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  • 1. CHAPTER 2: LIQUID LIQUID EXTRACTION

2. CHAPTER / CONTENT Definition & ApplicationLLE for Partially Miscible SolventLLE for Immiscible SolventLiquid liquid extraction equipment 3. Definition & ApplicationLLE= 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 4. Definition & Application 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 operationRaffinate -The residual liquid from which solutes has been removed. 5. Definition & Application 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. 6. Definition & Application Examples: Extraction of nitrobenzene after reaction of HNO3 with toluene in H2SO4 Extraction of perchlorethylenemethylacrylatefromorganicsolutionwithExtraction 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 7. Definition & Application 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. 8. Definition & ApplicationAdvantages of LLE over distillation process: Where distillation requires excessive amount of heat Presence of azeotropes or low relative volatilities are involved ( 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. 9. LLE for Partially Miscible SolventSINGLE STAGE CALCULATIONSMULTISTAGE COUNTER CURRENT SYSTEM 10. Single stage calculations 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. 11. Single stage calculationsLiquid-Liquid Extraction Extracting Solvent, S ys (A)Intermediate, MFeed Solution, FxM (A)xF (A)F E M yS x*Mass of feed solution Mass of extract phase Mass of intermediate Mass fraction of A in S Equilibrium mass fraction of A in RS R xF xM y*Extract phase, E y* (A) Raffinate phase, R x* (A)Mass of extracting solvent Mass of raffinate phase Mass fraction of A in F Mass fraction of A in M Equilibrium mass fraction of A in ENote: Intermediate shown just for purpose of demonstration. Dont have to draw it when answering the question 12. Single stage calculations 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: F+S =MEq. (1)Determine mass fraction of solute A in intermediate M using material balance for solute A : xF F + y S S = xM MUse both Eq. 1 and 2 to find xM valueEq. (2) 13. Single stage calculations 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. 14. Single stage calculations 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:F+S =R+EEq. (3)Using the material balance for solute A: x F F + y S S = x * R + y * EEq. (4)Solve those Eq 3 and 4 to determine the masses of E and R 15. Single stage calculations 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 layerWater rich layer% EG% water% furfural% EG% water% furfural0.05.095.00.092.08.08.54.587.02.089.68.414.54.581.05.586.08.521.06.073.07.084.48.629.07.064.08.083.38.742.08.549.514.077.28.850.014.036.031.060.09.051.033.016.051.033.016.0Use the following equilibrium tie line to construct the ternary phase diagram 16. Single stage calculations Solution 1 F = 100 kgS = 100 kgxF=0.4yS=0Calculate the mass of intermediate M using total material balance F+S =M100 + 100 = MM = 200 kgDetermine mass fraction of solute A in intermediate M using material balance for solute A: xF F + y S S = xM M0.4 100 + 0 100 = x M 200x M = 0 .2Locate point F & S, draw line FS. Locate point x M 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) 17. Single stage calculations Right angle methodF M RES 18. Single stage calculations Equilateral methodF EM RS 19. Single stage calculations Solution 1 (cont) Using the total material balance F+S =R+E100 + 100 = R + ER = 200 EEq. (i)Using the material balance for solute A: x F F + y S S = x * R + y * EInsert eq (i) into eq above 0.075( 200 E ) + 0.26E = 40 15 0.075E + 0.26E = 40 0.185E = 25 E = 135.14kg R = 200 E = 200 135.14 R = 64.86kgSolution for point 20.4 100 + 0 100 = 0.075 R + 0.26 E % of EG extracted = (Mass of EG in extract / Mass of EG in feed) x 100% % of EG extracted = y * E 0.26 135.14 100% = x100% = 87.8% xF F 0.40 100Solution for point 3 20. Tutorial 1. 12.5-2 2. 12.5-4 21. Tutorial 1. 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 composition of raffinate and extract phases the mass of extract and raffinate the percent acetic acid extracted 22. Tutorial 2. 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 composition of raffinate and extract phases the mass of extract and raffinate the percent acetone extracted 23. Multi stage counter current system 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. Final extract, EFeed solution, F xF (A) 1yE (A)2 3 n N-1 NExtracting solvent, S yS (A)Final raffinate, R XR (A) 24. Multi stage counter current system 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 speci