# Examples Solid-Liquid Extraction

Examples Solid - Liquid Extraction Lecturer: Dr.Gamse - 5 - The amount of the leaving flows L1 and V 1 can be calculated from the mass balance for com...

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

Examples Solid-Liquid Extraction 1. Rectangular Triangle Diagram

A ... inert material B ... extractable material C ... solvent C

D ve O ow rf l

F

(e xt ra ) ct

Y = C / (A + B + C)

1

c

E

a b 0 A

1 X = B / (A + B + C)

B

a ... constant underflow b ... variable underflow c ... constant ration solvent / inert material DE ... connode

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

2. Ponchon - Savarit Diagram

A ... inert material B ... extractable material C ... solvent

E

a

F

N = A / (B + C)

b

c

0

D

X,Y = B / (B + C)

a ... constant underflow b ... variable underflow c ... constant ration solvent / inert material DE ... connode

N=

inert material A = extractable substance + solvent 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

1

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

Example 1: Single Step Extraction In a single step solid-liquid extraction soybean oil has to be extracted from soybean flakes using hexane as solvent. 100 kg of the flakes with an oil content of 20 wt% are contacted with 100 kg fresh hexane. 1.5 kg of inert material hold back a constant value of 1 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.

V1

extract (overflow)

solvent extraction step

L0 feed

V2 L1

underflow

1. Rectangular Triangle Diagram

Total balance: L0 + V2 = M = L1 + V1 = 100 + 100 = 200 kg Balance for compound A:

L0 wA,L0 + V2 wA,V2 = M wA,M with the feed concentration wA,L0 = 0.8 and the suggestion, that no solid particles are included in the overflow, so wA,V2 = 0 follows: 100 * 0.8 + 100 * 0 = 200 * wA,M wA,M = 0.4

Examples Solid - Liquid Extraction

Lecturer: Dr.Gamse

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Balance for compound B: L0 wB,L0 + V2 wB,V2 = M wB,M with the feed concentration wB,L0 = 0.2 and with the knowledge, that pure hexane is used as solvent, wB,V2 = 0, follows 100 * 0.2 + 100 * 0 = 200 * wB,M wB,M = 0.1 The concentration of compound C (solvent) in the mixing point M can be determined either by a mass balance for compound C

L0 wC,L0 + V2 wC,V2 = M wC,M with wC,L0 = 0, because no solvent is included in the feed, and with wC,V2 = 1, pure hexane, follows

100 * 0 + 100 * 1 = 200 * wC,M wC,M = 0.5 or by the rule, that the sum of the mass percent of each compound in the point M has to be 1. wA,M + wB,M + wC.M = 1 0.4 + 0.1 + wC.M = 1 wC.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 1 kg inert material retains 1.5 kg solution (extractable substance + solvent = miscella = overflow). Therefore the concentration of the underflow is

w A,Underflow =

inert material A = inert material + extractable substance + solvent A + B + C

w A,Underflow = w A, L1 =

1 .5 = 0.6 1 .5 + 1

Examples Solid - Liquid Extraction

Lecturer: Dr.Gamse

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The amount of the leaving flows L1 and V1 can be calculated from the mass balance for compound A

M wA,M = V1 wA,V1 + L1 wA,L1 with wA,V1 = 0 (no solid material in the overflow) and wA,L1 = 0.6 (underflow)

L1 = M

w A,M w A,L1

= 200

0 .4 0 .6

L1 = 133.333 kg With the total balance

M = L1 + V1 follows V1 = M - L1 = 200 - 133.333 V1 = 66.666 kg The concentrations of B and C in the overflow V1 are calculated with the suggestion that no inert material A is included in the overflow.

w B, V1 =

B 20 = (A ) + B + C 0 + 20 + 100

wB,V1 = 0.1667 w C, V1 =

C 100 = (A ) + B + C 0 + 20 + 100

wC,V1 = 0,8333 The composition of the underflow can be calculated by mass balances for compound B and C. L1 wB,L1 + V1, wB,V1 = L0 wB,L0 + V2 wB,V2 with wB,V2 = 0

w B,L1 =

L 0 ∗ w B,Lo − V1∗ w B, V1

wB,L1 = 0.067

L1

=

100 ∗ 0.2 − 66.666 ∗ 0.1667 133.333

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

wA,L1 + wB,L1 + wC,L1 = 1 wC,L1 = 1 - 0.6 - 0.067 wC,L1 = 0.333

total mass [kg]

wt% A

wt% B

wt% C

feed L0

100

80

20

0

solvent V2

100

0

0

100

overflow V1

66.666

0

16.667

83.333

underflow L1

133.333

60

6.7

33.3

2. Ponchon - Savarit Diagram

Total balance: L0 + V2 = M = L1 + V1 L0 = B + C = 20 kg, no solvent is included in the feed material V2 = 100 kg, pure solvent C M = 20 + 100 = 120 kg

Compound balance: L0 XL0 + V2 XV2 = M XM XL0 = 1 =

B , no solvent in the feed material  C = 0 B+C

XV2 = 0, pure solvent C

20 * 1 + 100 * 0 = 120 * XM XM = 0.1667

N0 = ? N0 =

A L0

=

80 =4 20

NM = ? N0 * L0 = A = NM * LM LM = B + C = 20 + 100 = 120 kg NM =

A LM

=

80 = 0.667 120

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

The amount of the extract solution V1 and of the solution, retained by the solid material, L1 can be determined by law of balance or by calculation. M = 120 kg = L1 + V1 N1 = ? It is given, that 1.5 kg of inert material A retains 1 kg solution B+C

 N1 =

1 .5 = 1.5 = underflow, which is constant 1

A = N0 * L0 = N1 * L1 = NM * M L1 =

A L1

=

80 = 53.333 kg 1 .5

V1 = M - L1 = 120 - 53.333 = 66.666 kg

L

N

X

feed L0

20

4

1

solvent Vn

100

0

0

overflow V1

66.666

0

0.1667

underflow Ln

53.333

1.5

0.1667

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

Example 2: Continuous Countercurrent Solid - Liquid Extraction 10.000 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%. 1 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.

