Page 200 - Separation process principles 2

5.3 Single-Section, Liquid-Liquid Extraction Cascades 165

5.3 SINGLE-SECTION, LIQUID-LIQUID For the second stage, a material balance for B gives
!
EXTRACTION CASCADES

Three possible two-stage, single-section, liquid-liquid ex-
with
yaction cascades are the cocurrent, crosscurrent, and coun-
$ tercurrent arrangements in Figure 5.5. The countercurrent
: arrangement is generally preferred because, as will be
[
shown in this section, that arrangement results in a higher Combining (5-17) with (5-18), (5-13), and (5-16) to elimi-
1 degree of extraction for a given amount of solvent and num- nate x:), Yi'), and Y:) gives
her of equilibrium stages.
[ In Section 4.5, (4-25), for the fraction of solute, B, that is
E not extracted, was derived for a single liquid-liquid equilib-
rium extraction stage, assuming the use of pure solvent, and
I Comparison of (5-19) with (5-13) shows that xf) = x") B.
i a constant value for the distribution coefficient, KbB , for the Thus, no additional extraction takes place in the second
solute, B, dissolved in components A and C, which are mu-
[ tually insoluble. That equation is now extended to multiple the first stage are at equilibrium and when they are recon-
stage. This is as expected because the two streams leaving
I stages for each type of cascade shown in Figure 5.5. tacted in stage 2, no additional net mass transfer of B occurs.
! Cocurrent Cascade Accordingly, a cocurrent cascade of equilibrium stages has
/
no merit other than to provide increased residence time.

If additional stages are added in the cocurrent arrangement
in Figure 5.5a, the equation for the first stage is that of a sin- Crosscurrent Cascade
gle stage. That is, from (4-25) in mass ratio units,
For the crosscurrent cascade shown in Figure 5.5b, the feed
progresses through each stage, starting with stage 1 and fin-
ishing with stage N. The solvent flow rate, S, is divided into
portions that are sent to each stage. If the portions are equal,
where E is the extraction factor, given by
the following mass ratios are obtained by application of
E = KbBS/FA (5-14) (5-13), where S is replaced by SIN, so that E is replaced
by E/N:
Since Y;) is in equilibrium with x;) according to
x!'/x~' = 1/(1 + E/N)
KbB = Y~')/X:' (5-15)
X~"/X;) = 1/(1+ E/N)
(5-20)
the combination of (5- 15) with (5- 13) gives

Water and ) Water and )
( dip;;;e ( dip;;;e y(l)
(pur~~f;i:eni /~fi~L~~
Water and R
Extract

Stage Stage Stage
x(R 112 of Stage Extract 2

pure benzene
Stage
y;;;
)
I
Raffinate Extract Raffinate Raffinatn benzene )

X(2) X(R) y(2) X(2) X(Rl X(2) X(R)
B= 0 B B= B B= B
(a) (b) (c)
Figure 5.5 Two-stage arrangements: (a) cocurrent cascade; (b) crosscurrent cascade; (c) countercun-ent cascade.