#### Transcript What is Liquid-liquid extraction

Chapter 5 Liquid-Liquid Extraction Subject: 1304 332 Unit Operation in Heat transfer Instructor: Chakkrit Umpuch Department of Chemical Engineering Faculty of Engineering Ubon Ratchathani University 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 2 5.1 Introduction to Extraction Processes “When separation by distillation is ineffective or very difficult e.g. closeboiling 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) Raffinate is the residual feed solution containing little solute. 3 5.1 Introduction to Extraction Processes Liquid-liquid extraction principle When Liquid-liquid extraction is carried out in a test tube or flask the two immiscible phases are shaken together to allow molecules to partition (dissolve) into the preferred solvent phase. 4 5.1 Introduction to Extraction processes An example of extraction: Extract Acetic acid in H2O Organic layer contains most of acetic acid in ethyl acetate with a small amount of water. + Raffinate Ethyl acetate 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. 5 5.2 Single-stage liquid-liquid extraction processes 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 xC. The distance to the base CB represents xA, and the distance to base AC represents xB. xA + xB + xC = 0.4 + 0.2 + 0.4 = 1 Equilateral triangular diagram xB = 1.0 - xA - xC (A and B are partially miscible.) yB = 1.0 - yA - yC 6 Liquid-Liquid phase diagram where components A and B are partially miscible. 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. 7 8 Ex 5.1 Define the composition of point A, B, C, M, E, R, P and DEPRG in the ternary-mixture. 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. 9 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. xB = 1.0 - xA - xC yB = 1.0 - yA - yC Liquid-liquid phase diagram 10 EX 5.2 An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is equilibrated and the equilibrium phases separated. What are the compositions of the two equilibrium phases? Solution: Composition of original mixture is xc= 0.3, xA = 0.10, and xB = 0.60. 11 Liquid-liquid phase diagram 1. Composition of xC = 0.30, xA = 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 extract (ether) layer at g = 0.04, yC = 0.94, and 1.00 - 0.04 - 0.94 = mass fraction. the is yA yB = 0.02 4. The raffinate (water) layer composition at i is xA = 0.12, xC = 0.02, and xB = 1.00 – 0.12 – 0.02 = 0.86. 12 Phase diagram where the solvent pairs B-C and A-C are partially miscible. “The solvent pairs B and C and also A and C are partially miscible.” 13 5.3 Single-Stage Equilibrium Extraction Derivation of lever-arm rule for graphical addition An overall mass balance: A balance on A: V LM 5.1 Vy A LxA MxAM 5.2 Where xAM is the mass fraction of A in the M stream. A balance on C: VyC LxC MxC M 5.3 14 Derivation of lever-arm rule for graphical addition Sub 5.1 into 5.2 L y A x AM V x AM x A (5.4) Sub 5.1 into 5.3 L yC xC M V xC M xC (5.5) Equating 5.4 and 5.5 and rearranging xC xC M x A x AM xC M y C x AM y A (5.6) Eqn. 5.6 shows that points L, M, and V must lie on a straight line. By using the properties of similar right triangles, Lever arm’s rule L(kg ) V M V (kg ) L M (5.7) L(kg) V M M (kg) L V (5.8) 15 Ex 5.3 The compositions of the two equilibrium layers in Example 5.1 are for the extract layer (V) yA = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate layer (L) xA = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the amounts of V and L. Solution: Substituting into eq. 5.1 V L M 100 Substituting into eq. 5.2, where M = 100 kg and xAM = 0.10, V (0.04) L(0.12) 100(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, L L h g 4.2 M 100 gi 5.8 Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the material-balance method. 16 5.2 Single-stage liquid-liquid extraction processes Single-state equilibrium extraction We now study the separation of A from a mixture of A and B by a solvent C in a single equilibrium stage. An overall mass balance: L0 V2 L1 V1 M 5.9 A balance on A: L0 x A0 V2 y A2 L1 x A1 V1 y A1 Mx AM 5.10 A balance on C: L0 xC 0 V2 yC 2 L1 xC1 V1 yC1 MxC M 5.11 x A xB xC 1.0 17 To solve the three equations, the equilibrium-phase-diagram is used. 1. L0 and V2 are known. 