Chapter 14: Extraction of Partially Miscible Systems

Liquid-liquid extraction (LLE) was introduced in Chapter 13, and equipment was briefly discussed there. All extraction systems are partially miscible to some extent. When partial miscibility is very low, as for toluene and water, we can treat the system as if it were completely immiscible and use McCabe-Thiele analysis or the Kremser equation shown in Chapter 13. When partial miscibility becomes appreciable, it can no longer be ignored, and a calculation procedure that allows for variable flow rates must be used. In this case a different type of stage-by-stage analysis, which is very convenient for ternary systems, can be used. For multicomponent systems, computer calculations are required.

Leaching, or solid-liquid extraction, was also introduced in Chapter 13. When a large part of the solid dissolves, the calculation procedures are essentially the same as for partially miscible LLE. These leaching calculations will be discussed at the end of the chapter.

14.1 EXTRACTION EQUILIBRIA

Extraction systems are noted for the wide variety of equilibrium behavior that can occur in them. In the partially miscible range utilized for extraction, two liquid phases will be formed. At equilibrium the temperatures and pressures of the two phases will be equal and the compositions of the two phases will be related. The number of independent variables that can be arbitrarily specified (i.e., the degrees of freedom) for a system at equilibrium can be determined from the Gibbs phase rule

(14-1)

which for a ternary extraction is F = 3 − 2 + 2 = 3 degrees of freedom. In an extraction, temperature and pressure are almost always constant so only one degree of freedom remains. Thus, if we specify the composition of one component in either phase, all other compositions will be set at equilibrium.

Extraction equilibrium data are easily shown graphically as either right triangular diagrams or equilateral triangular diagrams. Figure 14-1 shows the data listed in Table 14-1 for the system water-chloroform-acetone at 25°C on a right triangular diagram. We have chosen chloroform as solvent, water as diluent, and acetone as solute. We could also call water the solvent and chloroform the diluent if the feed was an acetone-chloroform mixture. Curved line AEBRD represents the solubility envelope for this system. Any point below this line represents a two-phase mixture that will separate at equilibrium into a saturated extract phase and a saturated raffinate phase. Line AEB is the saturated extract line, while line BRD is the saturated raffinate line. Point B is called the plait point where extract and raffinate phases are identical. Remember that the extract phase is the phase with the higher concentration of solvent. Tie line ER connects extract and raffinate phases that are in equilibrium.

Figure 14-1. Equilibrium for water-chloroform-acetone at 25°C and 1 atm

TABLE 14-1 Equilibrium data for the system water-chloroform-acetone at 1 atm and 25°C (Alders, 1959; Perry and Green, 1997, p. 2-33)

Point N in Figure 14-1 is a single phase because the ternary system is miscible at these concentrations. Point M represents a mixture of two phases, since it is in the immiscible range for this ternary system. At equilibrium the mixture represented by M will separate into a saturated raffinate phase and a saturated extract phase in equilibrium with each other. Either of the conjugate lines shown in Figure 14-1 can be used to draw tie lines. Consider tie line ER, which was found by drawing a horizontal line from point E to the conjugate line (point C) and then a vertical line from point C to the saturated raffinate curve (point R). Points E and R are in equilibrium, so they are on the ends of a tie line. This construction is shown in Figure 14-2 for different equilibrium data. This procedure is analogous to the use of an auxiliary line on an enthalpy-composition diagram as illustrated in Figure 2-5.

Figure 14-2. Construction of tie line using conjugate line

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

The solubility envelope, tie lines, and conjugate lines shown on the triangular diagrams are derived from experimental equilibrium data. To obtain these data a mixture can be made up and allowed to separate in a separatory funnel. Then the concentrations of extract and raffinate phases in equilibrium are measured. This measurement will give the location of one point on the saturated extract line, one point on the saturated raffinate line, and the tie line connecting these two points. One point on the conjugate line can be constructed from this tie line by reversing the procedure used to construct a tie line when the conjugate line was known (Figure 14-2).

The equilibrium data represented by Figure 14-1 are often called a type I system, since there is one pair of immiscible binary compounds. It is also possible to have systems with zero, two, and three immiscible binary pairs (Alders, 1959; Sorenson and Arlt, 1979, 1980; Macedo and Rasmussen, 1987; Walas, 1985). It is possible to go from a type I to a type II system as temperature decreases. This is shown in Figure 14-3 (Fenske et al., 1955) for the methylcyclohexane-toluene-ammonia system. At 10°F this is a type II system.

Figure 14-3. Effect of temperature on equilibrium of methylcyclohexane-toluene-ammonia system from Fenske et al., AIChE Journal, 1, 335 (1955), copyright 1955, AIChE

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

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

14.2 MIXING CALCULATIONS AND THE LEVER-ARM RULE

Triangular diagrams can be used for mixing calculations. In Figure 14-4A a simple mixing operation is shown, where streams F₁ and F₂ are mixed to form stream M. Streams F₁, F₂, and M can be either single-phase or two-phase. Operation of the mixer is assumed to be isothermal. For ternary systems there are three independent mass balances. With right triangular diagrams it is convenient to use the diluent balance, the solute balance, and the overall mass balance. The solvent mass balance will be automatically satisfied if the three independent balances are satisfied. The nomenclature is the same as in Chapter 13 in fraction units (see Table 13-2).

Figure 14-4. Mixing operation; A) equipment, B) triangular diagram

For the mixing operation in Figure 14-4A the flow rates F₁ and F₂ would be given as well as the concentration of the two feeds: x_A,F1, x_D,F1, x_A,F2, x_D,F2. The three independent mass balances used to solve for M, x_A,M and x_D,M are

(14-2a)

(14-2b)

(14-2c)

The concentrations of the mixed stream M are

(14-3a)

(14-3b)

We will now show that points F₁, F₂, and M are collinear as shown in Figure 14-4B. We will first use Eq. (14-2a) to remove the mixed stream flow rate M from Eqs. (14-2b) and (14-2c). Next, we solve the resulting equations for the ratio F₁/F₂ and then set these two equations equal to each other. The manipulations are as follows:

Then

(14-4a)

(14-4b)

Finally, setting these equations equal to each other and rearranging, we have

(14-5)

Equation (14-5), the three-point form of a straight line, states that the three points (x_A,M, x_D,M) (x_A,F2, x_D,F2) and (x_A,F1, x_D,F1) lie on a straight line. The manipulations used to derive Eq. (14-5) are very similar to those used to develop difference points for countercurrent calculations, and we will return to them shortly.

It will often prove convenient to be able to determine the location of the mixing point on the line between F₁ and F₂ without having to solve the mass balances analytically. This can be done using Eqs. (14-4a) or 14-4b), which relate the ratio of the feed rates to differences in the ordinate and abscissa, respectively. With F₁/F₂, x_A,F1, x_A,F2 known, Eq. (14-3a) can be used to find x_A,M. Equation (14-3b) can be used in a similar way to find x_D,M.

