1. By extracting kerosene, 2 tons of waxed paper is to be dewaxed in a continuous
countercurrent extraction system that contains a number of ideal stages. The waxed paper
contains, by weight, 25% paraffin wax and 75% paper pulp. The extracted pulp is put
through a dryer to evaporate the kerosene. The pulp, which retains the unextracted wax
after evaporation, must not contain over 0.2 kg of wax per 100 kg of wax free pulp. The
kerosene used for the extraction contains 0.05 kg of wax paper per 100 kg of wax free
kerosene. Experiments show that the pulp retains 2.0 kg of kerosene per kg of kerosene
and wax free pulp as it is transferred from cell to cell. The extract from the battery is to
contain 5 kg of wax per 100 kg of wax free kerosene.
a. Find the overflow stream.
b. Find the underflow stream
c. Kg wax / kg kerosene in the underflow
d. No. of stages
Given:
V1:Overflow
5 𝑘𝑔 𝑤𝑎𝑥
100 𝑘𝑔 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒
S:
Vaya
V2y2
Vbyb
Solvent
0.05 𝑘𝑔 𝑤𝑎𝑥
100 𝑘𝑔 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒
First
Stage
Feed : 4 Tons Wax paper
25% wax
75% pulp
Lbxb
5 𝑘𝑔 𝑤𝑎𝑥
100 𝑘𝑔 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒
2 𝑘𝑔 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒
1 𝑘𝑔 𝑝𝑢𝑙𝑝
L1x1
L1: Underflow
0.2 𝑘𝑔 𝑤𝑎𝑥
100 𝑘𝑔 𝑓𝑟𝑒𝑒 𝑝𝑢𝑙𝑝
Required: Overflow and Underflow streamsN stages.
Solutions:
OMB : F + S = V1 + L1
Material balance for the wax
4 Ton x
1000 kg
1 Ton
= 4000 kg
4000 kg x .25 = 1000 kg
4000 kg x .75 = 3000 kg
Wax balance
1000 kg +
0.05 kg wax
5 kg wax
0.2 kg wax
(0.75 x 4000 kg)
Vb =
Va +
100 kg kerosene
100 kg kerosene
100 kg pulp
Solvent Balance
Vb = Va +
2 kg kerosene
(0.75 x 4000 kg) = Va + 6000 kg
1 kg pulp
Solving simultaneously we get the streams
𝐕𝐚 = 𝟐𝟎𝟏𝟒𝟏. 𝟒𝟏𝟒𝟏 𝐤𝐠 𝐊𝐞𝐫𝐨𝐬𝐞𝐧𝐞
𝐕𝐛 = 𝐕𝟐 = 𝟐𝟎𝟎𝟖𝟏. 𝟒𝟏𝟒𝟏 𝐤𝐠 𝐤𝐞𝐫𝐨𝐬𝐞𝐧𝐞
undeflow = 6000 kg kerosene
0.2 kg wax
x 0.75 x 4000 kg pulp
𝐤𝐠 𝐰𝐚𝐱
100 kg pulp
xb =
= 𝟎. 𝟎𝟎𝟏
6000 kg kerosene
𝐤𝐠 𝐤𝐞𝐫𝐨𝐬𝐞𝐧𝐞
Wax balance:
20081.4141 y2 + 1000 =
5
(20141.4141) + 6000x1
100
equilibrium condition x1 = ya = y2 = 0.05
kg wax
kg kerosene
y2 = 0.01529
No. of ideal stages
0.001 − 0.0005
)
0.05 − 0.01529 + 1 = 4 .5398 ≈ 𝟓
𝑁=
0.01529 − 0.0005
ln (
)
0.05 − 0.001
ln (
2. 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.
extract (overflow)
solvent
V1
V2
extraction
step
L0
L1
feed
underflow
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
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
inert material
w
A,Underflow =
A
=
inert material+extractable substance+solvent
A+B+C
1.5
w
A,Underflow
=w
A,L1
=
= 0.6
1.5 + 1
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
𝑊𝐴,𝑀
0.4
= 200
𝑊𝐴,𝐿1
. 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.
𝑊𝐵,𝑉1 =
𝑊𝐶,𝑉1 =
𝐵
𝐴+𝐵+𝐶
=
20
0+20+100
𝐶
100
=
𝐴 + 𝐵 + 𝐶 0 + 20 + 100
𝑊𝐵,𝑉1 = 0.1667
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
𝑊𝐵,𝐿1 =
𝐿𝑂 × 𝑊𝐵,𝐿𝑜 − 𝑉1 × 𝑊𝐵,𝑉1 100 × 0.2 − 66.666 × 0.1667
=
𝐿1
133.333
𝑊𝐵,𝐿1 = 0.067
𝑊𝐴,𝐿1 + 𝑊𝐵,𝐿1 + 𝑊𝐶,𝐿1 = 1
𝑊𝐶,𝐿1 = 1 − 0.6 − 0.67
𝑊𝐶,𝐿1 = 0.333
Feed LO
Solvent V2
Overflow V1
Underflow L1
Total mass (kg)
100
100
66.666
133.333
Wt% A
80
0
0
60
Wt% B
20
0
16.667
6.7
Wt%C
0
100
83.333
33.3
Situation for problems no. 23-28
By extraction with kerosene with 0.05 lb wax per 100 lb kerosene, 2 tons of waxed paper per day
is to be dewaxed in a continuous countercurrent extraction system that contains a number of
ideal stages. The waxed paper contains, by weight, 25 percent paraffin wax and 75 percent paper
pulp. The extracted pulp is put through a dryer to evaporate the kerosene. The pulp, which
retains the unextracted wax after evaporation, must not contain over 0.2 lbs of wax per 100 lbs of
wax-free kerosene-free pulp. Experiment show that the pulp retains 2.0 lb of kerosene per lb of
kerosene and wax-free pulp as it is transferred from cell to cell. The extract from the battery is to
contain 5 lb of wax paper per 100 lb of wax-free kerosene. Per 100 lb of wax-free kerosene-free
pulp,
23. The kerosene in the exhausted pulp is equal to
a. 150 lb
c. 117 lb
b. 200 lb
d. 212 lb
24. The kerosene in the strong solution is equal to
a. 561 lb
c. 761
b. 651 lb
d. 671 lb
25. The wax in the strong solution is equal to
a. 35.35 lb
c. 55.33 lb
b. 33.55 lb
d. 53.53 lb
26. The wax in the underflow to unit 2 is equal to
a. 8 lb
c. 12 lb
b. 10 lb
d. 14 lb
27. The wax in the overflow from the second cell to the first is
a. 10.22 lb
c. 12.11 lb
b. 11.12 lb
d. 13.19 lb
28. The total number of ideal stages is equal to
a. 3
c. 5
b. 4
d. 6
Given:
Solute = Wax
Solvent = Kerosene
Inert = Pulp
Y
YN
Y2
1
+1
1
2
N
X
X
1
N
F= 2 Tons
25%
Solute
75% Inert
Solution:
In Feed:
Inert = 100 lb
100 lb inert
Feed =
= 133.3333 lb
0.75
Solute = (0.25)(133.3333 lb) = 33.3333 lb
In Final Underflow:
Inert = 100 lb
0.2 lb solute
Solute = 100 lb inert x 100 lb inert = 0.2 lb
Solvent =
2 lb solvent
lb inert
x 100 lb inert = 𝟐𝟎𝟎 𝐥𝐛
Overall Solvent Balance
0 + VN+1 = V1 + 200
Equation 1
Overall Solute Balance
0.05 lb solute
5 lb solute
33.3333 + ( 100 lb solvent x VN+1) = (100 lb solvet x V1) + 0.2
Equation 2
From Equation 1:
VN+1 – V1 = 200
From Equation 2:
5𝑥10−4 VN+1 – 0.05V1 = -33.1333
Equate Equation 1 and Equation 2, solve for VN+1 and V1:
VN+1 = 871.3798 lb
V1 = 671.3798 lb
Solute in V1:
5 lb solute
100 lb solvent
x 671.3798 = 33.5690 lb
Solvent Balance in Stage 1:
0 + V2 = 671.3798 +200
V2 = 871.3798 lb
Solute Balance in Stage 2:
5 lb solute
5 lb solute
33.3333 + solute in V2 = (100 lb solvent x 200 lb solvent) + (100 lb solvent) 𝑥671.3798 lb solvent
Solute in V2 = 10.2357 lb
5 lb solute
Solute in Y2 = `00 lb solvent x 200 lb solvent = 𝟏𝟎 𝐥𝐛
Solving for Number of Stages:
N=1+
Y
−X
ln[ N+1 N ]
Y2 −X1
Y
−Y
ln[ N+1 2 ]
XN −X1
where:
0.05 lb solute
YN+1 = 100 lb solvent = 5x10−4
0.2 lb solute
XN = 200 lb solvent = 1x10−3
10.2357 lb solute
Y2=871.3798 lb solvent = 0.0117
10 lb solute
X1 = 200 lb solvent = 0.05
N=1+
5𝑥10−4 −1𝑥10−3
]
0.0117−0.05
5𝑥10−4 −0.0117
ln[
]
1𝑥10−3 −0.05
ln[
N = 3.9396 = 4 stages
Problems no. 29-31.
