12.10. COUNTERCURRENT MULTISTAGE LEACHING

12.10A. Introduction and Operating Line for Countercurrent Leaching

A process flow diagram for countercurrent multistage leaching is shown in Fig. 12.10-1 and is similar to Fig. 12.7-1 for liquid–liquid extraction. The ideal stages are numbered in the direction of the solids or underflow stream. The solvent (C)–solute (A) phase or V phase is the liquid phase that overflows continuously from stage to stage countercurrent to the solid phase, dissolving solute as it moves along. The slurry phase L composed of inert solids (B) and a liquid phase of A and C is the continuous underflow from each stage. Note that the composition of the V phase is denoted by x and the composition of the L phase by y, which is the reverse of that for liquid–liquid extraction.

It is assumed that solid B is insoluble and is not lost in the liquid V phase. The flow rate of the solids is constant throughout the cascade of stages. As in single-stage leaching, V is kg/h (lb_m/h) of overflow solution and L is kg/h of liquid solution in the slurry retained by the solid.

In order to derive the operating-line equation, an overall balance and a component balance on solute A is made over the first n stages:

Equation 12.10-1

Equation 12.10-2

Solving for x_n+1 and eliminating V_n+1,

Equation 12.10-3

The operating-line equation (12.10-3), when plotted on an x-y plot, passes through the terminal points x₁, y₀ and x_N+1, y_N.

In the leaching process, if the viscosity and density of the solution change appreciably with the solute (A) concentration, the solids from the lower-numbered stages where solute concentrations are high may retain more liquid solution than the solids from the higher-numbered stages, where the solute concentration is dilute. Then L_n, the liquid retained in the solids underflow, will vary, and the slope of Eq. (12.10-3) will vary from stage to stage. This condition of variable underflow will be considered first. The overflow will also vary. If the amount of solution L_n retained by the solid is constant and independent of concentration, then constant underflow occurs. This simplifies somewhat the stage-to-stage calculations. This case will be considered later.

12.10B. Variable Underflow in Countercurrent Multistage Leaching

The methods in this section are very similar to those used in Section 12.7B for countercurrent solvent extraction, where the L and V flow rates varied from stage to stage. Making an overall total solution (solute A + solvent C) balance on the process shown in Fig. 12.10-1,

Equation 12.10-4

where M is the total mixture flow rate in kg A + C/h. Next, making a component balance on A,

Equation 12.10-5

Making a total solids balance on B,

Equation 12.10-6

where N_M and x_AM are the coordinates of point M shown in Fig. 12.10-2, which is the operating diagram for the process. As demonstrated previously, L₀MV_N+1 must lie on a straight line and V₁ML_N must also lie on a straight line. Usually the flows and compositions of L₀ and V_N+1 are known and the desired exit concentration y_AN is set. Then the coordinates N_M and x_AM can be calculated from Eqs. (12.10-4)–(12.10-6) and point M plotted. Then L_N, M, and V₁ must lie on one line, as shown in Fig. 12.10-2.

In order to go stage by stage on Fig. 12.10-2, we must derive the operating-point equation. Making a total balance on stage 1 and then on stage n,

Equation 12.10-7

Equation 12.10-8

Rearranging Eq. (12.10-7) for the difference flows Δ in kg/h,

Equation 12.10-9

This value Δ is constant and also holds for Eq. (12.10-8) rearranged and for all stages:

Equation 12.10-10

This can also be written for a balance on solute A to give

Equation 12.10-11

where x_AΔ is the x coordinate of the operating point Δ. A balance made on solids gives

Equation 12.10-12

where N_Δ is the N coordinate of the operating point Δ.

As shown in Section 12.7B, Δ is the operating point. This point Δ is located graphically in Fig. 12.10-2 as the intersection of lines L₀V₁ and L_NV_N+1. From Eq. (12.10-10) we see that V₁ is on a line between L₀ and Δ, V₂ is on a line between L₁ and Δ, V_n+1 is on a line between L_n and Δ, and so on.

