15 extraction

UNZA Pharmacy Training Lecture notes

2. EXTRACTION

3. Extraction Theory

4. Extraction
•Itistheremovalofsolublematerialfrominsolublesolidor
liquidbytreatmentwithaliquidsolvent.
•Theextractionratedependson:
Thediffusionrateofsolutethroughliquidboundarylayer.
I. Solid - liquid extraction (leaching )
•Theextractionofasolubleconstituentfromasolidby
meansofasolvent.
1.Theextractionoffixedoilsfromseeds,
Alkaloidsas strychninefrom Nux vomica
or quininefrom cinchona bark,
2. The isolation of enzymes as rennin, hormonesas insulin.
•Leachingisusuallyoperatedasabatchprocess.. WHY?
•Itinvolvestheprocessingofhighcostmaterialspresentin
relativelysmallquantities.
•Frequentchangesofmaterialmaybemade,creatingproblems
ofcleaning&contamination.

5. Leaching is performed in one of two ways:
By percolation:
•Therawmaterialisplacedinavessel,forma
permeablebedthroughwhichthesolvent
(menstrum)percolates.
•Thedissolutionofthewantedconstituentsoccursand
thesolutionissuesfromthebottomofthebed.
•Thisliquidis(miscella) andtheexhaustedsolidsare
(marc) .
Byimmersion:
•Therawmaterialisimmersedinthesolventand
stirredfrequently.
•Afterasuitableperiodoftime,solidandliquidare
separatedbyfiltration.

6. Factors affecting the rate of leaching
1.ParticleSize:
•Small particle size is attained by milling higher extraction
rates.. WHY?
1. Millingcauses rupturing of cell wall (the main barrier to be crossed).
2. The smaller the size
The greater the interfacial area between solid and liquid.
The smaller the distance the solute must diffuse within solid.
2.TheSolvent:
A.Selective
B.Ofasufficientlylowviscositytocirculatefreely.
C.Arelativelypuresolventwillbeusedinitially,butastheextraction
proceeds,theconcentrationofsolutewillincreaseandtherateofextraction
willdecrease..WHY?
1.Theconcentrationgradientwillbereduced.
2.Thesolutionwillbecomemoreviscous.

7. 3.TheTemperature:
1.Theincreaseintemperaturewillleadto:
a.↑drugsolubility↑extractionrate.
b.↑diffusioncoefficient.
2.Theupperlimitoftemperatureisdeterminedby
secondaryconsiderations.
Eg.Thenecessityofpreventingenzymeactionduring
extractionofsoluteslikesugar.
4.TheAgitationofthesolvent:
↑Eddydiffusion↑transferofmaterialfromthesurfaceofthe
particlestobulkofsolution.
Beyondacertainlimit,increasingtherateofagitationwillnot
furtheraffecttherateofextraction..WHY?
•Becausethecontrollingresistancewillbefoundinthediffusion
ofthesolutefromthecellmatrixitselftooutside.

8. A. Leaching by percolation
•Whenthesolidsformanopen,permeable
massthroughouttheleachingoperation,
solventmaybepercolatedthroughan
unagitatedbedofsolids.
•The solid bed may be:
1. Stationary:batch operation
Eg. Batch percolator
2. Moving:continuous operation
Eg. Bollman Extractor).

9. Bollman Extractor
•Itisaverticalchamber
havinganumberofbaskets
withperforatedmetal
bottomsrotatingclockwise
inachainover2wheels.
Theascendingcolumnof
baskets:
1.Thebasketsmove
upwards(containingthe
herb)andthensprayedwith
thepuresolvent(counter-
current),
2.Thesolventpercolates
downwardstobecollectedat
thebottomashalfmiscella.

10. Thedescendingcolumnof
baskets:
•Asthebasketsmoveup,they
areinvertedtogetrideofthe
marcandfilledagainwithfresh
one.
•Thebasketsarethensprayedin
thissidewiththehalfmiscella
comingfromtheleftside(parallel
flow).
•Thehalfmiscellapercolate
downwardthroughthefreshherb
soitiscollectedatthebottomas
fullmiscella.
•Usedforextractionofoils

11. Batch percolator
•Itisatankwitha
perforatedfalsebottomto
supportsolidsandpermit
solventdrainage.
•Thesolidsareloadedinto
thetank,sprayedwithsolvent
untiltheirsolutecontentis
reducedtoeconomical
minimum,andexhausted.
•Theextractorbodymaybe
jacketedtogivecontrolofthe
extractiontemperature.

12. •Thepackingofthecoarselyground
materialmustbeevenotherwise,solvent
willpreferentiallyflowthroughalimited
volumeofbedandleachingwillbe
inefficient.
•Theleachliquidspasstoanevaporator
workingatareducedpressureThe
vapourleavingtheevaporatoris
condensedandisredirectedtothe
extractor.
•Extractionisstoppedwhenleachliquid
isfreefromwantedconstituents.
•Used for:extraction of alkaloids
Disadvantages:
1. Require large amounts of solvent.
2. Yield dilute extracts
extraction evaporation.

13. B. Leaching by Immersion
Iscarriedoutinsimpletanks, whichmaybe
agitatedbyturbinesorpaddles.
•Ifthesolidsareadequatelysuspended,intimate
contactbetweenphasespromoteefficientextraction.
•Advantages:
1.Incompleteextractionduetochannelingisavoided
2.Difficultiesduetoswellingwillnotarise.

14. •Separationofthephasesdependson:
Thenatureoftheextractedmaterials.
1.Finelydividedsolids:
Bycontinuouscentrifugation.
2.Coarse,compressiblesolids:
When agitation ceases,
Thesolidswillsettle
Theleachliquidissiphonedorpumpedoff.
Thesedimentwillcontaina↑volumeofmiscella
recoveredby:
1.re-suspendingsolidsinfreshsolvent.
2.re-separating.
*Alternatively, cakefiltrationmaybeused.

15. II. Liquid –Liquid Extraction
Principlesoftheextractiontowers:
1.Towersareoperatedwiththe2liquidphasesflow
counter-currenttoeachother.
Theheavyphaseintroducedattop&flowsdownward,
Thelighterphaseintroducedatbottom&flows
upward.
2.Thegravitationalforcecontrolsthecounter- current
motionandseparationofthetwophases.
Applications:
1-Refiningofvegetableoilsusingfurfural.
2-Manufactureofpharmaceuticalsasantibiotics.

16. Spray Towers
•Theyareemptytowers,
withoutpackingorbaffles.
•Oneliquidfillsthewholetower
asacontinuousphase,
•Otherphaseisdispersed
throughitbyspraying.
•Contactingeffectivenessis
low..WHY?
1.Re-circulationofcontinuous
phase.
2.Coalescenceofdispersed
phase.
•Highcolumnsarerequiredfor
efficientextraction

17. Baffle Plate column
•Thehorizontalbafflesare
usedtodirecttheflowof
liquidsinazigzagpathas
theypasseitherupordown
tothetower.
Advantages:
•Containnoholestobe
cloggedorenlargedby
corrosion.
•Suitablefordirtysolution
containsuspendedsolids.
•Valuableforliquidsthatcan
emulsifyeasily.

