Suppose that you are operating at five times the minimum fluidization velocity, uo = 5 What would be the effect of doubling the particle diameter on the catalyst weight for the same throughput and conversion? | |||
Solution | |||
Substitution for u0 into Equation (CD12-89) gives | |||
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(CDE12-7.1) (CDE12-7.2) |
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Since the temperature remains constant and there are no inter- and intraparticle resistances, kcat1 = kcat2 , the throughput (u01 = u02 ) and conversion (X1 =X 2 ) are the same for cases 1 and 2. The ratio of Equations (CDE12-7.1) and (CDE12-7.2) yields | |||
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(CDE12-7.3) | ||
Recalling Equation (CD12-25) gives | |||
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(CDE12-7.4) | ||
and neglecting the dependence of on dp, we see that the only parameters that vary between case 1 (dp ) and case 2 (dp2= 2dp1 ) are and W. Taking the ratio of 2to 1 and substituting for d p2 in terms of dp1 , we obtain | |||
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(CDE12-7.5) | ||
and therefore | |||
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(CDE12-7.6) | ||
Thus in the situation we have postulated, with a first-order reaction and reaction limiting the bed behavior, doubling the particle size will reduce the catalyst by approximately 75% and still maintain the same conversion. |