LiquidLiquid Extractions•Liquidliquid extraction (also known as solvent extraction) involves the separation of the constituents (solutes) of a liquid solution by contact with another insoluble liquid. •Solutes are separated based on their different solubilities in different liquids. •Separation is achieved when the substances constituting the original solution is transferred from the original solution to the other liquid solution. 1 The Figure showed a feed liquid (the "first" liquid) containing the desirable compound to be separated together with other compounds. Then an immiscible extraction liquid (the "second" liquid) is added and mixed with the feed liquid through agitation. The species redistribute themselves between the 2 liquid phases. Agitation of the 2 phases is continued until equilibrium, and then agitation is stopped and the liquids are allowed to settle until both phases are clear. The 2 phases can then be separated.that is The simplest liquidliquid extraction involves only a ternary (i.e. 3 components) system. The solution which is to be extracted is called the feed, and the liquid with which the feed is contacted is the solvent. The feed can be considered as comprising the solute A and the "carrier" liquid C. Solvent S is a pure liquid. During contact, mass transfer of A from the feed to the solvent S occurs, with little transfer of C to S. 2 Operasi Perpindahan Massa Ekstraksi Dalam Industri Pengolahan Minyak • The purpose of solvent extraction is to prevent corrosion, protect catalyst in subsequent processes, and improve finished products by removing unsaturated, aromatic hydrocarbons from lubricant and grease stocks. • The solvent extraction process separates aromatics, naphthenes, and impurities from the product stream by dissolving or precipitation. The feedstock is first dried and then treated using a continuous countercurrent solvent treatment operation. 3 Application of Liquid-Liquid Extraction 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. aliphatic, aromatic, naphthenic) rather than by molecular weight or vapour pressure. Other 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 solutions. Some examples are given below: •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 •Removing residual alkalis from dichlorohydrazobenzene with water •Extraction of methanol from LPG with water •Extraction of chloroacetic acid from methylchloroacetate with water •Extraction of acrylic acid from wastewater with butanol 4 evaporation. the feedstock is washed with a liquid in which the substances to be removed are more soluble than in the desired resultant product. selected solvents are added to cause impurities to precipitate out of the product.Solvent extraction • In one type of process. In the adsorption process. or fractionation. • The solvent is separated from the product stream by heating. and residual trace amounts are subsequently removed from the raffinate by steam stripping or vacuum flashing. In another process. 5 . highly porous solid materials collect liquid molecules on their surfaces. 6 . furfural.2' dichloroethyl ether.• • • • • Electric precipitation may be used for separation of inorganic compounds. The selection of specific processes and chemical agents depends on the nature of the feedstock being treated. and the finished product requirements. and cresylic acid. the contaminants present. The most widely used extraction solvents are phenol. Other solvents less frequently used are liquid sulfur dioxide. nitrobenzene. and 2. The solvent is regenerated for reused in the process. from a Depriester chart to determine the distribution of components between the two phases. A. A. Define the extract as the exiting phase rich in solvent. A. Now we’ll analyze the following system: Ternary LiquidLiquid Extraction In this case: • We create two liquid phases by introducing a solvent (C) (MSA) to a liquid mixture of a carrier (A) and a solute (B) • Solvent (C) and carrier (A) have very little solubility in each other LiquidLiquid Extraction Solvent Feed S. B. B. For that system we needed equilibrium data. C Liquid Feed F. B Solvent Rich Liquid Out Carrier Rich Liquid out R. C Define the raffinate as the exiting phase rich in carrier.LiquidLiquid Ternary Single Equilibrium Stage We last covered the Flash Calculations where a liquid phase and a vapor phase were in equilibrium. 