CHAPTER / COTET Definition & Application Solvent selectivit LLE for Partiall Miscible Solvent LLE for Immiscible Solvent Liquid liquid etraction equipment
Definition & Application The separation of constituents (solutes) of a liquid solution b 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 miture b treatment with a solvent in which one or more desired components is soluble. There are two requirements for liquid liquid etraction to be feasible: component (s) to be removed from the feed must preferentiall distribute in the solvent. the feed and solvent phases must be substantiall immiscible The two liquid phases must be either immiscible, or partiall miscible. usuall isothermal and isobaric can be done at low temperature (good for thermall fragile solutes, such as large organic molecules or biomolecules) can be ver difficult to achieve good contact between poorl miscible liquids (low stage efficienc) etracting solvent is usuall reccled, often b distillation (epensive and energ-intensive) can be single stage (mier-settler) or multistage (cascade) 2
Definition & Application The simplest LLE involves onl a ternar (i.e 3 component sstem) Important terms ou need to know: Feed - The solution which is to be etracted (denoted b component A) Solvent - The liquid with which the feed is contacted (denoted b component C) Diluent - Carrier liquid (denoted b component B) Etract (E) - The solvent rich product of the operation Raffinate (R) -The residual liquid from which solutes has been removed. Definition & Application In some operations, the solutes are the desired product, hence the etract stream is the desirable stream. In other applications, the solutes m be contaminants that need to be removed, and in this instance the raffinate is the desirable product stream. Etraction processes are well suited to the petroleum industr because of the need to separate heat sensitive liquid feeds according to chemical tpe (e.g aromatic, naphthenic) rather than b molecular weight or vapor pressure. Application: Major applications eist in the biochemical or pharmaceutical industr, where emphasis is on the separation of antibiotics and protein recover. In the inorganic chemical industr, the are used to recover high boiling components such as phosphoric acid, boric acid and sodium hdroide from aqueous solution. 3
Definition & Application Eamples: Etraction of nitrobenzene after reaction of HO 3 with toluene in H 2 SO 4 Etraction of methlacrlate from organic solution with perchlorethlene Etraction of benzlalcohol from a salt solution with toluene. Removing of H 2 S from LPG with MDEA Etraction of caprolactam from ammonium sulfate solution with benzene Etraction of acrlic acid from wastewater with butanol Removing residual alkalis from dichlorohdrazobenzene with water Definition & Application Eamples: Etraction of methanol from LPG with water Etraction of chloroacetic acid from methlchloroacetate with water. The difference between LLE and distillation process in the separation of liquid mitures: 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. 4
Definition & Application Advantages of LLE over distillation process: Where distillation requires ecessive amount of heat Presence of azeotropes or low relative volatilities are involved (α value near unit and distillation cannot be used) Dissolved or comple inorganic substances in organic or aqueous solution Removal of a component present in small concentrations, e.g hormones in animal oil. Recover of a high boiling point component present in small quantities in waste stream, e.g acetic acid from cellulose acetate. Recover of heat sensitive materials (e.g food) where low to moderate processing temperatures are needed. Thermal decomposition might occur. Solvent recover is eas and energ savings can be realized. Solvent selectivit The solvent is the ke to a successful separation b LLE. The several criteria: Distribution Coefficient. This is the ratio (at equilibrium) of the concentration of solute in the etract and raffinate phases. It gives a measure of the affinit of the solute for the two phases. Selectivit (Separation factor). If there are more than one solutes (sat two solutes A and B), then consideration should be given to the selectivit of the solvent for solutes A against B. The selectivit between two solutes A and B is defined as the ratio of the distribution coefficient of A to the distribution coefficient of B. For all useful etraction operation the selectivit must eceed unit. If the selectivit is unit, no separation is possible. 5
