Some Concepts regarding Multi Component Distillation

1. Multicomponent Distillation

2. Introduction to multicomponent distillation
• Most of the distillation processes deal with
multicomponent mixtures
• Multicomponent phase behaviour is much more
complex than that for the binary mixtures
• Rigorous design requires computers
• Short cut methods exist to outline the scope and
limitations of a particular process

3. Rigorous methods (Aspen)
Stage j
Short‐cut methods:
Fenske‐Underwood‐Gilliland
(+Kirkbride)
Introduction to multicomponent distillation

4. Multicomponent distillation in tray towers
• Objective of any distillation process is to recover pure products
• In case of multicomponent mixtures we
may be interested in one, two or more components
• Unlike in binary distillation, fixing mole
fraction of one of the components in a
product does not fix the mole fraction of other
components
• On the other hand fixing compositions of all
the components in the distillate and the bottoms
product, makes almost impossible to meet
specifications exactly
D
B
y1,y2,y3,y4…

5. Key components
• In practice we usually choose two components
separation of which serves as a good indication
that a desired degree of separation is achieved
These two components are called key components
‐ light key
‐ heavy key
• There are different strategies to select these key
components
• Choosing two components that are next to each other
on the relative volatility: sharp separation

6. Distributed and undistributed components
• Components that are present in both the distillate and
the bottoms product are called distributed components
‐ The key components are always distributed components
• Components with negligible concentration (<10‐6) in one
of the products are called undistributed
A B C D E G
key
components
heavy non‐distributed components
(will end up in bottoms product)
light non‐distributed components
(will end up in the overhead product)

7. Complete design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
a) Design a distillation process
F, zf
condenser
boiler
n‐pentane: 0.04
n‐hexane:0.40
n‐heptane: 0.50
n‐octane: 0.06
100kmol/h

8. Complete design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
‐ Design a distillation process
F, zf
condenser
boiler
n‐pentane: 0.04
n‐hexane:0.40
n‐heptane: 0.50
n‐octane: 0.06
100kmol/h
What is design of
a column?
‐ P (pressure)
‐ N (stages)
‐ R (reflux)
‐ D (diameter)
‐ auxilary
equipment
(condenser, boiler)

9. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
‐ Pressure consideration:
‐ what if you were not given P=1atm, how would you choose it?
‐ how do you validate that P=1atm is appropriate?
F, zf
condenser
boiler
‐ condenser uses cooling water
(20C). Let say the exit water
temperature is 30C.
‐ To maintain the temperature delta
at 10C, the dew point can not be
lower than 40C.
‐ Thus, the dew point of the distillate
has to be at least 40C.
‐ If not, will need higher pressure

10. Complete design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
‐ Design a distillation process
F, zf
condenser
boiler
n‐pentane: 0.04
n‐hexane:0.40
n‐heptane: 0.50
n‐octane: 0.06
100kmol/h
LK
HK
HNK
LNK

11. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 3.62
Hexane 0.4 40 1.39
Heptane 0.5 50 0.56
Octane 0.06 6 0.23
100
‐ Material balance conisderations
LK
HK
HNK
LNK

12. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 3.62
Hexane 0.4 40 39.2 1.39
Heptane 0.5 50 0.5 0.56
Octane 0.06 6 0.23
100
‐ Material balance considerations
LK
HK
HNK
LNK

13. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 4 3.62
Hexane
LK
0.4 40 39.2 1.39
Heptane
HK
0.5 50 0.5 0.56
Octane 0.06 6 0 0.23
100
‐ Sharp split: components lighter than the Light Key (LK) will end up completely
in the overheads
‐ Material balance considerations
LK
HK
HNK
LNK

14. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 4 0.092 3.62
Hexane
LK
0.4 40 39.2 0.897 1.39
Heptane
HK
0.5 50 0.5 0.011 0.56
Octane 0.06 6 0 0 0.23
100 D=43.7
‐ Material balance considerations
LK
HK
HNK
LNK

15. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 4 0.092 3.62
Hexane
LK
0.4 40 39.2 0.897 0.8 1.39
Heptane
HK
0.5 50 0.5 0.011 49.5 0.56
Octane 0.06 6 0 0 0.23
100 D=43.7
‐ Material balance considerations
LK
HK
HNK
LNK

16. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 4 0.092 0 3.62
Hexane
LK
0.4 40 39.2 0.897 0.8 1.39
Heptane
HK
0.5 50 0.5 0.011 49.5 0.56
Octane 0.06 6 0 0 6 0.23
100 D=43.7 B=56.3
‐ Sharp split: components heavier than the Heavy Key (HK) will
end up completely in the bottoms
‐ Material balance considerations
LK
HK
HNK
LNK

