120712ChE128 7 LiqLiq Extract

7/27/2019 120712ChE128 7 LiqLiq Extract

1/41

Liquid-liquid extraction


2/41

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)


3/41

Extraction equipment

Batch:

mixer-settler

column:

separatory funnel

rotating-disk contacter

a. agitator; b. stator disk

single-stage:

Continuous:


4/41

Mixer-settler column

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 anumber 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 Khni

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

Stage-wise phase contact


5/41


6/41

Agitated column


7/41


8/41

Packed extraction column

The ECP packed column is based on current state-of-the-a

Packing

The special Sulzer extraction packing reduces the back-mi

Liquid distributors

In order to create an even liquid flow velocity profile at eitMain 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/m3Low interfacial tension: < 2 mN/m

Tendency to form emulsions

Reliable scale-up


9/41

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/FSand xA,N.

Feed (F) contains solute A (xA) dissolved in

diluentD (xD= 1xA).

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 thedistribution ratio, Kd:

Kd= yA/xA

Note that yAdoes not refer to gas composition.

F Rdiluent flow rate

= FD= constan

S Esolvent flow rate

= FS= constant

feed mixtureextract

raffinate

mixer settler

solvent


10/41

McCabe-Thiele analysis:

Counter-current extraction with immiscible liquids

N = 3

(X0,Y1)

(XN,YN+1)

1

2

3

X0

(FD/FS)maxgives FS,minfor N = .

Can also use Kremser eqns, if solutions

are dilute andequil. line is straight.

For dilute solutions,

y=R

E

x+ (y1 R

E

x0)

Y=

FDFSX+ (Y1

FDFSX0)

Equation of the operating line:

(analogous to operating line for

stripper column).

N=

ln 1 mER

yN+1

y0

y1 y

0

+

mE

R

ln RmE( )


11/41

Cross-flow cascade

Increase overall efficiency by introducingfresh extracting solvent at each stage.

Each stage has its own mass balance andoperating line

Uses much more solvent than counter-current cascade (requires much moresolvent recovery)

A mixer-settler is just one cross-flow stage.

From Separation Process Engineering, Third Editionby Phillip C. Wankat

(ISBN: 0131382276) Copyright 2012 Pearson Education, Inc. All rights reserved.

Figure 13-8 Cross-flow cascade

N = 3

yj=

R

Ej

xi+ (y

j,in+R

Ej

xj 1

)

From mass balance around stagej:

(x0,y1,in)(x1,y2,in)(x2,y3,in)x3

(x1,y1)

(x2,y2)

(x3,y3)


12/41

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:

Aprefers solvent 1 (extract)

Bprefers solvent 2 (raffinate)

Kd,A= yA/xA> 1

Kd,B= yB/xB< 1

1

N

F

zAzB

solvent 1

yA,N+1= 0yB,N+1= 0

solvent 2xA,0 = 0xB,0= 0

extractyA,1yB,1

raffinate

xA,NxB,N

E R

E

R

abs

orbing

se

ction

s

tripping

section


13/41

One operating line for each solute i, in

each section of the column (i.e., 4 total).

McCabe-Thiele analysis: dilute fractional extraction

xA,N

5

3

yA,1

2

1

6

NF= 4,

feed stage

If yA,1and xA,Nare specified, and NFis

known, use M-T diagram to obtain N, then

use trial-and-error to find xB,0and xB,N+1

If yA,1and xB,Nare 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

Ex

i+ (y

i,1R

Ex

i,0)

Bottom operating lines (stripping section):

Equilibrium data is different for each

solute (use separate McCabe-Thiele

diagrams!)


14/41

Center-cut extraction

When there are 3 solutes: A, Band C,

and Bis desired

(Aand C may be > 1 component each)

solvent 1

solvent 2solvent 1+ A

solvent 2

+ B+ C solvent 3

solvent 2solvent 3+ B

solvent 2

+ C

F

zA, zB, zC

Requires twocolumns: column 1 separates Afrom B+C

column 2 separates Bfrom C

Requires threeextracting solvents:

Aprefers solvent 1 over solvent 2

B, Cprefer solvent 2 over solvent 1

Bprefers solvent 3 over solvent 2

Cprefers solvent 2 over solvent 3


15/41

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 3axes: 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 Editionby Phillip 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)


16/41

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 oppositethe

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 afurfural/water ratio in this range (otherwisesingle phase).

The mixture Mlies inside the miscibility

boundary, and will spontaneously separateinto two phases. Their compositions (Eand

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


17/41

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 Editionby Phillip C. Wankat(ISBN: 0131382276) Copyright 2012 Pearson Education, Inc. All rights

reserved.

