Liquid-liquid extraction

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Liquid-liquid extraction
Basic principles
In liquid-liquid extraction, a soluble component (the solute) moves
from one liquid phase to another. The two liquid phases must be
either immiscible, or partially miscible.
• usually isothermal and isobaric
• can be done at low temperature (good for thermally fragile
solutes, such as large organic molecules or biomolecules)
• can be very difficult to achieve good contact between poorly
miscible liquids (low stage efficiency)
• extracting solvent is usually recycled, often by distillation
(expensive and energy-intensive)
• can be single stage (mixer-settler) or multistage (cascade)
Extraction equipment
Batch:
Continuous:
single-stage:
column:
separatory funnel
mixer-settler
rotating-disk contacter
a. agitator; b. stator disk
Design
Mixer-settlers, both as stand-alone and as incolumn type, are offered for special
applications. As implied by the name, the
mixer-settler-column is a series of mixersettlers in the form of a column. It consists of a
stages installed one on top of the
Mixer-settlers operate withnumber
a purelyofstageother,there
eachishydraulically
separated, and each
wise contact. After every mixer
a
with
a mixing
ettler. Mixer-settlers can be
operated
inand
a settling zone (see below).
This design
enables the elimination of some of
multistage, co- or countercurrent
fashion.
the main disadvantages of conventional mixersettlers, whilst maintaining stage-wise phase
contact.
The mechanical design of the mixer-settlercolumn is comparable to the agitated ECR
Kühni column.
Key characteristics
For long residence times: >15 min
Extraction controlled by residence
time
Reactive extraction systems
Long phase separation
For extraction controlled by pH
(stage-wise pH adjustment)
Mixer-settler column
points
Separation of high-boiling products
or pollutants that are present in only low
concentrations
Separation of components with
similar boiling points or components forming
ction with high mass transfer and/or
azeotropes
ng physical properties, this is the
Separation of mixtures with
of choice. The geometry of the
thermally sensitive components
d compartments can be adapted for
Selective separation of single
ng hydrodynamic conditions. Other
components out of a complex mixture
atures are the special mixing turbines
Our portfolio includes a complete range of
perforated partition plates.
liquid-liquid-extraction equipment, enabling us
to provide you with the most appropriate
solution for your requirements. In addition to
agitated columns, it includes mixer-settlers and
packed columns.
Design
The agitated Kühni column has a simple and
robust design. The drive unit and the shaft are
supported at the top of the column, allowing
you to use all common types of shaft seals
(stuffing box, mechanical seals). In special
cases, the seal can be replaced by a magnetic
drive. Only radial slide-bearings are necessary
Agitated column
Packed extraction column
The ECP packed column is based on c
Packing
The special Sulzer extraction packing
Liquid distributors
In order to create an even liquid flow
Main benefits
High specific throughput fac
Small column diameters
Revamp of existing columns
Use in cases of difficult phy
Low density difference < 50
Low interfacial tension: < 2
Tendency to form emulsion
Reliable scale-up
Stream labeling
feed
solvent
raffinate
mixture
extract
mixer
E, yA,1
settler
F, xA,0
1
Feed (F) contains solute A (xA) dissolved in
diluent D (xD = 1 – xA).
S≠E
solvent flow rate
Solvent (S) extracts A (yA), creating the product = FS = constant
F≠R
diluent flow rate
= FD = constant
extract stream (E). The depleted feed becomes
the product raffinate stream (R).
Equilibrium (no longer VLE!) is defined by the
distribution ratio, Kd:
N
S, yA,N+1
R, xA,N
Kd = yA/xA
Note that yA does not refer to gas composition.
Usually specified:
yA,N+1, xA,0, FD/FS and xA,N.
McCabe-Thiele analysis:
Counter-current extraction with immiscible liquids
Equation of the operating line:
FD
FD
Y=
X + (Y1 X )
FS
FS 0
•X0
(analogous to operating line for
stripper column).
1•
(FD/FS)max gives FS,min for N = ∞.
For dilute solutions,
R
R
y = x + (y1 - x0 )
E
E
Can also use Kremser eqns, if solutions
are dilute and equil. line is straight.
