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:
mixer-settler
column:
separatory funnel
rotating-disk contacter
a. agitator; b. stator disk
single-stage:
Continuous:
Mixer-settler column
Mixer-settlers operate with a purely stage-
wise contact. After every mixer there is a
settler. Mixer-settlers can be operated in a
multistage, co- or countercurrent fashion.
Design
Mixer-settlers, both as stand-alone and as in-
column type, are offered for special
applications. As implied by the name, the
mixer-settler-column is a series of mixer-
settlers in the form of a column. It consists of a
number of stages installed one on top of the
other, each hydraulically separated, and each
with a mixing and settling zone (see below).
This design enables the elimination of some of
the main disadvantages of conventional mixer-
settlers, whilst maintaining stage-wise phase
contact.
The mechanical design of the mixer-settler-
column 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)
For batch extraction
Agitated column
In extraction with high mass transfer and/or
changing physical properties, this is the
column of choice. The geometry of the
agitated compartments can be adapted for
changing hydrodynamic conditions. Other
main features are the special mixing turbines
and the perforated partition plates.
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
azeotropes
Separation of mixtures with
thermally sensitive components
Selective separation of single
components out of a complex mixture
Our portfolio includes a complete range of
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
inside the column, which are accessible
Packed extraction column
The ECP packed column is based on current state
Packing
The special Sulzer extraction packing reduces the back
Liquid distributors
In order to create an even liquid flow velocity profile at either end of the packed bed, both liquid phases are distributed o
Main benefits
High specific throughput facilitation:
Small column diameters
Revamp of existing columns to increase capacity
Use in cases of difficult physical properties:
Low density difference < 50 kg/m3
Low interfacial tension: < 2
Tendency to form emulsions
Reliable scale-up
Stream labeling
1
N
F, xA,0
S, yA,N+1
E, yA,1
R, xA,N
Usually specified:
yA,N+1, xA,0, FD/FS and xA,N.
Feed (F) contains solute A (xA) dissolved in
diluent D (xD = 1 – xA).
Solvent (S) extracts A (yA), creating the product
extract stream (E). The depleted feed becomes
the product raffinate stream (R).
Equilibrium (no longer VLE!) is defined by the
distribution ratio, Kd:
Kd = yA/xA
Note that yA does not refer to gas composition.
F ≠ R
diluent flow rate
= FD = constant
S ≠ E
solvent flow rate
= FS = constant
feed mixture
extract
raffinate
mixer settler
solvent
McCabe-Thiele analysis:
Counter-current extraction with immiscible liquids
Y
X
N = 3
•(X0,Y1)
•(XN,YN+1)
•
•
• 1
• 2
• 3
•X0
(FD/FS)max gives FS,min for N = ∞.
Can also use Kremser eqns, if solutions
are dilute and equil. line is straight.
For dilute solutions,
y =
R
Ex + (y1 −
R
Ex0)
Y =
FDFSX + (Y1 −
FDFSX0)
Equation of the operating line:
(analogous to operating line for
stripper column).
N =
ln 1−mE
R
yN+1 − y
0
y1
− y0
+mE
R
ln RmE( )
Cross-flow cascade
• 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 counter-current cascade (requires much more solvent recovery)
• A mixer-settler is just one cross-flow stage.
From Separation Process Engineering, Third Edition by Phi l lip C. Wankat
(ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights reserved.
Figure 13-8 Cross-flow cascade
Y
X
N = 3
yj
= −R
Ej
xi+ (y
j ,in+R
Ej
xj−1)
From mass balance around stage j:
•(x0,y1,in) •(x1,y2,in) •(x2,y3,in)
• x3
• (x1,y1)
• (x2,y2)
• (x3,y3)
Dilute fractional extraction
A common situation:
the feed contains two important
solutes (A, B), and we want to
separate them from each other.
Choose two solvents:
A prefers solvent 1 (“extract”)
B prefers solvent 2 (“raffinate”)
Kd,A = yA/xA > 1
Kd,B = yB/xB < 1
1
N
F
zA
zB
solvent 1
yA,N+1 = 0
yB,N+1 = 0
solvent 2
xA,0 = 0
xB,0 = 0
extract
yA,1
yB,1
raffinate
xA,N
xB,N
E R
E R
ab
sorb
ing
sect
ion
st
rip
pin
g
se
ctio
n
One operating line for each solute i, in
each section of the column (i.e., 4 total).
