chapter 2 lle

7/23/2019 Chapter 2 Lle

1/59

CHAPTER 2: LIQUID LIQUID EXTRACTION

By : SYAFIZA ABDHASHIB

Credit to : Pn Siti

Wahidah


2/59

CHAPTER / CONTENT

Definition & Process Operation

LLE for Partially Miscible Solvent

LLE for Immiscible Solvent

Liquid liquid extraction equipment

Solvent selectivity & Phase Diagram


3/59

The separation of constituents (solutes) of a liquid solution by contact with

another insoluble liquid.

Solutes are separated based on their different solubilities in different liquid.

The separation process of the components of a liquid mixture by treatmentwith a solvent in which one or more desired components is soluble.

There are two requirements for liquid liquid extraction to be feasible:

component (s) to be removed from the feed must preferentially

distribute in the solvent.the feed and solvent phases must be substantially immiscible

Definition & Application


4/59

The simplest LLE involves only a ternary (i.e 3 component system)

Important terms you need to know:

Feed - The solution which is to be extracted(denoted by component A)

Solvent - The liquid with which the feed is contacted(denoted by component C)

Diluent - Carrier liquid(denoted by component B)

Extract - The solvent rich product of the operation

Raffinate - The residual liquid from which solutes has been removed.



5/59

In some operations, the solutes are the desired product, hence the extract stream isthe desirable stream. In other applications, the solutes my be contaminants that need tobe removed, and in this instance the raffinate is the desirable product stream.

Extraction processes are well suited to the petroleum industry because of the need toseparate heat sensitive liquid feeds according to chemical type (e.g aromatic,naphthenic) rather than by molecular weight or vapor pressure.

Application:

Major applications exist in the biochemical or pharmaceutical industry, whereemphasis is on the separation of antibiotics and protein recovery.

In the inorganic chemical industry, they are used to recover high boilingcomponents such as phosphoric acid, boric acid and sodium hydroxide fromaqueous solution.



6/59

Examples:

1. Extraction of nitrobenzene after reaction of HNO3with toluene inH2SO4

2. Extraction of methylacrylate from organic solution withperchlorethylene

3. Extraction of benzylalcohol from a salt solution with toluene.

4. Removing of H2S from LPG with MDEA



7/59

Examples:

5. Extraction of caprolactam from ammonium sulfate solutionwith benzene

6. Extraction of acrylic acid from wastewater with butanol

7. Removing residual alkalis from dichlorohydrazobenzene withwater

8. Extraction of methanol from LPG with water

9. Extraction of chloroacetic acid from methylchloroacetate withwater.



8/59


LLE VS distillation process :

LLE depends on solubilities between the liquid componentsand produces new solution which in turn has to be separated

again, whereas;

Distillation depends on the differences in relative volatilities /vapor pressures of substances. Furthermore, it requires heataddition.


9/59


Advantages of LLE over distillation process:

1. Where distillation requires excessive amount of heat

2. Presence of azeotropes or low relative volatilities are involved (valuenear unity and distillation cannot be used)

3. Removal of a component present in small concentrations, e.g

hormones in animal oil.


10/59


Advantages of LLE over distillation process:

4. Recovery of a high boiling point component present in small

quantities in waste stream, e.g acetic acid from cellulose acetate.

5. Recovery of heat sensitive materials (e.g food) where low tomoderate processing temperatures are needed. Thermaldecomposition might occur.

6. Solvent recovery is easy and energy savings can be realized.


11/59

Solvent selectivity

Selectivity (Separation factor).

If there are more than one solutes (sat two solutes A and B), then

consideration should be given to the selectivity of the solvent forsolutes A against B.

The selectivity between two solutes A and B is defined as the ratioof the distribution coefficient of A to the distribution coefficient of B.For all useful extraction operation the selectivity must exceed unity. Ifthe selectivity is unity, no separation is possible.


12/59

Solvent selectivity

Insolubility of Solvent.

The solvent should have low solubility in the feed solution,otherwise the separation is not clean. For example, if there issignificant solubility of solvent in the raffinate stream, anadditional separation step is required to recover the solvent.


13/59

Solvent selectivity

Recoverability.

It is always necessary to recover the solvent for re-use, and thismust ordinarily be done by other means, e.g distillation. If distillation isto be used, the solvent should form no azeotrope with the extracted

solute and mixtures should show high relative volatility. The solventshould also be thermally stable under the distillation temperature.

Density.

A large difference in density between extract and raffinate phases

permits high capacities in equipment. This is especially important forextraction devices utilizing gravity for phase separation.


