# Examples Solid- Liquid Extraction

```Examples Solid - Liquid Extraction
-1-
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
-2-
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
-3-
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
-4-
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
-5-
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
-6-
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
-7-
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
-8-
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
-9-
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
- 10 -
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
- 11 -
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
```