Chapter 15 Adsorption Page 15-1
Chapter 15 Adsorption
1. Dextran adsorption of IgG
The basic equation is Eq. 15.2-9, q = q0y
K + y
2. Phenol adsorption
a) Using the given isotherm, we have:
023
023 6
.
.kg phenol kg phenol

The carbon required is:


66
5300 730 10 10 10
0 0375
kg phenol
kg solution kg solution kg carbon
kg phenol
.kg carbon


b) Since the adsorbent is now in equilibrium with feed solution, we have:
023
023 6
.
.kg phenol kg phenol

The carbon required is:
Chapter 15 Adsorption Page 15-2


66
5300 730 10 10 10
kg phenol
kg solution kg solution . kg carbon


3. Nitrogen drying with molecular sieves
0
500
1500
0 5 10 15 20 25
c, ppm
The total amount adsorbed is obtained from the
shaded area, thus by numerical integration, we
have
Thus the weight of the adsorbed water is
wH2O = 1.2 mol/h-cm2*0.0255*18 g/mol
The weight of molecular sieve per unit area is
(the depth of the bed is L = 4.2 cm)
Therefore the saturation capacity is
The breakthrough point is the time when c = 0.05c0 = 75 ppm, thus from the Figure,
The exhaust point is the time when c = c0, thus
From Eq. 12.2-5, the unused fraction is
This is too large. If the concentration reaches c0 at some point between 18 and 24 h, then the
unused fraction would be much smaller.
4. Color Removal
a) percent saturation
0
140 0.667
y
Chapter 15 Adsorption Page 15-3
c) Unused bed length is the same, and the new bed volume must be 1000 times greater than the
old:
12.9
dcm
d)
23
4cm cm cm
5. Adsorption scale-up
Assuming a linear breakthrough the amount of the bed used is:

4052
2
1
40
40
00
0
yy
y
The unused bed length is:
We are told that ℓ’ scales linearly with the HTU. We know from its definition that the HTU
scales linearly with the mass transfer coefficient. Therefore we have:
The new unused bed length will be:
Chapter 15 Adsorption Page 15-4
The usable volume of the lab-scale column is:
3
2
27.1361.2201
4
4cmcmcmcmd
The diameter of the full-scale column will be:
cmcm
8.10130
‘
6. Compressible adsorbents
The specific cake resistance is defined as:
p
p
vv


For the lab column, we have:

 

2
3
2
600
2 3600 /
20 1.0
4
cm
cm
hr s hr
cm
cm











Since the cake resistance scales with the 0.43 power of pressure drop, we have:
0.43
0.43
022
1000 /
80
0.2
kPa s kPa m kPa s
cm cm
m









Using the cake resistance equation again, we have:
Chapter 15 Adsorption Page 15-5
2
3
2
1000
/100 0.179 /
55.9
2000 1000
7 3600
4
23.8
kPa
pcm
vc
kPa s
cm
Lcmhr
Qhr L s
dcm

 

ms
The used bed volume must scale linearly with the volumetric flow rate to give the required
capacity. The unused bed length remains constant.

3
2 2
3
2000 1000
7
1.0 20 8 23.8 8
2
28.2
Lcm
hr L
cm cm cm cm
hr
cm



 


 


7. Making adsorption more productive
As shown in the figure on the right, the total
amount adsorbed in the old operation is
0
0.5
1
012345
y
New Old
As for the new operation, the process ends at
certain time ts, when the concentration is ys,
The correlation between y and t is (because the
slope becomes half of the old slope)
Chapter 15 Adsorption Page 15-6
8. Elution of vaccine
The is a reversible adsorption2.