ADSORPTION

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ERT 313
BIOSEPARATION ENGINEERING
ADSORPTION
Prepared by:
Miss Hairul Nazirah Abdul Halim
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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 
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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 
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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
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Zeolite structure
Silica structure
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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
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Adsorption Equipment
Fixed-bed adsorbers
Gas-drying equipment
Pressure-swing adsorption
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Adsorption Isotherm
Adsorption isotherm – equilibrium relationship between the concentration in the fluid phase and the concentration in the adsorbent particles.
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Types of Isotherms
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Number of types of isotherm
1.	Linear - adsorption amount is proportional to the concentration in the fluid
Langmuir (favourable) 
 		W=Wmax [Kc/(1+Kc)]
Where:
	W = adsorbate loading
	c = the concentration in the fluid
	K = the adsorption constant
Freundlich (strongly favourable) – high adsorption at low fluid concentration	 
		W=bcm 	where b and m are constant
4.	Irreversible – independent of concentration
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FIGURE 25.3	Adsorption isotherms for water in air at 20 to 50 0C.
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Principles of Adsorption
Concentration profile in fixed beds
Figure 25.6(a)
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Breakthrough Curves
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 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*
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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
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Length of Unused Bed (LUB)
Determine the total solute adsorbed up to the break point by integration
The break point time, tb is calculated from the ideal time and the fraction of bed utilized:
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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)	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
b)	Predict the break point time for a bed length of 32 cm
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Data for n-butanol on Columbia JXC 4/6 carbon are as follows:
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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???
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 Use numerical integration (5-point quadrature formula)
From the graph plotted, the following data is obtained:
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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???
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 Use numerical integration (Trapezoidal rule)
From the graph plotted, the following data is obtained:
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