Chapter 13: Immiscible Extraction, Washing, Leaching and Supercritical Extraction

There are a number of other separation processes that can be analyzed, at least under limiting conditions, using McCabe-Thiele diagrams or the Kremser equation with an approach very similar to that used for absorption and stripping. These processes all require addition of a mass-separating agent.

13.1 EXTRACTION PROCESSES AND EQUIPMENT

Extraction is a process where one or more solutes are removed from a liquid by transferring the solute(s) into a second liquid phase. The two liquid phases must be immiscible (that is, insoluble in each other) or partially immiscible. In this chapter we will first discuss extraction equipment and immiscible extraction, while in Chapter 14 we will discuss partially miscible extraction. In both cases the separation is based on different solubilities of the solute in the two phases. Since vaporization is not required, extraction can be done at low temperature and is a gentle process suitable for unstable molecules such as proteins or DNA.

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 the paraffins by extraction with a solvent such as sulfolane. The mixture of sulfolane and aromatics is sent to a distillation column, where the sulfolane is the 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 Ritchy and Ashbrook (1979); and extraction of biological compounds, particularly proteins, is discussed by Belter et al. (1988) and Harrison et al. (2003).

As these commercial examples illustrate, the complete extraction process includes the extraction unit and the solvent recovery process. This is shown schematically in 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 (1995, 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 will discuss these later.

Figure 13-1. Complete extraction process

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 Godfrey and Slater (1994), Humphrey and Keller (1997), Lo (1997), Lo et al. (1983), Reissinger and Schroeter (1978), Robbins and Cusack (1997), Schiebel (1978), and Woods (1995). A summary of the features of the various types of extractors is presented in Table 13-1. Decision methods for choosing the type of extractor to use are presented by King (1980), Lo (1997), Reissinger and Schroeter (1978), and Robbins (1997). Woods (1995, Chapter 5) discusses the design of decanters (another name for settlers).

Figure 13-2. Extraction equipment

TABLE 13-1 Extractor types

The following heuristics are useful for making an initial decision on the type of extraction equipment to use:

1. If one or two equilibrium stages are required, use mixer-settlers.

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) have 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.

The number of equilibrium contacts required can be determined using the same stage-by-stage procedures for all the 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).

13.2 COUNTERCURRENT EXTRACTION

The most common type of extraction cascade is the countercurrent system shown schematically in Figure 13-3. In this cascade the two phases flow in opposite directions. Each stage is assumed to be an equilibrium stage so that the two phases leaving the stage are in equilibrium.

Figure 13-3. Mass balance envelope for countercurrent cascade

The 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 the extract. The nomenclature in both weight fraction and weight ratio units is given in Table 13-2.

TABLE 13-2. Nomenclature for extraction

13.2.1 MCCABE-THIELE METHOD FOR DILUTE SYSTEMS

The McCabe-Thiele analysis for dilute immiscible extraction is very similar to the analysis for dilute absorption and stripping discussed in Chapter 12. It was first developed by Evans (1934) and is reviewed by Robbins (1997). In order to use a McCabe-Thiele type of analysis we must be able to plot a single equilibrium curve, have the energy balances automatically satisfied, and have one operating line for each section.

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. 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:

4a. Diluent and solvent are totally immiscible.

When the fourth assumption is valid, then

(13-1a)

(13-1b)

These are flow rates of diluent only and solvent only and do not include the total raffinate and extract streams. Equations (13-1) are the diluent and solvent mass balances, and they are automatically satisfied when the phases are immiscible.

Most solvent-diluent pairs that are essentially completely immiscible become partially miscible as more solute is added (see Chapter 14). This occurs because appreciable amounts of solute make the two phases chemically more similar. Since Eq. (13-1) is usually valid only for very dilute systems, we can usually analyze immiscible extraction systems using constant total flow rates. Thus, the fifth assumption we usually make is

(13-2a)

(13-2b)

For the mass balance envelope shown in Figure 13-3, the mass balance becomes,

(13-3)

Solving for y_j+1 we obtain the operating equation

(13-4)

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

These equations 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.

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

(13-5)

in weight fractions or mole fracs. For very dilute systems K_d will be constant, while at higher concentrations K_d often becomes a function of concentration. Values of K_d are tabulated in Perry and Green (1997, pp. 15-10 to 15-15), Hartland (1970, Chap. 6), and Francis (1972). A brief listing is given in Table 13-3. 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. The ethylbenzene - p-xylene separation will be the most difficult of these, but is nonetheless feasible.

TABLE 13-3. Distribution coefficients for immiscible extraction

Extraction equilibria are dependent upon 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 the buffer salt is changed, and may have several orders of magnitude change in K_d when pH is varied (Harrison et al., 2003). Extraction processes often use shifts in temperature or pH to extract and then recover a solute from the solvent (Blumberg, 1988). For example, penicillin is extracted in a process with a pH swing.

Solvent selection is critical for the development of an economical extraction system. The 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. The solvent should be easy to separate from the diluent either as 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, 1980; Walas, 1985). The solubility parameter δ is defined as

(13-6)

where (ΔE_v)_i = (ΔH_v − pΔV)_i ~ (ΔH_v − RT)_i 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). The solubility parameter is useful for quick estimation of the miscibility of two liquids—the closer the values of δ_A and δ_B are available (Giddings, 1991; King, 1980; Walas, 1985). The solubility parameter is useful for quick estimation of the miscibility of two liquids—the closer the values of δ_A and δ_B are available (Giddings, 1991; King, 1980; Walas, 1985). 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. Blumberg (1988), King (1981), Lo et al. (1983) and Treybal (1963) discuss solvent selection in detail.

Students should be aware that the rules for selecting solvents are changing. Because of increased concern about the environment, global warming, 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, aqueous two-phase extraction, and the use of supercritical extraction (see section 13-10). 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.

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 5 (constant total flow rates) is valid. The operating line goes through the point (x_N, y_N+1), which are passing streams in the column. Since this procedure is very similar to stripping (e.g., Figure 12-5, but in wt frac units), we will consider a more challenging example problem.

EXAMPLE 13-1. Dilute countercurrent immiscible extraction

A feed of 100.0 kg/minute of a 1.2 wt % mixture of acetic acid in water is to be extracted with 1-butanol at 1 atm pressure and 26.7°C. We desire an outlet concentration of 0.1 wt % acetic acid in the exiting water. We have available solvent stream 1 that is 44.0 kg/minute of pure 1-butanol and solvent stream 2 that is 30.0 kg/minute of 1-butanol that contains 0.4 wt % acetic acid. Devise a scheme to do this separation, find the outlet flow rate and concentration of the exiting 1-butanol phase, and find the number of equilibrium contacts needed.

Solution

A. Define. A feasible scheme is one that will produce exiting water with 0.001 wt frac acetic acid. We will assume that we want to use all of the solvent available; thus, the outlet butanol flow rate is 74 kg/minute (the assumption that we want to use all of the butanol will be checked in Problem 13.D1). We need to find N and y₁.

B. Explore. The equilibrium data is given in Table 13-4, y = 1.613x where y and x are the acetic acid wt fracs in the solvent and diluent phases, respectively. We learned in previous chapters that mixing is the opposite of separation; thus, we probably want to keep the two solvent streams separate. The extraction column will have an aqueous feed at the top. Since x_N = 0.001 < y_s2 = 0.004, we probably want to input the pure solvent stream 1 at the bottom of the extractor and solvent 2 in the middle of the column (Figure 13-4A).

C. Plan. We can use the mass balance envelope in the bottom of the extractor to find the operating equation,

(13-7)

where the and the operating line goes through the point (x_N, y_N+1) = (0.001, 0.0). If we use the mass balance envelope in the top of the extractor, we obtain Eq. (13-4) with

slope = R/E = R/(E₁ + E₂) = 100/(44 + 30) = 1.35.

Figure 13-4. Dilute stripper with two feeds (Example 13-1);A) schematic, B) McCabe-Thiele diagram

This operating line goes through point (x₀, y₁). Although y₁ is not immediately known, we can find it from an overall mass balance. We could also use the horizontal line y = y_solvent2 = 0.004 as a feed line and use the point of intersection of the feed line and the bottom operating line to find a point on the top operating line.

Once we have plotted the operating lines, we step off stages until a stage crosses the feed line. At that point we switch operating lines.