In Figure 14-5 similar triangles F₁AM and MBF₂ have been drawn. Since the triangles are similar,

Figure 14-5. Development of lever-arm rule with similar triangles

(14-6a)

Rearranging this formulation, we have

(14-6b)

where the bar denotes distance. According to Eq. (14-4b), the right-hand side of this equation is equal to F₁/F₂. Thus, we have shown that

(14-7)

Equation (14-7) is the lever-arm rule, which was first introduced in Figure 2-10. It may be helpful to review that material now. By measuring along the straight line between F₁ and F₂ we can find point M so that the lever-arm rule is satisfied. When you use the lever-arm rule, you don’t need the individual values of the flow rates F₁ and F₂ to find the location of M.

Using Eq. (14-7), point M might be found by trial-and-error. Since this is a cumbersome procedure, it is worthwhile to develop the lever-arm rule in a different form. These alternative forms are (see Problem 14-C1)

(14-8)

In this form the lever-arm rule is useful for finding the location of a stream M that is the sum of the two streams F₁ and F₂.

14.3 SINGLE-STAGE AND CROSS-FLOW SYSTEMS

Single-stage extraction systems can easily be solved with the tools we have developed. A batch extractor would consist of a single vessel equipped with a mixer. The two feeds would be charged to the vessel, mixed, and then allowed to settle into the two product phases. A continuous single-stage system requires a mixer and a settler as shown in Figure 13-2. Here the feed and solvent are fed continuously to the mixer, and the raffinate and extract products are continuously withdrawn from the settler. Figure 14-6 shows this schematically. The calculation procedures for batch and continuous operation are the same, the only real difference being that in batch operations S, F, M, E, and R are measured as total weight of material, whereas in continuous operation they are flow rates.

Figure 14-6. Continuous mixer-settler

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

The calculation method proceeds as follows. 1) Plot the locations of S and F on the triangular equilibrium diagram. 2) Draw a straight line between S and F, and use the lever-arm rule or Eq. (14-3) to find the location of the mixed stream M. Now we know that stream M settles into two phases in equilibrium with each other. Therefore, 3) construct a tie line through point M to find the compositions of the extract and raffinate streams. 4) Find the ratio E/R using mass balances. We will follow this method to solve the following example.

EXAMPLE 14-1. Single-stage extractionSingle-stage extraction

A solvent stream containing 10% by weight acetone and 90% by weight chloroform is used to extract acetone from a feed containing 55 wt % acetone and 5 wt % chloroform with the remainder being water. The feed rate is 250 kg/hr, while the solvent rate is 400 kg/hr. Operation is at 25°C and atmospheric pressure. Find the extract and raffinate compositions and flow rates when one equilibrium stage is used for the separation.

Solution

A. Define. The equipment sketch is the same as Figure 14-6 with S = 400, y_A,S = 0.1, y_S,S = 0.9, y_D,S = 0 and F = 250, x_A,F = 0.55, x_S,F = 0.05, x_D,F = 0.40. Find x_A,R, x_D,R, y_A,E, y_D,E, R, and E.

B. Explore. Equilibrium data are obviously required. They can be obtained from Table 14-1 and Figure 14-1.

C. Plan. Plot streams F and S. Find mixing point M from the lever-arm rule or from Eqs. (14-3). Then a tie line through M gives locations of streams E and R. Flow rates can be found from mass balances.

D. Do it. The graphical solution is shown in Figure 14-7. After locating streams F and S, M is on the line SF and can be found from the lever arm rule,

or from Eq. (14-3a),

A tie line through M is then constructed by trial-and-error, and the extract and raffinate locations are obtained. Concentrations are

The flow rates can be determined from the mass balances

M = E + R and Mx_A,M = Ey_A,E + Rx_A,R

Solving for R, we obtain

(14-9)

or

and E = M − R = 650 −125.36 = 524.64.

Figure 14-7 Solution for single-stage extraction, Example 14-1

The lever arm rule can also be used but tends to be slightly less accurate.

E. Check. We can check the solute or diluent mass balances. For example, the solute mass balance is

Sy_A,S + Fx_A,F = Ey_A,E + Rx_A,R

which is

(400)(0.1) + (250)(0.55) = (524.64)(0.30) + (125.36)(0.16)

or 177.5 ~ 177.45, which is well within the accuracy of the calculation. The diluent mass balance also checks.

F. Generalize. This procedure is similar to the one we used for binary flash distillation in Figure 2-9. Thus, there is an analogy between distillation calculations on enthalpy-composition diagrams (Ponchon-Savarit diagrams) and extraction calculations on triangular diagrams.

From this example it is evident that a single extraction stage is sufficient to remove a considerable amount of acetone from water. However, quite a bit of solvent was needed for this operation, the resulting extract phase is not very concentrated, and the raffinate phase is not as dilute as it could be.

The separation achieved with one equilibrium stage can easily be enhanced with a cross-flow system as shown in Figure 14-8. Assume that a cross-flow stage is added to the problem given in Example 14-1 and another 400 kg/hr of solvent (stream S₂ with 10% acetone, 90% chloroform) is used in stage 2. The concentrations of E₂ and R₂ are easily found by doing a second mixing calculation with streams S₂ and R₁. During this mixing calculation, R_j–1 (the feed to stage j) is different for each stage. A tie line through the new mixing point M₂ (Figure 14-8B) gives the location of streams E₂ and R₂. Note that x_A,R2 < x_A,R1 as desired.

Figure 14-8. Cross-flow extraction; A) cascade, B) solution of triangular diagram

In Chapter 13 we found that cross-flow systems are less efficient than countercurrent systems. In the next section the calculations for countercurrent cascades will be developed.

14.4 COUNTERCURRENT EXTRACTION CASCADES

14.4.1 External Mass Balances

A countercurrent cascade allows for more complete removal of the solute, and the solvent is reused so less is needed. A schematic diagram of a countercurrent cascade is shown in Figure 14-9. All calculations will assume that the column is isothermal and isobaric and is operating at steady state. In the usual design problem, the column temperature and pressure, the flow rates and compositions of streams F and S, and the desired composition (or percent removal) of solute in the raffinate product are specified. The designer is required to determine the number of equilibrium stages needed for the specified separation and the flow rates and compositions of the outlet raffinate and extract streams. Thus, the known variables are T, p, R_N+1, E₀, x_A,N+1, y_A,0, y_D,0, and x_A,1, and the unknown quantities are E_N, R₁, x_D,1, y_A,N, y_D,N, and N.

