Lecture 6 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 6 - Tuesday 1/25/2011 Block 1: Block 2: Block 3: Block 4: Review of Blocks 1, 2 and 3 Examples : Undergraduate Reactor Experiments CSTR PFR BR Gas Phase Reaction with Change in the Total Number of Moles 2 Previous Lectures Reactor Differential Algebraic Integral X Batch N A0 dX dt r AV 0 V CSTR PFR t N A0 FA0 dX dV dX r AV t FA0 X rA X rA X V FA0 0 dX rA X PBR 3 FA0 dX dW X rA W FA0 0 dX r A W Previous Lectures Power Law Model rA kC A C B α order in A β order in B Overall Rection 4 Order α β 2 A B 3C A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written. e.g. If the above reaction follows an elementary rate law 2 rA k A C A C B 2nd order in A, 1st order in B, overall third order Previous Lectures 5 Previous Lectures 6 Previous Lectures 7 Today’s lecture Example for Liquid Phase Undergraduate Laboratory Experiment (CH2CO)2O + H2O A + B Entering Volumetric flow rate Acetic Anhydride Water Elementary with k’ Case I Case II 8 CSTR PFR 2CH3COOH 2C v0 = 0.0033 dm3/s 7.8% (1M) 92.2% (51.2M) 1.95x10-4 dm3/(mol.s) V = 1dm3 V = 0.311 dm3 Today’s lecture Example for Gas Phase : PFR and Batch Calculation 2NOCl 2NO + Cl2 2A 2B + C Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows an elementary rate law with k = 0.29 dm3/(mol.s) Case I Case II PFR with v0 = 10 dm3/s Find space time, with X = 0.9 Find reactor volume,V for X = 0.9 Batch constant volume Find the time, t, necessary to achieve 90% conversion. Compare and t. 9 Lecture 6 Algorithm for Isothermal Reactor Design 1. Mole Balance and Design Equation 2. Rate Law 3. Stoichiometry 4. Combine 5. Evaluate A. Graphically (Chapter 2 plots) 10 B. Numerical (Quadrature Formulas Chapter 2 and appendices) C. Analytical (Integral Tables in Appendix) D. Software Packages (Appendix- Polymath) Example: CH3CO2 + H20 2CH3OOH C A 0 1M C B 0 51 . 2 M X ? V 1 dm 3 0 3.3 10 3 dm 3 s 2C A+B 1) Mole Balance: 11 CSTR: V FA0 X rA 2) Rate Law: rA k A C A C B 3) Stoichiometry: 12 A FA0 -FA0X FA=FA0(1-X) B FA0ΘB -FA0X FB=FA0(ΘB-X) C 0 2FA0X FC=2FA0X CA CB B FA F A 0 1 X 0 F A 0 B X 0 51 . 2 C A 0 1 X C A 0 B X 51 . 2 1 C B C A 0 51 . 2 X C A 0 51 . 2 C B 0 13 rA k ' C B 0 C A 0 1 X kC A 0 1 X k V X 0 kC A 0 X C A 0 1 X k 1 k X 3 . 03 4 . 03 14 0 . 75 V 0 kX 1 X V 0 kX 1 X A + B 2C 0 . 00324 dm 3 0.311 dm s 1) Mole Balance: dX 2) Rate Law: 3) Stoichiometry: 15 dV X 3 rA FA0 r A kC A C B C A C A 0 1 X CB CB0 rA k ' C B 0 C A 0 1 X kC A 0 1 X 4) Combine: dX kC A 0 1 X dX ln 16 C A 0 0 dV 1 X 1 1 X k 0 dV kd k X 1 e k V 0 . 311 dm 0 0 . 00324 dm X 0 . 61 3 3 96 . 0 sec sec k 0 . 01 s 1 2 NOCl 2 NO + Cl2 2A 2B + C 0 10 dm 3 k 0 . 29 s T T0 2) Rate Law: 17 3 C A 0 0 .2 mol s P P0 1) Mole Balance: dm X 0.9 dX dV rA FA0 rA kC A 2 mol L 0 1 X 3) Stoich: Gas C A 0 1 X CA 1 X A B + ½C 4) Combine: rA 1 X 1 X 2 0 18`1 V dX 1 X 1 X 2 dV 2 kC 2 kC A 0 1 X 2 2 C A 0 0 1 X 2 dX X 2 A0 0 kC A 0 0 Da kC A 0V dV kC A 0 0 kC A 0 2 1 ln 1 X X 2 1 1 y A 0 1 2 2 kC A 0 17 . 02 17 . 02 294 sec kC A 0 V V 0 2940 L 19 1 2 X 1 X Gas Phase 2A 2B + C 1) Mole Balance: dX dt 2) Rate Law: 3) Stoich: rAV 0 N A0 N A0 V0 rA C A0 rA kC A 2 Gas, but: CA V V0 N A 0 1 X V0 r A kC A 0 1 X 2 20 rA C A 0 1 X 2 4) Combine: dX X kC A 0 1 X 2 kC dt 1 2 C A0 dX dt dX 1 X 1 1 X 21 2 A0 2 kC A 0 dt kC A 0 t t 155 sec kC A 0 1 X 2 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 22 End of Lecture 6 23