Triangular diagrams mixing point

When the solute is initially present as a solid, the amount that can be dissolved in a given amount of solvent is limited by the solubility of the material. A saturated solution of the solute will be represented by some point, such as A, on the hypotenuse of the triangular diagram (Figure 10.16). The line OA represents the compositions of all possible mixtures of saturated solution with insoluble solid, since xs/xA is constant at all points on this line. The part of the triangle above OA therefore represents unsaturated solutions mixed with the insoluble solid B. If a mixture, represented by some point N, is separated into solid and liquid, it will yield a solid, of composition represented by O, that is pure component B, and an unsaturated solution (N ). Again the lower part of the diagram represents mixtures... 

This nomenclature is shown with Example 14.3. On the triangular diagram, the proportions of feed and solvent locate the mix point... 

This problem is concerned with the properties of the triangular diagram used for representing ternary systems. Use the lever rule and notice that if the residue were mixed with the distillate at any moment it would give a mixture represented by point A, Note also that the composition of the distillate is alwa3rs somewhere on XF because Z is not volatile. 

The calculation method proceeds as follows. 1) Plot the locations of S and F on the triangular equilibrium diagram 2) Draw a straight line between S and F, and use the lever-arm rule or Eqs. tl3-31i to find the location of the mixed stream M. Now we know that stream M settles into two phases in equilibrium with each other. Therefore, 3) construct a tie line through point M to find the compositions of the extract and raffinate streams. 4) Find the ratio E/R using mass balances. We will follow this method to solve the following exanple. 

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