Chemical Reaction Engineering

Transcription

1 Lecture 4 hemical Reaction Engineering (RE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.

2 hapter 4 Lecture 4 Block 1 Mole Balances Size STRs and PFRs given r =f(x) Block 2 Rate Laws Reaction Orders rrhenius Equation Block 3 Stoichiometry Stoichiometric Table Definitions of oncentration alculate the Equilibrium onversion, X e 2

3 Review Lecture 2 Reactor Mole Balances Summary in terms of conversion, X Reactor Differential lgebraic Integral Batch STR N 0 dx dt r V V F 0 X r t N X 0 0 dx r V X t 3 PFR PBR dx F 0 dv r F 0 V F X 0 0 dx r dx dw dx r W F0 r 0 X X W

4 Review Lecture 2 Levenspiel Plots F 0 r X 4

5 Review Lecture 2 PFR 5

6 Review Lecture 2 Reactors in Series X i molesof reactedup to point i molesof fed to first reactor 6 Only valid if there are no side streams

7 Review Lecture 2 Reactors in Series 7

8 Review Lecture 2 Two steps to get r f X Step 1: Rate Law Step 2: Stoichiometry r g i X i h Step 3: ombine to get r f X 8

9 Review Lecture 3 Building Block 2: Rate Laws 9 Power Law Model: r k 2 B 3 B α β order in order in B Overall Rection Order 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 r 2nd order in, 1st order in B, overall third order k 2 B α β

10 Review Lecture 3 rrhenius Equation k e E RT E = ctivation energy (cal/mol) R = Gas constant (cal/mol*k) T = Temperature (K) k T = Frequency factor (same units as rate constant k) (units of, and k, depend on overall reaction order) T k T 0 k

11 Review Lecture 3 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another 11

12 Review Lecture 3 lgorithm How to find r f X Step 1: Rate Law r g i Step 2: Stoichiometry X i h Step 3: ombine to get r f X 12

13 Building Block 3: Stoichiometry hapter 4 We shall set up Stoichiometry Tables using species as our basis of calculation in the following reaction. We will use the stoichiometric tables to express the concentration as a function of conversion. We will combine i = f(x) with the appropriate rate law to obtain -r = f(x). b a B c a d a D 13 is the limiting reactant.

14 Stoichiometry N N 0 N 0 X For every mole of that reacts, b/a moles of B react. Therefore moles of B remaining: N B = N B0 - b a N 0 X = N 0 æ ç è N B0 N 0 - b a X ö ø hapter 4 14 Let Θ B = N B0 /N 0 Then: b N B N 0B X a N N 0 c a N X N c 0 0 a X

15 Batch System - Stoichiometry Table Species Symbol Initial hange Remaining N 0 -N 0 X N =N 0 (1-X) B B N B0 =N 0 Θ B -b/an 0 X N B =N 0 (Θ B -b/ax) N 0 =N 0 Θ +c/an 0 X N =N 0 (Θ +c/ax) D D N D0 =N 0 Θ D +d/an 0 X N D =N 0 (Θ D +d/ax) Inert I N I0 =N 0 Θ I N I =N 0 Θ I F T0 hapter 4 N T =N T0 +δn 0 X Where: i N N i0 0 i i0 0 y y i0 0 and d a c a b a 1 15 δ = change in total number of mol per mol reacted

16 hapter 4 Stoichiometry onstant 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 16 Then N V N 0 1 X B N B V N 0 etc. V 0 V 0 V V X B b a X b 0 B a X

17 Stoichiometry onstant Volume Batch hapter 4 Suppose r k 2 B Batch: V V0 -r = k X æ ( ) 2 çq B - b a X Equimolar feed: B 1 Stoichiometric feed: è B b a ö ø 17

18 Stoichiometry onstant Volume Batch hapter 4 If r k 2 B, then r 3 2 b 0 1 X B X onstant Volume Batch a and we have r f X 1 r 18 X

19 hapter 4 Batch Reactor - Example alculate the equilibrium conversion for gas phase reaction, X e. onsider the following elementary reaction with K =20 dm 3 /mol and 0 =0.2 mol/dm 3. Find X e for both a batch reactor and a flow reactor. 2 B r k 2 K B 19

20 hapter 4 Batch Reactor - Example 20 alculate X e Step 1: dx dt Step 2: rate law: r N 0 V mol dm 3 K 20 dm r r K k k k k B 2 mol 2 k B 3 K B B

21 hapter 4 Batch Reactor - Example Symbol Initial hange Remaining N 0 -N 0 X N 0 (1-X) B 0 ½ N 0 X N 0 X/2 Totals: N T0 =N 0 N T =N 0 -N 0 equilibrium: -r =0 0 2 e K Be K e Be 2 e N e e V 1 X 0 e 21 Be 0 X e 2

22 hapter 4 Batch Reactor - Example Solution: t equilibrium r 0 k 2 e K Be K Be 2 e Stoichiometry: onstant Volume: B/ 2 V V 0 Batch Species Initial hange Remaining N 0 -N 0 X N =N 0 (1-X) B 0 +N 0 X/2 N B =N 0 X/2 N T0 =N 0 N T =N 0 -N 0 X/2 22

23 hapter 4 Batch Reactor - Example K e 2 1 X 2 1 X X 2 e e 0 X e e Xe 2Ke0 2 1 X X eb e

24 hapter 4 Flow System Stoichiometry Table Species Symbol Reactor Feed hange Reactor Effluent F 0 -F 0 X F =F 0 (1-X) B B F B0 =F 0 Θ B -b/af 0 X F B =F 0 (Θ B -b/ax) 24 Where: i F i0 F 0 i i0 0 y i0 y 0

25 Flow System Stoichiometry Table Species Symbol Reactor Feed hange Reactor Effluent F 0 =F 0 Θ +c/af 0 X F =F 0 (Θ +c/ax) D D F D0 =F 0 Θ D +d/af 0 X F D =F 0 (Θ D +d/ax) Inert I F I0 = 0 Θ I F I =F 0 Θ I F T0 hapter 4 F T =F T0 +δf 0 X Where: i F F i0 0 i i0 0 y y i0 0 and d a c a b a 1 25 oncentration Flow System F

26 hapter 4 Flow System Stoichiometry Table Species Symbol Reactor Feed hange Reactor Effluent F 0 -F 0 X F =F 0 (1-X) B B F B0 =F 0 Θ B -b/af 0 X F B =F 0 (Θ B -b/ax) F 0 =F 0 Θ +c/af 0 X F =F 0 (Θ +c/ax) D D F D0 =F 0 Θ D +d/af 0 X F D =F 0 (Θ D +d/ax) Inert I F I0 =F 0 Θ I F I =F 0 Θ I 26 Where: F y F T0 and i0 i0 0 i0 i0 i 1 F y0 a a a F oncentration Flow System d c b F T =F T0 +δf 0 X

27 hapter 4 Stoichiometry oncentration Flow System: Liquid Phase Flow System: F 0 F F 0 1 X B N B N 0 0 etc X B b a X b 0 B a X Flow Liquid Phase 27 We will consider and B for gas phase reactions in the next lecture

28 lgorithm Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 28

29 29 End of Lecture 4