Figure 14-9. Countercurrent extraction cascade

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

(14-10a)

(14-10b)

(14-10c)

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

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

The procedure can also be carried out conveniently on a triangular diagram. Let us represent the cascade shown in Figure 14-9 as a mixing tank followed by a black box separation scheme that produces the desired extract and raffinate as shown in Figure 14-10A. In Figure 14-10A, streams E_N and R₁ are not in equilibrium as they were in Figure 14-6, but stream E_N is a saturated extract and stream R₁ is a saturated raffinate. The external mass balances for Figure 14-40A are

(14-11a)

(14-11b)

(14-11b)

Figure 14-10. External mass-balance calculation; A) mixer-separation representation, B) solution on triangular diagram

The coordinates of point M can be found from Eqs. (14-11):

(14-12a)

(14-12b)

Since Eq. (14-11) are the same type of mass balances as for a mixer, the points representing streams E_N, R₁, and M lie on a straight line given by

(14-13)

and the flow rates are related to the length of line segments by the lever-arm rule (see Problem 14-C2). We also know that E_N must lie on the saturated extract line and R₁ must lie on the saturated raffinate line. Since x_A,1 is known, the location of R₁ on the saturated raffinate line can be found. A straight line from R₁ extended through M (found from x_A,M and x_D,M or from the lever-arm rule) will intersect the saturated extract stream at the value of E_N. This construction is illustrated in Figure 14-10B. Note that this procedure is very similar to the one used for single equilibrium stages, but the line R₁ME_N is not a tie line. Mass balances can then be used to solve for the flow rates E_N and R₁.

The external mass balance and the equilibrium diagram in Figure 14-10B can be used to determine the effect of variation in the feed or solvent concentrations, the raffinate concentration, or the ratio F/S on the resulting separation. For example, if the amount of solvent is increased, the ratio of F/S will decrease. The mixing point M will move toward point S, and the resulting extract will contain less solute.

14.4.2 Difference Points and Stage-by-Stage Calculations

To determine the number of stages or flow rates and compositions inside the cascade, stage-by-stage calculations are needed. But first we use the external mass balances to find concentrations y_A,N and y_D,N and flow rates E_N and R₁. Starting at stage 1 (Figure 14-9) we note that streams R₁ and E₁ both leave equilibrium stage 1. Therefore, these two streams are in equilibrium and the concentration of stream E₁ can be found from an equilibrium tie line.

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

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

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

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

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

Referring to Figure 14-9, we can do a mass balance around the first stage. After rearrangement, this is

E₀ − R₁ = E₁ − R₂

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

(14-14a)

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

(14-14b)

and for diluent D,

(14-14c)

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

Equations (14-14) define a difference or Δ (delta) point. The coordinates of this point are easily found from Eqs. (14-14b) and (14-14c).

(14-15a)

(14-15b)

where Δ is given by Eq. (14-14a). x_A,Δ and x_D,Δ are the coordinates of the difference point and are not compositions that occur in the column. Note that x_A,Δ and x_D,Δ can be negative.

The difference point can be treated as a stream for mixing calculations. Thus, Eqs. (14-14a), (14-14b), (14-14c) show that the following points are collinear.

Δ(x_A,Δ, x_D,Δ), E₀(y_A,0, y_D,0), R₁(x_A,1, x_D,1)

Δ(x_A,Δ, x_D,Δ), E₁(y_A,1, y_D,1), R₂(x_A,2, x_D,2)

Δ(x_A,Δ, x_D,Δ), E_j(y_A,j, y_D,j), R_j+1(x_A,j+1, x_D,j+1)

Δ(x_A,Δ, x_D,Δ), E_N(y_A,N, y_D,N), R_N+1(x_A,N+1, x_D,N+1)

The existence of these straight lines and the applicability of the lever-arm rule can be proved by deriving Eq. (14-16) (see Problem 14-C3).

(14-16)

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

There are three methods for finding the location of Δ:

1. Graphical construction. Since the points Δ, E₀, and R₁; and Δ, E_N, and R_N+1 are on straight lines, we can draw these two straight lines. The point of intersection must be the Δ point. For the typical design problem (see Figure 14-9), points R_N+1, E₀, and R₁ are easily plotted. E_N can be found from the external balances (Figure 14-10B). Then Δ is found as shown in Figure 14-11.

Figure 14-11. Location of difference point for typical design problem

2. Coordinates. The coordinates of the difference point were found in Eq. (14-15). These coordinates can be used to find the location of Δ. It may be convenient to draw one of the straight lines in method 1 (such as line ) and use one of the coordinates (such as x_D,Δ) to find Δ. This procedure is useful since accurate graphical determination of Δ can be difficult.

3. Lever-arm rule. The general form of the lever-arm rule for two passing streams is (see Problem 14-C3)

(14-17)

The lever-arm rule can be used to find the Δ point. For instance, if flow rates R₁ and E₀ are known, then Δ can be found on the straight line through points R₁ and E₀ at a distance that satisfies Eq. (14-17) with j = 0.

Consider again the stage-by-stage calculation routine that was outlined previously. We start with the known concentration of raffinate product stream R₁ and use an equilibrium tie line (stage 1) to find the location of saturated extract stream E₁ (see Figure 14-12). The points representing Δ, E₁, and R₂ are collinear, since E₁ and R₂ are passing streams. If the location of Δ is known, the straight line from Δ to E₁ can be drawn and then be extended to the saturated raffinate curve. This has to be the location of stream R₂ (see Figure 14-12). Thus, the difference point allows us to solve the three simultaneous mass balances by drawing a single straight line—the operating line. The procedure may be continued by constructing a tie line (representing stage 2) to find the location of stream E₂ that is in equilibrium with stream R₂. Then the mass balances are again solved simultaneously by drawing a straight line from Δ through E₂ to the saturated raffinate line, which locates R₃. This process of alternating between equilibrium and mass balances is continued until the desired separation is achieved. The stages are counted along the tie lines, which represent extract and raffinate streams in equilibrium. To obtain an accurate solution, a large piece of graph paper is needed and care must be exercised in constructing the diagram.

Figure 14-12. Stage-by-stage solution

Before continuing you should carefully reread the preceding paragraph; it contains the essence of the stage-by-stage calculation method.

In Figure 14-12 we see that two equilibrium stages do not quite provide sufficient separation, and three equilibrium stages provide more separation than is needed. In a case like this, an approximate fractional number of stages can be reported.

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

If a mixer-settler system were used, where each mixer-settler combination is approximately an equilibrium stage, then we would have three choices: 1) Use three stages, and obtain more separation than desired; 2) use two stages, and obtain less separation than desired; or 3) change the feed-to-solvent ratio to obtain the desired separation with exactly two or exactly three equilibrium stages.

One further important point should be stressed with respect to Figure 14-12. If two equilibrium stages were used with F/S = 2 as in the original problem statement, the saturated extract would not be located at the value E₂ shown on the graph and saturated raffinate would not be at the value R₁ shown. The streams R₁ and E₂ do not satisfy the external mass balance for this system. The values R₁ and E_N do, but R₁ and E₂ do not. The exact compositions of the product streams for a two-stage system require a trial-and-error solution.

What do we do if flow rate E₀ is less than R₁? The easiest solution is to define Δ so that it is now positive.

