Map A distillation column, as shown in the figure to the right is a device used
ID: 484136 • Letter: M
Question
Map A distillation column, as shown in the figure to the right is a device used to separate components with different volatilities. In a specific process, a distillation column is used to separate a feed stream containing 0.4100 mol reflux overhead product ethanol/mol and 0.5900 mol water/mol into two product streams: an overhead stream that is ethanol-rich and a bottoms stream that is water-rich. Vapor containing feed Distillation 0.7520 mol ethanol/mol (and 0.2480 mol water/mol) Column leaves the top of the distillation column and enters a condenser, where the stream condenses completely boilup into a liquid. This stream is then split, the reflux returns to the column and the overhead product leaves the process. The ratio of the flow rates of reflux to overhead product is given by the reflux r reboiler flow rate of reflux bottoms product flow rate of overhead product and for this particular process, R 1.24. Liquid leaving the bottom of the distillation column enters a reboiler, where a fraction of the stream is vaporized and returns to the distillation column as boilup, while the liquid fraction leaves the process as bottoms product. The composition of the boilup and the bottoms product streams are governed by the relative volatility, a, of ethanol and water: E' E xn (1- x, where yE is the mole fraction of ethanol in the vapor stream, w is the mole fraction of water in the vapor stream, XE is the mole fraction of ethanol in the liquid stream, and xw is the mole fraction of water in the liquid stream. For this particular process a 7.273, and the ratio of ethanol leaving the reboiler in the boilup to ethanol leaving the reboiler in the bottoms product is 0.710. For a basis of 100 mol of feed, 47.82 mols of overhead product is produced. Given this basis, answer the following questions Label the process flow diagram below. Label any quantities as "unknown" if they are not given or implied above. Do not leave any blank spaces. Variables names are given for reference laterExplanation / Answer
Feed to the column =100 moles/hr
Writing overall balance , n1= n3+n6
Given n3= 47.82, n6= 100-n3= 100-47.82= 52.18 moles/hr
Since there is 100*0.41= n3*0.7520+n6*x6, x6 is the mole fraction of ethanol in the feed
41 = 47.82*0.7520+52.18*x6, x6= 0.0965, 1-x6= 1-0.0965=0.9035
R= 1.24 =molare flow rate of reflux/ molar flow rate of productss with drwan = molar flow rate of reflux/ 47.82
Molar flow rate of reflux, n4 = 48.72*1.24=59.29 moles/hr
Feed entering the condenser ,n2= n4+n3= 47.82+59.29=107.11 moles/hr
It is given that ethanol in the reboiler/ethanol in the bottoms = 0.710
Ethanol In the bottoms = 0.0965*n6= 0.0965*52.18=5.04 moles/hr
Ethanol in the reboiler = 5.04*0.710 =3.6 moles/hr
Since the streams leaving the reboiler are in equilibriu
Alpha ={ y7/(1-y7)}/{ x6/(1-x6)}=
7.273= {y7/(1-y7)}/(0.0965/0.9035) = {y7/(1-y7)//0.1068
Hecne y7/(1-y7)= 7.273*0.1068 =0.78, y7= 0.78*(1-y7)
Hence y7*(1+0.78)= 0.78, y7= 0.44