Solvent selectivit Insolubilit of Solvent. The solvent should have low solubilit in the feed solution, otherwise the separation is not clean. For eample, if there is significant solubilit of solvent in the raffinate stream, an additional separation step is required to recover the solvent. Recoverabilit. It is alwas necessar to recover the solvent for re-use, and this must ordinaril be done b other means, e.g distillation and the solvent should also be thermall stable under the distillation temperature. Densit. A large difference in densit between etract and raffinate phases permits high capacities in equipment. This is especiall important for etraction devices utilizing gravit for phase separation. Solvent selectivit Chemical Reactivit. The solvent should be stable chemicall and inert toward the other components of the sstem and toward the common materials of construction. Availabilit and Cost. Other Criteria. Toicit and flammabilit of the solvent are important occupational health and safet consideration. Stabilit of the solvent (i.e resistance to breakdown), particularl in the recover steps, is significant, especiall if the breakdown products might contaminate the products of the main separation. 6
(a) A homogeneous solution ma be formed and the selected solvent is then unsuitable. (b) The solvent ma be completel immiscible with the initial solvent. (c) The solvent ma be partiall miscible with the original solvent resulting in the formation of one pair of partiall miscible liquids. (d) The new solvent ma lead to the formation of two or three partiall miscible liquids. feed solvent miture raffinate etract E, A, F, A,0 mier settler Feed (F) contains solute A ( A ) dissolved in diluent D ( D = A ). Solvent (S) etracts A ( A ), creating the product solvent flow rate etract stream (E). The depleted feed becomes = F S = constant the product raffinate stream (R). Equilibrium (no longer VLE!) is defined b the distribution ratio, K d : S E F R diluent flow rate = F D = constant K d = A / A ote that A does not refer to gas composition. S, A,+ R, A, Usuall specified: A,+, A,0, F D /F S and A,. 7
LLE for Partiall Miscible Solvent/Immiscible SIGLE STAGE CALCULATIOS MULTISTAGE COUTER CURRET SYSTEM MULTIPLE STAGES WITH CROSSFLOW OF SOLVET There are two liquid phases Each phase is a ternar (3- component) miture of solute A, diluent D and solvent S Ternar equilibrium diagrams have 3 aes: usuall, mole or mass fractions of A, D, and S Literature data is commonl presentl on an equilateral triangle diagram (note O origin) Each ais is bounded 0 Miscibilit boundar = equilibrium line (depends on T, P) Figure 3-4 Effect of temperature on equilibrium of methlccloheane-toluene-ammonia sstem from Fenske et al., AIChE Journal,,335 (955), 955, AIChE From Separation Process Engineering, Third Edition b Phillip C. Wankat (ISB: 03382276) 202 Pearson Education, Inc. All rights reserved. 8
Consider the point M: water content ( A ) is? 0.9 ethlene glcol content ( B ) is? 0.20 furfural content ( C ) is? 0.6 check: A + B + C = Read the mole/mass fraction of each component on the ais for that component, using the lines parallel to the edge opposite the corner corresponding to the pure component. The miture M lies inside the miscibilit boundar, and will spontaneousl separate into two phases. Their compositions (E and R) are given b the tie-line through M. Raffinate region of partial miscibilit A-C The compositions of E and R converge at the plait point, P (i.e., no separation). A 2-component miture of furfural and water is partiall miscible over the compositio range from about 8 % furfural to 95 % furfural. Separation b etraction requires furfural/water ratio in this range (otherwise single phase). Etract Solvent What is the tie line? Connect the compositions of two phases in equilibrium with each other, and these compositions must be found b practical measurement. Plait Point: Represents a single phase that does not split into two phases and this must also be found b eperimental measurement 9
Equilibrium data on rectangular coordinates The sstem acetic acid (A) water (B) isopropl ether solvent (C). The solvent pair B and C are partiall miscible. B =.0 - A - C B =.0 - A - C Liquid-liquid phase diagram Liquid-liquid phase diagram. Composition of C = 0.30, A = 0.0 is plotted as point h. 2. The tie line g i is drawn through point h b trial and error. 3. The composition of the etract (ether) laer at g is A = 0.04, C = 0.94, and B =.00-0.04-0.94 = 0.02 mass fraction. 4. The raffinate (water) laer composition at i is A = 0.2, C = 0.02, and B =.00 0.2 0.02 = 0.86. 0
Each point on the conjugate line is composed of - one coordinate from the etract side of the equilibrium line - one coordinate from the raffinate side of the equilibrium line On this graph, which component is the diluent? which is the solute? Single stage calculations Solvent and the solution are in contact with each other onl once and thus the raffinate and etract are in equilibrium onl once. The solution normall binar solution containing solute (A) dissolved in a diluent or carrier (B). The etracting solvent can be either pure solvent C or ma content little A. Raffinate (R) is the eiting phase rich in carrier (B) while etract is eiting phase rich in solvent (C). When liquid solution mied with solvent (C), an intermediate phase M momentaril forms as the light liquid moves through the heav 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 etract are in equilibrium with each other.
