Chemical Reaction Engineering

Transcription

1 Lecture 22 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 Web Lecture 22 Class Lecture 18-hursday 3/28/2013 Review of Multiple Steady States (MSS) Reactor Safety (Chapter 13) Blowout Velocity CSR Explosion Batch Reactor Explosion 2

3 Review Last Lecture CSR with Heat Effects 3

4 Review Last Lecture Energy Balance for CSRs d dt = F A 0 N i C Pi [ G( ) R( ) ] R G( ) = ( r A V) [ ΔH ] Rx [ ] ( ) = C ( PS 1+κ) C κ = UA F A 0 C P 0 C = 0 +κ a 1+κ 4

5 Review Last Lecture Energy Balance for CSRs R [ ] ( ) = C ( PS 1+κ) C R() Increasing 0 5 Variation of heat removal line with inlet temperature.

6 Review Last Lecture Energy Balance for CSRs κ= R [ ] ( ) = C ( PS 1+κ) C R() κ=0 Increase κ 6 a 0 Variation of heat removal line with κ (κ=ua/c P0 F A0 )

7 Review Last Lecture Multiple Steady States (MSS) Variation of heat generation curve with space-time. 7

8 Review Last Lecture Multiple Steady States (MSS) Finding Multiple Steady States with 0 varied 8

9 Review Last Lecture Multiple Steady States (MSS) emperature ignition-extinction curve 9

10 Review Last Lecture Multiple Steady States (MSS) Bunsen Burner Effect (Blowout)

11 Review Last Lecture Multiple Steady States (MSS) R() G() G R Increase υ 0 Bunsen Burner Effect (Blowout)

12 Reversible Reaction Gas Flow in a PBR with Heat Effects A B υ0c A0X τk V =, X = X kc A0 1 X 1 k 1 K + τ + e ΔH Rxτk G = ΔH RxX = τ k 1 + K e R ( ) = C ( 1+ κ)[ ] R C P = 0 + κ 1+ κ = 310 ( ) = 400[ 310] a C 1 K e 12

13 Reversible Reaction Gas Flow in a PBR with Heat Effects A B UA = 73,520 UA = 0 13

14 Reversible Reaction Gas Flow in a PBR with Heat Effects A B K C = C Be C Ae = C A0 C A0 X e p 0 ( 1 X ) e p 0 8 ( ) X e = K C 1+ K C 14

15 Reversible Reaction Gas Flow in a PBR with Heat Effects A B Exothermic Case: K C X e K C Endothermic Case: X e ~1 15

16 Adiabatic Equilibrium Conversion Conversion on emperature Exothermic ΔH is negative Adiabatic Equilibrium temperature ( adia ) and conversion (X e,adia ) X X e adi a = 0 X + e ( ΔH ) C Rx PA KC = 1+ K C X 16 adia

17 Gas Phase Heat Effects rends: Adiabatic: X exothermic X endothermic Adiabatic and X e 17 0 = 0 ΔH + C + Θ PA Rx I X C PI 0

18 Gas Phase Heat Effects X K C X e = 1+ K C X eadia = 0 + ΔH Rx Θ i C Pi 18

19 Gas Phase Heat Effects Effect of adding inerts on adiabatic equilibrium conversion Adiabatic: X 0 Θ I = Adiabatic Equilibrium Conversion Θ I = 0 19 ( ) C PA + θ I C PI X = 0 [ ] ΔH Rx, = 0 + ( ΔH Rx ) C PA + θ I C PI

20 Q 1 Q 2 F A0 F A1 F A2 F A3 X 0 1 X X 2 20

21 X X e X 3 X 2 X EB X = θ i C Pi ΔH ( ) RX 0 X

22 Adiabatic Exothermic Reactions A B ΔH Rx = +15 kcal mol he heat of reaction for endothermic reaction is positive, i.e., Energy Balance : ΔH = 0 Rx X C PA + Θ I C PI ( and X = C P A + C PI Θ I ) 0 Isothermal ΔH Rx ( ) 22 Θ Ι V We want to learn the effects of adding inerts on conversion. How the conversion varies with the amount, i.e., Θ I, depends on what you vary and what you hold constant as you change Θ I.