McCabe-Thiele analysis: dilute fractional extraction
Y
X
•
xA,N
•
•
• 5
•
3
yA,1•
• •
•
•
2
1 6
•
• NF = 4,
feed stage
•
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
If yA,1 and xB,N are specified, vary NF (trial-and-error)
until N is the same for both solutes.
Operating lines intersect at feed
composition (not shown, may be
very large).
yi =
R
Exi + (yi,1 −
R
Exi,0 )
Top operating lines (absorbing section):
yi=R
Exi+ (y
i ,1−R
Exi ,0)
Bottom operating lines (stripping section):
Equilibrium data is different for each
solute (use separate McCabe-Thiele
diagrams!)
Center-cut extraction
When there are 3 solutes: A, B and C,
and B is desired
(A and C may be > 1 component each)
solvent 1
solvent 2 solvent 1
+ A
solvent 2
+ B + C solvent 3
solvent 2solvent 3
+ B
solvent 2
+ C
F
zA, zB, zC
Requires two columns:
• column 1 separates A from B+C
• column 2 separates B from C
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
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)
From Separation Process Engineering, Third Edition by Phi l lip C. Wankat
(ISBN: 0131382276) © 2012 Pearson Education, Inc. All rights reserved.
Figure 13-14 Effect of temperature on equilibrium of
methylcyclohexane-toluene-ammonia system from
Fenske et al., AIChE Journal, 1,335 (1955), ©1955, AIChE • Each axis is bounded 0 ≤ x ≤ 1
• Miscibility boundary = equilibrium line
(depends on T, P)
Consider the point M:
water content (xA) is ?
ethylene glycol content (xB) is ?
furfural content (xC) is ?
0.19
0.20
0.61
Reading ternary phase diagrams
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.
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).
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.
The compositions of E and R converge at the plait point, P (i.e., no separation). region of partial miscibility A-C
check: xA + xB + xC = 1
•
•
•
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 .
From Separation Process Engineering, Third Edition by Phi l lip C. Wankat
(ISBN: 0131382276) Copyright © 2012 Pearson Education, Inc. All rights
reserved.
Figure 13-12 Equilibrium for water-chloroform-acetone at 25°C and 1 atm
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)
• extract compositions are
represented by coordinates (yA, yB)
More tie-lines can be obtained by trial-and-
error, using the conjugate line.
Ex.: find the tie-line that passes through M.
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?
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 M E
R
mixer settler
S
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 (co-
linear with M).
4. Find flow rates of E and R.
mixing line
tie-line
• F
• S
• M
Flow rates of E and R are related
by mass balance.
Compositions of E and R are also
related by equilibrium.
Why does F appear on or
near the hypotenuse?
Why does S appear at or
near the origin?
E •
coord.:
(yD,yA)
•
R
coord.:
(xD,xA)
Co-linearity
Why are F, S and M co-linear on the Hunter-Nash diagram?
F M
mixer
S
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
solve for coordinates of M: (xA,M, xD,M)
xA,M
=Fx
A,F+Sx
A,S
F +S
xD ,M
=Fx
D ,F+Sx
D,S
F +S
F
S=xA,M
− xA,S
xA,F
− xA,M
=xD ,M
− xD,S
xD ,F
− xD ,M
slope from
M to S
slope from
F to M • F (xD,F, xA,F)
• S (xD,S ,xA,S)
• M (xD,M, xA,M) Therefore F, S and M are co-linear. To locate
M on the FS line: calculate either xA,M or xD,M.
xA,M
− xA,S
xD,M
− xD,S
=xA,F
− xA,M
xD,F
− xD,M
rearrange
CMBA CMBD
The lever-arm rule
S •
M •
F •
Another way to locate M:
MF
MS
F
M=xA,M
− xA,S
xA,F
− xA,S
=MS
FS
FxA,F + SxA,S = MxA,M
FxA,F + (M – F)xA,S = MxA,M
F(xA,F - xA,S) = M(xA,M - xA,S)
R
M=xA,M
− yA,E
xA,R
− yA,E
=ME
RE
M = R + E
Your choice! Use mass balances, or
measure distances and use lever-arm rule.
similar triangles
E •
M •
R •
To calculate flow rates E and R:
EM
MRsimilar triangles
Hunter-Nash analysis of cross-flow cascade
1 F = R0 R1
S1
E1
2
S2
E1
R2
F •
• S
E2 •
• R2
Treat each stage as a mixer-settler.
