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Lecture 24 (Ch. 13) 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. Web Lecture 24 Class Lecture 22-Thursday 3/24/2011 Review of Multiple Steady States Blowout Velocity CSTR Explosion Batch Reactor Explosion 2 3 dT FA 0 GT RT dt N iCPi GT rAV H Rx RT CPS 1 T TC 4 UA FA 0CP 0 T0 Ta TC 1 RT CPS 1 T TC R(T) Increasing T0 T 5 Variation of heat removal line with inlet temperature. κ=∞ RT CPS 1 T TC κ=0 R(T) Increase κ Ta 6 T0 T Variation of heat removal line with κ (κ=UA/CP0FA0) Variation of heat generation curve with space-time. 7 Finding Multiple Steady States with T0 varied 8 Temperature ignition-extinction curve 9 A <-> B 0 C A0 X k , X X 1 kCA 0 1 X 1 k1 K K e e H Rx k G H Rx X 1 1 k1 Ke V R T C P 1 T TC T0 Ta 310 1 R T 400T 310 TC 12 UA = 73,520 UA = 0 13 Example: PBR A ↔ B CBe CA 0 X e y T0 T KC CAe CA 0 1 X e y T0 T 8 14 KC Xe 1 KC Example: PBR A ↔ B Exothermic Case: KC Xe T T Endothermic Case: KC 15 ~1 Xe T T Conversion on Temperature Exothermic ΔH is negative Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia) X T T0 Xeadia H Rx X CPA KC Xe 1 KC 16 Tadia T Trends: -Adiabatic: X exothermic X endothermic Adiabatic T and Xe T0 17 T H R X T T0 C PA I C PI T0 T X Xe KC 1 K C Xeadia T T T0 18 H Rx iCPi Effect of adding inerts on adiabatic equilibrium conversion Adiabatic: X I Adiabatic Equilibrium Conversion I 0 T T0 19 X T T0 CPA I CPI H Rx , T T0 H Rx CPA I CPI Q1 20 FA0 FA1 T0 X1 Q2 FA2 T0 X2 FA3 T0 X3 X Xe X3 XEB C T T X i H RX X2 X1 T0 21 Pi T 0 A B HRx The heat of reaction for endothermic reaction is positive, i.e., X H Rx T T0 C PA I C P I Energy Balance and kcal 15 mol C P C P I T0 T X A I H Rx We want to learn the effects of adding inerts on conversion. How the conversion varies with the amount, i.e., I, depends on what you vary and what you hold constant as you change I. 22 dX rA dV FA0 A. First Order Reaction Combining the mole balance, rate law and stoichiometry dX kCA0 1 X k 1 X dV 0CA0 0 23 Two cases will be considered Case 1 Constant 0, volumetric flow rate Case 2: Variable 0, volumetric flow rate For liquids the volumetric flow rates are additive 0 A0 I0 A0 1 I 24 Effect of Adding Inerts to an Endothermic Adiabatic Reaction What happens when we add Inerts, i.e., vary Theta I??? It all depends what you change and what you hold constant!!! I I ISO ISO k T V 25 \ V k 1 y I k 1 yXI k k 1 1 II k 1 I I OPT 1 1 I I I I OPT I To keep 0 constant if we increase the amount of Inerts, i.e., increase I we will need to decrease the amount of A entering, i.e., A0. So I then A0 T T0 26 H Rx X CPA I CP I k1 X dX k1 X dV 0 A 1 I 27 Without Inerts CTA With Inerts and A FA0 P CA0 A A RTA CTI FTI FA0 FI0 P I I I RTI Taking the ratio of CTA to CTI FTI PI CTI I RTI PA CTA FTA A RTA Solving for I I A 28 FTI PA TI FTA PI TA We want to compare what happens when Inerts and A are fed to the case when only A is fed. Nomenclature note: Sub I with Inerts I and reactant A fed Sub A with only reactant A fed FTI = Total inlet molar flow rate of inert, I, plus reactant A, FTI = FA0 + FI0 FTA = Total inlet molar flow rate when no Inerts are fed, i.e., FTA = FA0 PI, TI = Inlet temperature and pressure for the case when both Inerts (I) and A are fed PA, TA = Inlet temperature and pressure when only A is fed CA0 = Concentration of A entering when no inerts are presents C A0 PA RTA PI CTI = Total concentration when both I and A are present RTI FA0 A CTA = Total concentration when no inerts are present A are entering FA0 CA0I = Concentration of A entering when inerts I (I) and reactant (A) I = Entering volumetric flow rate with both Inerts 29 1 1 FTI FA0 FI0 1 I FA0 y A0 FA0 FTA F F I0 A0 y A0 1 1 I PA TI I A 1 I T P I A 30 Maintain constant volumetric flow, 0, rate as inerts are added. I.e., 0 = I = A. Not a very reasonable situation, but does represent one extreme. Achieve constant 0 varying P, T to adjust conditions so term in brackets, [ ], is one. PA TI 1 I 1 PI T0 For example if I = 2 then I will be the same as A, but we need the entering pressures PI and PA to be in the relationship PI = 3PA with TA = TI 1 PA TA I A1 2 A3 A 0 3 3PA TA 31 That is the term in brackets, [ ], would be 1 which would keep 0 constant with I = A = 0. Returning to our combined mole balance, rate law and stoichiometry dX k1 X dV 0 32 I A FTI F F A A0 I 0 A 1 I FTA FA0 I A 1 I dX 1 k 1 X dV A 1 33 Pure A C A0 Inerts Plus A PA F A0 RTA A CTI 2 2 dX rA kCA0I 1 X dV FA0 FA0 34 FA0 1 I I PA TI I A 1 I PI TA 2 C 2A0I FA0 I F A0 2 FA0 FA0 I FA0 2 2 T 2 P A A 1 I A I PI TA 2 C A0 PI TA 2 P T A 1 I A I dX k dV 1 I 2 35 2 C A0 PI TA 2 1 X A PA TI For the same temperature and pressures for the cases with and without inerts, i.e., PI = PA and TI = TA, then dX k CA0 2 1 X dV 1 I 2 A 36 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 37 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 38 39