 # 1 - 17/04/2015 Department of Chemical Engineering Lecture 4 Kjemisk reaksjonsteknikk Chemical Reaction Engineering  Review of previous lectures  Stoichiometry.

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1 - 17/04/2015 Department of Chemical Engineering Lecture 4 Kjemisk reaksjonsteknikk Chemical Reaction Engineering  Review of previous lectures  Stoichiometry  Stoichiometric Table  Definitions of Concentration  Calculate the Equilibrium Conversion, X e

2 - 17/04/2015 Department of Chemical Engineering Reactor Mole Balances in terms of conversion ReactorDifferentialAlgebraicIntegral CSTR PFR Batch X t PBR X W 2

3 - 17/04/2015 Department of Chemical Engineering Last Lecture Relative Rates of Reaction 3

4 - 17/04/2015 Department of Chemical Engineering Last Lecture Rate Laws - Power Law Model 4 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written. e.g. If the above reaction follows an elementary rate law 2nd order in A, 1st order in B, overall third order

5 - 17/04/2015 Department of Chemical Engineering Last Lecture Arrhenius Equation k is the specific reaction rate (constant) and is given by the Arrhenius Equation. Where: k T 5

6 - 17/04/2015 Department of Chemical Engineering These topics build upon one another Mole BalanceRate LawsStoichiometry Reaction Engineering 6

7 - 17/04/2015 Department of Chemical Engineering How to find Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get 7

8 - 17/04/2015 Department of Chemical Engineering We shall set up Stoichiometry Tables using species A as our basis of calculation in the following reaction. We will use the stochiometric tables to express the concentration as a function of conversion. We will combine C i = f(X) with the appropriate rate law to obtain -r A = f(X). A is the limiting Reactant. 8

9 - 17/04/2015 Department of Chemical Engineering For every mole of A that react, b/a moles of B react. Therefore moles of B remaining: Let Θ B = N B0 /N A0 Then: 9

10 - 17/04/2015 Department of Chemical Engineering SpeciesSymbolInitialChangeRemaining Batch System Stoichiometry Table BBN B0 =N A0 Θ B -b/aN A0 XN B =N A0 (Θ B -b/aX) AAN A0 -N A0 XN A =N A0 (1-X) InertIN I0 =N A0 Θ I ----------N I =N A0 Θ I F T0 N T =N T0 +δN A0 X Where: and CCN C0 =N A0 Θ C +c/aN A0 XN C =N A0 (Θ C +c/aX) DDN D0 =N A0 Θ D +d/aN A0 XN D =N A0 (Θ D +d/aX) 10 δ = change in total number of mol per mol A reacted

11 - 17/04/2015 Department of Chemical Engineering Constant Volume Batch Note:If the reaction occurs in the liquid phase or if a gas phase reaction occurs in a rigid (e.g. steel) batch reactor Then etc. 11

12 - 17/04/2015 Department of Chemical Engineering Suppose Batch: Stoichiometry 12 Equimolar feed: Stoichiometric feed:

13 - 17/04/2015 Department of Chemical Engineering AAF A0 -F A0 XF A =F A0 (1-X) SpeciesSymbolReactor FeedChangeReactor Effluent BBF B0 =F A0 Θ B -b/aF A0 XF B =F A0 (Θ B -b/aX) Where: Flow System Stochiometric Table 13

14 - 17/04/2015 Department of Chemical Engineering SpeciesSymbolReactor FeedChangeReactor Effluent Where: Flow System Stochiometric Table InertIF I0 = A0 Θ I ----------F I =F A0 Θ I F T0 F T =F T0 +δF A0 X CCF C0 =F A0 Θ C +c/aF A0 XF C =F A0 (Θ C +c/aX) DDF D0 =F A0 Θ D +d/aF A0 XF D =F A0 (Θ D +d/aX) and Concentration – Flow System 14

15 - 17/04/2015 Department of Chemical Engineering SpeciesSymbolReactor FeedChangeReactor Effluent AAF A0 -F A0 XF A =F A0 (1-X) BBF B0 =F A0 Θ B -b/aF A0 XF B =F A0 (Θ B -b/aX) CCF C0 =F A0 Θ C +c/aF A0 XF C =F A0 (Θ C +c/aX) DDF D0 =F A0 Θ D +d/aF A0 XF D =F A0 (Θ D +d/aX) InertIF I0 =F A0 Θ I ----------F I =F A0 Θ I F T0 F T =F T0 +δF A0 X Where:and Concentration – Flow System Flow System Stochiometric Table 15

