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Solid - liquid extraction_3

Solid liquid extraction calculation : Ponchon-Savarit Diagram • Flow rates of streams : F‘ (feed) , S‘ ( solvent) , L' ( underflow ) and V' ( overflow) on solid free basis ( Carrier A is not consider in total mass)

• mass fraction of A :

mass of solid Z= mass of non solid ( solvent + solute)

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Solid liquid extraction calculation : Ponchon-Savarit Diagram • Material balance equations are: solid free material balance

F   S   L  V   M 

(9.6)

F '  solute   S '  solvent   L' ( solvent  solute )Underflow  V ' ( solvent  solute )Overflow  total mixture( without solid ) Solid balance

F Z F   S Z S   LZ L  V ZV   M Z M  (9.7)

solid    non-solid× non-solid   i i  feed , solvent , underflow , overflow

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Solid liquid extraction calculation : Ponchon-Savarit Diagram

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• From the above 9.6 and 9.7 equations,

ZM

F Z F   S Z S   F   S

(9.8)

Z M   Fraction of total solid in mixture M ' ZM =

total solid total kg solid  solvent+solute total kg non-solid

ZL vs XC

• Underflow and overflow curves are drawn from given data ZV vs YC

ZL vs XC

Underflow

ZV vs YC

Overflow

Solid liquid extraction calculation : Ponchon-Savarit Diagram • Point F' ( ZF’, XCF’ ) located Feed X CF' =

xCF' x  CF'  1 x BF'  = 0  + xCF' xCF'

xC XC = x B + xC

ZL =

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xA x B + xC

Feed is solvent free

( ZF’, XCF’=1) ZL vs XC

Z F' =

x BF'

x AF' x AF'  (= 0) + xCF' xCF'

 kg solid  Z F'    =  kg non-solid feed

 kg solid    kg solute  feed

ZV vs YC

( ZS’ =0 ,YCS’ =0) Pure solvent

Solid liquid extraction calculation : Ponchon-Savarit Diagram

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• Point S' ( ZS’ ,YCS’ ) located Solvent:

YCS'

( ZF’, XCF’=1)

solvent is pure; oil (solute) and solid free

yCS' ( 0) = 0 yBS' ( 1) + yCS' ( 0)

y AS' ( 0) Z S' = 0 yBS' ( 1) + yCS' ( 0)

ZL vs XC

yC YC = yB + yC

ZV =

yA yB + yC

ZV vs YC

( ZS’ =0 ,YCS’ =0) Pure solvent

Solid liquid extraction calculation : Ponchon-Savarit Diagram

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• point M' located on line S'F' using ZM’ value from Eq 9.8

ZM 

F Z F   S Z S  F   S

(9.8)

ZL vs XC

• Vertical tie line through M' is drawn

L’ • intersecting point of tie line with overflow curve gives point V’ and underflow curve gives L’ ZV vs YC

• solute concentrations in overflow and underflow

obtained from V’ and L’ points

V’

Solid liquid extraction calculation : Ponchon-Savarit Diagram

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• Point V’ gives in overflow • solute concentration = yCV’ • Point L’ gives in underflow • solute concentration = xCL’ ZL vs XC

• amounts of overflow and underflow can be obtained by material balance equations 9.6 and 9.7

L’

ZV vs YC

V’

EXAMPLE

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(Solid-liquid extraction in single - stage batch contactor )

• One thousand Kilograms of crushed oil seed ( 19.5% oil, 80.5% meal ) is extracted with 1500 Kg of 'pure' hexane in a batch extraction vessel. Calculate the fraction of the oil extracted using the 'Ponchon-Savarit diagram'. The equilibrium data given.

