Lecture 21 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.
Web Lecture 21 Class Lecture 17 uesday 3/19/2013 Gas Phase Reactions rends and Optimums 2
Review Last Lecture User Friendly Equations relate, X, or F i 1. Adiabatic CSR, PFR, Batch, PBR achieve this: W! = Δ ˆ S CP = 0 X EB = Θ i ĈP ( i ) 0 ΔH Rx 3 X = = 0 Θ i ĈP ( i ) 0 + ΔH Rx ( ΔH ) Θ Rx i C P i X
User Friendly Equations relate, X, or F i 2. CSR with heat exchanger, UA( a -) and a large coolant flow rate: X EB = F i0 " $ # UA F A0 a ( ) % '+ & ΔH Rx Θ! i CPi ( ) 0 4 m! C a X
User Friendly Equations relate, X, or F i 3. PFR/PBR with heat exchange: F A0 0 a Coolant 5 3A. In terms of conversion, X d dw = Ua ρ B F ʹ Δ ( ) + r H ( ) A0 a ~ Rx ( Θ C + ΔC X ) i P p i A
User Friendly Equations relate, X, or F i 6 3B. In terms of molar flow rates, F i d dw = Ua ρ B ( ) + r ʹ ΔH ( ) a 4. For multiple reactions Ua a d ρb = dv 5. Co-Current Balance d dv A ( ) Ua = m! C i A FC P i Rx ij + r Δ FC ( ) c P c a i P i ij H Rx ij
Reversible Reactions X e endothermic reaction exothermic reaction K P endothermic reaction exothermic reaction 7
Heat Exchange Example: Elementary liquid phase reaction carried out in a PFR F A0 F I m! c a A B Heat Exchange Fluid he feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. 8
Heat Exchange a) Adiabatic. Plot X, X e, and the rate of disappearance as a function of V up to V = 40 dm 3. b) Constant a. Plot X, X e,, a and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature is constant at 300 K for V = 40 dm 3. How do these curves differ from the adiabatic case. 9
Heat Exchange c) Variable a Co-Current. Plot X, X e,, a and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 40 dm 3. he coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)? 10 d) Variable a Countercurrent. Plot X, X e,, a and rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 20 dm 3. he coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?
Heat Exchange Example: PBR A B 5) Parameters For adiabatic: Ua = 0 Constant a : d a = dw 0 Co-current: Equations as is Counter-current: d dw ( 1) (or flip - a to a - ) 11
Reversible Reactions 1) Mole Balances dx dw = raʹ F A0 (1) W = ρ b V dx dv = rʹ Aρ F A0 B = r F A A0 12
Reversible Reactions 13 2) Rate Laws ) (2 = C B A A K C C k r (3) 1 1 exp 1 1 = R E k k (4) 1 1 exp 2 2 Δ = R H K K Rx C C
Reversible Reactions 3) Stoichiometry 5 ( ) C A = C A0 1 X ( ) p 0 ( ) Note: Nomenclature change for 5th edition p y ( 6) C B = C A0 Xp( 0 ) F = F 0 dp dw = α p F F 0! # " 0 $ & = α % 2 p! # " 0 $ & % W = ρv 14 dp dv = αρ b 2 p! # " 0 $ & %
Reversible Reactions Parameters F A0 ΔH, k Rx 1, E,, 2, C R, A0 1,, 0 K C 2,, α, ρ b (7) (15) 15
Reversible Reactions Gas Phase Heat Effects Example: PBR A B 3) Stoichiometry: Gas Phase v = v 0 ( 1+εX) P 0 P 0 ( 5) C A = F 1 X A0 ( ) v 0 ( 1+εX) ( 6) C B = C A0X 1+εX ( 7) dp dw = α 2p ( ) p 0 F F 0 ( ) ( ) P 0 P 0 = C 1 X A0 1+εX = α 0 2p 1+εX ( ) 0 p 0 16
Reversible Reactions Gas Phase Heat Effects Example: PBR 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 17
Reversible Reactions Gas Phase Heat Effects Example: PBR A B Exothermic Case: K C X e K C Endothermic Case: X e ~1 18
Reversible Reactions Gas Phase Heat Effects d dv = r A ( ) ΔH Rx ( ) Ua a F i C Pi ( ) F i C Pi = F A0 $ % Θ i C Pi + ΔC P X& ' Case 1: Adiabatic and ΔC P =0 ( = 0 + ΔH Rx)X Θ i C Pi (16A) Additional Parameters (17A) & (17B) 19 0, Θ i C Pi = C PA + Θ I C PI
Reversible Reactions Gas Phase Heat Effects Case 2: Heat Exchange Constant a Heat effects: d dw = ( r )( ΔH ) ( ) A F A Rx 0 Ua ρb θ C i Pi a ( 9) 20
Reversible Reactions Gas Phase Heat Effects Case 3. Variable a Co-Current da = dv ( ) Ua mc! P cool a, V = 0 a = ao (17C) Case 4. Variable a Countercurrent d dv a Ua mc! ( ) a = V a P cool = 0 =? Guess a at V = 0 to match a0 = a0 at exit, i.e., V = V f 21
22
23
24
25
26
Endothermic PFR A B dx dv = k 1 1+ 1 X K C, X e = K C υ 0 1+ K C X e X X EB X EB = Θ i C P i ( 0 ) ( ) = C P A + Θ I C PI 0 ΔH Rx ΔH Rx 27 = 0 + ( ΔH Rx)X C PA + Θ I C PI
Adiabatic Equilibrium Conversion on temperature Exothermic ΔH is negative Adiabatic Equilibrium temperature ( adia ) and conversion (Xe adia ) X X e adia = 0 + ( ΔH ) C Rx PA X X e = KC 1+ K C 28 adia
Q 1 Q 2 F A0 F A1 F A2 F A3 X 0 1 X 0 0 3 X 2 29
X X e X 3 X 2 X EB ( ) θ C = i Pi 0 ΔH Rx X 1 0 30
31
Gas Flow Heat Effects rends: Adiabatic X exothermic X endothermic Adiabatic and X e 32 0 = 0 ΔH + C + Θ PA Rx I X C PI 0
33 Effects of Inerts in the Feed
Endothermic First Order Irreversible 34 k 1 + Θ I As inert flow increases the conversion will increase. However as inerts increase, reactant concentration decreases, slowing down the reaction. herefore there is an optimal inert flow rate to maximize X.
Gas Phase Heat Effects Adiabatic: As 0 decreases the conversion X will increase, however the reaction will progress slower to equilibrium conversion and may not make it in the volume of reactor that you have. X X e X X 35 0 herefore, for exothermic reactions there is an optimum inlet temperature, where X reaches X eq right at the end of V. However, for endothermic reactions there is no temperature maximum and the X will continue to increase as increases.
Gas Phase Heat Effects Effect of adding inerts Adiabatic: X X Θ I = V 1 V 2 X e 0 Θ I = 0 X 36 ( ) C pa +θ I C pi X = 0 [ ] ΔH Rx
Exothermic Adiabatic k θ I As θ I increase, decrease and dx dv = k ( ) υ 0 Hθ I 37
38
39
Adiabatic Exothermic K C X e k X e X V V V V V Endothermic V K C V X e V k V X e X Frozen V OR X e X V 40
Heat Exchange Exothermic K C X e X e X V V V V Endothermic K C X e X e X V V V V 41
End of Web Lecture 21 End of Class Lecture 17 42