Problem 7.2.3

Experimental data on the retention of oil by liver is given (y_A, K), where y_A is mass ratio of oil to solution in the overflow, and K is mass ratio of solution retained to mass of inert (liver). Construct equilibrium diagram (x_S versus x_A) or (y_S versus y_A).

Solution:

Mass fraction of the solute, x_A, is given by y_A K over (K plus one). Mass fraction of the solvent, x_S, is given by (one minus y_A) times K over (K plus one). Finally, mass fraction of the inert, xI, is equal to one minus x_A minus x_S. This simplifies to one over (K plus one).

A typical calculation is as follows:

For the point (y_A, K) having values (0.2, 0.286), x_A is 0.044, x_S is 0.178. x_I can be calculated as 0.778.

Draw a graph between x_S and x_A. This graph represents underflow composition. A diagonal line represents overflow composition. A tie-line passes through origin and terminates on diagonal line.

Composition of the underflow could be read by drawing lines perpendicular to X and Y axes. For a point B, x_S is equal to 0.176 and x_A is 0.076. We can find x_I as one minus 0.176 minus 0.076. This equals to 0.748.

To determine the composition of the overflow in equilibrium with the underflow represented by point B, the tie line is drawn through point B to intersect the hypotenuse SA in C. The point C gives the required overflow composition which is y_A is equal to 0.3, and y_S as 0.7.