éæ mE ö æ y N+1 - y 0 ö mE ù
ln êç 1+
ú
ç
÷
÷
R
y
y
R
è
ø
è
ø
ê
úû
1
0
N= ë
ln R
mE
(
)
2•
3•
•(X0,Y1)
•
•
N=3
•(XN,YN+1)
Cross-flow cascade
From mass balance around stage j:
R
R
y j = - xi + (y j,in + x j -1)
Ej
Ej
Figure 13-8 Cross-flow cascade
From Separation Process Engineering, Third Edition by Phillip C. Wankat
(ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights reserved.
N=3
• Increase overall efficiency by introducing
fresh extracting solvent at each stage.
• Each stage has its own mass balance and
operating line
• Uses much more solvent than countercurrent cascade (requires much more
solvent recovery)
• A mixer-settler is just one cross-flow stage.
(x1,y1) •
(x2,y2)
•
(x3,y3)•
•
x3
•(x2,y3,in)•(x1,y2,in)
•(x0,y1,in)
Dilute fractional extraction
A prefers solvent 1 (“extract”)
1
F
zA
zB
E
R
E
R
B prefers solvent 2 (“raffinate”)
Kd,A = yA/xA > 1
Kd,B = yB/xB < 1
N
solvent 1
yA,N+1 = 0
yB,N+1 = 0
raffinate
xA,N
xB,N
absorbing
section
the feed contains two important
solutes (A, B), and we want to
separate them from each other.
Choose two solvents:
solvent 2
xA,0 = 0
xB,0 = 0
extract
yA,1
yB,1
stripping
section
A common situation:
McCabe-Thiele analysis: dilute fractional extraction
Operating lines intersect at feed
composition (not shown, may be
very large).
One operating line for each solute i, in
each section of the column (i.e., 4 total).
Top operating lines (absorbing section):
R
R
yi = xi + (yi,1 - xi,0 )
E
E
•
•5
R
R
y i = xi + (y i,1 - xi,0 )
E
E
Equilibrium data is different for each
solute (use separate McCabe-Thiele
diagrams!)
If yA,1 and xA,N are specified, and NF is
known, use M-T diagram to obtain N, then
use trial-and-error to find xB,0 and xB,N+1
3
•
•
Bottom operating lines (stripping section):
•
2
•
•
yA,1•
• N = 4,
F
feed stage
1 •
•6
•
•
xA,N
If yA,1 and xB,N are specified, vary NF (trial-and-error)
until N is the same for both solutes.
Center-cut extraction
When there are 3 solutes: A, B and C,
solvent 1
+A
solvent 2
solvent 3
+B
solvent 2
solvent 3
solvent 2
+C
and B is desired
(A and C may be > 1 component each)
Requires two columns:
• column 1 separates A from B+C
• column 2 separates B from C
F
zA, zB, zC
Requires three extracting solvents:
A prefers solvent 1 over solvent 2
B, C prefer solvent 2 over solvent 1
B prefers solvent 3 over solvent 2
C prefers solvent 2 over solvent 3
solvent 1
solvent 2
+B+C
Partially miscible solvents
• There are two liquid phases
• Each phase is a ternary (3-component)
mixture of solute A, diluent D and
solvent S
• Ternary equilibrium diagrams have 3
axes: usually, mole or mass fractions of
A, D, and S
• Literature data is commonly presently
on an equilateral triangle diagram (note
NO origin)
• Each axis is bounded 0 ≤ x ≤ 1
• Miscibility boundary = equilibrium line
(depends on T, P)
Figure 13-14 Effect of temperature on equilibrium of
methylcyclohexane-toluene-ammonia system from
Fenske et al., AIChE Journal, 1,335 (1955), ©1955, AIChE
From Separation Process Engineering, Third Edition by Phillip C. Wankat
(ISBN: 0131382276) © 2012 Pearson Education, Inc. All rights reserved.
Reading ternary phase diagrams
Consider the point M:
•
water content (xA) is ?
0.19
ethylene glycol content (xB) is ? 0.20
furfural content (xC) is ?
0.61
check: xA + xB + xC = 1
Read the mole/mass fraction of each
component on the axis for that component,
using the lines parallel to the edge opposite the
corner corresponding to the pure component.