2. We calculate M, xAM, and xCM by using equation 5.9-5.11. 3. Plot L0, V2, M in the Figure. 4. Using trial and error a tie line is drawn through the point M, which locates the compositions of L1 and V1. 5. The amounts of L1 and V1 can be determined by substitution in Equation 5.9-5.11 or by using leverarm rule. 18 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: L0 1000kg andV2 500kg Given: L0 V2 1000 500 M 1500kg xA0 0.235, xB0 0.765and y A2 1.0 L0 xA0 V2 y A2 MxAM (1000)(0.235) (500)(0) (1500) xAM x AM 0.157 Given: xc 0 1 xA0 xB0 1.0 0.235 0.765 0 L0 xC 0 V2 yC 2 MxC M (1000)(0) (500)(1) (1500) xC M xCM 0.33 19 V2 (0,1) = (yA2, yC2) V1 (0.1,0.89) = (yA1, yC1) M(0.157,0.33) = (xAM, xCM) L1(0.2,0.03) = (xA1, xC1) M L0(0.235,0) = (xA0, xC0) 20 From the graph: xA1 = 0.2 and yA1 = 0.1; L1 xA1 V1 y A1 MxAM L1 (0.2) V1 (0.1) (1500)(0.157) L1 0.5V1 1,177.5 (1) From the graph: xC1 = 0.03 and yC1 = 0.89; L1 xC1 V1 yC1 MxC M L1 (0.03) V1 (0.89) (1500)(0.33) L1 29.67V1 16,500 (2) Solving eq(2) and eq(3) to get L1 and V1; L1 914.86kg and V1 525.28kg xA1 0.2, y A1 0.1, xC1 0.03 and yC1 0.89 Answer 21 5.3 Equipment for Liquid-Liquid Extraction Introduction and Equipment Types As in the separation processes of distillation, the two phases in liquidliquid 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 22 Mixer-Settles for Extraction Separate mixer-settler Combined mixer-settler 23 Plate and Agitated Tower Contactors for Extraction Perforated plate tower Agitated extraction tower 24 Packed and Spray Extraction Towers Spray-type extraction tower Packed extraction tower 25 5.4 Continuous multistage countercurrent extraction Countercurrent process and overall balance An overall mass balance: A balance on C: L0 VN 1 LN V1 M 5.12 L0 xC 0 VN 1 yC N 1 LN xC N V1 yC1 MxC M Combining 5.12 and 5.13 Balance on component A gives xCM x AM L0 xC 0 VN 1 yCN 1 L0 VN 1 LN xCN V1 yC1 5.13 5.14 LN V1 L0 x A0 VN 1 y AN 1 LN x AN V1 y A1 L0 VN 1 LN V1 5.15 26 5.4 Continuous multistage countercurrent extraction Countercurrent process and overall balance 1. Usually, L0 and VN+1 are known and the desired exit composition xAN is set. 2. Plot points L0, VN+1, and M as in the figure, a straight line must connect these three points. 3. LN, M, and V1 must lie on one line. Also, LN and V1 must also lie on the phase envelope. 27 Ex 5.5 Pure solvent isopropyl ether at the rate of VN+1 = 600 kg/h is being used to extract an aqueous solution of L0=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 V1 and the aqueous raffinate LN. Use equilibrium data from the table. Solution: The given values are VN+1 = 600kg/h, yAN+1 = 0, yCN+1 = 1.0, L0 = 200kg/h, xA0 = 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.04. In figure below, VN+1 and L0 are plotted. Also, since LN is on the phase boundary, it can be plotted at xAN = 0.04. For the mixture point M, substituting into eqs. below, xCM x AM L0 xC 0 VN 1 yCN 1 L0 VN 1 200(0) 600(1.0) 0.75 200 600 L0 x A0 VN 1 y AN 1 200(0.30) 600(0) 0.075 L0 VN 1 200 600 28 Using these coordinates, 1) Point M is plotted in Figure below. 2) We locate V1 by drawing a line from LN through M and extending it until it intersects the phase boundary. This gives yA1 = 0.08 and yC1 = 0.90. 3) For LN a value of xCN = 0.017 is obtained. By substituting into Eqs. 5.12 and 5.13 and solving, LN = 136 kg/h and V1 = 664 kg/h. 29 Stage-to-stage calculations for countercurrent extraction. Total mass balance on stage 1 Total mass balance on stage n From 5.16 obtain difference Δ in flows Δ is constant and for all stages L0 V2 L1 V1 Ln1 Vn1 Ln Vn L0 V1 L1 V2 L0 V1 Ln Vn1 LN VN 1 .... x L0 x0 V1 y1 Ln xn Vn1 yn1 LN x N VN 1 y N 1 ... 