To graphically determine the number of stages, we start at L₀ and draw line L₀Δ to locate V₁. A tie line through V₁ locates L₁. Line L₁Δ is drawn given V₂. A tie line gives L₂. This is continued until the desired L_N is reached. In Fig. 12.10-2, about 3.5 stages are required.

EXAMPLE 12.10-1. Countercurrent Leaching of Oil from Meal A continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent (B3). The process is to treat 2000 kg/h of inert solid meal (B) containing 800 kg oil (A) and also 50 kg benzene (C). The inlet flow per hour of fresh solvent mixture contains 1310 kg benzene and 20 kg oil. The leached solids are to contain 120 kg oil. Settling experiments similar to those in the actual extractor show that the solution retained depends upon the concentration of oil in the solution. The data (B3) are tabulated below as N kg inert solid B/kg solution and yA kg oil A/kg solution: N yA N yA 2.00 0 1.82 0.4 1.98 0.1 1.75 0.5 1.94 0.2 1.68 0.6 1.89 0.3 1.61 0.7 Calculate the amounts and concentrations of the stream leaving the process and the number of stages required. Solution: The underflow data from the table are plotted in Fig. 12.10-3 as N versus yA. For the inlet solution with the untreated solid, L0 = 800 + 50 = 850 kg/h, yA0 = 800/(800 + 50) = 0.941, B = 2000 kg/h, N0 = 2000/(800 + 50) = 2.36. For the inlet leaching solvent, VN+1 = 1310 + 20 = 1330 kg/h and xAN+1 = 20/1330 = 0.015. The points VN+1 and L0 are plotted. Figure 12.10-3. Graphical construction for number of stages for Example 12.10-1. The point LN lies on the N-versus-yA line in Fig. 12.10-3. Also for this point, the ratio NN/yAN = (kg solid/kg solution)/(kg oil/kg solution) = kg solid/kg oil = 2000/120 = 16.67. Hence, a dashed line through the origin at yA = 0 and N = 0 is plotted with a slope of 16.67 and intersects the N-versus-yA line at LN. The coordinates of LN at this intersection are NN = 1.95 kg solid/kg solution and yAN = 0.118 kg oil/kg solution. Making an overall balance by substituting into Eq. (12.10-4) to determine point M, Substituting into Eq. (12.10-5) and solving, Substituting into Eq. (12.10-6) and solving, The point M is plotted with the coordinates xAM = 0.376 and NM = 0.918 in Fig. 12.10-3. The line VN+1ML0 is drawn, as is line LNM, which intersects the abscissa at point V1 where xA1 = 0.600. The amounts of streams V1 and LN are calculated by substituting into Eqs. (12.10-4) and (12.10-5), and solving simultaneously: Hence, LN = 1016 kg solution/h in the outlet underflow stream and V1 = 1164 kg solution/h in the exit overflow stream. Alternatively, the amounts could have been calculated using the lever-arm rule. The operating point Δ is obtained as the intersection of lines L0V1 and LNVN+1 in Fig. 12.10-3. Its coordinates can also be calculated from Eqs. (12.10-11) and (12.10-12). The stages are stepped off as shown. The fourth stage for L4 is slightly past the desired LN. Hence, about 3.9 stages are required.

12.10C. Constant Underflow in Countercurrent Multistage Leaching

In this case the liquid L_n retained in the underflow solids is constant from stage to stage. This means that a plot of N versus y_A is a horizontal straight line and N is constant. Then the operating-line equation (12.10-3) is a straight line when plotted as y_A versus x_A. The equilibrium line can also be plotted on the same diagram. In many cases the equilibrium line may also be straight, with y_A = x_A. Special treatment must be given the first stage, however, as L₀ is generally not equal to L_n, since it contains little or no solvent. A separate material and equilibrium balance is made on stage 1 to obtain L₁ and V₂ (see Fig. 12.10-1). Then the straight operating line can be used and the McCabe–Thiele method used to step off the number of stages.

Since this procedure for constant underflow requires almost as many calculations as the general case for variable underflow, the general procedure can be used for constant underflow by simply using a horizontal line of N versus y_A in Fig. 12.10-2 and stepping off the stages with the Δ point.