D. Do it. The overall mass balance is

Rx₀ + E₁y₁ + E₂y_s2 = Rx_N + Ey₁

Substituting in known values,

(100.0)(0.012) + (44.0)(0.0) + (30)(0.004) = (100.0)(0.001) + (74)y₁

and solving for y₁ we obtain y₁ = 0.01649.

The intersection point of the two operating lines occurs at the feed line, y = y_solvent2 = 0.004. Substituting this value of y into Eq. (13-7) and solving for x we obtain x_intersection = 0.00276. Plotting the equilibrium line and the two operating lines and stepping off stages, we obtain Figure 13-4B. The optimum feed stage is the third from the bottom, and we need 11.8 equilibrium contacts.

E. Check. The calculation is probably more accurate than the assumptions involved. Since there is some miscibility between the organic and aqueous phases, the flow rates will not be entirely constant. The methods discussed in Chapter 14 can be used for a more accurate calculation if sufficient data is available.

F. Generalization. The McCabe-Thiele procedure is quite general and can be applied to a number of different dilute extraction operations (sections 13.3 and 13.4) and to other separation methods (sections 13.7 and 13.8).

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

(13-8)

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. 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.

13.2.2 Kremser Method for Dilute Systems

If one additional assumption can be made, the Kremser equation can be used for dilute extraction of single or multicomponent systems.

7. Equilibrium is linear.

In this case, equilibrium has the form

(13-9)

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. (12-20) becomes

(13-10)

while if R/(mE) ≠ 1, Eqs. (12-29) (inverted) and (12-30) become

(13-11)

with = mx_o + b and

(13-12)

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. 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.

13.3 DILUTE FRACTIONAL EXTRACTION

Very often, particularly in bioseparation, extraction is used to separate solutes from each other (Belter et al., 1988; Schiebel, 1978). In this situation we can use fractional extraction with two solvents as illustrated in Figure 13-5. In fractional extraction the two solvents are chosen so that solute A prefers solvent 1 and concentrates at the top of the column, while solute B prefers solvent 2 and concentrates at the bottom of the column. In Figure 13-5 solvent 2 is labeled as diluent so that we can use the nomenclature of Table 13-2. The column sections in Figure 13-5 are often separate so that each section can be at a different pH or temperature. This will make the equilibrium curve different for the two sections. It is also common to have reflux at both ends (Schiebel, 1978) (see Chapter 14).

Figure 13-5. Fractional extraction

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 undesired chemicals are also produced.

Figure 13-6. Center-cut extraction

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 the solutes distribute between the two liquid phases. The ratio of solute flow rates in the column shown in Figure 13-5 is

(13-13)

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 13-6 we want

(13-14)

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 13-3; B goes down in the first column and up in the second column. Ranges of (K_DE/R) can be derived that will make the fractional extractor work (see Problem 13-A10). Since it is quite expensive to have a large number of equilibrium stages in a commercial extractor, the ratios in Eq. (13-14) should be significantly different.

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

(13-15a)

(13-15b)

(13-15c)

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

(13-16a)

(13-16b)

while if the feed is in solvent 2,

(13-17a)

(13-17b)

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

(13-18)

which is identical to the single component Eq. (13-7). Since the 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.

Figure 13-7. McCabe-Thiele diagram for fractional extraction; A) solute A, B) solute B

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 the 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. Some of the ways of specifying the four remaining degrees of freedom are illustrated below.

Case 1. Specify y_A,1, R, , and N_F. Calculate x_A,N from Eq. (13-15b), calculate from Eq. (13-16b) or (13-17b), and calculate E from Eq. (13-16a) or (13-17a). Since pairs of passing streams and slopes are known, both operating lines can be plotted for solute A. The total number of stages can be determined from the solute A diagram. Solution for solute B is trial-and-error.

Case 2. Specify , R, N_F, and N. This is trial-and-error for each solute, but the two solute problems are not coupled and can be solved separately.

Case 3. Specify R, , y_A,1, and x_B,N. After doing external mass balances, you can plot the operating lines as in Figure 13-7, but the problem is still trial-and-error. The feed stage must be varied until the total number of stages is the same for both solutes. Small changes in compositions or flow rates will probably be required to get an exact fit.

Because they are inherently trial-and-error, fractional extraction problems are naturals for computer solution.

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 (1980).

13.4 SINGLE-STAGE AND CROSS-FLOW EXTRACTION

Although countercurrent cascades are the most common, other cascades can be employed. One type occasionally used is the cross-flow cascade shown in Figure 13-8. Each stage is assumed to be an equilibrium stage. In this cascade, fresh extract streams are added to each stage and extract products are removed. For single-solute systems the same five assumptions made for countercurrent cascades are required for the McCabe-Thiele analysis. For dilute multicomponent systems, assumption 6 is again required.

Figure 13-8. Cross-flow cascade

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

(13-19)

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

(13-20)

Each stage will have a different operating equation. On a McCabe-Thiele diagram plotted as y vs. x, this 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, the intersection of the operating line and the equilibrium curve is at y₁ and x₁ (see Figure 13-9). This is the single-stage solution. For a cross-flow system, the raffinate input to stage 2 is x₁. Thus, the point (y_2,in, x₁) on the operating line is known and the operating line for stage 2 can be plotted. The procedure is repeated until x_N is reached or N stages have been stepped off.

Figure 13-9. McCabe-Thiele diagram for cross-flow

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 with the inlet concentrations (y_j,in, x_j−1) is on the operating line for each stage, which is easily proved. If we let y_j = y_j,in and x_j = x_j−1, Eq. (13-20) is satisfied. Thus, the point representing the inlet concentrations is on the operating line.

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.

The 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 the simulation problems do not require trial-and-error solution for cross-flow systems.

EXAMPLE 13-2. Single-stage and cross-flow extraction of a protein

We wish to extract a dilute solution of the protein alcohol dehydrogenase from a aqueous solution of 5 wt % poly (ethylene glycol) (PEG) with an aqueous solution that is 10 wt % dextran. Aqueous two-phase extraction system is a very gentle method of recovering proteins that is unlikely to denature the protein since both phases are aqueous (Albertsson et al., 1990; Harrison et al., 2003). The two phases can be considered to be essentially immiscible. The dextran phase is denser and will be the cross-flow solvent. The entering dextran phases contain no protein. The entering PEG phase flow rate is 20 kg/hr.

a. If 10 kg/hr of dextran phase is added to a single-stage extractor, find the total recovery fraction of alcohol dehydrogenase in the dextran solvent phase.

b. If 10 kg/hr of dextran phase is added to each stage of a cross-flow cascade with two stages, find the total recovery fraction of alcohol dehydrogenase in the dextran solvent phase.

The protein distribution coefficient is (Harrison et al., 2003)

K = (wt frac protein in PEG, x)/(wt frac protein in dextran, y) = 0.12

Solution

A. Define. The two-stage cross-flow process is shown in Figure 13-10. The single-stage system is the first stage of this cascade. For the single-stage extractor the fraction of the protein not extracted is (Rx)_out/(Rx)_in = x₁/x_F, and the fraction recovered = 1 − x₁/x_F. For the two stage system, the fraction of the protein not extracted is (Rx)_out/(Rx)_in = x₂/x_F, and the fraction recovered = 1 − x₂/x_F. If we find x₁/x_F and x₂/x_F, we have solved the problem.

B. Explore. Although the feed weight fraction is not specified, we can still plot the linear equilibrium with the abscissa x going from zero to x_F and the ordinate y from zero to an appropriate number of x_F units (see Figure 13-10B). This procedure works for linear systems since the slope of the equilibrium curve, y/x = 1/0.12 = 8.33333 is the same regardless of the value of x_F. The operating lines will also plot correctly regardless of the value of x_F.

C. Plan. Plot the equilibrium line, y = 8.33333 x, and plot the operating lines using Eq. (13-20) for each stage. Then calculate the recoveries.

D. Do it. Part a. Since y_in = 0, the operating line for stage 1 is

y = −(R/E₁)x + (R/E₁)x₀

Substituting in R = 20, E₁ = 10, and x₀ = x_F, the operating equation is

y = −2x+2x_F

The slope = −2 and if we set y = 0, we find x = x_F. Figure 13-10B shows that the operating and equilibrium lines intersect at y₁ = 1.613 x_F, x₁ = 0.194 x_F.

