Examples Solid-Liquid Extraction

Transcription

1 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse - - Examples Solid-Liquid Extraction. Rectangular Triangle Diagram C A... inert material B... extractable material C... solvent Y C / (A + B + C) F E D c Overflow (extract) a b 0 A X B / (A + B + C) B a... constant underflow b... variable underflow c... constant ration solvent / inert material DE... connode

2 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse Ponchon - Savarit Diagram A... inert material B... extractable material C... solvent F E a b N A / (B + C) c D 0 X,Y B / (B + C) a... constant underflow b... variable underflow c... constant ration solvent / inert material DE... connode inert material N extractable substance + solvent A B + C L solution extractable substance + solvent B + C N * L amount of inert material A L * X, L * Y amount of extractable substance B

3 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse Example : Single Step Extraction In a single step solid-liquid extraction soybean oil has to be extracted from soybean flakes using hexane as solvent. 00 kg of the flakes with an oil content of 20 wt% are contacted with 00 kg fresh hexane..5 kg of inert material hold back a constant value of kg solution. Determine in the rectangular triangle diagram and in the Ponchon - Savarit diagram the amount and composition of the flows leaving the extraction plant. V L 0 extract (overflow) feed extraction step solvent underflow V 2 L. Rectangular Triangle Diagram Total balance: L 0 + V 2 M L + V kg Balance for compound A: L 0 w A,L0 + V 2 w A,V2 M w A,M with the feed concentration w A,L0 0.8 and the suggestion, that no solid particles are included in the overflow, so w A,V2 0 follows: 00 * * * w A,M w A,M 0.4

4 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse Balance for compound B: L 0 w B,L0 + V 2 w B,V2 M w B,M with the feed concentration w B,L0 0.2 and with the knowledge, that pure hexane is used as solvent, w B,V2 0, follows 00 * * * w B,M w B,M 0. The concentration of compound C (solvent) in the mixing point M can be determined either by a mass balance for compound C L 0 w C,L0 + V 2 w C,V2 M w C,M with w C,L0 0, because no solvent is included in the feed, and with w C,V2, pure hexane, follows 00 * * 200 * w C,M w C,M 0.5 or by the rule, that the sum of the mass percent of each compound in the point M has to be. w A,M + w B,M + w C.M w C.M w C.M 0.5 With these concentrations the mixing point M can be drawn in the diagram, which has to be on the connection line of feed point F and solvent C. It is given, that kg inert material retains.5 kg solution (extractable substance + solvent miscella overflow). Therefore the concentration of the underflow is inert material w A, Underflow inert material + extractable substance + solvent.5 w A,Underflo w w A, L A A + B + C

5 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse The amount of the leaving flows L and V can be calculated from the mass balance for compound A M w A,M V w A,V + L w A,L with w A,V 0 (no solid material in the overflow) and w A,L 0.6 (underflow) w A, M L M 200 w A, L L kg With the total balance M L + V follows V M - L V kg The concentrations of B and C in the overflow V are calculated with the suggestion that no inert material A is included in the overflow. wb, V w B,V wc, V B w C,V 0, ( A) + B + C C 00 ( A) + B + C The composition of the underflow can be calculated by mass balances for compound B and C. L w B,L + V, w B,V L 0 w B,L0 + V 2 w B,V2 with w B,V2 0 wb,l L0 wb, Lo V L w B,L wb,v

6 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse w A,L + w B,L + w C,L w C,L w C,L total mass [kg] wt% A wt% B wt% C feed L solvent V overflow V underflow L Ponchon - Savarit Diagram Total balance: L 0 + V 2 M L + V L 0 B + C 20 kg, no solvent is included in the feed material V 2 00 kg, pure solvent C M kg Compound balance: L 0 X L0 + V 2 X V2 M X M X L0 B, no solvent in the feed material C 0 B + C X V2 0, pure solvent C 20 * + 00 * 0 20 * X M X M N 0? A 80 N 0 4 L 0 20 N M? N 0 * L 0 A N M * L M L M B + C kg A 80 N M LM 20

7 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse The amount of the extract solution V and of the solution, retained by the solid material, L can be determined by law of balance or by calculation. M 20 kg L + V N? It is given, that.5 kg of inert material A retains kg solution B+C N.5.5 underflow, which is constant A N 0 * L 0 N * L N M * M A 80 L kg L. 5 V M - L kg L N X feed L solvent V n overflow V underflow L n

