introduction to multicomponent distillation
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Introduction to multicomponent distillationIntroduction 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
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Multicomponent distillation in tray towersMulticomponent 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…
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Key componentsKey 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 scale often leads to all the components lighter then the light key components accumulating in the distillate and all the components heavier then the heavy key component accumulating in the bottoms product: sharp separation
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Distributed and undistributed componentsDistributed 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)
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Complete designComplete 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: 0.04n-hexane:0.40n-heptane: 0.50n-octane: 0.06
100kmol/h
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Complete designComplete 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: 0.04n-hexane:0.40n-heptane: 0.50n-octane: 0.06
100kmol/h
What is design ofa column?
- P (pressure)- N (stages)- R (reflux)- D (diameter)- auxilary equipment (condenser, boiler)
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Complete designComplete 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
Rigorous methods (Aspen)
Stage j
Fj
Fj
Fjjij TPhzF ,,,, , jQ
11
11,1
,
,,,
−−
−−−
jj
Ljjij
TP
hxL
11
11,1
,
,,,
++
+++
jj
Vjjij
TP
hyV
jj
Ljjij
TP
hxL
,
,,, ,
jj
Vjjij
TP
hyV
,
,,, ,
jU
jW
(See my notes on the web)
Short-cut methods:
Fenske-Underwood-Gilliland(+Kirkbride)
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Complete short cut designComplete 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
I. Material balance conisderations
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 1.39
HeptaneHK
0.5 50 0.5 0.56
Octane 0.06 6 0.23
100
I. Material balance considerations
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 1.39
HeptaneHK
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 completelyin the overheads
I. Material balance considerations
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 0.897 1.39
HeptaneHK
0.5 50 0.5 0.011 0.56
Octane 0.06 6 0 0 0.23
100 D=43.7
I. Material balance considerations
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 0.897 0.8 1.39
HeptaneHK
0.5 50 0.5 0.011 49.5 0.56
Octane 0.06 6 0 0 0.23
100 D=43.7
I. Material balance considerations
iFii DyFxBx −=
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 0.897 0.8 1.39
HeptaneHK
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 overheads
I. Material balance considerations
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Complete short cut designComplete 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
HexaneLK
0.4 40 39.2 0.897 0.8 0.014 1.39
HeptaneHK
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 overheads
I. Material balance considerations
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Complete short cut designComplete 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.
II. 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 distillatehas to be at least 40C.
- If not, will need higher pressure
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III. Gilliland correlation: Number of ideal III. Gilliland correlation: Number of ideal plates at the operating refluxplates at the operating reflux
+
−=+
−11
min
D
DmD
R
RRf
N
NN
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III. Gilliland correlation: Number of ideal III. Gilliland correlation: Number of ideal plates at the operating refluxplates at the operating reflux
+
−=+
−11
min
D
DmD
R
RRf
N
NN
Nmin
Rmin
R=1.5Rmin
Nmin
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IV. Fenske equation for multicomponentIV. Fenske equation for multicomponentdistillationsdistillations
Assumption: relative volatilities of components remain constantthroughout the column
1ln
ln
,
,
,
,
,
min −
=HKLK
HKD
HKB
LKB
LKD
x
x
x
x
Nα
LK – light componentHK – heavy component
)(
)()(, TK
TKT
HK
LKHKLK =α
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V. Minimum reflux ratio analysisV. 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
xzf
zf
xB xD
y1
yB
xN
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V. Minimum reflux ratio analysis:V. Minimum reflux ratio analysis:Underwood equationsUnderwood equations
∑ −=−
i HKi
iFHKi
A
xq
,
,,)1(αα
∑ −==+
i HKi
iDHKim A
x
D
VR
,
,,1αα
For a given q, and the feed composition we are looking for A satisfies this equation(usually A is between αLK and αHK)
Once A is found, we can calculate theminimum reflux ratio
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VI. Kirkbride equation: Feed stage locationVI. Kirkbride equation: Feed stage location
206.02
,
,
,
,
=
D
B
x
x
x
x
N
N
HKD
LKB
LKF
HKF
S
R
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Complete short cut design: Complete short cut design: Fenske-Underwood-Gilliland methodFenske-Underwood-Gilliland methodGiven a multicomponent distillation problem:Given a multicomponent distillation problem:
a) Identify light and heavy key components
b) Guess splits of the non-key components and compositionsof 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
HKLK ,α
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VII. Stage efficiency analysisVII. 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
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VII. Stage efficiency analysisVII. Stage efficiency analysis
Lin,xin
Lout,xout
Vout,yout
Vin,yin
Local efficiency
1*
1
+
+
−−′
=nn
nnmv yy
yyE
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 happenson the tray
Point efficiency
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VII. Stage efficiency analysisVII. Stage efficiency analysis
Depending on the location on the tray the point efficiency will vary
high concentrationgradients
low concentrationgradients
stagnation points
The overall plate efficiency can be characterized by the Murphreeplate efficiency:
1*
1
+
+
−−=
nn
nnmV yy
yyE
When both the vapour and liquidphases are perfectly mixed the plateefficiency is equal to the point efficiency
mvmV EE =
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VII. Stage efficiency analysisVII. Stage efficiency analysis
In general a number of empirical correlations exist that relate point and plate efficiencies
ce
LPe tD
ZN
2
=
Peclet number
length of liquid flow path
eddy diffusivity residence time of liquidon the tray
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VII. Stage efficiency analysis: O’Connell (1946)VII. Stage efficiency analysis: O’Connell (1946)
(Sinnott)
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VII. Stage efficiency analysis: Van Winkle (1972)VII. Stage efficiency analysis: Van Winkle (1972)
(Sinnott)
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VII. Stage efficiency analysisVII. Stage efficiency analysis
- AICHE method- AICHE method
- Fair-Chan- 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)
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VII. Stage efficiency analysisVII. Stage efficiency analysis
Finally the overall efficiency of the process defined as
actual
ltheoreticaO N
NE =
If no access to the data: E0=0.5 (i.e. double the number of plates)
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VIII. Column diameter, etcVIII. Column diameter, etc
Sinnott,
Jim Douglas, Conceptual design of chemical process