Examples Liquid- Liquid- Extraction

Examples Liquid- Liquid- Extraction Lecturer: Thomas Gamse ao.univ.prof.dipl.-ing.dr.techn. Department of Chemical Engineering and Environmental Technology Graz University of Technology Inffeldgasse 25, A-8010 Graz Tel.: ++43 316 873 7477 Fax: ++43 316 873 7472 email: Thomas.Gamse@TUGraz.at

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 1 - Examples Liquid - Liquid Extraction Nominating of the flos Solvent Feed Mixer Separator Raffinate Extract

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 2 - Example 1: Ternary Systems, Triangle Diagram To mixtures R and E, hich contain both the three compounds A, B, and C, have to mixed in the ratio 1:2. This ternary system has no miscibility gap so that all compounds are completely soluble each other. The mixture R has a composition of x AR, 07, and x BR, 02;, the mixture E exists of x AE, 01, and x BE, 05., Please determine: a) The points R and E in the triangle diagram and the concentration of the active agent C and b) the mixing point (calculation and graphical determination).

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 3 - a.) Mixtures in the triangle diagram For a partial solubility of substances A and B, hich is essential for extraction, all three compounds have to be taken into account for draing of the phase equilibrium. For this reason triangle co-ordinates are used, here each of the vertexes represents the pure compounds. Points at the triangle side represent the composition of the binary system and points inside the triangle the composition of the ternary system. The representation of a ternary point is based on the fact that the sum of the normal distances in a equal sided triangle is corresponding to the height of the triangle. If the height of the triangle is set 100% so result the concentrations of the single compounds from the normal distances (see figure). The given points R and E can therefore be dran in the diagram. From these points the concentration of C can be determined. x CR, 01, x CE, 04, (Control: The sum of the components A, B and C must be equal 1) The triangle diagram can also be given in eight percent t%.

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 4 - b) Mixing point If to mixtures ith given composition in the triangle diagram are mixed then the resulting mixing point lays on the connection line beteen these to points. The position of the mixing point can be calculated by a mass balance or graphically by the use of the la of balance. Calculating: Total balance: R+ E M Mass balance for compound C: R x + E x M x CR, CE, CM, x CM, R x + E x CR, CE, R+ E ith R E 1 2 the mass R and E can be eliminated, hich results in x CM, 05, x + x CR, CE, 15,

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 5 - or 05, 01, + 04,, 15, x CM x CM, 03, Analogous results therefore x AM, 03, and x BM, 04, Graphically: La of balance: RM ME E R ith R E 1 2 follos ME RM 1 2 From the diagram the length of the distance RE can be determined ith 77 mm. RE RM + ME 05, 77 RM RM RM 51, 3mm Draing this length in the diagram results the mixing point M and the concentrations of the compounds can be determined. x AM, 03, x BM, 04, x CM, 03,

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 6 - Triangle diagram

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 7 - Example 2: Ternary system ith mixing gap A aste ater from a process is loaded ith acetone, hich should be extracted ith chlorobenzene. The equilibrium data for the ternary system ater / acetone / chlorobenzene are given. composition of the coexisting phases in equilibrium in t% ater phase organic phase ater acetone chlorobenzene ater acetone chlorobenzene 99,89 00 11 18 00 99,82 89,79 100 21 49 179 88,72 79,69 200 31 79 22,23 76,98 69,42 300 58 1,72 37,48 680 58,64 400 1,36 3,05 49,44 47,51 46,28 500 3,72 7,24 59,19 33,57 27,41 600 12,59 22,85 61,07 15,08 25,66 658 13,76 25,66 658 13,76 You have to determine: a.) the triangle diagram including the phase equilibrium line and connodes. b.) The ater and chlorobenzene content of the aqueous phase (raffinate) ith an acetone concentration of 45 % and of the coexisting phase. c.) Which amount of acetone has to be added to a mixture, existing of 110 g chlorobenzene and 90 g ater? What is the composition of the mixing point? d.) What is the ater free composition of this mixing point? a) Construction of the phase equilibrium in the triangle diagram The given ternary system ha a mixing gap hich separates the system in a homogeneous one phase region and a heterogeneous to phase region. The boundary is the binodal curve.