100 kg of solid containing 50% of a soluble material were treated with 200 kg of a solvent
containing the same solute at 3% concentration in a vessel under the constant agitation. After a
long time, pressing separated in the solution and the solid. The solid analyzed 0.75 kg of solvent
per kg of inert solid.
29. The amount of solute in the final underflow is approximately equal to
a. 10.82 kg
c. 2.78 kg
b. 8.54 kg
d. 7.16 kg
30. The amount of solvent in the extract is approximately equal to
a. 106.2 kg
c. 178.3 kg
b. 216.0 kg
d. 156.5 kg
31. How much extract was collected?
a. 201.68 kg
c. 216.08 kg
b. 106.21 kg
d. 192.86 kg
Given:
Overflow, V1
Vo = 200 kg
(Extract)
3% solute
97% solvent
Feed, F = 100 kg
Underflow, L1
50% solute
solid =
0.75 kg solvent
kg inert solid
50% solid
Required: 29.) amount of solute in final underflow
30.) amount of solvent in the extract
31.) V1
Solution:
In the Feed, F: 𝑠𝑜𝑙𝑢𝑡𝑒 = (0.5)(100𝑘𝑔) = 50 𝑘𝑔
𝑖𝑛𝑒𝑟𝑡 𝑠𝑜𝑙𝑖𝑑 = (0.5)(100 𝑘𝑔) = 50 𝑘𝑔
𝑠𝑜𝑙𝑢𝑡𝑒 = (0.03)(200 𝑘𝑔) = 6 𝑘𝑔
In Vo:
𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = (0.97)(200 𝑘𝑔) = 194 𝑘𝑔
In overflow, V1.: 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑎
𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = Vi − a
Solute balance:
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐹 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉𝑜 = 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1
50 + 6 = 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 + 𝑎
In underflow, 𝐿1 :
𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = (50 𝑘𝑔 𝑖𝑛𝑒𝑟𝑡𝑠) (
𝑠𝑜𝑙𝑢𝑡𝑒 = 56 − 𝑎
0.75 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑘𝑔 𝑖𝑛𝑒𝑟𝑡𝑠
) = 37.5 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Continuation...
Inerts balance:
𝑖𝑛𝑒𝑟𝑡𝑠 𝑖𝑛 𝐹 = 𝑖𝑛𝑒𝑟𝑡𝑠 𝑖𝑛 𝐿1 = 50 𝑘𝑔
Solution balance:
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐹 + 𝑉𝑜 = 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿1 + 𝑉1
(50 + 0) + 200 = 93.5 − 𝑎 + 𝑉1
156.5 + 𝑎 = 𝑉𝑖
At Equilibrium:
𝑠𝑜𝑙𝑢𝑡𝑒
𝑠𝑜𝑙𝑢𝑡𝑒
(
) =(
)
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐿1
𝑎
56 − 𝑎
) =(
)
156.5 + 𝑎 𝑉1
37.5 + 56 − 𝑎 𝐿1
(
𝑎 = 45.1753 𝑘𝑔
29.) Amount of solute in underflow, L1:
𝒔𝒐𝒍𝒖𝒕𝒆 = 56 − 𝑎 = 56 − 45.1753 = 𝟏𝟎. 𝟖𝟐𝟒𝟕 𝒌𝒈
30.) Amount of solvent in Extract, Vi:
𝒔𝒐𝒍𝒗𝒆𝒏𝒕 = 𝑉1 − 𝑎 = 156.5 + 𝑎 − 𝑎 = 𝟏𝟓𝟔. 𝟓 𝒌𝒈
31.) V1:
𝑽1 = 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑉1 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑉1 = 156.5 + 45.1753 = 𝟐𝟎𝟏. 𝟔𝟕𝟓𝟑 𝒌𝒈
Situation for problems no. 32-34
A solid B, contains a soluble component, A, of mass fractions xA = 0.25, xB = 0.75 and is to be
recover A by a solvent extraction with C. Solid B and solvent C are mutually totally insoluble.
The extracted solid is to be screw passed to a 0.75 kg of solution/kg of B underflow. The
entrainment of B in the overflow can be neglected. Per kg of feed and to obtain 85% of A in the
extract overflow.
32. The composition of the solution in the underflow is
a. 0.04
c. 0.01
d. 0.10
b. 0.07
33. The amount of solvent in the underflow is
a. 0.44
b. 0.53
34. How much solvent C (A free) must be fed?
a. 3.5000 kg
b. 2.5712 kg
c. 1.7000 kg
d. 5.2311 kg
c. 0.88
d. 1.33
GIVEN:
REQUIRED:
32. x1 in L1
33. solvent in L1
34. solvent C
SOLUTION:
In Feed, F:
F = 1 kg
Solute (A) : (0.25)(1) = 0.25 kg
Inerts (B) : (0.75)(1) = 0.75 kg
In Underflow , L1 :
Inerts (B) : 0.75 kg
Solution (A + C) : (0.75)(0.75) = 0.5625 kg
Solute : (1 - 0.85)(0.25) = 0.0375 kg
Solvent : (0.5625 - 0.0375) = 0.5250 kg
x1 =
solute
solution
=
0.0375
0.5625
= 𝟎. 𝟎𝟔𝟔𝟕
In Overflow , V1 :
Solute (A) : (0.85)(0.25) = 0.2125 kg
V1 = ? ? ?
Solution Balance:
(0.25 + 0) + C = V1 + 0.5625
V1 = C - 0.3125
@ equilibrium:
solute
(solution)
solute
V1
0.2125
= (solution)
L1
0.0375
(C − 0.3125)
V1
= (0.5625)
C = 3.5000 kg
L1
Situation for problems no. 35-38
Seeds containing 30% weight oil are extracted in a countercurrent plant and 88% of the oil is
recovered in a solution containing 55% by weight of oil. The seeds are extracted with fresh
solvent and 1 kg of solution is removed in the underflow in association with every 1.5 kg of
insoluble material.
