Chapter 13. Liquid-Liquid Extraction

1. Explain what extraction is and outline the types of equipment used

2. Apply the McCabe-Thiele and Kremser methods to immiscible extraction

3. For partially miscible extraction, plot extraction equilibria on a triangular diagram. Find the saturated extract, saturated raffinate, and conjugate lines

4. For partially miscible extraction, find the mixing point and solve single-stage and cross-flow extraction problems

5. For partially miscible extraction for countercurrent systems, develop external mass balances, find the difference points, and step off the equilibrium stages

6. For partially miscible extraction, use difference points to plot operating line(s) on a McCabe-Thiele diagram

7. Use a computer simulation program to solve extraction problems

8. Do detailed design of mixer-settlers

Extraction is a common laboratory and commercial unit operation. For example, in commercial penicillin manufacturing, after the fermentation broth is sent to a centrifuge to remove cell particles, the penicillin is extracted from the broth. Then the solvent and the penicillin are separated from each other by one of several techniques. In petroleum processing, aromatic hydrocarbons such as benzene, toluene, and xylenes are separated from paraffins by extraction with a solvent such as sulfolane. The mixture of sulfolane and aromatics is sent to a distillation column, where sulfolane is recovered as bottoms product and is recycled back to the extractor. Flowcharts for acetic acid recovery from water, for the separation of aromatics from aliphatics, and uranium recovery are given by Robbins (1997). Reviews of industrial applications of extraction are presented by Lo et al. (1983) and Frank et al. (2008); and extraction of biological compounds, particularly proteins, is discussed by Belter et al. (1988) and Harrison et al. (2015). Although extraction is fairly common, the ratio of distillation to extraction units in industry is about 20 to 1 (Adler et al., 1998).

As these commercial examples illustrate, the complete extraction process includes the extraction unit and the solvent recovery process (Figure 13-1). In many applications the downstream solvent recovery step (often distillation or a chemical stripping step) is more expensive than the actual extraction step. Woods (1995a, Table 4-8) lists a variety of commercial processes and the solvent regeneration step employed. A variety of extraction cascades, including single-stages, countercurrent cascades, and cross-flow cascades, can be used; we discuss these later.

The variety of equipment used for extraction is much greater than for distillation, absorption, and stripping. Efficient contacting and separating of two liquid phases is considerably more difficult than contacting and separating a vapor and a liquid. In addition to plate and packed (random, structured and membrane) columns, many specialized pieces of equipment have been developed. Some of these are illustrated in Figure 13-2. Details of the different types of equipment are provided by Frank et al. (2008), Godfrey and Slater (1994), Humphrey and Keller (1997), Lo (1997), Lo et al. (1983), Robbins and Cusack (1997), Scheibel (1978), and Woods (1995a). A summary of the features of the various types of extractors is presented in Table 13-1. The following heuristics are useful for making an initial decision on the type of extraction equipment to use (Frank et al., 2008; King, 1981; Lo, 1997; Robbins, 1997):

1. If one or two equilibrium stages are required, use mixer-settlers. Detailed design of mixer-settlers is discussed in Section 13.14.

2. If three equilibrium stages are required, use a mixer-settler, a sieve tray column, a packed column (random or structured), or a membrane contactor.

3. If four or five equilibrium stages are required, use a sieve tray column, a packed column (random or structured), or a membrane contactor.

4. If more than five equilibrium stages are required, use one of the systems that apply mechanical energy to a column (Figure 13-2).

Godfrey and Slater (1994) present detailed chapters on all the major types of extraction equipment with considerable detail on mass transfer rates. Humphrey and Keller (1997) discuss equipment selection in considerable detail and have height equivalent to a theoretical plate (HETP) and capacity data. Frank et al. (2008) discusses equipment selection, holdup, flooding, and mass transfer.

The number of equilibrium contacts required can be determined using the same stage-by-stage procedures for all extractors. Determining stage efficiencies and hydrodynamic characteristics is more difficult. The more complicated systems are designed by specialists (e.g., see Lo, 1997, or Skelland and Tedder, 1987).

Solvent selection is critical for the development of an economical extraction system. Solvent should be highly selective for the desired solute and not very selective for contaminants. Both the selectivity, α = K_desired/K_undesired, and K_desired should be large [K is defined later in Eq. (13-6)]. The solvent should be easy to separate from diluent as either a totally immiscible system or a partially miscible system where separation by distillation is easy. In addition, the solvent should be nontoxic, noncorrosive, readily available, chemically stable, environmentally friendly, and inexpensive.

A useful, relatively simple approach to selecting a solvent with reasonably large selectivity is to use the solubility parameter (Giddings, 1991; King, 1981; Walas, 1985; Woods, 1995a). The solubility parameter δ is defined as

ΔE_v is the latent energy of vaporization, and V_i is the molal volume for component i. The solubility parameter has the advantage of being a property of only the pure components, it is easily calculated from parameters that are easy to measure and are often readily available, and tables of δ are available (Giddings, 1991; King, 1980; Walas, 1985; Woods, 1995b). The solubility parameter is useful for quick estimation of the miscibility of two liquids—the closer the values of δ_A and δ_B, the more likely the liquids are miscible. For example, the δ values in (cal/ml)^1/2 are: water = 23.4, ethanol = 12.7, and benzene = 9.2 (Giddings, 1991). Water and ethanol are miscible (although they are nonideal and the vapor-liquid equilibrium (VLE) shows an azeotrope), water and benzene are immiscible, and ethanol and benzene are miscible. Frank et al. (2008), King (1981), Lo et al. (1983), and Treybal (1963) discuss solvent selection in detail.

Rules for selecting solvents are changing. Because of increased concern about the environment, climate change, and the desire to use “green” processes, solvents that used to be quite acceptable may be unacceptable now or in the future (Allen and Shonnard, 2002). There is a tendency to stop using chlorinated solvents and to use fewer hydrocarbon solvents. This has led to increased interest in ionic liquids (Forsyth et al., 2004; Meindersma et al., 2012), aqueous two-phase extraction (Albertsson et al., 1990; Harrison et al., 2015), and the use of supercritical extraction (see Section 14.5). If a designer has a reasonable choice of solvents, consideration of life-cycle costs may lead to the use of a “greener” solvent even though it may be somewhat more expensive initially. However, a cautionary note: new solvents may have unexpected consequences that make them significantly less benign in the environment.

Solute, A, is initially dissolved in diluent, D, in the feed. Solute is extracted with solvent, S. The entering solvent stream is often presaturated with diluent. Streams with high concentrations of diluent are called raffinate, while streams with high concentrations of solvent are extract. The nomenclature in both weight fraction and weight ratio units is given in Table 13-2.

For equilibrium conditions, the Gibbs phase rule is F = C – P + 2. There are three components (solute, solvent, and diluent) and two phases. Thus, there are three degrees of freedom. In order to plot equilibrium data as a single curve, we must reduce this to one degree of freedom. The following two assumptions are usually made:

1. The system is isothermal.

2. The system is isobaric.

To have the energy balances automatically satisfied, we must also assume

3. The heat of mixing is negligible.

These three assumptions are usually true for dilute systems. For very dilute systems, we can usually analyze immiscible extraction systems using constant total flow rates. Thus, the fourth assumption we usually make is that total flow rates are constant:

As the system becomes more concentrated, the amount of solute transferred from the raffinate to the extract becomes significant, and total flow rates of R and E are no longer constant. If we switch to a ratio analysis similar to the analysis for absorption and stripping (Section 12.6), the operating line will be straight, which makes it easy to work with, and the solvent and diluent mass balances will be automatically satisfied if the following assumption is valid:

5. Diluent and solvent are totally immiscible.

When the fifth assumption is valid, then

These are flow rates of diluent only and solvent only and do not include the total raffinate and extract streams. Equations (13-3a and b) are the diluent and solvent mass balances, and they are automatically satisfied when the phases are immiscible. Concentrated immiscible systems will be studied in Section 13.5.

Most solvent-diluent pairs that are essentially completely immiscible become partially miscible as more solute is added (see Section 13.7). This occurs because appreciable amounts of solute make the two phases chemically more similar.

In this section we consider the simplest analysis—that for very dilute systems in which the total flow rates can be treated as constant (assumption 4). For the mass balance envelope shown in Figure 13-3, the mass balance is

Solving for y_j+1 we obtain the operating equation

Since R/E is assumed to be constant, this equation plots as a straight line on a y vs. x (McCabe-Thiele) graph.

Equilibrium data for dilute extraction are usually represented as a distribution ratio, K_d,

in weight fractions, mole fractions, or concentrations (e.g., kg/m³). For very dilute systems K_d will be constant, whereas at higher concentrations K_d often becomes a function of concentration. Values of K_d are tabulated in Frank et al. (2008, pp. 15–29 to 15–31), Robbins and Cusack (1997, pp. 15–10 to 15–15), Hartland (1970, Chap. 6), and Francis (1972). A brief listing is given in Table 13-3. The value of K_d in the table in Perry’s Handbook (Frank et al., 2008) ranges from 0.0012 to 181 for different solute-diluent-solvent combinations. Even for the same solute (phenol) in the same diluent (water), the value of K_d varies from 0.040 in isopropyl acetate (a nonselective solvent) to 39.8 in methyl isobutyl ketone.

It is common to think of K_d as dimensionless, but it has dimensions. For example, in weight fraction units the dimensions of K_d,wt are (g A/g extract)/(g A/g raffinate), and in mole fraction units the dimensions of K_d,mol are (mol A/mol extract)/(mol A/mol raffinate). If the molecular weights of extract and raffinate are not equal, then K_d,wt ≠ K_d,mol.

The equilibrium “constant” is temperature- and pH-dependent. The temperature dependence is illustrated in Table 13-3 for the distribution of acetic acid between water (the diluent) and benzene (the solvent). Note that there is an optimum temperature at which K_d is a maximum. Benzene is not a good solvent for acetic acid because the K_d values are low, but water would be a good solvent if benzene were the diluent. As shown in Table 13-3, 1-butanol is a much better solvent than benzene for acetic acid. In addition, benzene would probably not be used as a solvent because it is carcinogenic. The use of extraction to fractionate components requires that the selectivity, α₂₁ = K_d2/K_d1, be large. An example where fractional extraction (see Section 13.3) is feasible is the separation of ethylbenzene and xylenes illustrated in Table 13-3. Although feasible, this separation is not economical because the K_d values are too low.

Extraction equilibria are dependent on temperature, pH, and the presence of other chemicals. Although temperature dependence is small in the immiscible range (see Table 13-3), variations with temperature can be large when the solvents are partially miscible. Biological molecules, particularly proteins, can have an order of magnitude change in K_d when buffer salt is changed and may have several orders of magnitude change in K_d when pH is varied (Harrison et al., 2015). Extraction processes often use shifts in temperature or pH to extract and then recover a solute from the solvent (Frank et al., 2008). For example, penicillin is extracted in a process with a pH swing.

The McCabe-Thiele diagram for extraction can be obtained by plotting the equilibrium data, which is a straight line if K_d is constant, and the operating line, which is also straight if assumption 4 (constant total flow rates) is valid. The operating line goes through the point (x_N, y_N+1), which are the concentrations of passing streams in the column. Since this procedure is very similar to stripping (e.g., Figure 12-5, but in weight fraction units), we will consider a challenging example.

The McCabe-Thiele diagram shown in Figure 13-4 is very similar to the McCabe-Thiele diagrams for dilute stripping. This is true because the processes are analogous unit operations in that both contact two phases and solute is transferred from the x phase to the y phase. The analogy breaks down when we consider stage efficiencies and sizing the column diameter, since mass transfer characteristics and flow hydrodynamics are very different for extraction and stripping.

In stripping there was a maximum L/G. For extraction, a maximum value of R/E can be determined in the same way. This gives the minimum solvent flow rate, E_min, for which the desired separation can be obtained with an infinite number of stages.

For simulation problems the number of stages is specified, but the outlet raffinate concentration is unknown. A trial-and-error procedure is required in this case. This procedure is essentially the same as the simulation procedure used for absorption or stripping.

