ERT 313
BIOSEPARATION ENGINEERING
ADSORPTION
Prepared by:
Miss Hairul Nazirah Abdul Halim
Adsorption ≠ Absorption !
• Absorption – a fluid phase is transferred from one medium to
another
• Adsorption – certain components of a fluid (liquid or gas) phase
are transferred to and held at the surface of a solid (e.g. small
particles binding to a carbon bed to improve water quality)
• Adsorbent – the adsorbing phase (carbon, zeolite)
• Adsorbate – the material adsorbed at the surface of adsorbent
Application of Adsorption
• Used in many industrial processes:
– dehumidification
– odour/colour/taste removal
– gas pollutant removal (H2S)
– water softening and deionisation
– hydrocarbon fractionation
– pharmaceutical purification
Nature of Adsorbent
• Porous material - Large surface area per unit mass
- internal surface area greater than the external
surface area
- often 500 to 1000 m2/g.
• Separation occurs because differences in molecular weight,
shape or polarity of components
• Rate of mass transfer is dependent on the void fraction within
the pores
• Granular (50μm - 12 mm diameter)
• Suitable for packed bed use
• Activated carbon, silica gel, alumina, zeolites
Silica structure
Zeolite structure
Types of Adsorption
• Ion exchange
– Electrostatic attachment of ionic species to site of the opposite
charge at the surface of an adsorbent
• Physical Adsorption
– result of intermolecular forces causing preferential binding of
certain substances to certain adsorbents
– Van der Waal forces, London dispersion force
– reversible by addition of heat (via steam, hot inert gas, oven)
– Attachment to the outer layer of adsorbent material
• Chemisorption
– result of chemical interaction
– Irreversible, mainly found in catalysis
– change in the chemical form of adsorbate
Adsorption Equipment
• Fixed-bed adsorbers
• Gas-drying equipment
• Pressure-swing adsorption
Adsorption Isotherm
• Adsorption isotherm – equilibrium relationship between the
concentration in the fluid phase and the concentration in the
adsorbent particles.
Types of Isotherms
•
Number of types of isotherm
1. Linear - adsorption amount is proportional to the
concentration in the fluid
2. Langmuir (favourable)
W=Wmax [Kc/(1+Kc)]
Where:
W = adsorbate loading
c = the concentration in the fluid
K = the adsorption constant
3. Freundlich (strongly favourable) – high adsorption at low
fluid concentration
W=bcm
where b and m are constant
4. Irreversible – independent of concentration
FIGURE 25.3 Adsorption isotherms for water in air at 20 to 50 0C.
Principles of Adsorption
• Concentration profile in fixed beds
Figure 25.6(a)
Breakthrough Curves
• tb – time when the concentration reaches break point
• Break point – relative concentration c/co of 0.05 or 0.10
• Adsorption beyond the break point would rise rapidly to about
0.50
• Then, slowly approach 1.0
• t* is the ideal adsorption time for a vertical breakthrough curve
• t* is also the time when c/co reaches 0.50
• Amount of adsorbed is proportional to the rectangular area to the
left of the dashed line at t*
• Solute feed rate (FA) = superficial velocity (uo) X concentration (co)
Where:
Wo = initial adsorbate loading
Wsat = adsorbate at equilibrium with the fluid
L = length of the bed
ρb = bulk density of the bed
Length of Unused Bed (LUB)
•Determine the total solute adsorbed up to the break point by
integration

c
0 1  co dt
t
•The break point time, tb is calculated from the ideal time and
the fraction of bed utilized:
Tutorial 3
Example 25.2 (McCabe)
The adsorption of n-butanol from air was studied in a small fixed bed
(10.16 cm diameter) with 300 and 600 g carbon, corresponding
to bed lengths of 8 and 16 cm.
a)
b)
From the following data for effluent concentration, estimate the
saturation capacity of the carbon and the fraction of the bed used
at c/co = 0.05
Predict the break point time for a bed length of 32 cm
Data for n-butanol on Columbia JXC 4/6 carbon are as follows:
Solution
The total solute adsorbed is the area above the graph multiplied by
FA . For the 8-cm bed, the area is
How to solve this integration???
• Use numerical integration (5-point quadrature formula)
X4
h
X 0 f ( X )dX  3  f 0  4 f1  2 f 2  4 f 3  f 4 
X4  X0
h
4
•From the graph plotted, the following data is obtained:
t
c/co
f (X) = 1-c/co
0
0
1
2
0.027
0.973
4
0.29
0.71
6
0.78
0.22
8
0.99
0.01
80
h
2
4
t4
c
2


t0 1  c0 dt  3 1  4(0.973)  2(0.71)  4(0.22)  0.01  4.80 h
• The mass of carbon per unit cross-sectional area of bed is
= bed length x density of carbon
= 8 cm x 0.461g/cm3 = 3.69 g/cm2
How to solve this integration???
• Use numerical integration (Trapezoidal rule)
h
X 0 f ( X )dX  2  f ( X 0 )  f ( X1 )
h  X1  X 0
X1
•From the graph plotted, the following data is obtained:
t
c/co
f (X) = 1-c/co
0
0
1
2.4
0.05
0.95
h  2.4  0  2.4

c
2.4
t0 1  c0 dt  2 1  0.95  2.34 h
t1