11-Lect-Pressure Drop in Reactors0

11 lect pressure drop in reactors0

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Unformatted text preview: Outline: 1.  Pressure drop in reactors (Fogler 4.5) Next Time: (Fogler 4.6-4.8) 1.  Synthesizing the Design of a Chemical Plant (Fogler 4.6) 2.  Mole Balances (Fogler 4.7) 3.  Microreactors (Fogler 4.8) Fj0 Fj Pressure Drop in Reactors Chemical Reaction Engineering - Musgrave Lecture 11 - Page 1 Ethylene and oxygen are fed in stoichiometric proportions to a PBR operated isothermally at 260 C. Ethylene is fed at 0.30 lb mol/s at a pressure of 10 atm. 10 banks with 100 tubes/bank of 1.5 in diameter schedule 40 tubes packed with 120 lb/ft3 of 1/4 in diameter catalyst are to be used. k is 0.0141 (lb mol/atm•lbcat•hr) and the rate law is: 1 −rA + = kPA / 3 PB2 / 3 C2H4 + 1/2O2 → C2H4O A+ 1/2B → C F0=FC2H4,0 € FC2H4O FA0 dW FA This is similar to the reactor design problems we’ve been solving, except that we now have significant P drop in the reactor. We can still apply the CRE algorithm, but we need to take into account how changing P along the reactor affects X. So, the first step is to determine how P changes along the tube… Pressure Drop in Reactors Chemical Reaction Engineering - Musgrave Lecture 11 - Page 2 1 For liquid phase reactions, P has little effect on rA. However, for gas phase reactions carried out in PFRs and PBRs pressure both: • affects rA • can change significantly Remember from our gas phase stoichiometry table that: (1 − X ) T0 P CA = CA 0 (1 + εX ) P0T ȹ b ȹ ȹ Θ B − X ȹ ȹ a Ⱥ T0 P CB = CA 0 (1 + εX ) P0T So, evaluation of the combine step of the CRE algorithm requires evaluating P(X). € (1 − X ) y CA = CA 0 (1 + εX ) € ȹ b ȹ ȹ Θ B − X ȹ ȹ a Ⱥ CB = CA 0 y (1 + εX ) Where we’ve defined P/P0 as y and used the isothermal condition. € Pressure Drop in Reactors How does P vary along the reactor length? € Chemical Reaction Engineering - Musgrave Lecture 11 - Page 3 1. Mole Balance PBR CSTR Batch 1. Mole Balance dW = 2. Rate Law 2. Rate Law -rA=f(CA) -rA=g(CA) -rA=h(CA) FA 0 dX −rA +...
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