Dilute multicomponent extraction can be analyzed on a McCabe-Thiele diagram if we add one more assumption:

6. Each solute is independent.

When these assumptions are valid, the entire problem can be solved in mole or weight fractions. Then for each solute the mass balance for the balance envelope shown in Figure 13-3 is

where i represents the solute and terms are defined in Table 13-2. This equation is identical to Eq. (13-5) when there is only one solute. The operating lines for each solute have the same slopes but different y intercepts. The equilibrium curves for each solute are independent because of assumption 6. Then we can solve for each solute independently. This solution procedure is similar to the one for dilute multicomponent absorption and stripping (Section 12.8). We first solve for the number of stages, N, using the solute that has a specified outlet raffinate concentration. Then with N known, a trial-and-error procedure is used to find x_N+1 and y₁ for each of the other solutes.

Equation (13-5) or (13-8) can also be used, with care, if there is a constant small amount of solvent dissolved in the raffinate streams and a constant small amount of diluent dissolved in the extract streams. Flow rates R and E should include these small concentrations of the solvent or diluent. Unless entering streams are presaturated with solvent and diluent, R = R₁ > R₀ by the amount of solvent dissolved in the raffinate and R_N < R_N–1 = R by the amount of diluent that dissolves in entering solvent stream E_N+1. Similar adjustments need to be made to extract flow rates.

7. Equilibrium is linear.

In this case, equilibrium has the form

Usually b_i = 0 and m_i = K_d,i. The dilute extraction model now satisfies all the assumptions used to derive the Kremser equations in Chapter 12, so they can be used directly. Since we have used different symbols for flow rates, we replace L/V with R/E. Then Eq. (12-12) becomes

while if R/(mE) ≠ 1, Eqs. (12-21) (inverted) and (12-22) become

with y₁* = mx_o + b and

Other forms of the Kremser equation can also be written with this substitution. When the Kremser equation is used, simulation problems are no longer trial-and-error.

The grouping mE/R = K_dE/R is known as the extraction factor. Note that mE/R = (yE)/(Rx). If mE/R > 1, then there is more solute in the extract phase, and in Figure 13-3 net movement of solute is to the right. If mE/R < 1, then there is more solute in the raffinate phase, and net movement of solute is to the left in Figure 13-3. If the goal is to remove solute from the diluent, then to have a reasonable solute recovery, we must have mE/R > 1, and the absolute theoretical minimum value of the extraction factor is mE/R = 1. A practical minimum value that is often used for the preliminary design of extractors is (Frank et al., 2008)

As we will see when we discuss partially miscible extraction, there is also a maximum solvent to feed ratio when the solvent dissolves the entire feed and only one phase is formed.

Except for very dilute systems, the equilibrium will often be nonlinear and the various forms of the Kremser equation are not strictly valid. However, for modest curvature, a geometric average value of m (Frank et al., 2008)

often gives a good fit. For the best fit, the value of b must also be fit and will not be zero.

Application of the Kremser equation to extraction is considered in detail by Hartland (1970), who also gives linear fits to equilibrium data over various concentration ranges. As the feed becomes more concentrated and more solute is transferred into solvent, the total flow rate E increases while R decreases. The applicability of the McCabe-Thiele method and sometimes the Kremser equation can be extended by using solvent and diluent flow rates with ratio units (see Section 13.5).

A common problem in fractional extraction is the center cut. In Figure 13-6 solute B is the desired solute, while A represents a series of solutes that are more strongly extracted by solvent 1, and C represents a series of solutes that are less strongly extracted by solvent 1. Center cuts are common when pharmaceuticals are produced by fermentation, since a host of chemicals are produced.

Before looking at the analysis of fractional extraction, it will be helpful to develop a simple criterion to predict whether a solute will go up or down in a given column. At equilibrium solutes distribute between the two liquid phases. The ratio of solute flow rates in the column is

where we have used the equilibrium expression, Eq. (13-6). If (K_d,AE/R)_j > 1, the net movement of solute A is up at stage j, while if (K_d,AE/R)_j < 1, the net movement of solute is down at stage j. In Figure 13-5 we want

in both sections of the column. By adjusting K_D,i (say, by changing the solvents used or temperature) and/or the two solvent flow rates (E and R), we can change the direction of movement of a solute. This was done to solute B in Figure 13-6; B goes down in the first column and up in the second column. Ranges of (K_d,iE/R) can be derived that will make the fractional extractor work. Since it is quite expensive to have a large number of equilibrium stages in a commercial extractor, the ratios in Eq. (13-14b) should be significantly different.

Analysis of fractional extraction is straightforward for dilute mixtures with independent solutes and constant total flow rates in each section. External mass balances for the fractional extraction cascade shown in Figure 13-5 are

If the feed is contained in solvent 1, flow rates in the two sections are related by the expressions

and if the feed is in solvent 2,

The solute operating equations for the top section, which use the top mass balance envelope in Figure 13-5, is Eq. (13-8). For the bottom section the mass balances are

which is identical to single-component Eq. (13-7). Since feed is usually dissolved in one of the solvents, the phase flow rates are usually slightly different in the two sections. A McCabe-Thiele diagram can be plotted for each solute. Each diagram will have two operating lines and one equilibrium line, as shown in Figure 13-7.

Figures 13-7A and B show the characteristics of both absorber and stripper diagrams. The solutes are being “absorbed” in the top section (that is, the solute concentration is increasing as we go down the column) while the solute is being “stripped” in the bottom section (solute concentration is increasing as we go up the column). Thus, the solute is most concentrated at the feed stage and diluted at both ends (because we add lots of extra solvent). The two operating lines will intersect at a feed line, which is at the concentration of solute in the feed. This feed concentration is usually much greater than the solute concentration on the feed stage. If there is no solvent in the feed (the feed is liquid A + B), then the effective feed concentrations (y_i or x_i) are very large, and the operating lines are almost parallel.

The top section of the column in Figures 13-5 and 13-7 is removing (“absorbing”) component B from A. The more stages in this section, the purer the A product will be (smaller y_B,1), and the higher the recovery of B in the B product will be. The bottom section of the column is removing (“stripping”) component A from B. Extra stages in the bottom section increase the purity of the B product (reduce x_A,N) and increase the recovery of A in the A product. Specifications would typically include temperature, pressure, feed composition (A, B, and solvents), feed flow rate, and both solvent compositions.

Brian (1972, Chapt. 3) explores fractional extraction calculations in detail. He illustrates the use of extract reflux and derives forms of the Kremser equation for multisection columns. An abbreviated treatment of the Kremser equation for fractional extraction is also presented by King (1981).

To derive an operating equation, we use a mass balance envelope around a single stage, as shown around stage j in Figure 13-8. For dilute systems the resulting steady-state mass balance is

Solving for outlet extract mass fraction, y_j, we obtain the operating equation,

Each stage will have a different operating equation. On a McCabe-Thiele diagram plotted as y vs. x, each operating line is a straight line of slope –R/E_j and y intercept (x_j–1 R/E_j + y_j,in). In these equations y_j,in is the mass fraction of solute in the extract entering stage j and E_j is the flow rate of solvent entering stage j. The designer can specify all values of E_j and y_j,in as well as x₀, R, and either x_N or N. If the calculation is started at the first stage (j = 1), x_j–1 = x₀ is known and the operating equation can be plotted. Since the stage is an equilibrium stage, x₁ and y₁ are in equilibrium in addition to being related by operating equation (13-20). Thus, intersection of the operating line and the equilibrium curve is at y₁ and x₁ (see Figure 13-9). This is also the solution for a single-stage system. For a cross-flow system, the raffinate input to stage 2 is x₁. Thus, point (y_2,in, x₁) on the operating line for stage 2 is known and can be plotted. The procedure is repeated until x_N is reached or N stages have been stepped off.

In general, each stage can have different solvent flow rates, E_j, and different inlet solvent mass fractions, y_j,in, as illustrated in Figure 13-9. This figure also shows that the point representing the inlet concentrations (y_j,in, x_j–1) is on the operating line for each stage, which is easily proved by showing that if y_j = y_j,in and x_j = x_j–1, Eq. (13-20) is satisfied.

The operating lines in Figure 13-9 are similar to those we found for binary flash distillation. Both single-stage systems and cross-flow systems are arranged so that the two outlet streams are in equilibrium and on the operating line. This is not true of countercurrent systems.

Analysis for dilute multicomponent systems follows as a logical extension of the independent solution for each solute, as was discussed for countercurrent systems. Total flow rates and mole or weight fractions are used in these calculations. Note that simulation problems do not require trial-and-error solution for cross-flow systems.

The analysis procedure for cross-flow stripping and absorption systems is very similar to the extraction calculations developed here (see Problem 12.D12).

In countercurrent systems solvent is reused in each stage, while in cross-flow systems it is not. Because solvent is reused, countercurrent systems can obtain more separation with same total amount of solvent and same number of stages. They can also obtain both high purity (x_N+1 small) and high yield (high recovery of solute). Cross-flow systems can obtain either high purity or concentrated solvent streams but not both. For extraction cross flow may have an advantage when flooding or slow settling of phases is a problem. For absorption and stripping, pressure drop may be significantly lower in a cross-flow system. Cascades that combine cross-flow and countercurrent operation have been studied by Bogacki and Szymanowski (1990).

When diluent and solvent are immiscible, Eqs. (13-3) are valid. To use constant diluent and solvent flow rates in mass balances, we need to convert fractions to ratios. For immiscible phases, the weight ratio units are related to weight fractions as

where X is kg solute/kg diluent and Y is kg solute/kg solvent.

The operating equation can be derived with reference to the mass balance envelope shown in Figure 13-3. In weight ratio units, the steady-state mass balance is

where weight fractions in Figure 13-3 have been converted to weight ratios. Solving for Y_j+1 we obtain the operating equation,

When plotted on a McCabe-Thiele diagram of Y vs. X, this is a straight line with slope F_D/F_S and Y intercept (Y_l – (F_D/F_S)X₀). Since F_D and F_S are constant, the operating line is straight. For a typical design problem, F_D/F_S will be known, as will X₀, X_N and Y_N+1. Since X_N and Y_N+1 are the concentrations of passing streams, they represent the coordinates of a point on the operating line.

For the McCabe-Thiele diagram, if flow rates are given as in Eq. (13-3), equilibrium data must be expressed as weight or mole ratios. Equation (13-22) can be used to transform the equilibrium data, which is easy to do in tabular form. Usually equilibrium data are reported in fractions, not the ratio units given in the second part of Table 13-3.

With the equilibrium data and the operating equation known, the McCabe-Thiele diagram is plotted. First point (Y_N+1, X_N) is plotted, and the operating line passes through this point with a slope of F_D/F_S. Y₁ can be found from the operating line at the inlet raffinate concentration X₀. Then stages are stepped off. The procedure will become clearer by considering the following example.

When there is fairly significant partial miscibility of diluent and solvent, the McCabe-Thiele analysis can still be used if the following alternative assumption is valid:

5 Alternate. Solvent concentration in raffinate and diluent concentration in extract are both constant.

Flow rates of diluent and solvent streams are now defined as

The ratio units are defined as

The calculation procedure now follows Eqs. (13-23) to (13-24).

For all of these situations, as long as F_D and F_S are constant and the equilibrium in ratio units can be approximated as a straight line, Y = m_ratioX + b_ratio, the Kremser equations [e.g., Eq. (13-11a) or (13-11b)] are valid if written in terms of mass or mole ratios with F_D replacing R and F_S replacing E (Brian, 1972). Problems 13.D12 and 13.D17 compare the McCabe-Thiele analysis to the Kremser equation.

When phases are partially miscible and assumption 5 Alternate is not valid, the methods developed in sections 13.8 to 13.10 should be used.

For dilute immiscible phases, and = Ê. Solving for y, we obtain the operating equation

Except that this equation uses masses of raffinate and extract instead of flow rates, it is essentially the same as the continuous operating equation for a single-stage system [Eq. (13-19) with j = 1]. Thus the solution is identical to the continuous solution and can be obtained either graphically (Figures 13-9 and 13-10) or analytically [Eq. (13-21)] for linear isotherms with j = 1.

If solvent and diluent are immiscible and a concentrated solute is recovered, the methods developed in Section 13.5 can be used. Then the solvent and diluent balances are

The solute balance is

and the operating equation is

Equations (13-27) are also essentially the same as the continuous equations for a single-stage system (e.g., see problem 13.D16) except for the different units. If there is substantial mutual miscibility, then the methods developed in Sections 13.8 to 13.10 need to be used.