The mixture M lies inside the miscibility
•
boundary, and will spontaneously separate
into two phases. Their compositions (E and
R) are given by the tie-line through M.
region of partial miscibility A-C
The compositions of E and R converge at the plait point, P (i.e., no separation).
A 2-component mixture of furfural and water is partially miscible over the composition
range from about 8 % furfural to 95 % furfural. Separation by extraction requires a
furfural/water ratio in this range (otherwise – single phase).
•
Right-triangle phase diagrams
Raffinate (diluent-rich): xA + xB + xC = 1
Extract (solvent-rich): yA + yB + yC = 1
We need to specify only two of the
compositions in order to describe each
liquid phase completely .
This can be shown on a right-triangle phase
diagram, which is easy to plot and read.
• raffinate compositions are
represented by coordinates (xA, xB)
Figure 13-12 Equilibrium for water-chloroform-acetone at 25°C and 1 atm
• extract compositions are
represented by coordinates (yA, yB)
Vertical axis corresponds to both xA and yA.
Horizontal axis corresponds to both xB and yB
Q: Where does pure C appear on this diagram?
From Separation Process Engineering, Third Edition by Phillip C. Wankat
(ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights
reserved.
More tie-lines can be obtained by trial-anderror, using the conjugate line.
Ex.: find the tie-line that passes through M.
Obtaining the conjugate line
Each point on the conjugate line is composed of
- one coordinate from the extract side of the
equilibrium line
- one coordinate from the raffinate side of the
equilibrium line
•
•
•
•
•
•
•
On this graph, which component is the diluent? which is the solute?
Hunter-Nash analysis of mixer-settler
F
S
R
E
M
mixer
Why does F appear on or
near the hypotenuse?
settler
Flow rates of E and R are related
by mass balance.
Compositions of E and R are also
related by equilibrium.
Overview of solution using RT
diagram:
1. Plot F and S and join with a
line.
2. Find mixing point, M, which
is co-linear with F and S.
3. Find tie-line through M; find
E and R at either end (colinear with M).
4. Find flow rates of E and R.
Why does S appear at or
near the origin?
coord.:
(yD,yA)
mixing line
•F
E
•
•
M
tie-line
•S
coord.:
(xD,xA)
•
R
Co-linearity
F
S
M
Why are F, S and M co-linear on the Hunter-Nash diagram?
mixer
solve for coordinates of M: (xA,M, xD,M)
TMB:
F+S=M
CMBA: FxA,F + SxA,S = MxA,M = (F + S)xA,M
CMBD: FxD,F + SxD,S = MxD,M = (F + S)xD,M
F xA,M - xA,S xD,M - xD,S
=
=
S xA,F - xA,M xD,F - xD,M
rearrange
FxA,F + SxA,S
xA,M =
F +S
xD,M =
xA,M - xA,S
xD,M - xD,S
=
FxD,F + SxD,S
F +S
xA,F - xA,M
xD,F - xD,M
CMBD
CMBA
• F (xD,F, xA,F)
•
M (xD,M, xA,M)
• S (xD,S ,xA,S)
slope from slope from
M to S
F to M
Therefore F, S and M are co-linear. To locate
M on the FS line: calculate either xA,M or xD,M.
The lever-arm rule
Another way to locate M:
F
S
•
MS
M
•
MF •
similar triangles
To calculate flow rates E and R:
E
•
EM
similar triangles
M
• MR
FxA,F + SxA,S = MxA,M
M=R+E
FxA,F + (M – F)xA,S = MxA,M
R xA,M - y A,E ME
=
=
M xA,R - y A,E RE
F(xA,F - xA,S) = M(xA,M - xA,S)
F xA,M - xA,S MS
=
=
M xA,F - xA,S FS
R
•
Your choice! Use mass balances, or
measure distances and use lever-arm rule.
Hunter-Nash analysis of cross-flow cascade
S2
S1
F = R0
1
E1
R1
2
R2
E1
Treat each stage as a mixer-settler.