Figure 13-10. Single-stage and cross-flow extraction for Example 13-2; A) schematic, B) McCabe-Thiele diagram

Thus, x₁/x_F = 0.194 and the fractional recovery for the single-stage system = 1 − 0.194 = 0.806.

Part b. The operating line for stage 2 of the cross-flow system is

y = -(R/E₂)x + (R/E₂)x₁ = −2x + 2x₁

The slope again = −2, and if we set y = 0 we obtain x = x₁. Since x₁ = 0.194 x_F is known, we can plot this operating line (Figure 13-10B) and obtain y₂ = 0.312 x_F, x₂ = 0.037 x_F.

The fraction recovered = 1 − x₂/x_F = 1 − 0.037 = 0.963.

E. Check. This problem can also be solved analytically. We want to solve the linear equilibrium equation y = mx simultaneously with the linear operating equation for each stage, y = −(R/E_j)x + (R/E_j)x_j−1 + y_j,in. Do this sequentially starting with j = 1. This simultaneous solution is

(13-21)

For example, for stage 1, x₁ = [2x_F + 0]/[8.3333 + 2] = 0.19355 x_F which agrees with the graphical solution for part a. From the equilibrium expression, y₁ = 8.33333x₁ = 1.6129 x_F. For part b, stage 2 also agrees with the graphical solution.

Generalize. Although the problem statement involved a new type of system—aqueous two-phase extraction of proteins—we could apply the basic principles without difficulty. Since many of the details are often not necessary to solve the problem, it sometimes simplifies the problem to rewrite it as solute A being removed from diluent into solvent. Then see if you can solve the simplified version of the problem.

For linear equilibrium and operating lines an analytical solution is always relatively easy. The McCabe-Thiele diagram (even if only roughly sketched) is very useful to organize our thoughts and make sure we don’t make any dumb algebraic errors. If the equilibrium line is curved, the analytical solution becomes considerably more complicated, but the McCabe-Thiele solution doesn’t.

It is also interesting to compare the cross-flow system to a countercurrent system with 2 stages and a total flow rate of E = 20 kg/hr (see Problem 13.D2). The result shows that the countercurrent process has a higher recovery.

Cross-flow systems have also been explored for stripping (Wnek and Snow, 1972). 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 the 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 the same total amount of solvent and the 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 they may have an advantage when flooding or slow settling of the two phases is a problem. For absorption and stripping the pressure drop may be significantly lower in a cross-flow system. Cascades that combine cross-flow and countercurrent operation have been studied by Thibodeaux et al. (1977), and by Bogacki and Szymanowski (1990).

13.5 CONCENTRATED IMMISCIBLE EXTRACTION

If the extraction system is relatively concentrated but still immiscible (this is almost a contradiction), we can extend the ratio unit mass balances developed in Chapter 12 for a single solute to extraction. Assumption 5, Eq. (13-2), no longer needs to be valid. This ratio-unit analysis is easily extended to cross-flow systems.

When the diluent and solvent are immiscible, weight ratio units are related to weight fractions as

(13-22)

where X is kg solute/kg diluent and Y is kg solute/kg solvent. Note that these equations require that the phases be immiscible.

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

(13-23)

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

(13-24)

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 (Y₁ − (F_D/F_s)X₀). Note that since F_D and F_S are constant, the operating line is straight. For the usual design problem, F_D/F_S will be known as will be 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. 13-1, the 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 equlibrium data and the operating equation known, the McCabe-Thiele diagram is plotted. First the 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 X_o. Then the stages are easily stepped off.

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

4b. The concentration of solvent in the raffinate and the concentration of diluent in the extract are both constant.

The flow rates of the diluent and solvent streams are now defined as

(13-25a)

(13-25b)

The ratio units are defined as

(13-26a)

(13-26b)

The calculation procedure now follows Eqs. (13-23) to (13-24). When the phases are partially miscible and assumption 4b is not valid, the methods developed in Chapter 14 should be used.

13.6 BATCH EXTRACTION

In batch plants batch extraction will usually fit into the production scheme easier than continuous extraction. This is particularly true in the production of biochemicals (Belter et al., 1988). Batch extraction is very flexible, and there are a number of ways to do it. The simplest approach is to add solvent and diluent together in a tank, mix the two immiscible liquids, allow them to settle and then withdraw the solvent layer. If we define , , and as the mass (kg) of the streams, the resulting mass balance is

(13-24a)

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

(13-24b)

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 a repeated batch extraction is done, it is essentially identical to the continuous cross-flow system.

If we want to totally remove the solute from the diluent, the continuous solvent addition batch extraction shown in Figure 13-11 will use less solvent than repeated single-stage batch extractions. A solvent that is pre-saturated with diluent is added continuously to a mixed tank that contains the feed. The raffinate and extract phases are separated in a settler with withdrawal of the extract product and recycle of the diluent.

Figure 13-11. Continuous solvent addition batch extraction

This process is analogous to constant level batch distillation (section 9.3.) and the analysis for immiscible extraction is very similar to the analysis in that section. If we assume that the level in the mixed tank is constant, then the overall mass balance becomes in = out, and for d kg of entering solvent,

(13-25)

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

(13-26a)

where is the mass of raffinate phase in the tank and x_t is the mass fraction solute in the raffinate phase in the tank. If we assume that V_tank >> V_settler, then and , and Eq. (13-26a) simplifies to

(13-26b)

Substituting in Eq. (13-25) and integrating, we obtain

(13-27)

where y and x_t are in equilibrium and is the total mass of solvent added. In general, this equation can always be integrated numerically or graphically. If equilibrium is linear, y = Kx, analytical integration is straightforward and the result is

(13-28)

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

If the solvent and diluent are partially miscible, the methods in Chapter 14 need to be used. The single-stage and cross-flow systems are illustrated in section 14.3. Countercurrent columns can also be used and y in Eq. (13-27) must be related to x_t by the countercurrent analysis (see section 14.4).

13.7 GENERALIZED MCCABE-THIELE AND KREMSER PROCEDURES

The McCabe-Thiele procedure has been applied to flash distillation, continuous countercurrent distillation, batch distillation, absorption, stripping, and extraction. What are the common factors for the McCabe-Thiele analysis in all these cases?

All the McCabe-Thiele graphs are plots of concentration in one phase vs. concentration in the other phase. In all cases there is a single equilibrium curve, and there is one operating line for each column section. It is desirable for this operating line to be straight. In addition, although it isn’t evident on the graph, we want to satisfy the energy balance and mass balances for all other species.

In order to obtain a single equilibrium curve, we have to specify enough variables that only one degree of freedom remains. For binary distillation this can be done by specifying constant pressure. For absorption, stripping, and extraction we specified that pressure and temperature were constant, and if there were several solutes we assumed that they were independent. In general, we will specify that pressure and/or temperature are constant, and for multisolute systems we will assume that the solutes are independent.

To have a straight operating line for the more volatile component in distillation we assumed that constant molal overflow (CMO) was valid, which meant that in each section total flows were constant. For absorption, stripping, and extraction we could make the assumption that total flows were constant if the systems were very dilute. For more concentrated systems we assumed that there was one chemical species in each phase that did not transfer into the other phase; then the flow of this species (carrier gas, solvent, or diluent) was constant. In general, we have to assume either that total flows are constant or that flows of nontransferred species are constant.

These assumptions control the concentration units used to plot the McCabe-Thiele diagram. If total flows are constant, the solute mass balance is written in terms of fractions, and fractions are plotted on the McCabe-Thiele diagram. If flows of nontransferred species are constant, ratio units must be used, and ratios are plotted on the McCabe-Thiele diagram.

The McCabe-Thiele operating line satisfies the mass balance for only the more volatile component or the solute. In binary distillation the CMO assumption forces total vapor and liquid flow rates to be constant and therefore the overall mass balance will be satisfied. In absorption, when constant carrier and solvent flows are assumed, the mass balances for these two chemicals are automatically satisfied. In general, if overall flow rates are assumed constant, we are satisfying the overall mass balance. If the flow rates of nontransferred species are constant, we are satisfying the balances for these species.