35. The amount of solvent in final extract is approximately equal to
a. 26.4 kg
c. 46.67 kg
d. 43.07 kg
b. 21.6 kg
36. The amount of solvent in final underflow is approximately equal to
a. 26.4 kg
c. 46.67 kg
b. 21.6 kg
d. 43.07 kg
37. The concentration of oil in the solvent stream for stage 1 is approximately equal to
a. 0.55
c. 0.18
b. 0.08
d. 0.34
38. How many ideal stages are needed to attain the desired separation?
a. 4
b. 6
c. 8
d. 10
GIVEN:
REQUIRED:
35. Solvent in V1
36. Solvent in LN
37.Concentration of oil V2
38.N
SOLUTION:
Basis: 100 kg of Feed
In Feed, F:
Insoluble = 0.70(100 kg) = 70 kg
Oil = 0.30(100 kg) = 30 kg
In final Overflow, V1:
Oil = 0.88 (30 kg) = 26.4 kg
45
Solvent = 26.4 kg(55 ) = 21.6 kg
y1=x1=0.55
In final Underflow, LN=L1=46.6667 kg
Oil= 0.12(30 kg) = 3.6 kg
Insoluble = 70 kg
1 𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Solution = 70 kg(1.5 𝑘𝑔 𝑖𝑛𝑠𝑜𝑙𝑢𝑏𝑙𝑒)= 46.6667 kg
Solvent = 46.6667 kg– 3.6 kg= 43.0667 kg
3.6
xN= 46.6667 = 0.0771
In Fresh Solvent, VN+1:
Solvent = 21.6 kg + 43. 0667 kg = 64.6667 kg
Solute = 0
yN+1 = 0
Solute Balance around Stage 1:
30 kg + 64.6667 kg (y2) = 26.4 kg + 46.6667 kg(0.55)
y2= 0.3412 = x2
N= 1 +
0−0.0771
)
0.3412−0.55
0−0.3412
ln(
)
0.0771−0.55
ln(
= 4.05
Situation for problems 39-42
Calcium-carbonate precipitate can be produced by the reaction of an aqueous solution of
sodium carbonate and calcium oxide. The by-product is aqueous sodium hydroxide.
Following decantation, the slurry leaving the precipitation tank is 5 wt% calcium carbonate,
0.1 wt% sodium hydroxide, and the balance water. One hundred thousand lb/h of this slurry
is fed to a two-stage, continuous, countercurrent washing system to be washed with 20,000
lb/h of fresh water. The underflow from each thickener will contain 20 wt% solids.
39. The amount of extract
40. The amount of sodium hydroxide in final extract
41. The amount of sodium hydroxide in final underflow
42. The percent recovery of sodium hydroxide in the extract
Given:
20,000 lb/h
(V2)
(V1)
2
1
(F) 100,000 lb/h
(V3)
(L1)
5 wt% Calcium Carbonate
0.1 wt% Sodium Hydroxide
94.9 wt% Water
Solution:
In Feed:
𝐶𝑎𝑙𝑐𝑖𝑢𝑚 𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 = 0.05(100,000) = 5000 𝑙𝑏/ℎ
𝑆𝑜𝑑𝑖𝑢𝑚 𝐻𝑦𝑑𝑟𝑜𝑥𝑖𝑑𝑒 = 0.001(100,000) = 100 𝑙𝑏/ℎ
𝑊𝑎𝑡𝑒𝑟 = 0.949 (100,000) = 94900 𝑙𝑏/ℎ
Solid Balance:
𝑆𝑜𝑙𝑖𝑑 𝑖𝑛 𝐹 = 𝑆𝑜𝑙𝑖𝑑 𝑖𝑛 𝐿2
5000
𝑙𝑏
= 0.20 (𝐿2)
ℎ
𝐿2 = 25000
𝑙𝑏
ℎ
(L2)
20wt% solid
OMB:
𝐹 + 𝑉3 = 𝑉1 + 𝐿2
100,000 + 20,000 = 𝑉1 + 25,000
𝑉1 = 95000
𝑙𝑏
ℎ
Stage 1 (@ equilibrium)
𝑆𝑜𝑙𝑢𝑡𝑒
𝑆𝑜𝑙𝑢𝑡𝑒
(
)(𝑉1) = (
)
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐿1)
(
𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿2 = 25,000 (0.80) = 20,000
𝑉3 = 𝑉2 ; 𝐿2 = 𝐿1
𝑆𝑜𝑙𝑢𝑡𝑒
𝑆𝑜𝑙𝑢𝑡𝑒
)(𝑉1) = (
)
95,000
20,000 (𝐿1)
𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1 = 4.75 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 − − − 𝑒𝑞𝑛(1)
Stage 2 (@ equilibrium)
(
𝑆𝑜𝑙𝑢𝑡𝑒
𝑆𝑜𝑙𝑢𝑡𝑒
)(𝑉2) = (
)
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐿2)
(
𝑆𝑜𝑙𝑢𝑡𝑒
𝑆𝑜𝑙𝑢𝑡𝑒
)(𝑉2) = (
)
20,000
20,000 (𝐿2)
𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉2 = 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿2 − − − 𝑒𝑞𝑛(2)
Solute Balance in Stage 2
𝑆𝑜𝑙𝑢𝑡𝑒𝐿1 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝑉3 = 𝑆𝑜𝑙𝑢𝑡𝑒𝑉2 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝐿2
𝑆𝑜𝑙𝑢𝑡𝑒𝐿1 + 0 = 𝑆𝑜𝑙𝑢𝑡𝑒𝑉2 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝐿2
𝑆𝑜𝑙𝑢𝑡𝑒𝐿1 = 𝑆𝑜𝑙𝑢𝑡𝑒𝐿2 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝐿2
𝑆𝑜𝑙𝑢𝑡𝑒𝐿1 = 2𝑆𝑜𝑙𝑢𝑡𝑒 𝐿2 − − − 𝑒𝑞𝑛 (3)
Overall Solute Balance
𝑆𝑜𝑙𝑢𝑡𝑒𝐹 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝑉3 = 𝑆𝑜𝑙𝑢𝑡𝑒𝑉1 + 𝑆𝑜𝑙𝑢𝑡𝑒 𝐿2
100 + 0 = 𝑆𝑜𝑙𝑢𝑡𝑒𝑉1 + 𝑆𝑜𝑙𝑢𝑡𝑒𝐿2 − − − 𝑒𝑞𝑛 (4)
Substitute eqn (1) to eqn (4)
100 = 4.75𝑆𝑜𝑙𝑢𝑡𝑒𝐿1 + 𝑆𝑜𝑙𝑢𝑡𝑒𝐿2 − − − 𝑒𝑞𝑛 (5)
Substitute eqn (3) to eqn (5)
100 = 4.75(2)𝑆𝑜𝑙𝑢𝑡𝑒𝐿2 + 𝑆𝑜𝑙𝑢𝑡𝑒𝐿2
𝑆𝑜𝑙𝑢𝑡𝑒𝐿2 = 9.52
𝑙𝑏
ℎ
Using eqn 4:
100 + 0 = 𝑆𝑜𝑙𝑢𝑡𝑒𝑉1 + 9.52
𝑆𝑜𝑙𝑢𝑡𝑒𝑉1 = 90.48
𝑙𝑏
ℎ
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑅𝑒𝑐𝑜𝑣𝑒𝑟𝑦 (𝑅) =
𝑅=
(90.48 − 0 )
𝑥 100
100
𝑅 = 90.48 %
𝑆𝑜𝑙𝑢𝑡𝑒𝑉1 − 𝑆𝑜𝑙𝑢𝑡𝑒𝑉3
𝑆𝑜𝑙𝑢𝑡𝑒𝐹
PROBLEM 43-46
Ground roasted coffee contains 8% soluble solids, 2% water, and 90% inert insoluble solids.
