Multicomponent distillation - Department of Chemical Engineering

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8. Multi-component distillation
Prof. S. Scott, ChE 128, UC Santa Barbara
How many distillation columns are required?
One distillation column can be optimized to separate
one pair of volatile components.
D, xD,1
F
zF,1
zF,2
We can specify two fractional recoveries:
FR1 = (DxD,1)/(FzF,1) = 0.95
FR2 = (BxB,2)/(FzF,2) = 0.98
If the feed contains more than two volatile
components, we cannot specify the recoveries of the
additional components.
However, we can add more distillation columns.
B, xB,1
Distillation of multicomponent mixtures
1
Alternative?
1
1, 2
F
II
2
2,3,4
F
zF,1
zF,2
zF,3
zF,4
3
3,4
2
I
4
3
3, 4
III
Separation of C components requires (C1) distillation columns.
4
Key components
• Each column is designed to separate two components of adjacent relative
volatility. These components are the keys.
• All other components are non-keys.
design for
separation
component
a
1
1.5
Light non-key (LNK)
2
1.4
Light key (LK)
assume exclusively in distillate
specify recovery in distillate
3
1.3
Heavy key (HK)
specify recovery in bottoms
4
1.2
Heavy non-key (HNK)
5
1.0
Heavy non-key (HNK)
designation
assume exclusively in bottoms
• Distributions in the distillate and bottoms streams are specified for the two key
components.
• If we assume that the non-keys do not distribute, the overall mass balance is
easily solved.
Overall mass balance
with non-distribution of non-keys
If non-keys do not distribute, then
LNK:
Dxi,D = Fzi,F
HNK:
Bxi,B = Fzi,F
Define fi, di, and bi as molar flow rates of component i in feed, distillate and
bottoms, respectively.
Ex.: 100 mol/h of an equimolar mixture of four components.
Assume 80 % recovery of LK in D, and 80 % recovery of HK in B.
component i zF,i
fi = F zF,i
di
xD,i = di/D
bi
xB,i = bi/B
LNK
0.25
25
25
0.50
0
0
LK
0.25
25
20
0.40
5
0.10
HK
0.25
25
5
0.10
20
0.40
HNK
0.25
25
0
0
25
0.50
Σdi = D = 50
ΣxD,i = 1
Σbi = B = 50
ΣxB,i = 1
Stage-by-stage calculations
For a specified external reflux ratio R = L0/D and feed quality q, and assuming constant
molal overflow (CMO), first perform the overall column balance to obtain D. Then,
L = (L0/D) D
V=L+D
⇒
L/V
L̅ = L + q F
V̅ = V + (1 – q) F
⇒
L̅/V̅
Algebraic solution, starting at top of column:
1. yi,1 = xi,D
2. xi,1 = yi,1 / Ki,1
3. yi,2 = (L/V) xi,1 + (1 – L/V) xD
section)
(total condensor)
(VLE)
(mass balance, rectifying
Repeat steps (2) and (3) down to feed stage. Then, change mass balance equation to
yi,k+1 = (L̅/V̅ ) xi,k + (1 – L̅/V̅ ) xB
We just need to find Ki for each stage.
Obtaining Ki(T) for each stage
• Consider a ternary system with components A, B, C
• Assume constant relative volatility
• Choose B as the reference compound
a AB
K A y A / xA
=
=
K B y B / xB
xA,j =
K B,j =
y A,j
K A,j
y A,j
a AB
=
+
a BB = 1
y A,j
a ABK B,j
y B,j
a BB
+
yC,j
a CB
xA,j + xB,j + xC,j =
æ y i,j
ö
= åç
a iB ÷ø
i è
a CB =
y A,j
a ABK B,j
+
KC
KB
y B,j
a BBK B,j
y A,j
xA,j =
+
y C,j
a CBK B,j
a AB
æ y i,j
ö
å çè a ÷ø
iB
i
IF relative volatility is not constant, use geometric average, (a1aN)½. Or, find Tj
from dewpoint calculation (starting with xD,i), then find Ki(Tj) using VLE.