Figure 13-12 Equilibrium for water-chloroform-acetone at 25C 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 xAand yA.

Horizontal axis corresponds to both xBand yB

Q: Where does pure C appear on this diagram?


18/41

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?


19/41

Hunter-Nash analysis of mixer-settler

F ME

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.

Compositionsof E and R are alsorelated 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)


20/41

Co-linearity

Whyare 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,MCMBD: FxD,F+ SxD,S= MxD,M= (F+ S)xD,M

solve for coordinates of M: (xA,M, xD,M)

xA,M

=

FxA,F

+SxA,S

F+S

x

D,M=

FxD,F

+SxD,S

F+S

F

S=

xA,M

xA,S

xA,F

xA,M

=

xD,M

xD,S

xD,F

xD,M

slope from

Mto S

slope from

Fto M F (xD,F, xA,F)

S (xD,S ,xA,S)

M (xD,M, xA,M) ThereforeF, S and M are co-linear. To locate

M on the FS line: calculate either xA,Mor xD,M.

xA,M

xA,S

xD,M

xD,S

=

xA,F

xA,M

xD,F

xD,M

rearrange

CMBA CMBD


21/41

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+ (MF)xA,S= MxA,M

F(xA,F- xA,S) = M(xA,M- xA,S)

RM

=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


22/41

Hunter-Nash analysis of cross-flow cascade

1F = R0 R1

S1

E1

2

S2

E1

R2

F

S

E2

R2

Treat each stage as a mixer-settler.

each Ri, Sipair creates a mixing line

find each Ei, Ripair using a tie-line

E1

R1

M1

M2


23/41

Hunter-Nash analysis of counter-current cascade

FM

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,1on 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 atie-line

F

S

E and R are both points on the

equilibrium line. But they arenot

related by the same tie-line.

EN

MR1

xA,1


24/41

Stage-by-stage analysis

1

N

F = RN+1

xA,N+1

S = E0

yA,0

EN

yA,N+1

R1xA,1

R2 E1

stage 1 TMB: E0+ R2= E1+ R1

E0R1= E1R2= E2R3 etc.

constant difference in flow rates of passing streams

= EjRj+1= constant

stage 1 CMBA: E0yA,0+ R2xA,2= E1yA,1+ R1xA,1

E0yA,0R1xA,1= E1yA,1R2xA,2= etc.

constant difference in compositions of passing streams

net flow of A: xA,= EjyA,jRj+1xA,j+1

net flow of D: xD,= EjyD,jRj+1xD,j+1


25/41

The difference point

does not necessarily lie inside the RT graph.

All pairs of passing streams Ej, Rj+1are 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 R

2(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, =

E0yA,0 R1xA,1 xD, = E0yD,0 R1xD,1Define a difference point, , with coordinates (xA, , xD, ):


26/41

Procedure:

1. Plot F (= RN+1), S = (E0). Locate M.

2. Plot R1and locate EN.

3. Extend the lines joining E0-R1,

and EN-RN+1, to find at theintersection point.

Finding the -point

last mixing line

EN

M

R1 first mixing line

S = E0

F = RN+1

4. All intermediate mixing linesmust pass through .


27/41

Stepping off stages on the H-N diagram

Procedure:

1. Use R1and 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 E1and D-point to find R2


4. Use E2and D-point to find R3



28/41

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

yAxA

A

D

00

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

P1

xA

yA

0 0 1


29/41

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,1A

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 ourlabeling convention (feed enters at stage N).

(x1, y0)

(xN+1, yN)

R1M

WAIT! In general, operating line isnot straight.

Plot arbitrary intermediate mixing

lines to obtain more points.


30/41

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

Sminachieves the desired separation with N =

A

DS

F

M

Mmin

Mmax

as M moves towards S, (S/F) increases

(lever-arm rule)

when M reaches the equilibrium line, allfeed 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.