In order to obtain an extract with high soluble solids content without having to concentrate it
for spray drying, a countercurrent extraction process is to be used to prepare the extract. It is
desired that the final extract contain 0.15kg soluble/kg water and that the soluble of the spent
coffee grounds not to exceed 0.008 kg/kg dry inert solids. The coffee grounds carry 1 kg
water/kg of soluble-free inert solids and this quantity is constant with the solute concentration
in the extract.
REQD:
43) The amount of final extract is approximately equal to
a. 55.81 kg
b. 48.54 kg
c. 72.8 kg
d. 28.1 kg
44) The concentration of the solution adhering to the extracted solids is approximately equal
to
a. 0.0936
b. 0.0079
c. 0.1304
d. 0.0032
45) The water/coffee ratio to be used in the extraction is
a. 1.37
b. 2.88
c. 0.98
d. 1.87
46) The number of extraction stages needed for this process is
a. 5
b. 6
c. 7
d. 8
SOLUTION:
Overflow, V1
0.15 kg solute
kg H2O
V2
V3
V4
Y2
Y3
Y4
Solvent, Vn+1
Yn+1
1
2
3
N
Solution:
Feed, F
R1
R2
R3
L1
L2
L3
8% Solute
Final Underflow, Ln
Xn
Solute/Inerts = 0.008
2% H2O
90% Inerts
X1 R= 1 kg H2O/kg
X2Inerts
X3
In the feed: basis(100 kg)
Solute: 0.08(100)= 8kg
Solvent: 0.02(100)= 2kg
Inerts: 0.9(100)=90kg
In final underflow:
Inerts= inerts in F=90kg
Solute=0.008(90)=0.72kg
Solvent=90(1)=90 kg
XN= 0.008
𝑆𝑜𝑙𝑢𝑡𝑒
0.72
(𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛)in LN =90+0.72 = 𝟎. 𝟎𝟎𝟕𝟗(#44)
In final overflow, V1:
Solute balance: 8+0=0.72+Solute in V1
𝑆𝑜𝑙𝑢𝑡𝑒
Y1= X1=𝑆𝑜𝑙𝑣𝑒𝑛𝑡 = 0.15
Final Overflow(extract)=solute+solvent
= 7.28+48.5333kg=V1
= 55. 8133 kg(#43)
In solvent stream, Vn+1
Yn+1=0(pure water)
Solvent= Vn+1=?
Overall solvent bal:
2+Vn+1=90+48.5333
Vn+1=136.5333
Ratio:
𝑉𝑛+1
𝐹
=
136.5333
100
= 𝟏. 𝟑𝟔𝟓𝟑 (#45)
Solute for Y2 using Solute balance around stage 1
8+ V2Y2=L1X1+7.28
V2= Vn+1=136.5333
Y2=?
L1=LN=90
X1=Y1=0.15
8+136.5333=90(0.15)+7.28
Y2=0.0936
Solve for N:
N=1+
0.008
)
0.0926−0.15
0−0.0936
𝑙𝑛(
)
0.008−0.15
ln(
=5.69=6 stages(#46)
Problem 47
Given:
V1
V2
YN+1
1
Y1
2
Y2
Feed= 50 tons/hr
N
L1
LN
X1`
XN
48% H2O
40% Pulp
R=
3 𝑡𝑜𝑛𝑠 𝐻20
𝑡𝑜𝑛𝑠 𝑃𝑢𝑙𝑝
12%Sugar
Required: N = ?
Solutions:
In Feed: H2O = 0.48(50) = 24
𝑡𝑜𝑛𝑠
Pulp = 0.40(50) = 20
Sugar = 0.12(50) = 6
ℎ𝑟
𝑡𝑜𝑛𝑠
ℎ𝑟
𝑡𝑜𝑛𝑠
ℎ𝑟
In Final Overflow:
Sugar = 0.97(6) = 5.82
5.82
𝑡𝑜𝑛𝑠
Solution = 0.15 = 38.8
ℎ𝑟
𝑡𝑜𝑛𝑠
ℎ𝑟
H2O = V1 = 38.8 – 5.82 = 32.98
Y1 =
5.82
32.98
𝑡𝑜𝑛𝑠
ℎ𝑟
= 0.1765
X1 = Y1 = 0.1765
at equilibrium
In Final Underflow:
Sugar = 0.03(6) = 0.18
𝑡𝑜𝑛𝑠
ℎ𝑟
H2O = LN = 20(3) = 60
XN =
0.18
60
𝑡𝑜𝑛𝑠
ℎ𝑟
= 0.003
In Fresh Solvent:
OMB (Solvent): LN + V1 – Lo
VN+1 = 60 + 32.92 – 24
H20 = VN+1 = 68.98
𝑡𝑜𝑛𝑠
ℎ𝑟
YN+1 = 0 (pure solvent)
Sugar Balance Around Stage 1:
Sugar in F + V2Y2 = L1X1 + V1Y1
V2 = V3 = V4 = ….. = VN+1 = 68.98
L1 = L2 = L3 = L4 = ….. = LN = 60
6 + 68.98Y2 = 60(0.1765) + 5.82
Y2 = 0.1509
Solving for N:
N=
0 − 0.003
0.1509−0.1765
0−0.1509
𝑙𝑛
0.003−0.1765
ln
N = 16.36 = 17
+1
𝑡𝑜𝑛𝑠
ℎ𝑟
𝑡𝑜𝑛𝑠
ℎ𝑟
Problem 48
Constant Solution Retention: L8V – solution flowrates : X8Y – solute/solution
In Feed:
H2O = 0.48(50) = 24 tons/hr
Pulp = 0.40(50) = 20 tons/hr
Sugar = 0.12(50) = 6 tons/hr
In Final Overflow:
Sugar = 0.97(6) = 5.82 tons/hr
Y1 = 0.15
5.82
Solution = V1 = 0.15 = 38.8 tons/hr
X1 = Y1 = 0.15
(@ equilibrium)
In Final Underflow:
R=
3 𝑡𝑜𝑛𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑡𝑜𝑛 𝑑𝑟𝑦 𝑝𝑢𝑙𝑝
Sugar = 0.03(6) = 0.18 tons/hr
Solution = LN = 3(20) = 60 tons/hr
XR =
0,18
60
= 0.003
In Fresh Solvent
H2O = VN+1 = LN + V1 – L0 (overall solution balance)
H2O = VN+1 = 60 + 38.8 – (24 + 6)
VN+1 = 68.8 tons/hr
YN+1 = 0
(pure solvent)
Sugar Balance Around Stage 1:
Sugar in F + V2Y2 = L1X1 + V1Y1
V2 = V3 = V4 = . . . = VN+1 = 68.8 tons/hr
L1 + L2 + L3 = . . . = LN = 60 tons/hr
6 + 68.8Y2 = 60(0.15) + 38.8(0.15)
Y2 = 0.1282
Solving for N:
N=
0−0.003
]
0.1282−0.15
0−0.1282
ln[
]
0.003−0.15
ln[
N = 15.49 = 16
Situation for problems no. 49-52
A seashore sand contains 85% insoluble sand, 12% salt and 3% water. 1000 lb/hr of this
mixture is to be extracted in a countercurrent washing system with 2000 lb/hr of pure water
so that after drying it will contain only 0.2% salt. The sand retains 0.5 lb of water per pound
of insoluble sand.