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 8 - In the for the extraction interesting heterogeneous region a mixture splits in raffinate and extraction phase along a connode, hich connects the to coexisting phases. The higher the amount of the active agent (extractable substance C) is the shorter the connodes become until they melt to one point, the critical point K. By this critical point K the binodal curve is split into to parts. Normally the part on the left side represents the raffinate phase R, hich has a lo content of solvent B, and the right side represents the solvent rich extract phase E. According to the given table the coexisting phases (connodes) are given hich can no be dran in the triangle diagram. One line in the table corresponds to one connode. 1.connode point of the raffinate phase: AR, 9989 BR, 0011 CR, 00, 1.connode point of the extract phase: AE, 0018 BE, 9982 CE, 00, Connecting these to points gives the first connode and analogous for the other given data. The last ro corresponds to the critical point K. By connecting all raffinate and all extract points the result is the binodal curve. b) Raffinate phase / Extract phase Point in the raffinate phase Draing the acetone concentration of CR, 045, on the right side of the triangle for the active agent C and crossing this ith the binodal curve at the left side gives the point of the aqueous phase, so that the concentrations of ater and chlorobenzene can be determined. AR, 535 BR, 015

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 9 - Point of the extract phase The point of the extract phase has to be on a connode going through the already determined point on the raffinate side. But this connode is not given and has to be constructed. Possibility 1: By interpolation beteen the to connodes next to the point the connode through the given point can be constructed, but in a very inaccurate ay. Possibility 2: With the help of the conjugation line the connode can be determined better and ith higher accuracy. For this purpose the right and left triangle side has to be shifted parallel through the points of the connodes and the crossing of these to lines represents one point of the conjugation line. All these by this ay constructed point and the critical point have to be connected to the conjugation line. The searched coexisting phase can be constructed analogous: parallel shifting of the right triangle side through R, crossing ith the conjugation line and crossing of the parallel shifted left triangle side through the point on the conjugation line ith the right side of the binodal curve. The by this ay determined concentrations are: AE, 004, BE, 041, CE, 055, c) mixture near the phase boundary First the binary mixture has to dran at the basic side of the triangle diagram. mass [g] eight-% chlorobenzene 110 55 ater 90 45 Σ 200 100 This point of the binary mixture has to be connected ith the point C, pure acetone, and somehere on this line the mixing point must be. The boundary beteen one and to phase region is the binodal curve. Therefore the searched mixing point M

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 10 - can be determined by crossing the line GC ith the binodal curve. The necessary amount of acetone can be determine by the la of balance. CM 40 mm MG 69 mm CG 109 mm G CM C MG C 345 g 200 40 C 69 Composition of the mixing point M: AM, 016, BM, 021, CM, 063, d) ater free mixing point: To get the ater free mixing point, the edge point A and the mixing point M hove to be connected and this line has to be prolonged to the right side of the triangle diagram, the ater free side. The composition of the binary, ater free mixture of acetone and chlorobenzene is: BM, 025, CM, 075,

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 11 - Triangle diagram

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 12 - Example 3: Single Step Extraction The basic mixture of 100 kg exists of 40 mole% acetone and 60 mole% ater and has to be extracted ith trichloroethane, hich is preloaded ith 15 mole% acetone. Your have to determine: a) the phase diagram of the system acetone / ater / trichloroethane in the triangle diagram. b) the minimum and maximum amount of solvent, c) the necessary amount of solvent, if the raffinate contains 4,82 mole% acetone, d) the amount and composition of the produced raffinate and extract, e) the extraction process in the triangle diagram Phase equilibria data for the system ater / acetone / trichloroethane phase equilibria data for the coexisting phases in mole% extract phase raffinate phase trichloroethane acetone ater trichloroethane acetone ater 818 17,72 2, 07 1,9 97,99 59,01 35,96 5, 10 4,8 95,08 49,17 44,00 6, 12 6,8 93,03 35,99 53,78 12 17 136 89,47 25,04 58,34 16,6 29 14,98 84,73 14,56 56,96 28,4 78 21,98 77,24 9,94 52,48 37,5 1,50 27,38 71,12

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 13 - a) Phase equilibrium in the triangle diagram Draing of the single connode points analogous to example 2 and constructing the connodes and combining the single point to the binodal curve. The critical point as not dran because the composition is not given and therefore the exact position is not defined. b) minimal / maximal amount of solvent Draing of feed and solvent feed: x CF, 04, x AF, 06, The point F is on the left triangle side (binary mixture). solvent: x CL, 015, x BL, 085, The point L is on the right triangle side (binary mixture). The mixing point M has to be on the line beteen these to points F and L and M has to be in the to phase region, because for extraction the mixture has to separate in to phases. The minimal and maximal amount of solvent ( M min and M max ) are the to crossings of the connection line FL ith the binodal curve. By the length, hich can be determined from the diagram, the searched amounts can be calculated. Minimal amount of solvent: and FL 915, mm FM min 4mm la of balance: FM M min min L M min M F FM min min 100 F M L min, 4 915 4 M min 457kg,