The energy balance is automatically satisfied in distillation when the CMO assumption is valid. In absorption, stripping, and extraction the energy balances were satisfied by assuming constant temperature and a negligible heat of absorption, stripping or mixing. In general, we will assume constant temperature and a negligible heat involved in contacting the two phases.

The Kremser equation was used for absorption, stripping, and extraction. When total flows, pressure, and temperature are constant and the heat of contacting the phases is negligible, we can use the Kremser equation if the equilibrium expression is linear. When these assumptions are valid, the Kremser equation can be used for other separations.

Of course, the assumptions required to use a McCabe-Thiele analysis or the Kremser equation may not be valid for a given separation. If the assumptions are not valid, the results of the analysis could be garbage. To determine the validity of the assumptions, the engineer has to examine each specific case in detail. The more dilute the solute, the more likely it is that the assumptions will be valid.

In the remainder of this chapter these principles will be applied to generalize the McCabe-Thiele approach and the Kremser equation for a variety of unit operations. A listing of various applications is given in Table 13-4.

TABLE 13-4. Applications of McCabe-Thiele and Kremser procedures

13.8 WASHING

When solid particles are being processed in liquid slurries, the solids entrain liquid with them. The removal of any solute contained in this entrained liquid is called washing. To be specific, consider an operation that mines sand from the ocean. The wet sand contains salt, and this salt can be removed by washing with pure water. The entrained liquid is called underflow liquid, because the solids are normally removed from the bottom of a settler as shown in Figure 13-12A. Washing is done by mixing solid (sand) and wash liquor (water) together in a mixer and sending the mixture to a settler or a thickener (Perry and Green, 1997, p. 18-6). The solids and entrained underflow liquid exit from the bottom of the settler, and clear overflow liquid without solids is removed from the top. In washing, the solute (salt) is not held up or attached to the inert solid (sand). The salt is assumed to be at the same concentration in the underflow liquid as it is in the overflow liquid. Thus, it can be removed by displacing it with clear water. The separation can be done in single-stage, cross-flow, and countercurrent cascades.

Figure 13-12. Countercurrent washing; A) two-stage mixer-settler system, B) general system

The equilibrium condition for a washer is that solute concentration is the same in both the underflow and overflow liquid streams. This statement does not say anything about the solid, which changes the relative underflow and overflow flow rates but does not affect concentrations. Thus, the equilibrium equation is

(13-29)

where y = mass fraction solute in the overflow liquid and x = mass fraction solute in the underflow liquid.

For the mass balance envelope on the countercurrent cascade shown in Figure 13-12B, it is easy to write a steady state-mass balance.

(13-30)

where O_j and U_j are the total overflow and underflow liquid flow rates in kg/hr leaving stage j. The units for Eq. (13-30) are kg solute/hr. To develop the operating equation, we solve Eq. (13-30) for y_j+1:

(13-31)

In order for this to plot as a straight line the underflow liquid and overflow liquid flow rates must be constant.

Often the specifications will give the flow rate of dry solids or the flow rate of wet solids. The underflow liquid flow rate can be calculated from the volume of liquid entrained with the solids. Let ε be the porosity (void fraction) of the solids in the underflow. That is,

(13-32a)

and then

(13-32b)

We can now calculate the underflow liquid flow rate, U_j. Suppose we are given the flow rate of dry solids. Then the volume of solids per hour is

(13-33a)

and the total volume of underflow is

(13-33b)

Then the volume of underflow liquid in m₃ liquid per hour is

(13-33c)

and finally the kg/hr of underflow liquid

(13-33d)

In these equations ρ_f is the fluid density in kg/m³ and ρ_s is the density of dry solids in kg/m³.

If the solids rate, ε, ρ_f, and ρ_s are all constant, then from Eq. (13-33d) U_j = U = constant. If U is constant, then an overall mass balance shows that the overflow rate, O_j, must also be constant. Thus, to have constant flow rates we assume:

1. No solids in the overflow and solids do not dissolve. This ensures that the solids flow rate will be constant.

2. ρ_f and ρ_s are constant. Constant ρ_f implies that the solute has little effect on fluid density or that the solution is dilute.

3. Porosity ε is constant. Thus, the volume of liquid entrained from stage to stage is constant.

When these assumptions are valid, O and U are constant, and the operating equation simplifies to

(13-34)

which obviously represents a straight line on a McCabe-Thiele plot. Note that this equation is similar to all the other McCabe-Thiele operating equations we have developed. Only the nomenclature has changed.

An alternative way of stating the problem would be to specify the volume of wet solids processed per hour. Then the underflow volume is

(13-35a)

and

(13-35b)

If densities and ε are constant, volumetric flow rates are constant, and the washing problem can be solved using volumetric flow rates and concentrations in kg solute/m₃.

Note that we could have just assumed that overflow and underflow rates are constant and derived Eq. (13-34). However, it is much more informative to show the three assumptions required to make overflow and underflow rates constant. These assumptions show that this analysis for washing is likely to be invalid if the settlers are not removing all the solid, if for some reason the amount of liquid entrained changes, or if the fluid density changes markedly. The first two problems will not occur in well-designed systems. The third is easy to check with density data.

The McCabe-Thiele diagram can now be plotted as shown in Figure 13-13. This McCabe-Thiele diagram is unique, since temperature and pressure do not affect the equilibrium; however, temperature will affect the rate of attaining equilibrium and hence the efficiency, because at low temperatures more viscous solutions will be difficult to wash off the solid.

Figure 13-13. McCabe-Thiele diagram for washing

The analysis for washing can be extended to a variety of modifications. These include simulation problems, use of efficiencies, calculation of maximum U/O ratios, and calculations for cross-flow systems. The Kremser equation can also be applied to countercurrent washing with no additional assumptions. This adaptation is a straightforward translation of nomenclature and is illustrated in Example 13-3. Brian (1972) discusses application of the Kremser equation to washing in considerable detail.

Washing is also commonly done by collecting the solids on a filter and then washing the filter cake. This approach, which is often used for crystals and precipitates that may be too small to settle quickly and is usually a batch operation, is discussed by Mullin (2001).

EXAMPLE 13-3. Washing

In the production of sodium hydroxide by the lime soda process, a slurry of calcium carbonate particles in a dilute sodium hydroxide solution results. A four-stage countercurrent washing system is used. The underflow entrains approximately 3 kg liquid/kg calcium carbonate solids. The inlet water is pure water. If 8 kg wash water/kg calcium carbonate solids is used, predict the recovery of NaOH in the wash liquor.

Solution

A. Define. Recovery is defined as 1 – x_out/x_in. Thus, recovery can be determined even though x_in is unknown.

B and C. Explore and plan. If we pick a basis of 1 kg calcium carbonate/hr, then O = 8 kg wash water/hr and U = 3 kg/hr. This problem can be solved with the Kremser equation if we translate variables. To translate: Since y = overflow liquid weight fraction, we set O = V. Then U = L. this translation keeps y = mx as the equilibrium expression. It is convenient to use the Kremser equation in terms of x. For instance, Eq. (12-39) becomes

(13-36)

D. Do it. Equilibrium is y = x; thus, m = 1. Since inlet wash water is pure, y_N+1 = 0. Then , m O/U = (1)(8)/3, and N = 4. Then Eq. (13-36) is

and

Recovery = 1 = x_N/x₀ = 0.98755

E. Check. This solution can be checked with a McCabe-Thiele diagram. Since the x_N value desired is known, the check can be done without trial and error.

F. Generalize. Recoveries for linear equilibrium can be determined without knowing the inlet concentrations. This can be useful for the leaching of natural products because the inlet concentration fluctuates. The translation of variables shown here can be applied to other forms of the Kremser equation.

The washing analysis presented here is for a steady-state, completely mixed system where the wash water and the water entrained by the solid matrix are in equilibrium. When filter cakes are washed, the operation is batch, the system is not well mixed because flow is close to plug flow, and the operation is not at equilibrium since the entrained fluid has to diffuse into the wash liquid. This case is analyzed by Harrison et al. (2003).