(14-18)

Δ is still equal to the difference between the flow rates of any pair of passing streams, but it is now raffinate minus extract. The corresponding lever-arm rule for any pair of passing streams is still Eq. (14-17), but the Δ point will be on the opposite side of the triangular diagram. This situation is shown in Figure 14-13. The stage-by-stage calculation procedure is unchanged when the location of Δ is on the right side of the diagram.

Figure 14-13. Location of difference point when R₁ > E₀

14.3.3 Complete Extraction Problem

At this point you should be ready to solve a complete extraction problem.

EXAMPLE 14-2. Countercurrent extraction

A solution of acetic acid (A) in water (D) is to be extracted using isopropyl ether as the solvent (S). The feed is 1,000 kg/hr of a solution containing 35 wt % acid and 65 wt % water. The solvent used comes from a solvent recovery plant and is essentially pure isopropyl ether. Inlet solvent flow rate is 1475 kg/hr. The exiting raffinate stream should contain 10 wt % acetic acid. Operation is at 20°C and 1 atm. Find the outlet concentrations and the number of equilibrium stages required for this separation. The equilibrium data are given by Treybal (1968) and are reproduced in Table 14-2.

Solution

A. Define. The extraction will be a countercurrent system as shown in Figure 14-9. F = 1,000, x_A,F = 0.35, x_D,F = 0.65, S = 1475, y_A,S = 0, y_D,S = 0, x_A,1 = 0.1. Find x_D,1, y_A,N, y_D,N, and N.

B, C. Explore and plan. This looks like a straightforward design problem. Use the method illustrated in Figures 14-11 and 14-13 to find Δ. Then step off stages as illustrated in Figure 14-12.

D. Do it. The solution is shown in Figure 14-14.

Figure 14-14. Solution to Example 14-3

1. Plot equilibrium data and construct conjugate line.

2. Plot locations of streams E₀ = S, R_N+1 = F, and R₁.

3. Find mixing point M on line through points S and F at x_A,M value calculated from Eq. (14-12).

(14-14)

4. Line R₁M gives point E_N.

5. Find Δ point as intersection of straight lines E₀R₁ and E_NR_N+1

6. Step off stages, using the procedure shown in Figure 14-12. To keep the diagram less crowded, the operating lines, ΔE_j R_j+1, are not shown. You can use a straight edge on Figure 14-14 to check the operating lines. A total of 5.8 stages are required.

E. Check. Small errors in plotting the data or in drawing the operating and tie lines can cause fairly large errors in the number of stages required. If greater accuracy is required, a much larger scale and more finely divided graph paper can be used, or a McCabe-Thiele diagram (see the next section) or computer methods can be used.

14.5 RELATIONSHIP BETWEEN MCCABE-THIELE AND TRIANGULAR DIAGRAMS

Stepping off a lot of stages on a triangular diagram can be difficult and inaccurate. More accurate calculations can be done with a McCabe-Thiele diagram. Since total flow rates are not constant, the triangular diagram and the Δ point are used to plot a curved operating line on the McCabe-Thiele diagram. This construction is illustrated in Figure 14-15 for a single point. For any arbitrary operating line (which must go through Δ), the values of the extract and raffinate concentrations of passing streams (y_A,op, x_A,op) are easily determined. These concentrations must represent a point on the operating line in the McCabe-Thiele diagram. Thus, the values of y_A,op and x_A,op are transferred to the diagram. Since the raffinate value is an x, the y = x line is used to find x. A very similar procedure was used in Figure 2-6B to relate McCabe-Thiele to enthalpy-composition diagrams.

Figure 14-15. Use of triangular diagram to plot operating line on McCabe-Thiele diagram

When this construction is repeated for a number of arbitrary operating lines, a curved operating line is generated on the McCabe-Thiele diagram. The equilibrium data, y_A vs x_A, can also be plotted. Then stages can be stepped off on the diagram. This is shown in Figure 14-16 for Example 14-2. The equilibrium data were obtained from Table 14-2. The Δ point on the triangular diagram was used to find the operating line. The answer is 6 1/6 stages, which is reasonably close to the 5.8 found in Example 14-2.

Figure 14-16. Use of triangular and McCabe-Thiele diagrams to solve Example 14-2

TABLE 14-2. Equilibrium data for water-acetic acid-isopropyl ether at 20°C and 1 atm

Note that the operating line is close to straight. Thus, R_j+1/E_j is approximately constant. This occurs when the change in solubility of the solvent in the raffinate streams is approximately the same as the change in solubility of the diluent in the extract streams. When the changes in partial miscibility of the extract and raffinate phases are very unequal, R_j+1/E_j will vary significantly and the operating line will show more curvature. When the feed is not presaturated with solvent and the entering solvent is not presaturated with diluent, there can be a large change in flow rates on stages 1 and N. This is not evident in Figure 14-16, since the extract and raffinate phases are close to immiscible for the ranges of concentration shown.

The McCabe-Thiele diagrams are useful for more complicated extraction columns such as those with two feeds or extract reflux (Wankat, 1982, 1987).

14.6 MINIMUM SOLVENT RATE

As the solvent rate is increased, the separation should become easier, and the outlet extract stream, E_N, should become more diluted. The effect of increasing S/F is shown in Figure 14-17. As S/F increases, the mixing point moves toward the solvent and the solute concentration in stream E_N decreases. The difference point starts on the right-hand side of the diagram (R_j+1 > E_j) and moves away from the diagram. When R_j+1 = E_j, Δ is at infinity. A further increase in S/F puts Δ on the left-hand side of the diagram (R_j+1 < E_j). It now moves toward the diagram as S/F continues to increase. Some of these S/F ratios will be too low, and even a column with an infinite number of stages will not be able to do the desired separation.

Figure 14-17. Effect of increasing S/F; (S/F)₄ > (S/F)₃ > (S/F)₂ > (S/F)₁

It is often of interest to calculate the minimum amount of solvent that can be used and still obtain the desired separation. The minimum solvent rate (or minimum S/F) is the rate at which the desired separation can be achieved with an infinite number of stages. If less solvent is used, the desired separation is impossible; while if more solvent is used, the separation can be achieved with a finite number of stages. The corresponding Δ value, Δ_min, thus, represents the dividing point between impossible cases and possible solutions. This situation is analogous to minimum reflux in distillation.

To determine Δ_min and hence (S/F)_min, we note that to have an infinite number of stages, tie lines and operating lines must coincide (be parallel) somewhere on the diagram. This will require an infinite number of stages. The construction to determine Δ_min is shown in Figure 14-18 for Example 14-2, and is outlined here.

Figure 14-18. Determination of minimum solvent rate

1. Draw and extend line R₁S.

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

3. Δ_min is located at the point of intersection of a tie line that is closest to the diagram on the left-hand side or furthest from the diagram if on the right-hand side. Thus, Δ_min is at the largest S/F that requires an infinite number of stages. Often the tie line that when extended goes through the feed point is the desired tie line. This is not the case in Figure 14-18.