Overflow(100 kg), solution

Underflow(100 kg), slurry

WA (kg)

WB (kg)

WC(kg)

W'A (kg)

W'B(kg)

W'C(kg)

0.3 0.45 0.54 0.70 0.77 0.91 0.99 1.19 1.28 1.28 1.48

99.7 90.6 84.54 74.47 69.46 60.44 54.45 44.46 38.50 34.55 24.63

0.0 8.95 14.92 24.83 29.77 38.65 44.56 54.35 60.22 64.17 73.89

67.2 67.1 66.93 66.58 66.26 65.75 65.33 64.39 63.77 63.23 61.54

32.8 29.94 28.11 25.06 23.62 20.9 19.07 16.02 14.13 12.87 9.61

0.0 2.96 4.96 8.36 10.12 13.35 15.6 19.59 22.10 23.90 28.85

Calculation using the Ponchon –Savarit Diagram Given : For feed (on solid free basis)

F' = amount of non solid in feed  1000  0.195 = 195 kg XCF' = mass fraction of solute in feed on solid free basis

XCF' =

Z F' =

kg  solute  feed kg ( solvent + solute)feed kg  solid  feed

=

195 1 195

kg  non solid ( solvent + solute)feed

=

805  4.13 195

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Calculation using the Ponchon –Savarit Diagram Given For solvent (on solid free basis) S’=1500 YCS’ = 0 (because feed solvent is pure or oil free) ZS’ = 0

(solvent is free from solid)

• By material balance M’ (solid free basis) = solute + solvent= 195 + 1500 = 1695 kg • Solid fraction in mixture M’

Z M' Z M'

F ' ZF '  S ' ZS'  M' 195  4.13  1500  0   0.475 1695

Z F' = 4.13

ZS’ = 0

F' = 195 kg

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Calculation using the Ponchon –Savarit Diagram • Equilibrium data used to plot underflow (UF) and overflow (OF) curve • Points F’ (1, 4.13) and S’ (0,0) are located on diagram

• Point M’ located on line S’F’ at ZM’ = 0.475 • Vertical tie line L'V‘ drawn through point M'

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Calculation using the Ponchon –Savarit Diagram

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• Terminal points of vertical tie line L'V‘ give intersecting points with UF and OF curves at L' and V' • L' and V‘ points give ZL’ = 2.03

ZV’=0

XCM’ = 0.115 = XCL’ = YCV’

ZL’ XCL’

as M’ , L’, V’ are on same vertical line

ZV’ XCV’

Calculation using the Ponchon –Savarit Diagram

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• Putting values of different quantities in material balance equations 9.6 and 9.7 L  V   1695 kg

LZ L  V ZV   M Z M  L  2.03  V   0  1695  0.475 L 

F   S   L  V   M 

(9.6)

F Z F   S Z S   LZ L  V ZV   M Z M  (9.7)

1695  0.475  396.6 2.03

• Mass of underflow = non-solid + solid = L’ + L’ZL’ =L’ (1+ZL’) = 396.6 (1+ 2.03 ) = 1201.7 kg V’ = M’- L’= 1695 - 396.6 = 1298.4 kg

• Mass of overflow= non-solid + solid = V’ (1+ZV’) = 1298.4 × (1+0) = 1298.4 kg ( so overflow is free from solid)

V 'YCV ' 1298  0.115 Fraction of oil extracted =   0.765 F' 195

(76.5%)

Solid liquid extraction calculation : Crosscurrent contact Three stage cross current contact unit • Underflow (UF) from 1st stage goes to 2nd stage and UF from 2nd stage to 3rd stage • Solvent is added individually to each stage of same or different, composition and amount • Overflow from each stage is taken out separately

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EXAMPLE

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The extraction of 1000 kg of crushed oil seed ( 19.5% oil, 80.5% meal ) was carried out in a threestage cross-current unit using 500 kg 'pure' hexane solvent in each stage. Calculate the fraction of oil that can be extracted using the 'Ponchon-Savarit diagram'. The equilibrium data given.

