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Mass Transfer 2

Mass Transfer 2

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Mass Transfer 2

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  1. Mass Transfer 2 Section # 10

  2. Solids- Free coordinates in Leaching • Sometimes, we don’t use the triangle for representing the leaching ternary system. • Rectangular diagram is used in such cases. It is similar to (Ponchon-Savarit) used before in distillation. • B S A S A

  3. Solids- Free coordinates in Leaching • In this new form, the y-axis is called N. N = B / (A+S) • In this new form, the x-axis is called X or Y X,Y = A / (A+S) • The x-axis represents X if we are dealing with an underflow stream. The x-axis represents Y if we are dealing with an overflow stream. • In this new form, all the quantities are solid free. • So, the quantities we are dealing with are Ln’, Vn’,………etc. Where: Ln’ = Ln – B , Vn’ = Vn – B ( We only need A+S) • It is easy to prove that: Ln = Ln’ (1+NLn) , Vn = Vn’ (1+NVn)

  4. Solids- Free coordinates in Leaching • Where is the locus of Underflow?! - It can be given by any of the 4 methods taken in the 1st section. - The most accurate method is to have an experimental data table between N and X as follows:

  5. Solids- Free coordinates in Leaching • Where is the locus of Overflow?! - It can be given by any of the 2 methods taken in the 1st section. - Example: Overflow streams contain 1 lb oil-free meal per 9 lb of solution

  6. Solids- Free coordinates in Leaching • How is the ideal stage represented?! - Here, we don’t have the vertex (B) to match with ! - In this new form, the ideal stage is simply represented by a vertical line

  7. Solids- Free coordinates in Leaching • How is the actual stage represented for ηoverflow≠ 100% ?! - ηoverflow = yn - yn+1 (Measurement from the left) yn * - yn+1 Example: Yn+1 = zero Xn=0.2 ηoverflow = 80 %

  8. Solids- Free coordinates in Leaching • How is the actual stage represented for ηunderflow≠ 100% ?! - ηunderflow = xn-1-xn (Measurement from the Right) xn-1 - xn* Example: X2 = 0.4 Y3=0.2 ηunderflow = 80 %

  9. Solids- Free coordinates in Leaching • How to calculate number of ideal stages?! A) Conventional method 1) Match y1 with xo and extend the line. 2) Match yn+1 and xn till you meet the extended line between y1 and xo. The point got is (R) 3) Match y1vertically with the underflow locus to get x1. (y1-x1) is the first ideal stage. 4) Match x1 with R to cut the overflow locus in y2 5) Match y2vertically with the underflow locus to get x2. (y2-x2) is the first ideal stage. B) Op. line projection 6) Match x2 with R to cut the overflow locus in y3 We can also project 7) Repeat the steps till you reach xn. The operating curve 8) To calculate a fraction of a stage: in a downward Fraction of a stage = xn– x last complete stage diagram as done in xexcess – x last complete stage PonchonSavarit

  10. Solids- Free coordinates in Leaching

  11. Solids- Free coordinates in Leaching

  12. Solids- Free coordinates in Leaching

  13. Solids- Free coordinates in Leaching

  14. Solids- Free coordinates in Leaching • How to calculate number of actual stages? 1) There is a classic method that we will know later in the problems. 2)Another method which is similar to that used in (Ponchon-Savarit) can also be used here. - This method depends on having pseudo equilibrium relation between X and Y, so that we can directly calculate number of actual stages. The graph below shows an example for that relation

  15. Solids- Free coordinates in Leaching • How to calculate number of actual stages? 2nd method steps: - Locate the solids-free coordinates diagram above that of the pseudo-equilibrium relation. - Get point (R) by matching Y1 and Xo, and matching Xn and Yn+1. - Descend from Y1 to the pseudo-equilibrium curve. Then move horizontal to the 45 line. Finally, go vertically upwards to cut the underflow locus in X1actual. - Match X1actual with R to cut the overflow locus in Y2. - Descend from Y2 to the pseudo-equilibrium curve. Then move horizontal to the 45 line. Finally, go vertically upwards to cut the underflow locus in X2actual. - Match X2actual with R to cut the overflow locus in Y3. - Continue using the same steps till you reach Xn or exceed it.

  16. Solids- Free coordinates in Leaching

  17. Solids- Free coordinates in Leaching

  18. Solids- Free coordinates in Leaching

  19. Solids- Free coordinates in Leaching

  20. Solids- Free coordinates in Leaching

  21. Single stage problem • In a single stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybeans containing 20%wt oil is leached with 100 kg fresh hexane solvent. The value of N for the slurry underflow is essentially constant as 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow and the underflow products.

  22. Single stage problem Solution: • As nothing is stated about the overflow locus, we will take it as (B=0) • Steps: a) Plot the underflow locus b) locate xo, and yn+1on the diagram c) Apply lever arm between Lo’ and Vn+1’ to get point (M) d) Match B with M to cut the overflow locus in y1, and the underflow locus in x1 Why? a) From material balance, we know that (Lo’ + Vn+1’) = (L1’+ V1’) = M b) So, y1 and x1 lie on the same straight line with M c) But we have one single ideal stage, so y1 and x1 lie also on the same Vertical line • Locus of underflow: B = 1.5 xB = 0.6 A+S

  23. Single stage problem Solution • As nothing is given about the amount of solid in the solvent, and as it is also not one of the requirements, we have put it zero in the above table. • AS nothing is stated about the overflow locus, we have taken as (B=zero) • Degrees of freedom on the overall: Nv = 12 , Ne = 3, Ng = 8 , NAR = 1 D.f = zero no trial and error