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

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

6. From the lever-arm rule, . In Figure 14-18, (S/F)_min = 1.296 and S_min = 1296 kg/hr. The actual solvent rate for Example 14-2 is 1475 kg/hr, so the ratio S/S_min = 1.138. Use of more solvent in Figures 14-14 or 14-16 would decrease the required number of stages.

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

14.7 EXTRACTION COMPUTER SIMULATIONS

Partially miscible ternary and multi-solute extraction systems can be set up for computer calculations. If we redraw and renumber the countercurrent extraction column (Figure 14-9) as shown in Figure 14-19 and relate R to L and E to S, the extraction column is analogous to a stripping or absorption column. The mass balances will be identical to those for absorption and stripping [Eqs. (12-45) to (12-48)] and they can be arranged into a tridiagonal matrix, Eq. (6-13). The new flow rates L_j (R_j) and V_j (E_j) can be determined from the summation Eqs. (12-49) and (12-50) using convergence check Eq. (12-51). The energy balances can again be written as Eq. (12-52) and the multivariate Newtonian solution method shown in Eqs. (12-53) to (12-58) is again applicable. Conversion of these equations for a specific problem is explored in problem 14.D11.

Figure 14-19. Extraction column numbered for matrix calculations

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

Computer calculations for extraction can also be applied to more complex extraction systems such as two-feed extractors or extraction with extract reflux. The triangular diagrams for these systems are considered in detail by Wankat (1982, 1987).

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

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

Application of Aspen Plus for extraction is delineated in the appendix to this chapter.

14.8 LEACHING WITH VARIABLE FLOW RATES

Leaching, also known as solid-liquid extraction (SLE), and LLE will have very different hydrodynamic and mass transfer characteristics. However, equilibrium staged analysis is almost identical for the two processes because it is not affected by these differences. In this section we will briefly consider the analysis of leaching systems using triangular diagrams. In leaching the flow rates will generally not be constant. The variation in flow rates can be included by doing the analysis on a triangular diagram. A very similar technique uses ratio units in a Ponchon-Savarit diagram (Prabhudesai, 1997; Lydersen, 1983; McCabe et al., 2005; Miller, 1997; Treybal, 1980).

A schematic of a countercurrent leaching system is shown in Figure 14-20. Since leaching is quite similar to LLE, the same nomenclature is used (see Table 13-2). Even if flow rates E and R vary, it is easy to show that the differences in total and component flow rates for passing streams are constant. Thus, we can define the difference point from these differences. This was done in Eqs. (14-14) and (14-15) for the LLE cascade of Figure 14-9. Since the cascades are the same (compare Figures 14-9 and 14-20), the results for leaching are the same as for LLE.

Figure 14-20. Countercurrent leaching cascade and nomenclature

The calculation procedure for countercurrent leaching operations is exactly the same as for LLE:

1. Plot the equilibrium data.

2. Plot the locations of known points.

3. Find mixing point M.

4. Locate E_N.

5. Find the Δ point.

6. Step off stages.

This procedure is illustrated in Example 14-3.

The equilibrium data for leaching must be obtained experimentally since it will depend on the exact nature of the solids, which may change from source to source. If there is no entrainment, the overflow (extract) stream will often contain no inert solids (diluent). However, the raffinate stream will contain solvent. Test data for the extraction of oil from meal with benzene are given in Table 14-3 (Prabhudesai, 1997). In these data, inert solids are not extracted into the benzene. The data are plotted on a triangular diagram in Figure 14-3 (see Example 14-3). The conjugate line is constructed in the same way as for extraction.

TBALE 14-3. Test data for extraction of oil from meal with benzene

EXAMPLE 14-3. Leaching calculations

We wish to treat 1,000 kg/hr (wet basis) of meal (D) that contains 0.20 wt frac oil (A) and no benzene(S). The inlet solvent is pure benzene and flows at 662 kg/hr. We desire an underflow product that is 0.04 wt frac oil. Temperature and pressure are constant, and the equilibrium data are given in Table 14-3. Find the outlet extract concentration and the number of equilibrium stages needed in a countercurrent leaching system.

Figure 14-21. Solution to leaching problem, Example 14-3

Solution

A. Define. The system is similar to that of Figure 14-20 with streams E₀ and R_N+1 specified. In addition, weight fraction x_A,1 = 0.04 and is a saturated raffinate. We wish to find the composition of stream E_N and the number of equilibrium stages required.

B. Explore. This looks like a straightforward leaching problem, which can be solved like the corresponding extraction problem.

C. Plan. We will plot the equilibrium diagram on a scale that allows the Δ point to fit on the graph. Then we will plot points E₀, R_N+1, and R₁. We will use the lever-arm rule to find point M and then find point E_N. Next we find Δ and finally we step off the stages.

D. Do it. The diagram in Figure 14-21 shows the equilibrium data and the points that have been plotted. Point M was found along the line E₀R_N+1 from Eq. (14-12a).

Then points E_N and Δ were found as shown in Figure 14-21. Finally, stages were stepped off in exactly the same way as for a triangular diagram for extraction. The exit extract concentration is 0.305 wt frac oil, and 3 stages are more than enough. Two stages are not quite enough. Approximately 2.1 stages are needed.

E. Check. The outlet extract concentration can be checked with an overall mass balance. The number of stages could be checked by solving the problem using another method.

F. Generalize. Since the leaching example was quite similar to LLE, we might guess that the other calculation procedures developed for extraction would also be valid for leaching. Since this is true, there is little reason to reinvent the wheel and rederive all the methods. Some of these applications are explored in Problems 14.C4,14. D18, 14. D19, and 14. D20.

McCabe-Thiele diagrams (Chapter 13) can also be used for leaching if flow rates are close to constant. If flow rates are not constant, curved operating lines can be constructed on the McCabe-Thiele diagram with the triangular diagram.

14.9 SUMMARY—OBJECTIVES

In this chapter, methods for ternary partially miscible extraction systems are explored. At the end of this chapter, you should be able to satisfy the following objectives:

1. Plot extraction equilibria on a triangular diagram. Find the saturated extract, saturated raffinate, and conjugate lines

2. Find the mixing point and solve single-stage and cross-flow extraction problems

3. For countercurrent systems, do the external mass balances, find the difference points, and step off the equilibrium stages

4. Use the difference points to plot the operating line(s) on a McCabe-Thiele diagram

5. Apply the methods to leaching problems

6. Use a computer simulation program to solve extraction problems

REFERENCES

Alders, L. Liquid-Liquid Extraction, 2nd ed., Elsevier, Amsterdam, 1959.

Brown, G. G. and Associates, Unit Operations, Wiley, New York, 1950.

Fenske, M.R., R.H. McCormick, H. Lawroski, and R.G. Geier, “Extraction of Petroleum Fractions by Ammonia Solvents,” AIChE J., 1, 335 (1955).

