12.5. SINGLE-STAGE LIQUID–LIQUID EXTRACTION PROCESSES

12.5A. Introduction to Extraction Processes

In order to separate one or more of the components in a mixture, the mixture is brought into contact with another phase. The two-phase pair can be gas–liquid, which was discussed in Chapter 10; vapor–liquid, which was covered in Chapter 11; liquid–liquid; or fluid–solid. In this section liquid–liquid extraction separation processes are considered first. Alternative terms for the same process are liquid extraction or solvent extraction.

In distillation, the liquid is partially vaporized to create another phase, which is a vapor. The separation of the components depends on the relative vapor pressures of the substances. The vapor and liquid phases are similar chemically. In liquid–liquid extraction, the two phases are chemically quite different, which leads to a separation of the components according to physical and chemical properties.

Solvent extraction can sometimes be used as an alternative to separation by distillation or evaporation. For example, acetic acid can be removed from water by distillation or by solvent extraction using an organic solvent. The resulting organic solvent–acetic acid solution is then distilled. Choice of distillation or solvent extraction would depend on relative costs (C7). In another example, high-molecular-weight fatty acids can be separated from vegetable oils by extraction with liquid propane or by high-vacuum distillation, which is more expensive.

In the pharmaceutical industry, products such as penicillin occur in fermentation mixtures that are quite complex, and liquid extraction can be used to separate the penicillin. Many metal separations are being done commercially by extraction of aqueous solutions, such as copper–iron, uranium–vanadium, and tantalum–columbium.

12.5B. Equilibrium Relations in Extraction

1. Phase rule

Generally in a liquid–liquid system we have three components, A, B, and C, and two phases in equilibrium. Substituting into the phase rule, Eq. (10.2-1), the number of degrees of freedom is 3. The variables are temperature, pressure, and four concentrations. Four concentrations occur because only two of the three mass-fraction concentrations in a phase can be specified. The third must make the total mass fractions equal to 1.0: x_A + x_B + x_C = 1.0. If pressure and temperature are set, which is the usual case, then, at equilibrium, setting one concentration in either phase fixes the system.

2. Triangular coordinates and equilibrium data

Equilateral triangular coordinates are often used to represent the equilibrium data for a three-component system, since there are three axes. This is shown in Fig. 12.5-1. Each of the three corners represents a pure component, A, B, or C. The point M represents a mixture of A, B, and C. The perpendicular distance from the point M to the base AB represents the mass fraction x_C of C in the mixture at M, the distance to base CB the mass fraction x_A of A, and the distance to base AC the mass fraction x_B of B. Thus,

Equation 12.5-1

A common phase diagram where a pair of components A and B are partially miscible is shown in Fig. 12.5-2. Typical examples are methyl isobutyl ketone (A)–water (B)–acetone (C), water (A)–chloroform (B)–acetone (C), and benzene (A)–water (B)–acetic acid (C). Referring to Fig. 12.5-2, liquid C dissolves completely in A or in B. Liquid A is only slightly soluble in B and B slightly soluble in A. The two-phase region is included inside below the curved envelope. An original mixture of composition M will separate into two phases a and b which are on the equilibrium tie line through point M. Other tie lines are also shown. The two phases are identical at point P, the Plait point.

3. Equilibrium data on rectangular coordinates

Since triangular diagrams have some disadvantages due to the special coordinates, a more useful method of plotting the three-component data is to use rectangular coordinates. This is shown in Fig. 12.5-3 for the system acetic acid (A)–water (B)–isopropyl ether solvent (C). Data for this system are from Appendix A.3. The solvent pair B and C are partially miscible. The concentration of component C is plotted on the vertical axis and that of A on the horizontal axis. The concentration of component B is obtained by difference from Eq. (12.5-2) or (12.5-3):

Equation 12.5-2

Equation 12.5-3

The two-phase region in Fig. 12.5-3 is inside the envelope and the one-phase region outside. A tie line gi is shown connecting the water-rich layer i, called the raffinate layer, and the ether-rich solvent layer g, called the extract layer. The raffinate composition is designated by x and the extract by y. Hence, the mass fraction of C is designated as y_C in the extract layer and as x_C in the raffinate layer. To construct the tie line gi using the equilibrium y_A − x_A plot below the phase diagram, vertical lines to g and i are drawn.

