Introduction to multicomponent distillation

1
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

2
Introduction to multicomponent distillation
Rigorous methods (Aspen)
Short-cut methods Fenske-Underwood-Gilliland (K
irkbride)
(See my notes on the web)
3
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

4
Key components

• In practice we usually choose two components
• separation of which serves as an 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

5
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 (lt10-6)
  in one
• of the products are called undistributed

A
B
C
D
E
G
heavy non-distributed components (will end up in
bottoms product)
key components
light non-distributed components (will end up in
the overhead product)
6
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid. a) Design a distillation process
n-pentane 0.04 n-hexane0.40 n-heptane
0.50 n-octane 0.06
condenser
F, zf
100kmol/h
boiler
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid. - Design a distillation process
What is design of a column? - P (pressure) - N
(stages) - R (reflux) - D (diameter) - auxilary
equipment (condenser, boiler)
n-pentane 0.04 n-hexane0.40 n-heptane
0.50 n-octane 0.06
condenser
F, zf
100kmol/h
boiler
8
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Pressure consideration - what if you were
not given P1atm, how would you choose it? - how
do you validate that P1atm is appropriate?
- 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
condenser
F, zf
boiler
9
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid. - Design a distillation process
n-pentane 0.04 n-hexane0.40 n-heptane
0.50 n-octane 0.06
LNK
condenser
LK
HK
HNK
F, zf
100kmol/h
boiler
10
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance conisderations
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
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
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
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
- Sharp split components lighter than the
Light Key (LK) will end up completely in the
overheads
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
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
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 D43.7
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
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 D43.7
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
- Sharp split components heavier than the
Heavy Key (HK) will end up completely in the
bottoms
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 D43.7 B56.3
LNK
LK
HK
HNK
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 98of hexane and 1 of
heptane in the distillate. The feed is boiling
liquid.
- Material balance considerations
- Sharp split components heavier than the
Heavy Key (HK) will end up completely in the
bottoms
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 D43.7 B56.3
LNK
LK
HK
HNK
17
Gilliland correlation Number of ideal plates at
the operating reflux
18
Gilliland correlation Number of ideal plates at
the operating reflux
Nmin Rmin R1.5Rmin
N
19
Fenske equation for multicomponent distillations
Assumption relative volatilities of components
remain constant throughout the column
LK light component HK heavy component
20
Fenske equation for multicomponent distillations
T
Choices for relative volatility
D
B
1) Relative volatility at saturated feed condition
2) Geometric mean relative volatility
why geometric mean?
21
Non key component distribution from the Fenske
equation
Convince yourself and derive for
22
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

23
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

y1
zf
yB
zf
xB
xD
xN
24
Minimum reflux ratio analysis
condenser
F, zf
boiler
Invariant zones presence of heavy and light
non-distributed components
25
Minimum reflux ratio analysis
condenser
F, zf
boiler
Invariant zones only light non-distributed
components
26
Minimum reflux ratio analysis
condenser
F, zf
boiler
Invariant zones only heavy non-distributed
components
27
Minimum reflux ratio analysis
condenser
F, zf
boiler
Invariant zones no non-distributed components
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(No Transcript)
29
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 aLK and aHK)
Once is found, we can calculate the minimum
reflux ratio
30
Minimum reflux ratio analysis Underwood equations
31
Minimum reflux ratio analysis Underwood equations
1.48
32
Minimum reflux ratio analysis Underwood equations
33
Minimum reflux ratio analysis Underwood equations
2.33
1.48
34
Distribution of components in multicomponent
distillation process
Feed stage
xi
hexane LK
heptane HK
Non-distributed heavy non-key component
pentane
octane
1
2
3
4
5
6
7
8
9
10
Non-distributed Light non-key component
35
Kirkbride equation Feed stage location
36
Complete short cut design Fenske-Underwood-Gilli
land 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
37
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
38
Stage efficiency analysis
What are the sources of inefficiencies? For this
we need to look at what actually happens on the
tray
Local efficiency
Actual separation
Separation that would have been achieved on an
ideal tray
Point efficiency
39
Stage efficiency analysis
Depending on the location on the tray the point
efficiency will vary
stagnation points
The overall plate efficiency can be
characterized by the Murphree plate efficiency
low concentration gradients
high concentration gradients
When both the vapour and liquid phases are
perfectly mixed the plate efficiency is equal to
the point efficiency
40
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
41
Stage efficiency analysis OConnell (1946)
(Sinnott)
42
Stage efficiency analysis Van Winkle (1972)
(Sinnott)
43
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)
44
Stage efficiency analysis
Finally the overall efficiency of the process
defined as
If no access to the data E00.5 (i.e. double the
number of plates)
45
Column diameter, etc
Sinnott, Jim Douglas, Conceptual design of
chemical process
46
Column diameter, etc
47
Column diameter, etc