F9. Leaching Formulas

The leaching system generally consists of three components, a solute (A), a solvent (S), and an inert solid (I). The flow rate of inerts from stage to stage is constant. It is assumed that equilibrium is attained, thus the concentration of the solution leaving a stage is the same as the concentration of the solution adhering to the inerts. The equilibrium relationship is x_e = y_e.

Right-triangular diagram: Experiments are conducted to obtain y_A and K, where K is mass of solution retained per 1 lb of solute-free inert and y_A is mass of solute per lb of solution in the overflow. The mass fractions of the solute, the solvent, and the inerts are calculated from

Plot a graph between x_s or y_s and x_A or y_A. This graph represents underflow composition.

To determine the composition of the overflow in equilibrium with the underflow represented by point B(x_s, x_A), the tie line is drawn through point B to intersect the hypotenuse SA in C. The point C gives the required overflow composition (y_A, y_S).

Crosscurrent Leaching and Triangular diagrams:

Draw an underflow curve.

Locate point R₀ and draw a line R₀S. This represents a mixing operation.

Locate a point M₁ on this line such that x_M1

Draw a line IM₁ from origin passing through M₁.

Locate point R₁ on the underflow curve.

Composition of the underflow can be read from the point R₁.

Draw a line R₁S.

Locate a point M₂ with a composition x_M2.

Draw a line IM₂ from origin passing through M₂.

Locate point R₂ on the underflow curve.

Composition of the underflow can be read from the point R₂

Ponchon-Savarit Diagram:

In Ponchon-Savarit method, Y_I is plotted against X_A, where Y_I is the ratio of the inerts to the solution retained by the inerts and is given by the following relationship.

And X_A is the solute fraction in the solution and is given as

Countercurrent Leaching using Ponchon-Savarit Diagram:

Locate point R₀ representing condition of feed (X_A,X_I ).

Locate point V₁ representing condition of concentrated solvent, y₁

Draw a line R₀V₁ and extend to a point Δ representing a difference stream.

Locate point V_n+1 representing condition of the fresh solvent, y_n+1 .

Locate point R_n representing condition of the final underflow, y_n.

Draw a line V_n+1R_n and extend it to Δ.

Draw a vertical line from V₁ to intersect the underflow curve at R₁

Join R₁ with Δ and locate V₂.

Draw a vertical line from V₂ to intersect the underflow curve at R₂

Join R₂ with Δ and locate V₃.

Continue these steps till we reach R_n.