In leaching, the final saturated raffinate often contains a significant amount of solvent. For example, this occurs in Example 14-2. How do you recover this solvent?

Example 14-2

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We wish to treat 1000.0 kg/h (wet basis) of insoluble meal (D) that contains 20.0 wt% oil (A) and no solvent (S). Inlet solvent is pure solvent and flows at 662 kg/h. We desire an underflow product that is 4.0 wt% oil. Temperature and pressure are constant, and equilibrium data are given in Table 14-2. Find the outlet extract concentration and the number of equilibrium stages needed in a countercurrent leaching system. TABLE 14-2. Test data for extraction of oil from meal with solvent Mass Fraction Oil (Solute) in Solution Mass Fraction Underflow (Raffinate) XD XS 0.0 0.0 0.670 0.333 0.1 0.0336 0.664 0.302 0.2 0.0682 0.660 0.272 0.3 0.1039 0.6541 0.242 0.4 0.1419 0.6451 0.213 0.5 0.1817 0.6366 0.1817 0.6 0.224 0.6268 0.1492 0.7 0.268 0.6172 0.1148 Source: Prabhudesai (1997) .5 .4 EN 3 Conjugate Line -Saturated Raffinate F=RN+I R3 Saturated Extract R2 or XOXY .3 .2 E2 E E- R E-S .2 .4 .6 .8 1.0 yoor xp