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 14 - Maximal amount of solvent: M max la of balance: L 2mm FM M max max L M F max M F FM max, max 100 915 2 M L 2 max M max 4. 475kg c) effective amount of solvent The acetone concentration of the produced raffinate R, hich has to be on he binodal curve, must be 4,82 %. With the connode going through this point R the extract E is fixed. The mixing point M of feed F and solvent L is the crossing of the connode RE ith the connection line FL. With the la of balance the necessary amount of solvent L can be calculated. ML 45, 5mm la of balance: FM ML L F L F FM 100 ML 91, 5 45, 5 45, 5 L 1011, kg d) composition and amount of raffinate and extract Raffinate R: x AR, 9508 x BR, 001 x CR, 0482 Extract E: x AE, 0503 x BE, 5901 x CE, 3596

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 15 - amount of raffinate: total balance: E+ R F + L 100+ 1011, 2011, kg la of balance: RM ME E R RE 95mm ME 26 mm R 2011, 95 26 + 1 26 R 55kg amount of extract: E F + L R E 146, 1kg

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 16 - Triangle Diagram / Nernst Diagram

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 17 - Example 4: Multi Step Cross Flo Extraction From 2.000 kg/h of an acetic acid / ater mixture ith 45 t% acetic acid the acetic acid has to be extracted by a multi step cross flo extraction at an operation temperature of 20 C. The residual concentration of acetic acid has to be 10 t% and the used solvent isopropyl ether is free of acetic acid. You have to determine: a) the minimum amount of solvent for the first extraction step b) the necessary number of theoretical steps in the triangle diagram for the case that a solvent ration L & F & S of 1 is chosen and in every step the same amount of solvent is added. Phase equilibria data extract phase acetic acid ater isopropyl ether raffinate phase acetic acid ater isopropyl ether 002 005 993 007 981 012 004 007 989 014 971 015 008 008 984 029 955 016 019 010 971 064 917 019 048 019 933 133 844 023 114 039 847 255 711 034 216 069 715 367 589 044 311 108 581 443 451 106 362 151 487 464 371 165

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 18 - a) minimum amount of solvent For the liquid - liquid extraction at cross flo method the feed F & enters the first extraction step, here it is contacted ith solvent L &. The extraction results in a raffinate R & and a extract E &. The extract is ithdran hile the raffinate enters the next step here it is contacted ith fresh solvent again and so on. In the single steps equilibrium beteen raffinate and extract is reached so that the compositions can be determined in the triangle diagram. First the equilibrium data have to be dran and the binodal curve ith the given connodes has to be constructed. Then the point of the feed F and of the solvent L is dran. The mixing point M 1 has to be on the connection line FL. For the minimum amount of solvent the crossing point M min at the raffinate side of the binodal curve is significant. From the la of balance results : L& F& min FMmin M L min CF, C,min C,min C, L 45 415 415 0 0843 With the amount of feed results the minimum amount of solvent: L& F& 0843 2. 000 0843 168, 6 kg h min b) number of steps in the triangle diagram L& The ratio of solvent of feed is given ith F& 1 S 1. step ( ) ( ) F& F& 1 2. 000 1 45 1100. kg h S C, F With this follos: L&. &, F 1100 2. 000 055

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 19 - The mixing point M 1 of the first step can be determined by calculation or graphically. Calculation: L& F& FM1 M L 1 CF, CM, 1 CM, 1 CL, 045, CM, 1 CM, 1 0 055, 029, CM, 1 Graphically: The length of FL is 173 mm and ha to divided according the ratio L & F & 055., FM + M L 1 1 173 mm 55 M L+ M L 173 mm 1 1 173 M1L 111, 6 mm 155, Amount of the mixing point: M& & & 1 F + L 2. 000+ 1100. 3100. kg h The connode through the mixing point M 1 gives the extract E 1 and the raffinate R 1. The according compositions can be taken from the diagram. raffinate flo: M& 21 355 CE, 1 CR, 1 1 1 CM, 1 CE, 1 CR, 1 CE, 1 1 extract flo: ( CM, CE, ) ( ) M& 1 1 1 CR, 1 CE, 1 3100. 29 21 355 21 1710., 3 kg h E& & & 1 M1 R1 3100. 1710., 3 1389., 7 kg h 2. step For the second step the flo rate of the solvent isopropyl ether is also 1.100 kg/h. L& 1 RM 1 2 1100. 0643, M L 17103., 2