7 . for example. C E. XB (E). XC (E). T. T. P Raffinate out R. XA (E). XB. P The raffinate is the exiting phase rich in carrier. P Liquid Feed F. 8 . XA (R).T. XC (R). P E. then the raffinate will have a small amount of solvent in it and the extract will have a small concentration of carrier: LiquidLiquid Extraction Solvent Feed Extract out S. The extract is the exiting phase rich in solvent.LiquidLiquid Ternary Single Equilibrium Stage If the solvent and carrier have some solubility in each other. T. XC(S). XA(F). XB (R). P Extract out E. P All of carrier exits in the raffinate The raffinate is the exiting phase rich in carrier. XB (R). T. XB (E). XC(S). XB. P Liquid Feed F.T. XC (E).LiquidLiquid Ternary Single Equilibrium Stage If the solvent and carrier have no solubility in each other. 9 . P Raffinate out R. T. The extract is the exiting phase rich in solvent. T. XA(F). XA (R). then the raffinate will have no solvent in it and the extract will have no carrier in it: All of solvent exits in the extract LiquidLiquid Extraction Solvent Feed S. 10 . Note that the basis (choice of component) for the mass or mole ratio must be chosen. Mole ratio XB: The ratio of moles of component B to another component of the stream. XA(F). times feed rate of C. T. 11 . XB (E). times feed rate of A. This is generally done to simplify the expressions used in the analysis.Mass and Mole Ratios Often the concentrations are as mass or mole ratios. XB. T. T. Mass ratio XB: The ratio of mass of component B to another component of the stream. Rate of B in the solvent is the ratio of B to C. P Liquid Feed F. times rate of A. Mass Ratio Example: Solvent Feed S.T. Rate of B in the extract is the ratio of B to C. P Rate of B in the feed is the ratio of B to A. P FB XB F FA EB X BE EC XBE S RB XB R RA XB R FA S B X BS SC 0 Extract out E. XA (R). XC (E). P Raffinate out R. Rate of B in the raffinate is the ratio of B to A. rather than mass or mole fractions. times rate of C. XC(S). XB (R). XC (E) Liquid Feed Raffinate out F. FB SB RB EB S R E X B FA XB SC XB RA X B EC X BF FA XBR FA X BE S 12 . times feed rate of C. Rate of B in the raffinate is the ratio of B to A. XA (R). XA(F). XC(S) E. XB R. times rate of C. EB X BE EC XBE S R R RB XB RA XB FA S B X BS SC 0 Solute Material Balance: F Rate of B in the extract is the ratio of B to C. Rate of B in the solvent is the ratio of B to C. XB (E). times rate of A. times feed rate of A. XB (R) FB XB F FA Rate of B in the feed is the ratio of B to A.Material Balances Solute Material Balance: Solvent Feed Extract out S. XB (R) Note that concentrations of exiting streams from an equilibrium stage are always related by equilibrium. 13 . not a ratio of mole fractions. XA (R). Extract out B E. XB Note that the Kvalue is primed to signify that this is a ratio of mass or mole ratios.Equilibrium Distribution The way the solute will distribute itself between the extract and raffinate at equilibrium is given by the KValue: X BE K'DB XB R Solvent Feed S. XB (E). XC (E) Raffinate out R. XC(S) Liquid Feed F. XA(F). The Extraction Factor The degree of separation of the solute between the exiting streams is expressed as the extraction factor: Extraction Factor: The ratio of solute flow in the extract to solute flow in the raffinate. 