ADVANCED SEPARATION PROCESS

ADVANCED SEPARATION PROCESSAssignment 1Question 1:a)Explain in detail why the mixing of pure chemicals to form a homogeneous mixture is a spontaneous process, while the separation of that mixture into its pure species is not.Question 2:Based on the information from Figure 1.9 and Table 1.5, determine the SF (split factor) , SP (split ratio) ,and Overall Percent Recovery for each component by filling in the information Table 1.6.

Table 1.5: Operating Material Balance for Hydrocarbon Recovery Process

lbmol/h in Stream

Stream1234567

ComponentFeed to C1C5+ richFeed to C2C3Feed to C3iC4nC4-rich

C2H60.600.60.6000

C3H85705754.82.22.20

iC4H10171.80.1171.70.6171.1162.58.6

nC4H10227.30.7226.60226.610.8215.8

iC5H124011.928.1028.1028.1

nC5H1233.616.117.5017.5017.5

C6+205.3205.300000

TOTAL735.6234.1501.556445.5175.5270

Table 1.6: Split Fraction and Split ratios for Hydrocarbon Recovery ProcessUnitColumn 1Column 2Column 3Overall Percent Recovery(%)

ComponentSFSRSFSRSFSR

C2H6

C3H8

iC4H10

nC4H10

iC5H12

nC5H12

C6+

Question 3:The feed to Column 3 of the distillation sequence in Figure 1.9 is given in Table 1.5 (refer Question 2). However the separation is to be altered so as to produce distillate that is 95 mol% pure isobutene (iCH4) with recovery of isobutane in the (SF) of 96%. Because of the relatively sharp separation in Column C3 between iC4 and nC4, assume all propane goes to the distillate and all C5s goes to the bottoms.a) Compute the flowrates in lbmol/h of each component in each of the product leaving Column C3.b) What is the percent purity of the normal butane in the bottoms product?