23 A. First Order Reaction dx dv = r A F A0 Combining the mole balance, rate law and stoichiometry ( ) dx dv = kc A0 1 X υ 0 C A0 = k ( 1 X) υ 0 wo cases will be considered Case 1 Constant υ 0, volumetric flow rate Case 2: Variable υ 0, volumetric flow rate 23

24 A.1. Liquid Phase Reaction a F i0 0 a a V=0 V=V f ao For Liquids, volumetric flow rates are additive. υ 0 = υ A0 + υ I0 = υ A0 ( 1+ Θ I ) 24

25 Θ I ISO Effect of Adding Inerts to an Endothermic Adiabatic Reaction What happens when we add Inerts, i.e., vary heta I??? It all depends what you change and what you hold constant!!! k Θ I \ ISO 1 ( ) 1+ Θ I V V Θ I k 1+ Θ I X 25 Θ I OP Θ I Θ I OP Θ I

26 A.1.a. Case 1. Constant υ 0 o keep υ 0 constant if we increase the amount of Inerts, i.e., increase Θ I we will need to decrease the amount of A entering, i.e., υ A0. So Θ I then υ A0 = 0 ΔH Rx X ISO k ISO C PA + Θ I C P I Θ Ι Θ Ι V V k X 26 Θ Ι Θ Ι

27 A.1.a. Case 2. Constant υ A, Variable υ 0 ( ) ( ) ( ) dx dv = k 1 X = k 1 X υ 0 υ A 1+ Θ I V k V k k 1+Θ I Θ Ι Θ Ι X 27 Θ Ι

28 A.2. Gas Phase Without Inerts With Inerts and A F A0 C A0 υ A P A, A C A F I0 F A0 C A0I υ I P I, I C I C A = F A0 υ A = C A0 = P A R A C I = F I υ I = F A0 + F I0 υ I = P I R I aking the ratio of C A to C I C I C A = F I υ I = F A υ A P I R I P A R A Solving for υ I υ I = υ A F I F A P A P I I A 28 We want to compare what happens when Inerts and A are fed to the case when only A is fed.

29 Nomenclature note: Sub I with Inerts I and reactant A fed Sub A with only reactant A fed F I = otal inlet molar flow rate of inert, I, plus reactant A, F I = F A0 + F I0 F A = otal inlet molar flow rate when no Inerts are fed, i.e., F A = F A0 P I, I = Inlet temperature and pressure for the case when both Inerts (I) and A are fed P A, A = Inlet temperature and pressure when only A is fed C A0 = Concentration of A entering when no inerts are presents υ I = Entering volumetric flow rate with both Inerts (I) and reactant (A) C A0 = F A0 υ A C A = otal concentration when no inerts are present = P A R A C I = otal concentration when both I and A are present = P I R I C A0I = Concentration of A entering when inerts A are entering = F A0 υ I 29

30 F I F A = F A0 + F I 0 F A0 ( 1+ Θ I ) = p A0 = 1 ( 1+ Θ I ) 1 # F % A0 $ F I 0 + F A0 & ( ' = 1 p A0 ) υ I = υ A + 1+ Θ I * ( ) P A P I I A,. - 30

31 A.2.a. Case 1 Maintain constant volumetric flow, υ 0, rate as inerts are added. I.e., υ 0 = υ I = υ A. Not a very reasonable situation, but does represent one extreme. Achieve constant υ 0 varying P, to adjust conditions so term in brackets, [ ], is one. 1+ Θ I ( ) P A P I I 0 =1 For example if Θ I = 2 then υ I will be the same as υ A, but we need the entering pressures P I and P A to be in the relationship P I = 3P A with A = I υ I = υ A ( 1+ 2) P A A 3P A = υ A = υ A = υ 0 A 31

32 A.2.a. Case 1 hat is the term in brackets, [ ], would be 1 which would keep υ 0 constant with υ I = υ A = υ 0. Returning to our combined mole balance, rate law and stoichiometry ( ) dx dv = k 1 X υ 0 k X X Θ Ι Θ Ι Θ Ι 32

33 A.2.b. Case 2: Variable υ 0 Constant, P i.e., P I = P A, I = A υ I = υ A F I F A = υ A ( F A0 + F I0 ) = υ A ( 1+ Θ I ) F A0 υ I = υ A ( 1+ Θ I ) dx dv = 1 υ A k 1+ Θ 1 X ( ) k k k 1+Θ I V V Θ Ι Θ Ι X 33 Θ Ι

34 B. Gas Phase Second Order Reaction Pure A Inerts Plus A C A0 = P A R A = F A0 υ A dx dv = r A = kc 2 A0I F A0 F I0 F A0 C I = F A0( 1+ Θ I ) υ I ( 1 X) 2 F A0 C A0I, υ I P I, I, C I 34

35 B. Gas Phase Second Order Reaction 2 C A0I F A0 ( = F A0 υ I ) 2 F A0 υ I = υ A ( 1+ Θ I ) P A P I I A = F A0 2 υ = F A0 I υ A υ A 1+ Θ I ( ) 2 P A P I 2 I A 2 dx dv = = C A0 P I ( ) 2 υ A 1+ Θ I k C A0 ( ) 2 υ A 1+ Θ I A P A I P I A P A I 2 2 ( 1 X) 2 35

36 B. Gas Phase Second Order Reaction For the same temperature and pressures for the cases with and without inerts, i.e., P I = P A and I = A, then dx dv = k C A0 ( ) 2 υ A 1+ Θ I ( 1 X) 2 36

37 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 37

38 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 38

39 End of Web Lecture 24 Class Lecture 20 39