Additional purification can be attained by doing repeated batch extractions in the same tank. If the fresh feed is contacted with fresh solvent, the extract phase is removed, and then the raffinate is contacted again with fresh solvent, the operation is essentially identical to continuous cross-flow systems. The use of two-stage countercurrent contacting is more common (Frank et al., 2008). To start up this operation, the fresh feed is contacted first with fresh solvent. After removing the extract phase, the raffinate is contacted with fresh solvent again. The raffinate is removed as raffinate product, and the extract phase is saved in the tank. This completes the startup for the two-stage countercurrent batch contactor. For the next batch, the fresh feed is contacted with the saved extract. After settling and removal, the extract phase from this contacting is the extract product. The raffinate phase remaining in the tank is contacted with fresh solvent, and the mixture is allowed to settle. The raffinate phase is removed as the raffinate product, and the extract phase is left in the tank ready for the next batch. The operation can be repeated many times. Sketching the process will probably be helpful in understanding it (see Problem 13.A14).

If we want to totally remove solute from diluent and have it dissolved in a solvent, continuous solvent addition batch extraction, shown in Figure 13-12, will use less solvent than repeated single-stage batch extractions. A solvent that is presaturated with diluent is added continuously to a mixed tank that contains the feed. Raffinate and extract phases are separated in a continuous settler with withdrawal of extract product and recycle of diluent. This process is analogous to constant mole batch distillation and analysis for immiscible extraction is very similar to the analysis in Section 9.4. If we assume that the mass in the mixing tank is constant, then the overall mass balance becomes in = out, and for dkg of entering solvent,

For the component balance, since there is no entering solute, the equation is –out = accumulation,

where is mass of raffinate phase in the tank plus settler, and x_t is mass fraction solute in the raffinate phase. If we assume that raffinate holdup is much greater than solvent holdup Ê_t then , and Eq. (13-29b) simplifies to

Assuming is constant, substituting in Eq. (13-28a) and integrating, we obtain

where y and x_t are in equilibrium, and is total mass of solvent added. This equation can always be integrated numerically or graphically. If equilibrium is linear, y = K_dx, analytical integration is straightforward:

Numerical calculations for this process are in Problem 13.D20.

If partial miscibility of solvent and diluents is significant, the methods in Sections 13.8 to 13.10 need to be used. Development of Eqs. (13-28a) to (13-29b) is valid only after enough solvent has been added to the feed that two phases form. Since K_d is not usually constant, Eq. (13-29b) cannot be used. Integration of Eq. (13-29a) usually needs to be done numerically. The resulting amount of solvent added, _t, is the amount added after addition of enough solvent to form two phases.

Extraction systems are noted for the wide variety of equilibrium behavior that can occur. In the partially miscible range utilized for extraction, two liquid phases are formed. At equilibrium, temperatures and pressures of the two phases will be equal and compositions of the two phases will be related. The number of independent variables that can be arbitrarily specified (i.e., the degrees of freedom) for a system at equilibrium can be determined from Gibbs phase rule F = C – P + 2, which for a ternary extraction is F = 3 – 2 + 2 = 3 degrees of freedom. Temperature and pressure are almost always constant, so only one degree of freedom remains. Thus, if we specify composition of one component in either phase, all other compositions will be set at equilibrium even if there is significant partial miscibility.

Ternary extraction equilibrium data are usually shown graphically as either right triangular diagrams or equilateral triangular diagrams. Figure 13-13 shows the data listed in Table 13-5 for the system water-chloroform-acetone at 25ºC on a right triangular diagram. We have chosen chloroform as solvent, water as diluent, and acetone as solute. We could also call water the solvent and chloroform the diluent if the feed was an acetone-chloroform mixture. Curved line AEBRD represents the solubility envelope for this system. Any point below this line represents a two-phase mixture that will separate at equilibrium into a saturated extract phase and a saturated raffinate phase. Line AEB is the saturated extract line, and line BRD is the saturated raffinate line. Point B is called the plait point where extract and raffinate phases are identical. Remember that the extract phase is the phase with the higher concentration of solvent. Tie line ER connects extract and raffinate phases that are in equilibrium.

Point N in Figure 13-13 is a single phase because the ternary system is miscible at these concentrations. Point M represents a mixture of two phases, since it is in the immiscible range for this ternary system. At equilibrium the mixture represented by M will separate into a saturated raffinate phase and a saturated extract phase in equilibrium with each other. Either of the conjugate lines shown in Figure 13-13 can be used to draw tie lines. Consider tie line ER, which was found by drawing a horizontal line from point E to the conjugate line (point C) and then a vertical line from point C to the saturated raffinate curve (point R). Points E and R are in equilibrium, so they are on the ends of a tie line. This construction is shown in Figure 13-14 for different equilibrium data. This procedure is analogous to the use of an auxiliary line on an enthalpy-composition diagram, as illustrated in Figure 2-5.

To find the raffinate and extract phases that result when mixture M separates into two phases, we need a tie line through point M. This requires a simple eyeball trial-and-error calculation. Guess the location of the end point of the tie line on the saturated extract or raffinate curve, construct a tie line through this point, and check whether the line passes through point M. If the first guess does not pass through M, repeat the process until you find a tie line that does. This is not too difficult because tie lines that are close to each other are approximately parallel.

The solubility envelope, tie lines, and conjugate lines shown on the triangular diagrams are derived from experimental equilibrium data. To obtain these data, a mixture can be made up and allowed to separate in a separatory funnel. Then the concentrations of extract and raffinate phases in equilibrium are measured. This measurement gives location of one point on the saturated extract line, one point on the saturated raffinate line, and the tie line connecting these two points. One point on the conjugate line can be constructed from a tie line by reversing the procedure used to construct a tie line when the conjugate line was known (Figure 13-14).

Equilibrium data represented by Figure 13-13 are often called a type I system, since there is one pair of immiscible binary compounds. It is also possible to have systems with zero, two, and three immiscible binary pairs (Alders, 1959; Macedo and Rasmussen, 1987; Sorenson and Arlt, 1979, 1980; Walas, 1985). It is possible to go from a type I to a type II system as temperature decreases. This is shown in Figure 13-15 (Fenske et al., 1955) for the methyl cyclohexane-toluene-ammonia system. At 10ºF this is a type II system.

We use right triangular diagrams exclusively in the remainder of this chapter because they are easy to read, they don’t require special paper, the scales of the axes can be varied, and portions of the diagram can be enlarged. Although equilateral diagrams have none of these advantages, they are used extensively in the literature for reporting extraction data; therefore it is important to be able to read and use this type of extraction diagram.

Equilibrium data can be correlated and estimated with thermodynamic models that calculate activity coefficients. Although these calculations are similar to those for VLE, they are more complicated and generally less accurate. An extensive compilation of data and UNIQUAC and NRTL parameters is given by Sorenson and Arlt (1979, 1980) and Macedo and Rasmussen (1987).

For the mixing operation in Figure 13-16A, the flow rates F₁ and F₂ would be known as well as the concentration of the two feeds: x_A,F1, x_D,F1, x_A,F2, x_D,F2. The three independent mass balances used to solve for M, x_A,M, and x_D,M are

The concentrations of the mixed stream M are

Points F₁, F₂, and M are shown as collinear in Figure 13-16B. Problem 13.C5 asks you to show that

Equation (13-32), the three-point form of a straight line, proves that the three points (x_A,M, x_D,M) (x_A,F2, x_D,F2) and (x_A,F1, x_D,F1) lie on a straight line.

Although analytical solution of the mass balances for mixing can be more accurate, the lever arm rule illustrated in Figure 13-17 is often convenient for determining location of mixing point M. Noting that the superscript bar denotes distance, the lever-arm rule is

By measuring along the straight line between F₁ and F₂ we can find point M so that the lever-arm rule is satisfied. When you use the lever-arm rule, you need the ratio of flow rates, not the individual values of F₁ and F₂. Using Eq. (13-33a), point M can be found by trial-and-error. Since this is cumbersome, it is worthwhile to develop the lever-arm rule in different forms.

Usually solvent and feed streams are completely specified in addition to temperature and pressure. Thus, known variables are S, F, y_A,S, y_D,S, x_A,F, x_D,F, T, and p. Values of E, R, y_A,E, y_D,E, x_A,R, and x_D,R are usually desired. If we make the usual assumption that the mixer-settler combination acts as one equilibrium stage, then streams E and R are in equilibrium with each other.

The calculation proceeds as follows: (1) Plot locations of S and F on the triangular equilibrium diagram. (2) Draw a straight line between S and F, and use Eqs. (13-31) or the lever-arm rule to find location of mixed stream M. Stream M settles into two phases in equilibrium with each other. Therefore, (3) construct a tie line through point M to find compositions of extract and raffinate streams. (4) Find ratio E/R using mass balances. We follow this method in Example 13-4.

For an isothermal ternary extraction problem, outlet compositions and flow rates can be calculated from external mass balances used in conjunction with equilibrium. Mass balances around the entire cascade are

Since five variables are unknown (actually, there are seven, but x_S,1 and y_S,N are easily found once x_A,1, x_D,1, y_A,N, and y_D,N are known), a total of five independent equations are needed.

To find two additional relationships, note that streams R₁ and E_N are both leaving equilibrium stages; thus, the compositions of stream R₁ must be related in such a way that R₁ is on the saturated raffinate curve. This gives a relationship between x_A,1 and x_D,1. Similarly, since stream E_N must be a saturated extract, y_A,N and y_D,N are related. If saturated extract and saturated raffinate data have been fit by equations, these two equations can be added to the three mass balances, and the resulting five equations can be solved simultaneously for the five unknowns.

The solution can also be conveniently obtained on a triangular diagram. Let us represent the cascade in Figure 13-21 as a mixing tank followed by a black box separation scheme that produces desired extract and raffinate, as shown in Figure 13-22A. Streams E_N and R₁ are not in equilibrium, but stream E_N is a saturated extract and stream R₁ is a saturated raffinate. The external mass balances for Figure 13-22A are

The coordinates of point M can be found from Eqs. (13-36):

Since Eqs. (13-36) are the same type of mass balances as for a mixer, the points representing streams E_N, R₁, and M lie on a straight line. We also know that E_N must lie on the saturated extract line and R₁ must lie on the saturated raffinate line. Since x_A,1 is known, the location of R₁ on the saturated raffinate line can be found. A straight line from R₁ extended through M (found from x_A,M and x_D,M) will intersect the saturated extract stream at the value of E_N. This construction is illustrated in Figure 13-22B. Note that this procedure is very similar to the one used for single equilibrium stages, but the line R₁ME_N is not a tie line. Once compositions of E_N and R₁ have been determined, mass balance Eqs. (13-35) or (13-36) are solved for flow rates E_N and R₁.

The external mass balance and the equilibrium diagram in Figure 13-22B can be used to determine the effect of variation in the feed or solvent concentrations, the raffinate concentration, or the ratio F/S on the resulting separation. For example, if the amount of solvent is increased, the ratio of F/S will decrease. The mixing point M will move toward point S, and the resulting extract will be less concentrated.

Streams E₁ and R₂ pass each other in the diagram and are called passing streams. These streams can be related to each other by mass balances around stage 1 and the raffinate end of the extraction train. The unknown variables for these mass balances are concentrations x_A,2, x_D,2, and x_S,2 and flow rates E₁ and R₂. Concentration x_S,2 can be determined from the stoichiometric relation x_S,2 = 1.0 – x_A,2 – x_D,2. Taking this equation into account, there are four unknowns (E₁, R₂, x_A,2, and x_D,2) but only three independent mass balances. What is the fourth relation that must be used?

To develop a fourth relation we must realize that stream R₂ is a saturated raffinate stream. Thus, it will be located on the saturated raffinate line, and x_A,2 and x_D,2 are related by the relationship describing the saturated raffinate line. With four equations and four unknowns, we can now solve for the variables E₁, R₂, x_A,2, and x_D,2.

To continue along the column, we repeat the procedure for stage 2. Since streams E₂ and R₂ are in equilibrium, a tie line gives concentration of stream E₂. Streams E₂ and R₃ are passing streams; thus they are related by mass balances. It will prove to be convenient if we write mass balances around stages 1 and 2 instead of around stage 2 alone. The fourth required relationship is that stream R₃ must be a saturated raffinate stream. The stage-by-stage calculation procedure is then continued for stages 3, 4, and so on. When the calculated solute concentration in the extract is greater than or equal to the specified concentration, that is, y_A,jcalc ≥ y_A,Nspecified, the problem is finished.

