Chapter 5: Isothermal Reactor Design: Conversion
Chapter 5 Example
Batch Reactor Optimization
| The following irreversible liquid phase reaction follows an elementary rate law | ||
| A+B → C | ||
| Determine the minimum number of batch reactors 1.0 cubic meter in size to produce 10,000 moles of C in a 300 day period. The processing time to fill, empty and clean the reactor between batches is 4.5 hours. The feed is equal molar in A and B and the initial concentration of A is 1.0 moles per cubic meter. In a trial run it was found that 50% conversion was achieved in 2 hours. | ||
Solution |
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| Part 1. Find X as a function of t and then find k. | ||
| Batch | ||
| A + B → C | Gas Phase |
Equil Molar Feed |
| Mole Balance |  |
(1) |
| Rate Law |  |
(2) |
| Stoichiometry | Gas, but  |
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(3) |
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(4) |
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| The number of moles C formed in a batch is | ||
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(5) |
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| Combine |  |
(6) |
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(7) |
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(8) |
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| Integrating | ||
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(9) |
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| Evaluate k from information given | ||
| Evaluate | ||
| At t = 2 hr then X = 0.5 | ||
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| The conversion for a reaction time tR | ||
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| Part 2. Find the minimum number of reactors | ||
| The total time for carry out one batch t is the reaction time tR plus the processing time tP which is the sum of the times to empty, te, to clean, tclean, and to fill, tf | ||
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| The total processing time, tP, between reactions is tP, = 4.5 hrs. How many 1 m3 reactors do you need to make 10,000 moles of C in 300 days? Consider a single reactor. The number of moles of C formed in one batch | ||
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| Total number of C moles made in one reactor for a 24 hour work day over 300 days is | ||
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n = number of batches in 300 days |
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n = 300 days x no. of batch per day |
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n  |
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| Rearranging | ||
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| For max production rate | ||
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| Solving for tR | ||
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| The number of moles 1 reactor can make in 300 days is | ||
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| The number of reactors, m, to make 10,000 moles of C in 300 days is | ||
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