18. Packed Tower
•Theuseofpacking
increasetheareaof
contactbetweenphases.
•Emptycolumnisfilled
withamaterial(metal,
stoneandporcelain)and
cutintheformofrings
called“Raschigrings”
alargesurfacearea
wettedbytheliquid.
•The packing is
supportedonanopen
screennotofferany
resistancetoflow.

19. Liquid-liquid extractionis a useful method to separate components
(compounds) of a mixture

20. Let's see an example.
Suppose that you have a mixture of sugarin vegetable oil(it tastes
sweet!) and you want to separate the sugar from the oil. You
observe that the sugar particles are too tiny to filter and you
suspect that the sugar is partially dissolved in the vegetable oil.
What will you do?

21. How about shaking the mixture
with water
Will it separate the sugar from the
oil? Sugar is much more soluble in
water than in vegetable oil, and,
as you know, water is immiscible
(=not soluble) with oil.
Did you see the result?The water
phase is thebottom layerandthe oil
phase is the top layer, because
water is denserthan oil.
*You have not shaken the mixture
yet, so sugar is still in the oil phase.

22. By shaking the layers (phases) well, you
increase thecontact area between the
two phases.The sugar will move to the
phase in which it is most soluble: the
water layer
Now the water phase tastes
sweet,because the sugar is moved to
the water phase upon shaking.**You
extractedsugarfrom the oil with
water.**In this example,water was
the extraction solvent ;the original
oil-sugar mixture was the solution to
be extracted; and sugar was the
compound extracted from one phase
to another. Separating the two layers
accomplishes the separation of the
sugar from the vegetable oil

23. Did you get it?.....the concept of liquid-liquid extraction?
Liquid-liquid extraction is based on the transfer of a solute
substance from one liquid phase into another liquid phase according
to the solubility.Extraction becomes a very useful tool if you choose
a suitable extraction solvent.You can use extraction to separate a
substance selectively from a mixture, or to remove unwanted
impurities from a solution.In the practical use, usually one phase is a
water or water-based (aqueous) solution and the other an organic
solvent which is immiscible with water.
The success of this method depends upon the difference in solubility
of a compound in various solvents. For a given compound, solubility
differences between solvents is quantified as the "distribution
coefficient"

24. Partition Coefficient Kp (Distribution Coefficient Kd)
When a compound is shaken in a separatory funnel with two immiscible
solvents, the compound will distribute itself between the two solvents.
Normally one solvent is water
and the other solvent is a
water-immiscible organic
solvent.
Most organic compounds are
more soluble in organic solvents,
while some organic compounds
are more soluble in water.

25. Here is the universal rule:
At a certain temperature, the ratio of concentrations of a solute
in each solvent is always constant.ﾊAnd this ratio is called the
distribution coefficient, K.
(when solvent
1and solvent
2are immiscible liquids
For example,Suppose the
compound has a distribution
coefficient K = 2between
solvent
1and solvent
2
By convention the organic
solvent is (1) and waater is
(2)

26. (1)If there are 30particles
of compound , these are
distributed between equal
volumesof solvent
1and solvent
2..
(2) If there are 300
particles of compound , the
same distribution ratio is
observed in solvents 1 and 2
(3) When you doublethe
volume of solvent
2(i.e., 200
mLof solvent
2and 100 mLof
solvent
1),the 300particles of
compound distribute as shown
If you use a larger amount of extraction solvent, more soluteis
extracted

27. What happens if you extract twicewith 100 mLof solvent
2?
In this case, the amount of extraction solvent is the same volume as was
used in Figure 3, but the total volume is divided into two portions and
you extract with each.
As seen previously, with 200 mL
of solvent
2you extracted 240
particles of compound . One
extraction with 200 mL gave a
TOTAL of 240 particles
You still have 100 mL of solvent
1,
containing 100particles. Now you
add a second 100 mL volume of
fresh solvent
2. According to the
distribution coefficient K=2, you
can extract 67more particles
from the remaining solution

28. An additional 67particles are
extracted with the secondportion
of extraction solvent
(solvent
2).The total number of
particles extracted from the first
(200 particles) and second(67
particles) volumes of extraction
solvent is 267.This is a greater
number of particles than the
single extraction (240particles)
using one 200 mL portion of
solvent
2!
It is more efficient to carry out
two extractions with 1/2 volume
of extraction solvent than one
large volume!

29. If you extract twice with 1/2 the volume, the extraction is more
efficient than if you extract once with a full volume. Likewise,
extraction three times with 1/3 the volume is even more efficient….
four times with 1/4 the volume is more efficient….five times with 1/5
the volume is more efficient…
ad infinitum
The greater the number of small extractions, the greater the
quantity of solute removed. However for maximum efficiency the
rule of thumb is to extract three times with 1/3 volume

30. Chemically active (acid-base) extraction
Can you change the solubility property of a compound? How?
Most organic compounds are more
soluble in organic solventsthan in
water,usually by the distribution
coefficientK > 4
However, specific classes of
organic compounds can be
reversibly alteredchemically to
become more water-soluble.
This is a powerful technique and allows you to separate organic
compounds from a mixture --if they belong to different solubility
classes

31. What type of organic compounds can be made water-soluble?
Compounds belonging to the following solubility classes can be
converted to their water-soluble salt form
(1)Organic acidsinclude carboxylic acids (strong organic acids)
and phenols (weak organic acids).
(2)Organic basesincludes amines

32. How can organic acids or bases be converted to a water-solubleform?
1.Organic Acidscan be converted to their saltform when treated with an
aqueous solution of inorganic base (
e.g.,NaOH (sodium hydroxide) and
NaHCO
3(sodium bicarbonate)).Salts are ionic, and in general, ions are
soluble in waterbut
not soluble in water-immiscible organic
solvents.Remember: water is a very polarsolvent thus salts (i.e.,ionic
species) are well dissolved in it.
A.Carboxylic Acidsare
converted to the salt form
with 5% NaOHaqueous
solution. NaOH is a strong
inorganic base.
Carboxylic acids are strong
organic acids (pKa = 3 to 4),
so they can also be ionized
with weakinorganic bases
(e.g.,NaHCO
3(sodium
bicarbonate)) aqueous
solution
.

