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1.  An 11.5 wt% mixture of acetic acid in water is to be extracted with 1-butanol at 1 atm pressure and 26.7 oC. We desire outlet concentration of 0.5 wt% in the water and 9.6 wt% in the 1-butanol. Inlet butanol is pure. Find the number of equilibrium stages required and the ratio between the actual 1-butanol flow rate and the minimal 1-butanol flow rate using the McCabe-Thiele method. The equilibrium data for dilute extraction are usually represented as a distribution ratio, Kd,

Kd=y/x

in weight fractions or mole fractions. For very dilute systems, Kd will be constant. Values of Kd are tabulated in Perry and Green’ Chemical Engineering Handbook. For the current system at 26.7 oC, Kd=1.613, with x and y in weight fractions. 

2.  The 11.5 wt% aqueous solution of acetic acid processed in problem 1 is now treated in a three-stage cross-flow system. The same amount of solvent (pure 1-butanol) is to be used, but now it will be divided equally among the three stages. Operation is at 1 atm and 26.7 oC. Find the outlet concentration of all streams. 

3. A water solution containing 0.005 mole fraction benzoic acid is to be extracted using pure benzene as the solvent in a counter-current cascade of 20 actual stages, with an overall cascade/stage efficiency of 0.5. If the feed rate is 100 mol/hr and the solvent flow rate is 10 mol/hr, find the recovery efficiency of benzoic acid. Operation is isothermal at 6 oC, with the equilibrium equation given by: Y=0.0446X. 

4.A water solution containing X0 mole fraction benzoic acid is to be extracted using pure benzene as the solvent in a counter-current cascade of N equilibrium stages. The solute- free feed rate is Rs kmol/hr and the solvent flow rate is chosen as 1.2 times of the minimum flow rate, Es,min. Operation is isothermal, with the equilibrium equation given by: Y=mX, with m as a constant.

a.        Derive the following relationship for the minimum solvent flow rate:

Es,min  = Rs (1 - XN / X 0 ) / m

b.        Derive the following relationship for XN/X0

 

X    /  =1.2(1- X N  / X 0 ) -1N             0         [1.2(1- X/ X 0)]N +1 -1

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