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 20

2. 2 Substituting for

3. User friendly Energy Balance Derivations Adiabatic Heat Exchange Constant T a Heat Exchange Variable T a Co-current Heat Exchange Variable T a Counter Current 3

4. The feed consists of both - Inerts I and Species A with the ratio of inerts I to the species A being 2 to 1. Elementary liquid phase reaction carried out in a CSTR 4

5. a)Assuming the reaction is irreversible for CSTR, A  B, (K C = 0) what reactor volume is necessary to achieve 80% conversion? b)If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume? c)Make a Levenspiel Plot and then determine the PFR reactor volume for 60% conversion and 95% conversion. Compare with the CSTR volumes at these conversions. d)Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K. 5

6. Mole Balance: 6

7. Rate Law: Stoichiometry: 7

8. Energy Balance - Adiabatic, ∆C p =0: 8

9. Irreversible for Parts (a) through (c) (a) Given X = 0.8, find T and V (if reverible) 9

10. Given X, Calculate T and V 10

11. (b) (if reverible) Given T, Calculate X and V 11

12. (c) Levenspiel Plot 12

13. (c) Levenspiel Plot 13

14. CSTR X = 0.95 T = 395 CSTR X = 0.6 T = 360 14

15. PFR X = 0.6 PFR X = 0.95 15

16. CSTRX = 0.6T = 360V = 2.05 dm 3 PFRX = 0.6T exit = 360V = 5.28 dm 3 CSTRX = 0.95T = 395V = 7.59 dm 3 PFRX = 0.95T exit = 395V = 6.62 dm 3 16

17. 17

18. 18

19. 19 Need to determine T a

20. 20 Need to determine T a

21. Adiabtic (Ua=0) and Δ C P =0 21

22. A.Constant Ta e.g., Ta = 300K B. Variable T a Co-Current C. Variable T a Counter Current Guess T a at V = 0 to match T a0 = T a0 at exit, i.e., V = V f 22

23. 23

24. In - Out + Heat Added = 0 24

25. Elementary liquid phase reaction carried out in a PFR The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. 25 F A0 F I TaTa Heat Exchange Fluid T

26. 26

27. 27

28. 28

29. Heat removed Heat generated 29

30. Energy Balance: Adiabtic and Δ C P =0 Ua=0 Additional Parameters (17A) & (17B) 30 Mole Balance:

31. 31

32. Find conversion, X eq and T as a function of reactor volume V rate V T V X X X eq 32

33. 33 Need to determine T a

34. A. Constant Ta (17B) Ta = 300K Additional Parameters (18B – (20B): B. Variable T a Co-Current C. Variable T a Counter Current Guess T a at V = 0 to match T a0 = T a0 at exit, i.e., V = V f 34

35. All equations can be used from before except T a parameter, use differential T a instead, adding m C and C PC 35

36. In - Out + Heat Added = 0 All equations can be used from before except dT a /dV which must be changed to a negative. To arrive at the correct integration we must guess the T a value at V=0, integrate and see if T a0 matches; if not, re-guess the value for T a at V=0 36

37. Differentiating with respect to W: 37

38. Mole Balance on species i: Enthalpy for species i: 38

39. Differentiating with respect to W: 39

40. Final Form of the Differential Equations in Terms of Conversion: A: 40

41. Final Form of terms of Molar Flow Rate: B: 41

42. The rate law for this reaction will follow an elementary rate law. Where K e is the concentration equilibrium constant. We know from Le Chaltlier’s law that if the reaction is exothermic, K e will decrease as the temperature is increased and the reaction will be shifted back to the left. If the reaction is endothermic and the temperature is increased, K e will increase and the reaction will shift to the right. 42

43. Van’t Hoff Equation: 43

44. For the special case of Δ C P =0 Integrating the Van’t Hoff Equation gives: 44

45. endothermic reaction exothermic reaction KPKP T endothermic reaction exothermic reaction XeXe T 45

46. 46