31/41

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 SR1mixing line

4. Locate several tie-lines

5. Extend tie-lines to the SR1

mixing line

6. Tie-line with furthest intersection

from Slocates Dmin

7. Mixing line from Dminthrough Flocates EN,min

8. Connecting R1and EN,min

completes the mass balance

9. Mminis located at the intersection

of SF and R1

EN,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


32/41

Minimum solvent flow rate

Strategy:

1. Plot S = E0, F = RN+1, R12. Join S and F

3. Extend SR1mixing line

4. Locate several tie-lines

5. Extend tie-lines to SR1

mixing line

6. Find tie-line which givesclosest intersection to S;

this locates Dmin7. Draw mixing line from Dmin

through F to locate EN,min8. Connect R1and EN,minto

complete mass balance

S

F

R1

min

EN,min

Mmin

9. Mminis at the intersection of SF and R1EN,min

10. (S/F)min= (FMmin)/(SMmin)

On H-N diagram whose tie-lines have positive slopes:


33/41

Two feed counter-current column

1

N

F1= RN+1

E0= SR1

EN

F2

ER

ER

Feed balance: F1+ F2 = FT

S

F2

F1

FT

MEN

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 F2enters the column

upper and lower sections have different sets of operating

linesdifferent D-points


34/41

Hunter-Nash analysis of 2-feed column

Overall balance:

1. Plot F1and F2. Locate FT(co-

linear with F1and F2).

2. Plot S . Locate M (co-linear with

S and FT).

3. Plot R1. Locate EN(co-linear with

R1and M).

1. Calculate flow rates R1and EN.

EN

M

R1

S = E0

FT

F2

F1


35/41

Stage-by-stage analysis

Balance around top of column:

R1E0= Rj+1Ej= D1 R1, E0, D1are co-

linear

Note: D1and D2may be on different sides of the phase diagram.

1

N

F2

F1= RN+1

E0= SR1

EN

ER

j

E

R

k

Balance around bottom of column:

ENRN+1= EkRk+1= D2 RN+1, EN, D2are co-linear

Overall balance:

F2+ RN+1+ E0= EN+ R1F2= (ENRN+1) + (R1E0) = D1+ D2

F2, D1, D2are co-linear feed-line

D2 is located at the intersection of two mixing lines:

RN+1

, EN, D

2 and F

2, D

1, D

2

Need another line to locate D1:

TMB: FT= F1+ F2= EN+ (R1E0) = EN+ D1 FT, EN, D1are co-linear

D1 is located at the intersection of two mixing lines:

R1, E0, D1 and FT, EN, D2

U i h f d li


36/41

2

3. Step off stages, initially using D1to

generate the first mixing lines

1

Using the feed-line

feed lineM

EN

S

FT

R1

F2

F1

1. Locate D1at intersection of R1E0

and ENFT

2. Locate D2at intersection of F2D1

and ENRN+1

5. When the tie-line crosses the feed

line, the next mixing line will begenerated using D2

4. Identify the optimum feed stage

when the mixing line crosses the feed

line, F2D1D2

E1

E2

R2

C t t li id li id t ti ith fl


37/41

Countercurrent liquid-liquid extraction with reflux

1

N

F1= RN+1xA,N+1

E0= SR1

ENyA,N

ER

In a conventional liquid-liquid extraction column:

yA,Nis related by equilibrium to xA,N

xA,Ndepends on xA,N+1

dilute feed gives dilute extracthighest yA,Nobtained with S Smin, but this requires very large N

How to increase yA,N?

need to increase xA,N+1make RN+1an reflux stream

1

N

F

RN+1reflux

E0R1

EN

ER

E

R

PEproduct 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


38/41

Analogy to distillation reflux

1

V1

L0 D

Saturated liquid reflux stream isobtained 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

RN+1reflux

EN

PEproduct 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


39/41

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, R1and RN+1are specified, same stage-by-stage analysis

But RN+1is an internal stream, usually notspecified.

Usually specified:F, xA,F, xD,F plot FyA,0, yD,0 plot E0xA,1 plot R1on satdraffinate curve

xA,PE, xD,PE plot PE(same location as RN+1and Q, different flowrates)

yA,SR, yD,SR plot SRRN+1/PE

FT= F + R

N+1 cant locate F

T(or E

N) because we dont know R

N+1


40/41

Mass balance: solvent separator

PE

Q

SR

solventseparator

RN+1

EN

ENis co-linear with Q and SR.

ENalso lies on satdextract 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/SRfrom lever-arm rule.

We will also need RN+1/SR:

= RN+1+ PE+ SR

dont know

A

DE0

F

SR R1

EN

PE, Q, RN+1


41/41

Finding the D-points

D2

= EN

- RN+1

D2xA,D2 = ENyA,N- RN+1xA,N+1

R1

Locate D1at the intersectionof two mixing lines:

D1 = E0- R1

F = D1+ D2

xA, 2 =

ENyA,N

RN+1xA,N+1

EN R

N+1

We dont know the individualflow rates EN, RN+1, but we know

EN/SRand RN+1/SR. We can

calculate xA,D2and thereby locate

D2on the ENRN+1line.

D1

F

PE, Q, RN+1

D2

E0

EN

SR

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120712ChE128 7 LiqLiq Extract