14/59

Solvent selectivity

Chemical Reactivity.

The solvent should be stable chemically and inert toward theother components of the system and toward the commonmaterials of construction.

Other Criteria.

Toxicity and flammability of the solvent are importantoccupational health and safety consideration.

Stability of the solvent (i.e resistance to breakdown), particularlyin the recovery steps, is significant, especially if the breakdownproducts might contaminate the products of the main separation.


15/59

Design ofLLE tower


16/59

1. SINGLE STAGE CALCULATIONS

2. MULTISTAGE COUNTER CURRENT SYSTEM

LLE for Partially Miscible Solvent


17/59

Solvent and the solution are in contact with each other only once andthus the raffinate and extract are in equilibrium only once.

The solution normally binary solution containing solute (A) dissolvedin a diluent or carrier (B). The extracting solvent can be either puresolvent C or may content little A. Raffinate (R) is the exiting phase rich incarrier (B) while extract is exiting phase rich in solvent (C).

Single stage calculations


18/59

When liquid solution mixed with solvent (C), an intermediate phaseM momentarily forms as the light liquid moves through the heavyliquid in the form of bubbles. These bubbles provide a large surface

area for contact between the solution and the solvent that speed upmass transfer process.

The raffinate and extract are in equilibrium with each other.



19/59


Liquid-Liquid Extraction

y* (A)ys(A) Intermediate, M

Raffinate phase, R

x* (A)

Extract phase, E

xM (A)

F Mass of feed solution S Mass of extracting solventE Mass of extract phase R Mass of raffinate phaseM Mass of intermediate xF Mass fraction of A in FyS Mass fraction of A in S xM Mass fraction of A in Mx* Equilibrium mass fraction y* Equilibrium mass fraction of A in E

of A in R

Note: Intermediate shown just for purpose of demonstration. Dont have to draw it whenanswering the question

Feed Solution, F

xF(A)

Extracting Solvent, S


20/59

In most single extraction, we are interested to determine theequilibrium composition and masses of raffinate and extract phases byusing ternary phase diagram and simple material balances.

Using material balance,


Calculate the mass of intermediate M using total material balance:MSF Eq. (1)

Determine mass fraction of solute A in intermediate M using materialbalance for solute A :

MxSyFx MSF Eq. (2)

Use both Eq. 1 and 2 to find xM value


21/59

Single stage graphical steps

1. Locate point F (xF) and S (yS)

2. Draw a straight line from F to S

3. Using the calculated value of xM, locate point M (xM) on the FS

line. Note that point M must be on FS line.

4. Draw a new tie line that pass through point M. This new tie linemust take shape of the nearest given tie lines.

5. From the new tie line, you can locate point E and R and hence

you can determine the composition of raffinate, R and extract, Ethat are in equilibrium.


22/59

Once you have determine composition of R and E, you can determine themasses of E and R using the material balance as follows:


Using the total material balance:

Eq. (3)ERSF

Using the material balance for solute A:

EyRxSyFx SF ** Eq. (4)

Solve those Eq 3 and 4 to determine the masses of E and R


23/59

Example 1


100 kg of a solution containing 0.4 mass fraction of ethylene glycol (EG) in water isto be extracted with equal mass of furfural 250C and 101 kPa. Using the ternaryphase equilibrium diagram method, determine the followings:

the composition of raffinate and extract phases

the mass of extract and raffinate

the percent glycol extracted

Furfural rich layer Water rich layer

% EG % water % furfural % EG % water % furfural

0.0 5.0 95.0 0.0 92.0 8.0

8.5 4.5 87.0 2.0 89.6 8.4

14.5 4.5 81.0 5.5 86.0 8.5

21.0 6.0 73.0 7.0 84.4 8.6

29.0 7.0 64.0 8.0 83.3 8.7

42.0 8.5 49.5 14.0 77.2 8.8

50.0 14.0 36.0 31.0 60.0 9.0

51.0 33.0 16.0 51.0 33.0 16.0

Use the followingequilibrium tie lineto construct theternary phasediagram


24/59

Solution 1


Calculate the mass of intermediate M using total material balance

F = 100 kg S = 100 kg xF=0.4 yS=0

MSF M100100 200M kg

Determine mass fraction of solute A in intermediate M using materialbalance for solute A:

MxSyFx MSF 200x100010040 M . 20xM .