These stage-by-stage calculations can be done analytically and can be programmed for spreadsheet solution if equations are available for tie lines and the saturated extract and saturated raffinate curves. If equations are not readily available, either equilibrium data must be fitted to an analytical form or a data matrix with a suitable interpolation routine must be developed (Teppaitoon, 2016). Graphical techniques can be employed and have the advantage of giving a visual interpretation.

In a graphical procedure for countercurrrent systems, equilibrium calculations are done by constructing tie lines. The relation between x_A,j and x_D,j is already shown as the saturated raffinate curve. All that remains is to develop a method for representing mass balances graphically.

Referring to Figure 13-21, we can do a mass balance around the first stage. After rearrangement, this is

E₀ – R₁ = E₁ – R₂

If we now do mass balances around each stage and rearrange each balance as the difference between passing streams, we obtain

Thus, differences in flow rates of passing streams is constant even though both extract and raffinate flow rates are varying. The same difference calculation can be repeated for solute A,

and for diluent D,

The differences in flow rates (which is the net flow) of solute and diluent are constant.

Equations (13-38) define a difference or Δ (delta) point. Coordinates of this point are easily found from Eqs. (13-38b) and (13-38c):

where “flow rate” Δ is given by Eq. (13-38a). x_AΔ and x_DΔ are coordinates of the difference point and are not compositions that occur in the column. Note that x_AΔ and x_DΔ can be negative.

The difference point can be treated as a stream for mixing calculations. Thus Eqs. (13-38a), (13-38b), (13-38c) show that the following points are collinear.

Δ(x_AΔ, x_DΔ), E₀(y_A0, y_D0), R₁(x_A1, x_D1)

Δ(x_AΔ, x_DΔ), E₁(y_A1, y_D1), R₂(x_A2, x_,2)

Δ(x_AΔ, x_DΔ), E_j(y_Aj, y_Dj), R_j+1(x_Aj+1, x_Dj+1)

Δ(x_AΔ, x_DΔ), E_N(y_AN, y_DN), R_N+1(x_AN+1, x_DN+1)

The existence of these straight lines can be proved by deriving Eq. (13-40), which is left as an exercise (Problem 13.C4)

Since all pairs of passing streams lie on a straight line through the Δ point, the Δ point is used to determine operating lines for mass balances. A difference point in each section replaces the single operating line in each section on a McCabe-Thiele diagram. Procedure for stepping off stages is illustrated after we discuss finding the location of the Δ point.

There are three methods for finding the location of Δ:

1. Graphical construction. Since points Δ, E₀, and R₁; and Δ, E_N, and R_N+1 are on straight lines, we can draw these two straight lines. The point of intersection must be the Δ point. For a typical design problem (see Figure 13-21), points R_N+1, E₀, and R₁ are easily plotted, and point E_N can be found from external balances (Figure 13-22B). Then Δ is found as shown in Figure 13-23.

2. Coordinates. Coordinates of the difference point were found in Eq. (13-39). These coordinates can be used to find the location of Δ. It may be convenient to draw one of the straight lines in method 1 (such as line ) and use one of the coordinates (such as x_DΔ) to find Δ. This procedure is useful, since accurate graphical determination of Δ can be difficult.

3. Lever-arm rule. The general form of the lever-arm rule for two passing streams is

The lever-arm rule can be used to find the Δ point. For instance, if flow rates R₁ and E₀ are known, then Δ can be found on the straight line through points R₁ and E₀ at a distance that satisfies Eq. (13-41) with j = 0.

Consider again the stage-by-stage calculation outlined previously. We start with known concentration of raffinate product stream R₁ and use an equilibrium tie line (stage 1) to find the location of saturated extract stream E₁ (see Figure 13-24). Points representing Δ, E₁, and R₂ are collinear, since E₁ and R₂ are passing streams. If location of Δ is known, the straight line from Δ to E₁ can be drawn and be extended to the saturated raffinate curve. This intersection has to be the location of stream R₂ (see Figure 13-24). Thus the difference point allows us to solve the three simultaneous mass balances by drawing a single straight line––the operating line. The procedure is continued by constructing a tie line (representing stage 2) to find the location of stream E₂ that is in equilibrium with stream R₂. Then mass balances are again solved simultaneously by drawing a straight line from Δ through E₂ to saturated raffinate line, which locates R₃. This process of alternating between equilibrium and mass balances is continued until desired separation is achieved. Stages are counted along tie lines, which represent extract and raffinate streams in equilibrium. To obtain an accurate solution, a large piece of graph paper is needed and care must be exercised in constructing the diagram.

Before continuing you should carefully reread the preceding paragraph; it contains the essence of stage-by-stage calculations on triangular diagrams.

In Figure 13-24 two equilibrium stages do not quite provide sufficient separation, and three equilibrium stages provide more separation than is needed. In a case like this, an approximate fractional number of stages can be reported.

This fraction can be measured along the curved saturated extract line. It should be stressed that the resulting number of stages, 2.2, is only approximate. The fractional number of stages is useful when the actual stages are not equilibrium stages. Thus, if a sieve-plate column with an overall plate efficiency of 25% were being used, the actual number of plates required would be

If a mixer-settler system were used, where each mixer-settler combination probably has an 80% to 95% efficiency (see Section 13.14), we would probably use three stages, since even 80% efficiency provides more separation than required.

One further important point should be stressed with respect to Figure 13-24. If two equilibrium stages were used with F/S = 2, as in the original problem statement, the saturated extract would not be located at the value E₂ shown on the graph, and saturated raffinate would not be at the value R₁ shown. The streams R₁ and E₂ do not satisfy the external mass balance for this system. The values R₁ and E_N do, but R₁ and E₂ do not. The exact compositions of the product streams for a two-stage system require a trial-and-error solution.

What do we do if flow rate E₀ is less than R₁? The easiest method is to redefine Δ:

Δ is still the difference between flow rates of any pair of passing streams, but it is now raffinate minus extract. The corresponding lever-arm rule for any pair of passing streams is still Eq. (13-41), but the Δ point will be on the opposite side of the triangular diagram. The stage-by-stage calculation procedure is unchanged when the location of Δ is on the right side of the diagram.

When this construction is repeated for a number of arbitrary operating lines, a curved operating line is generated on the McCabe-Thiele diagram. The equilibrium data, y_A vs. x_A, can also be plotted. Then stages can be stepped off. This procedure is shown in Figure 13-27 for Example 13-5. Equilibrium data were obtained from Table 13-6. The answer is 6 1/6 stages, which is reasonably close to the 5.8 found in Example 13-5.

Note that the operating line is close to straight; thus R_j+1/E_j is approximately constant. This occurs when the change in solubility of the solvent in the raffinate streams is approximately the same as the change in solubility of the diluent in the extract streams. When changes in partial miscibility of extract and raffinate phases are very unequal, R_j+1/E_j will vary significantly and the operating line will show more curvature. When feed is not presaturated with solvent and entering solvent is not presaturated with diluent, large changes in flow rates can occur on stages 1 and N. This is not evident in Figure 13-27, since the extract and raffinate phases are close to immiscible for the ranges of concentration shown.

McCabe-Thiele diagrams are useful for more complicated extraction columns such as those with two feeds or extract reflux (Wankat, 1982, 1988).

The minimum solvent rate (or minimum S/F) is the rate at which the desired separation would be achieved if an extractor with an infinite number of stages could be built. If less solvent is used, the desired separation is impossible; if more solvent is used, the separation can be achieved with a finite number of stages. The corresponding Δ value, Δ_min, is the dividing point between impossible cases and possible solutions. Minimum S/F is analogous to minimum reflux in distillation.

To determine Δ_min and hence (S/F)_min, we note that to have an infinite number of stages, tie lines and operating lines must coincide (be parallel) somewhere on the diagram. The construction to determine Δ_min is shown in Figure 13-29 for Example 13-5 and is outlined here.

1. Draw and extend line R₁S.

2. Draw a series of arbitrary tie lines in the range between points F and R₁. Extend these tie lines until they intersect line R₁S.

3. Δ_min is located at the point of intersection of line R₁S with a tie line that is closest to diagram on the left-hand side or furthest from diagram if on the right-hand side. Often the tie line that, when extended, goes through the feed point is the desired tie line. This is not the case in Figure 13-29.

4. Draw the line Δ_minF. Intersection of this line with the saturated extract curve is E_N,min.

5. Draw the line from E_N,min to R₁. Intersection of this line with line from S to F gives M_min.

6. From the lever-arm rule, . In Figure 13-29, (S/F)_min = 1.296 and S_min = 1296 kg/h. Actual solvent rate for Example 13-4 is 1475 kg/h, so the ratio S/S_min = 1.138. Use of more solvent in Figures 13-25 or 13-27 will decrease required number of stages.

On a McCabe-Thiele diagram the behavior will appear simpler. At low S/F the operating line will intersect the equilibrium curve. As S/F increases, the operating line will eventually just touch the equilibrium curve (minimum solvent rate). Then as S/F increases further, the operating line will move away from the equilibrium curve. Unfortunately, since the operating line is curved, it may be difficult to find an accurate value of (S/F)_min from a McCabe-Thiele diagram. An approximate value can easily be estimated. Another way to find an approximate value of (S/F)_min is to use the ratio analysis procedure in Section 13.5.

The flowsheet for the calculation shown in Figure 12-13 is applicable except that because of the nonideal behavior of extraction equilibrium, a convergence loop for x_i,j and y_i,j must be added. This loop may cause convergence problems.

The analogy between stripping and extraction breaks down when one considers equilibrium (also hydraulics and efficiencies, but they do not affect the computer calculation). Understanding of VLE is more advanced than understanding of liquid-liquid equilibrium (LLE), and data banks available in process simulators are much more complete and accurate for VLE than for LLE. Many correlations that work well for VLE do not work for LLE.

The correlations that are suggested for LLE are UNIQUAC and NRTL (Sorenson and Arlt, 1979, 1980; Macedo and Rasmussen, 1987; Walas, 1985). To obtain useful fits with experimental data, specific parameters for the liquid-liquid system, not general parameters used for VLE, should be used (see Appendix B at end of book). If an extraction system will be used for which equilibrium data are unavailable, simulations can be used to determine if the system is worth investigating experimentally (Walas, 1985); however, LLE must be measured experimentally before an extractor can be designed with confidence.

Application of computer simulators for extraction is delineated in the appendix to this chapter.

Mixer-settlers (Figure 13-2) are one of the more common types of extraction equipment. Advantages of mixer-settlers are that they

• can be designed for any capacity.

• are useful for batch or continuous operation.

• can be stopped and then restarted easily.

• have high stage efficiency (typically in the range from 0.8 to 1.0).

• are flexible and have a high turndown ratio (can be operated with very different feed rates).

• can be designed based on simple laboratory experiments.

• can be staged for countercurrent (illustrated for washing in Figure 14-1A—extraction will be similar) or cross-flow operation.

• can be designed as more compact multicompartment units that incorporate mixer and settler in the same shell.

• work for a large variety of systems.

The disadvantages of mixer-settlers are that

• they have a relatively long residence time that is a problem for unstable compounds.

• the mixer may cause emulsions to form that the settler may be unable to separate.

• they are not as compact as many other designs.

• they require a large footprint (large floor area).

• they are relatively expensive for high feed rates.

• they are expensive if many stages are required.

In this section detailed design of mixers and settlers is discussed. Mass transfer characteristics are discussed in Chapter 16.

Because of buoyancy or sedimentation forces, dispersed phases tend to move faster than continuous phases. Thus

Q_D and Q_C are volumetric flow rates of dispersed and continuous phases in the feed. The value of ϕ_D is easy to determine experimentally. Run the mixer continuously until steady state is obtained, close feed and exit lines, turn the mixer off, and allow phases to settle. The value of ϕ_D can then be determined from Eq. (13-43a) by measuring dispersed phase and total volumes.

Which phase is dispersed is often indicated by these rules of thumb (Frank et al., 2008),

These rules of thumb use volume fractions in the mixer, not volume fraction in the feed. In the ambivalent region, which phase is dispersed depends on startup procedures, rates of mixing, mixer geometry, and materials of construction. The phase that fills the mixer first during startup often becomes the dispersed phase. In general, since metals are wet by aqueous phases, use of metals in the mixer promotes the aqueous phase becoming the continuous phase. Polymers tend to be wetted by organics and promote the organic phase becoming the continuous phase. In the ambivalent region, the dispersed phase may not be stable, and phase inversion can occur particularly if operation changes (e.g., stirrer speed or flow rates change). As phase inversion is approached, entrainment increases (Godfrey, 1994); thus best practice is to not operate too close to phase inversion.