13.9 LEACHING

Leaching, or solid-liquid extraction, is a process in which a soluble solute is removed from a solid matrix using a solvent to dissolve the solute. The most familiar examples are making coffee from ground coffee beans and tea from tea leaves. The complex mixture of chemicals that give coffee and tea their odor, taste, and physiological effects are leached from the solids by the hot water. An espresso machine just does the leaching faster into a smaller volume of water. Instant coffee and tea are made by leaching ground coffee beans or tea leaves with hot water and then drying the liquid to produce a solid. There are many other commercial applications of leaching such as leaching soybeans to recover soybean oil (a source of biodiesel), leaching ores to recover a variety of minerals, and leaching plant leaves to extract a variety of pharmaceuticals (Rickles, 1965; Schwartzberg, 1980, 1987).

The equipment and operation of washing and leaching systems are often very similar. In both cases a solid and a liquid must be contacted, allowed to equilibrate, and then separated from each other. Thus, the mixer-settler type of equipment shown in Figure 13-12 is also commonly used for leaching easy-to-handle solids. A variety of other specialized equipment has been developed to move the solid and liquid countercurrently during leaching. Prabhudesai (1997), Miller (1997), Lydersen (1983), and Schwartzberg (1980, 1987) present good introductions to this leaching equipment.

In leaching, the solute is initially part of the solid and dissolves into the liquid. In washing, which can be considered as a special case of leaching, the solute is initially retained in the pores of the solid and the solid itself does not dissolve. In leaching, the equilibrium equation is usually not y = x, and the total solids flow rate is usually not constant. Since diffusion rates in a solid are low, mass transfer rates are low. Thus, equilibrium may take days for large pieces such as pickles, where it is desirable to leach out excess salt, or even years for in-situ leaching of copper ores (Lydersen, 1983). A rigorous analysis of leaching requires that the changing solid and liquid flow rates be included. This situation is very similar to partially miscible extraction and is included in Chapter 14. In this section we will look at simple cases where a modified McCabe-Thiele or Kremser equation can be used.

A countercurrent cascade for leaching is shown in Figure 13-14. We will consider the (idealized) case where entrainment of liquid with the solid underflow can be ignored. The assumptions are:

1. The system is isothermal.

2. The system is isobaric.

3. No solvent dissolves into solid.

4. No solvent entrained with the solid.

5. There is an insoluble solid backbone or matrix.

6. The heat of mixing of solute in solvent is negligible.

7. The stages are equilibrium stages.

Figure 13-14. Countercurrent leaching; A) cascade, B) McCabe-Thiele diagram

With these assumptions the energy balance is automatically satisfied. A straight operating line is easily derived using the mass balance envelope shown in Figure 13-14. Defining ratios and flow rates

(13-37)

the operating equation is

(13-38)

This represents a straight line as plotted in Figure 13-14B. The equilibrium curve is now the equilibrium of the solute between the solvent and solid phases. The equilibrium data must be measured experimentally. If the equilibrium line is straight, the Kremser equation can be applied.

In the previous analysis, assumptions 4 and 7 are often faulty. There is always entrainment of liquid in the underflow (for the same reason that there is an underflow liquid in washing). Since diffusion in solids is very slow, equilibrium is seldom attained in real processes. The combined effects of entrainment and nonequilibrium stages are often included by determining an “effective equilibrium constant.” This effective equilibrium depends on flow conditions and residence times and is valid only for the conditions at which it was measured. Thus, the effective equilibrium constant is not a fundamental quantity. However, it is easy to measure and use. The McCabe-Thiele diagram will look the same as Figure 13-14B. Further simplification is obtained by assuming that the effective equilibrium is linear, y = m_Ex.

13.10 SUPERCRITICAL FLUID EXTRACTION

There has been an increasing amount of interest in the use of supercritical fluids (SCF) for extracting compounds from solids or liquids in the food and pharmaceutical industries because of the nontoxic nature of the primary SCF––carbon dioxide. In this section we will briefly consider the properties of SCFs that make them interesting for extraction. Then a typical process for SCF extraction will be explored and several applications will be discussed.

First, what is an SCF? Figure 13-15A shows a typical pressure-temperature diagram for a single component. Above the critical temperature T_c, it is impossible to liquefy the compound. The critical pressure p_c is the pressure required to liquefy the compound at the critical temperature. The critical temperatures and pressures for a large number of compounds have been determined (Paulaitis et al., 1983; Poling et al., 2001). SCFs of interest include primarily carbon dioxide (p_c = 72.8 atm, T_c = 31° C, ρ = 0.47 g/mL), but also propane (p_c = 41.9 atm, T_c = 97° C, ρ = 0.22 g/mL), and water (p_c = 217.7 atm, T_c = 374° C, ρ = 0.32 g/mL). An SCF behaves like a gas in that it will expand to fill the confines of the container.

Figure 13-15. Thermodynamics of SCF extraction; A) pressure-temperature diagram for pure component, B) solubility of naphthalene in CO₂

As can be seen from this very short list, SCFs have densities much greater than those of typical gases and less than those of liquids by roughly a factor of 2 to 3. The viscosities of SCFs are about one-tenth those of liquids. This leads to low-pressure drops. The high diffusivities, roughly ten times those of liquids, plus the lack of a phase boundary leads to very high mass transfer rates and low HETP values in packed beds. The SCFs can often dissolve almost the same amount of solute as a good solvent. Extraction can often be carried out at low temperatures, particularly when CO₂ is the SCF. This is particularly advantageous for extraction of foods and pharmaceuticals. Many SCFs are also completely natural or “green” and thus, are totally acceptable as additives in foods and pharmaceuticals (Allen and Shonnard, 2002). This is a major advantage to the use of supercritical CO₂ without additives.

The solubility of a solute in an SCF is a complex function of temperature and pressure. This is commonly illustrated with the solubility of naphthalene in CO₂, which is illustrated in Figure 13-15B (Hoyer, 1985; Paulaitis et al., 1983). As pressure is increased the solubility first decreases and then increases. At both high and low pressures the naphthalene is more soluble at high temperatures than at low temperatures. This is the expected behavior, because the vapor pressure of naphthalene increases with increasing temperature. Immediately above the critical pressure the solute is more soluble at the lower temperature. This is a retrograde phenomenon. If naphthalene solubility is plotted vs. CO₂ density, the retrograde behavior does not appear. In addition to having high solubilities, the SCF should be selective for the desired solutes. Solute-solvent interactions can affect the solubility and the selectivity of the SCF, and therefore entrainers are often added to the SCF to increase solubility and selectivity.

Figure 13-15B also shows how the solute can be recovered from the CO₂. If the pressure is dropped, the naphthalene solubility plummets, and naphthalene will drop out as a finely divided solid. A typical process using pressure reduction is shown in Figure 13-16. Note that this will probably be a batch process if solids are being processed because of the difficulty in feeding and withdrawing solids at the high pressures of supercritical extraction. Regeneration can also be achieved by changing the temperature, distilling the CO₂-solute mixture at high pressure or absorbing the solute in water (McHugh and Krukonis, 1994). In many cases one of these processes may be preferable because it will decrease the cost of compression.

Figure 13-16. Batch SCF extraction; regeneration is by pressure swing

Several current applications have been widely publicized. Kerr-McGee developed its ROSE (Residuum Oil Supercritical Extraction) process in the 1950s (Humphrey and Keller, 1997; Johnston and Lemert, 1997; McHugh and Krukonis, 1994). When oil prices went up, the process attracted considerable attention, since it has lower operating costs than competing processes. The ROSE process uses an SCF such as propane to extract useful hydrocarbons from the residue left after distillation. This process utilizes the high temperatures and pressures expected for residuum treatment to lead naturally to SCF extraction.

Much of the commercial interest has been in the food and pharmaceutical industries. Here, the major driving force is the desire to have completely “natural” processes, which cannot contain any residual hydrocarbon or chlorinated solvents (Humphrey and Keller, 1997). Supercritical carbon dioxide has been the SCF of choice because it is natural, nontoxic, and cheap, is completely acceptable as a food or pharmaceutical ingredient, and often has good selectivity and capacity. Currently, supercritical CO₂ is used to extract caffeine from green coffee beans to make decaffeinated coffee. Supercritical CO₂ is also used to extract flavor compounds from hops to make a hop extract that is used in beer production. The leaching processes that were replaced were adequate in all ways except that they used solvents that were undesirable in the final product.