Single stage calculations Liquid-Liquid Etraction Etracting Solvent, S s (A) Feed Solution, F F (A) Intermediate, M M (A) Etract phase, E * (A) Raffinate phase, R * (A) F Mass of feed solution S Mass of etracting solvent E Mass of etract phase R Mass of raffinate phase M Mass of intermediate F Mass fraction of A in F S Mass fraction of A in S M Mass fraction of A in M * Equilibrium mass fraction * Equilibrium mass fraction of A in E of A in R ote: Intermediate shown just for purpose of demonstration. Don t have to draw it when answering the question Single stage calculations In most single etraction, we are interested to determine the equilibrium composition and masses of raffinate and etract phases b using ternar phase diagram and simple material balances. Using material balance, Calculate the mass of intermediate M using total material balance: F S M Eq. () Determine mass fraction of solute A in intermediate M using material balance for solute A : F S M Eq. (2) F S Use both Eq. and 2 to find M value M 2
Single stage calculations On a right angle triangular diagram or equilateral triangular diagram for A-B-C sstem: Locate point F ( F ) and S ( S ) Draw a straight line from F to S Using the calculated value of M, locate point M ( M ) on the FS line. ote 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, ou can locate point E and R and hence ou can determine the composition of raffinate, R and etract, E that are in equilibrium. Single stage calculations Once ou have determine composition of R and E, ou can determine the masses of E and R using the material balance as follows: Using the total material balance: F S R E Eq. (3) Using the material balance for solute A: F S * R * E Eq. (4) F S Solve those Eq 3 and 4 to determine the masses of E and R 3
Single stage calculations Eample 00 kg of a solution containing 0.4 mass fraction of ethlene glcol (EG) in water is to be etracted with equal mass of furfural 25 0 C and 0 kpa. Using the ternar phase equilibrium diagram method, determine the followings: the composition of raffinate and etract phases the mass of etract and raffinate the percent glcol etracted Furfural rich laer Water rich laer % 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 4.5 4.5 8.0 5.5 86.0 8.5 2.0 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 4.0 77.2 8.8 50.0 4.0 36.0 3.0 60.0 9.0 5.0 33.0 6.0 5.0 33.0 6.0 Use the following equilibrium tie line to construct the ternar phase diagram Single stage calculations Solution F = 00 kg S = 00 kg F =0.4 S =0 Calculate the mass of intermediate M using total material balance F S M 00 00 M M 200 kg Determine mass fraction of solute A in intermediate M using material balance for solute A: F S M 0. 4 00 0 00 M 200 M 0. 2 F S M Locate point F & S, draw line FS. Locate point M on FS line. Draw new tie line that pass through point M. From that tie line, locate point E and R hence ou can determine the composition of R (*) and E (*) which is in equilibrium. From the graph, * = 0.26, * = 0.075 (Solution for point ) 4
Single stage calculations Right angle method F M E R S Single stage calculations Equilateral method F E M S R 5
Single stage calculations Solution (cont ) Using the total material balance F S R E 00 00 R E R 200 E Eq. (i) Using the material balance for solute A: F S * R * E 0. 4 00 000 0. 075 R 0. 26 E F S Insert eq (i) into eq above 0. 075 200 E 0. 26E 40 5 0. 075E 0. 26E 40 0. 85E 25 E 35. 4kg R 200 E 200 35. 4 R 64. 86kg % of EG etracted = (Mass of EG in etract / Mass of EG in feed) 00% % of EG etracted = * E 0. 26 35. 4 00% 00% 87. 8% F 0. 40 00 F Solution for point 2 Solution for point 3 Multi stage counter current sstem Solvent and solution which flow opposite (countercurrent) to each other, come into contact more than once and mi on stages inside the reactor. ormall numbering of the stages begin at the top down to the bottom. Thus the top most stage is named as stage, stage directl below stage is stage 2 and so on. Feed solution, F F (A) 2 Final etract, E E (A) 3 n - Etracting solvent, S S (A) Final raffinate, R X R (A) 6
Multi stage counter current sstem The analsis of multistage etraction 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: F S M Eq. () Determine mass fraction of solute A in intermediate M using material balance for solute A : F S M Eq. (2) F S Use both Eq. and 2 to find M value M Multi stage counter current sstem On a right angle triangular diagram or equilateral triangular diagram for A-B-C sstem: Locate point F ( F ) and S ( S ) Draw a straight line from F to S Using the calculated value of M, locate point M ( M ) on the FS line. ote that point M must be on FS line. Locate point E (Point M must be on E R line). Operating Points and Lines. Locate the Operating Point b finding the intersection of operating lines for the left most and right most stage. Draw a line through E and F. Draw a line through S and R. Locate the intersection P. This point is the operating point P. 7
