Problem 4.9 Distillation with Reflux When operation of a distillation column beg
ID: 536022 • Letter: P
Question
Problem 4.9 Distillation with Reflux When operation of a distillation column begins, vapor is generated at the bottom of the column, flows upward through the column, and leaves the top of the column to be condensed in a heat exchanger. During the startup phase, all of the condensate is allowed to flow back into the top of the column as reflux where it contacts the upward-flowing vapor. When conditions in the column begin to stabilize, a valve controlling the flow of reflux to the column is opened, and a fraction of the condensate is allowed to leave the system as product instead of re-entering the column as reflux. The ratio of the flow rate of condensate returned to the column (reflux) to the product withdrawal rate is 2.45. A mixture of 89.0 mol% ethanol and the remainder water enters the condenser at a rate of 100.0 lb-mol/h. condensate has density of 49 lbm/ft3. When operated at steady-state conditions, the tank collecting the condensate is half full, and the The ethanol-water a mean residence time in the tank is 10.0 minutes. Using the following figure as a guide, write a general mass balance around a system consisting of the condenser, condensate tank, and splitter.Explanation / Answer
writing balance over the system containing condenser ,condensate tank and splitter,
let F= flow rate into the condenser, P= flow rate of product, R= Reflux
Flow rate in = flow rate of product+ flow rate of reflux= P+R, 100= P+R (1)
given R/P=2.45, R= 2.45P (2)
Substituting the value of R in Eq.1, 2.45P+P=100, P=100/3.45= 28.98 lbmoles/hr
composition of Vapor = 0.89 ethanol and 1-0.89=0.11 water
molar mass of ethanol(C2H5OH)=46, water= 18
Basis :1 lbmoles of mixture, for 1 lbmole, mass =molar mass
hence mass= 0.89*46+0.11*18= 42.92 lb/mole
so mass of vapor entering the condenser=molar flow rate* molar mass= 100lbmoles/hr*42.92 lb/mole= 4292 lb/hr
density of the mixture= 49 lbm/ft3
volumetric flow rate= mass flow rate/density= 4292/49= 87.6 ft3/hr=0.0243 ft3/min
residence time in the tank= 10min, volume of tank= 0.0243ft3/min*10min= 0.243 ft3.
but only 50% of the tank is full, hence actual volume required= 0.243/0.5= 0.486 ft3
since 1ft3=7.48 gallons, volume of tank= 0.486*7.48 galloons = 3.63 gallons