tugas#3 - liquid-liquid extraction (tayangan#)

8/10/2019 TUGAS#3 - Liquid-Liquid Extraction (Tayangan#)

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TUGAS KELOMPOK:

MEMBUAT TAYANGAN

(PPT)

Kalian diminta untuk membuat TAYANGAN yang berbasis dari contoh tayangan berikut ini.

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terdapat pada referensi-referensi lainnya, termasuk yang dimiliki oleh kelompok-kelompok

lainnya (PAGI atau pun SIANG).


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11 Liquid-liquid Extraction

A liquid mixture (e. g. A+B) is treated

with a solvent S (extractant) in which

one or more of the desired components is

preferentially soluble.

The separation of compounds is based on

differences in solubilities

[distribution of components between

(partially) immiscible liquid phases].

11.1 Introduction

11.1.1 Extraction processes

physical operation

depends on the solubilities

in two (partially) immiscible liquids

chemical operation

chemical reaction occurs,

with higher selectivity

Requirement for the solvent in physical

extractions:

(1) the solvent S and liquid-mixture are at

least partially if not completely

immiscible;

(2) its solubilities for components aredifferent.

A+B S

A

A single stage batch extraction:

Subsequent separation process:

distillation, crystallization, reextraction,

solvent S extract E

(S+A+B)

yA, yB

feed F

(solute A +

diluent B )

raffinate R

(B+A+S)xA, xB

B

A

B

A

x

x

y

y>

Extraction is preferable in the following

cases:

(1) when the relative volatility is near unity,

or when the azeotrope is formed;

(2) when the concentration of less volatile

components is low;

(3) when heating must be avoided; and

(4) when the components to be separated

are quite different in nature.

Factors to be considered for selecting

solvent:

(1) solubility

(2) selectivity

(3) subsequent separations


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11.1.2 Contact between two phases

differential contacting equipment

e. g., packed columns, pulsed columns,

spray towers.

stagewise equipment

e. g., mixer-settler units

11.2 Equilibrium of Ternary Systems and

Principle of Extraction

xS = 0.3

xA+xB +xS = 1

x mass fraction S0.2 0.4 0.6 0.8

xS

xA

xB

M

A

0.8

0.6

0.4

0.2

B

11.2.1 Triangular diagram

1. Triangular diagram

0.2

0.4

0.8

0.6

xA = 0.4

xB = 0.3

2. Lever rule

Two liquid mixtures

R kg (xA

,xB

,xS

) and

E kg (yA

,yB

,yS

) are

mixed to form a

mixture

Mkg (zA

,zB

,zS

). S

A

B xS

RxA

yA E

yS

M

zS

zA

Lever rule:

(1) M, E, and R are along the same line;

(2) the mass ratio of mixtures equals to the

length ratio of

line segments.

ME

MR

R

E

=

,RE

MR

M

E=

RE

ME

M

R=

S

A

B xS

RxA

yA E

yS

M

zS

zA

When the solvent S is mixed with the

binary mixture F (A+B) , point M must be

on line SF, and its position is determined by

the lever rule.

S

A

B

MS

MF

F

S=

(11-26)

On the line SF, the mass

ratio of A and B is the same.

FM

11.2.2 Equilibrium of partially misciblesystems

Liquid mixture: solute A + solvent B

New solvent S

Types of ternary mixture:

(1) S may be completely immiscible with

the original solvent B.

(2) S may be partially miscible with B.

(3) S and B, S and A form two partially

miscible liquids.

I

II


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1. Solubility curve, tie line, and plait point

plait point P

conjugate phasesE and R areconnected by thetie line.

P

A

SB

solubility curve

homogeneous

phase

tie line

R0 E0

R1E1

M1

two-phase region

Solubility curve:

Extract on the RHS of pint P

ys = (yA ) (11-9)

Raffinate on the LHS of pint P

xs = (xA ) (11-10)

P

A

SB R0 E0

2. Distribution coefficient and

distribution curve

Distribution curve

yA =f (xA ) (11-8)

Distribution coefficient

A

A

x

y=

(11-6)

kA =mass fraction of A in extract .

mass fraction of A in raffinate

yA

xA0

y = x

P

11.2.3 Phase equilibrium and extraction

operation

1. Single-stage extraction process

solventrecovery

purified extract

E,yA

purified raffinate

R,xA

feed F

solvent S

SB

A

ERR

E

F

extract E

raffinate R

M

2. Selectivity ratio of solvent

Binary systems:

B

A

BA

BA

k

k

xx

yy=

/

/

(11-11)

oo

oo

BA

BA

xx

yy

/

/=

or (11-12)

o

o

o

A

AA

x

xy

)1(1 (11-13)

, = 1, separation by

extraction is impossible.

