CHBE 345 – Unit Operations IIAssignment #2Due: February 4 20161. (25%) An 11.5 wt% mixture of acetic acid in water is to be extracted with 1-butanol at 1 atmpressure and 26.7 oC. We desire outlet concentration of 0.5 wt% in the water and 9.6 wt% inthe 1-butanol. Inlet butanol is pure. Find the number of equilibrium stages required and theratio between the actual 1-butanol flow rate and the minimal 1-butanol flow rate using theMcCabe-Thiele method. The equilibrium data for dilute extraction are usually represented asa distribution ratio Kd Kd=y/xin weight fractions or mole fractions. For very dilute systems Kd will be constant. Values ofKd are tabulated in Perry and Green’ Chemical Engineering Handbook. For the currentsystem at 26.7 oC Kd=1.613 with x and y in weight fractions.2. (25%) The 11.5 wt% aqueous solution of acetic acid processed in problem 1 is now treated ina 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. (25%) A water solution containing 0.005 mole fraction benzoic acid is to be extracted usingpure benzene as the solvent in a counter-current cascade of 20 actual stages with an overallcascade/stage efficiency of 0.5. If the feed rate is 100 mol/hr and the solvent flow rate is 10mol/hr find the recovery efficiency of benzoic acid. Operation is isothermal at 6 oC with theequilibrium equation given by: Y=0.0446X.4. (25%) A water solution containing X0 mole fraction benzoic acid is to be extracted usingpure benzene as the solvent in a counter-current cascade of N equilibrium stages. The solutefree feed rate is Rs kmol/hr and the solvent flow rate is chosen as 1.2 times of the minimumflow 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 X N / X 0 ) / mb. Derive the following relationship for XN/X01.2(1 X N / X 0 ) 1XN / X0[1.2(1 X N / X 0 )] N 1 1