Both mass transfer rates (see Chapter 16) and settling depend on which phase is dispersed. Mass transfer tends to be higher if the phase with the controlling resistance is the continuous phase (this may not be possible). A predictive test that is somewhat more conclusive than Eq. (13-44) can be developed by defining χ (Frank et al., 2008; Jacobs and Penney, 1987),

φ_L and φ_H are volume fractions of light and heavy phases, respectively, in the mixer.

If χ < 0.3, light phase is always dispersed.

If 0.3 < χ < 0.5, light phase is probably dispersed.

If 0.5 < χ < 2.0, either phase is dispersed, phase inversion possible; design for worst case.

If 2.0 < χ < 3.3, heavy phase is probably dispersed.

If χ > 3.3, heavy phase is always dispersed.

If we have experimental data for φ_L and φ_H, Eq. (13-45) and the test will give us an idea of the stability of the current arrangement of dispersed and continuous phases.

To predict system behavior without experimental data, we must first estimate values of φ_L and φ_H. An approximate method is to assume φ_L and φ_H are both equal to their volumetric fractions in the feed. This assumption will assign a value that is too large for dispersed phase φ_D and too low for continuous phase φ_C. Thus, if light phase is dispersed, the χ value calculated from Eq. (13-45) will be too large. If the heavy phase is dispersed, the χ value calculated from Eq. (13-45) will be too small. As a result, the preceding test is conservative if feed values are used. More accurate methods of predicting φ_D are considered later.

Mixer tanks (Figure 13-31) usually have baffles to improve mixing. Enclosed tanks operated full of liquid do not require baffles. If there is a liquid-vapor interface, baffles are required to prevent vortex formation. Baffles typically have a width in the range 1/12 to 1/10 times tank diameter (Treybal, 1980). Usual flow direction in a continuous mixer is to have both liquids enter together near the mixer bottom and exit near the top. Typical residence times in mixers, V_liq-tank/(Q_D + Q_C), where V_liq-tank is volume of liquid in the tank, is in range of 1 to 3 minutes, although longer times may be required for reactive systems (Frank et al., 2008). Static mixers are also used, but the residence time may be too short for adequate mass transfer.

Fasano (2015) describes many different types of impellers used for mixing and dispersion, and he recommends impellers for different applications. For flat-blade impellers, the typical ratio of impeller diameter d_i to tank diameter d_tank ranges from d_i/d_tank = 0.25 to 0.35, and the typical ratio of impeller width w_i to diameter is w_i/d_i = 1/5 to 1/8 (Treybal, 1980). If high shear is not required, a pump-mix combination that does both mixing and pumping will reduce costs (Godfrey, 1994).

The key variable in mixer design is the power number N_Po,

In this equation, P is the power typically in Watts, ω is the revolutions per time typically in s^–1, and ρ_m is mixture density calculated as

where ρ_C and ρ_D are densities of continuous and dispersed phases, respectively.

For preliminary estimates, the power number for mixing can be estimated from correlations (Frank et al., 2008; Treybal, 1980). A commonly used correlation is shown in Figure 13-32 (Treybal, 1980). This correlation was originally developed for mixing of homogeneous liquids. Propellers (curve a in Figure 13-32) are typically used for liquid blending, not extraction. To use Figure 13-32 for extraction with two immiscible or partially miscible liquids, replace ρ_L with ρ_M from Eq. (13-47) and replace µ_L with µ_M from

Viscosity of a two-phase mixture can be greater than viscosity of both pure phases.

If experimental measurements are not available, an estimate of φ_d is needed (if φ_d is estimated, then φ_c = 1 – φ_d) to use Eqs. (13-47) and (13-48) and Figure 13-32. As a first approximation, we can assume φ_d = φ_d,feed. A more accurate value of φ_d for mixers with a vapor-liquid interface and baffles can be estimated from the following empirical equation (Frank et al., 2008; Treybal, 1980).

This equation is valid for power/volume > 105 W/m³. σ is interfacial tension, and Δρ = |ρ_c – ρ_D|. Units are selected so that all terms in parenthesis are dimensionless. For unbaffled vessels run full of liquid with no vapor-liquid interface and no vortex, a different correlation is recommended (Frank et al., 2008; Treybal, 1980),

For both correlations, if calculated value of φ_D/φ_D,feed > 1.0, use φ_D = φ_D,feed. Treybal (1980) notes that water dispersed in hydrocarbons may not follow these correlations.

An estimate of the power P is needed to determine φ_D from Eq. (13-49) or (13-50), and an estimate of φ_D is needed to determine ρ_M and µ_M from Eqs. (13-47) and (13-49) to determine P from Figure 13-32. Thus, prediction of φ_D and power in mixers is a classical trial-and-error problem. A solution is illustrated later in Example 13-6.

Unfortunately, almost all literature data are on propellers or flat-blade impellers. If data are not available, the mixer can be studied in a miniplant equipped with a variable-speed mixer. Power P (Watts) required to completely disperse the system and operating conditions that provide reasonable stage efficiencies are determined by experiment. Murphree efficiencies are defined as

where x* is the raffinate mass or mole fraction in equilibrium with the exiting extract mass or mole fraction. Efficiencies of at least 0.8 and preferably in the range from 0.9 to 0.95 are desirable.

Because this scale-up procedure keeps N_Po constant, which often results in a slight increase in efficiency, this is a conservative scale-up method (Frank et al., 2008). Scale-up also assumes that entering feed and solvent concentrations and solvent-to-feed ratio, S/F, used in miniplant and large unit are identical. In addition, we need geometric similarity by keeping ratios d_i/d_tank, H_t/d_tank, and (baffle width)/d_tank constant. Finally, residence time V_t/(Q_D + Q_C) must be equal in the two units. Based on equal power numbers and geometric similarity for the two units,

When a new system is scaled up, the mixer on the large-scale unit should be equipped with a variable-speed drive so that rpm can be tuned for optimum operation of both mixer and settler. Frank et al. (2008) briefly outline other scale-up approaches.

Best design practice is to use experimental data for design of settlers. First, settling of the actual feed/solvent mixture, not a mixture of pure compounds, is characterized with a simple shaker test, which can be done with a 1-inch-diameter graduated cylinder (Frank et al., 2008). The cylinder is shaken and then placed on a bench, and the settling behavior is timed. Initially, the entire cylinder contains an emulsion. Then clear layers form at the top and bottom of the cylinder. These clear layers grow as the dispersion band becomes smaller. In type I systems the dispersion band disappears in less than 5 minutes, resulting in two clear liquids with a clean interface. Systems showing this type of behavior typically have moderate to high interfacial tensions (> ∼10 dyne/cm), density difference > 0.1 g/cm³, viscosities of each phase < 5 cP, and negligible fine solids and surfactants. Type I systems are controlled by the settling rate of drops.

In type II systems the dispersion band disappears in less than 20 minutes, resulting in two clear liquids with a clean interface. Systems showing this type of behavior typically have moderate interfacial tensions (∼10 dyne/cm), density difference > 0.1 g/cm³, viscosities of each phase < 20 cP, and negligible fine solids and surfactants. Type II systems are typically controlled by slow coalescence and are designed using experimental coalescence data.

Frank et al. (2008) also define systems with solids or surfactants (type III), or high viscosity (type IVa), or low interfacial tension (IVb), or low density difference (IVc), or formation of a stable emulsion stabilized by solids/surfactants (IVd). Successful settling of type III systems that have solids or surfactants present may be possible if solids are removed first by filtration and surfactants by adsorption (see Chapter 18). After feed and solvent are pretreated, repeat the shaker tests. Type IV systems probably require a different phase separation system, such as coalescers, centrifuges, hydrocyclones, ultrafiltration or electrotreatment; and pretreatment may also be necessary.

The design of rapid-settling type I systems is usually done by determining the settling velocity of drops with Stokes law or with a modified Stokes law. Stokes law is remarkably simple. If g is the acceleration of gravity and d_d is the drop diameter, the terminal settling velocity u_t,Stokes is

However, its usefulness is quite restricted. Stokes law applies for dilute, low Reynolds number (Re_drop = d_dρ_C u_t /µ_C < 0.3) systems in a quiescent fluid with no interaction between highly dispersed rigid spherical drops, with gravity being the only body force (e.g., no electrostatic forces).

When Stokes law is applied outside its range of validity, care must be taken in using results. The “settling” will be downwards if ρ_D > ρ_C and upwards if ρ_D < ρ_C.

Use of Eq. (13-53) requires knowledge of the diameter of the smallest drops, which, because they have the smallest velocity, are most difficult to remove. If reliable experimental data on smallest drop size are available, they can be used in Eq. (13-53). Although correlations for largest and average drop size are available, reliable correlations for smallest drop size are not available (Godfrey, 1994; Hartland and Jeelani, 1994). An empirical design procedure is to use a drop diameter d_d = 150 µm, which is well less than typical minimum drop diameter of ∼300 µm observed in industrial practice; thus it includes a safety factor (Frank et al., 2008; Jacobs and Penney, 1987).

In addition, the assumption that drops do not interact is seldom valid. Frank et al. (2008) recommend the following empirical expression ignoring coalescence during settling (a conservative assumption) to calculate reduction in terminal settling velocity u_t:

If u_t,hindered > u_t,Stokes, set u_t,hindered = u_t,Stokes for remaining calculations.

In a settler, as the clear layers are formed, the continuous phase is assumed to move vertically upward or downward from inlet to outlet in a uniform plug flow. At the interface, continuous phase velocity is Q_c/A_i, where A_i = L × (interface width) is the interface area. Interface width depends on its location in the settler. If droplets settle (or rise) faster than the continuous phase velocity, droplets will be collected and phases will separate. Thus, design so that

A safety factor of about 0.2 (20%) is often used (Jacobs and Penney, 1987).

However, fluid in a settler is not quiescent, since continuous phase also flows to the outlet end. The settler Reynolds number is defined as

To estimate Re_settler we first estimate continuous phase velocity. In a horizontal settler continuous phase is flowing horizontally, which is perpendicular to the upward or downward drop movement. Continuous phase velocity is equal to volumetric flow rate divided by flow area

where flow area A_f for the continuous phase depends on the dispersion band location (Figure 13-33B). The hydraulic diameter d_hydraulic for continuous phase flow is

Flow perimeter Per_f includes the interface (Figure 13-33B). Combining Eqs. (13-56a) to (13-58), if

there will be minimal interference from turbulence (Jacobs and Penney, 1987). Dispersion band location needs to be known at least approximately to do this design check.

Type II slowly coalescing systems may be underdesigned if Stokes law, Eq. (13-53), or modified Stokes law, Eq. (13-54), are used. Instead, data obtained with a miniplant coalescer on steady-state height ΔH of the dispersion band are preferable. With constant mixture composition and phase ratio, ΔH can be related to changes in Q/A [A = L × (settler width) is the settler cross-sectional area]. Jacobs and Penney (1987) recommend that ΔH < 0.1 d_settler, while Frank et al. (2008) recommend ΔH < 0.15 d_settler. These heuristics will prevent flooding of the settler if the flow rate increases, which can cause ΔH to increase drastically. As a rule of thumb, residence time of the dispersed phase in the dispersion band should be greater than about 2 to 5 minutes (Jacobs and Penney, 1987). Because the dispersed phase occupies about half of the dispersion band,

Several methods have been developed to speed up coalescence (Frank et al., 2008; Hartland and Jeelani, 1994; Jacobs and Penney, 1987). Increasing operating temperature usually increases coalescence mainly because viscosity decreases. Coalescence-enhancing structures such as mesh or baffles are often employed. Coalescence can be further encouraged by using construction materials that are wetted by the dispersed phase (i.e., metals for aqueous and plastics for organics). Coalescence is also faster if mass transfer (assuming the mixer does not have an efficiency of 100%) is from drops to continuous phase; thus, one may want the raffinate phase to be dispersed. Settler performance can be improved by using two diffuser plates with increasing hole area at the settler inlet and by using overflow and underflow weirs to remove the light and heavy phases, respectively. If φ_d is high, settling will be hindered. Recycling continuous phase to the settler inlet will decrease φ_d and may increase coalescence. Recycling needs to be done with care, since Eq. (13-56b) may be violated by increased velocity of the continuous phase.