Locate point F & S, draw line FS. Locate point xM on FS line. Draw new tieline that pass through point xM. From that tie line, locate point E and R henceyou can determine the composition of R (x*) and E (y*) which is in equilibrium.From the graph, y* = 0.26, x* = 0.075(Solution for point 1)


25/59


Right angle method

F

S

M E

R


26/59


Equilateral method

S

ME

R

F


27/59

Solution 1 (cont)


Using the total material balance

Using the material balance for solute A:

ERSF ER100100 E200R Eq. (i)

EyRxSyFx SF ** E260R0750100010040 ...

Insert eq (i) into eq above

kg8664R

14135200E200R

kg14135E

25E1850

40E260E075015

40E260E2000750

.

.

.

.

..

..

Solution for point 2

% of EG extracted = (Mass of EG in extract /Mass of EG in feed) x 100%

%.%.

..%

*887100x

10040014135260100

FxEy

F

% of EG extracted =



28/59

Solvent and solution which flow opposite (countercurrent) to each other,come into contact more than onceand mix on stages inside the reactor.

Normally numbering of the stages begin at the top down to the bottom.Thus the top most stage is named as stage 1, stage directly below stage 1 isstage 2 and so on.

Multi stage counter current system

XR (A)

Final raffinate, RExtracting solvent, S

yS (A)

Final extract, E

xF (A)

Feed solution, F

1

2

3

n

N-1

N

yE (A)


29/59

The analysis of multistage extraction can be performed using right angleor equilateral triangular diagram to determine the number of ideal stagesrequired for a specified separation.

Using material balance,


Calculate the mass of intermediate M using total material balance:

MSF Eq. (1)

Determine mass fraction of solute A in intermediate M using material balancefor solute A :

MxSyFx MSF Eq. (2)

Use both Eq. 1 and 2 to find xMvalue


30/59

On a right angle triangulardiagram or equilateral triangulardiagram forA-B-C system:

Locate point F (xF) and S (yS)

Draw a straight line from F to S

Using the calculated value of xM, locate point M (xM) on the FS line. Note that

point M must be on FS line.

Locate point E1(Point M must be on E1RNline).


Operating Points and Lines.

Locate the Operating Point by finding the intersection of operating linesfor the left most and right most stage.

Draw a line through E1 and F.Draw a line through S and RN.Locate the intersection P. This point is the operating point P.


31/59


Acetone

TCE Water

Plait Point

Carrier

Solute

Feed

RN

M

E1

S

Operating Point

P


32/59


Operating Lines and Tie Lines: Stepping Off Stages:

Locate point R1from the tie line intersecting E1

Draw a line from the operating point P through R1 to the extract side ofthe equilibrium curve. The intersection locates E2

Locate point R2from a tie line.

Repeat Steps 2 and 3 until RNis obtained

Summary: E1 R1: Tie line, R1 E2: Operating line. Stop until E valueis slightly below RNvalue


33/59


Acetone

TCE Water

Plait Point

Solvent C

Carrier

Solute

E1

R1

Feed

RN

E2

E3

E4

E5

E6

M

Operating Point

P


34/59

Minimum solvent amount / minimum solvent flow rate

Minimum solvent flow rate is the lowest rate / amount at which solvent could betheoretically used for a specified extraction.

Occurs when operating line touches the equilibrium curve at which the

separation requires infinite numberof ideal stages.

Point M is dependent upon the solvent flow rate / amount. The larger the rate /amount, the closer is point M to point S on the FS line.



35/59


On a right angle triangulardiagram or equilateral triangulardiagram for A-B-C system:

Locate point F (xF) and S (yS)

Draw a best tie linethat originate from F. The intersection of this line withextract half dome is point Emin (minimum extract flow rate / amount).

Draw a straight line from Emin to point R. The intersection of this line withFS gives point Mmin. From point Mminyou can read the value of xmin.

Use the value of xminand material balance to calculate the Smin.

% Overall efficiency of multi stage extraction column:

% Overall efficiency = (number of ideal stage / number of real stage) x 100%


36/59

Example 2

5300 kg/h of a solution containing 30% by weight of ethylene glycol (EG) in wateris to be reduced to 4.5% (solvent free) by a continuous extraction in acountercurrent column using recycled furfural that contains 1.5% EG as theextracting solvent:

Determine the minimum solvent flow rate for the extraction above

If the solvent enters at 1.25 times the minimum solvent rate, how many idealstages are required?

