A solution of equimolar amounts of methanol, ethanol, isopropanol, and water is
ID: 532531 • Letter: A
Question
A solution of equimolar amounts of methanol, ethanol, isopropanol, and water is fed at 100 kmoles/hr, to a distillation column shown schematically to the right. There is no water in the overhead stream and no methanol in the bottoms. The ratio of ethanol in the overhead to the bottoms is 60/40-i.e. 60 mole percent of the ethanol goes overhead and 40 mole percent goes to the bottoms. The same ratio for isopropanol is 30/70. Calculate the flow rate and composition (as mole fractions) of the overhead and bottoms streams or, If there is no unique solution, set values for the number of variables needed to solve the problem.Explanation / Answer
Flow rate of feed= 100 kmoles/hr. Since equimolar amounts of methanol, ethanol, isopropanol and water enter, flow rate of each component= 100/4= 25 moles/hr(since there are four components)
Their mole fractions= 25/100=0.25
All the water goes to the bottom. Hence flow rate of water at the bottom= 25 moles
Similarly , all the methanol goes to the top. So moles of methanol at the top= 25 mole
Ethanol gets distributed in the ratio of 60/40. 60% of ethanol goes to the top. Moles of ethanol at the top= 25*0.6= 15, moles of ethanol at the bottom= 25-15=10 moles.
Isopropanol at the top similarly is = 25*0.3= 7.5 moles isopropanol at the top is =25-7.5= 17.5 moles
Top product from distillation column : Water=0, methanol= 25 moles, ethanol= 15 and isopropanol= 7.5
Total flow at the top= 0+25+15+7.5= 47.5, composition : water=0, methanol= 25/49.5= 0.505, ethanol= 15/49.5= 0.303, isopropanol= 7.5/49.5= 0.1515
Bottom product contains (moles) : water= 25, methanol=0, ethanol= 10, isopropanol= 17.5
Bottom total flow = 25+10+17.5= 52.5, composition: water= 25/52.5= 0.476, methanol=0, ethanol= 10/52.5= 0.19, isopropanol= 17.5/52.5=0.333
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.