Suppose that you are operating at five times the minimum fluidization
velocity, uo = 5![]() |
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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 ![]() ![]() ![]() ![]() |
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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. |