Determine the number of real stages if the overall efficiency of the column is60%



37/59

Solution 2

F = 5300 kg/hr xF= 0.30 RSF= 0.045 yS= 0.015


F

S

Emin

RSF

Mmin


38/59


From the graph above, XMmin= 0.25

From material balance:

minmin MSF

minmin MS5300

minmin minMxSyFx MsF

minmin ... M250S0150530030 Smin=1127.66 kg/h


S = 1.25 x Smin=1.25 x 1127.66 kg/h S = 1409.58 kg/h

Calculate the mass of intermediate M using total material balance :

MSF M1409.585300 kgM 58.6409

Determine mass fraction of solute A in intermediate M using material

balance for solute A:

From material balance:


39/59



MxSyFx MSF 6709.58x1409.580.01553000.3 M 0.24xM

F

S

E1

RSF

M

P

E2

E3

E4

E5

From figure above, noof ideal stages = 5

Number of real stages = 5 / 0.60 = 8.33 = 9 stages.Solution for point 3


40/59


Sometimes extraction use a solvent C that is only slightly soluble in B or thesolvent C used is in range where the solubility in B is so low that for all practicepurpose, it can be assumed to be completely insoluble / immiscible in B and viceversa.

Bancroft weight fractions or mass ratio, x and y are defined as follows:

x (in raffinate phase) = mass of solute A / mass of diluent B

y (in extract phase) = mass of solute A / mass of solvent C


41/59

SINGLE STAGE CALCULATIONS

MULTISTAGE COUNTER CURRENT SYSTEM



42/59

Solvent C is used in such a range that it is considered insoluble in B.

Material balance of the solute (A) are:

Feed solutionM kg A in feed

S kg solvent C in Extracty kg A/kg solvent C

SolventN kg A in feed

B kg diluent B in Raffinatex kg A/kg diluent B

BxSyNM ''

S

NMx

S

By

''

Eq. (3)



43/59

Example 3


An aqueous solution of acetic acid is to be extracted in a single extractor withisopropyl ether. The solution contains 24.6 kg of acetic acid and 80 kg of water.

If 100 kg of isopropyl ether is added to the solution, what weight of acetic acidwill be extracted by isopropyl ether if equilibrium is attained?

Water and isopropyl ether may be considered as completely immiscible under

the condition of extraction. The equilibrium data as follows:

x (kg acid/kg isopropyl ether) 0.030 0.046 0.063 0.070 0.078 0.086 0.106

y (kg acid/kg water) 0.10 0.15 0.20 0.22 0.24 0.26 0.30


44/59

A = Acetic Acid B = Water C = Isopropyl ether


From mass balance,

Feed solution24.6 kg A

100 kg solvent C in Extracty kg A/kg solvent C

Solvent0 kg A in feed

80 kg diluent B in Raffinatex kg A/kg diluent B

BxSyNM '' '80100'06.24 xy 2460x800y .'.'

The equilibrium data and equation above is plotted in figure next page.

From the intersection of lines, x = 0.062, y = 0.195

Amount of acetic acid extracted = 19.5 kg

Solution 3


45/59


Solution 3

x' - y' diagram for s ystem acetic acid-water-isopro pyl ether

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

x'

y'


46/59

The principle is quite same with the partially miscible solvent, but in this casethe extraction process involved the immiscible solvent.


B kg/h pure diluent B in feed solution

x2kg A/kg pure B

B kg/h pure diluent B in raffinate

xnkg A/kg pure B

S kg/h pure solvent C

yn+1 kg A/kg pure solvent

S kg/h pure solvent in extract phase

y2 kg A/kg pure solvent1

2

3

n

N-1

N


47/59

For immiscible solvents, analysis of the extraction process become much simplersince the flow rate of pure solvents (extracting solvent and feed solvent (diluent) )are constants).

The operating line equation for multi stage liquid liquid extraction:


S

BxSyx

S

By 22

''''

Form of y = mx + c

The operating line gives a relationship between mass fraction of A (xn) inraffinate phase coming out of the nth stage, with mass fraction of A (yn+1) in extractphase entering the nth stage.

The operating line can be plotted by locating points (x1,y1) and (x2,y2) anddraw a straight line through these points.

Using the method similar to Mc Cabe Thiele diagram, the number of idealstages can be determined by drawing the triangular steps connecting equilibriumline and the operating line.


48/59


Example 4

8000 kg/h of an acetic acid-water solution containing 20% acid by mass is to becounter currently extracted with isopropyl ether to reduce the concentration of acidto 2% in the solvent-free raffinate product.

Determine the number of theoretical stages if 20,000 kg/h solvent is used.

