Lecture 19 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.
oday s lecture Gas Phase Reactions rends and Optimums 2
Last Lecture User Friendly Equations Relate and X or F i 1. Adiabatic CSR, PFR, Batch, PBR achieve this: 3
User Friendly Equations Relate and X or F i 2. CSR with heat exchanger, UA( a -) and a large coolant flow rate a 4
User Friendly Equations Relate and X or F i 3. PFR/PBR with heat exchange F A0 0 a Coolant 3A. In terms of conversion, X 5
User Friendly Equations Relate and X or F i 3B. In terms of molar flow rates, F i 4. For multiple reactions 5. Coolant Balance 6
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 a Heat Exchange Fluid F A0 F I 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 Counter 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 = 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)?
Reversible Reactions Gas Phase Heat Effects Example: PBR A B 5) Parameters. For adiabatic: Constant a : Co-current: Counter-current: Equations as is 11
Reversible Reactions Mole Balance: 12
Reversible Reactions Rate Law: 13
Reversible Reactions Stoichiometry: ( 5) C A = C ( A 0 1 X)y( 0 ) ( 6) C B = C A 0 Xy( 0 ) dy dw = α y F F 0 0 = α 2y 0 W = ρv dy dv = αρ b 2y 0 14
Reversible Reactions Parameters: 15
Reversible Reactions Gas Phase Heat Effects Example: PBR A B 3) Stoich (gas): v = v 0 ( 1+εX) P 0 P 0 ( 5) C A = F A 0( 1 X) v 0 ( 1+εX) ( 6) C B = C A 0X 1+εX ( 7) dy dw = α 2y ( ) y 0 F F 0 ( ) ( ) P 0 P 0 = C A 0 1 X 1+εX = α 0 2y 1+εX ( ) 0 y 0 16
Reversible Reactions Gas Phase Heat Effects Example: PBR A B K C = C Be C Ae = C A 0 X e y 0 C ( A 0 1 X e )y 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 Endothermic Case: K C X e ~1 18
Adibatic Equilibrium Conversion on emperature Exothermic H is negative Adiabatic Equilibrium temperature ( adia ) and conversion (Xe adia ) X X e adia 19 adia
Q 1 Q 2 F A0 F A1 F A2 F A3 X 0 1 X 0 0 3 X 2 20
X X e X EB X 3 X 2 X 1 0 21
Example A B Gas Phase Heat Effects d dv = r A ( )( ΔH ) Rx Ua( ) a F i C Pi F i C Pi = F [ A 0 Θ i C Pi + ΔC P X] Case 1: Adiabtic and C P =0 ( = 0 + ΔH Rx)X Θ i C Pi Additional Parameters (17A) & (17B) (16A) 22 0, Θ i C Pi = C PA + Θ I C PI
Example A B Gas Phase Heat Effects Case 2: Heat Exchange Constant a Heat effects: d dw = ( r )( a ΔH ) R U a ρ b F A 0 θ i C Pi ( ) a ( 9) 23
Example A B Case 3. Variable a Co-Current Case 4. Variable a Counter Current Guess a at V = 0 to match a0 = a0 at exit, i.e., V = V f 24
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31 Effects of Inerts In he Feed
Endothermic First Order Irreversible What happens when we vary 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. 32
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Gas Phase Heat Effects rends: -Adiabatic: X exothermic X endothermic Adiabatic and X e 0 0 34
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 V 1 V 2 X e 0 X 36 ( ) C pa +θ I C pi X = 0 [ ] ΔH Rx
Exothermic Adiabatic k I As I increase, decrease and dx dv = υ 0 k ( ) Hθ I 37
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Adiabatic Exothermic Endothermic 40
Heat Exchange Exothermic Endothermic 41
Derive the Steady State Energy Balance (w/o Work) Differentiating with respect to W: 42
Derive the Steady State Energy Balance (w/o Work) Mole Balance on species i: Enthalpy for species i: 43
Derive the Steady State Energy Balance (w/o Work) Differentiating with respect to W: 44
Final Form of the Differential Equations in erms of Conversion: A: 45
Final Form of terms of Molar Flow Rate: B: 46
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48 End of Lecture 19
Gas Phase Heat Effects Example: PBR A B 1) Mole balance: 2) Rates: 49