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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 19

2 Today’s lecture Gas Phase Reactions Trends and Optimums 2

3 User Friendly Equations Relate T and X or F i 1. Adiabatic CSTR, PFR, Batch, PBR achieve this: Last Lecture 3

4 2. CSTR with heat exchanger, UA(T a -T) and a large coolant flow rate T TaTa User Friendly Equations Relate T and X or F i 4

5 3. PFR/PBR with heat exchange F A0 T 0 Coolant TaTa User Friendly Equations Relate T and X or F i 3A. In terms of conversion, X 5

6 User Friendly Equations Relate T and X or F i 3B. In terms of molar flow rates, F i 4. For multiple reactions 5. Coolant Balance 6

7 Reversible Reactions endothermic reaction exothermic reaction KPKP T endothermic reaction exothermic reaction XeXe T 7

8 Heat Exchange 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

9 Heat Exchange 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

10 Heat Exchange 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

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

12 Mole Balance: 12 Reversible Reactions

13 Rate Law: 13 Reversible Reactions

14 Stoichiometry: 14 Reversible Reactions

15 Parameters: 15 Reversible Reactions

16 Example: PBR A ↔ B 3) Stoich (gas): Gas Phase Heat Effects 16 Reversible Reactions

17 Example: PBR A ↔ B 17 Gas Phase Heat Effects Reversible Reactions

18 Example: PBR A ↔ B Gas Phase Heat Effects 18 Exothermic Case: XeXe T KCKC T KCKC T T XeXe ~1 Endothermic Case: Reversible Reactions

19 Adibatic Equilibrium Conversion on Temperature Exothermic Δ H is negative Adiabatic Equilibrium temperature (T adia ) and conversion (Xe adia ) X X e adia T adia T 19

20 X2X2 F A0 F A1 F A2 F A3 T0T0 X1X1 X3X3 T0T0 T0T0 Q1Q1 Q2Q2 20

21 X T X3X3 X2X2 X1X1 T0T0 XeXe X EB 21

22 Gas Phase Heat Effects 22 Case 1: Adiabtic and Δ C P =0 Additional Parameters (17A) & (17B) Example A ↔ B

23 Heat effects: 23 Case 2: Heat Exchange – Constant T a Gas Phase Heat Effects Example A ↔ B

24 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 24 Example A ↔ B

25 25

26 26

27 27

28 28

29 29

30 30

31 31 Effects of Inerts In The Feed

32 What happens when we vary 32 Endothermic 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

33 33

34 T X Adiabatic T and X e T0T0 exothermic T X T0T0 endothermic Gas Phase Heat Effects Trends: -Adiabatic: 34

35 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. Gas Phase Heat Effects 35 X T XeXe T0T0 X T X T

36 Adiabatic: Gas Phase Heat Effects 36 Effect of adding Inerts X T V1V1 V2V2 X T T0T0 XeXe X

37 37 Exothermic Adiabatic As θ I increase, T decrease and k θIθI

38 38

39 39

40 Adiabatic Endothermic Exothermic 40

41 Heat Exchange Endothermic Exothermic 41

42 Derive the Steady State Energy Balance (w/o Work) Differentiating with respect to W: 42

43 Mole Balance on species i: Enthalpy for species i: Derive the Steady State Energy Balance (w/o Work) 43

44 Differentiating with respect to W: Derive the Steady State Energy Balance (w/o Work) 44

45 Final Form of the Differential Equations in Terms of Conversion: A: 45

46 Final Form of terms of Molar Flow Rate: B: 46

47 47

48 End of Lecture 19 48

49 Example: PBR A ↔ B 2) Rates: 1) Mole balance: Gas Phase Heat Effects 49


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