  24. Single stage problem

  25. Single stage problem

  26. Single stage problem

  27. Single stage problem Solution • Calculation steps: 1) MB on B 2) Calculate Ln 3) Get V1 4) From NLn and BLn get (A+S)Ln 5) MB on (A+S), get (A+S)V1 6) From the steps shown in the previous slides, get X1, and Y1, and put them in the table

  28. Single stage problem Solution • Calculation steps: 7) From X1, and (A+S)Ln get ALn and SLn 8) From Y1, and (A+S)V1 get AV1 and SV1

  29. Leaching Sheet , Problem#2’ • Givens: Lo : B=2000 lb, A=800 lb, S=50 lb Vn+1 : A=20 lb, S=1310 lb Ln: A=120 lb Locus of underflow: Experimental data is given (the same data shown in slide 4) Basis: 1 hr Counter current • Required: a) The composition of the strong solution b) The composition of the solution adhering to the extracted solid c) The weight of the solution leaving with the extracted meal d) The weight of the strong solution e) The number of units required

  30. Leaching Sheet , Problem#2’ Solution • As nothing is given about the amount of solid in the solvent, and as it is also not one of the requirements, we have put it zero in the above table. • AS nothing is stated about the overflow locus, we have taken as (B=zero) • Degrees of freedom on the overall: Nv = 12 , Ne = 3, Ng = 8 , NAR = 1 D.f = zero no trial and error [As the underflow locus isn’t given directly as a composition]

  31. Leaching Sheet , Problem#2’ Solution • Steps: a) Plot the underflow locus b) Calculate xo, and yn+1 , then locate them on the diagram c) Apply lever arm between Lo’ and Vn+1’ to get point (M) d) Perform MB on B to get BLn e) I know that Xn will be on the underflow locus, so I want to use the 2 givens about Ln together with the underflow locus to locate Xn.

  32. Leaching Sheet , Problem#2’ Solution • Steps: f) Nn / Xn = B / (A+S) = B / A = 2000 / 120 = 16.666667 A / (A+S) LnLn (There is only 1 line having the above slope that can pass by the underflow locus) g) From the origin, construct a line having the slope (16.66667) to cut the underflow locus in Xn

  33. Leaching Sheet , Problem#2’ Solution • Steps: h) Match Xn with M to cut the overflow locus in Y1 i) Read Y1 and Xn, and out them in the table j) Complete the table normally

  34. Leaching Sheet , Problem#2’ Solution • To calculate number of ideal stages, follow the slides from 9 to 13 • NTS=4 • The results are of course the same as number 2.

  35. Leaching Sheet , Problem#5’ • Givens: Lo : B=2000 lb, A=800 lb, S=50 lb Vn+1 : A=20 lb, S=1310 lb Ln: A=120 lb Locus of underflow: Experimental data is given (4th method taken before in plotting the underflow locus) [ηunderflow = 80%] Basis: 1 hr Counter-current • Required: a) The composition of the strong solution b) The composition of the solution adhering to the extracted solid. c) The weight of the solution leaving with the extracted meal d) The weight of the strong solution e) The number of units required

  36. Leaching Sheet , Problem#5’ Solution • The same material balance table of problem 2’

  37. Leaching Sheet , Problem#5’ To get number of actual stages: 1) Match Xn with Yn+1 and extend. 2) Match Xo with Y1 and extend till you meet the extended line between xn and yn+1 3) Match Y1vertically with the underflow locus to get X1* 4) As we know that ηunderflow = Xn-1-Xn = 0.8 Xn-1 - Xn* Then for the 1st stage: X0-X1 = 0.8 locate X1 on the underflow locus. X0- X1* 5) Match X1 with Y1. This is the 1st actual stage 6) Match R with X1 to cut the underflow locus in Y2 7) Match Y2vertically with the underflow locus to get X2* 8) X1-X2 = 0.8 Locate X2 on the underflow locus. X1- X2*

  38. Leaching Sheet , Problem#5’ To get number of actual stages (Cont’d): 9) Match X2 with Y2. This is the 2nd actual stage 10) Match R with X2 to cut the underflow locus in Y3 11) Repeat the steps till you reach Xn or exceed it.

  39. Leaching Sheet , Problem#5’

  40. Leaching Sheet , Problem#5’

  41. Leaching Sheet , Problem#5’

  42. Leaching Sheet , Problem#5’ • Continue using the same sequence to find that number of actual stages = 5 • This is of course the same result as that got from number 5

  43. May 2010 • Givens: Lo : Lo=1000 kg/hr, %A = 28%, %S = 2.5% Vn+1 : Vn+1 = 500 kg/hr, %A = 1.5% Ln: %A = 5% Locus of underflow: Experimental data is given Counter current • Required: a) The amount of extract, and its oil content b) The amount of solution retained by the meal and the oil content in it c) The number of stages required

  44. May 2010 (Cont’d) • Where is the locus of Underflow?!

  45. May 2010 (Cont’d) Solution • As nothing is given about the amount of solid in the solvent, and as it is also not one of the requirements, we have put it zero in the above table. • Degrees of freedom on the overall: Nv = 12 , Ne = 3, Ng = 8 , NAR = 1 D.f = zero no trial and error [As the underflow locus isn’t given directly as a composition]

  46. May 2010 (Cont’d) Solution • Steps: a) Plot the underflow locus b) Calculate Xo, and Yn+1 , then locate them on the diagram c) Perform MB on B to get BLn

  47. May 2010 (Cont’d) Solution Steps: d) Nn / Xn = B / (A+S) = B / A = 695 / 14 = 49.64 A / (A+S) LnLn (There is only 1 line having the above slope that can pass by the underflow locus) e) From the origin, construct a line having the slope (49.64) to cut the underflow locus in Xn f) Continue normally to get the requirements