PROCESS EQUIPMENT DESIGN GAS ADSORPTION COLUMN

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

The silica gel has an average particle size of 0 The silica gel has an average particle size of 0.068 inches in diameter and external surface of particle is 10.58 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.

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.

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

Regeneration of Adsorbent Thermal Pressure based Using Inert gas Steam condensation

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

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 1-1.5 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

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

Evaluation of height of tower (Z)

Results Minimum solid to gas ratio = 0.76 Minimum adsorbent required = 1.009 Kg/m2.sec Actual adsorbent rate = 1.52 Kg/m2.sec NtOG = 4.6 HtOG= 2.54 m Z = 11.6 m

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

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