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Lecture 21 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 21 Class Lecture 17 – Tuesday 3/19/2013 Gas Phase Reactions Trends and Optimums 2 Review Last Lecture User Friendly Equations relate T, X, or Fi 1. Adiabatic CSTR, PFR, Batch, PBR achieve this: W S Cˆ P 0 X EB Cˆ T T X T T0 3 i Pi 0 H Rx ~ C PA T T0 H Rx H Rx X C i Pi User Friendly Equations relate T, X, or Fi 2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow rate: X EB m C T Ta 4 UA ~ T T a iCPi T T 0 FA0 H Rx User Friendly Equations relate T, X, or Fi 3. PFR/PBR with heat exchange: FA0 T0 Coolant Ta 3A. In terms of conversion, X 5 Ua Ta T rAH Rx T dT B ~ dW FA0 i C Pi C p X User Friendly Equations relate T, X, or Fi 3B. In terms of molar flow rates, Fi Ua Ta T rAH Rxij T dT B dW FiCPi 4. For multiple reactions Ua Ta T rij H Rxij dT B dV FiCPi 5. Coolant Balance 6 dTA UaT Ta dV m c C Pc Reversible Reactions Xe KP endothermic reaction endothermic reaction exothermic reaction exothermic reaction T 7 T Heat Exchange Example: Elementary liquid phase reaction carried out in a PFR m c FA0 FI Ta Heat Exchange Fluid A B The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. 8 Heat Exchange a) Adiabatic. Plot X, Xe, T and the rate of disappearance as a function of V up to V = 40 dm3. b) Constant Ta. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature is constant at 300 K for V = 40 dm3. How do these curves differ from the adiabatic case. 9 Heat Exchange c) Variable Ta Co-Current. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 40 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? 10 d) Variable Ta Countercurrent. Plot X, Xe, T, Ta and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 20 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? Heat Exchange Example: PBR A ↔ B 5) Parameters • For adiabatic: • Constant Ta: • Co-current: • Counter-current: 11 Ua 0 dTa 0 dW Equations as is dT (1) (or flip T - Ta to Ta - T) dW Reversible Reactions 1) Mole Balances dX rA FA 0 dW (1) W b V dX rA B rA dV FA 0 FA 0 12 Reversible Reactions CB 2) Rate Laws rA k C A KC (2) E 1 1 k k1 exp R T1 T H Rx K C K C 2 exp R 13 (3) 1 1 T2 T (4) Reversible Reactions 3) Stoichiometry 5 CA CA 0 1 X yT0 T 6 CB CA 0 XyT0 T 14 FT FT 0 dy FT T T dW y FT0 T0 2 y T0 W V b T dy dV 2 y T0 Reversible Reactions Parameters FA0 , k1 , E , R, T1 , K C 2 , H Rx , T2 , C A0 , T0 , , b 15 (7) (15) Reversible Reactions Gas Phase Heat Effects Example: PBR A ↔ B P0 T 3) Stoichiometry: v v 0 1 X P T0 Gas Phase FA 0 1 X P T0 CA 0 1 X T0 y 5 CA v 0 1 X P0 T 1 X T 16 6 CA 0 X T0 CB y 1 X T 7 dy FT T T 1 X dW 2y FT 0 T0 2y T0 Reversible Reactions Gas Phase Heat Effects Example: PBR A ↔ B CBe CA 0 X e y T0 T KC CAe CA 0 1 X e y T0 T 8 17 KC Xe 1 KC Reversible Reactions Gas Phase Heat Effects Example: PBR A ↔ B Exothermic Case: KC Xe T T Endothermic Case: KC 18 ~1 Xe T T Reversible Reactions Gas Phase Heat Effects dT rA HRx UaT Ta dV FiCPi FiCPi FA 0 iCPi CP X Case 1: Adiabatic and ΔCP=0 T T0 HRx X iCPi (1 6A) Additional Parameters (17A) & (17B) 19 T0, iCPi CPA I CPI Reversible Reactions Gas Phase Heat Effects Case 2: Heat Exchange – Constant Ta Heat effects: dT dW 20 rA H Rx Ua b FA0 i C Pi T Ta 9 Reversible Reactions Gas Phase Heat Effects Case 3. Variable Ta Co-Current dTa UaT Ta ,V0 CPcool dV m Ta Tao (17C) Case 4. Variable Ta Countercurrent dTa UaTa T dV m C Pcool V 0 Ta ? Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf 21 22 23 24 25 26 Endothermic PFR A B 1 k1 1 X dX KC KC , Xe dV 0 1 KC Xe X XEB i CPi T T0 C PA I CP I T T0 X EB H Rx H Rx T0 T T0 27 H Rx X CPA I CPI Adiabatic Equilibrium 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 28 Tadia T Q1 29 FA0 FA1 T0 X1 Q2 FA2 T0 X2 FA3 T0 X3 X Xe X3 X EB X2 X1 T0 30 T C T T i Pi H Rx 0 31 Gas Flow Heat Effects Trends: Adiabatic X exothermic X endothermic Adiabatic T and Xe T0 32 T H Rx X T T0 C PA I C PI T0 T Effects of Inerts in the Feed 33 34 Endothermic First Order Irreversible k 1 I As inert flow increases the conversion will increase. However as inerts increase, reactant concentration decreases, slowing down the reaction. Therefore there is an optimal inert flow rate to maximize X. Gas Phase Heat Effects Adiabatic: As T0 decreases the conversion X will increase, however the reaction will progress slower to equilibrium conversion and may not make it in the volume of reactor that you have. X Xe T 35 X T0 X T T Therefore, for exothermic reactions there is an optimum inlet temperature, where X reaches Xeq right at the end of V. However, for endothermic reactions there is no temperature maximum and the X will continue to increase as T increases. Gas Phase Heat Effects Effect of adding inerts Adiabatic: X X V1 V2 Xe I I 0 T0 36 X T T T0 C pA I C pI H Rx X T Exothermic Adiabatic k θI As θI increase, T decrease and dX k dV 0 H I 37 38 39 Adiabatic Exothermic V k Xe KC T V Xe X V V V Endothermic k Xe KC T Xe Frozen Xe OR X X V 40 V V V V V Heat Exchange Exothermic Xe KC T V V Xe X V V Endothermic V 41 Xe KC T V Xe X V V End of Web Lecture 21 End of Class Lecture 17 42