Page 166 - Separation process principles 2

4.5 Ternary Liquid-Liquid Systems 131

and
~-~~lues enthalpies estimated from the P-R equation of state, compositions of the solute as mass or mole ratios instead of
the following results are obtained: mass or mole fractions.
krnolh Let:

Feed 120°F Vapor 112°F Liquid 112°F FA = feed rate of carrier A
Component 485 psia 165 psia 165 psia S = flow rate of solvent C
Hydrogen 1 .O 0.7
0.3 XB = ratio of mass (or moles) of solute B, to mass (or
Methane 27.9 15.2
12.7 moles) of the other component in the feed (F), raffi-
Benzene 345.1 0.4 344.7 nate (R), or extract (E)
113.4
Toluene - 0.04 113.36
Total 487.4 16.34 47 1.06 Then, the solute material balance is
Enthalpy, kT/h - 1,089,000 362,000 - 1,45 1,000
xF) F* = xF) s + xF) FA (4-21)
This case involves a wide-boiling feed, so the procedure involv-
and the distribution of solute at equilibrium is given by
ing (4-17) is the best choice. The above results show that only a
small amount of vapor (* = 0.0035), predominantly H2 and xp) = KbBxF) (4-22)
CHc is produced by the adiabatic flash. The computed flash tem-
perature of 112°F is 8°F below the feed temperature. The enthalpy where KbB is the distribution coefficient defined in terms of
of the feed is equal to the sum of the vapor and liquid product mass or mole ratios. Substituting (4-22) into (4-21) to elimi-
enthalpies for this adiabatic operation. nate XB (El gives

4.5 TERNARY LIQUID-LIQUID SYSTEMS

Temary mixtures that undergo phase splitting to form two It is convenient to define an extraction factor, EB, for the
separate liquid phases can differ as to the extent of sol~~bility solute B:
of the three components in each of the two liquid phases.
The simplest case is shown in Figure 4.13a, where only the
solute, component B, has any appreciable solubility in either The larger the value of E, the greater the extent to which the
the carrier, A, or the solvent, C, both of which have negligi- solute is extracted. Large values of E result from large values
ble (although never zero) solubility in each other. In this of the distribution coefficient, KbB, or large ratios of solvent
case, the equations can be derived for a single equilibrium to carrier. Substituting (4-24) in (4-23) gives the fraction of
stage, using the variables F, S, L('), and L(*) to refer, re- B that is not extracted as
spectively, to the flow rates (or amounts) of the feed, solvent,
exiting extract, and exiting raffinate. By definilion, the ex-
tract is the exiting liquid phase that contains the solvent and where it is clear that the larger the extraction factor, the
the extracted solute; the raffinate is the exiting liquid phase smaller the fraction of B not extracted.
that contains the carrier, A, of the feed and the portion of the Values of mass (mole) ratios, X, are related to mass (mole)
solute, B, that is not extracted. Although the extract is shown fractions, x, by
in Figure 4.13a as leaving from the top of the stage, this will
only be so if the extract is the lighter (lower-density) exiting
phase. Assuming that the entering solvent contains no Values of the distribution coefficient, Kb, in terms of ratios,
solute, B, it is convenient to write material balance and are related to KD in terms of fractions as given in (2-20) by
phase-equilibrium equations for the solute, B. These two
equations may be written in terms of molar or mass flow
rates. To obtain the simplest result, it is preferable to express

Solvent, S Solvent, S
Extract, E
Extract, E
component C components B, C component C components A, B, C
I
I

Figure
I I - 4.13 Phase splitting of ternary
Raffinate, R
Feed, F components A, B Feed, F components A, B, C mixtures: (a) components A and C mutually
components A, B components A, B
insoluble; (b) components A and C partially
(b) soluble.