14 . the larger the extraction factor. and the larger the ratio of solvent to feed. EB X BE EC XBE S RB XB R RA XB R FA XB E S EB B R RB X B FA Combining this definition with the equilibrium relationship: X BE K'DB XB R results in another expression for the extraction factor: B K'DB S FA The larger the equilibrium driving force to separate B. XB R XB R FA F K' S F XB DB A 1 1 F K' S B 1 XB DB 1 FA The amount not extracted increases with the feed rate. and smaller ratio distribution between extract and raffinate and less solvent. 15 .Extraction Efficiency We can determine the amount not extracted starting with the material balance of the solute: X BF FA XBE S X BR FA We substitute in the Kvalue ratio: X BF FA K'DB XB R S XB R FA And simplify: This ratio gives the amount of solute left in the raffinate to the amount originally in the feed stream. Note that the variables for these diagrams are only composition and that pressure and temperature are held constant (that is that these diagrams are slices through a four dimensional space with constant T and P). A C B 16 .Ternary Phase Diagrams It is convenient to construct ternary phase diagrams on a Gibbs Triangle (shown at right). 60% A] C [94% C. Note: Only two mole fractions are needed (use the third as a check). 3% B. Read the compositions off of the three axes. 20% B. 33% B. Compositions can be mole fractions or mass fractions. 70% B] [100% A] A B 17 . 3% A] [33% C. 33% A] [30% C.Ternary Phase Diagrams Compositions are read as follows: Draw three lines from the composition point parallel to the composition lines. [20% C. Construction of Ternary Phase Diagrams G 18 . Ternary Eutectic Phase Diagrams G G L L 19 . Ternary Eutectic Phase Diagrams C T above T of the ternary eutectic. but below the binary eutectics. T L +L + + + + +L +L L +L + +L 0 XB 1 +L + +L A + B 20 . Ternary Phase Diagrams C +L L A B 21 . Equilibrium data can be obtained graphically.Partially Soluble Ternary Systems If the two phases both have a partial solubility of the other component. The ternary phase diagram is a typical way of representing the equilibrium compositions of the two phases: Solute Ethylene Glycol Water 50 % EthGly 50% Furfural 17% EthGly 27% Furfural 56 % Water A composition where two liquid phases coexist. then the analysis is somewhat more complicated: The difficulty is that now equilibrium data must be obtained for the ternary which relates the partial solubilities. Furfural Carrier 22 . or from tables. 100% Furfural Solvent 66% EthGly 7% Furfural 27 % Water A composition where only a single liquid exists. Specification of LiquidLiquid Equilibrium For two phase equilibrium (either complete insolubility. Furfural 23 . Ethylene Glycol Water TieLines: Show the compositions of the equilibrium phases. or partially solubility): • the equilibrium is between two liquids phases ( = 2) • three components (ternary) distribute between the two phases (C = 3) For the static equilibrium case we can specify 3 variables: If we specify T and P we are left with one additional variable: Thus if we specify the concentration of one component in either of the phases this completely defines the state of the system. 24 .Tie Line • Tie line adalah garis yang menghubungkan satu titik pada rafinat dan pada extraktan • Campuran yang komposisinya terletak dalam tie line akan memiliki kesetimbangan sama yaitu pada kedua titik diujung tie line • Tie line dibuat dengan cara membiarkan suatu campuran yang diketahui komposisinya mencapai kesetimbangan 2 fasa. Namun berguna dalam memahami neraca masa perhitungan selanjutnya L0 + V2 = L1 + V1 = M L.yC = M.xAM L.xAM L.xCM Ekstrak E V1 V2 L0 L1 Campuran Rafinat R 25 .xA + V.yA = M.xC + V.Ekstraksi Kesetimbangan Satu tahap • Counter current satu tahap seperti dilukiskan pada gambar sebelah kiri bawah ini tidaklah ada dalam aplikasinya (bisa anda jelaskan mengapa?).xCM L+V=M L.xAM L.xAM L.xA + V.yA = M.xC + V.yA = M.yA = M.xA + V.yC = M.xA + V. 9. 