Sample Problem #3

Sample Problem #3:

A solution contains a valuable material M in water. M is recovered from the solution using solvent S. If 9 kg of  S is used per kg solution, and the distribution equilibrium X/Z = 3, where X = kg M/ kg water and Z= kg M / kg S. What will remain in the solution after extraction using (a) single batch equilibrium stage and (b) two successive batch equilibrium stages using fresh solvent for each step?

Answer: Basis: 10 kg feed

Extract 1 kg M/ 9 kg S

Solvent 9 kg S

Raffinate Extractor 

Solution 5 kg (M+water)

Raffinate 3 kg M/ kg water

 

Distributive Equilibrium= 3 = kg M/ kg Water  Kg M/ kg S Required: % remaining in the solution after solvent extraction Solution:

Solution + solvent = M From distributive equilibrium: 3 kg M + 1 kg Water from raffinate + 1 kg water from extract= 5 kg feed (Lo) 5 kg (water + M) + 9 kg solvent = 14 kg mixture (Lo) Mass of raffinate = 1 kg Water + 3 kg M= 4kg raffinate (L1) Mass of extract = 9 kg S + 1 kg M = 10 kg extract ( V1) Fraction of M in extract: 1 kg / 10 kg = 0.1 ( Ya1) Fraction of M in raffinate: 3 kg / 4 kg = 0.75 ( Xa1) Mass fraction of solute solute M in feed = required (Xao) (Xao) From the formula K = ya = L1 Lo Xa V1 L1

= 0.1 0.1 0.75

Xao - 1 xa1

= 4 kg (M + wate water) r) [ 14kg 14kg / 4kg 4kg ( Xao/ Xao/ 0.75 0.75)) –1] –1] 10 kg (M+S)

Xa1 = 82.14 % ( fraction of M in the feed) Fraction remaining in the solution= 0.8214-.75= 0.071 4 = 7.14% fraction remaining in the feed