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Multiple Reaction Conversion Notes

# 6 06 04 04 02 02 0 0 0 0 2 4 2 4 6 6 titimeaau me

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Unformatted text preview: [aa.u.] ChE ChE 400 - Reactive Process Engineering Series Rctns - Maximum Yield A - PFR A - PFR B - PFR B - PFR C - PFR C - PFR A - CS TR A - CS TR B - CS TR B - CS TR C - CS TR C - CS TR 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0 0 2 4 2 4 6 6 titime[aa.u.] me [ .u.] PFR: CSTR: (log mean rate constant)-1 (geometric mean rate constant)-1 What if C and not B is the desired product? 8 8 10 10 ChE ChE 400 - Reactive Process Engineering Example: L11-17 L11- We wish to produce a product B from a reactant A in a PFR with V = 4 l/min and CA0 = 2 mol/l. However, another reaction is also occurring, forming an undesired product C. (Both reactions are irreversible, 1st order, with kB= 0.5 min-1 and kC = 0.1 min-1). (a) Assuming a series reaction A -> B -> C, calculate the maximum achievable yield of B, as well as the necessary reactor volume. (b) Assuming parallel reactions A -> B and A -> C, calculate the reactor volume necessary to achieve the same conversion of A as in (a). What is the yield of B in this case? Procedure: (a) Calculate τopt, from there CB(τopt) and CA(τopt). From these you obtain SB,max and XA,max. (b) Calculate τ(XA=0.865), from there: V = 13.36 l. With τ from the equation for CB in a PFR/series reactions… Check it in LDS, examples 4-2 and 4-3, incl. the CSTR case! Check it in LDS, examples 4-2 and 4-3, incl. the CSTR case!...
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