Differential distillation

1. 1

Differential Distillation

Rayleigh Equation
()
i
ii
dxdL
L y x



……… (1)
Integrating: 11
()
i
o io
xL
i
iiLx
dxdL
L y x




1
1
ln
()
i
io
x
i
o i ix
dxL
L y x



……… (2)
Where:
oL Initial amount of liquid in pot, ,moles
1L Remaining amount of liquid in pot,,moles
dL Amount of liquid vaporized in timed
,
iidx dy Change in concentration for time intervald

Solution of the relation depends on the form of the equilibrium relationship: ()
iiy f x

a) When i i iy m x 1 1 1
1 1
ln
( ) ( ) ( 1)
i i i
io io io
x x x
i i i
o i i i i i i ix x x
dx dx dxL
L y x m x x m x
  
  
  
11 1
ln ln
( 1)
i
o i io
xL
L m x


……… (3)
Rewritten in another form: 1
1
11
im
i
o io
xL
Lx




……… (4)

b) For i i i iy m x c
1 1 1
1
ln
( ) ( ) ( 1)
i i i
io io io
x x x
i i i
o i i i i i i i i ix x x
dx dx dxL
L y x m x c x m x c
  
    
  

11
( 1)1
ln ln
( 1) ( 1)
i i i
o i i io i
m x cL
L m m x c


  
……… (5)

2. 2

c) When the relative volatilityij
 is constant and the equilibrium expressed is as: 1 ( 1)
ij i
i
ij i
x
y
x





11
11
1
ln
()
()
1 ( 1)
(1 ( 1) ) (1 ( 1) )
( 1 ) ( 1)(1 )
ii
io io
ii
io io
xx
ii
ij io i ixx
i
ij i
i
xx
ij i i ij i i
i ij ij i i i ij ixx
L dx dx
xL y x
x
x
x dx x dx
x x x x x



  




   

    




11
( 1)
( 1)(1 ) ( 1)(1 )
ii
io io
xx
ij i ii
i ij i i ij ixx
x dxdx
x x x x




   


11
1
( 1) (1 ) (1 )
ii
io io
xx
ii
ij i i ixx
dx dx
x x x

  


The integration becomes:
11
11
111
ln ln ln
( 1) 1 1
io i io
o ij i io i
x x xL
L x x x
   
    
  
   


1
11
111
ln ( 1)ln
( 1) 1 1
io i io
ij
ij i io i
x x x
x x x


   
     
  
   

1
1
11
ln ln
( 1) 1
i io
ij
ij io i
xx
xx


 
  



And finally:
111
11
ln ln ln
( 1) 1
ii
ij
o ij io io
xxL
L x x


 
  


……… (6)

3. 3

d) Graphical integration is applied when the equilibrium data( , )
iixy is given in tabular
form:
For each equilibrium point ( , )
iixy the value 1
iiyx



 is calculated:
1
ii
yx
iiyx iy ix
- - - -
- - - -
- - - -

A plot of1
iiyx



 versus[]
ix is then made and the area under the curve between 1[]
ix and []
iox
gives1
ln
o
L
L .