• each Ri, Si pair creates a mixing line
• find each Ei, Ri pair using a tie-line
F
•
E1
•
•
M1
E2
•
•
S
•
M2
•
R1
•
R2
Hunter-Nash analysis of counter-current cascade
F
R1
EN
M
S
mixer
separator
(column)
E and R are both points on the
equilibrium line. But they are not
related by the same tie-line.
Overview of solution using RT
diagram:
1. Plot F and S and join with a line.
2. Find mixing point, M, which is colinear with F and S.
3. Plot specified xA,1 on raffinate
side of equilibrium line to find R1.
4. Extrapolate R1M line to find EN.
5. Find flow rates of E and R.
mixing line
EN
•
xA,1
•
•S
•
M
NOT a
tie-line
•F
•
R1
Stage-by-stage analysis
R1
xA,1
S = E0
yA,0
stage 1 TMB:
E0 – R1 = E1 – R2 = E2 – R3 etc.
1
E1
R2
constant difference in flow rates of passing streams
= Ej – Rj+1 = constant
stage 1 CMBA:
N
F = RN+1
xA,N+1
E 0 + R2 = E 1 + R1
E0yA,0 + R2xA,2 = E1yA,1 + R1xA,1
E0yA,0 – R1xA,1 = E1yA,1 – R2xA,2 = etc.
EN
yA,N+1
constant difference in compositions of passing streams
net flow of A:
xA,  = EjyA,j – Rj+1xA,j+1
net flow of D:
xD,  = EjyD,j – Rj+1xD,j+1
The difference point
Define a difference point, , with coordinates (xA, , xD, ):
xA,D =
E0 y A,0 - R1xA,1
D
xD,D =
E0 y D,0 - R1xD,1
D
 does not necessarily lie inside the RT graph.
All pairs of passing streams Ej, Rj+1 are co-linear with .
Using the -point to step off stages on Hunter-Nash diagram:
• using the specified location of R1 (as xA,1), can find E1 (use tie-line);
• given the location of E1, can find R2 (use );
• given the location of R2, can find E2 (use tie-line);
• given the location of E2, can find R3 (use );
• and so on, until desired separation is achieved.
First, need to locate . It may be on either side of the Hunter-Nash diagram.
Finding the -point
Procedure:
1. Plot F (= RN+1), S = (E0). Locate M.
2. Plot R1 and locate EN.
last mixing line


F = RN+1
•
EN
•
•
M
•
S = E0
3. Extend the lines joining E0-R1,
and EN-RN+1, to find at the
intersection point.
•
R1
first mixing line
4. All intermediate mixing lines
must pass through .
Stepping off stages on the H-N diagram
Procedure:
N=3
1. Use R1 and conjugate line to find E1
2. Use E1 and D-point to find R2
3. Use R2 and conjugate line to find E2
4. Use E2 and D-point to find R3
EN
•
E2
•
E1
•
•
S = E0
3. Use R3 and conjugate line to find E3
•
M
F = RN+1
•
•
R3
•
R2


•
R1
Stop when you reach or pass EN.
Using McCabe-Thiele diagram instead of Hunter-Nash
• M-T diagram can be used with much greater accuracy than H-N diagram
• Need to transfer ternary equilibrium data from RT diagram
• Need to obtain the operating line
Transferring equilibrium data from RT diagram
Each tie-line represents a pair of equilibrium streams
• extract composition represented by yA
• raffinate composition represented by xA
Each (xA, yA) pair is a point on the M-T equilibrium line
1
extract
compositions
A
yA
xA
•
•
0 •
0
•
P
•
•
raffinate
compositions
•
•
•
•
D
1
•
P
equilibrium line ends at P
•
yA
•
1
•
0•
0
xA
1
Obtaining the M-T operating line
R1
xA,1
Mixing lines represent passing streams.
All mixing lines lie between the limits:
(x1, y0) and (xN+1, yN)
S = E0
yA,0
A
1
yN
xN+1
x1
y0•
E0
EN
•
•
M
RN+1
•
•R
WAIT! In general, operating line is
not straight.
Plot arbitrary intermediate mixing
lines to obtain more points.
1
D
N
F = RN+1
xA,N+1
1
EN
yA,N+1
•
•
yA
•
Note: passing streams are (xj+1, yj) instead of
(xj, yj+1) as in distillation, simply due to our
labeling convention (feed enters at stage N).
