sample problem #3
TRANSCRIPT
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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
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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.
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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) = ?
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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
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• 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
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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
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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
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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
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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
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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
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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
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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
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• 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:
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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
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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