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 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

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 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) 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

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) From the graph plotted, the following data is obtained: t c/co f (X) = 1-c/co 1 2 0.027 0.973 4 0.29 0.71 6 0.78 0.22 8 0.99 0.01

How to solve this integration??? 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) From the graph plotted, the following data is obtained: t c/co f (X) = 1-c/co 1 2.4 0.05 0.95