1. PROCESS EQUIPMENT DESIGN
GAS ADSORPTION COLUMN

2. Problem Statement Partial pressure NO2, mm Hg 2 4 6 8 10 12
Nitrogen dioxide is produced by a thermal process is to be removed from a dilute mixture with air by adsorption on silica gel in a continuous counter current adsorber. The gas flow rate is 1000 lb/m2hr and contains 1.5% NO2 by volume. It is desired to remove 90% of the NO2. Experimental data for equilibrium adsorption isotherm at 25 °C and 1 atm is given
Partial pressure NO2, mm Hg 2 4 6 8 10 12
Kg NO2/ 100 kg gel
0.4
0.9
1.65
2.6
3.65
4.85

3. The silica gel has an average particle size of 0
The silica gel has an average particle size of inches in diameter and external surface of particle is sq ft/lb. the individual resistance to mass transfer in fluid and within solid during adsorption of water vapor from air by silica gel are
KyaP = 188G0.55 lbH20/ hr cu ft
KsaP = 217 lbH20/ hr cu ft where G is the mass velocity of gas.

4. Adsorption
Process that occurs when a gas or liquid solute accumulates on the surface of a solid or more rarely, a liquid (adsorbent), forming a molecular or atomic film (the adsorbate). The term sorption encompasses both processes, while desorption is the reverse process.
Silica gel is a highly porous solid adsorbent material. It has a very large internal surface composed of myriad microscopic cavities and a vast system of capillary channels.

5. Conveying of Silica Gel
Silica Gel is characterized as a B28 material
Possible methods
Belt conveyor: Costlier
Bucket elevator
Screw Conveyor
Screw dia: 12 in.
Approximate area covered by materials: 15%
Costing depends on correlation of manufacture

6. Regeneration of Adsorbent
Thermal
Pressure based
Using Inert gas
Steam condensation

7. Brief Theory Solute balance about the entire tower:
Gs (Y1-Y2) = Ss(X1-X2)
And the upper part
Gs(Y – Y2) = Ss(X-X2)
Where:
Gs = NO2 free air flow rate
Ss = NO2 free solid flow rate
X = Kg NO2/Kg silica Gel
Y= Kg NO2 / Kg NO2 free air

8. Solution Procedure
Fix (X2, Y2) by the specifications of the problem and the ordinate Y1
Draw a tangent through (X2, Y2) to the equilibrium curve. This gives the minimum adsorbent rate required
We operate at around times the minimum rate
We then assume, for ease of calculations, the equilibrium curve to be a straight line for the range of values that are encountered in the problem.
Calculate NtOG and HtOG and hence Z

10. Resistances to mass transfer
Gas phase resistance
Liquid phase resistance = 217 lbNO2/ hr cu ft

11. Evaluation of height of tower (Z)

12. Results Minimum solid to gas ratio = 0.76
Minimum adsorbent required = Kg/m2.sec
Actual adsorbent rate = 1.52 Kg/m2.sec
NtOG = 4.6
HtOG= 2.54 m
Z = 11.6 m

13. References
Mass Transfer Operations, by Robert E. Treybal, Third edition, McGraw Hill Book Company.
Chemical Engineers Handbook, Perry, R.H., and C.H. Chilt5th Ed. McGraw-Hill, NY, USA
Analysis of the Adsorption Process and of Desiccant Cooling Systems -A Pseudo-steady state Model for Coupled Heat and Mass Transfer, Robert S Barlow,1982
Gas purification processes, Nonhebel G, George Newnes Limited, London

14. Thank You