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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 21 (Chapter 12)

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1. Adiabatic CSTR, PFR, Batch, PBR achieve this: 2

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Gas Phase Reactions Trends and Optimums 3

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2. CSTR with heat exchanger, UA(T a -T) and a large coolant flow rate: T TaTa 4

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3. PFR/PBR with heat exchange: F A0 T 0 Coolant TaTa 3A. In terms of conversion, X 5

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3B. In terms of molar flow rates, F i 4. For multiple reactions 5. Coolant Balance 6

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endothermic reaction exothermic reaction KPKP T endothermic reaction exothermic reaction XeXe T 7

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Example: Elementary liquid phase reaction carried out in a PFR F A0 F I TaTa Heat Exchange Fluid The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. 8

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a)Adiabatic. Plot X, X e, T and the rate of disappearance as a function of V up to V = 40 dm 3. b)Constant T a. Plot X, X e, T, T a 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 dm 3. How do these curves differ from the adiabatic case. 9

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c)Variable T a Co-Current. Plot X, X e, T, T a 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 dm 3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? d)Variable T a Counter Current. Plot X, X e, T, T a 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 dm 3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? 10

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Example: PBR A ↔ B 5) Parameters…. For adiabatic: Constant T a : Co-current: Equations as is Counter-current: 11

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Example: PBR A ↔ B 16

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Example: PBR A ↔ B 17

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Example: PBR A ↔ B 18 Exothermic Case: XeXe T KCKC T KCKC T T XeXe ~1 Endothermic Case:

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19 Case 1: Adiabtic and Δ C P =0 Additional Parameters (17A) & (17B)

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Heat effects: 20 Case 2: Heat Exchange – Constant T a

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Case 3. Variable T a Co-Current Case 4. Variable T a Counter Current Guess T a at V = 0 to match T a0 = T a0 at exit, i.e., V = V f 21

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Conversion on Temperature Exothermic Δ H is negative Adiabatic Equilibrium temperature (T adia ) and conversion (Xe adia ) X X e adia T adia T 28

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X2X2 F A0 F A1 F A2 F A3 T0T0 X1X1 X3X3 T0T0 T0T0 Q1Q1 Q2Q2 29

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X T X3X3 X2X2 X1X1 T0T0 XeXe X EB 30

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T X Adiabatic T and X e T0T0 exothermic T X T0T0 endothermic Trends: -Adiabatic: 32

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What happens when we vary 34 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. First Order Irreversible

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Adiabatic: As T 0 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. Therefore, for exothermic reactions there is an optimum inlet temperature, where X reaches X eq 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. 35 X T XeXe T0T0 X T X T

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Adiabatic: 36 Effect of adding Inerts X T V1V1 V2V2 X T T0T0 XeXe X

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37 As θ I increase, T decrease and k θIθI

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Endothermic Exothermic 40

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Endothermic Exothermic 41

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