The selection criteria listed in Section 13.1 indicate that mixer-settlers are a good choice if only a few equilibrium stages are needed; however, if, as is often the case, a few more stages are required, then a nonagitated (static) column such as a spray, sieve tray, or packed column should be used. These columns are commonly used for relatively large-scale systems in the chemical and petrochemical industries. Design methods for static columns are discussed in detail by Frank et al. (2008). However, if more than five stages are required, an agitated column system is often used.

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A1. What is the designer trying to do in the extraction equipment shown in Figure 13-2 and listed in Table 13-1? Why are there so many types of extraction equipment and only two major types of equipment for vapor-liquid contact?

A2. Write your key relations chart for this chapter.

A3. For Figure 13-24 suppose the raffinate concentration had to be obtained with exactly two equilibrium stages. This can be accomplished by changing amount of solvent used. Would we want to increase or decrease amount of solvent? Explain the effect this change will have on M, E_N, Δ, and number of stages required.

A4. Extractors are analogous to strippers and absorbers; however, we ignore heat effects for concentrated extractors and assume they are isothermal, but concentrated absorbers and strippers typically have large ΔH effects that cause large changes in temperature. Explain the difference.

A5. If extract and raffinate phases are totally immiscible, triangular diagrams can still be used. Explain how and describe what equilibrium diagrams will look like.

A6. What can be done to increase concentration of solute in the extract (i.e., increase y_A,N)?

A7. What situation in analysis of countercurrent extraction on triangular diagrams is superficially analogous to total reflux in distillation? How does it differ?

A8. Study Figure 13-28. Explain what happens as S/F increases. What happens to M? What happens to E_N? What happens to Δ? How do you find Δ_min if it lies on the left-hand side? How do you find Δ_min if it lies on the right-hand side?

A9. Five analysis procedures were developed for extraction in Chapter 13: Kremser, McCabe-Thiele with mass or mole fractions, McCabe-Thiele with mass or mole ratios, triangular diagrams, and computer simulation. If you have an extraction problem, how do you decide which method to use? (In other words, explain when each is applicable.)

A10. Compare K_dE/R to KV/L used in distillation, absorption, and stripping. Do these quantities have the same significance?

A11. Explain why Kremser, McCabe-Thiele with mass or mole fractions, McCabe-Thiele with mass or mole ratios, and triangular diagrams are not appropriate methods to analyze a concentrated, immiscible extraction problem with four components (diluent, solvent, and two solutes).

A12. In fractional extraction what happens to solute C if

a.

b.

c. How would you adjust the extractor so that the conditions in parts a or b would occur?

A13. Figure 13-6 implies that a center-cut can be done using two columns either with three solvents or with two solvents. If the same two solvents are used in both columns explain how component B can be made to go down in the first column and up in the second.

A14. Sketch the two-stage countercurrent batch extraction process discussed in Section 13.6.

B. Generation of Alternatives

B1. For fractional extraction, list possible problems other than the three in the text. Outline a solution to these problems.

B2. How would you couple together cross-flow and countercurrent cascades? What might be advantages of this arrangement?

B3. The extract stream typically contains product (solute) dissolved in solvent. List as many ways as you can for recovering product and preparing solvent for recycle.

C. Derivations

C1. Derive Eq. (13-10) starting with a McCabe-Thiele diagram (follow procedure used to develop Kremser equation in Chapter 12).

C2. Derive Eq. (13-18).

C3. Single-stage systems (N = 1) can be designed as countercurrent systems, Figure 13-4, or as cross-flow systems, Figure 13-9. Develop the methods for both these designs. Which is easier? If the system is dilute, how can the Kremser equation be used?

C4. Define Δ and the coordinates of Δ from Eqs. (13-38) and (13-39). Prove that points Δ, E_j, and R_j+1 (passing streams) lie on a straight line by developing Eq. (13-40).

C5. Prove that the locations of streams M, F₁, and F₂ in Figure 13-17 lie on a straight line.

D. Problems

*Answers to problems with an asterisk are at the back of the book.

D1. In Example 13-1 we assumed we were going to use all available solvent. There are other alternatives. Determine if the following alternatives are capable of producing outlet water of desired acetic acid concentration.

a. Use only pure solvent at bottom of extractor.

b. Mix all of pure and all of impure solvent together and use them at bottom of column.

c. Mix all of pure and part of impure solvent together and use them at bottom of column.

D2. A countercurrent extraction system extracts furfural from water into methyl-isobutyl ketone (MIBK) at 25°C. Aqueous feed contains 0.11 wt% furfural and has a feed rate of 150 kg/min. Recovery of the furfural that enters in the aqueous feed is 90.0%. Entering solvent stream contains 0.03 wt% furfural and remainder is MIBK. Assume flow rates of raffinate and extract streams are constant. Equilibrium data are in Table 13-3.

a. How many equilibrium stages are required for a solvent flow rate of 25.0 kg/min?

b. Compare the result in part a with solution to problem 13.D8.

c. What is minimum solvent flow rate?

D3. Acetic acid (solute) is extracted from water (diluent) into butanol (solvent). A single mixer-settler (one equilibrium contact) is used. Feed flow rate is 100 kg/h, and feed is 1.3 wt% acetic acid with remainder being water saturated with butanol. Entering solvent stream is 0.1 wt% acetic acid with remainder being butanol saturated with water. We want weight fraction acetic acid in exiting water stream (raffinate) to be x = 0.7. Assume total flow rates of extract (butanol phase) and raffinate (water phase) are constant. Note complete immiscibility is not required as long as total flow rates are constant, which occurs if there is little change in amount of water in extract phase and little change in amount of butanol in raffinate. Since these streams are presaturated, assumption of constant flow rates is reasonable. Equilibrium data are in Table 13-3. Find required solvent flow rate S in kg/h.

D4.* 400 kg/h of an aqueous feed that is 0.5 wt% acetic acid mixture is extracted with 3-heptanol at 25°C. Assume water and heptanol are immiscible. Equilibrium is

Solvent flow rate is 560 kg/h. Entering solvent contains 0.01 wt% acetic acid.

a.* Find outlet raffinate and extract mole fractions for 10 equilibrium stages.

b. Find outlet raffinate and extract mole fractions for 11 equilibrium stages.

D5. The equilibrium for extraction of acetic acid from water into 3-heptanol at 25°C is y = 0.828 x, where y is weight fraction acetic acid in 3-heptanol and x = weight fraction acetic acid in water. 400 kg/h of feed with x₀ = 0.5 wt% acetic acid and 99.5 wt% water is contacted in a countercurrent extractor with E = 560 kg/h of solvent that is y_N+1 = 0.01 wt% acetic acid and 99.99 wt% 3-heptanol. Outlet raffinate concentration is x_N = 0.03 wt% acetic acid and 99.97 wt% water. Assume water and 3-heptanol are immiscible and that R and E are constant.

a. Determine number of equilibrium stages N required.

b. What is minimum solvent flow rate, E_min?

D6. Equilibrium for extraction of acetic acid from 3-heptanol into water at 25°C is y = 1.208 x, where y = weight fraction acetic acid in water and x = weight fraction acetic acid in 3-heptanol. 100 kg/h of feed with x₀ = 0.005 is contacted in a countercurrent extractor with solvent with y_N+1 = 0.0002. Outlet raffinate concentration is x_N = 0.0005. Assume water and 3-heptanol are immiscible and that R and E are constant. Note that parts a, b, and c can be solved independently, or value of y₁ from part a can be used in part b.

a. If solvent flow rate E = 140 kg/h, calculate exiting extract weight fraction y₁.

b. If solvent flow rate E = 140 kg/h, determine number of equilibrium stages N.

c. What are minimum solvent flow rate, and maximum exiting extract weight fraction?

d. Problems 13.D5 and 13.D6 are for same system, but y = 1.208 x in one problem and y = 0.828 x in the other problem. Explain why.

D7.* We are extracting acetic acid from water into 3-heptanol at 25°C in an extraction column with 30 equilibrium stages. Equilibrium is given in Problem 13.D4. The aqueous feed rate is 500 kg/h. The feed is 1.1 wt% acetic acid, and the exit water should be 0.037 wt% acetic acid. Inlet 3-heptanol contains 0.02 wt% acetic acid. What solvent flow rate is required? Assume total flow rates are constant.

D8. A cross-flow extraction system is being used to extract furfural from water into methyl-isobutyl ketone (MIBK) at 25°C. The 150 kg/min of aqueous feed contains 0.11 wt% furfural. We desire a 90.0% recovery of the furfural that enters the system in the aqueous feed. Entering solvent stream contains 0.03 wt% furfural in MIBK. Assume flow rates of aqueous raffinate stream and extract streams are constant. Equilibrium data are given in Table 13-3. If flow rate of each solvent stream is 50.0 kg/min, how many equilibrium stages are required?

a. Solve graphically.

b. Solve each stage analytically (see part E of Example 13-2).

c. Compare your answers for parts a and b.

D9.* We have a mixture of linoleic and oleic acids dissolved in methylcellosolve and 10% water. Feed is 0.3 wt% linoleic acid and 0.25 wt% oleic acid. Feed flow rate is 1500 kg/h. A simple countercurrent extractor will be used with 750 kg/h of pure heptane as solvent. We desire a 99% recovery of the oleic acid in extract product. Equilibrium data are given in Table 13-3. Find N and recovery of linoleic acid in extract product.

D10. Two feed solutions of dioxane in water are being extracted with benzene in a four-stage cross-flow system. Entering benzene solvent to each stage is pure (y_dioxane = 0), and solvent flow rate to each stage is 50 kg/h. Entering feed to stage 1 total flow rate is 100 kg/h and is x_dioxane,F1 = 1.5 wt% dioxane. Entering feed 2 total flow rate is 70 kg/h and is x_dioxane,F2 = 0.5 wt% dioxane. Feed 2 is mixed with the raffinate leaving the second stage, and this mixture is fed to the third stage. Assume benzene and water are completely immiscible and total flow rates of raffinate and extract entering and leaving each stage are constant. Equilibrium for dioxane distributing between benzene and water is y_dioxane = 1.02 x_dioxane where y_dioxane is weight fraction of dioxane in extract (benzene phase) and x_dioxane is weight fraction of dioxane in raffinate (water phase). Find dioxane weight fractions of following: raffinate leaving stage 2, raffinate entering stage 3, raffinate leaving stage 4, and extract leaving stage 4.

D11.* A fractional extraction system (Figure 13-5) is separating abietic acid from other acids. Solvent 1, heptane, enters at and is pure. Solvent 2, methylcellosolve + 10% water, is pure and has a flow rate of R = 2500 kg/h. Feed is 5 wt% abietic acid in solvent 2 and flows at 1 kg/h. There are only traces of other acids in the feed. We desire to recover 95% of the abietic acid in the bottom raffinate stream. Feed is on stage 6. Assume solvents are completely immiscible, and system can be considered to be very dilute. Equilibrium data are given in Table 13-3. Find N.

D12. 200 kg/h of a water stream containing 26.3 wt% acetone is to be extracted with chloroform at 1.0 atm and 25°C. Exiting raffinate stream should contain 1.4 wt% acetone. The entering chloroform contains 0.2 wt% acetone. Operate at a solvent flow rate that is 1.4 × (minimum solvent rate). Determine number of stages needed and outlet concentration of extract stream. Use a McCabe-Thiele diagram in weight ratios. Compare answer to problem 13.D17.

D13. Extract p-xylene and o-xylene from n-hexane diluent using β, β′-Thiodipropionitrile as solvent. Solvent and diluent can be assumed to be immiscible. Feed rate is 1000.0 kg/h. The feed contains 0.3 wt% p-xylene and 0.5 wt% o-xylene in n-hexane. We desire at least a 90% recovery of p-xylene and at least 95% recovery of o-xylene. Entering solvent is pure. Operation is at 25°C, and equilibrium data are in Table 13-3. Use a simple countercurrent cascade.

a. Calculate value of (R/E)_max for both p-xylene and for o-xylene to just meet recovery requirements.

b. The smaller (R/E)_max value represents the controlling or key solute. Operate at E = 1.5(E)_{min,controlling}. Find number of stages and solvent inlet flow rate.

c. Determine outlet raffinate concentration and percent recovery of noncontrolling solute.

D14.* Equilibrium data for the system water-acetic acid-isopropyl ether are given in Table 13-6. 15 kg of feed that is 30 wt% acetic acid and 70 wt% water is extracted with pure isopropyl ether. A batch extraction is done in a mixed tank.

a. If 10 kg of solvent is used, find outlet extract and raffinate compositions.

b. If a raffinate composition of 10 wt% is desired, how much solvent is needed?