King, C.J., Separation Processes, 2nd ed., McGraw-Hill, New York, 1981.

Laddha, G.S. and T.E. Degaleesan, Transport Phenomena in Liquid Extraction, McGraw-Hill, New York, 1978.

Lyderson, A. L., Mass Transfer in Engineering Practice, Chichester, UK, Wiley, 1983.

Macedo, M. E. A. and P. Rasmussen, Liquid-Liquid Equilibrium Data Collection, Supplement 1, vol. V, part 4, 1987, DECHEMA, Frankfurt (Main), Germany.

McCabe, W. L., J. C. Smith and P. Harriott, Unit Operations of Chemical Engineering, 7^th ed., McGraw-Hill, New York, 2005.

Miller, S. A., “Leaching,” in Perry, R. H. and D. W. Green, Eds., Perry’s Chemical Engineers’ Handbook, 7^th edition, McGraw-Hill, New York, pp. 18-55 to 18-59, 1997.

Perry, R.H. and D. Green (Eds.), Perry’s Chemical Engineer’s Handbook, 6th ed., McGraw-Hill, New York, 1984.

Prabhudesai, R.K., “Leaching,” in P.A. Schweitzer (Ed.), Handbook of Separation Techniques for Chemical Engineers, McGraw-Hill, New York, 1997, Section 5.1.

Smith, B.D., Design of Equilibrium Stage Processes, McGraw-Hill, New York, 1963.

Sorenson, J. M. and W. Arlt, Liquid-Liquid Equilibrium Data Collection, Binary Systems, vol. V, part 1, 1979, Ternary Systems, vol. V, part 2, 1980, Ternary and Quaternary Systems, vol. 5, part 3, 1980, DECHEMA, Frankfurt (Main), Germany.

Treybal, R.E., Mass Transfer Operations, 3rd ed., McGraw-Hill, New York, 1980.

Varteressian, K.A. and M.R. Fenske, “The System Methylcyclohexane - Aniline - N-Heptane,” Ind. Eng. Chem., 29, 270 (1937).

Walas, S. M., Phase Equilibria in Chemical Engineering, Butterworths, Boston, 1985.

Wankat, P. C., Equilibrium-Staged Separations, Prentice Hall, Upper Saddle River, New Jersey, Chapter 18, 1987.

Wankat, P.C., “Advanced Graphical Extraction Calculations,” in J.M. Calo and E.J. Henley (Eds.), Stagewise and Mass Transfer Operations, Vol. 3, Extraction and Leaching, AIChEMI, Series B, AICHE, New York, 1982. p. 17.

HOMEWORK

A. Discussion Problems

A1. Write your key relations chart for this chapter.

A2. Construct a conjugate line on Figure 14-3 at 10°C.

A3. When E₀ > R₁ the difference point is on the left side of the diagram; and when R₁ > E₀ the difference point is on the right of the diagram. What happens when R₁ = E₀? Answer this question using a logical argument.

A4. For Figure 14-12 suppose we desired to obtain the desired raffinate concentration with exactly two equilibrium stages. This can be accomplished by changing the amount of solvent used. Would we want to increase or decrease the amount of solvent? Explain the effect this change will have on M, E_N, Δ, and the number of stages required.

A5. You have the mixer in Figure 14-4A and know F₁ and M and the values x_A,F1, x_D,F1, x_A,M, and x_D,M. Explain how you would use the lever-arm rule and a triangular diagram to find x_A,F2 and x_D,F2.

A6. All the calculations in this chapter use right triangle diagrams. What differences would result if equilateral triangle diagrams were used to solve the extraction problem?

A7. How will the diagram change if we switch axes and plot weight fraction solvent vs. weight fraction diluent?

A8. If the extract and raffinate phases are totally immiscible, extraction problems can still be solved using triangular diagrams. Explain how and describe what the equilibrium diagram will look like.

A9. What can be done to increase the removal of solute in a countercurrent extractor (i.e., decrease x_A,1)?

A10. What can be done to increase the concentration of solute in the extract (i.e., increase y_A,N)?

A11. What situation in countercurrent extraction is analogous to minimum reflux in distillation?

A12. What situation in countercurrent extraction is superficially analogous to total reflux in distillation? How does it differ?

A13. Study Figure 14-17. Explain what happens as S/F increases. What happens to M? What happens to E_N? What happens to Δ? How do you find Δ_min if it lies on the left-hand side? How do you find Δ_min if it lies on the right-hand side?

A14. Three analysis procedures were developed for extraction in Chapters 13 and 14: McCabe-Thiele, Kremser, and triangular diagrams. If you have an extraction problem, how do you decide which method to use? (In other words, explain when each is applicable.)

A15. Would you expect stage efficiencies to be higher or lower in leaching than in LLE? Explain.

C. Derivations

C1. Derive Eqs. (14-8).

C2. Prove that the locations of streams M, E_N, and R₁ in Figure 14-10A lie on a straight line as shown in Figure 14-10B. In other words, derive Eq. (14-13).

C3. Define Δ and the coordinates of Δ from Eqs. (14-14) and (14-15). Prove that points Δ, E_j, and R_j+1 (passing streams) lie on a straight line by developing Eq. (14-16). While doing this, prove that the lever-arm rule is valid.

C4. Develop the procedures for single-stage and cross-flow systems for leaching.

D. Problems

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

D1.* The equilibrium data for extraction of methylcyclohexane (A) from n-heptane (D) into aniline (S) is given in Table 14-4. We have 100 kg/hr of a feed that is 60% methylcyclohexane and 40% n-heptane and 50 kg/hr of a feed that is 20% methylcyclohexane and 80% n-heptane. These two feeds are mixed with 200 kg/hr of pure aniline in a single equilibrium stage.

TABLE 14-4. Equilibrium data for methylcyclohexane, n-heptane-aniline

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

b. What is the flow rate of the extract product?

D2. We contact 100 kg/hr of a feed mixture that is 50 wt % methylcyclohexane and 50 wt % n-heptane with a solvent stream (15 wt % methylcyclohexane and 85 wt % aniline) in a mixer-settler that acts as a single equilibrium stage. We obtain an extract phase which is 10 wt % methylcyclohexane and a raffinate phase which is 61 wt % methylcyclohexane. Equilibrium data are in Table 14-4. What flow rate of the solvent stream was used?

D3. 100.0 kg/hr of a feed which is 10 wt % acetic acid and 90 wt % water is contacted with pure isopropyl ether solvent in a countercurrent extractor. Solvent flow rate is 500 kg/hr. The outlet raffinate should be 2 wt % acetic acid. Equilibrium data are in Table 14-2. Find.

a. Mixing point M and outlet extract point E_N.

b. Δ point.

c. Number of equilibrium stages required.

D4. * The equilibrium data for the system water-acetic acid-isopropyl ether are given in Table 14-2. We have a feed that is 30 wt % acetic acid and 70 wt % water. The feed is to be treated with pure isopropyl ether. A batch extraction with a single mixer-settler will process 15 kg of feed.

