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11-Lect-Pressure Drop in Reactors0

11-Lect-Pressure Drop in Reactors0 - Outline 1 Pressure...

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1 Chemical Reaction Engineering - Musgrave Pressure Drop in Reactors Lecture 11 - Page 1 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) F j0 F j Chemical Reaction Engineering - Musgrave Pressure Drop in Reactors Lecture 11 - Page 2 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/ft 3 of 1/4 in diameter catalyst are to be used. k is 0.0141 (lb mol/atm•lb cat •hr) and the rate law is: 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 F 0 =F C2H4,0 F C2H4O r A ʹஒ = kP A 1/ 3 P B 2/ 3 C 2 H 4 + 1/2O 2 C 2 H 4 O A+ 1/2B C F A0 F A dW
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2 Chemical Reaction Engineering - Musgrave Pressure Drop in Reactors Lecture 11 - Page 3 For liquid phase reactions, P has little effect on r A . However, for gas phase reactions carried out in PFRs and PBRs pressure both: • affects r A • can change significantly C A = C A 0 1 X ( ) 1 + ε X ( ) T 0 P P 0 T C B = C A 0 Θ B b a X Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 1 + ε X ( ) T 0 P P 0 T Remember from our gas phase stoichiometry table that: So, evaluation of the combine step of the CRE algorithm requires evaluating P(X). C A = C A 0 1 X ( ) 1 + ε X ( ) y C B = C A 0 Θ B b a X Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 1 + ε X ( ) y Where we’ve defined P/P 0 as y and used the isothermal condition. How does P vary along the reactor length? Chemical Reaction Engineering - Musgrave Pressure Drop in Reactors Lecture 11 - Page 4 PBR CSTR Batch 1. Mole Balance 2. Rate Law -r A =f(C A ) -r A =g(C A ) -r A =h(C A ) 3. Stoichiometry Flow Batch 4. Combine 5. Evaluate dV=dV(X) dt=dt(X) V=V(X) t=t(X) Liquid Gas Liquid Gas dW = F A 0 dX r A ʹஒ C A = F A υ υ = υ 0 1 + ε X ( ) P 0 T PT 0 1. Mole Balance 2. Rate Law 3. Stoichiometry Flow r A ʹஒ = kP A 1/ 3 P B 2/ 3 = kRTC A 1/ 3 C B 2/ 3 Gas C A = C A 0 1 X ( ) 1 + ε X ( ) P P 0 C B = C A 0 Θ B b a X Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 1 + ε X ( ) P P 0 4. Combine 5. Evaluate dW = F A 0 dX r A ʹஒ dW = F A 0 dX kRT C A 0 1 X ( ) 1 + ε X ( ) P P 0 Ρਟ Σਿ ΢ਯ Τ੏ Φ੯ Υ੟ 1/ 3 C A 0 Θ B b a X Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 1 + ε X ( ) P P 0 Ρਟ Σਿ ΢ਯ ΢ਯ ΢ਯ ΢ਯ Τ੏ Φ੯ Υ੟ Υ੟ Υ੟ Υ੟ 2/ 3 dW = F A 0 dX kRTC A 1/ 3 C B 2/ 3 How does P vary along the reactor length?
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3 Chemical Reaction Engineering - Musgrave Pressure Drop in Reactors Lecture 11 - Page 5 dP dz = G ρ g c D p 1 φ φ 3 Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 150 1 φ ( ) μ D p + 1.75 G Ρਟ Σਿ ΢ਯ Τ੏ Φ੯ Υ੟ P = pressure, kPa φ = porosity g c = conversion factor (32.174 lbft/s) D p = particle diameter (ft) A mass transport analysis of the pressure drop of a gas through a
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