33. Let's try a sample problem.
Here is a mixture of naphthaleneand benzoic acid, dissolved in
dichloromethane.
You want to separate
these two compounds.
What will you do?
You may use an aqueous solution of
either 5% NaOH or sat. NaHCO
3, to
extract benzoic acidas a saltform

34. B. Phenols are considered to be weakorganic acids. Phenol, the parent
compound, is partially water-soluble (1 g will dissolve in 15 mL of water),
whereas substituted phenols are not.Sodium bicarbonate (NaHCO
3) aqueous
solution, a weakinorganic base, will not deprotonate phenols to make it
ionic, because it is not strong enough.However, treatment with NaOH, a
stronginorganic base, can change phenol to its ionic (salt) form.

35. Let's try a another sample problem.
Here is a mixture of benzoic acid and p-methoxyphenol, dissolved in
dichloromethane.
You want to separate
these two compounds.
What will you do?
You cannot use 5% NaOH to separate these two
compounds. NaOH will react with both benzoic acidand p-
methoxyphenol, thus both compounds will be extracted into the
aqueous layer.

36. Let's try this problem again.
Here is another mixture of benzoic acid and p-methoxyphenol, dissolved in
dichloromethane.
Strong organic acids such as benzoic acid would be
deprotonated and ionized, while weak organic acids such
as phenols would NOT be deprotonated
NaOH was too strong a base,
thus it does not differentiate
the strong and weak organic
acids. Use of weak inorganic
base such as NaHCO
3will
differentiate between the
compounds

37. 2. Organic Bases(amines)can be converted to their salt form when treated
with an aqueous solution of an inorganic acid such as HCl (hydrochloric acid).
Recall that salts are ionicand generally soluble in waterbut not soluble in
water-immiscible organic solvents.

38. Let's try a third sample problem.
Here is a mixture of benzoic acid and p-chloroaniline, dissolved in
dichloromethane.
You want to separate
these two compounds.
What will you do?
You may use an aqueous solution of
either 5% HCl, to extract the amineas
a saltform and benzoic acid has
remained in the organic layer

39. You can separate four different classes of compounds from a mixture based on
differing solubility properties. The four classes are:1.Amines(organic
base)2.Carboxylic acids(strong acid)3.Phenols(weak acid)4.Neutral
compounds
.

40. After the separationof the mixture of four components, we will have four
solutions: each solution contains one component.
The first three compounds are chemically altered, existing in their salt form
dissolved in aqueous solution. The fourth compound is not chemically altered,
but it is dissolved in an organic solvent.We now want to recovereach compound
in its original state (
i.e.,in the non-ionic form) to complete the experiment. We
call this step isolationor recovery.
Let's see, one by one, how to recover each compound obtained from the
separation process

41. Isolation (Recovery) of amines
An amine is a basic compound. It is protonated in the presence of excess HCl
forming a salt that is soluble in aqueous solution. This is how you separated the
amine from the original mixture containing it.
An amine is soluble in acidicaqueous
solution because it forms a salt, an
ionic form.
However, if you change the pH
of the solution tobasicthe
amine can no longer stay
dissolved because it is no longer
ionic! This process is called
basification.
Basificationis done by carefully
adding concentrated NaOH
solution to the solution containing
the amine salt until it becomes
basic.

42. ･In the basification step, you use
concentratedNaOH solution to
minimize the volume of the final
solution. Recall that a dilute
solution of HCl was used to
extract the amine as its water-
soluble salt (see the picture on
the right side).
･Basification must be done
carefully, portion by portion, with
swirling each time because the
acid-base neutralization reaction
is exothermic.
･Check the pH of the solution to
ensure that it is basic. (~pH 10)

43. Isolation (Recovery) of Acids
There are two different groups of organic acids: carboxylic acids (strong acids) and
phenols(weak acids).In the separation procedure, acids were extracted using (weak or
strong) basic aqueous solutions
Both acids can be returned to the original form in the same
manner! Organic acids are currently dissolved in a basic aqueous
solution, because the acid forms a salt, an ionic form. When you
make the aqueous solution acidic, the organic acids no longer
remain dissolved because they are no longer ionic and usually
precipitate out of solution. This process is called acidification.
Acidificationis done by carefully adding
concentrated HCl solution until the mixture
becomes acidic,
When the weak base, NaHCO
3, was the
extracting solution, CO
2gas will evolve
during acidification.

44. ･The recovery of organic acids
requires acidification with
concentratedHCl solution. Recall
that in the extraction step for
the separation of an organic acid
either dilute NaHCO
3or NaOH
was used. Concentrated HCl will
now help minimize the volume of
the final aqueous solution
･Acidification must be done
carefully, portion by portion, with
swirling each time because the
acid-base neutralization reaction
is exothermic.
･Check the pH of the solution to
ensure that it is acidic. (~pH 3)

45. Separatory Funnel Extraction Procedure
Separatory funnels are designed to facilitate the mixing of immiscible liquids

46. Separatory Funnel Extraction Procedure
1. Support the
separatory funnel in a
ring on a ringstand.
Make sure stopcock is
closed
2. Pour in liquid to
be extracted
3. Add extraction
solvent
4. Add ground glass
Stopper (well greased)

47. Separatory Funnel Extraction Procedure
Pick up the separatory
funnel with the stopper in
palce and the stopcock
closed, and rock it once
gently.
Then, point the stem up and slowly open the
stopcock to release excess pressure. Close the
stopcock. Repeat this procedure until only a small
amount of pressure is released when it is vented
Shake the separatory funnel.

48. Separatory Funnel Extraction Procedure
Shake the separatory funnel vigorously.
Now, shake the funnel vigorously for a few seconds. Release
the pressure, then again shake vigorously. About 30 sec
total vigorous shaking is usually sufficient to allow solutes to
come to equilibrium between the two solvents.
Vent frequently to prevent pressure buildup,
which can cause the stopcock and perhaps
hazardous chemicals from blowing out. Take
special care when washing acidic solutions with
bicarbonate or carbonate since this produces a
large volume of CO
2gas

49. Separatory Funnel Extraction Procedure
Separate the layers.
Let the funnel rest
undisturbed until the layers
are clearly separated
While waiting, remove the
stopper and place a beaker
or flask under the sep
funnel.
Carefully open the stopcock and
allow the lower layer to drain
into the flask. Drain just to the
point that the upper liquid
barely reaches the stopcock

50. LIQUID EXTRACTION
•Liquid extraction, sometimes called solvent extraction, is
the separation of the constituents of a liquid solutionby
contact with another insoluble liquid
.
•If the substances constituting the original solution
distribute themselves differently between two liquid
phases, a certain degree of separationwill result.

51. The separationcan be enhancedby the use of
multistage contactsor their equivalentsin the
manner of gas absorption & distillation.
•Feed: Solution which is to be extracted.
(F=A+C)
•Solvent:Liquid with which feed is contacted.
(B)
•Extract: Solvent reach product. (B+C)
•Raffinate: Residual liquid from which solute
has been removed. (A+C)

52. M
FEE
D
SOLVE
NT
MIXER
RAFFINAT
E
EXTRACT
SETTL
ER

53. IMPORTANT REQUIREMENTS
A) Insolubility of extract & raffinate
phases.
B) Density difference between extract &
raffinate phases.
•Limitations:
A) Like gas absorption, pure products are
not obtained. Hence, other separation
techniques like distillation are to be
used for obtaining pure products.