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

b. If a raffinate composition of x_A = 0.1 is desired, how much solvent is needed?

D5. 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 1,000 kg/hr of a 10 wt % acetone, 90 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. Compare results to problem 13.D19.

D6. * A countercurrent system with three equilibrium stages is to be used for water-acetic acid-isopropyl ether extraction (see Table 14-2). Feed is 40 wt % acetic acid and 60 wt % water. Feed flow rate is 2000 kg/hr. Solvent added contains 1 wt % acetic acid but no water. We desire a raffinate that is 5 wt % acetic acid. What solvent flow rate is required? What are the flow rates of E_N and R₁?

D7. We have 500 kg/hr of a feed that is 30 wt % pyridine and 70 wt % water. This will be extracted with pure chlorobenzene. Operation is at 1 atm and 25°C. Equilibrium data are in Table 14-5.

a. A single mixer-settler is used. Chlorobenzene flow rate is 300 kg/hr. Find the outlet extract and raffinate flow rates and weight fractions.

b. We now want to convert the single-stage system in part a into a two-stage cross flow system. The raffinate from part a is fed to the second mixer-settler and is contacted with an additional 300 kg/hr of pure chlorobenzene. Find the raffinate and extract flow rates and weight fractions leaving the second stage.

D8. We will feed 500 kg/hr of a feed that is 30 wt % pyridine and 70 wt % water to a counter current extractor. The feed is contacted with pure chlorobenzene. Solvent flow rate is 600 kg/hr. The desired outlet raffinate wt. fraction is 4 wt % pyridine. Find the outlet wt frac of the exiting extract stream, the flow rates of the exiting extract and raffinate streams, and the number of equilibrium contacts needed. Equilibrium data is in Table 14-5.

D9. * We wish to remove acetic acid from water using isopropyl ether as solvent. The operation is at 20° C and 1 atm (see Table 14-2). The feed is 0.45 wt frac acetic acid and 0.55 wt frac water. Feed flow rate is 2000 kg/hr. A countercurrent system is used. Pure solvent is used. We desire an extract stream that is 0.20 wt frac acetic acid and a raffinate that is 0.20 wt frac acetic acid.

TABLE 14-5. Equilibrium data for pyridine, water, chlorobenzene at 25°C and 1 atm. (Treybal, 1980)

a. How much solvent is required?

b. How many equilibrium stages are needed?

D10. We plan to remove methylcyclohexane from n-heptane using aniline as the solvent (see Table 14-4 for equilibrium data). The feed is 200 kg/hr total. It is 40 wt % methylcyclohexane and 55 wt % n-heptane. The solvent is 95 wt % aniline and 5 wt % n-heptane.

a. With a single mixer-settler, set the total entering flow rate of solvent stream as 600 kg/hr. Find the outlet wt fracs and the outlet flow rates.

b. Suppose we decide to use a two stage cross flow system with 300 kg/hr of the solvent stream entering each stage. Find the outlet flow rates and wt fracs.

Note that the answers are very sensitive to how one draws the tie lines. This is typical of type II systems.

D11. For Example 14-2 set up the tridiagonal matrix for the mass balances assuming there are 6 stages in the column and the exit raffinate stream concentration is unknown. Develop the values for A, B, C, and D in only the acetic acid matrix using K values determined from Table 14-2. Do not invert the matrix.

D12. We have a total feed of 100 kg/hr to a countercurrent extraction column. The feed is 40 wt % acetic acid and 60 wt % water. We plan to extract the acetic acid with isopropyl ether. Equilibrium data are in Table 14-2. The entering isopropyl ether is pure and has a flow rate of 111.2 kg/hr. We desire a raffinate that is 20 wt % acetic acid.

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

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

c. Find the number of equilibrium contacts.

D13. We are extracting acetic acid from water using isopropyl ether as the solvent at 20°C and 1 atm (see Table 14-2 for equilibrium data). The feed is 20 wt % acetic acid and 80 wt % water. The total flow rate of the feed is 1 kg/hr. We use pure isopropyl ether as the solvent. The extraction is done in a special laboratory system that has 200 equilibrium contacts. The outlet raffinate wt frac of acetic acid is measured as 0.02. What solvent flow rate is used?

D14. * We are extracting acetic acid from water with isopropyl ether at 20°C and 1 atm pressure. Equilibrium data are in Table 14-2. The column has three equilibrium stages. The entering feed rate is 1,000 kg/hr. The feed is 40 wt % acetic acid and 60 wt % water. The exiting extract stream has a flow rate of 2500 kg/hr and is 20 wt % acetic acid. The entering extract stream (which is not pure isopropyl ether) contains no water. Find:

a. The exit raffinate concentration.

b. The required entering extract stream concentration.

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

Trial-and-error is not needed.

D15. The equilibrium data for extraction of acetic acid (A) from water (D) into isopropyl ether (S) is given in Table 14-2. We have 1,000 kg/hr of a feed that is 40 % A and 60 % D. This feed is extracted with a total of 1,000 kg/hr of pure isopropyl ether in a two-stage system.

a. If a cross-flow system is used with 500 kg/hr of isopropyl ether into each stage, find the extract compositions, the final raffinate composition and the raffinate flow rate leaving the system.

b. If a two-stage counter current system is used with a solvent flow rate of 1,000 kg/hr, find the extract composition, the raffinate composition and the raffinate flow rate. If the difference point is off the page, tape another sheet of paper to the first sheet.

c. Compare the cross-flow and countercurrent systems.

D16. We are recovering pyridine from water using chlorobenzene as the solvent in a countercurrent extractor. The feed is 35 wt % pyridine and 65 wt % water. Feed flow rate is 1,000 kg/hr. The solvent used is pure. The desired outlet extract is 20 wt % pyridine, and the desired outlet raffinate is 4 wt % pyridine. Operation is at 25°C and 1 atm. Equilibrium data are in Table 14-5.

a. Find the number of equilibrium stages needed.

b. Determine the solvent flow rate required in kg/hr.

D17. We are extracting acetic acid from water using isopropyl ether in a counter-current cascade at 20°C and 1 atm. Equilibrium data is in Table 14-2. The entering extract stream is pure isopropyl ether. The exiting raffinate stream is 10 wt % acetic acid and the exiting raffinate flow rate is R₁=100 kg/hr. The exiting extract stream is 9 wt % acetic acid at a flow rate of E_N=500 kg/hr. The entering feed contains no isopropyl ether. NOTE: This problem is not trial-and-error.

a. Find the feed composition x_Afeed, x_wfeed.

b. Find the flow rates of feed, F, and of entering extract stream E₀.

c. Find the number of equilibrium stages needed.

D18. * Repeat Example 14-3 except for a single-stage system and unknown underflow product concentration.

D19. * Repeat Example 14-3 except for a three-stage countercurrrent system and unknown underflow product concentration.

