Sample Problem #3:
An aqueous solution containing a valuable solute is colored by small amounts of an impurity. Before crystallization, the impurity is to be removed by adsorption on a decolorizing carbon, which adsorbs only insignificant amounts of the principal solute. A series of laboratory tests was made by stirring various amounts of the adsorbent into batches of the original solution until equilibrium was established, yielding the following data at constant temperature:
kg carbon/kg kg carbon/kg solutionsolution
00 0.0010.001 0.0040.004 0.0080.008 0.020.02 0.040.04
Equilibrium colorEquilibrium color 9.69.6 8.68.6 6.36.3 4.34.3 1.71.7 0.70.7
The color intensity was measured on an arbitrary scale, proportional to the concentration of the colored substance. It is desired to reduce the color to 10% of its original value, 9.6. Determine the quantity of fresh carbon required per 1000 kg of solution for a single-stage operation, for a two-stage crosscurrent process using the minimum total amount of carbon, and for a two-stage countercurrent operation.
A. SINGLE STAGE ADSORPTION
Adsorber Solution with impurity Lean solution
Spent Carbon
Co = 9.6 S = 1000 kg
C = 0.1(9.6) = 0.96
M = ?qo = 0
q (per 1000 kg of solution) = ?
SOLUTION
kg carbon/ kg sol’n
C* = equilibrium color,
Units/kg sol’n
Q = adsorbate concentration,
units/kg carbon
0 9.6
0.001 8.6
0.004 6.3
0.008 4.3
0.02 1.7
0.04 0.7
(9.6-8.6)/0.001 = 1000
(8.6-6.3)/0.001 = 825
(6.3-4.3)/0.001 = 663
(4.3-1.7)/0.001 = 395
(1.7- 0.7)/0.001 = 223
• Using Material Balance
S(Co - C) = M(q - qo)
1000(9.6 – 0.96) = M(q – 0)8640 = Mq -------------- eqn 1
• Using the equilibrium data given, the Freundlich equation applies for the system
c q log c log q
9.6 0
8.6 1000 0.934498 3
6.3 825 0.799341 2.916454
4.3 663 0.633468 2.821514
1.7 395 0.230449 2.596597
0.7 223 -0.1549 2.348305
y = 0.60x + 2.4634
R 2 = 0.9996
2.55
2.65
2.75
2.85
2.95
3.05
0 0.2 0.4 0.6 0.8 1
log c
log
q• Graphical Representation
From the equation of the line of the equilibrium data
Slope is 0.60 at empirical constant (n)
• Using Freundlich Equation:nKcq
nc
qK
(0.60)4.3
663K
in order to find K, we use the equation;
thus; = 276.3
Therefore , the resulting Freundlich isotherm is: (0.60)276.3c q
to find q at the spent adsorbent stream, use co equal to 0.96
(0.60) )276.3(0.96 q = 269.61
To solve for M, use equation 1 from the material balance
8640 = Mq
8640 = M (269.61)
Therefore
M = 32.04 kg Carbon
where q from the solved data is 269.61
B. Two-stage crosscurrent process using the minimum total amount of carbon
M1 = ?qo = 0
M2 = ?qo = 0
q1 = ? q2 = ?
Co = 9.6 S = 1000
kg
Lean solutionAdsorber Adsorber
C=0.1(9.6) = 0.96
Since the Freundlich equation applies, use Fig.11.19
0.19.6
0.96
Y
Y
0
2
0.1Y
Y
0
2
0.275C
C
0
1
From Fig. 11.19 w/ n=0.6 and
therefore 2.64 C1
• Using Freundlich equation: nKcq
0.6(2.64)*(277.06) q = 496.060.6 (0.96)*(277.06) q = 270.38
From the Material Balance: S(Co - C) = M(q - qo)
21T MMM 14.03kgM1
Solving for M2
(270.38)M10000.962.64 2
6.21kgM2
Solving for the Total Mass:
21T M M M
kg 6.21 kg 14.03 MT
solution kg 1000
carbon kg 20.24 MT
C. Two-stage countercurrent operation
M1 = ?qo = 0
M2 = ?qo = 0
q1 = ? q2 = ?Co = 9.6 S = 1000 kg
1 2
C = 0.1(9.6) = 0.96
• Using the Mc Cabe Thiele Method, locate the operating line and draw the equilibrium curve given the following data:
C0 = 9.6C = 0.96QN+1 = 0
Note: The operating line is located by Trial and error until two stages can be drawn between operating line and equilibrium curve:
Initial color, 0.96
Final color, 0.96
0 200 400 600 800 1000
q
10
9
8
7
6
5
4
3
2
1
0
c
Operating line Equilibrium
Curve
Graphical representation
1
2
From the Material Balance:
S(Co - C) = M(q - qo)
We could solve for the Mass of Carbon present
0675 M0.969.61000
tion1000kgsolu
carbon kg12.80M