17. Complete short cut design
A mixture of 4% n‐pentane, 40% n‐hexane, 50% n‐heptane and 6%
n‐octane is distilled at 1 atm. The goal is to recover 98%of hexane and
1% of heptane in the distillate. The feed is boiling liquid.
xF F xF Moles in D xD Moles in B xB Ki
Pentane 0.04 4 4 0.092 0 0 3.62
Hexane
LK
0.4 40 39.2 0.897 0.8 0.014 1.39
Heptane
HK
0.5 50 0.5 0.011 49.5 0.879 0.56
Octane 0.06 6 0 0 6 0.107 0.23
100 D=43.7 B=56.3
‐ Sharp split: components heavier than the Heavy Key (HK) will
end up completely in the bottoms
‐ Material balance considerations
LK
HK
HNK
LNK

18. Minimum reflux ratio analysis
• At the minimum reflux ratio
condition there are invariant zones
that occur above and below the
feed plate, where the number of
plates is infinite and the liquid and
vapour compositions do not
change from plate to plate
• Unlike in binary distillations, in
multicomponent mixtures these
zones are not necessarily adjacent
to the feed plate location
y
x
zf
zf
xB xD
y1
yB
xN

19. Minimum reflux ratio analysis
• At the minimum reflux ratio condition
there are invariant zones that occur
above and below the feed plate, where
the number of plates is infinite and the
liquid and vapour compositions do not
change from plate to plate
• Unlike in binary distillations, in
multicomponent mixtures these zones
are not necessarily adjacent to the feed
plate location
y
x
zf
zf
xB xD
y1
yB
xN

20. Minimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: presence of heavy and light non‐distributed
components

21. Minimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: only light non‐distributed
components

22. Minimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: only heavy non‐distributed
components

23. Minimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: no non‐distributed
components

24. xi
1 2 3 4 5 6 7 8 9 10
hexane LK
heptane HK
octane
pentane
Feed stage
Distribution of components in
multicomponent distillation process
Non‐distributed
heavy non‐key
component
Non‐distributed
Light non‐key
component

25. Gilliland correlation: Number of ideal
plates at the operating reflux GP‐A
Fenske equation for multicomponent
Distillations GP‐B
Non key component distribution from
the Fenske equation GP‐C
Minimum reflux ratio analysis:
Underwood equations GP‐D
Kirkbride equation: Feed stage location GP‐E
Stage efficiency analysis GP‐F
Stage efficiency analysis: O’Connell (1946)
Stage efficiency analysis: Van Winkle (1972)
Column diameter GP‐G

26. Gilliland correlation: Number of ideal
plates at the operating reflux

27. Gilliland correlation: Number of ideal
plates at the operating reflux
Nmin
Rmin
R=1.5Rmin
N

28. Fenske equation for multicomponent
distillations
Assumption: relative volatilities of components remain constant
throughout the column
LK – light component
HK – heavy component

29. Fenske equation for multicomponent
distillations
Choices for relative volatility:
D
B
T
1) Relative volatility at saturated feed condition
2) Geometric mean relative volatility
why geometric mean?

30. Non key component distribution from
the Fenske equation
Convince yourself and
derive for

31. Minimum reflux ratio analysis:
Underwood equations
For a given q, and the feed composition
we are looking for A satisfies this equation
(usually is between αLK and αHK)
Once is found, we can calculate the
minimum reflux ratio

32. Minimum reflux ratio analysis:
Underwood equations

33. Minimum reflux ratio analysis:
Underwood equations
1.48

34. Minimum reflux ratio analysis:
Underwood equations

35. Minimum reflux ratio analysis:
Underwood equations
1.48
2.33

36. xi
1 2 3 4 5 6 7 8 9 10
hexane LK
heptane HK
octane
pentane
Feed stage
Distribution of components in
multicomponent distillation process
Non‐distributed
heavy non‐key
component
Non‐distributed
Light non‐key
component

37. Kirkbride equation: Feed stage location

38. Complete short cut design:
Fenske‐Underwood‐Gilliland method
Given a multicomponent distillation problem:
a) Identify light and heavy key components
b) Guess splits of the non‐key components and compositions
of the distillate and bottoms products
c) Calculate
d) Use Fenske equation to find Nmin
e) Use Underwood method to find RDm
f) Use Gilliland correlation to find actual number of ideal stages
given operating reflux
g) Use Kirkbride equation to locate the feed stage