Note:

distribution coefficientk

phase equilibrium

selectivity ratio

separability by extraction

o

o

o

A

AA

x

xy

)1(1

(11-13)

SB

A

R

E

FM

oo

AA xy=When

E

R


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3. Effect of miscibility of original solvent Band extractant S

small miscibility

larger two-phase region

, better result of separation

lower temperature

miscibility

, surface tension .

11.3 Calculation for an Extraction Process

11.3.1 Mathematical description

1. Material balance

Stage m:

Rm1,xm1, A,xm1, S Rm ,xm, A,xm, S

mEm ,ym, A,ym, S Em+1,ym+1, A,ym+1, S

Total material balance:

Rm1 +Em+1 =Rm +Em

Material balance for solute A:

Rm1xm1, A +Em+1ym+1, A =Rmxm, A +Emym, A

Material balance for solvent S:

Rm1xm1, S +Em+1ym+1, S =Rmxm, S +Emym, S

Rm1,xm1, A,xm1, S Rm ,xm, A,xm, S


2. Equilibrium stage and stage efficiency

Equilibrium stage:

The compositions of the two streams

leaving the stage are in equilibrium,

no mater what composition they have

when they enter the stage.Rm1,xm1, A,xm1, S Rm ,xm, A,xm, S


A problem of extraction may beconsidered from two aspects:

equilibrium stage

it is independent of equipment and

may be calculated first.

stage efficiency

it depends on the equipment, determined

experimentally.

actual number of stages

= equilibrium stage/efficiency

11.3.2 Single-stage extraction

For design:

given:F,xFA

specify:xA

select:zA,zS

calculate: S,E,yA,yS,

R,xS.

For operation:

given:F,xFA

S,zA,zS

calculate:E,yA,yS,

R,xA,xS.

F, xFA

S, zA, zS

R, xA, xS

E, yA, yS


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1. Analytical method

Six unknowns are solved from 6

equations.

Phase equilibrium relationship:

yA =f (xA ) (11-8)

ys = (yA ) (11-9)

xs = (xA ) (11-10)


F + S =R +E =M (11-20)


xFA+SzA =RxA+EyA =MxMA (11-21)

Material balance for solvent S:

SzS =RxS+EyS (11-22)

2. Graphical method

(1) Draw the solubility

curve.

(2) Locate points F and

S, and the mixture

M is on the line FS.(3) Locate point R byxA,

find the tie line RE by interpolation,

M is the intersection of lines RE and FS.

R

A

SB

E

M

F

xA

?

A

P E3R3

R2R1

SB

E2

E1

Find the tie line RE:

(a) By estimation;

(b) By auxiliary

curve.

E

R

?

(4) Determine the flowrates of streamsby material balance and lever rule


M=F + S

R

A

SB

E

M

F

xA

MS

MF

F

S=

R =ME

RE

MRME

=

(5) Find the composition in the extractyAandyS , and the concentration of solvent

in the raffinatexS.

R =F E

are the amount of

purified extract and

raffinate, respectively. xS

E

R

yS

yA

=

ER

FRFE

R

A

SB

E

M

F

xA


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Or: the flowrates of streams may be

obtained by material balance only,

in whichxFA is given,xA is specified,zA is

selected, andxMA andyA are found from the

phase diagram.

AMA

MAFA

zx

xxFS

=

AA

AMA

xy

xxME

=

3. Range of separation in a single-stage

extraction and the limit in the amount of

extractant

Point c:

the maximum amount

of solvent Smax .

Correspondingly,

in raffinate:xA, min

in purified raffinate:xA,min

A

SB

E

M

FxF,A

R co

min,Axo

min,Ax

Point d:

the minimum amount of solvent Smin .