49. The mass of salt in the final underflow is equal to
a. 1.7 lb/hr
c. 2.3 lb/hr
b. 1.2 lb/hr
d. 2.5 lb/hr
50. The concentration of salt in the final overflow is equal to
a. 0.03
c. 0.07
b. 0.05
d. 0.09
51. The concentration of salt in the solvent stream for stage 1 is approximately equal to
a. 0.023
c. 0.07
b. 0.015
d. 0.19
52. The number of washing is approximately equal to
a. 3
c. 5
b. 4
d. 6
Given:
Y
2000
lb/hrY
Y2
N
1
+1
1
2
L
N
LN
1
F= 1000 lb/hr
12% Solute
85% Inert
3% Solvent
Solution:
In Feed :
Inert = 1000 lb/hr (0.85) = 850 lb/hr
Solute = 1000 lb/hr (0.12) = 120 lb/hr
Solvent = 1000 lb/hr (0.03) = 30 lb/hr
In Final Underflow :
after drying = 0.2%
salt
Inert = 850 lb/hr
0.5 𝑙𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Solvent = 𝑙𝑏 𝑖𝑛𝑒𝑟𝑡 × 850 𝑙𝑏 𝑖𝑛𝑒𝑟𝑡 = 425 𝑙𝑏 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Solute :
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁 =
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁 =
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁 =
0.2
(𝑖𝑛𝑒𝑟𝑡 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁 )
100
0.2
0.2
𝑖𝑛𝑒𝑟𝑡 +
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁
100
100
0.2
𝑙𝑏
0.2
(850 ) +
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿𝑁
100
ℎ𝑟
100
𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝑳𝑵 = 𝟏. 𝟕𝟎𝟑𝟒 𝒍𝒃/𝒉𝒓
Final underflow = inert +solvent + solute
𝑙𝑏
= (850 + 425 + 1.7034) ℎ𝑟
𝐹𝑖𝑛𝑎𝑙 𝑈𝑛𝑑𝑒𝑟𝑓𝑙𝑜𝑤 = 1276.7034 𝑙𝑏/ℎ𝑟
Overall Material Balance (OMB) :
𝐹 + 𝑉𝑁+1 = 𝐿𝑁 + 𝑉1
1000
𝑙𝑏
𝑙𝑏
𝑙𝑏
+ 2000
= 1276.7034 + 𝑉1
ℎ𝑟
ℎ𝑟
ℎ𝑟
𝑉1 = 1723.2966
𝑙𝑏
ℎ𝑟
Overall Solute Balance :
𝐹𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑉𝑁+1 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑉1𝑆𝑜𝑙𝑢𝑡𝑒 + 𝐿𝑁 𝑠𝑜𝑙𝑢𝑡𝑒
120
𝑙𝑏
𝑙𝑏
+ 0 = 𝑉1 𝑠𝑜𝑙𝑢𝑡𝑒 + 1.7034
ℎ𝑟
ℎ𝑟
𝑉1 𝑠𝑜𝑙𝑢𝑡𝑒 = 118.2966
𝑙𝑏
ℎ𝑟
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1 =
118.2966
1723.2966
𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝑽𝟏 = 𝟎. 𝟎𝟔𝟖𝟔𝟓 ≈ 𝟎. 𝟎𝟕
𝑙𝑏
𝑙𝑏
𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑉1 = (1723.2966 − 118.2966) ℎ𝑟 = 1605 ℎ𝑟
In Stage I :
V1=1605 lb/hr
V2 = 2000 lb/hr
LN= 425 lb/hr
F= 30 lb/hr
(
𝑠𝑜𝑙𝑢𝑡𝑒
𝑠𝑜𝑙𝑢𝑡𝑒
)𝑉1 = (
)𝐿
𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑠𝑜𝑙𝑣𝑒𝑛𝑡 1
𝑙𝑏
ℎ𝑟 = 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1
𝑙𝑏
𝑙𝑏
1605
425
ℎ𝑟
ℎ𝑟
118.2966
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 = 31.3246
𝑙𝑏
ℎ𝑟
Solute Balance in Stage I :
𝐹𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑉2 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑉1 𝑠𝑜𝑙𝑢𝑡𝑒 + 𝐿1 𝑠𝑜𝑙𝑢𝑡𝑒
120
𝑙𝑏
𝑙𝑏
𝑙𝑏
+ 𝑉2 𝑠𝑜𝑙𝑢𝑡𝑒 = 118.2966 + 31.3246
ℎ𝑟
ℎ𝑟
ℎ𝑟
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉2 = 29.6212
𝑙𝑏
ℎ𝑟
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉2 =
29.6212
2000
𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝑽𝟏 = 𝟎. 𝟎𝟏𝟒𝟖𝟏 ≈ 𝟎. 𝟎𝟏𝟓
Solving for Number of Stages:
N=1+
Y
−X
ln[ N+1 N ]
Y2 −X1
YN+1 −Y2
ln[
]
XN −X1
where:
YN+1 = 0
XN =
1.7034 lb solute
29.6212 lb solute
Y2=
X1 =
N=1+
= 4.008x10−3
425 lb solvent
2000 lb solvent
= 0.01481
31.3246 lb solute
425 lb solvent
0−4.008𝑥10−3
]
0.01481−0.0737
0.01481−0
ln[
]
0.07370−4.008𝑥10−3
ln[
N = 2.7352 = 3 stages
= 0.07370
55. A slurry of flaked soybeans weighing 100 kg contains 75 kg inert solids and 25 kg of
solution 10 weight % oil and 90 weight % solvent hexane. This slurry is contacted with 100
kg pure hexane in a single stage so that the value of retention for the outlet underflow is 1.5
kg on insoluble solid per kg solvent in the adhering solution. The composition of underflow
leaving the extraction stage in percent by weight oil is
GIVEN:
V1
V0 = 100 kg hexane
y1
y0
F = 100 kg
Inert = 75 kg
Sol’n = 25 kg
L1
x1
𝑅 = 1.5
𝑘𝑔 𝑖𝑛𝑒𝑟𝑡
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
REQUIRED:
The composition of underflow leaving the extraction
SOLUTION:
In Feed:
F = 100 kg
Inert = 75 kg
Sol’n = 25 kg
Oil (solute) = .10(25 kg) = 2.5 kg
Inert balance:
Inert in feed = Inert in L1
Inert in L1= 75 kg
In Underfeed (L1):
Inert = 75 kg
Solvent = ? = 50 kg
solvent =
75 kg inert
kg inert
1.5
kg solvent
Solute = ?
