11-Lect-Pressure Drop in Reactors0

# Mole balance pbr cstr batch 1 mole balance dw 2 rate

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Unformatted text preview: ȹȺ ρ 0 gc Dp ȹ φ 3 ȺȺ Dp Ⱥ FA 0 dX 1/ 2 kCA 0 (1 − X )[1 − αW ] Separating variables: € € € [1 − αW ] 1/ 2 dW = FA 0 dX kCA 0 (1 − X ) Integrating gives: € 2 −F 32 (1 − αW ) = A 0 ln(1 − X ) α2 kCA 0 23 When ε = 0, we can solve the pressure drop DFQ by separation of variables, substitute y into the dW/dF DFQ and solve Solving for W or X gives: Ⱥ 1 1 Ⱥ−α 2 FA 0 W = − Ⱥ ln(1 − X )Ⱥ €α α Ⱥ 2kCA 0 Ⱥ Pressure Drop in Reactors Ⱥ 2kCA 0 3 2 Ⱥ X = 1 − expȺ 1 − αW ) Ⱥ 2( Ⱥ− FA 0α Ⱥ Lecture 11 - Page 10 Chemical Reaction Engineering - Musgrave € € 5 We can plot W vs X using: We can plot X vs W using: 23 W= W Ⱥ 1 1 Ⱥ−α 2 FA 0 − Ⱥ ln(1 − X )Ⱥ α α Ⱥ 2kCA 0 Ⱥ Ⱥ 2 kCA 0 3 2 Ⱥ X = 1 − expȺ− 1 − αW ) Ⱥ 2( Ⱥ FA 0α Ⱥ X € € X W Pressure Drop in Reactors Chemical Reaction Engineering - Musgrave Lecture 11 - Page 11 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 + 2 −rA + = kCA 3. Stoichiometry Flow Batch 3. Stoichiometry € Flow Gas € CA = FA υ Gas υ = υ 0 (1 + εX ) P0T PT0 Liquid Gas Liquid Because ε = 0 € CA = CA 0 (1 − X ) € 4. Combine dV=dV(X) dt=dt(X) P P0 5. Evaluate V=V(X) t=t(X) 4. Combine F dX dW = A 0 € −rA + dW = dW = FA 0 dX 2 kCA • 2nd order • Equimolar • Irreversible • Gas phase • PBR Pressure Drop in Reactors FA 0 dX € 5. Evaluate €Ⱥ 2 P Ⱥ kȺCA 0 (1 − X ) Ⱥ P0 Ⱥ Ⱥ Chemical Reaction Engineering - Musgrave € Lecture 11 - Page 12 6 5. Evaluate dW = FA 0 dX 2 Ⱥ P Ⱥ kȺCA 0 (1 − X ) Ⱥ P0 Ⱥ Ⱥ εX Negligible y= α≡ P 1/ 2 = [1 − αW ] P0 2β 0 1 − φ ) Ac ρ c P0 ( Because ε is zero we can use: y€ = Which gives: P 1/ 2 = [1 − αW ] P0 dW = α≡ 2β 0 (1 − φ ) Ac ρc P0 € β0 ≡ Ⱥ G ȹ 1 − φ ȹȺ150(1 − φ )µ + 1.75GȺ ȹ ȹȺ ρ 0 gc Dp ȹ φ 3 ȺȺ Dp Ⱥ Separating variables: € € € FA 0...
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## This note was uploaded on 04/25/2010 for the course CHEN 4330 taught by Professor Staff during the Spring '08 term at Colorado.

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