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In the regeneration of catalyst, the SR of catalytic reforming unit fixed bed reactor.
Could any one tell me why we keep the O2 between 0.3-0.8mol % in the carbon burn step. Also, in the oxidation step should the O2 be kept at10mol%? Another thing, what will happen to the catalyst if the sulfur in the feed is more than 0.5ppm?
13/12/2017 A: Muhammad Akhtar, Orpic, mbadgk@gmail.com
0.3% to 0.8% is kept to have controlled burning. Since the coke level on the catalyst is high and thermal runaway can happen if you give more oxygen to the system during coke burning step.
Basically it is not Oxidation Step, it is Oxychlorination Step. After coke burning the metal on the catalyst become sintered (i.e. the dispersion of metal is effected). So in order to re-disperse the metal on catalyst we need high oxygen concentration in the presence of chlorinating agent. This step recovers the chloride content of the catalyst as well as disperse the metal on the catalyst to improve its metal function.
Sulfur is the temporary poison for the catalyst. higher the sulphur lesser will be the activity of the catalyst. In SR type reformer, you might have to shutdown the reactor for regeneration if it is too much contaminated with sulphur.