Solute balance:
Solute in F + Solute inV0 = Solute in V1 + Solute in L1
2.5 kg + 0 = Solute in V1 + Solute in L1
Eq. 1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1 = 2.5 − 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1
Solvent balance:
Solvent in F + Solvent in V0 = Solvent in V1 + Solvent in L1
22.25 kg + 100 kg = solvent in V1 + 50 kg
Solvent in V1 = 72.5 kg
At Equilibrium:
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1
=
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝐿1
Eq. 2
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿2
=
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1 + 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿2 + 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝐿2
Subs. Eq. 1 to Eq. 2:
2.5 − 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1
=
(2.5 − 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 ) + 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑉1
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 + 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝐿1
Solute in L1 = 1.0204 kg
Subs to Eq. 1
Solute in V1 = 1.4795
Composition on underflow leaving:
𝑉1
𝐿1
1.4795 𝑘𝑔
=
1.0204 𝑘𝑔
=
= 1.45
56. Tung meal containing 55% oil is to be extracted at a rate of 4000 kg per hour using nhexane containing 5% wt oil as solvent. A counter current multiple stage extraction system is
to be used. The meal retains 2 kg of solvent per kg of oil free meal while the residual charge
contains 0.11 kg oil per kg oil free meal while the product is composed of 15 weight percent
oil. The theoretical number of ideal stages is
(A) 3
(C) 5
(B) 4
(D) 6
Given:
𝑉1
𝑉𝑛+1
15% oil
5% oil
1
2
3
𝑘𝑔
𝐹 = 4000 ℎ𝑟
n
𝐿𝑛
55% oil
0.11 𝑘𝑔 𝑜𝑖𝑙
𝑘𝑔 𝑜𝑖𝑙 𝑓𝑟𝑒𝑒 𝑚𝑒𝑎𝑙
2 𝑘𝑔𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑅 = 𝑘𝑔 𝑓𝑟𝑒𝑒 𝑚𝑒𝑎𝑙
Required:
Theoretical number of ideal stages
Solution:
Basis: 1 hr
In the Feed,
𝑘𝑔 𝑜𝑖𝑙: 0.55𝑥 4000 = 2200 𝑘𝑔
𝑘𝑔𝑚𝑒𝑎𝑙: 0.45 𝑥 4000 = 1800 𝑘𝑔
In the final underflow,
0.11 𝑘𝑔 𝑜𝑖𝑙
𝑘𝑔𝑜𝑖𝑙:
𝑥 1800 𝑘𝑔 = 198 𝑘𝑔
𝑘𝑔 𝑜𝑖𝑙 𝑓𝑟𝑒𝑒 𝑚𝑒𝑎𝑙
2 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡:
𝑥 1800 𝑘𝑔 = 3600 𝑘𝑔
𝑘𝑔 𝑓𝑟𝑒𝑒 𝑚𝑒𝑎𝑙
𝑘𝑔 𝑚𝑒𝑎𝑙: 1800 𝑘𝑔
Overall Solution Balance: 2200 + 𝑉𝑛+1 = 𝑉1 + 3600 + 198
𝑉𝑛+1 = 𝑉1 + 1598  eq. 1
Overall Solute Balance: 2200 + 0.05𝑉𝑛+1 = 198 + 0.15𝑉1 eq 2
𝑉1 = 20819 𝑘𝑔
𝑉𝑛+1 = 22417 𝑘𝑔
At equilibrium condition,
𝑠𝑜𝑙𝑢𝑡𝑒
𝑠𝑜𝑙𝑢𝑡𝑒
(
) 𝑉1 = (
)𝐿
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1
(
0.15𝑥20819
𝑘𝑔𝑜𝑖𝑙 𝑖𝑛 𝐿1
)=(
)
20819
3600 + 𝑘𝑔 𝑜𝑖𝑙 𝑖𝑛 𝐿1
Kg oil in L1= 635.29 kg
Solute balance in stage 1: 2200 + 𝑘𝑔 𝑜𝑖𝑙 𝑖𝑛 𝑉2 = 0.15𝑥20819 + 635.29
Kg oil in V2= 1558.14 kg
Solvent balance in stage 1:
0 + 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑉2 = 3600 + 0.85𝑥20819
Kg solvent in V2= 21296.15 kg
𝑦2 :
1558.14
= 0.0682
1558.14 + 21296.15
198
𝑥𝑛 :
= 0.0521
3600 + 198
635.29
𝑥1 :
= 0.15
635.29 + 3600
At constant underflow,
𝑦𝑛+1−𝑥
𝑙𝑜𝑔 𝑦 − 𝑥 𝑛
2
1
𝑁−1=
𝑦𝑛+1 − 𝑦2
log 𝑥 − 𝑥
𝑛
1
0.05 − 0.0521
𝑙𝑜𝑔
0.0682 − 0.15
𝑁−1=
0.05 − 0.0682
𝑙𝑜𝑔
0.0521 − 0.15
N= 3.1665 ≈ 4 stages
57. Coconut oil is to be produced from dry copra in two stages. First, through expellers to
squeeze out part of the coconut oil and then through a counter current multi stage solvent
extraction process. After expelling, the dry copra cake contains 20% residual oil. In the
solvent extraction operation, 90% of the residual oil in the expeller cake is extracted as a
solution containing 50% by weight oil. If fresh solvent is used and on kg of solution with
every 2 kg of insoluble cake is removed with the underflow, the number of ideal stages is
(A) 4
(C) 6
(B) 5
(D) 7
Given:
𝑉1
𝑉𝑛+1
90% recovery
50% oil
𝑉𝑛+1
1
2
3
𝐿𝑛
F
Copra
𝑅=
20% oil
1 𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2 𝑘𝑔 𝑐𝑎𝑘𝑒
Required:
Number of Ideal Stages
Solution:
Basis: 100 kg Copra
In the Feed,
F= 100 kg
Kg oil: 0.20 x 100= 20 kg
Kg inert: 0.80 x 100= 80 kg
In the Solvent, V(n+1)
𝑦𝑛+1 =
𝑠𝑜𝑙𝑢𝑡𝑒
=0
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
In the Final Underflow, Ln,
Kg inert= 80 kg
Kg solution:
n
1 𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2 𝑘𝑔 𝑖𝑛𝑒𝑟𝑡
𝑥 80 𝑘𝑔 𝑖𝑛𝑒𝑟𝑡 = 40 𝑘𝑔
𝐿𝑛 : 80𝑘𝑔 + 40 𝑘𝑔 = 120 𝑘𝑔
𝑠𝑜𝑙𝑢𝑡𝑒: 0.10 𝑥 20 = 2 𝑘𝑔
𝑠𝑜𝑙𝑢𝑡𝑒 2
:
= 0.05
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 40
In the Final Overflow, V1
𝑦1 = 0.50
𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑒: 0.90 𝑥 20 = 18 𝑘𝑔
18𝑘𝑔
𝑉1 :
= 36 𝑘𝑔
0.50
In the first undeflow, L1
𝑠𝑜𝑙𝑢𝑡𝑒
= 𝑦1 = 0.50
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Kg inert: 80 kg
Kg solution: 40 kg
𝐿1 : 80𝑘𝑔 + 40𝑘𝑔 = 120 𝑘𝑔
𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑒: 0.50 𝑥 40 𝑘𝑔 = 20 𝑘𝑔
In the Ovreflow 2, V2,
OMB on stage 1,
𝐹 + 𝑉2 = 𝐿1 + 𝑉1
100 + 𝑉2 = 120 + 36
𝑉2 = 56 𝑘𝑔
Solute balance on stage 1,
𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐹 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉2 = 𝑆𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝐿1 + 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑉1
20 + 𝑦2 𝑥 56 = 20 + 18
𝑦2 = 0.3214
For constant underflow,
𝑦𝑛+1−𝑥
𝑙𝑜𝑔 𝑦 − 𝑥 𝑛
2
1
𝑁−1=
𝑦𝑛+1 − 𝑦2
log 𝑥 − 𝑥
𝑛
1
0 − 0.05
𝑙𝑜𝑔
0.3214 − 0.50
𝑁−1=
0 − 0.3214
𝑙𝑜𝑔
0.05 − 0.50
N= 5 stages
58. Roasted copper ore containing the copper as CuSO4 is to be extracted in countercurrent
stage extractor. Each hour, a charge consisting of 10 tons gangue, 1.2 tons CuSO4 and 0.5 ton
water is to be treated. The strong solution produced is to consist of 90% wt. water and 10%
wt. CuSO4. The recovery of CuSO4 is to be 98% of that in the ore. Pure water is to be used
as fresh solvent. After each stage, one ton inert gangue retained 2 tons of water plus the
copper sulfate dissolve in that water. Equilibrium is attained in each stage. The number of
stages required is.
