Chemical Reaction Engineering Overview: Rates, Mechanisms, Reactor Design

lecture 20 n.w
1 / 46
Embed
Share

Chemical Reaction Engineering (CRE) delves into the study of chemical reaction rates, mechanisms, and the design of reactors for these reactions. Explore topics such as energy balances, adiabatic operations, and reactor volume calculations for various conversions. Dive into the world of CSTR and PFR systems to understand the dynamics of chemical processes.

  • Chemical Engineering
  • Reactor Design
  • Adiabatic Operation
  • CSTR
  • PFR
rayaanacharya
jcumm Follow

Uploaded on | 0 Views

Download Presentation

Please find below an Image/Link to download the presentation.

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.

You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.

Download Presentation

The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.

E N D

Presentation Transcript

  1. Lecture 20 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.

  2. Last Lecture Energy Balance Fundamentals dE Q sys + = F E F E W 0 0 i i i i dt Substituting for Q + = 0 W F H F H 0 0 S i i i i 2

  3. Web Lecture 20 Class Lecture 16-Thursday 3/14/2013 Reactors with Heat Exchange User friendly Energy Balance Derivations Adiabatic Heat Exchange Constant Ta Heat Exchange Variable TaCo-current Heat Exchange Variable TaCounter Current 3 3

  4. Adiabatic Operation CSTR FA0 FI A B Elementary liquid phase reaction carried out in a CSTR The feed consists of both - Inerts I and Species A with the ratio of inerts I to the species A being 2 to 1. 4

  5. Adiabatic Operation CSTR Assuming the reaction is irreversible for CSTR, A B, (KC= 0) what reactor volume is necessary to achieve 80% conversion? If the exiting temperature to the reactor is 360K, what is the corresponding reactor volume? 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. Now assume the reaction is reversible, make a plot of the equilibrium conversion as a function of temperature between 290K and 400K. 5 5

  6. CSTR: Adiabatic Example ??? ??? ?(exothermic) ??0= 5??? ????= 20000 ??? FA0 FI ?0= 300 ? ??= 10??? A B ? = ? ? = ? ??? F X = 1) Mole Balances: A r 0 V A exit 6 6

  7. CSTR: Adiabatic Example 2) Rate Laws: C = B r k C A A K C E 1 1 R T T = k k e 1 1 H 1 1 = Rx K K exp C C 1 R T T 2 3) Stoichiometry: ( ) = C C 1 X A A 0 = C C X B A 0 7 7

  8. CSTR: Adiabatic Example 4) Energy Balance Adiabatic, Cp=0 ( ) ( ) C H X H + X = + = + Rx Rx T T T 0 0 C C i P P I P i A I ( ) 20 + 000 , 20 000 , + = + = + T 300 X 300 X ( )( ) 18 2 164 164 36 = + T 300 100 X 8 8

  9. CSTR: Adiabatic Example Irreversible for Parts (a) through (c) ( ) = = r kC 1 X (i.e., K ) A A 0 C (a) Given X = 0.8, find T and V Calc T Calc k Calc rA Calc V Given X CalcKC (if reversible) 9

  10. CSTR: Adiabatic Example Given X, Calculate T and V F X F X = = A r 0 A 0 1 V ( ) kC X A A 0 exit ( 8 . 0 ) = + = T 300 100 380 K 10 000 , 1 1 = = k 1 . 0 exp . 3 81 . 1 989 298 380 ( )( )( )( 2 81 ) F X 5 8 . 0 = = = 3 A 0 r V . 2 82 dm ( . 3 ) 8 . 0 1 A 10

  11. CSTR: Adiabatic Example Given T, Calculate X and V Calc Calc Calc Calc (b) Given X T k r V A Calc (if reversible) ble) K C ( ) = 1 (Irreversi r = kC X 0 A A 360 T = T K 300 = 6 . 0 X 100 . 1 = 1 83 min k ( )( 83 ) 5 6 . 0 = = 3 . 2 05 V dm ( . 1 )( )( 2 ) 4 . 0 11

  12. CSTR: Adiabatic Example (c) Levenspiel Plot F T F = A 0 A 0 ( X ) r = kC + 1 X A A 0 300 100 FA0 rA Calc T Calc k Calc rA Calc Choose X 12