D15.* The system shown in the figure is extracting acetic acid from water using benzene as the solvent. A temperature shift is used to regenerate solvent and return acid to water phase.

a.* Determine y₁ and y_N+1 (units are weight fractions) for column at 40°C.

b.* Determine R′ and x_N′ for column at 25°C.

c. Is this a practical way to concentrate acid?

Data are in Table 13-3. Note: A similar scheme is used commercially for citric acid concentration using a more selective solvent.

D16. 500 kg/h of a 30 wt% pyridine, 70 wt% water feed is extracted with 300 kg/h of pure chlorobenzene at 1 atm and 25°C. Assume chlorobenzene and water are immiscible. A single mixer-settler is used. Equilibrium data are in Table 13-4.

a. Find outlet extract and raffinate total flow rates and weight fractions.

b. Compare answer to part a with answer to problem 13.D23, which solves the same problem but without assuming water and chlorobenzene are immiscible.

D17. Repeat problem 13.D12, but find N using the Kremser equation in weight ratio units. Compare answer with problem 13.D12.

D18. We contact 100 kg/h of a 50 wt% methylcyclohexane and 50 wt% n-heptane feed mixture with a solvent stream (15 wt% methylcyclohexane and 85 wt% aniline) in a mixer-settler that acts as a single equilibrium stage. Extract phase is 10 wt% methylcyclohexane and raffinate phase is 61 wt% methylcyclohexane. Equilibrium data are in Table 13-7. What flow rate of solvent stream was used?

D19. Many extraction systems are partially miscible at high concentrations of solute but close to immiscible at low solute concentrations. At relatively low solute concentrations both McCabe-Thiele and trianglar diagram analyses are applicable. Use chloroform to extract acetone from water. Equilibrium data are given in Table 13-5. Find number of equilibrium stages required for a countercurrent cascade if feed is 1000.0 kg/h of a 10.0 wt% acetone, 90.0 wt% water mixture. Solvent is chloroform saturated with water (no acetone). Flow rate of stream E₀ = 1371 kg/h. We desire an outlet raffinate concentration of 0.50 wt% acetone. Assume immiscibility and use a weight ratio graphical analysis. Convert equilibrium to weight ratios. Compare results with Problem 13.D43.

D20. The aqueous two-phase system in Example 13-2 will be used in a batch extraction. 7.5 kg of PEG solution contains protein at mass fraction x_F. Use 6.0 kg of pure dextran solution to extract the protein from the solution. Equilibrium data are in Example 13-2.

a. Find fractional recovery of protein in dextran phase if two solutions are mixed together and then allowed to settle.

b. Find fractional recovery of protein in dextran phase if continuous solvent addition batch extraction system shown in Figure 13-11 is used.

D21.* We have 100 kg/h of a feed that is 60 wt% methylcyclohexane (A) and 40 wt% n-heptane (D) and 50 kg/h of a feed that is 20 wt% methylcyclohexane and 80 wt% n-heptane. These two feeds are mixed with 200 kg/h of pure aniline (S) in a single equilibrium stage. Equilibrium data for extraction of methylcyclohexane (A) from n-heptane (D) into aniline (S) are given in Table 13-7. Equilibrium data are in Table 13-7.

a. What are extract and raffinate compositions leaving the stage?

b. What is the flow rate of extract product?

D22. The horizontal settler calculation in Example 13-6 was done for a settler diameter of D_s = 1.023 m with the dispersion band assumed to be at the center of the circle. The conclusion was that there was likely to be a limited amount of interference due to flow. Repeat the calculation of Re_settler and determine whether interference is likely if the interface of the dispersion band is 0.1 m below the center of the circle.

D23. 500 kg/h of a 30 wt% pyridine and 70 wt% water feed is extracted with pure chlorobenzene at 1 atm and 25°C. Equilibrium data are in Table 13-4.

a. A single mixer-settler is used. Chlorobenzene flow rate is 300 kg/h. Find outlet extract and raffinate flow rates and weight fractions. Do NOT assume that raffinate and extract are immiscible.

b. Convert single-stage system in part a into a two-stage cross-flow system. Raffinate from part a is fed to second mixer-settler and is contacted with an additional 300 kg/h of pure chlorobenzene. Find raffinate and extract flow rates and weight fractions leaving second stage. Do NOT assume raffinate and extract are immiscible.

D24. Remove methylcyclohexane from n-heptane using aniline as solvent (see Table 13-7). Feed consists of 80 kg/h methylcyclohexane, 110 kg/h n-heptane, and 10 kg/h aniline. Solvent is 95 wt% aniline and 5 wt% n-heptane.

a. With a single mixer-settler, set total entering flow rate of solvent stream at 600 kg/h. Find the outlet weight fractions and outlet flow rates.

b. Suppose we decide to use a two-stage cross-flow system with 300 kg/h of solvent stream entering each stage. Find outlet flow rates and weight fractions.

Note: Answers are very sensitive to how one draws tie lines. This is typical of type II systems.

D25.* We wish to remove acetic acid from water using isopropyl ether as solvent at 20°C and 1 atm (see Table 13-6). Feed is 45.0 wt% acetic acid and 55.0 wt% water. Feed flow rate is 2000 kg/h. A countercurrent system with pure solvent is used. Extract stream is 20.0 wt% acetic acid and raffinate is 20.0 wt% acetic acid.

a. How much solvent is required?

b. How many equilibrium stages are needed?

D26.* A countercurrent system with three equilibrium stages is to be used for water-acetic acid-isopropyl ether extraction (see Table 13-6). Feed is 40 wt% acetic acid and 60 wt% water. Feed flow rate is 2000 kg/h. Solvent added contains 1 wt% acetic acid but no water. We desire a 5 wt% acetic acid raffinate. What solvent flow rate is required? What are flow rates of E_N and R₁?

D27. 100 kg/h of a 40 wt% acetic acid and 60 wt% water feed is sent to a countercurrent extraction column where acetic acid is extracted with isopropyl ether. Equilibrium data are in Table 13-6. Entering isopropyl ether is pure and has a flow rate of 111.2 kg/h. Raffinate is 20 wt% acetic acid.

a. Find weight fraction acetic acid in the outlet extract stream.

b. Determine the flow rates of outlet raffinate and extract streams.

c. Find number of equilibrium contacts.

D28.* Acetic acid is extracted from water with isopropyl ether at 20°C and 1 atm pressure in a column with three equilibrium stages. Equilibrium data are in Table 13-6. Entering feed rate is 1000 kg/h, and feed is 40 wt% acetic acid and 60 wt% water. Exiting extract stream has a flow rate of 2500 kg/h and is 20 wt% acetic acid. Entering extract stream (which is not pure isopropyl ether) contains no water. Trial and error is not needed. Find

a. Exit raffinate concentration.

b. Required entering extract stream concentration.

c. Flow rates of exiting raffinate and entering extract streams.

D29. Recover pyridine from water using chlorobenzene as solvent in a countercurrent extractor. Feed is 35 wt% pyridine and 65 wt% water. Feed flow rate is 1000 kg/h. Solvent is pure. Outlet extract is 20 wt% pyridine, and outlet raffinate is 4 wt% pyridine. Operation is at 25°C and 1 atm. Equilibrium data are in Table 13-4.

a. Find number of equilibrium stages needed.

b. Determine solvent flow rate required in kg/h.

D30. Suppose in Example 13-6 that we decide to build the settler with a diameter of 1.0 m and a length of 4.0 m. What safety factor are we employing?

D31. A mixer-settler operates as one equilibrium stage for extraction of a 1.0 wt%% acetic acid and 99.0 wt% water feed. Total feed flow rate is 50 kg/h. Feed is mixed with 100 kg/h of pure solvent (1-butanol). Equilibrium data are in Table 13-3. Assume 1-butanol and water are immiscible and system is dilute (constant total flow rates). Find acetic acid weight fractions in extract, y₁, and raffinate, x₁, products. There are multiple solution paths.

D32. For toluene-water system in Example 13-6 we found toluene is the dispersed phase if Q_solvent/Q_feed = 0.2. Which phase is dispersed if

a. Q_solvent/Q_feed = 0.6?

b. Q_solvent/Q_feed = 1.0?

c. Q_solvent/Q_feed = 2.0?

d. Q_solvent/Q_feed = 5.0?

D33. For the extraction in Example 13-6, suppose we decide to have H_tank = 2d_tank and want a 1.5 minute residence time. Find the tank dimensions.

D34. Estimate value of φ_d in the tank and power P required for Example 13-6 with H_tank = d_tank = 0.8279 m, and a 1.0-minute residence time, a six-blade flat turbine impeller with impeller diameter d_i = 0.20d_tank and impeller width w_i = d_i/5 is operated at 500 rpm.

D35. We want to solve Example 13-5 with a multicomponent computer program similar to absorption and stripping programs. Set up the tridiagonal matrix for the mass balances assuming there are six stages in the column and exit raffinate stream concentration is unknown. Develop the values for A, B, C, and D for only the acetic acid matrix using K values determined from Table 13-5. Do not invert the matrix.

D36. A 60 vol% tributyl phosphate (TBP) in kerosene solvent extracts Zr(NO₃)₄ from an aqueous solution. Entering solvent is recycled from a solvent recovery system and contains 0.008 mol Zr(NO₃)₄/L. Aqueous feed contains 0.10 mol Zr(NO₃)₄/L and outlet aqueous solution should contain 0.008 mol Zr(NO₃)₄/L. Feed rate of aqueous solution is 200L/h. Equilibrium is mol Zr(NO₃)₄/L (Benedict and Pigford, 1957).

a. Find minimum entering solvent rate F_solvent,min in L/h.

b. If F_solvent,min = 1.4 F_solvent,min, find number of equilibrium stages required.

c. Calculate mole fraction and mass fraction of Zr(NO₃)₄ in feed and determine if system is dilute. Assume density of feed is 1.0 g/ml.

Assume that aqueous and organic phases are completely immiscible and that densities of the two phases are both constant (volumetric flow rates are constant).

D37. Repeat Example 13.4, but use a McCabe-Thiele diagram with mass ratio units.

D38.* We are extracting benzoic acid from water into toluene in a single equilibrium stage system. Entering toluene is pure and entering water contains 0.023 mol% benzoic acid. Feed flow rate is 1.0 kmol/h and solvent flow rate is 0.06 kmol/h. Find outlet mole fractions of benzoic acid in exiting raffinate and extract phases. Data are in Example 13-6. Assume water and toluene are completely immiscible.

D39. Plot equilibrium data from Table 13A-1 for tri-ethylamine (solvent), carbon tetrachloride (solute), acetic acid (diluent) on a right triangle diagram with ordinate = mole fraction CCl₄ and abscissa = mole fraction acetic acid. Solve Standard Extraction Problem in Lab 12 for a single-stage extraction graphically. Find extract and raffinate mole fractions and flow rates.

D40. Plot equilibrium data from Table 13A-1 for tri-ethylamine (solvent), carbon tet- rachloride (solute), acetic acid (diluent) on a right triangle diagram with ordinate = mole fraction CCl₄ and abscissa = mole fraction acetic acid. Solve following two stage cross-flow problem graphically. F = 10 kmol/h, and feed is 10 mol% CCl₄ and 90 mol% acetic acid. Entering solvent is pure and 10 kmol/h are added to each stage. Find mole fractions and flow rates of extract streams and raffinate stream.

D41. Find number of stages needed for a countercurrent extractor if 10 kmol/h feed that is 10 mol% CCl₄ and 90 mol% acetic acid is processed at 293K and 1 atm. Solvent is pure tri-ethylamine with a flow rate of 14.5 kmol/h. Exiting extract is 9.1 mol% CCl₄. Also find flow rates and mole fraction of exiting extract and raffinate streams. Equilibrium data in Table 13A-1.

D42. Recover pyridine from water using chlorobenzene as solvent in a countercurrent extractor. Feed is 25.9 wt% pyridine and 74.1 wt% water. Solvent is pure chlorobenzene, and solvent flow rate is 1000.0 kg/h. Outlet extract is 26 wt% pyridine, and outlet raffinate is 4 wt% pyridine. Operation is at 25°C and 1.0 atm. Equilibrium data are in Table 13-4.

a. Find number of equilibrium stages needed.

b. Determine feed flow rate required in kg/h.

NEATNESS COUNTS!