Given:
OverFlow
water)
90% water, 10% CuSO4
Solvent(Pure
Feed
Underflow
1 𝑡𝑜𝑛 𝑔𝑎𝑛𝑔𝑢𝑒
𝑅 = 2 𝑡𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
10 tons gangue
1.2 tons CuSO4
0.5 tons water
Solution:
Inert in feed = inert in underflow
Amount of solution in underflow
10 𝑡𝑜𝑛 𝑖𝑛𝑒𝑟𝑡 𝑥
2 𝑡𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝑡𝑜𝑛 𝑖𝑛𝑒𝑟𝑡
= 20 𝑡𝑜𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Amount of overflow
1.2 𝑡𝑜𝑛𝑠 𝑥 0.98
0.10
= 11.76 𝑡𝑜𝑛𝑠
Overall balance of solute and solvent
11.76 + 20 = 0.5 + solvent stream
Solvent stream = 30.06 tons
Composition of final underflow
Solute in underflow = 1.2 – 1.2x0.98
=0.024 tons
0.024
%wt.
=
20
= 1.2 𝑥 10−3
For stage 1
11.76 tons
30.06 tons
1.2 tons CuSO4
0.5 tons Water
At equilibrium
𝑠𝑜𝑙𝑢𝑡𝑒
20 tons
𝑠𝑜𝑙𝑢𝑡𝑒
(𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)𝑜𝑣𝑒𝑟𝑓𝑙𝑜𝑤 = (𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)𝑢𝑛𝑑𝑒𝑟𝑓𝑙𝑜𝑤 = 0.10
Solute balance ; let x = fraction of solute at solvent stream
1.2 = 30.06 x =1.176 + 2
X = 0.065735
Number of stages
=1+
0−1.2𝑥10−3
]
0.065735− .10
0−0.065735
ln[
]
1.2𝑥10−3 −.10
ln[
= 9.226 = 10 stages
Situation for Problems 59-63
Oil is to be extracted from meal by means of benzene using a continuous countercurrent
extractor. The unit is to be treat 1000 kg of meal (based on completely exhausted solid) per
hour. The untreated meal contains 400 kg of oil and is contaminated with 25 kg of benzene.
The fresh solvent mixture contains 10 kg of oil and 655 kg of benzene. The exhausted solids
are to contain 60 kg of unextracted oil. Experiments carried out under conditions identical
with those of the projected battery show that the solution retained depends on the
concentration of the solution, as shown in table below. All quantities are given in an hourly
basis.
Concentration, kg
oil/kg solution
0.0
0.1
0.2
0.3
Solution retained,
kg/kg solid
0.500
0.505
0.515
0.530
Concentration, kg
oil/kg solution
0.4
0.5
0.6
0.7
Solution retained,
kg/kg solid
0.550
0.571
0.595
0.620
59. The concentration of the strong solution or extract is approximately equal to
a. 0.56
b. 0.58
c. 0.60
d. 0.62
60. The concentration of the solution adhering to the extracted solids is approximately equal
to
a. 0.193
b. 0.218
c. 0.021
d. 0.118
61. The mass of the solution leaving with the extracted meal is approximately equal to
a. 507 kg/h
b. 306 kg/h
c. 418 kg/h
d. 621 kg/h
62. The mass of the extract is approximately equal to
a. 583 kg/h
b. 512 kg/h
c. 536 kg/h
d. 571 kg/h
63. The number of stages required is
a. 3
b. 4
c. 6
d. 7
Given:
Final Overflow, Vi
V2,
1
y2
VN,
2
yN
Solvent, VN+1
N
10 kg oil
655 kg benzene
Feed, F = 1000 kg meal/hr
L1,
L2,
400 kg oil
x1
x2
LN
60 kg unextracted oil
25 kg benzene
575 kg solid
Solution:
In the feed: F = 1000 kg meal/hr
10
bVn+1 = 665 = 0.015
Solute: 400 kg oil
Solvent: 25 kg benzene
Inert Solid: 1000 – (400+25) = 575 kg
Solution: 400 + 25 = 425 kg/h solution
400
af = 425 = 0.941
By trial and error,
Assume aVn+1 = 0.1,
𝑘𝑔
from table, Solution in Ln = 0.505 𝑘𝑔 𝑠𝑜𝑙𝑖𝑑
LN = 0.505 (1000) = 505 kg/hr
60
In the Solvent: VN+1 = 10+655 = 665 kg
Solute: 10 kg oil
avn+1 = 505 = 0.119
@ avn+1 = 0.119,
𝑘𝑔
Solvent: 655 kg benzene
from table, Solution Ln = 0.507 𝑘𝑔 𝑠𝑜𝑙𝑖𝑑
Ln = 0.507(1000) = 507 kg/h
60
In the Final Underflow: LN
avn+1 =
Solute: 60 kg unextracted oil
@ Final Underflow, Ln:
Benzene: Ln – 60
Benzene: 507 – 60 = 447 kg/h
507
= 0.118
@ Final Overflow
Let:
a = mass fraction of oil in final underflow
OMB Solute: Feed + Solvent = Final
(Underflow + Overflow)
b = mass fraction of oil in final overflow
Oil: 400 + 10 = 60 + Final Overflow
Final Overflow Solute: 350 kg/h
OMB Solvent: Feed + Solvent = Final
(Underflow + Overflow)
Benzene: 25 + 655 = 447 + Final Overflow
Final Overflow Solvent: 233 kg/h
Vi = 350 + 233 = 583 kg/h extracted
350
b = 583 = 0.60
At equilibrium:
a = bvi = 0.60,
𝑘𝑔
from table, Solution = 0.595 𝑘𝑔 𝑠𝑜𝑙𝑖𝑑
At stage 1:
MB: Feed + V2 = V1 + L1
425 + V2 = 583 + 595
V2 = 953 kg
Oil Balance:
595 (0.60) + 583 (0.6) = 425 (0.) + 753Y2
y2 = 0.408
N=1+
0.015−0.118
]
0.408−0.6
0.015−0.408
ln[
]
0.118−0.6
ln[
= 4.05 = 4
64. An oil-sand mixture that is 25% (by mass) oil and 75% (by mass) sand is to be extracted or
leached with 75 tons/day of naphtha in a countercurrent extractor. The feed consists of 100
tons/day of mixture. The final extract (overflow) produced contains 35% (by mass) oil and 65%
(by mass) naphtha, and the underflow from each unit consists of 32% (by mass) oil and 68% (by
mass) sand. The overall efficiency of the extraction is 80% (by mass). Assume the solvent is
miscible with the oil in all portions and the extractor has reached equilibrium conditions in each
stage. Assume there is no sand in the overflow. The number stages required to effect the desired
separation of oil from sand is
a. 3
c. 5
b. 4
d. 6
Given:
75 tons/day
of naphtha
OverFlow
Yoil = 0.35
Ynaphtha = 0.65
Feed, F
100 tons/day
Xsolution = 0.32
Xoil =0.25
Xsand = 0.68
Xsand = 0.75
Overall efficiency = 80%
Required:
Number of stages
Solution:
Assume: 1 day
*in the feed*
(100
𝑡𝑜𝑛𝑠
𝑑𝑎𝑦
of mixture)(1 day) = 100 tons of mixture
*amount of raffinate*
75 𝑡𝑜𝑛𝑠 𝑜𝑓 𝑠𝑎𝑛𝑑
0.68
= 110.29
Sand = 100(0.75) = 75
Oil = 100(0.25) = 25
*OMB*
Feed + naphtha = raffinate + extract
100 + 75 = 110.29 + extract
Extract = 64.71
*Naphtha balance*
Amount of naphtha entering = amount of naphtha extract + amount of naphtha raffinate
Let X = mass fraction of naphtha in raffinate
75 = (64.71) (0.65) + (110.29) (X)
X= 0.2986
Mass fraction of oil in raffinate = 1- 0.2986 – 0.68 = 0.0214
The further solution will be subjected to a graphical method
65. A copper ore containing 10.3% by mass copper sulfate, 85.4% by mass inert and 4.3 % by
mass water is to be extracted with pure water in a counter current extractor. The daily feed
consist of 281 tons. The final extract produced contains 10% by mass copper sulfate and 90% by
mass water. The underflow from each stage consist of 66.7% by mass solution and 33.3% by
mass inert. The process is to recover 92% of the copper sulfate from the ore. Assume the
extractor has reached equilibrium conditions in each stage the minimum number of stages
required to effect the desired separation of copper sulfate from the inert.