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 20 - RM + M L 1 2 2 167 mm 643 M L+ M L 167 mm 2 2 167 M2 L 101, 6 mm 1, 643 amount of mixing point: M& & & 2 R1+ L 1710., 3+ 1100. 2. 81 3kg h concentrations: CE, 2 CM, 2 CR, 2 raffinate flo: 2 extract flo: 014, 215 029, ( CM, CE, ) ( ) M& 2 2 2 CR, 2 CE, 2 2. 81 3 215 14 029, 014, 1405., 15 kg h E& & & 2 M2 R2 2. 81 3 140515., 140515., kg h 3. step L& 2 RM 2 3 1100. M L 1405., 15 3 783 RM + M L 2 3 3 171mm 783 M L+ M L 171mm 3 3 171 M3L 95, 9 mm 1, 783 amount of mixing point: M& & & 3 R2+ L 1405., 15+ 1100. 2. 505, 15kg h concentrations: 097 16 225 raffinate flo: CE, CM, CR, 3 3 3 3 ( CM, CE, ) ( ) M& 3 3 3 CR, 3 CE, 3 2. 505, 15 16 097 225 097 1233 kg h

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 21 - extract flo: E& & & 3 M3 R3 2. 505, 15 1233. 1272, 15kg h 4. step L& 3 RM 3 4 1100. M L 1233 4 892 RM + M L 3 4 4 175 mm 892 M L+ M L 175mm 4 4 175 M4 L 92, 5 mm 1892, amount of mixing point: M& & & 4 R3+ L 1233. + 1100. 2. 333kg h concentrations: 07 117 173 raffinate flo: CE, 4 CM, 4 CR, 4 4 ( CM, CE, ) ( ) M& 4 4 4 CR, 4 CE, 4 2. 333 117 07 173 07 extract flo: E& & & 4 M4 R4 2. 333 1064., 6 1268., 4 kg h 1064., 6 kg h 5. step L& 4 RM 4 5 1100. M L 1064., 6 5 1, 033 RM + M L 4 5 5 180 mm 1, 033 M L+ M L 180 mm 5 5 180 M5L 88, 54 mm 2, 033 amount of mixing point: M& & & 5 R4+ L 1064., 6+ 1100. 2164., 6 kg h

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 22 - concentrations: 045 084 128 raffinate flo: CE, 5 CM, 5 CR, 5 5 ( ) ( ) M& 5 CM, 5 CE, 5 CR, 5 CE, 5 2. 164, 6 084 045 128 045 extract flo: E& & & 5 M5 R5 2. 164, 6 1017, 1 1147, 5kg h 1017, 1kg h 6. step L& 5 RM 5 6 1100. M L 1017, 1 6 1, 082 RM + M L 5 6 6 183mm 1, 082 M L+ M L 183mm 6 6 183 M6L 87, 9 mm 2, 082 amount of mixing point: M& & & 6 R5+ L 1017, 1+ 1100. 2117, 1kg h concentrations: 033 06 094 raffinate flo: CE, 6 CM, 6 CR, 6 6 ( CM, CE, ) ( ) M& 6 6 6 CR, 6 CE, 6 2. 117, 1 06 033 094 033 extract flo: E& & & 6 M6 R6 2117., 1 937, 1 1180 kg h 937, 1kg h The concentration of the raffinate of this 6. step is loer than the necessary concentration so that the extraction can be stopped. necessary number of steps: : N th 6

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 23 - Ternary triangle diagram

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 24 - Example 5: Multi Step Countercurrent Extraction The acetic acid / ater mixture of example 4 has to be extracted in a multi step countercurrent extraction cascade ith isopropyl ether as solvent. The residual acetic acid concentration is also given ith 10 t%. You have to determine: a) the necessary amount of theoretical extraction steps in the triangle diagram for the case that the effective amount of solvent is 2649 kg/h. Phase equilibria data see example 4. a) number of theoretical steps in the triangle diagram The number of theoretical steps can be determined in the triangle diagram by a method developed by Hunter and Nash. It has to be considered that: points of coexisting phases in equilibrium are on a connode points of phases, hich contact at a cross section of the extractor, have to be on a pole line. construction: 45 227 291 CF, CE, CE, 1 max The mixing point M is given by the ratio of feed F and solvent B and has to be on the connection line balance: F B. The amount of the mixing point M can be determined by a total

