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10-Lect-Design Tubular0.Reactor

10-Lect-Design Tubular0.Reactor - Outline 1 Design of...

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1 Chemical Reaction Engineering - Musgrave Isothermal Tubular Reactor Design Lecture 10 - Page 1 Outline: 1. Design of tubular reactors (Fogler 4.4) Next Time: (Fogler 4.5) 1. Pressure drop in reactors (Fogler 4.5) F j0 F j Chemical Reaction Engineering - Musgrave Isothermal Tubular Reactor Design Lecture 10 - Page 2 What PFR volume is necessary to produce 1 billion kg of ethylene per year by cracking a pure stream of ethane gas? Given: The reaction C 2 H 6 C 2 H 4 + H 2 follows an elementary rate law The reactor is isothermal at T=1050K Pressure is nearly constant at P=5 atm The specific rate (k A ) is 3.07 sec -1 at 1100K and E A =82 kcal/mol. Our target is 75% conversion Let’s first apply the CRE algorithm to both liquid and gas reactions carried out in isothermal tubular reactors with negligible pressure drop that follow either 1 st or 2 nd order rate laws. We will then use the appropriate resulting expression to solve this problem… F 0 =F ethane,0 F Ethane
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2 Chemical Reaction Engineering - Musgrave Isothermal Tubular Reactor Design Lecture 10 - Page 3 For a CSTR , none of the conditions vary within the reactor: X, C A , P, T, -r A , υ are all constant within the reactor volume. For a tubular reactor (PFR or PBR) all of these variables can vary along the length of the reactor. Varying : X - always, unless there is no reaction. C A - usually, although balances in X and V could lead to constant C A P - pressure drop, usually due to frictional losses T - often, if the reaction is not thermal neutral and T not controlled. υ - usually due to changes in T, P and number of moles of gas F j0 F j F j0 F j V V=0 V 1 Chemical Reaction Engineering - Musgrave Isothermal Tubular Reactor Design Lecture 10 - Page 4 PFR 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 dV = F A 0 dX r A V = F A 0 X r A dt = N A 0 dX r A V r A = k A C A r A = kC A 1 + k A C A r A = k C A 1 k C A C B Ρਟ Σਿ ΢ਯ Τ੏ Φ੯ Υ੟ C A = F A υ C A = N A V υ = υ 0 υ = υ 0 1 + ε X ( ) P 0 T PT 0 V = V 0 1 + ε X ( ) P 0 T PT 0 V = V 0 C A = C A 0 1 X ( ) C A = C A 0 1 X ( ) 1 + ε X ( ) T 0 P P 0 T C A = C A 0 1 X ( ) dV = F A 0 dX r A dV = F A 0 dX k A C A dV = F A 0 1 + ε X ( ) P 0 TdX k A C A 0 1 X ( ) T 0 P V = P 0 F A 0 T 0 C A 0 T 1 + ε X ( ) dX Pk A 1 X ( ) 0 X = ?
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