Mass Transfer

Section # 10Mass Transfer 21Solids- Free coordinates in LeachingSometimes, we dont 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 A2Solids- Free coordinates in LeachingIn this new form, the y-axis is called N. N = B / (A+S) In this new form, the x-axis is called X or YX,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) 3Solids- Free coordinates in LeachingWhere 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:

NX201.980.11.940.21.890.31.820.41.750.51.680.61.610.7

4Solids- Free coordinates in LeachingWhere 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

5Solids- Free coordinates in LeachingHow is the ideal stage represented?!- Here, we dont have the vertex (B) to match with !- In this new form, the ideal stage is simply represented by a vertical line

6Solids- Free coordinates in LeachingHow is the actual stage represented for overflow 100% ?!- overflow = yn - yn+1 (Measurement from the left) yn * - yn+1Example:Yn+1 = zero Xn=0.2 overflow = 80 %

7Solids- Free coordinates in LeachingHow is the actual stage represented for underflow 100% ?!- underflow = xn-1-xn (Measurement from the Right) xn-1 - xn*Example:X2 = 0.4 Y3=0.2 underflow = 80 %

8Solids- Free coordinates in Leaching9How 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 y1 vertically 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 y2 vertically 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 y3We 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 stagediagram as done in xexcess x last complete stagePonchon Savarit

Solids- Free coordinates in Leaching

10Solids- Free coordinates in Leaching

11Solids- Free coordinates in Leaching

1212Solids- Free coordinates in Leaching

13Solids- Free coordinates in LeachingHow 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

14Solids- Free coordinates in LeachingHow 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.15Solids- Free coordinates in Leaching

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Solids- Free coordinates in Leaching

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Solids- Free coordinates in Leaching

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Solids- Free coordinates in Leaching

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Solids- Free coordinates in Leaching

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Single stage problemIn 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.

21Single stage problemSolution: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+1 on 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

22Single stage problemSolution

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 = zerono trial and error

LoVn+1LnV1A200B80 00.6 Ln0SZero100Total100100(A+S)20100N = B / (A+S)401.50X,Y = A / (A+S)1023Single stage problem

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Single stage problem

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Single stage problem

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Single stage problemSolution

Calculation steps: 1) MB on B2) Calculate Ln 3) Get V1 4) From NLn and BLn get (A+S)Ln5) 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

LoVn+1LnV1A200B80 00.6 Ln = 800SZero100Total100100133.3333366.666(A+S)2010053.33333N = B / (A+S)401.50X,Y = A / (A+S)100.1680.16827Single stage problemSolution

Calculation steps: 7) From X1, and (A+S)Ln get ALn and SLn8) From Y1, and (A+S)V1 get AV1 and SV1

LoVn+1LnV1A2008.9611.04B80 00.6 Ln = 800SZero10044.373355.6266Total100100133.3333366.666(A+S)2010053.3333366.666N = B / (A+S)401.50X,Y = A / (A+S)100.1680.16828Leaching Sheet , Problem#2Givens: 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 currentRequired: 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

29Leaching Sheet , Problem#2Solution

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 = zerono trial and error[As the underflow locus isnt given directly as a composition]

LoVn+1LnV1A80020120B2000 00S501310Total28501330(A+S)8501330N = B / (A+S)2.352900X,Y = A / (A+S)0.940.01530Leaching Sheet , Problem#2Solution

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 BLne) 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.

LoVn+1LnV1A80020120B2000 00S501310Total28501330(A+S)8501330N = B / (A+S)2.352900X,Y = A / (A+S)0.940.01531Leaching Sheet , Problem#2Solution

Steps: f) Nn / Xn = B / (A+S) = B / A= 2000 / 120 = 16.666667 A / (A+S) Ln Ln(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

LoVn+1LnV1A80020120700B2000 020000S501310Total28501330(A+S)8501330N = B / (A+S)2.352900X,Y = A / (A+S)0.940.01532Leaching Sheet , Problem#2Solution

Steps: h) Match Xn with M to cut the overflow locus in Y1i) Read Y1 and Xn, and out them in the tablej) Complete the table normally

LoVn+1LnV1A80020120700B2000 020000S501310893.333466.66Total285013303013.331166.666(A+S)85013301013.331166.666N = B / (A+S)2.352901.97360X,Y = A / (A+S)0.940.0150.118420.633Leaching Sheet , Problem#2SolutionTo calculate number of ideal stages, follow the slides from 9 to 13NTS=4 The results are of course the same as number 2.

34Leaching Sheet , Problem#5Givens: 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) [underflow = 80%] Basis: 1 hr Counter-currentRequired: 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

35Leaching Sheet , Problem#5Solution

The same material balance table of problem 2

LoVn+1LnV1A80020120700B2000 020000S501310893.333466.66Total285013303013.331166.666(A+S)85013301013.331166.666N = B / (A+S)2.352901.97360X,Y = A / (A+S)0.940.0150.118420.636Leaching Sheet , Problem#5To 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+13) Match Y1 vertically with the underflow locus to get X1*4) As we know that underflow = 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 stage6) Match R with X1 to cut the underflow locus in Y27) Match Y2 vertically with the underflow locus to get X2*8) X1-X2 = 0.8 Locate X2 on the underflow locus. X1- X2*37Leaching Sheet , Problem#5To get number of actual stages (Contd):9) Match X2 with Y2. This is the 2nd actual stage10) Match R with X2 to cut the underflow locus in Y311) Repeat the steps till you reach Xn or exceed it.38Leaching Sheet , Problem#5

39Leaching Sheet , Problem#5

40Leaching Sheet , Problem#5

41Leaching Sheet , Problem#5Continue using the same sequence to find that number of actual stages = 5This is of course the same result as that got from number 542May 2010Givens: 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 currentRequired: a) The amount of extract, and its oil content b) The amount of solution retained by the meal and the oil content in itc) The number of stages required

43May 2010 (Contd)Where is the locus of Underflow?!

NX201.980.11.9420.21.8870.31.8180.41.7510.51.6810.61.6130.7

44May 2010 (Contd)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 = zerono trial and error[As the underflow locus isnt given directly as a composition]

LoVn+1LnV1A2807.50.05*280 = 14273.5B695 06950S25492.5Total1000500(A+S)305500N = B / (A+S)2.278690 0X,Y = A / (A+S)0.9180.01545May 2010 (Contd)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

LoVn+1LnV1A2807.514273.5B695 06950S25492.5Total1000500(A+S)305500N = B / (A+S)2.278690 0X,Y = A / (A+S)0.9180.01546May 2010 (Contd)Solution

Steps:

d) Nn / Xn = B / (A+S) = B / A= 695 / 14 = 49.64 A / (A+S) Ln Ln(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

LoVn+1LnV1A2807.514273.5B695 06950S25492.5Total1000500(A+S)305500N = B / (A+S)2.278690 0X,Y = A / (A+S)0.9180.01547