CENG 3210 Separation Processes
Tutorial 10 solution
The ternary equilibrium data of trichloroethane (C)-acetone (A)-water (B) at 25oC, 1 atm, are given
in the following table.
Water phase (mass fraction)
Trichloroethane phase (mass fraction)
Acetone
(xA)
Water
(xB)
Trichloroethane Acetone
(xC)
(yA)
Water
(yB)
Trichloroethane
(yC)
0
0.9956
0.0044
0
0.0011
0.9989
0.0596
0.9352
0.0052
0.0875
0.0032
0.9093
0.1000
0.894
0.0060
0.1500
0.0060
0.8440
0.1397
0.8535
0.0068
0.2078
0.0090
0.7832
0.1905
0.8016
0.0079
0.2766
0.0133
0.7101
0.2763
0.7133
0.0104
0.3939
0.0240
0.5821
0.3573
0.6267
0.0160
0.4821
0.0426
0.4753
0.4605
0.5020
0.0375
0.5740
0.0890
0.3370
0.58
0.22
0.2
0.58
0.22
0.2
(1) Plot the equilibrium curve on the right-angle triangle diagram.
(2) Draw the tie lines
(3) Construct the conjugate line.
In a single stage extractor, pure trichloroethane solvent is used to extract acetone from an acetonewater solution containing 54 kg acetone and 66 kg water. It is required to reduce the acetone
content in the raffinate to below 10% (mass %), calculate
(4) The amount of trichloroethane required.
(5) The mass in the extract phase and its acetone mass fraction
1
(1)
equilibrium curve
C
1.0
0.8
x C, y C
0.6
0.4
0.2
0.0
B 0.0
(2)
0.2
0.4
0.6
xA, yA
tie lines
2
0.8
1.0
A
(3)
conjugate line
(4)
In the feed, F = 54 + 66 = 120 kg,
xAF = 54/120 = 0.45, xBF = 66/120 = 0.55, xCF = 0.
This is point F in the figure. The entering solvent is
pure trichloroethane at the corner C. Draw line FC.
In the raffinate (x) phase, xA = 0.1. Locate point R at
the equilibrium boundary curve at xA = 0.1, xC = 0.006.
Draw a vertical line from R to intersect the conjugate
3
line at H, then draw a horizontal line from H to
intersect the boundary curve at E, which is the
composition of the extract phase, yA = 0.15, yC = 0.844.
Link line RE to intersect line FC at point M, which is
the composition of the mixture of extract and raffinate
phases, xAM = 0.14.
Let S be the solvent entering, yAS = 0 (pure solvent),
from the material balance, we have
F+S=M
FxAF + SyAS = MxAM = (F+S)xAM
120(0.45) + 0 = (120 + S)(0.14)
S = 266 kg
Or using the lever rule,
𝑆=
̅̅̅̅̅
𝑀𝐹
𝐹 ̅̅̅̅̅
𝑀𝐶
= 120
4
15.2
6.85
= 266 𝑘𝑔
C
1.0
E
H
0.8
M
x C, y C
0.6
Conjugate line
0.4
0.2
A
0.0
B
0.0
R
0.2
0.4 F
0.6
0.8
1.0
xA, yA
(5)
M = F + S = 120 + 266 = 386 kg
From part (4) we have known that acetone
concentrations are yA = 0.15 in the extract phase and
xA = 0.1 in the raffinate phase. Let E be the amount of
the extract phase and R the amount of the raffinate
phase. From the material balance, we have
E + R = M = 386 kg
EyA + RxA = MxAM
E(0.15) + R(0.1) = 386(0.14)
Solving to obtain
E = 77 kg
R = 309 kg
5