EXAMPLE 12.5-1. Material Balance for Equilibrium Layers An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is equilibrated and the equilibrium phases separated. What are the compositions of the two equilibrium phases? Solution: The composition of the original mixture is xC = 0.30, xA = 0.10, and xB = 0.60. This composition of xC = 0.30 and xA = 0.10 is plotted as point h on Fig. 12.5-3. The tie line gi is drawn through point h by trial and error. The composition of the extract (ether) layer at g is yA = 0.04, yC = 0.94, and yB = 1.00 − 0.04 − 0.94 = 0.02 mass fraction. The raffinate (water)-layer composition at i is xA = 0.12, xC = 0.02, and xB = 1.00 − 0.12 − 0.02 = 0.86.

Another common type of phase diagram is shown in Fig. 12.5-4, where the solvent pairs B and C and A and C are partially miscible. Examples are the system styrene (A)–ethylbenzene (B)–diethylene glycol (C) and the system chlorobenzene (A)–methyl ethyl ketone (B)–water (C).

12.5C. Single-Stage Equilibrium Extraction

1. Derivation of lever-arm rule for graphical addition

This will be derived for use with the rectangular extraction-phase-diagram charts. In Fig. 12.5-5a two streams, L kg and V kg, containing components A, B, and C, are mixed (added) to give a resulting mixture stream M kg total mass. Writing an overall mass balance and a balance on A,

Equation 12.5-4

Equation 12.5-5

where x_AM is the mass fraction of A in the M stream. Writing a balance for component C,

Equation 12.5-6

Combining Eqs. (12.5-4) and (12.5-5),

Equation 12.5-7

Combining Eqs. (12.5-4) and (12.5-6),

Equation 12.5-8

Equating Eqs. (12.5-7) and (12.5-8) and rearranging,

Equation 12.5-9

This shows that points L, M, and V must lie on a straight line. By using the properties of similar right triangles,

Equation 12.5-10

This is the lever-arm rule, which states that kg L/kg V is equal to the length of line /length of line . Also,

Equation 12.5-11

These same equations also hold for kg mol and mol frac, lb_m, and so on.

EXAMPLE 12.5-2. Amounts of Phases in Solvent Extraction The compositions of the two equilibrium layers in Example 12.5-1 are for the extract layer (V), yA = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate layer (L), xA = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the amounts of V and L. Solution: Substituting into Eq. (12.5-4), Substituting into Eq. (12.5-5), where M = 100 kg and xAM = 0.10, Solving the two equations simultaneously, L = 75.0 and V = 25.0. Alternatively, using the lever-arm rule, the distance hg in Fig. 12.5-3 is measured as 4.2 units and gi as 5.8 units. Then, by Eq. (12.5-11), Solving, L = 72.5 kg and V = 27.5 kg, which is in reasonably close agreement with the material-balance method.

2. Single-stage equilibrium extraction

We now study the separation of A from a mixture of A and B by a solvent C in a single equilibrium stage. The process is shown in Fig. 12.5-6a, where the solvent, as stream V₂, enters together with the stream L₀. The streams are mixed and equilibrated and the exit streams L₁ and V₁ leave in equilibrium with each other.

The equations for this process are the same as those given in Section 10.3 for a single equilibrium stage, where y represents the composition of the V streams and x the L streams:

Equation 12.5-12

Equation 12.5-13

Equation 12.5-14

Since x_A + x_B + x_C = 1.0, an equation for B is not needed. To solve the three equations, the equilibrium-phase diagram in Fig. 12.5-6b is used. Since the amounts and compositions of L₀ and V₂ are known, we can calculate values of M, x_AM, and x_CM from Eqs. (12.5-12)–(12.5-14). The points L₀, V₂, and M can be plotted as shown in Fig. 12.5-6b. Then, using trial and error, a tie line is drawn through point M, which locates the compositions of L₁ and V₁. The amounts of L₁ and V₁ can be determined by substitution into Eqs. (12.5-12)–(12.5-14) or by using the lever-arm rule.