• each Ri, Si pair creates a mixing line
• find each Ei, Ri pair using a tie-line
E1 •
• R1
• M1
• M2
Hunter-Nash analysis of counter-current cascade
F M
EN
R1
mixer separator
(column)
S
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. 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
NOT a tie-line
• F
• S
E and R are both points on the
equilibrium line. But they are not
related by the same tie-line.
EN •
• M
• R1
xA,1 •
Stage-by-stage analysis
1
N
F = RN+1
xA,N+1
S = E0
yA,0
EN
yA,N+1
R1
xA,1
R2 E1
stage 1 TMB: E0 + R2 = E1 + R1
E0 – R1 = E1 – R2 = E2 – R3 etc.
constant difference in flow rates of passing streams
⊗ = Ej – Rj+1 = constant
stage 1 CMBA: E0yA,0 + R2xA,2 = E1yA,1 + R1xA,1
E0yA,0 – R1xA,1 = E1yA,1 – R2xA,2 = etc.
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
⊗⊗⊗⊗ 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.
xA,∆
=E
0yA,0
− R1xA,1
∆ xD ,∆
=E
0yD ,0
−R1xD,1
∆
Define a difference point, ⊗, with coordinates (xA, ⊗, xD, ⊗):
Procedure:
1. Plot F (= RN+1), S = (E0). Locate M.
2. Plot R1 and locate EN.
3. Extend the lines joining E0-R1,
and EN-RN+1, to find ⊗ at the
intersection point.
Finding the ⊗-point
last mixing line
EN •
• M •
R1 first mixing line
⊗
•
S = E0
F = RN+1 •
4. All intermediate mixing lines
must pass through ⊗.
Stepping off stages on the H-N diagram
Procedure:
1. Use R1 and conjugate line to find E1
• M
⊗
•
S = E0
F = RN+1 •
Stop when you reach or pass EN.
N = 3
EN •
E1 •
E2 •
• R1
• R2
• R3
2. Use E1 and ∆-point to find R2
3. Use R2 and conjugate line to find E2
4. Use E2 and ∆-point to find R3
3. Use R3 and conjugate line to find E3
equilibrium line ends at P
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
yA
xA
A
D
0 0
1
1 • •
•
•
•
• •
P
• •
raffinate
compositions
extract
compositions
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
•
•
•
•
• •
P 1
xA
yA
0 0 1
xA
yA
0 0
1
1 •
•
•
•
• •
P
Obtaining the M-T operating line
1
N
F = RN+1
xA,N+1
S = E0
yA,0
EN
yA,N+1
R1
xA,1
A
D
x1
• EN
y0
yN RN+1
E0
•
• xN+1
Mixing lines represent passing streams.
All mixing lines lie between the limits:
(x1, y0) and (xN+1, yN)
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).
•
(x1, y0) •
(xN+1, yN)
• R1
M •
WAIT! In general, operating line is
not straight.
Plot arbitrary intermediate mixing
lines to obtain more points.
•
•
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
D S •
F •
M • • Mmin
•
Mmax
• 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, ∆min (corresponding to Mmin) occurs when a mixing
line and a tie-line coincide.
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Acetone-water-trichloroethane at 25 °C and 1 atm
weight fraction acetone
weight fraction water
EN,min•
•
S
F
R1
•
•
•Minimum solvent flow rate
•
∆min
•
Mmin
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 the SR1
mixing line
6. Tie-line with furthest intersection
from S locates ∆min
7. Mixing line from ∆min 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
10. (S/F)min = (FMmin)/(SMmin)
On H-N diagram whose tie-lines have negative slopes:
Rule-of-thumb: (S/F)act ~ 1.5 (S/F)min
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
Portion of right triangle phase diagram for water-acetic acid-isopropyl ether at 20 °C, 1 atm
weight fraction water
weight fraction acetic acid
Minimum solvent flow rate
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 ∆min
7. Draw mixing line from ∆min
through F to locate EN,min
8. Connect R1 and EN,min to
complete mass balance
•
S
F
R1
•
•
•
∆min
EN,min
• •
Mmin
9. Mmin is at the intersection of SF and R1EN,min
10. (S/F)min = (FMmin)/(SMmin)
On H-N diagram whose tie-lines have positive slopes:
Two feed counter-current column
1
N
F1 = RN+1
E0 = S R1
EN
F2
E R
E R
Feed balance: F1 + F2 = FT
S
F2
F1
FT
M
EN
R1
mixer 1 mixer 2 separator
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 ∆-points
Hunter-Nash analysis of 2-feed column
Overall balance:
1. Plot F1 and F2. Locate FT (co-
linear with F1 and F2).
2. Plot S . Locate M (co-linear with
S and FT).