54. FIELDS OF USEFULNESS
A) Competition with other mass transfer
operations—relative costs* will decide
the separation technique, i.e.,
i) Dilute aqueous solutions, e.g., acetic
acid-water mixtures, e.g., selected solvent
will
reduce distillation load and/or improve
relative
volatility.
ii) Substitution of high vacuum distillation
operation
*Cost includes both direct & indirect costs

55. B) Substitution for chemical methods—
avoids consumption of chemicals & disposal
problems, e.g., metal separations such as
uranium-vanadium, hafnium-zirconium, etc.
Also, purification of copper, phosphoric acid,
boric acid, etc.

56. C) Separation where other methods do
not
give desired products:
i) Separation of mixture with same
molecular weight (near B.P.)
II) Pharmaceutical products like
penicillin

57. Choice of solvent
There is usually wide choice of liquids to be used as solvent. It is unlikely
that any particular liquid will possess all the properties considered
desirable for extraction & some compromise is usually necessary.
Following properties should be considered while making a choice:
1.SELECTIVITY: IS MEASURED BY COMPAIRING THE RATIO OF C TO A IN THE B -
RICH PHASE (EXRACT) TO THAT IN THE A -RICH PHASE(RAFFINATE) AT
EQUILIBRIUM. ANALOGOUS TO RELATIVE VOLATILITY :
β
= (Wt. fraction C in E/ Wt. fraction A in E)
(Wt. fraction C in R/ Wt. fraction A in R )
= (Wt. fraction A in R) ×y*
E
(Wt. fraction A in E)×x
R1
>1,higher value desirable
=1-no separation

58. 2. Distribution coefficient: y*/x at equilibrium. Not
necessary that y*/x should be > 1. However, higher
values are desirable since less solvent will be
required for extraction.
3
. Insolubility of solvent: For both diagrams shown, only
those A-C mixtures between D & A can be separated
by use of solvent B or B‘. Since mixtures richer in C
will not form two liquid phases with the solvent and
no separation will take place. Clearly, solvent B is
more useful.

60. 4)Recoverability: It is always necessary to recover the
solvent for reuse. It is usually recovered by
distillation. This requires:
A) high relative volatility between solvent & solute
(less
no. of theoretical stages) .
B) Smaller quantity of material should be more volatile
(less vapor rate* & consequently less heat load**).
C) Low heats of vaporization (less heat load).
D) No azeotrope formation with solute.
* Small column diameter ** Small reboiler & condenser
& also low utility requirements.

61. 5) Density: differences are a must between
extract & raffinate phases. Larger the
difference, better & quicker is the separation.
6) Interfacial Tension: Larger the interfacial
tension, the more readily coalescence of
emulsion takes place.
7) Chemical Reactivity: Solvent should be
stable
chemically & inert towards other components
of
the system & common materials of
construction.

62. 8) Viscosity, Vapor Pressure & Freezing
Point: Low value will help in ease of
handling & storage.
9) Solvent should be non toxic, non
inflammable & of low cost.
Liquid extraction requires two
immiscible
phases to intimately mix to accomplish
mass transfer & then separate
efficiently

63. System of three liquids-
one pair partially miscible.
This is the most common type of system
in extraction. Typical examples are:
•Water (A)- Chloroform (B)- Acetone (C)
•Benzene (A)- Water (B)- Acetic acid (C)
Triangular co ordinates are used as isotherms.
Liquid C dissolves completely in A & B,
respectively. But A & B dissolve only to a
limited extent in each other to give rise to the
saturated
liquid solution at L (A rich) & K (B rich).

64. System of three liquids, A and B
partially miscible

65. More insoluble A & B, L & K will be near
apex points of triangle. A binary mixture
at J will separate into two liquids of
composition at L & K.
Curve LRPEK is the binodal solubility
curve, indicating the change of solubility
of A rich & B rich phases upon addition
of C.

66. •Any mixture outside this curve would be
a homogeneous liquid of one phase
(i.e., no separation). Any ternary
mixture, such as M, will form two
insoluble, saturated liquid phases of
equilibrium composition as indicated by
R(A rich) and E(B rich). The line RE
joining these equilibrium compositions is
a tie line, which must necessarily pass
through point M representing the
mixture as a whole.

67. •There are an infinite number of tie lines
in the two phase region & only a few are
shown. They are rarely parallel &
usually
•change slope in one direction as shown.
Point P, the plait point, the last of the tie
lines & the point where the A-rich & B-
rich
•
solubility curves merge, is ordinarily not
at the maximum value of C on the
equilibrium curve

68. •The % of C in solution E is clearly greater
than that in R, and it is said that in this case
the distribution of C favors the B rich phase.
This is conveniently shown on the distribution
diagram (Fig. 10.3), where the point (E,R) lies
above the diagonal y=x. The ratio y*/x, the
distribution coefficient, is in this case greater
than unity. The concentrations of C at ends of
the tie lines, when plotted against each other
, gives rise to the distribution curve shown.

69. EFFECT OF TEMPERATURE
•For most systems, the mutual solubility
of A & B increases with temperature,
and above certain critical temperature, t
4
, they dissolve completely. The
increased solubility at higher
temperature influences the ternary
equilibrium considerably. At higher
temperatures,
•i) Area of heterogeneity decreases
ii) Slope of tie line may change.

70. •Liquid extraction operations, which
depend upon the formation of insoluble
liquid phases, must be carried out at
temperatures below t
4.

71. SINGLE STAGE EXTRACTION

72. F+S
1 = M
1 =E
1+ R
1----(1)
Fx
F+S
1y
S= Mx
M1 ----(2)
x
M1= (Fx
F+S
1y
S)/M
1---(3)
WHERE F+S
1 = M
1.
This allows computation of x
M1.
Fx
F+S
1y
S=(F+S
1) x
M1
F(x
F-x
M1) = S
1 (x
M1-y
S)
S
1/F = (x
F-x
M1)/ (x
M1-y
S)---(4)
This allows computation of solvent for given
location of M
1.

73. E
1y
1+R
1x
1=M
1x
M1---(5)
E
1y
1 = M
1x
M1-x
1(M
1-E
1)
E
1= M
1 (x
M1-x
1)/ (y
1-x
1)--(6)

74. •Since two immiscible phases must form for an
extraction operation, point M
1 must lie within
the heterogeneous liquid area as shown. The
minimum amount of solvent is thus found by
locating M
1at D, which would then provide an
infinitesimal extract at G, and the maximum
amount of solvent is found by locating M
1at
K, which provides an infinitesimal amount of
raffinate at L. Point L also represents the
raffinate with the lowest possible C
concentration.