5.16 5.17 5.18 5.19 5.20 30 Stage-to-stage calculations for countercurrent extraction. Δx is the x coordinate of point Δ x L0 x0 V1 y1 Ln xn Vn1 y n1 LN x N VN 1 y N 1 L0 V1 Ln Vn1 LN VN 1 5.21 5.18 and 5.19 can be written as L0 V1 Ln Vn1 LN VN 1 5.22 31 Stage-to-stage calculations for countercurrent extraction. 1. Δ is a point common to all streams passing each other, such as L0 and V1, Ln and Vn+1, Ln and Vn+1, LN and VN+1, and so on. 2. This coordinates to locate this Δ operating point are given for x cΔ and x AΔ in eqn. 5.21. Since the end points VN+1, LN or V1, and L0 are known, xΔ can be calculated and point Δ located. 3. Alternatively, the Δ point is located graphically in the figure as the intersection of lines L0 V1 and LN VN+1. 4. In order to step off the number of stages using eqn. 5.22 we start at L0 and draw the line L0Δ, which locates V1 on the phase boundary. 5. Next a tie line through V1 locates L1, which is in equilibrium with V1. 6. Then line L1Δ is drawn giving V2. The tie line V2L2 is drawn. This stepwise procedure is repeated until the desired raffinate composition LN is reached. The number of stages N is obtained to perform the extraction. 32 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 VN+1 = 450, yAN+1 = 0, yCN+1 = 1.0, L0 = 150, xA0 = 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.10. 1. The points VN+1, L0, and LN are plotted in Fig. below. For the mixture point M, substituting into eqs. 5.12 and 5.13, xCM = 0.75 and xAM = 0.075. 2. The point M is plotted and V1 is located at the intersection of line LNM with the phase boundary to give yA1 = 0.072 and yC1 = 0.895. This construction is not shown. 3. The lines L0V1 and LNVN+1 are drawn and the intersection is the operating point Δ as shown. 33 1. Alternatively, the coordinates of Δ can be calculated from eq. 5.21 to locate point Δ. 2. Starting at L0 we draw line L0 Δ, which locates V1. Then a tie line through V1 locates L1 in equilibrium with V1. (The tie-line data are obtained from an enlarged plot.) 3. Line L1 Δ is next drawn locating V2. A tie line through V2 gives L2. 4. A line L2 Δ is next drawn locating V2. A tie line through V2 gives L2. 5. A line L2 Δ gives V3. 6. A final tie line gives L3, which has gone beyond the desired LN. Hence, about 2.5 theoretical stages are needed. 34 5.4 Continuous multistage countercurrent extraction Countercurrent-Stage Extraction with Immiscible Liquids If the solvent stream VN+1 contains components A and C and the feed stream L0 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 L0 to VN+1. x y x L 0 V N 1 L N 1 x0 1 y N 1 1 xN y V 1 1 y1 x y x L 0 V n 1 L n 1 x0 1 y n 1 1 xn y V 1 1 y1 5.23 5.24 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. 35 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 36 Solution: The given values are L0 = 100 kg/h, x0 = 0.010, VN+1 = 200 kg/h, yN+1 = 0.0005, xN = 0.0010. The inert streams are L L(1 x) L0 (1 x0 ) 100(1 0.010) 99.0kg water / hr V / V (1 y) VN 1 (1 yN 1 ) 200(1 0.0005) 199.9kgkerosene/ hr Making an overall balance on A using eq. 5.23 and solving, y1 = 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. 37 38 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 raffinate layers. Solve for the amounts both algebraically and by the lever-arm rule. What percent of the acetic acid is removed? 39 Homework No.10 1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of a feed solution containing 25 wt% acetic acid in water. (a) If 400 kg of isopropyl is used, calculate the percent recovery in the isopropyl solution in a one-stage process. (b) If a multiple four-stage system is used and 100 kg fresh isopropyl is used in each stage, calculate the overall percent recovery of the acid in the total outlet isopropyl ether. (Hint: First, calculate the outlet extract and raffinate streams for the first stage using 400 kg of feed solution and 100 kg of isopropyl ether. For the second stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the first stage. For the third stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the first stage. For the third stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the