A variety of other SCF extraction processes have been explored (Hoyer, 1985; McHugh and Krukonis, 1994; Paulaitis et al., 1983). These include extraction of oils from seeds such as soybeans, removal of excess oil from potato chips, fruit juice extraction, extraction of oxygenated organics such as ethanol from water, dry cleaning, removal of lignite from wood, desorption of solutes from activated carbon, and treatment of hazardous wastes. Not all of these applications were successful, and many that were technically successful are not economical.

The main problems in applying SCF extraction on a large scale have been scaling up for the high pressure required. The high-pressure equipment becomes quite heavy and expensive. In addition, methods for charging and discharging solids continuously have not been well developed for these high-pressure applications. Another problem has been the lack of design data for supercritical extraction. Supercritical extraction is not expected to be a cheap process. Thus, the most likely applications are extractions for which existing separation methods have at least one serious drawback and for which SCF extraction does not have major processing disadvantages.

13.11 APPLICATION TO OTHER SEPARATIONS

The McCabe-Thiele and Kremser methods can be applied to analyze other separation processes. Adsorption, chromatography, and ion exchange are occasionally operated in counter current columns. In some situations crystallization can be analyzed as an equilibrium stage separation. The application of the McCabe-Thiele procedure in these cases is explored by Wankat (1990).

The McCabe-Thiele and Kremser methods have also been applied to analyze less common separation methods. A modification of the McCabe-Thiele method has been applied to parametric pumping, which is a cyclic adsorption or ion exchange process (Grevillot and Tondeur, 1977). A similar modification can be used to analyze cycling zone adsorption (Wankat, 1986). McCabe-Thiele and Kremser methods can also be used to analyze three-phase separations (Wankat, 1988).

13.12 SUMMARY—OBJECTIVES

In this chapter we looked at the general applicability of the McCabe-Thiele and Kremser analysis procedures. The methods are reviewed in Table 13-4. At the end of this chapter you should be able to achieve the following objectives:

1. Explain in general terms how the McCabe-Thiele and Kremser analyses can be applied to other separation schemes and delineate when these procedures are applicable

2. Explain what extraction is, outline the types of equipment used, and apply the McCabe-Thiele and Kremser methods to immiscible extraction problems including fractional extraction and cross-flow extraction

3. Explain what washing is and apply the McCabe-Thiele and Kremser procedures to washing problems

4. Explain what leaching is and apply both McCabe-Thiele and Kremser methods to leaching problems

5. Explain how supercritical extraction works, and discuss its advantages and disadvantages

REFERENCES

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Allen, D. T. and D. R. Shonnard, Green Engineering: Environmentally Conscious Design of Chemical Processes, Prentice Hall, Upper Saddle River, New Jersey, 2002.

Belter, P. A., E. L. Cussler and W.-S. Hu, Bioseparations. Downstream Processing for Biotechnology, Wiley-Interscience, New York, 1988.

Benedict, M. and T. Pigford, Nuclear Chemical Engineering, McGraw-Hill, New York, 1957.

Bogacki, M. B. and J. Szymanowski, “Modeling of Extraction Equilibrium and Computer Simulation of Extraction-Stripping Systems for Copper Extraction by 2-Hydroxy-5-Nonylbenzaldehyde Oxime,” Ind. Engr. Chem. Research, 29, 601 (1990).

Blumberg, R., Liquid-Liquid Extraction, Academic Press, London, 1988.

Brian, P. L. T., Staged Cascades in Chemical Processes, Prentice Hall, Upper Saddle River, New Jersey, 1972.

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HOMEWORK

A. Discussion Problems

A1. Develop your key relations chart for this chapter. Remember that a key relations chart is not a core dump but is selective.

A2. In your own words, describe the McCabe-Thiele analytical procedure in general terms that could be applied to any separation.

A3. How do the ideas of a general McCabe-Thiele procedure and the concept of unit operation relate to each other?

A4. 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?

A5. Compare the advantages and disadvantages of the McCabe-Thiele and Kremser design procedures.

A6. Explain the similarities (analogies) between extraction and stripping (or absorption). How do the separations differ?

A7. How does the solid enter into washing calculations? Where does solids flow rate implicitly appear in Figure 13-12?

A8. Referring to Table 13-4, list similarities and differences between absorption, stripping, extraction, washing, and leaching.

A9. Compare K_DE/R to KV/L used in distillation, absorption, and stripping. Do these quantities have the same significance?

A10. What are the appropriate ranges of K_D,jE/R for solutes A, B, C in the two columns in Figure 13-6?

A11. 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?

A12. Show how Figure 13-16 could be modified to use a temperature swing instead of a pressure swing. What might be the advantage and disadvantage of doing this?

A13. What are some of the properties you would look for in a good solvent for extraction, leaching and supercritical extraction?

B. Generation of Alternatives

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

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

C. Derivations

C1. Chapters 12 and 13 cover separations that use a mass-separating agent. Derive general operating and equilibrium equations for the separations in these chapters.

C2. Derive Eq. (13-10) starting with a McCabe-Thiele diagram (follow the procedure used to develop the Kremser equation in Chapter 12).

C3. Derive Eq. (13-11) starting with a McCabe-Thiele diagram (follow the procedure used to develop the Kremser equation in Chapter 12).

C4. Derive Eq. (13-12) from Eq. (13-11).

C5. Derive Eq. (13-18).

C6. Prove that the two operating lines in fractional extraction (Figure 13-7) intersect at a feed line.

C7. For fractional extraction outline in detail a solution procedure for (a) case 1, (b) case 2, and (c) case 3.

C8. For the cross-flow cascade, show that the point (y_j,in, x_j–1) is on the operating line. Also show that if the two entering feeds are combined as a mixed feed the operating equation is the same as for flash distillation.

C9. Develop the solution method for a dilute multicomponent extraction in a cross-flow cascade. Sketch the McCabe-Thiele diagrams.

C10. 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?

C11. Explain the derivation of Eq. (13-33d). Write out the units for this equation.

C12. Adapt the Kremser equation to leaching.

C13. Derive the operating equations and sketch the McCabe-Thiele analytical procedure for cross-flow and single-stage washing systems.

C14. Derive Eq. (13-38).

D. Problems

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

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

a. Use only the pure solvent at the bottom of the extractor.

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

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

D2. Repeat Example 13-2 but use a countercurrent cascade with all of the solvent (E = 20 kg/hr) flowing countercurrent to the feed through the two stages.

D3. A feed that is 0.008 wt frac acetone in water is fed to a single mixer-settler that has an efficiency of 100 %. The total flow rate of the feed is 150.0 kg/hr. We contact this with 500.0 kg/hr of pure chloroform. Determine the outlet wt fracs of acetone in the extract and raffinate. You may assume that water and chloroform are immiscible and that the equilibrium expression for acetone is

K = y/x = 1.816.

a. Solve this graphically like a flash calculation (operating line has a negative slope).

b. Solve using the Kremser equation with N = 1.

D4.* We have a mixture of acetic acid in water and wish to extract this with 3-heptanol at 25° C. Equilibrium is

The inlet water solution flows at 550 lb/hr and is 0.0097 wt frac acetic acid. We desire an outlet water concentration of 0.00046 wt frac acetic acid. The solvent flow rate is 700 lb/hr. The entering solvent contains 0.0003 wt frac acetic acid. Find the outlet solvent concentration and the number of equilibrium stages required (use the Kremser equation). Is this an economical way to extract acetic acid?

D5. We are extracting acetic acid from benzene (diluent) into water (solvent) at 25°C. 100.0 kg/hr of a feed that is 0.0092 wt frac acetic acid and 0.9908 wt frac benzene is fed to a column. The inlet water (solvent) is pure and flows at 25.0 kg/hr. We desire an outlet weight fraction of 0.00040 acetic acid in benzene. Assume total flow rates are constant, water and benzene are completely immiscible, equilibrium is linear,

a. The outlet wt frac of acetic acid in the water.

b. The number of equilibrium stages required.

c. The minimum solvent rate.

D6. We plan to recover acetic acid from water using 1-butanol as the solvent in a batch extraction. Operation is at 26.7°C. The feed is 10.0 kg of an aqueous solution that contains 0.0046 mole frac acetic acid. Add 5.0 kg pure solvent. This operation will be done in a single mixer-settler, which can be assumed to be an equilibrium stage. Equilibrium data are available in Table 13-3.

a. Find the outlet mole fracs of solvent and diluent using a McCabe-Thiele diagram. Note that this can be done either in a form similar to a flash distillation or as a counter current process with one stage.

b. Check your McCabe-Thiele solution using the Kremser equation.