•
•
P
•(xN+1, yN)
•
•
0•
•
0 (x1, y0)
xA
1
Choice of extracting solvent flow rate
• As S increases, separation improves, but extract becomes more dilute
• As S decreases, N must increase to maintain desired separation
• Smin achieves the desired separation with N = ∞
A
F
•
M• •
Mmin
•
S
•
Mmax
D
• as M moves towards S, (S/F) increases
(lever-arm rule)
• when M reaches the equilibrium line, all
feed dissolves in extracting solvent (Mmax)
• as M moves towards F, (S/F) decreases
• before reaching the equilibrium line, there is
usually a pinch point (Mmin)
It is not easy to locate this pinch point on a McCabe-Thiele diagram, since the
operating line curvature changes as S changes.
On a Hunter-Nash diagram, Dmin (corresponding to Mmin) occurs when a mixing
line and a tie-line coincide.
Minimum solvent flow rate
On H-N diagram whose tie-lines have negative slopes:
E
N,min
•
•
•
•
min
•
R
•
S
Plot S = E0, F = RN+1, R1
Join S and F
Extend SR1 mixing line
Locate several tie-lines
Extend tie-lines to the SR1
mixing line
6. Tie-line with furthest intersection
from S locates Dmin
7. Mixing line from Dmin through F
locates EN,min
8. Connecting R1 and EN,min
completes the mass balance
9. Mmin is located at the intersection
of SF and R1EN,min
F
M
1.
2.
3.
4.
5.
D
min
1
10. (S/F)min = (FMmin)/(SMmin)
Rule-of-thumb: (S/F)act ~ 1.5 (S/F)min
Minimum solvent flow rate
On H-N diagram whose tie-lines have positive slopes:
•
F
•
min
M
min
R
1
•
S
D
•
N,min
•
E
•
Strategy:
1. Plot S = E0, F = RN+1, R1
2. Join S and F
3. Extend SR1 mixing line
4. Locate several tie-lines
5. Extend tie-lines to SR1
mixing line
6. Find tie-line which gives
closest intersection to S;
this locates Dmin
7. Draw mixing line from Dmin
through F to locate EN,min
8. Connect R1 and EN,min to
complete mass balance
9. Mmin is at the intersection of SF and R1EN,min
10. (S/F)min = (FMmin)/(SMmin)
Two feed counter-current column
S
E0 = S
R1
F1
FT
F2
1
R
mixer 1
R1
M
mixer 2
EN
separator
E
F2
Feed balance: F1 + F2 = FT
R
E
N
F1 = RN+1
EN
Overall balance:
• hypothetical mixed feedstream FT is co-linear with F1, F2
Stage-by-stage analysis:
• mass balance changes where F2 enters the column
• upper and lower sections have different sets of operating
lines ➙ different D-points
Hunter-Nash analysis of 2-feed column
Overall balance:
1. Plot F1 and F2. Locate FT (colinear with F1 and F2).
2. Plot S . Locate M (co-linear with
S and FT).
F1
•
FT
•
EN
•
•
M
•
S = E0
3. Plot R1. Locate EN (co-linear with
R1 and M).
F2
•
•
R1
1. Calculate flow rates R1 and EN.
Stage-by-stage analysis
E0 = S
R1
1
E
F2
E
N
F1 = RN+1
R1, E0, D1 are co-
Balance around bottom of column:
EN – RN+1 = Ek – Rk+1 = D2 ➙ RN+1, EN, D2 are co-linear
➙
k
R
➙
Overall balance:
F2 + RN+1 + E0 = EN + R1
F2 = (EN – RN+1) + (R1 – E0) = D1 + D2
j
R
Balance around top of column:
R1 – E0 = Rj+1 – Ej = D1
linear
EN
F2, D1, D2 are co-linear
“feed-line”
D2 is located at the intersection of two mixing lines:
RN+1, EN, D2 and
F2 , D 1 , D 2
Need another line to locate D1:
TMB:
FT = F1 + F2 = EN + (R1 – E0) = EN + D1
➙ FT, EN, D1 are co-linear
D1 is located at the intersection of two mixing lines:
R1, E0, D1
and
FT, EN, D2
Note: D1 and D2 may be on different sides of the phase diagram.