Overflow(100 kg), solution WA (kg) 0.3 0.45 0.54 0.70 0.77 0.91 0.99 1.19 1.28 1.28 1.48

WB (kg) 99.7 90.6 84.54 74.47 69.46 60.44 54.45 44.46 38.50 34.55 24.63

WC(kg) 0.0 8.95 14.92 24.83 29.77 38.65 44.56 54.35 60.22 64.17 73.89

Underflow(100 kg), slurry W'A (kg) 67.2 67.1 66.93 66.58 66.26 65.75 65.33 64.39 63.77 63.23 61.54

W'B(kg) 32.8 29.94 28.11 25.06 23.62 20.9 19.07 16.02 14.13 12.87 9.61

W'C(kg) 0.0 2.96 4.96 8.36 10.12 13.35 15.6 19.59 22.10 23.90 28.85

For stage 1

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• Input streams are: F = 1000 kg,

xCF = 0.195;

S = 500 kg,

yCS = 0.

• On solid-free basis are:

F' = 195 kg

XCF'  1

solid 805 Z F' =  = 4.13 non  solid 195 S' = 500 ZS' = 0 (no solid in feed solvent)

X CF' =

xCF' x  CF'  1 x BF'  = 0  + xCF' xCF'

For stage 1

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• mass and solid content of mixed stream in stage 1,

M'1 = F' + S' = 195 + 500 = 695

Z M' Z M'

F ' ZF '  S ' ZS'  M' 195  4.13  500  0   1.16 695

• points F‘ (1, 4.13) and S‘ (0,0) are located • point M'1 is located on line S'F' at ZM'1 = 1.16.

ZM'1 = 1.16

For stage 1 • Vertical Tie line L'1V'1 through M'1 is drawn.

• From the graph, ZL'1 = 1.96, ZV'1 ≈ 0

• From solid balance

L'1 Z L'1 + V1' Z V'1 = M1' Z M'1 L'1  1.96 + V1'  0 = 695  1.16 L'1 = 411 kg V1' = M'-L'= 695-411=284 kg

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For stage 2

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• inputs to stage 2 : L'1 and S'.

L'1 = 411 kg (feed)

S' = 500 kg

• Locate Feed point at L’1 (ZL'1 = 1.96), for stage 2

• S point remains same at (0,0) • Feed line obtained joining L’1 and S’

For stage 2

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L'1 = 411 kg (feed)

S' = 500 kg

• M'2 = amount of mixed stream M'2 = L'1 + S' = 411+500 = 911

• ZM'2 = concentration of solid in mixed stream

ZM'2 =

𝟒𝟏𝟏 𝟏.𝟗𝟔 + 𝟓𝟎𝟎 𝟎 𝟗𝟏𝟏

= 0.884

For stage 2 • point M'2 located on Feed line L'1S’ at ZM’2 = 0.884 • Vertical tie line L'2V'2 passing through M'2 is drawn. • From graph, ZL'2 = 2.02, ZV'2 ≈ 0.

• by a 'solid balance', L'2ZL'2 + V'2ZV'2 = M'2ZL'2 = 805 L'2 = 398.5 kg V'2 = 911 - 398.5 = 512.5 kg

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For stage 3

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L'2 = 398.5 kg

inputs to stage 3 : L‘2 and S'.

S’ = 500 kg

M'3 = L'2 + S' = 398.5 + 500 = 898.5 kg

ZM'3 =

𝟑𝟗𝟖.𝟓 𝟐.𝟎𝟐 + 𝟓𝟎𝟎 𝟎 𝟖𝟗𝟖.𝟓

= 0.896

• point M'3 is located on feed line L'2 S'. • Vertical tie line L'3V'3 is drawn through M'3. • solute concentrations (on solid-free basis) in

streams : XCL'3 = YCV'3 = 0.055

and

ZL'3 = 2.04

For stage 3

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from material balance, L'3 × ZL'3 + V'3 × ZV'3 = M'3 × ZM'3 L'3 × 2.04 + V'3 ×0 = 898.5 × 0.896 L'3 = 394.6; V'3 = 898.5 - 394.6 = 504

Mass of oil leaving with underflow from stage 3 = L'3 × XCL'3 = 394.6 × 0.055 = 21.7 kg.

195-21.7  0.89 Fractional oil recovery = 195

i.e. 89 %