The amount of

solvent used for

operation:

Smin < S < Smax

In general,

S = (1.1 2.0)Smin xA,min

A

SB

E

M

FxF,A

R co

min,Ax

d

Draw a line tangent

to the solubility curve

through point S, we

obtain the maximum

concentration in the

purified extract,o

max,Ay

o

max,Ay

xA,min

A

SB

E

M

FxF,A

R co

min,Ax

d

11.3.3 Multi-stage crosscurrent extraction

for single-stage extraction,

but the amount of liquid

mixture in each stage is

usually different.

Features:

the concentration of solute

is low in the raffinate, and

the amount of solvent is

large.

1

2

3

S E1

S E3

S E2

F

R1

R3

R2

The calculation is the repeat of that 11.3.4 Multistage countercurrent extraction

Higher extract rate may be obtained with

less extractant.

Calculation:

amount of solvent, number of equilibrium

stages, amount and concentration of

streams leaving each stage.

Method of stage-to-stage:

analytical or graphical


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11.3.5 Calculation for extraction with

completely immiscible solvents

1. Representation of composition and phase

equilibrium

X mass ratio in raffinate

(kg solute/kg diluent )

Y mass ratio in extract

(kg solute/kg solvent )

Equilibrium relationship:

Y=KX (11-35)

K distribution coefficient, generally

dependent upon concentration.

2. Single-stage extraction

B amount of original solvent

S amount of pure solvent

Z

mass ratio of A in extractant


S(YZ) =B(XF X) (11-36)

feed B, XF

solvent S, Z

extract S, Y

raffinate B, X

Equilibrium relationship:

Y=KX (11-35)


S(YZ) =B(XF X) (11-36)

For givenB,XF andZ, specify one variable

among S, YandX, other two may be

calculated.

3. Multistage crosscurrent extraction and

countercurrent extraction

11.3.6 Extraction with reflux

High purity may be achieved.

11.3.7 Differential contacting

countercurrent extraction

The diameter of column depends on the

flowrates and velocities of two phases.

The height of column may be obtained by

calculating the number of transfer unit or

HETS.

Exercises: 11 1, 3, 5


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11.4 Extraction Equipment

Requirement:

bringing the feed mixture and the solvent

into intimate contact

separation of the resulting two phases

11.4.1 Types of extraction equipment

stagewise contactors

differential contactors

Both may be with or without the addition

of external energy.

Stagewise contactors:

mixer-settler unit

sieve-plate column

Differential contactors:

1. spray tower

2. packed column

3. pulsed packed column

pulsed sieve column

4. vibrational sieve plate column

5. rotary disc column

6. centrifugal extractor

11.4.2 Flooding in extraction equipment

and limit of velocity

The performance of equipment is mainly

determined by:

allowable flowrates of two phases (D)

rate of mass transfer (H)

The superficial velocities of two phases at

which the flooding occurs are the limit for

the velocities.

11.4.3 Rate of mass transfer in extraction

equipment

Factors affecting mass transfer:

(1) drop size and size distribution of

dispersed phase

(2) circulation inside drops and interfacial

disturbance

(3) axial mixing in the equipment

11.4.4 Selection of extraction equipment

Following factors are to be considered:

the number of stages required

production capacity

properties of the materials

residence time


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11.5 Supercritical Fluid Extraction and

Liquid Membrane Extraction

11.5.1 Supercritical fluid extraction

1. Principle

It is an extraction process in which a

supercritical fluid is used as the solvent.

Supercritical fluid:

viscosity that of gas

density and solubility

those of liquid

diffusivity >> that in liquid

S

criticalpoint

t

pSFL

G

2. A process of supercritical extraction

CO2 is compressed as a supercritical

fluid, and used as an extractant.

reducing valveextractfeed

extraction

productraffinate SF

compressor

11.5.2 Liquid membrane extraction

(

separation )

The extraction and reverse extraction

are carried out simultaneously.

(a) make an emulsion by dispersingaqueous drops (in m) into an organic

phase (W/O);

(b) disperse the

emulsion into the

aqueous phase

(in 0.11 mm)

(W/O/W).

liquid

membrane

external

phase

internal

phase

A

B

Or: with O/W/O system.

Extraction with supported membrane:

The liquid

membrane phase

is retained in a

porous medium.liquid

membrane

lyophilic

porous

membrane

recovery

phase

original

liquid

A

tugas#3 - liquid-liquid extraction (tayangan#)

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