liquid-liquid extraction presentation 1 -...

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)


(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 .


(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


4. Use E2 and ∆-point to find R3


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 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




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




• 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

liquid-liquid extraction presentation 1 -...

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