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 25 -. M. F +. B. Rn +. E1 The point E 1 has to be on the binodal curve and on the connection line Rn M. Attention: The line Rn E1 represents no connode but only a balance line!! The points F and E 1 are connected by the upper pole line. F represents the feed, hich enters the first extraction step and E 1 is the extract hich leaves this first step. The loer pole line is given by the connection of the solvent L and the raffinate R n leaving the extraction plant. The pole is fixed by crossing the to pole lines F E 1 and B R n. For the case that the first extraction step is a theoretical step the leaving phases have to be in equilibrium. R 1 as point of the leaving raffinate phase & R 1, has to be on the binodal curve and has to be on a connode through the extract E 1. The raffinate phase & R 1 and the extract phase & E 2 contact in the next extraction step. The point E 2 of the extract phase & E 2 has to be on the binodal curve and further on the pole line, hich goes through R 1. Doing the construction for all points R 2, R 3 and E 3, E 4 by this method finally the necessary number of theoretical steps N th for the extraction can be determined. N th 6,5 For the calculation of the amounts of E i and R i the concentrations are determined from the triangle diagram.

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 26 - step 1 2 3 4 5 6 7 CRi, 383 335 285 233 171 119 029 CEi, 227 175 134 100 069 036 008 balance: F& + L& E& + CF, CL, 1 CE, 1 n CR, n M& E& + F& + L& 2. 000+ 2. 649 4. 649 kg h ges 1 n ( ) F& + L& M& + n CF, CL, ges n CE, 1 n CR, n F& + L& M& CF, CL, ges CE, 1 CR, n CE, 1 2. 000 45+ 2. 649 0 4. 649 227 1 227 E& M& 4. 649 1223. 3. 426 kg h 1 ges n 1223. kg h 1. step F& + E& + E& 2 1 1 F& + E& + E& CF, 2 CE, 2 1 CR, 1 1 CE, 1 ( ) F& + E& F& + E& E& + E& E& 2 CF, 2 CE, 2 2 1 CR, 1 1 CE, 1 ( ) & ( ) F& + E CF, CR, 1 1 CR, 1 CE, 1 CR, 1 CE, 2 & 2. 000 ( 45 383) + 3. 426 ( 383 227) E2 383 175 2. 000+ 3. 214 3. 426 1788. kg h 1 The folloing steps are calculated analogous 3. 214 kg h 2. step E& 3 ( ) & ( ) + E 1 CR, 1 CR, 2 2 CR, 2 CE, 2 CR, 2 CE, 3 & 1788. ( 383 335) + 3. 376 ( 335 175) E3 335 134 1788. + 3114. 3. 214 1688. kg h 2 3114. kg h

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 27-3. step E& 4 ( ) & ( ) + E 2 CR, 2 CR, 3 3 CR, 3 CE, 3 CR, 3 CE, 4 & 1688. ( 335 285) + 3114. ( 285 134) E4 285 100 1688. + 2998 3114. 1572. kg h 3 2998 kg h 4. step E& 5 ( ) & ( ) + E 3 CR, 3 CR, 4 4 CR, 4 CE, 4 CR, 4 CE, 5 & 1572. ( 285 233) + 2998 ( 233 100) E5 233 069 2. 930 kg h 1572. + 2. 930 2998 1504. kg h 4 5. step E& 6 ( ) & ( ) + E 4 CR, 4 CR, 5 5 CR, 5 CE, 5 CR, 5 CE, 6 & 1504. ( 233 171) + 2. 930 ( 171 069) E6 171 036 1504. + 2. 905 2. 930 1479. kg h 5 2. 905 kg h 6. step E& 7 ( ) & ( ) + E 5 CR, 5 CR, 6 6 CR, 6 CE, 6 CR, 6 CE, 7 & 1479. ( 171 119) + 2. 905 ( 119 036) E7 119 008 1479. + 2. 865 2. 905 1439. kg h 6 2. 865 kg h

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 28 - Ternary triangle diagram

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 29 - Appendix: Triangle Diagrams

Examples LiquidLiquid Extraction Lecturer: Dr. Gamse - 30 -