16 - 17/04/2015 Department of Chemical Engineering Concentration Flow System: Liquid Phase Flow System: Flow Liquid Phase etc. 16 Liquid Systems

17 - 17/04/2015 Department of Chemical Engineering If the rate of reaction were then we would have This gives us 17 Liquid Systems

18 - 17/04/2015 Department of Chemical Engineering For Gas Phase Flow Systems We obtain: Combining the compressibility factor equation of state with Z = Z 0 Stoichiometry: 18

19 - 17/04/2015 Department of Chemical Engineering For Gas Phase Flow Systems 19

20 - 17/04/2015 Department of Chemical Engineering The total molar flow rate is: For Gas Phase Flow Systems Substituting F T gives: 20

21 - 17/04/2015 Department of Chemical Engineering 21 For Gas Phase Flow Systems

22 - 17/04/2015 Department of Chemical Engineering Gas Phase Flow System: Concentration Flow System: 22 For Gas Phase Flow Systems

23 - 17/04/2015 Department of Chemical Engineering For Gas Phase Flow Systems C j =f(F j, T, P) =f(x, T,P)

24 - 17/04/2015 Department of Chemical Engineering If –r A =kC A C B This gives us F A0 /-r A X 24 For Gas Phase Flow Systems

25 - 17/04/2015 Department of Chemical Engineering Consider the following elementary reaction with K C =20 dm 3 /mol and C A0 =0.2 mol/dm 3. Calculate Equilibrium Conversion or both a batch reactor (X eb ) and a flow reactor (X ef ). Example: Calculating the equilibrium conversion for gas phase reaction in a flow reactor, X ef 25

26 - 17/04/2015 Department of Chemical Engineering 26

27 - 17/04/2015 Department of Chemical Engineering Consider the following elementary reaction with K C =20 m 3 /mol and C A0 =0.2 mol/m 3. X e ’ for both a batch reactor and a flow reactor. Calculating the equilibrium conversion for gas phase reaction,X e 27 Batch Reactor Example

28 - 17/04/2015 Department of Chemical Engineering Step 1: Step 2: rate law, Calculate X e 28 Batch Reactor Example

29 - 17/04/2015 Department of Chemical Engineering SymbolInitialChangeRemaining B0½ N A0 XN A0 X/2 AN A0 -N A0 XN A0 (1-X) Totals:N T0 =N A0 N T =N A0 - N A0 X/2 @ equilibrium: -r A =0 29 Calculate X e Batch Reactor Example

30 - 17/04/2015 Department of Chemical Engineering SpeciesInitialChangeRemaining AN A0 -N A0 XN A =N A0 (1-X) B0+N A0 X/2N B =N A0 X/2 N T0 =N A0 N T =N A0 -N A0 X/2 Solution: At equilibrium Stoichiometry Constant volume Batch Calculating the equilibrium conversion for gas phase reaction 30 Batch Reactor Example

31 - 17/04/2015 Department of Chemical Engineering BR Example X eb 31

32 - 17/04/2015 Department of Chemical Engineering Gas Phase Example X ef Rate law: 32 Solution:

33 - 17/04/2015 Department of Chemical Engineering SpeciesFedChangeRemaining AF A0 -F A0 XF A =F A0 (1-X) B0+F A0 X/2F B =F A0 X/2 F T0 =F A0 F T =F A0 -F A0 X/2 Gas Flow Example X ef 33

34 - 17/04/2015 Department of Chemical Engineering AF A0 -F A0 XF A =F A0 (1-X) B0F A0 X/2F B =F A0 X/2 Stoichiometry: Gas isothermal T=T 0, isobaric P=P 0 Gas Flow Example X ef 34

35 - 17/04/2015 Department of Chemical Engineering Pure A  y A0 =1, C A0 =y A0 P 0 /RT 0, C A0 =P 0 /RT 0 @ eq: -r A =0 Gas Flow Example X ef 35

36 - 17/04/2015 Department of Chemical Engineering Gas Flow Example X ef Flow: Recall Batch: 36

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