=1
Trial-and-error solution
Perform stage-by-stage analysis, from top to bottom of column.
xHK,N+1 ≥ xHK,B
xLK,N+1 ≤ xLK,B
must be satisfied simultaneously
• if not true, choose different starting concentrations at the top of the column,
and repeat
• cannot assume xD,HNK = 0, but no way to guess
• works best if there are no HNKs. If there is no LNK, can start at bottom of
column and work to the top
• not obvious where the optimum feed stage is
• very tedious, convergence is difficult
Approximate shortcut methods
for multicomponent distillation
• Suitable for preliminary designs
• Three-part F-U-G method:
1. Use Fenske equation to estimate Nmin (at total reflux)
2. Use Underwood equations (2) to estimate Rmin (with N = ∞)
3. Use Gilliland correlation to estimate Nactual and Nfeed (can also
use Kirkbride equation)
Operating at total reflux
• designate the keys A and B, and assume constant aAB
• according to the definition of equilibrium:
yi,1
1
xi,0
é yA ù
é xA ù
ê ú = a AB ê ú
ë y B ûN+1
ë xB ûN+1
(even though there is no liquid
stream with composition xi,N+1)
• at total reflux, the operating line is y = x
yi,j+1 = xi,j
so
yi,N+1 = xi,N
yi,N+1
N
xi,N
N+1
é xA ù
é xA ù
ê ú = a AB ê ú
ë xB ûN
ë xB ûN+1
• move up to stage N:
é yA ù
é xA ù
é xA ù
2
ê ú = a AB ê ú = a AB ê ú
ë y B ûN
ë xB ûN
ë xB ûN+1
• continue to the top of the column
Estimating Nmin
• at the top of the column
Fenske
equation
• solve for Nmin
é xA ù é y A ù
Nmin é xA ù
ê ú = ê ú = a AB ê ú
ë xB û0 ë y B û1
ë xB ûN+1
Nmin
éæ x ö
æ xA ö ù
A
ln êç ÷
ú
ç
÷
êëè xB ø 0 è xB ø N+1 úû
=
lna AB
• can also write in terms of fractional recoveries (FR)
Nmin
éæ FR ö
ù
æ 1- FRB ö
A
ln êç
ú
÷
ç FR ÷
1FR
è
êëè
A ø distillate
B ø bottoms ú
û
=
ln a AB
Other uses of the Fenske equation
Once Nmin is known,
• check non-distribution of non-keys, or estimate their distribution:
FRC,distillate =
N
a CB
min
æ FRB ö
Nmin
+
a
CB
ç 1- FR ÷
è
ø
B bottoms
• estimate optimum feed stage at total reflux:
NF,min
éæ x ö
æ zA ö ù
A
ln êç ÷
çz ÷ ú
x
êëè B ø 0 è B ø F úû
=
lna AB
Pinch points in binary distillation
Recall for binary distillation:
Rmin
Binary composition profile
•xD
xD • •
•
•
pinch
point
•
x
zF•
•zF
• • • •
•
plateau of
•
constant
•
composition
•
•
• xB
Using Rmin generates
a pinch point at the
feed stage
xB•
•
condensor
•
•
feed
•
reboiler
stage number
• near the pinch point, composition changes little from stage to stage
- passing streams are very close to equilibrium
- no change in temperature between stages
Behavior in ternary systems
• in the presence of a third component whose
composition does change through the
feed stage, the pinch point moves
Ternary composition profile
LK-HK-HNK
1
stripping
rectifying
Consider LK
• LK is the most volatile component in the
system on every stage
• there is almost no LK at the reboiler stage
xi
Consider HNK
• HNK is the least volatile component in the
system on every stage
• almost all HNK is found at the reboiler stage
• finite HNK at feed stage, drops off rapidly above
0
C
F
stage number
R
Behavior of HK
Consider HK:
Ternary composition profile
LK-HK-HNK
1
pinch
point
• HK behavior is most complex
(HK-LK), with HK the less volatile component xi
• near reboiler, distillation is almost binary
~binary
~binary
• above feed stage, distillation is almost binary
(HNK-HK), but HK is the more volatile
component
• HNK goes through a maximum in the stripping
section, creating a pinch point there
0
C
F
stage number
R
Pinch points in multicomponent distillation
pinch
F
LK, HK,
HNK
pinch
F
LNK, LK,
HK
pinch
F
LNK, LK,
HK, HNK
pinch
• LK-HK-HNK system
• LNK-LK-HK system
• LNK-LK-HK-HNK
has a pinch point in
has a pinch point in
has pinch points in
the stripping section
the rectifying section
both sections
Graphical methods are not helpful in finding these pinch points.
