Introduction to multicomponent distillation 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
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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
Rigorous methods (Aspen)
Stage j
F
j
F
j
F
jjij TPhzF ,,,, , jQ
11
11,1
,
,,,
−−
−−−
jj
L
jjij
TP
hxL
11
11,1
,
,,,
++
+++
jj
V
jjij
TP
hyV
jj
L
jjij
TP
hxL
,
,,, ,
jj
V
jjij
TP
hyV
,
,,, ,
jU
jW
(See my notes on the web)
Short-cut methods:
Fenske-Underwood-Gilliland
(+Kirkbride)
Introduction to multicomponent distillationIntroduction to multicomponent distillation
MulticomponentMulticomponent distillation in tray towersdistillation 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�
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: sharp separation
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)
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: 0.04
n-hexane:0.40
n-heptane: 0.50
n-octane: 0.06
100kmol/h
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.
- 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)
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.
- 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
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.
- 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
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.
0.2360.06Octane
100
0.56500.5Heptane
1.39400.4Hexane
3.6240.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Material balance conisderations
LK
HK
HNK
LNK
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.
0.2360.06Octane
100
0.560.5500.5Heptane
1.3939.2400.4Hexane
3.6240.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Material balance considerations
LK
HK
HNK
LNK
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.
0.23060.06Octane
100
0.560.5500.5Heptane
HK
1.3939.2400.4Hexane
LK
3.62440.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Sharp split: components lighter than the Light Key (LK) will end up completely
in the overheads
- Material balance considerations
LK
HK
HNK
LNK
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.
0.230060.06Octane
D=43.7100
0.560.0110.5500.5Heptane
HK
1.390.89739.2400.4Hexane
LK
3.620.092440.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Material balance considerations
LK
HK
HNK
LNK
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.
0.230060.06Octane
D=43.7100
0.5649.50.0110.5500.5Heptane
HK
1.390.80.89739.2400.4Hexane
LK
3.620.092440.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Material balance considerations
i
F
ii DyFxBx −=
LK
HK
HNK
LNK
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.
0.2360060.06Octane
B=56.3D=43.7100
0.5649.50.0110.5500.5Heptane
HK
1.390.80.89739.2400.4Hexane
LK
3.6200.092440.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Sharp split: components heavier than the Heavy Key (HK) will
end up completely in the bottoms
- Material balance considerations
LK
HK
HNK
LNK
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.
0.230.10760060.06Octane
B=56.3D=43.7100
0.560.87949.50.0110.5500.5Heptane
HK
1.390.0140.80.89739.2400.4Hexane
LK
3.62000.092440.04Pentane
KixBMoles in BxDMoles in DF xFxF
- Sharp split: components heavier than the Heavy Key (HK) will
end up completely in the bottoms
- Material balance considerations
LK
HK
HNK
LNK
Gilliland correlation: Number of ideal Gilliland correlation: Number of ideal
plates at the operating refluxplates at the operating reflux
+−
=+
−11
min
D
DmD
R
RRf
N
NN
Gilliland correlation: Number of ideal 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
N
FenskeFenske equation for multicomponentequation for multicomponent
distillationsdistillations
Assumption: relative volatilities of components remain constant
throughout the column
1ln
ln
,
,
,
,
,
min −
=HKLK
HKD
HKB
LKB
LKD
x
x
x
x
Nα
LK – light component
HK – heavy component
)(
)()(,
TK
TKT
HK
LKHKLK =α
FenskeFenske equation for multicomponentequation for multicomponent
distillationsdistillations
)(
)()(,
TK
TKT
HK
LKHKLK =α
Choices for relative volatility:
D
B
T
1) Relative volatility at saturated feed condition
)(,, F
F
HKLK THKLK
αα =
2) Geometric mean relative volatility
)()(,,, B
B
D
D
HKLK TTHKLKHKLK
ααα =
3, )()()(
,,, B
B
D
D
F
F
HKLK TTTHKLKHKLKHKLK
αααα =
why geometric mean?
Non key component distribution from Non key component distribution from
the the FenskeFenske equationequation
= +
HKB
HKDN
HKi
iB
iD
x
x
x
x
,
,1
,
,
, minαHK
iHKi
K
K=,α
⋅+
⋅⋅
=+
+
HKB
HKDN
HKi
HKB
HKDN
HKiiF
iD
Bx
Dx
Bx
DxFx
Dx
,
,1
,
,
,1
,,
,
min
min
1 α
α
Convince yourself and
derive for
iBBx ,
Minimum reflux ratio analysisMinimum 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
Minimum reflux ratio analysisMinimum 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
Minimum reflux ratio analysisMinimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: presence of heavy and light non-distributed
components
Minimum reflux ratio analysisMinimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: only light non-distributed
components
Minimum reflux ratio analysisMinimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: only heavy non-distributed
components
Minimum reflux ratio analysisMinimum reflux ratio analysis
F, zf
condenser
boiler
Invariant zones: no non-distributed
components
Minimum reflux ratio analysis:Minimum reflux ratio analysis:
Underwood equationsUnderwood equations
∑ −=−
i HKi
iFHKi xq
φαα
,
,,)1(
∑ −==+
i HKi
iDHKi
m
x
D
VR
φαα
,
,,1
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
φ
φ
Minimum reflux ratio analysis:Minimum reflux ratio analysis:
Underwood equationsUnderwood equations∑ −
=−i HKi
iFHKi xq
φαα
,
,,)1(
Minimum reflux ratio analysis:Minimum reflux ratio analysis:
Underwood equationsUnderwood equations∑ −
=i HKi
iFHKi x
φαα
,
,,0
1.48
Minimum reflux ratio analysis:Minimum reflux ratio analysis:
Underwood equationsUnderwood equations∑ −
==+i HKi
iDHKi
m
x
D
VR
φαα
,
,,1
Minimum reflux ratio analysis:Minimum reflux ratio analysis:
Underwood equationsUnderwood equations∑ −
==+i HKi
iDHKi
m
x
D
VR
φαα
,
,,1
1.48
2.33
xi
1 2 3 4 5 6 7 8 9 10
hexane LKheptane HK
octanepentane
Feed stage
Distribution of components in Distribution of components in
multicomponentmulticomponent distillation processdistillation process