Using the feed-line
1. Locate D1 at intersection of R1E0
and ENFT
2. Locate D2 at intersection of F2D1
and ENRN+1
3. Step off stages, initially using D1 to
generate the first mixing lines
•
4. Identify the optimum feed stage
when the mixing line crosses the feed
line, F2D1D2
EN
F1
feed line
•
•
•
R2
R1
D1
•
F2
•
•
•
S
FT
•
•
E2 •
•
E1
M
D2
5. When the tie-line crosses the feed
line, the next mixing line will be
generated using D2
Countercurrent liquid-liquid extraction with reflux
E0 = S
R1
1
makeup solvent
How to increase yA,N?
need to increase xA,N+1
make RN+1 an reflux stream
1
“extract reflux”
R
E
E0
R1
(no benefit to raffinate reflux)
R
E
R
E
F
N
F1 = RN+1
xA,N+1
EN
yA,N
Turning extract into raffinate :
• extract is mostly solvent
• raffinate is mostly diluent
We need to remove solvent,
e.g., distillation, stripping
recovered
solvent
N
RN+1
In a conventional liquid-liquid extraction column:
reflux
yA,N is related by equilibrium to xA,N
xA,N depends on xA,N+1
dilute feed gives dilute extract
highest yA,N obtained with S ≈ Smin , but this requires very large N
SR
EN
solvent
separator
Q
PE
product extract
Analogy to distillation reflux
V1
1
recovered
solvent
N
L0
SR
D
RN+1
Saturated liquid reflux stream is
obtained by condensing V1 (vapor
stream rich in A) to give L0 (liquid
stream rich in A)
External reflux ratio = L0/D
Internal reflux ratio = L/V
reflux
EN
solvent
separator
Q
PE
product extract
Extract reflux stream is obtained by removing
solvent from EN (extract stream rich in A and
solvent) to give RN+1 (raffinate stream rich in A
and depleted in solvent)
External reflux ratio = RN+1/PE
Internal reflux ratio = RN+1/EN
Stage-by-stage balances
Similar to 2-feed liq-liq extraction column:
- two D-points (mass balance above and below feed stage)
- if F, E0, R1 and RN+1 are specified, same stage-by-stage analysis
But RN+1 is an internal stream, usually not specified.
Usually specified:
F, xA,F, xD,F
yA,0, yD,0
xA,1
xA,PE, xD,PE
rates)
yA,SR, yD,SR
RN+1/PE
➙ plot F
➙ plot E0
➙ plot R1 on sat’d raffinate curve
➙ plot PE (same location as RN+1 and Q, different flow
➙ plot SR
Mass balance: solvent separator
SR
EN
RN+1
EN = Q + SR
solvent
separator
= RN+1 + PE + SR
Q
PE
EN is co-linear with Q and SR.
EN also lies on sat’d extract line.
Obtain EN/SR from lever-arm rule.
We will also need RN+1/SR:
EN RN+1 PE
=
+
+1
SR
SR SR don’t know
EN SR RN+1
S
´
=
+ 1+ R
SR PE
PE
PE
A
•
EN
•
SR •
•
E0
PE, Q, RN+1
F
•
•
R1
D
æ RN+1 ö
+1
PE ÷ø
R
E
P
SR çè
➙ N+1 = N - E - 1
=
SR
SR SR
æ EN ö
PE
1
çè S ÷ø
R
Finding the D-points
D2 = EN - RN+1
D2xA,D2 = ENyA,N - RN+1xA,N+1
xA,D 2 =
EN y A,N - RN+1xA,N+1
EN - RN+1
We don’t know the individual
flow rates EN, RN+1, but we know
EN/SR and RN+1/SR. We can
calculate xA,D2 and thereby locate
D2 on the ENRN+1 line.
Locate D1 at the intersection
D2
of two mixing lines:
•
D1 = E0 - R1
•
D1
F = D1 + D2
Proceed to step off stages.
PE, Q, RN+1
•
F
•
••
SREN
E0•
•
R1

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