chapter 2 lle edited

CHAPTER 2: LIQUID – LIQUID EXTRACTION

CHAPTER / CONTENT

Definition & Application

LLE for Partially Miscible Solvent

LLE for Immiscible Solvent

Liquid – liquid extraction equipment

LLE= 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.

LLE= Separation process of the components of a liquid mixture by treatment with 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


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.


In some operations, the solutes are the desired product, hence the extract stream is the desirable stream. In other applications, the solutes my be contaminants that need to be 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 to separate 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, where emphasis is on the separation of antibiotics and protein recovery.

In the inorganic chemical industry, they are used to recover high – boiling components such as phosphoric acid, boric acid and sodium hydroxide from aqueous solution.


Examples:

Extraction of nitrobenzene after reaction of HNO3 with toluene in H2SO4

Extraction of methylacrylate from organic solution with perchlorethylene

Extraction of benzylalcohol from a salt solution with toluene.

Removing of H2S from LPG with MDEA

Extraction of caprolactam from ammonium sulfate solution with benzene

Extraction of acrylic acid from wastewater with butanol

Removing residual alkalis from dichlorohydrazobenzene with water


Examples:

Extraction of methanol from LPG with water

Extraction of chloroacetic acid from methylchloroacetate with water.


The difference between LLE and distillation process in the separation of liquid mixtures:

LLE depends on solubilities between the liquid components and 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 heat addition.


Advantages of LLE over distillation process:

Where distillation requires excessive amount of heat

Presence of azeotropes or low relative volatilities are involved (α value near unity and distillation cannot be used)

Removal of a component present in small concentrations, e.g hormones in animal oil.

Recovery of a high – boiling point component present in small quantities in waste stream, e.g acetic acid from cellulose acetate.

Recovery of heat – sensitive materials (e.g food) where low to moderate processing temperatures are needed. Thermal decomposition might occur.

Solvent recovery is easy and energy savings can be realized.

SINGLE STAGE CALCULATIONS

MULTISTAGE COUNTER CURRENT SYSTEM

LLE for Partially Miscible Solvent

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

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

When liquid solution mixed with solvent (C), an intermediate phase M momentarily forms as the light liquid moves through the heavy liquid in the form of bubbles. These bubbles provide a large surface area for contact between the solution and the solvent that speed up mass transfer process.

The raffinate and extract are in equilibrium with each other.

Single – stage calculations


Liquid-Liquid Extraction

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

Raffinate phase, Rx* (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. Don’t have to draw it when answering the question

Feed Solution, FxF (A)

Extracting Solvent, S

In most single – extraction, we are interested to determine the equilibrium composition and masses of raffinate and extract phases by using 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 material balance for solute A :

MxSyFx MSF Eq. (2)

Use both Eq. 1 and 2 to find xM value

On a right angle triangular diagram or equilateral triangular diagram for A-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.

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

From the new tie line, you can locate point E and R and hence you can determine the composition of raffinate, R and extract, E that are in equilibrium.

Once you have determine composition of R and E, you can determine the masses 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

Example 1


100 kg of a solution containing 0.4 mass fraction of ethylene glycol (EG) in water is to be extracted with equal mass of furfural 250C and 101 kPa. Using the ternary phase 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 following equilibrium tie – line to construct the ternary phase diagram

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 material balance for solute A:

MxSyFx MSF 200x100010040 M . 20xM .

Locate point F & S, draw line FS. Locate point xM on FS line. Draw new tie line that pass through point xM. From that tie line, locate point E and R hence you 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)


Right angle method

F

S

M E

R


Equilateral method

S

ME

R

F

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

100400

14135260100

Fx

Ey

F

% of EG extracted =


Tutorial

1. 12.5-2

2. 12.5-4

Tutorial1. 12.5-2 (Textbook-page 832)A single stage extraction is performed in which 400kg of a

solution containing 35 wt% acetic acid in water is

contacted with 400 kg of pure isopropyl ether. By using

equilateral diagram, determine



the percent acetic acid extracted

Tutorial2. 12.5-4 (Textbook-page 832)A mixture weighing 1000kg contains 23.5 wt% acetone

and 75.5 wt% water and is to be extracted by 500 kg

methyl isobutyl ketone in a single stage extraction. By

using equilateral diagram, determine



the percent acetone extracted

Solvent and solution which flow opposite (countercurrent) to each other, come into contact more than once and 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 is stage 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)

The analysis of multistage extraction can be performed using right – angle or equilateral triangular diagram to determine the number of ideal stages required 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 balance for solute A :

MxSyFx MSF Eq. (2)

Use both Eq. 1 and 2 to find xM value



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 E1RN line).


Operating Points and Lines.

Locate the Operating Point by finding the intersection of operating lines for 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.


Acetone

TCE Water

Plait Point

Carrier

Solute

Feed

RN

M

E1

S

Operating PointP


Operating Lines and Tie Lines: Stepping Off Stages:

Locate point R1 from the tie line intersecting E1

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

Locate point R2 from a tie line.

Repeat Steps 2 and 3 until RN is obtained

Summary: E1 → R1 : Tie line, R1 → E2 : Operating line. Stop until E value is slightly below RN value


Acetone

TCE Water

Plait Point

Solvent C

Carrier

Solute

E1

R1

Feed

RN

E2E3

E4

E5

E6

M

Operating PointP

Minimum solvent amount / minimum solvent flow rate

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

Occurs when operating line touches the equilibrium curve at which the separation requires infinite number of 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.





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

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

Use the value of xmin and 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%

Example 2

5300 kg/h of a solution containing 30% by weight of ethylene glycol (EG) in water is to be reduced to 4.5% (solvent free) by a continuous extraction in a countercurrent column using recycled furfural that contains 1.5% EG as the extracting 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 ideal stages are required?