3. Plot R1. Locate EN (co-linear with
R1 and M).
1. Calculate flow rates R1 and EN.
EN •
• M •
R1
•
S = E0
FT •
F2 •
F1 •
Stage-by-stage analysis
Balance around top of column:
R1 – E0 = Rj+1 – Ej = ∆1 ➙ R1, E0, ∆1 are co-
linear
Note: ∆1 and ∆2 may be on different sides of the phase diagram.
1
N
F2
F1 = RN+1
E0 = S R1
EN
E R
j
E R
k
Balance around bottom of column:
EN – RN+1 = Ek – Rk+1 = ∆2 ➙ RN+1, EN, ∆2 are co-linear
Overall balance:
F2 + RN+1 + E0 = EN + R1
F2 = (EN – RN+1) + (R1 – E0) = ∆1 + ∆2
➙ F2, ∆1, ∆2 are co-linear “feed-line”
∆2 is located at the intersection of two mixing lines:
RN+1, EN, ∆2 and F2, ∆1, ∆2
Need another line to locate ∆1:
TMB: FT = F1 + F2 = EN + (R1 – E0) = EN + ∆1
➙ FT, EN, ∆1 are co-linear
∆1 is located at the intersection of two mixing lines:
R1, E0, ∆1 and FT, EN, ∆2
∆2
•
3. Step off stages, initially using ∆1 to
generate the first mixing lines
∆1
•
Using the feed-line
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Acetone-water-trichloroethane at 25 °C and 1 atm
weight fraction acetone
weight fraction water
feed line
•M
•
EN
•
S
FT
R1
•
••F
2
•F1
1. Locate ∆1 at intersection of R1E0
and ENFT
2. Locate ∆2 at intersection of F2∆1
and ENRN+1
5. When the tie-line crosses the feed
line, the next mixing line will be
generated using ∆2
4. Identify the optimum feed stage
when the mixing line crosses the feed
line, F2∆1∆2
• E1
E2 •
• R2
Countercurrent liquid-liquid extraction with reflux
1
N
F1 = RN+1
xA,N+1
E0 = S R1
EN
yA,N
E R
In a conventional liquid-liquid extraction column:
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
How to increase yA,N?
need to increase xA,N+1
make RN+1 an reflux stream 1
N
F
RN+1
reflux
E0 R1
EN
E R
E R
PE product extract
makeup solvent
Turning extract into raffinate :
• extract is mostly solvent
• raffinate is mostly diluent
Q
recovered
solvent
SR
solvent
separator
“extract reflux” (no benefit to raffinate reflux)
We need to remove solvent,
e.g., distillation, stripping
Analogy to distillation reflux
1
V1
L0 D
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
N N
RN+1
reflux
EN
PE product extract
Q
recovered
solvent
SR
solvent
separator
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 ∆-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 ➙ plot F
yA,0, yD,0 ➙ plot E0
xA,1 ➙ plot R1 on sat’d raffinate curve
xA,PE, xD,PE ➙ plot PE (same location as RN+1 and Q, different flow
rates)
yA,SR, yD,SR ➙ plot SR
RN+1/PE
FT = F + RN+1 ➙ can’t locate FT (or EN) because we don’t know RN+1
Mass balance: solvent separator
PE
Q
SR
solvent
separator
RN+1
EN
EN is co-linear with Q and SR.
EN also lies on sat’d extract line.
EN
SR
×SR
PE
=RN+1
PE
+1+SR
PE
EN
SR
=RN+1
SR
+PE
SR
+1
EN = Q + SR
SR
PE
=
RN+1
PE
+1
EN
SR
−1
RN+1
SR
=EN
SR
−PE
SR
−1➙
Obtain EN/SR from lever-arm rule.
We will also need RN+1/SR:
= RN+1 + PE + SR
don’t know
A
D • E0
F •
SR • • R1
EN •
PE, Q, RN+1 •
Finding the ∆-points
∆2 = EN - RN+1
∆2xA,∆2 = ENyA,N - RN+1xA,N+1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Acetone-water-trichloroethane at 25 °C and 1 atm
weight fraction acetone
weight fraction water
• R1
Locate ∆1 at the intersection
of two mixing lines:
∆1 = E0 - R1
F = ∆1 + ∆2
xA,∆2
=ENyA,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,∆2 and thereby locate
∆2 on the ENRN+1 line.
Proceed to step off stages.
• ∆1
F •
PE, Q, RN+1 •
∆2 •
E0•
• EN
• SR