Transcript Lecture 6
Lecture 6 Kjemisk reaksjonsteknikk Chemical Reaction Engineering Review of previous lectures Pressure drop in fixed bed reactor PFR reactor design with pressure drop (ε=0) Department of Chemical Engineering 1 - 29/04/2020 Reaction Engineering Stoichiometry Rate Laws Department of Chemical Engineering 2 Mole Balance Isothermal reactor design These topics build upon one another 2 - 29/04/2020 Reactor Mole Balances in terms of conversion Reactor Differential Algebraic Integral X X Batch N A0 Department of Chemical Engineering 3 PFR 0 V CSTR FA 0 dX rA dV dX rA V t N A0 dX r AV dt t FA 0 X rA X V FA 0 0 dX rA X PBR FA 0 dX rA dW X W FA 0 0 dX rA W 3 - 29/04/2020 Concentration Flow System: Gas Phase Flow System: Department of Chemical Engineering 4 FA CA 0 1 X T P0 T0 P FA 0 1 X C A 0 1 X T0 P FA CA P T 1 X T P0 0 0 1 X T0 P b b FA 0 B X C A 0 B X FB a a T0 P CB P T 1 X 1 X T P0 0 0 T0 P 4 - 29/04/2020 Pressure Drop in Packed Bed Reactors Note: Pressure drop does NOT affect liquid phase reactions Sample Question: Department of Chemical Engineering 5 Analyze the following second order gas phase reaction that occurs isothermally in a PBR: 2AB Mole Balance: Must use the differential form of the mole balance to separate variables: dX FA 0 rA dW Rate Law: 2 r kC Second order in A and irreversible: A A 5 - 29/04/2020 Pressure Drop in Packed Bed Reactors 1 X P T0 CA CA 0 1 X P0 T FA Stoichiometry: 1 X P CA CA 0 1 X P0 Isothermal, T=T0 Department of Chemical Engineering 6 Combine: 2 A0 dX kC dW FA 0 1 X P 1 X 2 P0 2 2 Need to find (P/P0) as a function of W (or V if you have a PFR) 6 - 29/04/2020 Pressure Drop in Packed Bed Reactors Ergun Equation: Department of Chemical Engineering 7 dP G 1 1501 3 1 .75 G dz g c D p D p TURBULENT LAMINAR P pressure, Φ porosity (volume of void/total bed volume) 1- Φ (volume of solid/total bed volume) gc conversion factor. 1.0 for metric system Dp diameter of particle in bed μ viscosity of gas passing through the bed Z length down the packed bed u, superficial velocity ρ gas density G= ρu superficial mass velocity kPa m kg/m.s m m/s kg/m3 kg/m2,s 7 - 29/04/2020 Pressure Drop in Packed Bed Reactors dP G 1 1501 3 1 .75 G dz g c D p Dp TURBULENT LAMINAR Ergun Equation: Constant mass flow: m 0 m 0 0 Department of Chemical Engineering 8 0 0 FT P0 T 0 FT 0 P T0 P0 T 0 (1 X) P T0 8 - 29/04/2020 Pressure Drop in Packed Bed Reactors Variable Density dP G dz 0 g c D p Department of Chemical Engineering Let P T0 FT 0 0 P0 T FT P0 T FT 1 1501 3 1.75G Dp P T0 FT 0 G 0 0 g c D p 1 1501 3 1.75G Dp 9 9 - 29/04/2020 Pressure Drop in Packed Bed Reactors Catalyst Weight W zA zA 1 c Where b c c b bulk density c solid catalyst density porosity (a.k.a., void fraction) Department of Chemical Engineering Let Ac, cross section area 0 P0 T FT dP dW A c 1 c P T0 FT 0 2 0 1 A c 1 c P0 10 10 - 29/04/2020 Pressure Drop in Packed Bed Reactors dy T FT dW 2 y T0 FT 0 P y P0 We will use this form for single reactions: d P P0 1 T 1 X dW 2 P P0 T0 Department of Chemical Engineering 11 dy T 1 X dW 2 y T0 dy 1 X dW 2y Isothermal case 11 - 29/04/2020 Pressure Drop in Packed Bed Reactors dX kC2A 0 1 X 2 y 2 dW FA 0 1 X 2 Department of Chemical Engineering dX dP dy f X, P and f X, P or f y, X dW dW dW The two expressions are coupled ordinary differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution. Polymath will combine the mole balance, rate law and stoichiometry. 12 12 - 29/04/2020 PBR AB 1) Mole Balance: dX rA dW FA 0 Department of Chemical Engineering 1 X P 1 X CA 0 CA CA 0 y 1 X P0 1 X 1 X 2 1 X y 2 2) Rate Law: rA kC 2 A0 13 13 - 29/04/2020 PBR For dy T 1 X dW 2 y T0 0 dy dW 2 y When W0 y 1 Initial condition Department of Chemical Engineering dy 2 dW y 2 (1 W ) y (1 W ) 1/ 2 2 0 1 A c 1 c P0 14 14 - 29/04/2020 1 P y 1 W 12 Department of Chemical Engineering 15 W 15 - 29/04/2020 2 CA P CA CA 0 1 X P0 Department of Chemical Engineering 16 No P P W 16 - 29/04/2020 3 -rA rA kCA2 No P Department of Chemical Engineering 17 P W 17 - 29/04/2020 4 X No P P W 18 Department of Chemical Engineering 18 - 29/04/2020 P0 T 0 (1 X) P T0 P0 T T0 , y P 0 1 f (1 X) y Department of Chemical Engineering 19 19 - 29/04/2020 5 P 1.0 No P Department of Chemical Engineering W 20 20 - 29/04/2020 Example 1: Gas Phase Reaction in PBR for δ = 0 Gas Phase Reaction in PBR with δ = 0 (Polymath Solution) A + B 2C Repeat the previous one with equil molar feed of A and B and kA = 1.5dm9/mol2/kg/min α = 0.0099 kg-1 Find X at 100 kg Department of Chemical Engineering 21 21 - 29/04/2020 Example 1: Gas Phase Reaction in PBR for δ = 0 A + B 2C dm6 k 1.5 mol kg min Case 1: Department of Chemical Engineering Case 2: W 100 kg DP 2DP1 0.0099 kg 1 X? 1 P02 P01 2 P? X? P? 22 22 - 29/04/2020 Example 1: Gas Phase Reaction in PBR for δ = 0 Department of Chemical Engineering 1) Mole Balance: dX r 'A dW FA 0 2) Rate Law: r'A kCACB 3) CA CA 0 1 X y 4) CB CA 0 1 X y W0 y 1 23 23 - 29/04/2020 Example 1: Gas Phase Reaction in PBR for δ = 0 5) dy dW 2y 2 ydy dW y 2 1 W y 1 W 12 Department of Chemical Engineering rA kC2A0 1 X y2 kC2A0 1 X 1 W 2 2 dX kC 2A 0 1 X 1 W dW FA 0 2 24 24 - 29/04/2020 Example 1: Gas Phase Reaction in PBR for δ = 0 kC 2A 0 dX 1 W dW 2 FA 0 1 X kC2A 0 X 1 X FA 0 W 2 W 2 Department of Chemical Engineering W 0, X 0, W W, X X X 0.6 with pressure drop X 0.75 without pressure drop, i.e. 0 25 25 - 29/04/2020 Example A + B → 2C Department of Chemical Engineering 26 26 - 29/04/2020 Example A + B → 2C Department of Chemical Engineering 27 27 - 29/04/2020