the

equation for y0:

N — 1

У0 = DxnJ2 fl. (9.25)

i = 0

Mass conservation implies

IFx0 = IFxN + IE У0 (9.26)

which allows us to write

eN -1

У0 = Dx0 flN + 1 _ 1 (9.27)

One sees that for large N and P > 1 the efficiency tends towards one. For

P < 1,p! P for N! 1.

The preceding calculation assumes that yN = 0. This condition can easily be dropped and one gets an overall recovery of the extractable component

Note that, because of the definition of p = IEy0//Fx0, the term depending on yN is partly trivial since it gives a finite contribution even when there is no transfer from the aqueous to the organic phase (P = 0). If it is the extraction efficiency that is of interest, it might be more informative to use

IE yN yN

7—— = P — P —

IF x0 Dx0

N

For yN = 0, p = reff. When yN = 0, the extraction efficiency decreases, which seems natural since the final value xN = yN _ 1 /D has a finite minimum value yN/D.

We have also assumed that the value of D does not depend on the stage number. This is only true for rather dilute solutions, as we have seen above. When this is not the case we refer to the discussion in reference [138].[48]

It is important that the treatment is completely symmetric with respect to the nature of the phases. Thus it is possible to transfer a component from an organic phase to an aqueous phase. This makes recovering the expensive organic solvent possible, as shown in figure 9.5.