A novice manufacturer of alcohol for gasohol is having a bit difficulty with a d

Question

A novice manufacturer of alcohol for gasohol is having a bit difficulty with a distillation column. The operation is shown in the Figure.

5. A novice manufacturer of alcohol for gasohol is having a bit difficulty with a distillation column. The operation is shown in the Figure. Technicians think too much 90% H2O alcohol is lost in the bottoms (waste). If 1000 kg/min is theHeaBottoms (Waste) feed rate, calculate the composition of the bottoms and the mass of alcohol lost in the bottoms. Heat Exchanger ooling Water Vapor Reflux Distillate (Product) 60% EtOH 40% H2O Wt-1/10 Feed Feed 10% EtOH-1 Column Distillation EtoH? H20?

Explanation / Answer

Feed, F = 1000 kg/min

Feed contains 10% EtOH and 90% H2O

EtOH in feed =1000*0.1=100kg/hr and water =100-10= 90 kg/hr

The outlet of distillation column contains two streams one distillate (D) and bottoms (B)

Distillate is 1/10 of feed , D = 1000/10= 100 kg/min

F= D+ B , 1000= 100+ B , B=1000-100= 900 kg/min

Writing alcohol balance for the oveall plant

1000*0.1= Alcohol entering

Alcohol leaving the top from D = 100*0.6= 60 kg/hr

Alcohol entering = Alcohol in Distllate + Alcohol in bottom= 60 + Alcohol in bottom

100= 60 + Alcohol in bottom

Alcohol in bottom B=100-60 =40 kg/min

But B= 900 kg/min and B= Alcohol + Water

Water in bottoms =900-40= 860 kg/min

Alcohol lost from the bottoms= 40 kg/hr

Composition of bottoms ; 100*40/900= 4.4% Alcohol and Water= 100-4.4= 95.6%

So around 40% of alcohol enteing the column is lost.

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