Second Underwood equation
CMB in rectifying section at minimum reflux:
at the pinch point:
xi,j = xi,j +1
VLE:
y i,j +1 = K i xi,j +1
relative volatility:
a i = K i Kref
Lmin
Vmin y i,j +1 =
y i,j +1 + Dxi,D
a i Kref
Vmin y i,j +1 =
a i Dxi,D
Lmin
ai VminK ref
Vmin y i,j +1 = Lmin xi,j + Dxi,D
y i,j +1 = a i Kref xi,j +1
æ
Lmin ö
Vmin y i,j +1 ç 1= Dxi,D
÷
è a iVminK ref ø
define: f =
a i Dxi,D
2nd Underwood equation: Vmin = åVmin y i,j +1 = å
ai - f
i
i
Lmin
VminK ref
First Underwood equation
in the stripping section:
assume
a i Bxi,B
-Vmin = åVmin y i,k +1 = å
i
i ai - f
f =f
DVfeed
é a i Dxi,D a i Bxi,B ù
= Vmin -Vmin = å ê
+
ú
a
f
a
f
i ê
úû
i
ë i
é a i Fzi,F ù
= åê
ú
i ê
ë a i - f úû
1st Underwood equation:
DVfeed
feed quality definition:
DVfeed = F(1- q)
searching forφ:
with C components, the 1st Underwood equation has C roots, one between
each adjacent pair of a-values, and one between 0 and the smallest a-value.
We need the root aHK < φ < aLK
• binary system: use quadratic equation to solve forφ
• trial-and-error: use ½(aHK + aLK) as initial guess
Estimating Rmin
Onceφis known, obtain Vmin using the 2nd Underwood equation.
a i Dxi,D
Vmin = åVmin y i,j +1 = å
ai - f
i
i
To get xi,D values for non-keys:
• assume non-distribution
OR
• assume distribution is the same at minimum reflux and total reflux
i.e., use Fenske equation
Lmin = Vmin – D
Rmin = Lmin/D
Gilliland correlation
We have Nmin (Fenske) and Rmin (Underwood).
R is often specified as a multiple of Rmin, e.g., R = 1.5 Rmin
y=
y = 1- e
éæ 1+54.4 x ö æ x-1ö ù
êç
÷ø çè 0.5 ÷ø ú
11+117.2x
è
x
ë
û
x=
Figure 7-3 Gilliland correlation as modified by Liddle (1968);
Reprinted with permission from Chemical Engineering, 75(23), 137 (1968), copyright 1968, McGraw-Hill.
From Separation Process Engineering, Third Edition by Phillip C. Wankat
(ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights reserved.
Optimum feed stage
• obtain Nmin and NF,min from Fenske eqn
• obtain N from Gilliland correlation
D
Assume relative position of the feed stage
is the same as it was at total reflux:
NF,min
Nmin
=
NF
N
D
F
F
NF
NF,min
Remember! Nmin and N include partial
reboiler (and partial condensor).
Works well for symmetrical columns
(feed stage close to middle).
B
B
Kirkbride equation
Better feed stage estimation for
unsymmetrical columns:
Nrectifying
Nstripping
éæ z ö æ x ö 2 æ ö ù
B
= êç HK ÷ ç LK ,B ÷ ç ÷ ú
êè zLK ø è xHK ,D ø è D ø ú
F
ë
û
F
D
0.206
D
where Nrect and Nstrip do not include the
feed stage:
B
F
N = Nrect + Nstrip + 1
B
Sandwich components
At the beginning of this section, we specified that the key components should
have adjacent relative volatilities. What if there is a non-key with intermediate
volatility (i.e., between the two key components)?
This component is called a “sandwich” component.
component
a
1
1.5
Light non-key (LNK)
2
1.4
Light key (LK)
3
1.3
Sandwich non-key
4
1.2
Heavy key (HK)
5
1.0
Heavy non-key (HNK)
designation
Sandwich components tend to concentrate in the
middle of the column, and can cause flooding even
when present at trace concentrations.
If unavoidable, use side-streams to withdraw them.
D
F
S
B

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