75. SINGLE STAGE CONTACTOF
IMMISCIBLE
LIQUIDS
•If A & B are completely immiscible, then there
is no need to use triangular diagrams. C will
distribute between B & A
•and equilibrium concentrations will be given
by plot of y( concentration of solute in extract
) vs. x ( concentration of solute in raffinate ).
•Ax
F=Ax
1+ Sy
1 Alt. y
1 / (x
1 -x
F) = -A/S
•This line passes through (x
F,0) & has slope of
•-A/S. It intersects equilibrium curve at point
(x,y)

76. Other Co-0rdinates
•Because the equilibrium relations can rarely
be expressed algebraically with any
convenience, extraction computation must
usually be made graphically on a phase
diagram.
The co- ordinates scales of an equilateral
triangles are necessarily always the same,
and in order to be able to expand one
concentration scale relative to other,
rectangular co ordinates could be used.

77. •One of these is formed by plotting
concentration of B as abscissa against
concentration of C (x &y) as ordinate as
shown in fig.10.9. Unequal scales can
be used in order to expand the plot as
desired. Equation given below applies,
regardless of inequality of the scales.
•R/E= (x
E-x
M)/ (x
M-x
R)

79. Multistage cross –current
operations
1 2 3
Solvent
S
1, y
s
S
2
,y
s
S
3,
y
s
R
1,
x
1
R
2
x
2
R
3
x
3
E
1,
y
1
E
2,
y
2
E
3,
y
3
fee
dF,
x
F
Extract of
all stages
Fig. shows a
three stage
cross current
extraction
operations
where the
raffinate
from any stage is
the feed for the
next stage

80. For any stage n,
Overall mass balance:
R
n-1+S
n= R
n+ E
n= M
n
Solute balance:
R
n-1 x
n-1 + S
n y
n= x
n R
n + y
n E
n
= x
Mn M
n
For stage 1, R
n-1= F, x
n-1 = x
F
From 1 and 2
x
Mn =( R
n-1 x
n-1 + S
n y
n)/ (R
n-1 + S
n)
(1)
(2)
(3
)

81. •E
n= {M
n(x
Mn -x
n)}/( y
n –x
n)
Graphical representation of various
streams is shown in fig.
(4)

82. •The solvent S
1, S
2 and S
3 have the
same compositions and are represented
by point S. The points F, M
1, S
1 and R
1
are located as in the case of single
stage operation. The point R
1 now
represents the feed to stage2. M
2lies on
line R
1S at mass fraction of C =x
M2 .R
2.
and E
2arethe ends.

83. Completely immiscible A & B
•If the feed contain “A” Kg of A and the solvent
contain “B” Kg of B, then all raffinate contain “A”
Kg of A and all the extract will contain “B” Kg of
B. thus solute balance for any stage n:
Ax’
n-1+ B
ny’
s= B
ny’
n + Ax’
n ----------(5)
or -(A/ B
n) = { (y’
s-y’
n) }/ {x’
n-1 -x’
n} ---(6)
•Equation(6) represent the operating line of stage
“n” of slope –A/B
nand passing through (x’
n-1 ,y’
s)
and (x’
n,y’
n). Graphical representation is as
shown:

85. •Ex. Water –dioxane solutions
forms a minimum boiling
azeotrope at atmosphere
pressure and cannot be
separated by ordinary
distillation methods. Benzene
forms no azeotropic with
dioxane and may be used as
an extraction solvent. At 25
deg c , equilibrium distribution
of dioxane between water and
benzene is as follow:
•
Wt
%in
H
2O
5.118.925.2
Wt
%in
C
6H
6
5.222.532

86. At these concentrations, water and benzene are
substantially insoluble 1000Kg of a 25%
dioxane –water solution is to be extracted
with benzene to remove 95% of dioxane.
Benzene is dioxane free.Calculate:
•Benzene requirement for single batch
operation
•if extraction were done with equal amount of
solvent in 3 cross- current stages, how much
solvent would be required?

87. •Basis:1000Kg dioxane-water
solution.
A-water B- Benzene C-Dioxane
Feed: A-750Kg , C=250 Kg
x’
F= (250/750) = 0.333
Since water and benzene are
substantially insoluble, the solute
conc. In the extract and raffinate
phases are expressed on a ‘c’
free basis
•The raffinate will contain750 Kg
water and all extract will contain
‘B’ Kg of B
X’0.05370.2330.337
Y’0.05480.2900.470

88. •A) since 95% dioxane is recovered, the final raffinate
will contain 5% of dioxane in feed.
Dioxane in raffinate= 0.05 x 250
= 12.5 Kg
X’ = (12.5/ 750)=0.01557
Now extract and reffinate phase leave in equilibrium
y’ = 0.0170
Now, (-A/B) = {(y’
s-y’
s)/x’
F-x’
(-750/B)={(0- 0.0170)/(0.333-0.01557)}
B=14,004Kg

89. •Since equal amount of solvent in
each stage the operating line for all
stages have the same slope. The
equilibrium values for stage 1(x’
1
,y’
1), for the stage 2 through (x’
2,
y’
2) and for the stage 3 through (x’
3,
y’
3) respectively, which lies on the
equilibrium curve

90. •x’
F
= 0.333, y’
s
= 0
x’
3
= 0.01557, y’
3
=0.017
Slope = -(A/B) = 0.5
B= 2 A = 1500 Kg
}by trial and
error method
Solvent required = 3x1500=4500Kg
conc. Of dioxane = {(250- 12.5)/(4500)} =0.05277

91. Multi stage counter-current
extraction
•For immiscible liquids, equilibrium curve is st.
line and solvent is pure, i.e.
y’
s
=0 and y’=mx’, s
1= s
2= s
3= B
(-A/ B
n) = {(y’
s–y’
n)/(x’
n-1–x’
n)}
Becomes (- A/ B) = -{(y’
n)/(x’
n-1–x’
n)}
B y’
n = A (x’
n-1–x’
n)
mBx
1’ = A (x’
F–x’
1)
(A+ mB) x’
1= A x’
F

92. •x’
1= {A/(A+mB)} x’
F
Similarly x’
2= {A/(A+mB)} x’
1
x’
2 = {A/(A+mB)}
2
x’
F
By extension, x’
n= {A/(A+mB)}
n
x’
F
n= {ln (x’
n/x’
F)/ln(A/mB)} recast earlier
example & find solvent requirement

131. •Another rectangular coordinate system
involves plotting as abscissa the weight
fraction C on a B free basis, X and Y in
the A-rich and B-rich phases,
respectively, against N, the B-
concentration on a B-free basis, as
ordinate, as shown in the upper part of
fig. 10.10. This has been plotted for a
system of two partly miscible pairs, such
as that of fig. 10.5