  13. CSTR: Adiabatic Example (c) Levenspiel Plot 13

  14. CSTR: Adiabatic Example CSTR X = 0.6 T = 360 K 30 25 20 -Fa0/Ra 15 10 5 CSTR 60% 0 0 0.1 0.2 0.3 0.4 0.5 X 0.6 0.7 0.8 0.9 1 CSTR X = 0.95 T = 395 K 30 25 20 -Fa0/Ra 15 10 5 CSTR 95% 14 0 0 0.1 0.2 0.3 0.4 0.5 X 0.6 0.7 0.8 0.9 1

  15. CSTR: Adiabatic Example PFR X = 0.6 30 25 20 -Fa0/Ra 15 10 PFR 60% 5 0 0 0.1 0.2 0.3 0.4 0.5 X 0.6 0.7 0.8 0.9 1 PFR X = 0.95 PFR 95% 15

  16. CSTR: Adiabatic Example - Summary X = 0.6 X = 0.6 X = 0.95 X = 0.95 T = 360 Texit= 360 T = 395 Texit= 395 V = 2.05 dm3 V = 5.28 dm3 V = 7.59 dm3 V = 6.62 dm3 CSTR PFR CSTR PFR 16

  17. Energy Balance in terms of Enthalpy ( ) + = 0 F H F H Ua T T V i i i i a + V V V dV d F H ( ) i i + = 0 Ua T T a dV d F H dH dF i i = + i i F H i i dV dV 17

  18. PFR Heat Effects ( i i r dV dF ) = = i r A ( ) = + 0 H H C T T i i Pi R dH dT i= C Pi dV dV d F H dT ( ) i i = + i F C H r i Pi i A dV dV = H H i i R x 18

  19. PFR Heat Effects +Ua Ta-T dT dV+DHRx-rA ( ) ( )=0 - CPiFi dT dV=DHRxrA-Ua T -Ta CPi ( ) Fi ( ) -rA CPi ( )-Ua T -Ta Fi ( ) DHRx dT dV= Need to determine Ta 19

  20. Heat Exchange: ( )( ) C ( ) r H Ua T T dT = A Rx F a dV i iP ( ) -DHRx FA0 QiCPi ( )-Ua T -Ta ( ) -rA dT dV= Need to determine Ta 20

  21. Heat Exchange Example: Case 1 - Adiabatic Energy Balance: Adiabatic (Ua=0) and CP=0 ( ) H X = + Rx T T 16 ( A ) 0 C i iP 21

  22. User Friendly Equations A. Constant Ta e.g., Ta = 300K B. Variable Ta Co-Current ( ) , dT Ua T T = = = a a 0 17 ( ) V T T C a ao dV m C P cool C. Variable Ta Counter Current ( = P C m dV cool Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = V ) dT Ua T T = = 0 Guess ? a a V T a 22

  23. Heat Exchanger Energy Balance Variable Ta Co-current Coolant Balance: In - Out + Heat Added = 0 ( ) + = 0 m H m H Ua V T T C C dH C C a + V V V ( ) + = C 0 m Ua T T C a dV H ( ) = + 0 C H C T T C PC a r dH dT = C a C PC dV dV ( ) dT Ua T T = = = a a , 0 V T T 0 a a dV m C 23 C PC

  24. Heat Exchanger Energy Balance Variable Ta Counter-current In - Out + Heat Added = 0 ( ) + = 0 m H m H Ua V T T C C C C a + V V V dH ( ) + = C 0 m Ua T T C a dV dT ( ) Ua T T = a a C dV m C PC 24

  25. Heat Exchanger Example Case 1 Constant Ta Elementary liquid phase reaction carried out in a PFR Heat Exchange Fluid m c Ta FA0 FI T 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. 25

  26. Heat Exchanger Example Case 1 Constant Ta dX 1) Mole Balance: = ) 1 ( r F A A 0 dV C = ) 2 ( r k C B 2) Rate Laws: A A K C E 1 1 = ) 3 ( k k exp 1 R T T 1 H 1 1 = ) 4 ( K K exp Rx C C 2 R T T 2 26

  27. Heat Exchanger Example Case 1 Constant Ta ( ) ( ) ( ) 6 = C C 1 X 5 3) Stoichiometry: A A 0 = C C X B A 0 ( )( F ) ( )( ) 7 H r Ua T T dT 4) Heat Effects: = R A a 0 dV C A i Pi ( ) 0 CP= k + ( ) 8 = C X eq 1 k C ( ) 9 = + C C C i Pi PA I PI 27

  28. Heat Exchanger Example Case 1 Constant Ta , , , , , H , 1 E , 2 R T T Parameters: 1 2 R , , , k k , Ua , T F , 0 C a , A C C C 0 A PA r PI I = rate A 28

