12.9. EQUILIBRIUM RELATIONS AND SINGLE-STAGE LEACHING

12.9A. Equilibrium Relations in Leaching

1. Introduction

To analyze single-stage and countercurrent-stage leaching, an operating-line equation or material-balance relation and the equilibrium relations between the two streams are needed, as in liquid–liquid extraction. It is assumed that the solute-free solid is insoluble in the solvent. In leaching, assuming there is sufficient solvent present so that all the solute in the entering solid can be dissolved into the liquid, equilibrium is reached when the solute is dissolved. Hence, all the solute is completely dissolved in the first stage. There usually is sufficient time for this to occur in the first stage.

It is also assumed that there is no adsorption of the solute by the solid in the leaching. This means that the solution in the liquid phase leaving a stage is the same as the solution that remains with the solid matrix in the settled slurry leaving the stage. In the settler in a stage it is not possible or feasible to separate all the liquid from the solid. Hence, the settled solid leaving a stage always contains some liquid in which dissolved solute is present. This solid–liquid stream is called the underflow or slurry stream. Consequently, the concentration of oil or solute in the liquid or overflow stream is equal to the concentration of solute in the liquid solution accompanying the slurry or underflow stream. Hence, on an x-y plot the equilibrium line is on the 45° line.

The amount of solution retained with the solids in the settling portion of each stage may depend upon the viscosity and density of the liquid in which the solid is suspended. This, in turn, depends upon the concentration of the solute in the solution. Hence, experimental data showing the variation of the amount and composition of solution retained in the solids as a function of the solute composition must be obtained. These data should be obtained under conditions of concentrations, time, and temperature similar to those in the process for which the stage calculations are to be made.

2. Equilibrium diagrams for leaching

The equilibrium data can be plotted on the rectangular diagram as wt fraction for the three components: solute (A), inert or leached solid (B), and solvent (C). The two phases are the overflow (liquid) phase and the underflow (slurry) phase. This method is discussed elsewhere (B2). Another convenient method of plotting the equilibrium data will be used, instead, which is similar to the method discussed in the enthalpy–concentration plots in Section 11.6.

The concentration of inert or insoluble solid B in the solution mixture or the slurry mixture can be expressed in kg (lb_m) units:

Equation 12.9-1

There will be a value of N for the overflow where N = 0, and for the underflow N will have different values, depending on the solute concentration in the liquid. The compositions of solute A in the liquid will be expressed as wt fractions:

Equation 12.9-2

Equation 12.9-3

where x_A is the wt fraction of solute A in the overflow liquid and y_A is the wt fraction of A on a solid-B-free basis in the liquid associated with the slurry or underflow. For the entering solid feed to be leached, N is kg inert solid/kg solute A and y_A = 1.0. For pure entering solvent N = 0 and x_A = 0.

In Fig. 12.9-1a a typical equilibrium diagram is shown, where solute A is infinitely soluble in solvent C, which would occur, for example, in the system soybean oil (A)–soybean inert solid meal (B)–hexane solvent (C). The upper curve of N versus y_A for the slurry underflow represents the separated solid under experimental conditions similar to the actual stage process. The bottom line of N versus x_A, where N = 0 on the axis, represents the overflow liquid composition where all the solid has been removed. In some cases small amounts of solid may remain in the overflow. The tie lines are vertical, and on a y-x diagram, the equilibrium line is y_A = x_A on the 45° line. In Fig. 12.9-1b the tie lines are not vertical, which can result from insufficient contact time, so that all the solute is not dissolved; adsorption of solute A on the solid; or the solute being soluble in the solid B.

If the underflow line of N versus y is straight and horizontal, the amount of liquid associated with the solid in the slurry is constant for all concentrations. This would mean that the underflow liquid rate is constant throughout the various stages as well as the overflow stream. This is a special case which is sometimes approximated in practice.

12.9B. Single-Stage Leaching

In Fig. 12.9-2a a single-stage leaching process is shown where V is kg/h (lb_m/h) of overflow solution with composition x_A and L is kg/h of liquid in the slurry solution with composition y_A based on a given flow rate B kg/h of dry, solute-free solid. The material-balance equations are almost identical to Eqs. (12.5-12)–(12.5-14) for single-stage liquid–liquid extraction and are as follows for a total solution balance (solute A + solvent C), a component balance on A, and a solids balance on B, respectively:

Equation 12.9-4

Equation 12.9-5

Equation 12.9-6

where M is the total flow rate in kg A + C/h and x_AM and N_M are the coordinates of this point M. A balance on C is not needed, since x_A + x_C = 1.0 and y_A + y_C = 1.0. As shown before, L₁MV₁ must lie on a straight line and L₀MV₂ must also lie on a straight line. This is shown in Fig. 12.9-2b. Also, L₁ and V₁ must lie on the vertical tie line. The point M is the intersection of the two lines. If L₀ entering is the fresh solid feed to be leached with no solvent C present, it will be located above the N-versus-y line in Fig. 12.9-2b.

EXAMPLE 12.9-1. Single-Stage Leaching of Flaked Soybeans In a single-stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybeans containing 20 wt % oil is leached with 100 kg of fresh hexane solvent. The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage. Solution: The process flow diagram is the same as given in Fig. 12.9-2a. The known process variables are as follows: for the entering solvent flow, V2 = 100 kg, xA2 = 0, xC2 = 1.0; for the entering slurry stream, B = 100(1.0 − 0.2) = 80 kg insoluble solid, L0 = 100(1.0 − 0.8) = 20 kg A, N0 = 80/20 = 4.0 kg solid/kg solution, yA0 = 1.0. To calculate the location of M, substituting into Eqs. (12.9-4), (12.9-5), and (12.9-6) and solving, Hence, xAM = 0.167. The point M is plotted in Fig. 12.9-3 along with V2 and L0. The vertical tie line is drawn locating L1 and V1 in equilibrium with each other. Then N1 = 1.5, yA1 = 0.167, xA1 = 0.167. Substituting into Eqs. (12.9-4) and (12.9-6) and solving or using the lever-arm rule, L1 = 53.3 kg and V1 = 66.7 kg. Figure 12.9-3. Graphical solution of single-stage leaching for Example 12.9-1.