134. Liquid-Liquid Extraction

135. Here is what you will learn in this chapter.
5.1 Introduction to Extraction Processes
5.2 Equilibrium Relations in Extraction
5.3 Single- Stage Equilibrium Extraction
5.4 Equipment for Liquid- Liquid Extraction
5.5 Continuous Multistage Countercurrent
Extraction
135

136. 5.1 Introduction to Extraction Processes
“When separation by distillation is ineffective or very difficult e.g.close-
boiling mixture, liquid extraction is one of the main alternative to
consider.”
What is Liquid-liquid extraction (or solvent extraction)?
Liquid-Liquid extraction is a mass transfer operation in which a liquid
solution (feed) is contacted with an immiscible or nearly immiscible liquid
(solvent) that exhibits preferential affinity or selectivity towards one or
more of the components in the feed. Two streams result from this contact:
a) Extract is the solvent rich solution containing the
desired extracted solute.
b) Raffinateis the residual feed solution containing little solute.
136

137. Liquid-liquid extraction principle
WhenLiquid-liquid extractioniscarriedoutinatesttubeorflaskthe
twoimmisciblephasesareshakentogethertoallowmoleculesto
partition(dissolve)intothepreferredsolventphase.
5.1 Introduction to Extraction Processes
137

138. An example of extraction:
5.1 Introduction to Extraction processes
Acetic acid in H
2O
+
Ethyl acetate
Extract Organic layer contains most of acetic acid in
ethyl acetate with a small amount of water.
Raffinate
Aqueous layer contains a weak acetic acid
solution with a small amount of ethyl
acetate.
The amount of water in the extract and ethyl acetate in the raffinate
depends upon their solubilites in one another.
138

139. Equilateral triangular diagram
(A and B are partially miscible.)
Triangular coordinates and equilibrium data
Each of the three corners
represents a pure component A,
B, or C.
Point M represents a mixture of
A, B, and C.
The perpendicular distance from
the point M to the base AB
represents the mass fraction x
C.
The distance to the base CB
represents x
A, and the distance
to base AC represents x
B.
x
A+ x
B+ x
C= 0.4 + 0.2 + 0.4 = 1
x
B= 1.0 -x
A-x
C
y
B= 1.0 -y
A-y
C
5.2 Single- stage liquid-liquid extraction processes
139

140. Liquid C dissolves completely in A or in B.
Liquid A is only slightly soluble in B and B slightly soluble in A.
The two-phase region is included inside below the curved envelope.
An original mixture of composition M will separate into two phases a and b which are on
the equilibrium tie line through point M.
The two phases are identical at point P, the Plait point.
Liquid-Liquid phase diagram where components A and B are partially
miscible.
140

141. 141

142. Point A = 100% Water
Point B = 100% Ethylene Glycol
Point C = 100% Furfural
Point M = 30% glycol, 40% water, 30% furfural
Point E = 41.8% glycol, 10% water, 48.2% furfural
Point R = 11.5% glycol, 81.5% water, 7% furfural
The miscibility limits for the furfural-water binary
system are at point D and G.
Point P (Plait point), the two liquid phases have
identical compositions.
DEPRG is saturation curve; for example, if feed
50% solution of furfural and glycol, the second
phase occurs when mixture composition is 10%
water, 45% furfural, 45% glycol or on the
saturation curve.
Liquid-Liquid equilibrium, ethylene glycol-furfural-water, 25ºC,101 kPa.
Ex 5.1
Define the composition of point A, B, C, M, E, R, P and DEPRG in
the ternary-mixture.
142

143. Equilibrium data on rectangular coordinates
The system acetic acid (A) –
water (B) – isopropyl ether
solvent (C). The solvent pair B
and C are partially miscible.
x
B= 1.0 -x
A-x
C
y
B= 1.0 -y
A-y
C
Liquid-liquid phase diagram
143

144. EX5.2 Anoriginalmixtureweighing100kgandcontaining30kgof
isopropylether(C),10kgofaceticacid(A),and60kgwater(B)is
equilibratedandtheequilibriumphasesseparated. Whatarethe
compositionsofthetwoequilibriumphases?
Solution:
Composition of original mixture is x
c= 0.3, x
A= 0.10, and x
B= 0.60.
144

145. Liquid-liquid phase diagram
1.Composition of x
C= 0.30, x
A
= 0.10 is plotted as point h.
2.The tie line gi is drawn
through point h by trial and
error.
3.The composition of the
extract (ether) layer at g is y
A
= 0.04, y
C= 0.94, and y
B=
1.00 -0.04 -0.94 = 0.02
mass fraction.
4.The raffinate(water) layer
composition at iis x
A= 0.12,
x
C= 0.02, and x
B= 1.00 –
0.12 –0.02 = 0.86.
145

146. “The solvent pairs B and C and also A and C are partially miscible.”
Phase diagram where the solvent pairs B-C and A-C are partially miscible.
146

147. Derivation of lever-arm rule for graphical addition
5.3 Single- Stage Equilibrium Extraction
MLV=+An overall mass balance:
MCCCMxLxVy =+
AMAAMxLxVy =+
A balance on A:
A balance on C:
5.1
5.2
5.3
Where x
AMis the mass fraction of A in the M stream.
147

148. Derivation of lever-arm rule for graphical addition
AAM
AMA
xx
xy
V
L
−
−
=
AMC
MCC
xx
xy
V
L
−
−
=
AAM
CMC
AMA
MCC
yx
yx
xx
xx
−
−
=
−
−
ML
MV
kgV
kgL
=
)(
)(
VL
MV
kgM
kgL
=
)(
)(
(5.4)
(5.5)
(5.6)
Sub 5.1 into 5.2
Sub 5.1 into 5.3
Sub 5.1 into 5.3
(5.7)
(5.8)
Lever arm’s rule
Eqn. 5.6 shows that points L, M, and V must lie on a straight line.
148

149. Ex 5.3 The compositions of the two equilibrium layersin Example 5.1 are for
the extract layer (V) y
A= 0.04, y
B= 0.02, and y
C= 0.94, and for the raffinate
layer (L) x
A= 0.12, x
B= 0.86, and x
C= 0.02. The original mixture contained
100 kg and x
AM= 0.10. Determine the amounts of V and L.
Solution: Substituting into eq. 5.1
Substituting into eq. 5.2, where M = 100 kg and x
AM= 0.10,
Solving the two equations simultaneously, L = 75.0 and V = 25.0. Alternatively, using
the lever-arm rule, the distance hg in Figure below is measured as 4.2 units and gi
as 5.8 units. Then by eq. 5.8,
Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the
material-balance method.
100==+MLV
)10.0(100)12.0()04.0( =+LV
8.5
2.4
100
===
ig
ghL
M
L
149