Determine the number of real stages if the overall efficiency of the column is 60%



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 kg M 58.6409

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

From material balance:



MxSyFx MSF 6709.58x1409.580.01553000.3 M 0.24xM

F

S

E1

RSF

M

P

E2

E3

E4

E5

From figure above, no of ideal stages = 5

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

Exercise1. 12.7-3 (Textbook, pages 833)An aqueous feed solution of 1000 kg/hr containing

23.5 wt% acetone and 76.5 wt% water is being

extracted in a countercurrent multistage extraction

system using pure methylisobutyl ketone solvent at

298-299 K. The outlet water raffinate will contain 2.5

wt% acetone. By using equilateral method,

a) Determine the minimum solvent that can be used (Answer: Xmin = 0.18, Smin = 305.56 kg/hr , Mmin = 1305.56 kg/hr)

b) Using a solvent flow rate of 1.5 times the minimum, determine the number of theoretical stages. (Answer: S= 458.34 kg/hr, Xm=0.16 , 5 stages)


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

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

SINGLE STAGE CALCULATIONS

MULTISTAGE COUNTER CURRENT SYSTEM


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)


Example 3


An aqueous solution of acetic acid is to be extracted in a single extractor with isopropyl 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 acid will 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

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


Solution 3

x' - y' diagram for system acetic acid-water-isopropyl 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'

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


B kg/h pure diluent B in feed solution

x’2 kg A/kg pure B

B kg/h pure diluent B in raffinate

x’n kg A/kg pure B

S kg/h pure solvent C

yn’ kg A/kg pure solvent

S kg/h pure solvent in extract phase

y’2 kg A/kg pure solvent1

2

3

n

N-1

N

For immiscible solvents, analysis of the extraction process become much simpler since 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 nn

22 ''''

Form of y = mx + c

The operating line gives a relationship between mass fraction of A (x’n) in raffinate phase coming out of the nth stage, with mass fraction of A (yn) in extract phase entering the nth stage.

The operating line can be plotted by locating points (x’1,y’1) and (x’2,y’2) and draw a straight line through these points.

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


Example 4

8000 kg/h of an acetic acid-water solution containing 20% acid by mass is to be counter currently extracted with isopropyl ether to reduce the concentration of acid to 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/kg water)

0.000

0.025

0.050

0.100

0.150

0.200

0.220

0.240

0.260

0.300

y’ (kg acid/kg isopropyl

ether)

0.000

0.005

0.013

0.030

0.046

0.063

0.070

0.078

0.086

0.106


Solution 4

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

BOTTOM (1)

TOP (2)

0.8(8000) kg/h of water in

x’2

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

y’1

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

x’1

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

y’2


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 (x’1,y’1) = (0.0204, 0)


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 - x’1x B = 1600 – 0.0204 x 6400= 1469.44 kg/h

Mass of solvent C = 20 000 kg/h


Solution 4

For location 2,

Extract

25006400

1600x2 .' 07350

20000

441469y2 .

.'

Coordinates for (x’2,y’2) = (0.250, 0.0735)

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

No. of theoretical stages = 14 stages


Solution 4

TOP (0.250,0.0735)

BOTTOM(0.0204,0)

Minimum solvent amount / flow rate: immiscible solvent

Minimum solvent flow rate, Smin is defined as the flow rate of solvent at which the number of stages approaches infinity.

Smin can be determined graphically by drawing a straight line originating from the bottom (x’1,y’1) until it becomes tangential to the equilibrium curve. The slope of this line corresponds to minimum slope (mmin):


minmin S

Bm

minmin m

BS


The slope of the tangential line = 0.343 3430

800080

m

BS

.

.

minmin

= 18 658.89 kg/h.


Example 5

Determine the minimum flow rate of isopropyl ether for the problem in Example 4.


Solution 4

TOP (0.250,0.0735)

BOTTOM(0.0204,0)


The slope of the tangential line = 0.343 3430

800080

m

BS

.

.

minmin

= 18 658.89 kg/h.

Tutorial12.7-6 (Textbook)A water solution of 1000kg/hr containing 1.5 wt%

nicotine in water is stripped with a kerosene stream

of 2000kg/hr containing 0.05 wt% nicotine in a

countercurrent stage tower. The exit water is to

contain only 10% of the original nicotine, that is,

90% is removed.

1.Determine the number of theoretical stages needed.

2. Determine the minimum solvent(kerosene) flow rate


Two main classes of solvent – extraction equipment:

1) Vessels in which mechanical agitation is provided for mixing

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

The extraction equipment can be operated batch or continuous.


Mixer – Settlers for ExtractionA mechanical mixer is often used to provide intimate contact between the two 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 entirely separate 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 extraction of uranium salts or copper salts from aqueous solution.


Spray Extraction Towers

In Figure 12.6 – 2 the heavy liquid enters at the top of the spray tower, fills the tower as the continuous phase, and flows out through the bottom.

The light liquid enters through a nozzle distributor at the bottom, which disperses or sprays the droplets upward.

The light liquid coalesces at the top and flows out.

In some cases the heavy liquid is sprayed downward into a rising, light continuous phase.


Packed Extraction TowersMore effective tower – made by packing the column with random packing such as Raschig rings, Berl saddles, Pall rings and so on

Packings cause the droplets to coalesce and redisperse at frequent intervals through the tower.

Packed tower is more efficient than spray tower.

Table 12.6 – 1 shows typical performance for several types of commercial extraction towers.

chapter 2 lle edited

Technology

liquid solution

liquid components

residual liquid

different liquid

heavy liquid

insoluble liquid

solvent recovery

single extraction