Chemical Reaction Engineering: Studying Reaction Rates and Reactor Design
"Chemical Reaction Engineering (CRE) involves the study of chemical reaction rates, mechanisms, and reactor design. Topics covered include Mole Balances, Rate Laws, Stoichiometry, Reactors in Series, Levenspiel Plots, Power Law Model, and the Arrhenius Equation."
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Lecture 4 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.
Chapter 4 Lecture 4 Lecture 4 Block 1 Mole Balances Size CSTRs and PFRs given rA=f(X) Block 2 Rate Laws Reaction Orders Arrhenius Equation Block 3 Stoichiometry Stoichiometric Table Definitions of Concentration Calculate the Equilibrium Conversion, Xe 2 2
Review Lecture 2 Reactor Reactor Mole Balances Mole Balances Summary in terms of conversion, X Summary Reactor Differential Algebraic Integral X X dX dX 0 = Batch t N = N r V A A r V 0 A dt A 0 t V =FA0X CSTR rA X dX dX dV= rA 0 = V F FA0 PFR 0 A r A X X dX dX dW 0 = PBR W F = r A FA0 0 A A r 3 3 W
Review Lecture 2 Levenspiel Levenspiel Plots Plots FA0 rA X 4 4
Review Lecture 2 PFR PFR 5 5
Review Lecture 2 Reactors in Series Reactors in Series moles of reacted A point to up i Xi= moles of A fed first to reactor Only valid if there are no side streams 6 6
Review Lecture 2 Reactors in Series Reactors in Series 7 7
Review Lecture 2 rA= f X ( ) Two steps to get ( ) C r = g Step 1: Rate Law A i ( ) Ci= ( ) X h Step 2: Stoichiometry ( ) X rA= f Step 3: Combine to get 8 8
Review Lecture 3 Building Block 2: Building Block 2: Rate Laws Power Law Model: r = Rate Laws order in A A B kC C A order in B Overall = + Rection Order + 2 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 C k r = 3 A B C 2 C A A A B 2nd order in A, 1st order in B, overall third order 9 9
Review Lecture 3 Arrhenius Equation Arrhenius Equation Ae k = T k A T 0 k 0 A 1013 E RT k E = Activation energy (cal/mol) R = Gas constant (cal/mol*K) T = Temperature (K) A = Frequency factor (same units as rate constant k) (units of A, and k, depend on overall reaction order) T 10 10
Review Lecture 3 Reaction Engineering Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another 11 11
Review Lecture 3 Algorithm Algorithm How to find ( ) X rA= f ( ) C r = g Step 1: Rate Law A i ( ) Ci= ( ) X h Step 2: Stoichiometry ( ) X rA= f Step 3: Combine to get 12 12
Chapter 4 Building Block 3: Building Block 3: Stoichiometry Stoichiometry We shall set up Stoichiometry Tables using species A 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 Ci = f(X) with the appropriate rate law to obtain -rA = f(X). b A + c d + B C D a a a A is the limiting reactant. 13 13
Chapter 4 Stoichiometry Stoichiometry NA= NA0 NA0X For every mole of A that reacts, b/a moles of B react. Therefore moles of B remaining: NB= NB0-b NB0 NA0 -b aNA0X = NA0 aX Let B = NB0/NA0 Then: = N N A B b X 0 B a NC= NC0+c aNA0X = NA0 C+c aX 14 14
Chapter 4 Batch System Batch System - - Stoichiometry Species Symbol Stoichiometry Table Initial Change Table Remaining A A NA0 -NA0X NA=NA0(1-X) NB0=NA0 B NB=NA0( B-b/aX) B B -b/aNA0X NC0=NA0 C NC=NA0( C+c/aX) C C +c/aNA0X ND0=NA0 D ND=NA0( D+d/aX) D D +d/aNA0X NI0=NA0 I FT0 NI=NA0 I NT=NT0+ NA0X c a Inert I ---------- N C C y d b = = = = Where: = + 0 0 0 0 0 i i i i 1 and i N C C y a a 0 0 0 0 0 A A A A = change in total number of mol per mol A reacted 15 15
