Differential Distillation

Home -> ChE Learning Resources -> Solved Problems -> Mass Transfer->


A liquid feed consisting of 1200 gmoles of mixture containing 30% naphthalene and 70% dipropylene glycol is differentially distilled at 100 mm Hg pressure and final distillate contains 55% of the feed solution. The VLE data are

x

8.4

11.6

28.0

50.6

68.7

80.6

88

y

22.3

41.1

62.9

74.8

80.2

84.4

88

  1. determine the amount of distillate
  2. determine the concentration of naphthalene in residue and distillate

Calculations:

F = 1200 gmol

F = D + W

D = 0.55 x 1200 = 660 gmol

Amount of distillate, D = 660 gmol

W = 1200 - 660 = 540 gmol

From Rayleigh's equation

xW is calculated from the graphical method, for which the integral term is the above equation is balanced:

ln (F/W) = ln (1200/540) = 0.7985

 

The area of 0.7985 is obtained for xW = 0.07

 F xF = D yD,avg + W xW

660 x  yD,avg = 1200 x 0.3 - 540 x 0.07

yD,avg = 0.4882

Concentration of naphthalene in residue = 7%

Concentration of naphthalene in distillate = 48.82%


[Index]     [Learn More from Our Online Course...]


Last Modified on: 04-Feb-2022

Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in