Jw

Figure 8. Graphical solution for multiple contact.

Since liquid-liquid extraction frequently involves only a few stages, the above equation can be used for an analytical solution.

The desired concentration of extract YE is set equal to Yx, and the raffinate in equilibrium with the first stage, Xx, is determined from the equilibrium curve. With this value ofZl5 Y2 is calculated from the above operating equation; then X2 is determined from the equilibrium line and the calculation procedure is continued until X„ < Xr.

A graphical solution is also readily obtainable. The operating line, with slope F/S, is drawn from the inlet and outlet concentrations. The number of stages is then stepped off in the same fashion as with a McCabe Thiele diagram in distillation, as shown in Fig. 8.

With a ternary equilibrium diagram, such as Fig. 5, the process result can be determined graphically. In Fig. 9, the addition of solvent to a feed containing XF solute will be along the straight line connecting S with XF. From an overall mass balance, the composition Mof the mixture of feed and

solvent is determined. With M in the two-phase zone, the overall mixture M separates along a tie line to end points YE an&XR on the equilibrium curve. The relative quantities of each phase can be calculated using the inverse lever-arm rule.

Solute

Solute

Figure 9. Graphical solution for single contact with ternary equilibrium data.

With more than one contact, an operating point Q is located outside the ternary diagram, as shown in Fig. 10. With a specified solvent/feed ratio and a desired raffinate purity, X1, with the given feed, XF the composition of the final extract, Y„, is fixed by material balance. Point Q is formed by the intersection of the line drawn from Yn through XF, with the line drawn from the fresh solvent Ys through X1.

Solute

Solute

Point Min Fig. 9 represented the material balance: F+ S = E + R = M

Point Q in Fig. 10 represents a hypothetical quantity obtained by rearrangement of the above equation:

The material balance for each stage is:

Thus, a line through Q represents the operating line between stages. The number of stages is obtained by sequentially stepping off first the equilibrium distribution along a tie line, and then to the next stage by a line drawn from point Q through the raffinate to locate the next extract.

4.1 Simplified Solution

If the distribution coefficient is constant, and if there is essentially no mutual solubility, the fraction not extracted, VP, can be calculated directly as a function of the extraction factor, E, and the number of stages, n.

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