Use the following equilibrium data:

x (kg acid/kgwater)

0.000 0.025 0.050 0.100 0.150 0.200 0.220 0.240 0.260 0.300

y (kg acid/kgisopropyl ether)

0.000 0.005 0.013 0.030 0.046 0.063 0.070 0.078 0.086 0.106


49/59


Solution 4

Since acid is to be extracted, acid is the absorbable component A, while water iscomponent B and isopropyl ether is component C.

BOTTOM (1)

TOP (2)

0.8(8000) kg/h of water in

x2

20,000 kg/h isopropyl ether (C) in

y1

0.8(8000) kg/h of water (B) in raffinate0.02 kg A/kg (solvent free)

x1

20,000 kg/h isopropyl ether (C) in extract phase

y2


50/59


From the definition:

x (in raffinate phase) = mass of solute A / mass of diluent B

y (in extract phase) = mass of solute A / mass of solvent C

Solution 4

F = 8000 kg/h xF= 0.20 xBF= 0.80 S = 20,000 kg/h

Solvent feed

Mass of solute A = 0 kg/h

Mass of solvent C = 20,000 kg/h.

For location 1,

Raffinate phase

Solvent-free composition of A = 0.02, B = 0.98

02040980

020

B

BA

BA

A

BA

B

BA

A

B

Ax1 .

.

.'

020000

0y1 'Coordinates for (x1,y1) = (0.0204, 0)


51/59


Solution 4

Feed

Mass of solute A = 0.20 x 8000 = 1600 kg/h

Mass of diluent B = 0.80 x 8000 = 6400 kg/h.

For location 2,

Extract

Mass of solute A = Mass of acid in Mass of acid in raffinate

Mass of solute A = 1600 - x1x B = 1600 0.0204 x 6400= 1469.44 kg/h

Mass of solvent C = 20 000 kg/h


52/59


Solution 4

For location 2,

Extract

25006400

1600x2 .'

0735020000

441469y2 .

.'

Coordinates for (x2,y2) = (0.250, 0.0735)

The points for operating line are (0.0204, 0) and (0.250, 0.0735).

No. of theoretical stages = 14 stages


53/59


Solution 4


54/59

Minimum solvent amount / flow rate: immiscible solvent

Minimum solvent flow rate, Sminis defined as the flow rate of solvent at whichthe number of stages approaches infinity.

Smincan be determined graphically by drawing a straight line originating fromthe bottom (x1,y1) until it becomes tangential to the equilibrium curve. The slope

of this line corresponds to minimum slope (mmin):


min

minS

Bm

min

minm

BS

Example 5

Determine the minimum flow rate of isopropyl ether for the problem in

Example 4.


55/59


Solution 5

The line tangential to the equilibrium curve and originates from (x1,y1)which is (0.0204, 0) is drawn in figure below:

The slope of the tangential line = 0.3433430

800080

m

BS

.

.

min

min

= 18 658.89 kg/h.


56/59


Two main classes of solvent extraction equipment:

1) Vessels in which mechanical agitation is provided for mixing

2) Vesselsin which the mixing is done by the flow of the fluid themselves.

The extraction equipment can be operated batch or continuous.


57/59


Mixer Settlers for Extraction

A mechanical mixer is often used to provide intimate contact between thetwo liquid phases to provide efficient mass transfer.

One phase is usually dispersed into the other in the form of small droplet.

In figure 12.6 1(a) for typical mixer settler, mixer or agitator is entirelyseparate from the settler. The feed of aqueous phase and organic phase are

mixed in the mixer, and then the mixed phases are separated in the settler.

In figure 12.6 1(b) for combined mixer settler, sometimes used in extractionof uranium salts or copper salts from aqueous solution.


58/59


Spray Extraction Towers

In Figure 12.6 2 the heavy liquid enters atthe top of the spray tower, fills the tower asthe continuous phase, and flows out throughthe bottom.

The light liquid enters through a nozzledistributor at the bottom, which disperses orsprays the droplets upward.

The light liquid coalesces at the top andflows out.

In some cases the heavy liquid is sprayed

downward into a rising, light continuousphase.


59/59


Packed Extraction Towers

More effective tower madeby packing the column withrandom packing such asRaschig rings, Berl saddles,Pall rings and so on

Packings cause the droplets tocoalesce and redisperse atfrequent intervals through thetower.

Packed tower is more efficientthan spray tower.

Table 12.6 1 shows typical

performance for several typesof commercial extractiontowers.

chapter 2 lle

Documents