D7.* We have an extraction column with 30 equilibrium stages. We are extracting acetic acid from water into 3-heptanol at 25° C. Equilibrium is given in Problem 13-D4. The aqueous feed flows at a rate of 500 kg/hr. The feed is 0.011 wt frac acetic acid, and the exit water should be 0.00037 wt frac acetic acid. The inlet 3-heptanol contains 0.0002 wt frac acetic acid. What solvent flow rate is required? Assume that total flow rates are constant.

D8. At low concentrations of acetone, chloroform and water are essentially immiscible and the distribution coefficient is essentially constant (at constant temperature). At 25°C and 1.0 atm this distribution coefficient is,

(wt frac acetone in extract phase)/(wt frac acetone in raffinate phase) = y/x = 1.816

We have 100.0 kg/hr of a feed that is x₀ = 0.007 wt frac acetone and the remainder is water. We wish to recover 98%of the acetone in the chloroform. Use a countercurrent extractor operating at 25°C and 1.0 atm. The entering extract stream is pure chloroform, y_N+1 = 0. The system has 18 equilibrium stages. Find the wt frac of acetone in the outlet raffinate, x_N, the solvent flow rate, E, needed and the wt frac of acetone in the outlet extract, y₁.

D9.* We have a mixture of linoleic and oleic acids dissolved in methylcellosolve and 10% water. Feed is 0.003 wt frac linoleic acid and 0.0025 wt frac oleic acid. Feed flow rate is 1500 kg/hr. A simple countercurrent extractor will be used with 750 kg/hr of pure heptane as solvent. We desire a 99% recovery of the oleic acid in the extract product. Equilibrium data are given in Table 13-3. Find N and the recovery of linoleic acid in the extract product.

D10. We are extracting acetic acid from benzene (diluent) into water (solvent) at 25°C and 1.0 atm. 100.0 kg/hr of a feed that is 0.00092 wt frac acetic acid and 0.99908 wt frac benzene is fed to a column. The inlet water (solvent) is pure and flows at 25.0 kg/hr. We have an extractor that operates with 2 equilibrium stages. Data are in Problem 13.D5. Find:

a. The outlet wt frac of acetic acid in the benzene, x_out.

b. The outlet wt frac of acetic acid in the water, y_out.

D11.* The fractional extraction system shown in Figure 13-5 is separating abietic acid from other acids. Solvent 1, heptane, enters at = 1000 kg/hr and is pure. Solvent 2, methylcellosolve + 10% water, is pure and has a flow rate of R = 2500 kg/hr. Feed is 5 wt % abietic acid in solvent 2 and flows at 1 kg/hr. 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 that the solvents are completely immiscible and that the system can be considered to be very dilute. Equilibrium data are given in Table 13-3. Find N.

D12. We plan to recover acetic acid from water using 1-butanol as the solvent. Operation is at 26.7°C. The feed flow rate is 10.0 kg moles per hour of an aqueous solution that contains 0.0046 mole frac acetic acid. The entering solvent is pure and flows at 5.0 kg moles per hour. This operation will be done with three mixer-settlers arranged as a countercurrent cascade. Each mixer-settler can be assumed to be an equilibrium stage. Equilibrium data are available in Table 13-3. Find the exiting raffinate and extract mole fractions.

D13. We wish to extract p-xylene and o-xylene from n-hexane diluent using β, β′ -Thiodipropionitrile as the solvent. The solvent and diluent can be assumed to be immiscible. The feed is 1000.0 kg/hr. The feed contains 0.003 wt frac p-xylene and 0.005 wt frac o-xylene in n-hexane. We desire at least a 90% recovery of p-xylene and at least 95% recovery of o-xylene. The entering solvent is pure. Operation is at 25°C and equilibrium data are in Table 13-3. Use a simple countercurrent cascade.

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

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

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

D14. We have 100.0 kg/hr of a feed that is 0.007 wt frac acetone and the remainder is water. We wish to recover 98% of the acetone in the chloroform. Use a countercurrent extractor operating at 25°C and 1.0 atmosphere. The entering extract stream is pure chloroform. Use a solvent flow rate that is 1.1 times the minimum solvent rate. Equilibrium data are in Problem 13.D8. Find the solvent flow rate needed, the wt fracs of acetone in the outlet extract and raffinate streams, and the number of stages required.

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

a. Determine y₁ and y_N+1 (units are wt fracs) for the column at 40° C.

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

c. Is this a practical way to concentrate the acid?

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

D16. Acetic acid is being extracted from water with butanol as the solvent. Operation is at 26.7°C and equilibrium data are in Table 13-3. The feed is 10.0 kg/minute of an aqueous feed that contains 0.01 wt frac acetic acid. The entering solvent stream is butanol with 0.0002 wt frac acetic acid. The flow rate of the solvent stream is 8.0 kg/minute. The column has 6 equilibrium stages. Find the outlet weight fractions. Assume butanol and water are immiscible.

a. Solve with the Kremser equation.

b. Check your answer with a McCabe-Thiele diagram (done in weight fractions). Note that if used as a check, the McCabe-Thiele diagram is not trial-and-error even when the number of stages is known.

D17. The feed in Problem 13.D16 is to be processed in a 6 stage cross-flow extractor. The same solvent as in Problem 13.D16 is increased to a total flow rate of 12.0 kg/minute that is equally divided among the 6 stages, so that flow rate of solvent to each stage is 2.0 kg/minute. Find the outlet diluent weight fraction.

D18. Tributyl phosphate (TBP) in kerosene is used as a solvent for extracting metal ions from aqueous solution in a countercurrent cascade. The inlet feed is an aqueous solution of nitric acid and sodium nitrate that contains 0.110 gmoles/liter of Zr(NO₃)₄. Feed rate is F_Aq = 100.0 liters/hr. The entering TBP solution contains no Zr(NO₃)₄. Assume that the aqueous and organic phases are completely immiscible and that the densities of the two phases are both constant (volumetric flow rates are constant). An outlet concentration of 0.010 gmole Zr(NO₃)₄ /liter is desired. For zirconium nitrate, Zr(NO₃)₄, the equilibrium data are listed below for 60 volume % TBP in kerosene (Benedict and Pigford, 1957):

a. Find the number of equilibrium stages required if the rate of the entering TBP/kerosene solvent is F_Org = 85.0 liter/hr.

b. Find the minimum entering solvent rate, F_{Org, Min} (liters/hr) if an infinite number of stages are available.

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 the analyses in both Chapters 13 and 14 are applicable. This problem explores this. We wish to use chloroform to extract acetone from water. Equilibrium data is given in Table 14-1. Find the number of equilibrium stages required for a countercurrent cascade if we have a feed of 1000.0 kg/hr of a 10.0 wt % acetone, 90.0 wt % water mixture. The solvent used is chloroform saturated with water (no acetone). Flow rate of stream E₀ = 1371 kg/hr. We desire an outlet raffinate concentration of 0.50 wt % acetone. Assume immiscibility and use a weight ratio units graphical analysis. Compare results with Problem 14.D5.

Note: Use the lowest acetone wt % in Table 14-1 to estimate the distribution coefficient for acetone. Then convert this equilibrium to weight ratios.

D20. The aqueous two-phase system in Example 13-2 will be used in a batch extraction. We have 5.0 kg of PEG solution containing protein at mass fraction x_F. We will use 4.0 kg of pure dextran solution to extract the protein from the solution. Equilibrium data are in Example 13-2.

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

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

D21. Check the solution to Example 13-3 with a McCabe-Thiele diagram.

D22.* You are working on a new glass factory near the ocean. The sand is to be mined wet from the beach. However, the wet sand carries with it seawater entrained between the sand grains. Several studies have shown that 40% by volume seawater is consistently carried with the sand. The seawater is 0.035 wt frac salt, which must be removed by a washing process.