D20. * Repeat Example 14-3 except for a three-stage cross-flow system, with pure solvent at the rate of 421 kg/hr added to each stage and unknown underflow product concentration.

D21. A slurry of pure NaCl crystals, NaCl in solution, NaOH in solution and water is sent to a system of thickener(s) at a rate of 100 kg/min. The feed slurry is 45 wt % crystals. The mass fractions of the entire feed (including crystals and solution) are: x_NaCl = 0.5193, x_NaOH = 0.099, and x_water = 0.3187. We desire to remove the NaOH by washing with a saturated NaCl solution. The thickener(s) are operated so that the underflow is 80% solids (crystals) and 20% liquid. There are no solids in the overflow. Solubility of NaCl in caustic solutions is listed in the table (the y values) (Brown et al., 1950). The first two rows of x values (wt frac in underflow including both crystals and solution) have been calculated so that you can check your calculation procedure. The x values are specific for this operation where the underflow is 80% solids and the crystals are pure NaCl.

a. Complete the table and plot the saturated extract (y) and saturated raffinate (x) curves and construct a conjugate line.

b. If the feed is mixed with 20.0 kg/min of a saturated NaCl solution (y_NaCl = 0.27, y_NaOH = 0.0) in a single thickener, find the flow rates of the underflow and overflow, the compositions of these streams, and the weight of crystals in the underflow.

c. If a countercurrent cascade of thickeners is used with S = 20 kg/min of a saturated NaCl solution (y_NaCl = 0.27, y_NaOH = 0.0) and we desire a raffinate product that has x_NaOH = 0.01, determine the flow rates of R₁ and E_N, and the number of stages required.

Note: Although this is a washing problem, there is not a constant flow rate of solids. Thus, it needs to be solved like the leaching problems in this chapter.

E. More Complex Problems

E1. You are processing halibut livers that contain approximately 25.2 wt % fish oil and 74.8 wt % insoluble solids. The following data for leaching fish oil from halibut livers using diethyl ether solvent is given by Brown and Associates (1950):

where y_oil is the mass fraction oil in the overflow, which was found to contain no insoluble solids, and Z is the pounds of solution per pound of oil free solids in the underflow.

a. Convert the data to mass fraction oil and solvent in the overflow and to mass fraction oil, solvent and solids in the underflow. Plot the data as a saturated extract curve and a saturated raffinate curve with appropriate tie lines on a graph with y_oil or x_oil as the ordinate and y_solids or x_solids as the abscissa. Develop the conjugate line.

b. We have a total of 1,000 pounds of fish (oil + solids). We mix the fish with 500 pounds of pure diethyl ether, let the mixture settle and draw off the solvent layer. Calculate the weight fraction of oil in the extract and raffinate layers, and the amounts of the extract and the raffinate.

c. The raffinate from step b is now mixed with 500 pounds of pure diethyl ether. After settling, calculate the weight fraction of oil in the extract and raffinate layers, and the amounts of the extract and the raffinate.

d. We have a total of 1,000 pounds of fish (oil + solids)/hr. We continuously mix the fish with 500 pounds of pure diethyl ether/hr in a thickener and draw off the solvent layer. Calculate the weight fraction of oil in the extract and raffinate layers, and the flow rates of the extract and the raffinate.

e. The raffinate from step d is now mixed with 500 pounds/hr of pure diethyl ether in a second thickener. Calculate the weight fraction of oil in the extract and raffinate layers, and the flow rates of the extract and the raffinate.

f. We have a total of 1,000.0 pounds of fish (oil + solids)/hr which we are going to continuously process in a countercurrent cascade of thickeners. Use 500 kg/hr of pure diethyl ether. We want the outlet raffinate to contain 0.02 weight fraction fish oil. Find the number of stages required.

CHAPTER 14 APPENDIX COMPUTER SIMULATION OF EXTRACTION

Lab 10. The major purpose of this lab assignment is to have you learn how to simulate LLE. It will be helpful if you become familiar with the information resources available in Aspen Plus. The sources useful for this lab are: help, the Aspen Plus User Guide, and the Unit Operations document. The latter two are available as PDF files when you go to Documents instead of the Aspen User Interface. Find the documents, and read the few pages you need.

We will separate a feed that is 10 mole % carbon tetrachloride (CCl₄) and 90 mole % acetic acid (C₂H₄O₂) using a solvent that is pure triethylamine (C₆H₁₅N). Operation is at 293 K and 1 atm. Data and NRTL parameters are available for this system in the DECHEMA data bank (Sorensen and Arlt, 1980). The DECHEMA parameters for NRTL are in Table 14-A1.

TABLE 14-A1 NRTL parameters and experimental tie line data for triethylamine (1), carbon tetrachloride (2), and acetic acid (3) at 293K and 1 atm (Sorensen and Arlt, 1980); alpha = 0.2, units are K

Hints on running Aspen Plus for extraction:

1. In Setup, choose liquid-liquid-vapor as the allowable phases.

2. For a mixer-settler, all you need in the flowsheet is a decanter with the feed and solvent streams input into the decanter feed port and two product streams. Select acetic acid as the key component for the first liquid phase and triethylamine as the key component for the second liquid phase.

3. For inputting equilibrium data, select NRTL as the property method. Aspen Plus has its own database for parameters, but the Aspen Plus method of inputting the parameter values is different than the DECHEMA method. On the input page for NRTL, select one of the columns for a binary pair and click on the DECHEMA button. Then input the Aij and Aji values from DECHEMA in the Aspen Plus locations for Bij and Bji, respectively. The alpha value from DECHEMA is the Cij in the Aspen Plus Table. This column in the table should now list “user” for source and units should be K. Repeat this procedure for the other binary pairs.

4. For the countercurrent extractor, select “column” then “extract” in the bottom menu bar. The acetic acid feed should be input at the top of the column and the pure solvent at the bottom of the column. For thermal option select “specify temperature profile” and use 293 on the stages. Select acetic acid as the key component for the first liquid phase and triethylamine as the key component for the second liquid phase.

Do the following extraction problems using the NRTL parameters in Table 14-A1 and report your answer.

1. Feed flow rate is 10 kmoles/hr. The feed is mixed with 10 kmoles/hr of pure solvent in a decanter. All streams and the column are at 293 K and 1 atm.

a. Determine the K value for carbon tetrachloride and compare to the data.

b. Find the outlet flow rates and mole fracs in the two outlet streams.

c. Try changing the ratio of S/F and the feed mole frac to determine what happens in this single-stage system. Note that it is possible for one of the phases to disappear.

2. Repeat Problem 1, but in a countercurrent extraction column with two equilibrium stages.

a. Determine the K value for carbon tetrachloride and compare to the data.

b. Find the outlet flow rates and mole fracs in the two outlet streams.

c. Try changing the ratio of S/F and the feed mole frac to determine what happens in this single-stage system. Note that it is possible for one of the phases to disappear.

d. Try increasing the number of stages to determine what happens.