V1

extract (overflow)

L0

extraction steps 1,2,..n

feed

solvent

Vn Ln

underflow

1. Rectangular triangle diagram

90% sugar (B) have to be extracted and the extract solution must contain 40 wt% sugar

V1 * xB,V1 = 0.9 * L0 * xB,L0 with xB,V1 = 0.4, L0 = 10.000 kg and xB,L0 = 0.32

V1 =

0.9 * 10,000 * 0.32 = 7.200 kg 0 .4

Balance for inert material A Ln * xA,Ln + V1 * xA,V1 = Vn * xA,Vn + L0 * xA,L0 with xA,Ln = xA,Underflow =

A 1 = = 0.25 A + B + C 1+ 3

and xA,V1 = 0 with the suggestion that no solid material is included in the overflow and with xA,Vn = 0 because of pure solvent water C

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

follows Ln = L 0

x A,L0

= 10,000

x A,Ln

0 .4 = 16,000 kg 0.25

Balance for sugar B:

V1 * xB,V1 + Ln xB,Ln = L0 * xB,L0 + Vn * xB,Vn with xB,Vn = 0 because the solvent is pure water C follows

xB,Ln =

L 0 * x B,L0 − V1 * x B, V1 Ln

=

10,000 * 0.32 − 7,200 * 0.4 = 0.02 16,000

The amount of necessary solvent water C can be calculated by a total mass balance

L0 + Vn = Ln + V1



Vn = Ln + V1 - L0 = 16,000 + 7,200 - 10,000 = 13,200 kg

total mass [kg]

wt% A

wt% B

wt% C

feed L0

10,000

40

32

28

solvent Vn

13,200

0

0

100

overflow V1

7.200

0

40

60

underflow Ln

13.200

25

2

73

Determination of the number of ideal steps First of all the constant underflow with xA,Ln = 0.25 and the given points L0 (xA,L0 = 0.4, xB,L0 = 0.32, xC,L0 = 0.28), V1 (xA,V1 = 0, xB,V1 = 0.4, xC,V1 = 0.6), Vn (xC,Vn = 1) and Ln (xA,Ln = 0.25, xB,Ln = 0.02, xC,Ln = 0.73) are drawn in the diagram. The one pole line is the connection of V1 with L0 and the other one the connection of Vn with Ln. Crossing these pole lines results in the pole point ∆.

Examples Solid - Liquid Extraction

Lecturer: Dr.Gamse

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Construction of the connode (= connection line with point A) through V1 gives the underflow L1 at the underflow line. Connecting L1 with the pole point ∆ give the extract composition V2, and so on. Finally the number of ideal steps results with Nth = 10 2. Ponchon - Savarit Diagram

Determination of the feed point A 0. 4 = 0.666 = B + C 0.32 + 0.28

N0 =

B 0.32 = 0.5333 = B + C 0.32 + 0.28

XL0 =

L0 = (0.32 + 0.28) * F = (0.32 + 0.28) * 10.000 = 6,000 kg

concentration of the overflow (extract solution) XV1 =

B 0 .4 = = 0.4 B + C 0 . 4 + 0 .6

90% extraction efficiency: V1 * XV1 = 0.9 * L0 * XL0



V1 =

0 .9 * L 0 * X L 0 X V1

=

0.9 * 6000 * 0.533 = 7.200 kg 0 .4

Balance for solid material: N0 * L0 = Nn * Ln Nn = NUnderflow =



Ln =

A 1 = = 0.333 B+C 3

N0 * L 0 0.666 * 6,000 = = 12.000 kg 0.333 NN

Total balance: L0 + Vn = LM = Ln + V1



LM = 12,000 + 7,200 = 19,200 kg



Vn = 19,200 - 6,000 = 13,200 kg

Examples Solid - Liquid Extraction

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Lecturer: Dr.Gamse

Balance for sugar B: L0 * XL0 + Vn * XVn = LM * XM = V1 * XV1 + Ln * XLn with XVn = 0 (pure solvent) follows XLn =

L 0 * XL0 − V1 * X V1 6,000 * 0.533 − 7,200 * 0.4 = = 0,0267 12,000 Ln

L

N

X

feed L0

6,000

0.666

0.533

solvent Vn

13,200

0

0

overflow V1

7,200

0

0.4

underflow Ln

12,000

0.333

0.0267

Determination of the ideal number of steps: Drawing of the points L0 (NL0 = 0.666, XL0 = 0.533), V1 (NV1 = 0, XV1 = 0.4), Vn (NVn = 0, XVn = 0) and Ln (NLn = 0.333, XVn = 0.0267). The connection of L0 and V1 gives the first pole line and connection of Ln and Vn the second one. Crossing these two pole lines gives the pole point ∆. The first connode is a vertical line through V1 which gives at the underflow the point L1. Connecting this point L1 with the pole point ∆ give the next extract composition V2 and so on.

Finally the number of ideal steps results with Nth = 10