5. Hubungkan L1 dengan . perpotongan pada satu sisi amplop menjadi V2 7. Buat garis yg menghubungkan LN dan VN+1 3. Buat pertemuan kedua garis tsb (1 & 2) > 4. Dst sampai melewati LN. Buat tie line yg mengenai M. Buat tie line kedua dari V2.Ekstraksi jamak 1. M dan L0 VN+1 2. Pada sisi amplop fasa > L1 & V1 6. Jumlah tahap = jumlah garis yg menemui V2 V1 M L0 LN L1 26 . Buatlah garis yang menghubungkan VN+1. 8. 27 . 28 . 29 . Add 300kg of pure furfural solvent. XB (R) Ethylene Glycol Water Furfural 30 . XB R. XA (R). XC (E) Liquid Feed Raffinate out F. Solvent Feed Extract out S. XB (E). XC(S) E. XA(F).Partially Soluble Ternary Systems Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Step 1: Locate the Solvent and Feed points Ethylene Glycol S 300 Kg Water F 60 kg EG 140 kg water Furfural 31 . Add 300kg of solvent which is pure furfural.Partially Soluble Ternary Systems Example: Consider a feed of 200 kg of 30% ethylene glycol in water. 12 FS 500kg Ethylene Glycol Water F 60 kg EG 140 kg water M S 300 Kg Furfural 32 .Partially Soluble Ternary Systems Step 2: Locate the mixing point M: X BF FA XBS S 0.3 200kg 0 300kg 0. Extract (4% water. 82% furfural) Raffinate (87% water. 5%EG. 8% furfural) Ethylene Glycol Water F 60 kg EG 140 kg water M E R S 300 Kg Furfural 33 . 14%EG.Partially Soluble Ternary Systems Step 3: Use the tieline to get the raffinate and extract compositions. 82 R f R E 0.4kg E 1 0. 5%EG. 14%EG.08 0. 8% furfural) R 0.Partially Soluble Ternary Systems Step 4: Determine the amount of extract and raffinate (can use lever rule) XC M XC E 0.257 500kg 128.63 0. 82% furfural) Raffinate (87% water.6kg Ethylene Glycol Water F 60 kg EG 140 kg water M E R S 300 Kg Furfural 34 .257 XC XC Extract (4% water.82 0.257 500kg 371. Partially Soluble Ternary Systems Step 5: Determine the solvent free extract: Mixtures of E and S. Extend line from S through E to solvent free point at H. 80% EG) X BR XB H Ethylene Glycol Water F 60 kg EG 140 kg water M E F R S 300 Kg Furfural 35 . Solvent free extract H (20% water. Differential extractors Extract Extract Feed Extract Feed Feed Extracting solvent Raffinate Extract Feed Extracting solvent Raffinate Extracting solvent Raffinate Raffinate 36 . Ekstraksi Diferensial • • • • • Continuous process Feed and extracting solvent flow past one another One phase is dispersed in the other Contacting and phase separation takes place within one unit Phases are not in equilibrium except locally i.e at the interface Extract Feed Extracting solvent Raffinate 37 . Transport zat terlarut pada ekstraksi Raffinate side film yi Concentration profile y Raffinate Raffinate Feed Extract x xi Extract side film Raffinate side film Extracting solvent Extract side film Interface 38 . tekanan uap dan titik beku Toksisitas. nonflammable dan murah 39 .Pemilihan Pelarut (Solvent) • • • • • • • • • Selektivitas Adalah kemampuan pelarut untuk melarutkan suatu zat yang dikehendaki. yaitu Koefisien distribusi Ketidaklarutan solven Rekoverabilitas Densitas Tegangan permukaan Reaktivitas kimia Viskositas. Selektivitas ditentukan dengan menghitung . 40 . • Ternary phase diagrams. • A procedure to determine the product compositions and flow rates of a liquidliquid extraction separation.Summary This lecture covered: • Ternary LiquidLiquid extractions. Next lecture will cover: • Leaching • Crystallization 41 . ALTERNATE VERSION • USING SOLVENTFREE DIAGRAM 42 . MASS BALANCE EQUATIONS • OVERALL & COMPONENT La Vb Lb Va M xa La y bVb x b Lb y aVa x m M • CAN BE REARRANGED TO SOLVE FOR Va AND Lb ON THE BASIS OF La AND Vb FED TO PROCESS La y b xm Vb xm xa FROM LEVER RULE 43 . Va]1 [ya.M]1 [xm.Va]2 [xb.La [ya.M]2 xA.Vb]1.Lb]2 B S [yb.La]2 [xb.Lb]1 = [xa.MULTIPLE CROSSCURRENT EXTRACTION A [xm.2 44 . COUNTERCURRENT MULTISTAGE EXTRACTION • MINIMUM SOLVENT DETERMINATION BASED ON EQUILIBRIUM LINE THROUGH xa A xj.min S yb xb B 45 .min xa P.min ya. DESIGN SOLVENT RATES • GRAPHICAL CALCULATION OF NUMBER OF STAGES 46 . 1518.TRANSFER TO McCABETHIELE • • OPERATING LINE SLOPE BASED ON L/V PERRY’S P. 47 .