Chapter 4 Stoichiometry Stoichiometry Constant Volume Constant Volume Batch 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 V =V0 ) = CA01 X Then ( V0 B b =NA01 X CA=NA ( ) V CB=NB =NA0 V0 = CA0 B b aX aX V etc. 16 16
Chapter 4 Stoichiometry Stoichiometry Constant Volume Constant Volume Batch Batch rA= kACA V = 2CB Suppose V Batch: 0 2QB-b 31-X ( ) -rA=kACA0 aX B = 1 Equimolar feed: b B= Stoichiometric feed: a 17 17
Chapter 4 Stoichiometry Stoichiometry Constant Volume Constant Volume Batch Batch = 2 If , then r k C C A A A B b ( ) 3 2 = 01 Constant Volume Batch r C X X A A B a ( ) X rA= f and we have 1 Ar X 18 18
Chapter 4 Batch Reactor Batch Reactor - - Example Example Calculate the equilibrium conversion for gas phase reaction, Xe . Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Find Xe for both a batch reactor and a flow reactor. A 2 B C 2 = B r k C A A A K C 19 19
Chapter 4 Batch Reactor Batch Reactor - - Example Example CA= KC 3 2 . 0 mol dm Calculate Xe 0 = 3 20 dm mol dX r V Step 1: = A dt N 0 A = 2 A r k C k C Step 2: rate law: A A B B C 2 = B r k C A A A K C k K = A C k B 20 20
Chapter 4 Batch Reactor Batch Reactor - - Example Example Symbol A Initial NA0 Change -NA0X Remaining NA0(1-X) B 0 NA0X NA0 X/2 Totals: NT0=NA0 NT=NA0 -NA0 X/2 C @ equilibrium: -rA=0 = 2 0 Be C Ae K C Ke=CBe CAe=NAe =CA01 Xe ( ) 2 CAe V Xe 2 CBe=CA0 21 21
Chapter 4 Batch Reactor Batch Reactor - - Example Solution: r At equilibrium Example C C 2 = = Be 0 k C K = Be A A Ae C K 2 Ae C C Stoichiometry: Constant Volume: A B / 2 V = V 0 Batch Species A B Initial NA0 0 NT0=NA0 Change -NA0X +NA0X/2 Remaining NA=NA0(1-X) NB=NA0X/2 NT=NA0-NA0X/2 22 22
Chapter 4 Batch Reactor Batch Reactor - - Example Example X C e X A 0 2 = = e K ( ) ( )2 e 2 C 1 X 2 C 1 X A 0 e A 0 e X ( )( 20 2 ) 8 = = = e 2 K C 2 . 0 ( ) e A 0 2 1 X e Xeb=0.703 23 23
Chapter 4 Flow System Flow System Stoichiometry Stoichiometry Table Table Species Symbol Reactor Feed Change Reactor Effluent A A FA0 -FA0X FA=FA0(1-X) FB0=FA0 B FB=FA0( B-b/aX) B B -b/aFA0X =Ci0 0 CA0 0 i=Fi0 =Ci0 CA0 =yi0 yA0 Where: FA0 24 24
Chapter 4 Flow System Flow System Stoichiometry Stoichiometry Table Table Species C Symbol C Reactor Feed FC0=FA0 C FD0=FA0 D Change +c/aFA0X +d/aFA0X Reactor Effluent FC=FA0( C+c/aX) FD=FA0( D+d/aX) D D Inert I FI0=A0 I FT0 ---------- FI=FA0 I FT=FT0+ FA0X d c b F C C y Where: and = + 1 = = = = 0 i 0 i 0 0 i 0 i i a a a F C C y A 0 A 0 0 A 0 A 0 F = Concentration Flow System A C A 25 25
Chapter 4 Flow System Flow System Stoichiometry Stoichiometry Table Table Species Symbol Reactor Feed Change Reactor Effluent A A FA0 -FA0X -b/aFA0X +c/aFA0X +d/aFA0X ---------- FA=FA0(1-X) FB=FA0( B-b/aX) FC=FA0( C+c/aX) FD=FA0( D+d/aX) FI=FA0 I FT=FT0+ FA0X B B FB0=FA0 B FC0=FA0 C FD0=FA0 D FI0=FA0 I FT0 y C C D D Inert I F C C d c b Where: and = = = = = + 0 i 0 i 0 0 i 0 i 1 i a a a F C C y A 0 A 0 0 A 0 A 0 F = Concentration Flow System A C A 26
Chapter 4 Stoichiometry Stoichiometry Concentration Flow System: F = A C A = Liquid Phase Flow System: 0 ( 0 B b ) =FA01 X CA=FA =CA01 X ( ) Flow Liquid Phase CB=NB =NA0 0 = CA0 B b aX aX etc. We will consider CAand CBfor gas phase reactions in the next lecture 27
Algorithm Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 28
End of End of Lecture Lecture 4 4 29 29