D43. Use chloroform to extract acetone from water. Equilibrium data are given in Table 13-5. Find number of equilibrium stages required for a countercurrent cascade if feed is 1000.0 kg/h of a 10.0 wt% acetone, 90.0 wt% water mixture. Solvent is chloroform saturated with water (no acetone). Flow rate of stream E₀ = 1371 kg/h. Outlet raffinate concentration is 0.50 wt% acetone. Use a triangle diagram and compare results with Problem 13.D19.

D44. In a countercurrent extraction system we are recovering 92% of the solute (acetaldehyde) from the aqueous feed (0.9 wt% acetaldehyde, 99.1 wt% water, feed flow rate =1500 kg/h). The solvent is benzene and the entering solvent contains 0.007 wt% acetaldehyde with remainder benzene. Assume that water and benzene are immiscible. The distribution coefficient for acetaldehyde between water and benzene is

K_d = 1.119 = y/x = wt% acetaldehyde in benzene/wt% acetaldehyde in water.

a. If there are eight equilibrium stages, what solvent flow (kg/h) is required?

b. What is the minimum solvent flow rate (kg/h)?

D45. We are extracting acetic acid from benzene into water at 25°C and 1.0 bar. 100 kg/h of a feed that is 0.092 wt% acetic acid and 99.908 wt% benzene is fed to the column. The inlet solvent is water with 0.007 wt% acetic acid, and total flow rate of inlet solvent is 20.0 kg/h. Assume water and benzene are completely immiscible. The extractor has 5.0 equilibrium stages. Equilibrium is

a. Find the outlet weight fraction of acetic acid in the benzene, x₅.

b. Find the outlet weight fraction of acetic acid in the water, y₁.

D46. We are extracting pyridine from 500 kg/h of a feed that is 15.0 wt% pyridine and 85.0 wt% water using 225 kg/h of pure chlorobenzene as the solvent. Assume water and chlorobenzene are immiscible. The equilibrium is y_pyr = 2.2 x_pyr with y = wt% pyr in extract, and x = wt% pyr in raffinate. A counter-cascade is used. If we desire a final concentration of x_N = 4.0 wt% pyridine in the exiting raffinate (x_N),

a. What is the outlet wt% of the pyridine in the extract phase, y₁?

b. How many equilibrium stages are required?

E. More Complex Problems

E1.* A liquid feed that is 48 wt% m-xylene and 52 wt% o-xylene is separated in a fractional extractor (Figure 13-5) at 25°C and 101.3 kPa. Solvent 1 is β,β′-thiodipropionitrile, and solvent 2 is n-hexane. Equilibrium data are in Table 13-3. For each kilogram of feed, 200 kg of solvent 1 and 20 kg of solvent 2 are used. Both solvents are pure when they enter the cascade. We desire a 92% recovery of o-xylene in solvent 1 and a 94% recovery of m-xylene in n-hexane. Find outlet composition, N, and N_f. Adjust recovery of m-xylene if necessary to solve this problem.

E2. Extract meta-, ortho-, and para-xylenes from n-hexane using β,β′-thiodipropionitrile as solvent. Solvent and diluent (n-hexane) are immiscible. Feed flow rate is 1000.0 kg/h. Feed is x_m–xy = 0.5 wt% m-xylene, x_o–xy = 0.6 wt% o-xylene, and x_p–xy = 0.4 wt% p-xylene in n-hexane. Solvent flow rate is 20,000 kg/h. Entering solvent is pure. We desire 96% recovery of p-xylene in solvent. Operation is at 25°C and 1.0 atm. Equilibrium data are in Table 13-3. Use a simple countercurrent cascade.

a. Find outlet p-xylene weight fraction, x_N,p–xy.

b. Find N.

c. Find x_N,o–xy.

d. Find x_N,m–xy.

e. What is minimum solvent flowrate (N approaches infinity)?

E3. Equilibrium data for extraction of methylcyclohexane (A) from n-heptane (D) into aniline (S) are given in Table 13-7. Compare batch extractions of 20 kg of 40 wt% methylcyclohexane and 60 wt% n-heptane feed. Solvent added is first presaturated with diluent n-heptane so that, according to data in Table 13-7, it is 93.8 % aniline and 6.2 % n-heptane.

a. Do a normal batch extraction adding 20 kg of presaturated solvent. Find flow rates and compositions of extract and raffinate products.

b. Process same feed with presaturated solvent in a continuous solvent addition batch extraction. First, presaturated solvent is added with no removal of extract until a two-phase mixture is obtained. How much presaturated solvent is added to obtain two phases? Second, continuous solvent addition batch extraction is conducted until raffinate has a methylcyclohexane weight fraction of 0.292. How much solvent is added during this step? What is final raffinate composition? How much extract is collected, and what is its average composition? Assume R_t is constant and Eq. (13-27) is valid.

G. Computer Simulation Problems

G1. Do Lab 12 in the Appendix to this chapter, but for three-stage systems. Operation is at 293K, 1.0 atm, F = 10 kmol/h and is 10 mol% carbon tetrachloride and 90 mol% acetic acid. Entering solvent is pure triethylamine.

a. Simulate a 3-stage cross-flow system with 30 kmol/h total pure solvent with 10.0 kmol/h fed to each stage. Find total and component flow rates (kmol/h) in the four outlet streams. Calculate fraction of entering carbon tetrachloride that is extracted.

b. Simulate a 3-stage countercurrent system using 3 decanters. 10 kmol/h of pure solvent is used. Find total and component flow rates (kmol/h) in the two outlet streams. Calculate fraction of entering carbon tetrachloride that is extracted.

c. Repeat 3-stage countercurrent system with different flow rates of pure solvent until fraction of entering carbon tetrachloride extracted is same as in part a. Which process uses less solvent?

Preparation

• As needed, review computer assignments 1 to 11.

• Read Appendix B in the back of the textbook.

Goals:

1. Learn how to input LLE correlation parameters into Aspen Plus.

2. Learn how to do regression analysis of LLE correlation parameters to fit data.

3. Investigate Aspen Plus simulation of extractors.

4. Explore the behavior of partially miscible extraction cascades.

I. Getting Started

We use pure triethylamine (C₆H₁₅N) solvent to separate 10kmol/h of a feed that is 10 mol% carbon-tetrachloride (CCl₄) (the hyphen is necessary when directly inputting the component name in Aspen Plus, or you can use the Find button and search for the component by keyword) and 90 mol% acetic-acid (C₂H₄O₂). Extractor, solvent, and feed are at 293K and 1 bar. Data and NRTL parameters are available for this system in the DECHEMA data bank (Sorensen and Arlt, 1980). The DECHEMA parameters for NRTL and the experimental tie line data are in Table 13-A1. Note: diluent = acetic-acid, solute = carbon-tetrachloride.

1. In Setup, choose Vapor-Liquid-Liquid as the allowable phases.

2. For inputting equilibrium data, select NRTL as the property method. Aspen Plus has its own database for parameters, but the Aspen Plus method of inputting the parameter values is different than the DECHEMA method. On the input page for NRTL (Figure 13-A1), click on the top of one of the columns for a binary pair and then click the DECHEMA button (upper right side of Input tab). You should get a DECHEMA pop-up menu (Figure 13-A2). Input the Aij and Aji and alpha values from DECHEMA in Aspen Plus. Be sure that the button for LLE is clicked, not the button for VLE. Click OK. The column in the Aspen Plus table should now list “user” for source and units should be K. Select the components i and j for the next binary pair, and input the Aij and Aji and alpha values from DECHEMA. Repeat for the third binary pair.

3. To look at the predicted equilibrium data, go to Analysis→Property→Ternary, select Liquid-Liquid, input the temperature, Enter, and then click Go. Compare the predicted results (the Table is probably easier to read than the graph) with Table 13-A1 experimental tie lines. The fit should be reasonable. If the fit is poor, go back to your NRTL parameters and input the values again, making sure that the LLE button is pressed for each binary pair. If you have difficulty finding where to input the NRTL parameters, the All Items list on the left side of Figure 13-A1 shows how to navigate to the desired input page.

4. Go to Simulation in Aspen Plus.

II. Simulations

The following standard problem will be used with different equipment:

Standard Extraction Problem: Feed is 10 mol% carbon tetrachloride (CCl₄) and 90 mol% acetic acid (C₂H₄O₂) with flow rate of 10.0 kmol/h. Solvent is 10.0 kmol/h of pure solvent. All streams and the decanters and column are at 293K and 1 bar.

1. Mixer-Settler (Decanter):

a. All you need in the flowsheet is a decanter (listed under separators) with feed and solvent streams input into decanter feed port and two product streams. Select triethylamine as key component for second liquid phase, and use default value for second liquid threshold.

b. Input information for the Standard Extraction Problem (listed above after II). Run the simulation.

c. From the results determine K_d value [= (y carbon tetra chloride)/(x carbon tetra chloride)] and compare to data in Table 13.A1.

d. Find the outlet flow rates and mole fractions in the two outlet streams.

e. Try changing S/F ratio and feed mole fraction to determine what happens in this single-stage system. Find conditions that make one phase disappear.

2. Countercurrent Extractor—Column:

a. To build a countercurrent extractor flowsheet, select Column, then Extract, in the bottom menu bar. Acetic acid feed should be input at the top of the column and pure solvent at the column bottom. For thermal option select Specify temperature profile, and use 293K on all stages. Select acetic acid as the key component for the first liquid phase and triethylamine as the key component for second liquid phase. Use two equilibrium stages.

Input the conditions for the Standard Extraction Problem, run the simulation using the NRTL parameters in Table 13-A1, and report your answer.

b. Determine the K_d value for carbon tetrachloride and compare to the data in Table 13-A1. Find the flow rates and mole fractions in the two outlet streams.

c. Try changing the ratio of S/F and the feed mole fraction to determine what happens in this two-stage system.

d. Increase the number of stages to 3, and determine the separation.

e. Change conditions so that one of the phases disappears and see what happens.

3. Cross-Flow Extractor—Decanters:

a. A cross-flow system can be simulated with two decanters. Set up a two-stage cross-flow system for the Standard Extraction Problem except there are two solvent streams (pure triethylamine solvent with S₁ = S₂ = 10 kmol/h).

b. Run the simulation. Try changing S₁ and S₂, noting that flow rates do not have to be equal. Compare results with the two-stage countercurrent extractor that uses half as much solvent.

4. Countercurrent Extractor—Decanters:

a. Countercurrent systems can also be simulated with two decanters. Start with a two-stage cross-flow system from Part 3 set up for the Standard Extraction Problem S₁ = S₂ = 10, and run the simulation. (We do this first to give Aspen Plus initial values for streams that will enter decanter 1 when extract 2 is fed to decanter 1.)

b. Click in the flowsheet on extract stream from stage 2, then right-click and choose Reconnect Destination, and put the destination at feed to decanter 1. Keep both S₁ and S₂ at 10 kmol/h. Run the simulation.

c. Reduce the amount of S₁ to 1, and run a simulation of the Standard Extraction Problem again. Reduce the amount of S₁ to zero, and run the simulation again. Because Aspen Plus is programmed to flag zero values of flow rates, you will get a warning that S₁ flow rate is not specified. Ignore this warning. Decanter system is now a two-stage countercurrent system that should match previous results from step 2.

d. Repeat runs done with the countercurrent extraction column and compare results.

III. Regression of LLE Correlations against Data

1. LLE correlations in AspenPlus are not always as accurate as possible. If data are available, AspenPlus will find parameter values for any of VLE and LLE correlations by doing a regression against data you input. The purpose of this part is to obtain an improved fit for the NRTL correlation for ternary system water, chloroform, and acetone. Open a new AspenPlus file, in Setup list valid phases as Vapor-Liquid-Liquid, and input components water, chloroform, acetone. Use NRTL as LLE correlation. Go to Analysis, list ternary and valid phases as Liquid-Liquid, and obtain predicted equilibrium results. Compare predicted equilibrium data to experimental data in Table 13-5—also save Aspen Plus predictions for later comparison.

2. To regress NRTL parameters to fit data in Table 13-5, follow instructions in Appendix B (at the end of the book) as modified for LLE in the last paragraph.

3. When finished, look at the ternary graph from Analysis (list valid phases as Liquid-Liquid). Compare to experimental data and to predicted equilibrium for non-regressed parameters.

4. Go to Simulation and set up a decanter system for Example 13-4. Run a simulation for this single-stage system. Compare your results with Example 13-4.