8 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse Example 2: Continuous Countercurrent Solid - Liquid Extraction kg of wet sugar beet chips with a composition of 28 wt% water, 32 wt% sugar and 40 wt% inert material have to be extracted in a continuous countercurrent extraction plant using hot water as solvent. The produced extract must contain 40 wt% sugar and the total extraction efficiency for sugar has to be 90%. kg inert material retains 3 kg solution and this value is constant. Determine in the rectangular triangle diagram and in the Ponchon - Savarit diagram the number of ideal steps for this separation problem. V L 0 extract (overflow) feed extraction steps,2,..n solvent underflow V n L n. Rectangular triangle diagram 90% sugar (B) have to be extracted and the extract solution must contain 40 wt% sugar V * x B,V 0.9 * L 0 * x B,L0 with x B,V 0.4, L kg and x B,L V 0.9 *0,000 * kg Balance for inert material A with L n * x A,Ln + V * x A,V V n * x A,Vn + L 0 * x A,L0 x A,Ln x A,Underflow A A + B + C and x A,V 0 with the suggestion that no solid material is included in the overflow and with x A,Vn 0 because of pure solvent water C

9 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse follows xa,l0 L n L 0 x A, Ln 0, ,000 kg Balance for sugar B: V * x B,V + L n x B,Ln L 0 * x B,L0 + V n * x B,Vn with x B,Vn 0 because the solvent is pure water C follows x B,Ln * L0 xb,l0 V * Ln xb,v 0,000 * ,200 * 0.4 6, The amount of necessary solvent water C can be calculated by a total mass balance L 0 + V n L n + V V n L n + V - L 0 6, ,200-0,000 3,200 kg total mass [kg] wt% A wt% B wt% C feed L 0 0, solvent V n 3, overflow V underflow L n Determination of the number of ideal steps First of all the constant underflow with x A,Ln 0.25 and the given points L 0 (x A,L0 0.4, x B,L0 0.32, x C,L0 0.28), V (x A,V 0, x B,V 0.4, x C,V 0.6), V n (x C,Vn ) and L n (x A,Ln 0.25, x B,Ln 0.02, x C,Ln 0.73) are drawn in the diagram. The one pole line is the connection of V with L 0 and the other one the connection of V n with L n. Crossing these pole lines results in the pole point.

10 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse Construction of the connode ( connection line with point A) through V gives the underflow L at the underflow line. Connecting L with the pole point give the extract composition V 2, and so on. Finally the number of ideal steps results with N th 0 2. Ponchon - Savarit Diagram Determination of the feed point A 0.4 N B + C B 0.32 X L B + C L 0 ( ) * F ( ) * ,000 kg concentration of the overflow (extract solution) B 0.4 X V 0.4 B + C % extraction efficiency: V V * X V 0.9 * L 0 * X L0 0.9 * L0 * XV XL0 0.9 * 6000 * kg Balance for solid material: L n N 0 * L 0 N n * L n N n N Underflow A B + C N0 * L * 6, kg NN Total balance: L 0 + V n L M L n + V L M 2, ,200 9,200 kg V n 9,200-6,000 3,200 kg

11 Examples Solid - Liquid Extraction Lecturer: Dr.Gamse - - Balance for sugar B: L 0 * X L0 + V n * X Vn L M * X M V * X V + L n * X Ln with X Vn 0 (pure solvent) follows X Ln * L0 XL0 V * Ln XV 6,000 * ,200 * 0.4 0,0267 2,000 L N X feed L 0 6, solvent V n 3, overflow V 7, underflow L n 2, Determination of the ideal number of steps: Drawing of the points L 0 (N L , X L ), V (N V 0, X V 0.4), V n (N Vn 0, X Vn 0) and L n (N Ln 0.333, X Vn ). The connection of L 0 and V gives the first pole line and connection of L n and V n the second one. Crossing these two pole lines gives the pole point. The first connode is a vertical line through V which gives at the underflow the point L. Connecting this point L with the pole point give the next extract composition V 2 and so on. Finally the number of ideal steps results with N th 0