Question

A stream containing toluene (T), heptane (H) and benzene (Z) is fed into a distillation column at a rate of 200kmol/hr. The composition of the feed has molar fractions for T of 0.4, H of 0.2, Z of 0.4. The amount of toluene in the bottoms is 60% the amount of toluene in the distillate. The molar fraction of hexane in the bottom stream is 1/21 and the molar fraction of benzene in the distillate is 2/19. Solve for the distillate and bottoms and the molar composition of both outlet streams.

Answer 1

Basis : 200 kmol/hr of feed (F). It contains 0.4 toulene (T), 0.2 Heptane (H) and0.4 benzene(Z).

Moles/hr of    T= 200*0.4= 80 mol/hr H= 200*0.2= 40 mol/hr and Z= 200*0.4= 80 mol/hr

Distillation produces two streams, one is distillate (D) and bottoms (B).

F= D+B ( Overall material balance) (1) ,200= D+B (1A)

Let y1, y2 and y3 be mole fractions of T, H and Z in the distillate and x1 x2, x3 be mole fractions in the bottoms (B)

Writing Toluene balance 80= D*y1+B*x1 (2)    and it is given that   B*x1= 0.6* D*y1 (3)

1.6Dy1= 80 and Dy1=80/1.6= 50 y1= 50/D

Writing H balance 40= y2*D+x2*B   (4)

Given x₂=1/21=0.047619

40= y2*D+0.047619*B (5), Dy2= 40-0.047619*B= 40-0.0471619*(200-D)= 40-9.52+0.047619D=30.448-0.047619D   and y2= (30.448-0.047169D)/D (5A)

Writing Z balance 80= D*y3+B*x3

and y3=2/19= 0.105, 80= D*0.105+B*x3 (6)

we know that y1+y2+y3=1 and from (3A), 5A and y2=0.105

50/D+( 30.448-0.047619D)/D+0.105 =1

50/D + (30.448-0.047619)/D =1-0.105=0.895

Using solver of excel Solving by assuming D and matching LHS and RHS D= 85.3

B= 200-85.3= 114.7 moles/hr

And hence y1=50/85.3 =0.586 y3= 0.105 and y2= 1-0.586-0.105=0.309

From (3) x1= D*y1/B= 85.3*0.586/114.7 =0.435, x2= 0.047619 and x3= 1-x1-x2= 0.5165