Densities: Water, 1.0 g/cm₃ (assume constant); Dry sand, 1.8 g/cm₃ (including air in voids); Dry sand without air, 1.8/0.6 = 3.0 g/cm₃.

a. We desire a final wet sand product in which the entrained water has 0.002 wt frac salt. For each 1000 cm₃ of wet sand fed we will use 0.5 kg of pure wash water. In a countercurrent washing process, how many stages are required? What is the outlet concentration of the wash water?

b. In a cross-flow process we wish to use seven stages with 0.2 kg of pure wash water added to each stage for each 1000 cm₃ of wet sand fed. What is the outlet concentration of the water entrained with the sand?

D23. Your boss has looked at the results of Example 13-3. He now asks you what if a four-stage cross-flow system is used instead of the countercurrent system. The pure wash water will be divided equally between the four stages. Thus, overhead flow is 2 kg wash water/kg calcium carbonate on each stage. Determine the recovery of NaOH and compare to the result in Example 13-3.

D24.* We wish to wash an alumina solids to remove NaOH from the entrained liquid. The underflow from the settler tank is 20 vol % solid and 80 vol % liquid. The two solid feeds to the system are also 20 vol % solids. In one of these feeds, NaOH concentration in the liquid is 5 wt %. This feed’s solid flow rate (on a dry basis) is 1000 kg/hr. The second feed has a NaOH concentration in the liquid of 2 wt %, and its solids flow rate (on a dry basis) is 2000 kg/hr. We desire the final NaOH concentration in the underflow liquid to be 0.6 wt % (0.006 wt frac) NaOH. A countercurrent operation is used. The inlet washing water is pure and flows at 4000 kg/hr. Find the optimum feed location for the intermediate feed and the number of equilibrium stages required.

Data: ρ_w = 1.0 kg/liter (constant), ρ_alumina = 2.5 kg/liter (dry crushed)

D25. After reading the solution to Problem 13.D24, your boss thinks it will be just as good to take the two feeds and combine them rather than keeping the feeds separate. Calculate the number of equilibrium stages required to achieve the same outlet concentrations with the same flow rates if the two feeds are combined before being fed to the washing cascade. Compare with the answer to 13.D24.

D26. We wish to wash 1000.0 kg/hr of wet sand to remove salt from the entrained water. A counter-flow cascade with 5 equilibrium stages is used. The inlet wash water is pure. The inlet concentration of the salt in the entrained liquid is 0.028 wt frac. The outlet concentration of the salt in the entrained liquid (the underflow) should be 0.005 wt frac.

a. Find the ratio of U/O = (Underflow liquid rate)/(Overflow liquid rate) required.

b. Find the kg/hr of Underflow liquid.

Densities: Water = 1000 kg/ m₃(assume constant). Dry crushed sand = 2900 kg/m₃.

Void fraction = 0.38 (this is the porosity ρ).

D27.* You are working on a new glass factory near the ocean. The sand is to be mined wet from the beach. However, the wet sand carries with it seawater entrained between the sand grains. Data are given in Problem 13-D22. The salt must be removed by a washing process. A cross-flow process will be employed, with 0.2 kg of wash water added to each stage for each 1000 cm₃ of wet sand fed. The wash water outlet from the last stage will be used as the wash water inlet for stage 3. Wash water outlet from stage N–1 will be used as wash inlet for stage 2, and wash water outlet from stage N–2 as wash water inlet to stage 1. All other stages have pure wash water inlet (see figure). We desire an outlet concentration of less than 0.002 wt frac salt in the entrained liquid. What is the minimum number of stages required to obtain this concentration? (This is not a minimum wash water flow rate problem. This specification means you don’t have to obtain exactly 0.002 with an integer number of stages.) Note: This problem is not trial-and-error.

D28.* In the leaching of sugar from sugar cane, water is used as the solvent. Typically about 11 stages are used in a countercurrent Rotocel or other leaching system. On a volumetric basis liquid flow rate/solid flow rate = 0.95. The effective equilibrium constant is m_E = 1.18, where m_E = (concentration, g/liter, in liquid)/(concentration, g/liter, in solid) (Schwartzberg, 1980). If pure water is used as the inlet solvent, predict the recovery of sugar in the solvent.

D29. We plan to wash dilute sulfuric and hydrochloric acids from crushed rock in a counter current system. Operation is at 25°C and one atmosphere. 100.0 m₃/day of wet rock are to be washed. After settling, the porosity is constant at 0.40. Thus, the underflow rate is 40.0 m₃/day. The initial concentration of the underflow liquid is 1.0 kg sulfuric acid/m₃ and 0.75 kg hydrochloric acid/m₃. We desire an outlet underflow sulfuric acid concentration of 0.09 kg/m₃. The wash liquid (overflow) rate is 50.0 m₃/day, and the inlet wash water is pure. For these dilute solutions assume that the solution densities are constant, and are the same as pure water, 1000.0 kg/m₃. Find:

a. The number of equilibrium stages required

b. The outlet concentration (kg/m₃) of hydrochloric acid in the underflow liquid

D30.* The use of slurry adsorbents has received some industrial attention because it allows for countercurrent movement of the solid phase. Your manager wants you to design a slurry adsorbent system for removing methane from a hydrogen gas stream. The actual separation process is a complex combination of adsorption and absorption, but the total equilibrium can be represented by a simple equation. At 5° C, equilibrium can be represented as

Weight fraction CH₄ in gas = 1.2 × (weight fraction CH₄ in slurry)

At 5° C, no hydrogen could be detected in the slurry and the heat of sorption was negligible. We wish to separate a gas feed at 5° C that contains 100 lb/hr of hydrogen and 30 lb/hr of methane. An outlet gas concentration of 0.05 wt frac methane is desired. The entering slurry will contain no methane and flows at a rate of 120 lb/hr. Find the number of equilibrium stages required for this separation and the mass fraction methane leaving with the slurry.

D31. 2000.0 kg/hr of wet sand needs to be washed to remove the salt from the water entrained with the sand. The inlet concentration of the salt in the entrained water is 0.033 wt frac. We desire an outlet concentration of salt in the entrained underflow liquid of 0.004 wt frac. The inlet wash water (Overflow) is pure and its flow rate is the same as the Underflow liquid flow rate. Densities: Water = 1000 kg/m₃ (assume constant). Dry crushed sand = 3000 kg/m₃. Void fraction = 0.40.

Find the number of equilibrium stages required.

Find the kg/hr of Underflow liquid.

D32.* A countercurrent leaching system is recovering oil from soybeans. The system has five stages. On a volumetric basis, liquid flow rate/solids flow rate = 1.36. 97.5% of the oil entering with the nonsoluble solids is recovered with the solvent. Solvent used is pure. Determine the effective equlibrium constant, m_E, where m_E is (kg/m₃ of solute in solvent)/(kg/m₃ of solute in solid) and is given by the equation y = m_Ex.

D33. Batch leaching will be similar to a batch extraction, and the equations developed in section 13-6can be adapted when the solution is dilute or there is an insoluble solid matrix. We have 12.5 liters of pure water that we will use to leach 10.0 liter of wet sugar cane solids. Equilibrium data are in Problem 13.D28.

a. Find the fractional recovery of the sugar in the water if the water and wet sugar cane solids are mixed together and after settling, the water layer is removed.

b. Find the fractional recovery of the sugar in the water if a continuous solvent addition batch leaching system analogous to Figure 13-11 is used.

D34. Barium sulfide is produced by reacting barium sulfate ore with coal. The result is barium black ash, which is BaS plus insoluble solids. Since BaS is soluble in water, it can be leached out with water. In thickeners the insoluble solids in the underflow typically carry with them 1.5 kg liquid per kg insoluble solids. At equilibrium the overflow and underflow liquids have the same BaS concentrations (Treybal, 1980). We want to process 350 kg of insoluble solids plus its associated underflow liquid containing 0.20 mass fraction BaS. Use a countercurrent system with 2075 kg/hr of water as solvent. The entering water is pure. We desire the outlet underflow liquid to be 0.00001 mass fraction BaS. Find:

a. The BaS mass fraction in the exiting overflow liquid.

b. The number of equilibrium stages required.

E. More Complex Problems

E1.* We have a liquid feed that is 48 wt % m-xylene and 52 wt % o-xylene, which are to be 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 the recovery of m-xylene if necessary to solve this problem.

F. Problems Requiring Other Resources

The article by Lo (1997) is an excellent review of commercial liquid-liquid extraction equipment. Read it, and write a critique (maximum of two pages, typed, double-spaced).