39. Stage efficiency analysis
In general the overall efficiency will depend:
1) Geometry and design of contact stages
2) Flow rates and patterns on the tray
3) Composition and properties of vapour and
liquid streams

40. Stage efficiency analysis
Lin,xin
Lout,xout
Vout,yout
Vin,yin
Local efficiency
Actual separation
Separation that
would have been
achieved on an
ideal tray
What are the sources of inefficiencies?
For this we need to look at what actually happens
on the tray
Point efficiency

41. Stage efficiency analysis
Depending on the location on the tray
the point efficiency will vary
high concentration
gradients
low concentration
gradients
stagnation points
The overall plate efficiency can
be characterized by the Murphree
plate efficiency:
When both the vapour and liquid
phases are perfectly mixed the plate
efficiency is equal to the point
efficiency

42. Stage efficiency analysis
In general a number of
empirical correlations exist
that relate point and plate
efficiencies
Peclet number
length of liquid
flow path
eddy diffusivity residence time of liquid
on the tray

43. Stage efficiency analysis: O’Connell (1946)
(Sinnott)

44. Stage efficiency analysis: Van Winkle (1972)
(Sinnott)

45. Stage efficiency analysis
‐ AICHE method
‐ Fair‐Chan
Chan, H., J.R. Fair,” Prediction of Point Efficiencies for Sieve Trays, 1. Binary Systems”,
Ind Eng. Chem. .Process Des. Dev., 23, 814‐819 (1984)
Chan, H., J.R. Fair, ,” Prediction of Point Efficiencies for Sieve Trays, 1. Multi‐component Systems”,
Ind Eng. Chem. .Process Des. Dev., 23, 820‐827 (1984)
(Sinnott)

46. Stage efficiency analysis
Finally the overall efficiency of the process defined as
If no access to the data: E0=0.5 (i.e. double the number of plates)

47. Column diameter, etc
Sinnott,
Jim Douglas, Conceptual design of chemical process

48. Column diameter, etc

49. Column diameter, etc

50. Multi‐Component Distillation
Phase equilibrium For multi‐components

51. Bubble point calculations
Given the total pressure and liquid mole
fraction ‐‐> T and y
Iterate over temperature till solution is found  calculate the vapor mole fraction from

53. component xi Ki αι α ιxi
Butane
Pentane
Hexane
Heptane
0.4
0.25
0.2
0.15
1.68
0.63
0.245
0.093
6.857
2.571
1.0
0.38
2.743
0.643
0.2
0.057
Total 3.643
A liquid stream at 405.3 kPa with 40% butane, 25% pentane, 30% hexane and 15%
heptane is fed to a distillation column. Find the bubble point and the corresponding
vapor composition in equilibrium with the liquid.
Using the K-value diagram we build this Table:
Let K-hexane be the reference component
Let T = 65 oC

54. A liquid stream at 405.3 kPa and 100 mole/h with 40% butane (A), 25%
pentane (B), 30% hexane (C) and 15% hepatne (D) is fed to a distillation
column. 90% of B is recovered in distillate and 90% of C is the bottom,
calculate
(a) mole and composition of distillate and bottom
(b) Dew point of the distillate and bubble point of the bottom
(c) minimum stages for total reflux
(d) minimum reflux ratio
(e) number of theoretical stage working at 1.5Rm
(f) location of feed tray
(a) solve the material balance
mole balance on butane
25 = nBD + nBB
Q nBD = 0.9(25) = 22.5 mole ‐‐> nDD = 2.5 mole
mole balance on hexane
30 = nCD + nCB

55. • Q ncB = 0.9(20) = 18 mole ‐‐> nCD = 2 mole
• Assume all A in distillate ‐‐> nAD = 40 mole ‐‐> nAB = 0
• Assume all D in bottom ‐‐> nAD = 40 mole ‐‐> nAB = 0

56. • Part b)

57. • part ( c)
• let B  light component; let C  heavy component
• αLD = 2.5; αLB = 2.04

59. part (d)
Solve by trial‐and‐error
 θ = 1.2096

60. part (d)
R = 1.5Rm = 1.5(0.395) = 0.593
R/(R+1) = 0.3723
Rm/(Rm+1) = 0.283
Using the chart  Nm/N = 0.49  N = Nm/0.49  N = 5.4/0.49 = 11
part (e)
 Ne/Ns = 1.184
 Ne+Ns = 11
Ne = 6 ; Ns = 5