Multi stage counter current sstem Solute Plait Point E M Feed Operating Point P R S Carrier Multi stage counter current sstem Operating Lines and Tie Lines: Stepping Off Stages: Locate point R from the tie line intersecting E Draw a line from the operating point P through R to the etract side of the equilibrium curve. The intersection locates E 2 Locate point R 2 from a tie line. Repeat Steps 2 and 3 until R is obtained Summar: E R : Tie line, R E 2 : Operating line. Stop until E value is slightl below R value 8
Multi stage counter current sstem Solute Plait Point E E5 E4 E2 E3 M R Feed Operating Point P E6 R Carrier Solvent C Multi stage counter current sstem Minimum solvent amount / minimum solvent flow rate Minimum solvent flow rate is the lowest rate / amount at which solvent could be theoreticall used for a specified etraction. 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. 9
Multi stage counter current sstem On a right angle triangular diagram or equilateral triangular diagram for A-B- C sstem: Locate point F ( F ) and S ( S ) Draw a best tie line that originate from F. The intersection of this line with etract half dome is point E min (minimum etract flow rate / amount). Draw a straight line from E min to point R. The intersection of this line with FS gives point M min. From point M min ou can read the value of min. Use the value of min and material balance to calculate the S min. % Overall efficienc of multi stage etraction column: % Overall efficienc = (number of ideal stage / number of real stage) 00% Multi stage counter current sstem Eample 2 5300 kg/h of a solution containing 30% b weight of ethlene glcol (EG) in water is to be reduced to 4.5% (solvent free) b a continuous etraction in a countercurrent column using reccled furfural that contains.5% EG as the etracting solvent: Determine the minimum solvent flow rate for the etraction above If the solvent enters at.25 times the minimum solvent rate, how man ideal stages are required? Determine the number of real stages if the overall efficienc of the column is 60% 20
Multi stage counter current sstem Solution 2 F = 5300 kg/hr F = 0.30 R SF = 0.045 S = 0.05 E min F M min S R SF Multi stage counter current sstem From the graph above, X Mmin = 0.25 From material balance: F S min M min 0. 35300 0. 05S 0 25M min. min 5300 S min M F S F s min min Mmin M min Solution for point S min =27.66 kg/h S =.25 S min =.25 27.66 kg/h S = 409.58 kg/h Calculate the mass of intermediate M using total material balance : F S M 5300 409.58 M M 6409. 58 kg Determine mass fraction of solute A in intermediate M using material balance for solute A: F F S S M M From material balance: 0.3 5300 0.05 409.58 M 6709.58 0.24 M 2
Multi stage counter current sstem F S M 0.3 5300 0.05 409.58 M 6709.58 M 0.24 F S M Solution for point 2 E E 3 E 2 M F From figure above, no of ideal stages = 5 E 4 E 5 R SF P S umber of real stages = 5 / 0.60 = 8.33 = 9 stages. Solution for point 3 LLE for Immiscible Solvent Sometimes etraction use a solvent C that is onl slightl soluble in B or the solvent C used is in range where the solubilit in B is so low that for all practice purpose, it can be assumed to be completel insoluble / immiscible in B and vice versa. Bancroft weight fractions or mass ratio, and are defined as follows: (in raffinate phase) = mass of solute A / mass of diluent B (in etract phase) = mass of solute A / mass of solvent C 22
LLE for Immiscible Solvent SIGLE STAGE CALCULATIOS MULTISTAGE COUTER CURRET SYSTEM Single stage calculations Feed solution M kg A in feed S kg solvent C in Etract kg A/kg solvent C Solvent kg A in feed B kg diluent B in Raffinate kg A/kg diluent B Solvent C is used in such a range that it is considered insoluble in B. Material balance of the solute (A) are: M ' S ' B B M ' ' Eq. (3) S S 23