second stage, and so on.) 40 Homework No.9 (Solution) 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 raffinate layers. Solve for the amounts both algebraically and by the lever-arm rule. What percent of the acetic acid is removed? Solution Given: L0 V2 400 400 M 800kg xA0 0.35, xB0 0.65 and y A2 1.0 L0 xA0 V2 y A2 MxAM (400)(0.35) (400)(0) (800) xAM x AM 0.175 Given: xc0 1 xA0 xB0 1.0 0.35 0.65 0 L0 xC 0 V2 yC 2 MxC M (400)(0) (400)(1) (800) xC M xCM 0.5 41 V2 (0,1) = (yA2, yC2) V1 (0.12,0.87) = (yA1, yC1) M(0.175,0.5) = (xAM, xCM) M L1(0.22,0.03) = (xA1, xC1) L0(0.35,0) = (xA0, xC0) 42 From the graph: xA1 = 0.22 and yA1 = 0.12; L1 xA1 V1 y A1 MxAM L1 (0.22) V1 (0.12) (800)(0.175) L1 0.54V1 636.36 (1) From the graph: xC1 = 0.03 and yC1 = 0.87; L1 xC1 V1 yC1 MxC M L1 (0.03) V1 (0.87) (800)(0.5) L1 29V1 1,333.33 (2) Solving eq(1) and eq(2) to get L1 and V1; L1 623.12kg and V1 24.49kg xA1 0.23, y A1 0.12, xC1 0.03 and yC1 0.88 Answer 43 V1 (0.12,0.88) = (yA1, yC1) M(0.175,0.5) = (xAM, xCM) M L1(0.23,0.03) = (xA1, xC1) Lever arm’s rule L1(kg ) V M M (kg ) L V L1(kg) 0.38 0.44 M (kg) 0.86 L(kg ) V M V (kg ) L M L(kg ) 0.38 0.81 V (kg) 352/ 0.81 434.57kg V (kg ) 0.47 L1(kg) 0.44x800 352kg 44 The amount and composition of extract are L1 325kg * xA1 0.23, xC1 0.03 and xB1 0.74 Answer The amount and composition of raffinate are V1 434.57kg * y A1 0.12, yC1 0.88 and yB1 0 Answer *The correction of amount of extract and raffinate depends on the identification of points on the graph. The value obtained from Lever’s arm rule is more reliable personally. The percent of acetic acid removed is about =(400x0.35-325x0.23)x100% = 46.6% 400x0.35 Answer 45 Homework No.10 1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of a feed solution containing 25 wt% acetic acid in water. (a) If 400 kg of isopropyl is used, calculate the percent recovery in the isopropyl solution in a one-stage process. (b) If a multiple four-stage system is used and 100 kg fresh isopropyl is used in each stage, calculate the overall percent recovery of the acid in the total outlet isopropyl ether. (Hint: First, calculate the outlet extract and raffinate streams for the first stage using 400 kg of feed solution and 100 kg of isopropyl ether. For the second stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the first stage. For the third stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the first stage. For the third stage, 100 kg of isopropyl ether contacts the outlet aqueous phase from the second stage, and so on.) 46 Solution a): Given: L0 V2 400 400 M 800kg xA0 0.25, xB0 0.75 and y A2 1.0 L0 xA0 V2 y A2 MxAM (400)(0.25) (400)(0) (800) xAM x AM 0.125 Given: xc0 1 xA0 xB0 1.0 0.35 0.75 0 L0 xC 0 V2 yC 2 MxC M (400)(0) (400)(1) (800) xC M xCM 0.5 47 V2 (0,1) = (yA2, yC2) V1 (0.09,0.9) = (yA1, yC1) M(0.125,0.5) = (xAM, xCM) M L1(0.17,0.03) = (xA1, xC1) L0(0.25,0) = (xA0, xC0) 48 From the graph: xA1 = 0.17 and yA1 = 0.09; L1 xA1 V1 y A1 MxAM L1 (0.17) V1 (0.12) (800)(0.125) L1 0.71V1 588.24 (1) From the graph: xC1 = 0.03 and yC1 = 0.9; L1 xC1 V1 yC1 MxC M L1 (0.03) V1 (0.9) (800)(0.5) L1 30V1 1,333.33 (2) Solving eq(1) and eq(2) to get L1 and V1; L1 570.18kg and V1 25.43kg xA1 0.17, y A1 0.09, xC1 0.03 and yC1 0.9 Answer 49 Solution b): Given: L0 V2 400 100 M 500kg xA0 0.25, xB0 0.75 and y A2 1.0 First stage L0 xA0 V2 y A2 MxAM (400)(0.25) (100)(0) (500) xAM x AM 0.2 Given: xc0 1 xA0 xB0 1.0 0.35 0.75 0 L0 xC 0 V2 yC 2 MxC M (400)(0) (100)(1) (500) xC M xCM 0.2 50 V2 (0,1) = (yA2, yC2) V1 (0.12,0.86) = (yA1, yC1) M(0.2,0.2) = (xAM, xCM) L1(0.17,0.03) = (xA1, xC1) L0(0.25,0) = (xA0, xC0) M 51 From the graph: xA1 = 0.17 and yA1 = 0.09; L1 xA1 V1 y A1 MxAM L1 (0.17) V1 (0.12) (800)(0.125) L1 0.71V1 588.24 (1) From the graph: xC1 = 0.03 and yC1 = 0.9; L1 xC1 V1 yC1 MxC M L1 (0.03) V1 (0.9) (800)(0.5) L1 30V1 1,333.33 (2) Solving eq(1) and eq(2) to get L1 and V1; L1 570.18kg and V1 25.43kg xA1 0.17, y A1 0.09, xC1 0.03 and yC1 0.9 % recovery in isopropyl solution is 96.93% Answer 52