4.9. Inside a distillation column (see Problem 4.8), a downward-flowing liquid a

Question

4.9. Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter 6, the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product, and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: CONDENSER Vapor Condensate CONDENSATE TANK TOP OF Overhead Product DISTILLATION COLUMN Reflux A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (ess volatile). A vapor mixture containing 89.0 mole% ethanol and the balance water enters the overhead condenser at a rate of 100 lb-mole/h. The liquid condensate has a density of49.01bm/ft, and Problems 177 the reflux ratio is 3 lbm reflux/lbm overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft /min and the condenser tank Volume (galo

Explanation / Answer

Let L= Flow rate of condensate in to the column, D- flow rate of overhead product

flow rate of liqud

L/D= 3 or L=3D , let the volume of liquid in the tank = V ft3 .its density is 49 lb/ft3, mass of liquid in the tank= V*49/2 =V*24.5ft3

Residence time = 24.5V/Volumetrif flow rate of liquid , Volumetric flow rate of liquid = 24.5V/10 =2.45V

Molar masses= ethanol (C2H5OH)= 46 and water= 18

molar mass of overhead producs = 0.89*46+0.11*18 = 43

1 lb mole occupies 43 lb

100 lb moles occupy 100*43 lb/hr = 4300 lb/hr

Volumetric flow rate= 4300/49 ft3/hr = 88 ft3/hr, =83/60 f3/min = 1.383 ft3/min

Holding time is 10 min , volume of the tank in the liquid = 1.383*10 =13.83 ft3

but the tank is half full. hence volume of tank=13.83*2= 27.66 ft3

1ft3= 7.48 gallons, 27.66 ft3= 27.66*7.48 gal=207 gal

From overall balance V ( Vapor ) = L+D but L=3D

100 = 3D+D, D= 25 moles/hr and L= 75 moles/hr

flow rate of product condensste= 25*43 lb/hr/ 49 lb/ft3= 21.93 ft3/hr

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