  29. PFR Heat Effects Heat generated Heat removed Q Q dT g r = dV F C i Pi qiCPi+DCPX FA0qi+uiX ( )CPi=FA0 FiCPi= ( ) rA ( )-Ua T -Ta qiCPi+DCPX ( ) DHR FA0 dT dV= 29

  30. Heat Exchanger Example Case 2 Adiabatic dX dV= rA Mole Balance: FA0 Energy Balance: Adiabatic and CP=0 Ua=0 ( ) H X = + Rx T T 16 ( A ) 0 C i iP Additional Parameters (17A) & (17B) = + T , C C C 30 0 i P P I P i A I

  31. Adiabatic PFR 31 31

  32. Example: Adiabatic Find conversion, Xeqand T as a function of reactor volume Xeq rate X T X V V V 32

  33. Heat Exchange ( )( ) C ( ) r H Ua T T dT = A Rx F a dV i iP ( )( ) ( ) r H Ua T T dT = A Rx a 16 ( B ) dV F C A 0 i iP Need to determine Ta 33 33

  34. User Friendly Equations A. Constant Ta (17B) Ta = 300K Additional Parameters (18B (20B): T , C , Ua a i iP B. Variable Ta Co-Current ( ) dT Ua T T = = = 0 17 ( ) a a V T T C a ao dV m C P cool C. Variable Ta Countercurrent ( = C m dV ) dT Ua T T = = 0 ? a a V T a P cool Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf 34

  35. Heat Exchange Energy Balance Variable TaCounter-current Coolant balance: In - Out + Heat Added = 0 ( ) + = 0 m H m H Ua V T T C C dH C C a + V V V ( ) + = 0 C m Ua T T C a dV All equations can be used from before except Taparameter, use differential Ta instead, adding mC and CPC ( ) = + 0 C H H C T T C PC a r dH dT = C a C PC dV dV ( ) dT Ua T T = = = , 0 a a V T T 0 a a dV m C 35 35 C PC

  36. Heat Exchange Energy Balance Variable TaCo-current In - Out + Heat Added = 0 ( ) + = 0 m H m H Ua V T T C C C C a + V V V ( ) dH dT Ua T T ( ) + = = 0 C a a C m Ua T T C a dV dV m C PC All equations can be used from before except dTa/dV which must be changed to a negative. To arrive at the correct integration we must guess the Tavalue at V=0, integrate and see if Ta0matches; if not, re-guess the value for Taat V=0 36

  37. Derive the user-friendly Energy Balance for a PBR W Ua ( ) + = T T dW F H F H 0 a 0 i 0 i i i B 0 Differentiating with respect to W: Ua dF dH ( ) + = i i T T 0 H F 0 a i i dW dW B 37

  38. Derive the user-friendly Energy Balance for a PBR Mole Balance on species i: dF = = i r r i i A dW Enthalpy for species i: T ( ) = + H H T C dT i i R Pi T R 38 38

  39. Derive the user-friendly Energy Balance for a PBR Differentiating with respect to W: dH dT = + i 0 C Pi dW dW Ua dT ( ) + = T T r H F C 0 a A i i i Pi dW B 39 39

  40. Derive the user-friendly Energy Balance for a PBR Ua F dT ( ) + = T T r H F C 0 a A i i i Pi dW B ( ) T = H H i i R ( ) = + F X i A 0 i i Final Form of the Differential Equations in Terms of Conversion: A: 40 40

  41. Derive the user-friendly Energy Balance for a PBR Final form of terms of Molar Flow Rate: Ua ( ) + T T r H a A dT = B dW F C i Pi B: dX r ( ) = A= g X , T dW F A 0 41

  42. Reversible Reactions + + A B C D The rate law for this reaction will follow an elementary rate law. C C = C K D r k C C A A B C Where Ke is the concentration equilibrium constant. We know from Le Chaltlier s law that if the reaction is exothermic, Ke 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, Ke will increase and the reaction will shift to the right. 42

  43. Reversible Reactions K = K P ( ) C RT Van t Hoff Equation: ( ) 2 ( ) ( ) C + d ln K H T H T T T = = P R R R P R 2 dT RT RT 43

  44. Reversible Reactions For the special case of CP=0 Integrating the Van t Hoff Equation gives: ( ) H T 1 1 ( ) T ( ) T R R = K K exp R P 2 P 1 T T 1 2 44

  45. Reversible Reactions Xe KP endothermic reaction endothermic reaction exothermic reaction exothermic reaction T T 45

  46. End of Lecture 20 46 46

Related


No related presentations.

More Related Content