Single stage calculations Eample 3 An aqueous solution of acetic acid is to be etracted in a single etractor with isopropl ether. The solution contains 24.6 kg of acetic acid and 80 kg of water. If 00 kg of isopropl ether is added to the solution, what weight of acetic acid will be etracted b isopropl ether if equilibrium is attained? Water and isopropl ether ma be considered as completel immiscible under the condition of etraction. The equilibrium data as follows: (kg acid/kg isopropl ether) 0.030 0.046 0.063 0.070 0.078 0.086 0.06 (kg acid/kg water) 0.0 0.5 0.20 0.22 0.24 0.26 0.30 Single stage calculations Solution 3 A = Acetic Acid B = Water C = Isopropl ether Feed solution 24.6 kg A 80 kg diluent B in Raffinate kg A/kg diluent B Solvent 0 kg A in feed 00 kg solvent C in Etract kg A/kg solvent C From mass balance, M ' S ' B 24.6 0 ' 00 80' ' 0. 80' 0. 246 The equilibrium data and equation above is plotted in figure net page. 24
' Single stage calculations Solution 3 0.35 ' - ' diagram for sstem acetic acid-water-isopropl ether 0.3 0.25 0.2 0.5 0. 0.05 0 0 0.02 0.04 0.06 0.08 0. 0.2 0.4 ' From the intersection of lines, = 0.062, = 0.95 Amount of acetic acid etracted = 9.5 kg (How?) Multi stage counter current sstem The principle is quite same with the partiall miscible solvent, but in this case the etraction process involved the immiscible solvent. B kg/h pure diluent B in feed solution 2 kg A/kg pure B 2 S kg/h pure solvent in etract phase 2 kg A/kg pure solvent 3 n - S kg/h pure solvent C n+ kg A/kg pure solvent B kg/h pure diluent B in raffinate n kg A/kg pure B 25
26 Countercurrent-Stage Etraction with Immiscible Liquids 0 0 V L V L 0 0 V L V L n n n n If the solvent stream V + contains components A and C and the feed stream L 0 contains A and B and components B and C are relativel immiscible in each other, the stage calculations are made more easil. The solute A is relativel dilute and is being transferred from L 0 to V +. Where Lʹ = kg inert B/h, Vʹ = kg inert C/h, = mass fraction A in V stream, and = mass fraction A in L stream. (5.24) is an operating-line equation whose slope Lʹ/Vʹ. If and are quite dilute, the line will be straight when plotted on an diagram. 5 Eample: An inlet water solution of 00 kg/h containing 0.00 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 essentiall immiscible in each other. It is desired to reduce the concentration of the eit water to 0.000 wt fraction nicotine. Determine the theoretical number of stages needed. The equilibrium data are as follows (C5), with the weight fraction of nicotine in the water solution and in the kerosene. 52 Solution: The given values are L 0 = 00 kg/h, 0 = 0.00, V + = 200 kg/h, + = 0.0005, = 0.000. The inert streams are: hr kgwater L L L / 99.0 0.00) 00( ) ( ) ( 0 0 hr osene kg V V V / ker 99.9 0.0005) 200( ) ( ) ( /
27 Making an overall balance on A using the below equation and solving, = 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 = 3.8 theoretical stages. 5 3 0 0 V L V L Solution: The given values are L 0 = 00 kg/h, 0 = 0.00, V + = 200 kg/h, + = 0.0005, = 0.000. The inert streams are hr kgwater L L L / 99.0 0.00) 00( ) ( ) ( 0 0 hr osene kg V V V / ker 99.9 0.0005) 200( ) ( ) ( / 99.9 0.00 0.00 99 0.0005 0.0005 99.9 0.0 0.0 99 = 0.005 = 3.7 5 4 99.9 0.00 0.00 99 0.0005 0.0005 99.9 0.0 0.0 99 = 0.005 0 0 V L V L
Liquid liquid etraction equipment Two main classes of solvent etraction equipment: ) Vessels in which mechanical agitation is provided for miing 2) Vessels in which the miing is done b the flow of the fluid themselves. The etraction equipment can be operated batch or continuous. Liquid liquid etraction equipment Mier Settlers for Etraction A mechanical mier is often used to provide intimate contact between the two liquid phases to provide efficient mass transfer. One phase is usuall dispersed into the other in the form of small droplet. In figure 2.6 (a) for tpical mier settler, mier or agitator is entirel separate from the settler. The feed of aqueous phase and organic phase are mied in the mier, and then the mied phases are separated in the settler. In figure 2.6 (b) for combined mier settler, sometimes used in etraction of uranium salts or copper salts from aqueous solution. 28
Liquid liquid etraction equipment Spra Etraction Towers In Figure 2.6 2 the heav liquid enters at the top of the spra 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 spras the droplets upward. The light liquid coalesces at the top and flows out. In some cases the heav liquid is spraed downward into a rising, light continuous phase. Liquid liquid etraction equipment Packed Etraction Towers More effective tower made b 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 spra tower. Table 2.6 shows tpical performance for several tpes of commercial etraction towers. 29