150. 5.2 Single- stage liquid-liquid extraction processes
Single-state equilibrium extraction
MVLVL =+=+
1120
AMAAAA
MxyVxLyVxL =+=+
11112200
MCCCCCMxyVxLyVxL =+=+
11112200
We now study the separation of A from a mixture of A and B by a solvent C in a single equilibrium
stage.
0.1=++
CBAxxx
An overall mass balance:
A balance on A:
A balance on C:
5.9
5.10
5.11
150

151. To solve the three equations, the equilibrium-phase-diagram is used.
1. L
0and V
2are known.
2. We calculate M, x
AM, and x
CMby
using equation 5.9-5.11.
3. Plot L
0, V
2, M in the Figure.
4. Using trial and error a tie line is
drawn through the point M, which
locates the compositions of L
1and V
1.
5. The amounts of L
1and V
1can be
determined by substitution in
Equation 5.9-5.11 or by using lever-
arm rule.
151

152. Ex 5.4 A mixture weighing 1000 kg contains 23.5 wt% acetic acid (A) and
76.5 wt% water (B) and is to be extracted by 500 kg isopropyl ether (C) in a
single-stage extraction. Determine the amounts and compositions of the
extract and raffinate phases.
Solution Given:
kgVandkgL 5001000
20==
AM
x)1500()0)(500()235.0)(1000( =+
kgMVL 15005001000
20
==+=+
0.1765.0,235.0
200===
ABAyandxx
Given:
157.0=
AM
x
MCCC
MxyVxL =+
2200
Given: 0765.0235.00.11
000
=−−=−−=
BAc
xxx
AMAA MxyVxL =+
2200
MCx)1500()1)(500()0)(1000( =+
33.0=
CM
x 152

153. M
V
2 (0,1) = (y
A2, y
C2)
V
1(0.1,0.89) = (y
A1, y
C1)
L
1(0.2,0.03) = (x
A1, x
C1)
L
0(0.235,0) = (x
A0, x
C0)
M(0.157,0.33) = (x
AM, x
CM)
153

154. (1)
AMAA
MxyVxL =+
1111
MCCC MxyVxL =+
1111
)157.0)(1500()1.0()2.0(
11
=+VL
)33.0)(1500()89.0()03.0(
11 =+VL
From the graph: x
C1= 0.03 and y
C1= 0.89;
From the graph: x
A1= 0.2 and y
A1= 0.1;
(2)
5.177,15.0
11
=+VL
500,1667.29
11
=+ VL
Solving eq(2) and eq(3) to get L
1and V
1;
kgVandkgL 28.52586.914
11==
89.003.0,1.0,2.0
1111
====
CCAA
yandxyx Answer
154

155. 5.3 Equipment for Liquid- Liquid Extraction
Introduction and Equipment Types
As in the separation processes of distillation, the two phases in liquid-
liquid extraction must be brought into intimate contact with a high
degree of turbulence in order to obtain high mass-transfer rates.
Distillation: Rapid and easy because of the large difference in
density (Vapor-Liquid).
Liquid extraction: Density difference between the two phases is not
large and separation is more difficult.
Liquid extraction equipment
Mixing by mechanical
agitation
Mixing by fluid flow
themselves
155

156. Mixer-Settles for Extraction
Separate mixer-settler Combined mixer-settler
156

157. Plate and Agitated Tower Contactors for Extraction
Perforated plate tower Agitated extraction tower
157

158. Packed and Spray Extraction Towers
Spray- type extraction tower Packed extraction tower
158

159. 5.4 Continuous multistage countercurrent extraction
Countercurrent process and overall balance
MVLVL
NN
=+=+
+ 110
MCCNCNNCNCMxyVxLyVxL =+=+
++ 111100
1
11
10
1100VL
yVxL
VL
yVxL
x
N
CNCN
N
NCNC
MC+
+
=
+
+
=
+
++
1
11
10
1100
VL
yVxL
VL
yVxL
x
N
AANN
N
ANNA
MA+
+
=
+
+
=
+
++
An overall mass balance:
A balance on C:
Combining 5.12 and 5.13
Balance on component A gives
5.12
5.13
5.14
5.15
159

160. 5.4 Continuous multistage countercurrent extraction
Countercurrent process and overall balance
1. Usually, L
0and V
N+1are known and
the desired exit composition x
ANis set.
2. Plot points L
0, V
N+1, and M as in the
figure, a straight line must connect these
three points.
3. L
N, M, and V
1must lie on one line.
Also, L
Nand V
1must also lie on the
phase envelope.
160

161. Ex 5.5 Pure solvent isopropyl ether at the rate of V
N+1= 600 kg/h is being
used to extract an aqueous solution of L
0=200 kg/h containing 30 wt% acetic
acid (A) by countercurrent multistage extraction. The desired exit acetic acid
concentration in the aqueous phase is 4%. Calculate the compositions and
amounts of the ether extract V
1and the aqueous raffinate L
N. Use equilibrium
data from the table.
Solution: The given values are V
N+1= 600kg/h, y
AN+1= 0, y
CN+1= 1.0, L
0= 200kg/h,
x
A0= 0.30, x
B0= 0.70, x
C0= 0, and x
AN= 0.04.
In figure below, V
N+1and L
0are plotted. Also, since L
Nis on the phase
boundary, it can be plotted at x
AN= 0.04. For the mixture point M,
substituting into eqs. below,
75.0
600200
)0.1(600)0(200
10
1100
=
+
+
=
+
+
=
+
++
N
NCNC
MC
VL
yVxL
x
075.0
600200
)0(600)30.0(200
10
1100
=
+
+
=
+
+
=
+
++
N
ANNA
MAVL
yVxL
x
161

162. Usingthesecoordinates,
1)PointMisplottedinFigurebelow.
2)WelocateV
1bydrawingalinefromL
NthroughMandextendingituntil
itintersectsthephaseboundary.Thisgivesy
A1=0.08andy
C1=0.90.
3)ForL
Navalueofx
CN=0.017isobtained. BysubstitutingintoEqs.5.12
and5.13andsolving,L
N=136kg/handV
1=664kg/h.
162

163. Stage-to -stage calculations for countercurrent extraction.
1120VLVL +=+
nnnnVLVL +=+
+− 11
∆=−=−
2110
VLVL
....
1110
=−=−=−=∆
++ NNnn
VLVLVL
...
11111100
=−=−=−=∆
++++∆ NNNNnnnn
yVxLyVxLyVxLx
Total mass balance on stage 1
Total mass balance on stage n
From 5.16 obtain difference Δin flows
5.16
5.17
5.18
5.19
5.20
Δis constant and for all stages
163

164. Stage-to -stage calculations for countercurrent extraction.
1
11
1
11
10
1100
+
++
+
++
∆
−
−
=
−
−
=
−
−
=
NN
NNNN
nn
nnnn
VL
yVxL
VL
yVxL
VL
yVxL
x
10
VL +∆=
1++∆=
nnVL
1++∆=
NNVL
5.21
5.22
Δx is the x coordinate of point Δ
5.18 and 5.19 can be written as
164