Given:
Overflow
10% CuSO4, 90% water
Solvent
%recovery = 92
Feed
281 tons
10.3 % CuSO4
solution
85.4 % inert
4.3 % water
Solution:
Basis: 281 tons feed
.103(281) = 28.943 kg CuSO4
.854(281) = 239.974 kg inert
.043(2810 = 12.083 kg water
Inert in feed = inert in underflow
Amount of solution in underflow
239.974
.333
(. 667) = 480.6686 tons solution
Amount of overflow
0.92(28.943)
0.10
= 266.2756
66.7 %
33.3 % inerts
%solute in underflow
28.943−(28.943 𝑥 0.92)
480.6636
= 4.817 𝑥 10−3
Solvent balance
12.083 + solvent stream = 0.90(266.2756) + 480.6636 – 2.31544
Solvent stream =705.9182
At stage 1
266.2756 tons solution
28.943 tonsCuSO4
12.083 tons water
705.9182 tons solution
480.6686 tons solution
At equilibrium
𝑠𝑜𝑙𝑢𝑡𝑒
𝑠𝑜𝑙𝑢𝑡𝑒
(𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)𝑜𝑣𝑒𝑟𝑓𝑙𝑜𝑤 = (𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)𝑢𝑛𝑑𝑒𝑟𝑓𝑙𝑜𝑤 = 0.10
Solute balance; let x be fraction of solute at solvent stream
.10(266.2756) + .10(480.6686) = 28.943 + 705.9182x
X = 0.0648
Number of stages
=1+
0−4.817𝑥10−3
]
0.0648− .10
0−0.0648
ln[
]
4.817𝑥10−3 −.10
ln[
= 6.1727 = 7 stages
1. 60 tons per day oil sand (25 wt% oil and 75 wt % sand) is to be extracted with 40 tons per
day of naphthalene in a counter current extraction battery. The final extract from the
battery is to contain 40 wt% oil and 60 wt% naphthalene and the underflow from each
unit is expected to consists of 35 wt% solution and 65 wt% sand. If the overall efficiency
of the battery is 50%, how many stages will be required?
GIVEN:
Final Vo
Vn+1
F
Final Ln
Final Vo ( Final Overflow)
Final Ln ( Final Underflow)
X oil= 0.40
X naphthalene+ X oil= 0.35
X naphthalene= 0.60
X sand= 0.65
Feed= 60 tons/day
Vn+1= 40 tons/day
X oil= 0.25
X sand= 0.75
Required:
N (Number of Stages) =?
X naphthalene= 1
Detailed Solution:
Let A= Oil (Solute)
B= Sand (Insoluble Solid)
C= Naphthalene (Solvent)
In the Feed
F= 60 tons/day
A= 0.25(60) = 15 tons/day
B= 0.75(60) = 45 tons/day
Overall Insoluble Solid Balance:
(B)FEED=(B)UNDERFLOW
(B)UNDERFLOW= 45 tons/day
In the underflow
(B)UNDERFLOW= 45 tons/day=0.65 (underflow)
Underflow= 69.53 tons/day
Ln=(69.23)(0.35)= 24.23 tons/day
Liquid Balance: ( Solute+Solvent)
15+40=24.23+Vo
Vo= 30.67 tons/day
Solvent Balance:
40= (C)UNDERFLOW + (0.60)(30.67)
(C)UNDERFLOW= 21.598 tons/day
In the Underflow:
Xc = ((C)UNDERFLOW/ Ln)= 21.598/24.23= 0.89
Xn= Xa= 1-0.89= 0.11
No. of Stages:
N THEO= 1+ ln( (Yn+1- XN)/ (Y2-X1))/ln(((Yn+1- Y2)/ (XN-X1)))
Balance at Stage 1:
0.25(60)+ Y2(40)= 0.40( 30.77)+0.40(24.23)
Y2= 0.175
Substitute:
Yn+1= 0
XN= 0.11
Y2= 0.175
X1= 0.40
N THEO = 1+ ln( (Yn+1- XN)/ (Y2-X1))/ln(((Yn+1- Y2)/ (XN-X1)))
N THEO= 2.42 stages
N ACTUAL= N THEO/EFFECIENCY= 2.42/ 0.50
N ACTUAL= 4.84= 5 STAGES
(Principles of Mass Transfer and Separation Processes by Binay K. Dutta)
A solid feed containing 22% of solute, 3% water and 75% inerts (insoluble) is to be leached a
rate of 1 ton per hour with water in a countercurrent leaching cascade. The strong leachate
leaving the unit should have 16% of the solute in it. Desired recovery of the solute in the feed is
99%. The overflow does not have any entrained inert in it, and the amount of solution retained in
the sludge is 0.45 kg solution per kg inert. Analytically determine the number of stages required
for the separation.
Given:
Final V1
Vn+1
F
Final Ln
Solution:
Basis: 1 hour operation
1 ton = 1000 kg
In Feed
Solute: 1000(0.22) = 220 kg
Water: 1000(0.03) = 30 kg
Inert: 1000(0.75) = 750 kg
In Underflow:
0.45 𝑘𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 750 𝑘𝑔 𝑖𝑛𝑒𝑟𝑡 = 337.5 𝑘𝑔
𝑘𝑔 𝑖𝑛𝑒𝑟𝑡
LN = 337.5 kg solution
Mass of solute leaving with the sludge (99% recovery) = (220)(0.01) = 2.2 kg
Solute = 2.2 kg
Solvent = 335.3 kg
𝑥𝑁 =
2.2
= 0.00652
337.5
In Overflow:
Solvent balance:
Solvent in
F + VN+1 = LN + V1
30 + VN+1 = 335.3 + V1
V1 = VN+1 – 305.3
Solute balance :
Solute in
F + VN+1 = LN + V1
220 + 0 = 2.2 + (VN+1 – 305.3) (0.16/0.84)
VN+1 = 1448.75 kg
V1 = 1143.45 kg
Solute in V1 = 182.95 kg
Solvent in V1 = 960.5 kg
Solute Balance at Stage 1:
VN+1 = 1448.75 = V2
X1 = Y1 = 0.16
YN+1 = 0 Pure solvent
220 + V2y2 = L1x1 + 182.95
220 + 1448.75y2 = 337.5 (0.16) + 182.95
Y2 = 0.0117
𝑥𝑁 = 0.00652
Using the equation:
𝑦
− 𝑥𝑁
ln( 𝑁+1
𝑦2 − 𝑥1 )
𝑁−1=
𝑦
− 𝑦2
ln( 𝑁+1
𝑥𝑁 − 𝑥1 )
𝑁−1=
0 − 0.00652
ln(0.0117 − 0.16)
ln(
N= 2.2
N= 3 STAGES
N = 3 stages
0 − 0.0117
)
0.00652 − 0.16