165. Stage-to-stage calculations for countercurrent extraction.
1. Δis a point common to all streams passing each
other, such as L
0and V
1, L
nand V
n+1, L
nand V
n+1,
L
Nand V
N+1, and so on.
2. This coordinates to locate this Δoperating point
are given for x
cΔand x
AΔin eqn. 5.21. Since the
end points V
N+1, L
Nor V
1, and L
0are known, x
Δcan
be calculated and point Δlocated.
3. Alternatively, the Δ point is located graphically in
the figure as the intersection of lines L
0V
1and L
N
V
N+1.
4. In order to step off the number of stages using
eqn. 5.22 we start at L
0and draw the line L
0Δ,
which locates V
1on the phase boundary.
5. Next a tie line through V
1locates L
1, which is in
equilibrium with V
1.
6. Then line L
1Δis drawn giving V
2. The tie line
V
2L
2is drawn. This stepwise procedure is
repeated until the desired raffinate composition L
N
is reached. The number of stages N is obtained to
perform the extraction.
165

166. Ex 5.6 Pure isopropyl ether of 450 kg/h is being used to extract an aqueous
solution of 150 kg/h with 30 wt% acetic acid (A) by countercurrent multistage
extraction. The exit acid concentration in the aqueous phase is 10 wt%.
Calculate the number of stages required.
Solution: The known values are V
N+1= 450, y
AN+1= 0, y
CN+1= 1.0, L
0= 150, x
A0=
0.30, x
B0= 0.70, x
C0= 0, and x
AN= 0.10.
1. The points V
N+1, L
0, and L
Nare plotted in Fig. below. For the mixture point M,
substituting into eqs. 5.12 and 5.13, x
CM= 0.75 and x
AM= 0.075.
2. The point M is plotted and V
1is located at the intersection of line L
NM with the
phase boundary to give y
A1= 0.072 and y
C1= 0.895. This construction is not shown.
3.ThelinesL
0V
1andL
NV
N+1aredrawnandtheintersectionistheoperatingpointΔ
asshown.
166

167. 167
1.Alternatively,thecoordinatesofΔcan
becalculatedfromeq.5.21tolocate
pointΔ.
2.StartingatL
0wedrawlineL
0Δ,which
locatesV
1.ThenatielinethroughV
1
locatesL
1inequilibriumwithV
1.(The
tie-linedataareobtainedfroman
enlargedplot.)
3.LineL
1ΔisnextdrawnlocatingV
2.Atie
linethroughV
2givesL
2.
4.AlineL
2ΔisnextdrawnlocatingV
2.A
tielinethroughV
2givesL
2.
5.AlineL
2ΔgivesV
3.
6.AfinaltielinegivesL
3,whichhasgone
beyondthedesiredL
N.Hence,about
2.5theoreticalstagesareneeded.

168. 5.4 Continuous multistage countercurrent extraction
Countercurrent-Stage Extraction with Immiscible Liquids








−
′+








−
′=








−
′+








−
′
+
+
1
1
1
1
0
0
1111 y
y
V
x
x
L
y
y
V
x
x
L
N
N
N
N








−
′+








−
′=








−
′+








−
′
+
+
1
1
1
1
0
0 1111 y
y
V
x
x
L
y
y
V
x
x
L
n
n
n
n
If the solvent stream V
N+1contains components A and C and the feed stream L
0
contains A and B and components B and C are relatively immiscible in each other, the
stage calculations are made more easily.The solute A is relatively dilute and is being
transferred from L
0to V
N+1.
Where L
/
= kg inert B/h, V
/
= kg inert C/h, y = mass fraction A in V stream, and x =
mass fraction A in L stream. (5.24) is an operating-line equation whose slope ≈ L
/
/V
/
.
If y and x are quite dilute, the line will be straight when plotted on an xy diagram.
5.23
5.24
168

169. Ex 5.7 An inlet water solution of 100 kg/h containing 0.010 wt fraction
nicotine (A) in water is stripped with a kerosene stream of 200 kg/h
containing 0.0005 wt fraction nicotine in a countercurrent stage tower. The
water and kerosene are essentially immiscible in each other. It is desired to
reduce the concentration of the exit water to 0.0010 wt fraction nicotine.
Determine the theoretical number of stages needed. The equilibrium data are
as follows (C5), with x the weight fraction of nicotine in the water solution and
y in the kerosene.
X y x y
0.001010 0.000806 0.00746 0.00682
0.00246 0.001959 0.00988 0.00904
0.00500 0.00454 0.0202 0.0185
169

170. Solution: The given values are L
0= 100 kg/h, x
0= 0.010, V
N+1= 200 kg/h, y
N+1=
0.0005, x
N= 0.0010. The inert streams are
hrwaterkgxLxLL /0.99)010.01(100)1()1(
00
=−=−=−=′
hrosenekgyVyVV
NN /ker9.199)0005.01(200)1()1(
11
/ =−=−=−=
++
Making an overall balance on A using eq. 5.23 and solving, y
1= 0.00497.
These end points on the operating line are plotted in Fig. below. Since the
solutions are quite dilute, the line is straight. The equilibrium line is also
shown. The number of stages are stepped off, giving N = 3.8 theoretical
stages.
170

171. 171

172. Any Questions or Additions

173. THANK YOU

174. Definethefollowingterms:
[Extraction, etc]
Respondtothefollowingquestions:
Giveadetailedaccountof………………
Explainindetailsthe processof…………..
Describeindetailswithexamplesthe…………
Withexamples,illustratethe pharmaceuticalapplicationsof……………

175. Groupworkdiscussionalquestions:
Explainindetailstheprocessof………
Describewithexamplesindetailsthe…………..
Withexamples,illustratethepharmaceuticalapplicationsof…….

176. 176
Homework No.9
1.A single-stage extraction is performed in which 400 kg of a solution
containing 35 wt% acetic acid in water is contacted with 400 kg of pure
isopropyl ether. Calculate the amounts and compositions of the extract
and raffinatelayers. Solve for the amounts both algebraically and by
the lever-arm rule. What percent of the acetic acid is removed?

177. 177
Homework No.10
1.Pure water is to be used to extract acetic acid from 400 kg of a feed
solution containing 25 wt% acetic acid in isopropyl ether.
(a)If 400 kg of water is used, calculate the percent recovery in the water
solution in a one-stage process.
(b)If a multiple four-stage system is used and 100 kg fresh water is used
in each stage, calculate the overall percent recovery of the acid in the
total outlet water. (Hint: First, calculate the outlet extract and raffinate
streams for the first stage using 400 kg of feed solution and 100 kg of
water. For the second stage, 100 kg of water contacts the outlet organic
phase from the first stage. For the third stage, 100 kg of water contacts
the outlet organic phase from the first stage. For the third stage, 100 kg
of water contacts the outlet organic phase from the second stage, and
so on.)