Thermochemistry for Nonisothermal Reactor Design FA L12-1 Review

Explore the application of thermochemistry in designing nonisothermal reactors for liquid-phase exothermic reactions. Understand the energy balance, rate laws, and stoichiometry involved. Learn about terms in energy balance, steady-state conditions, and relating temperature to conversion in reactor systems. Dive into the fundamentals of heat and work interactions in reactor design.

• Thermochemistry
• Reactor Design
• Energy Balance
• Exothermic Reactions
• Rate Laws

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1. Review: Thermochemistry for Nonisothermal Reactor Design FA L12-1 FA0 XA = 0.7 Consider an exothermic, liquid-phase reaction operated adiabatically in a PFR (adiabatic operation- temperature increases down length of PFR): k A B E 1 R T 1 T dX dV r ERT A A = 1 = k k exp Mole balance: = k Ae 1 F A0 Arrhenius Equation E 1 1 R T 1 k C (1 X ) dX dV dX dV (1 X ) T A0 C A A = A A r kC = Rate law: = k exp A A 1 A0 0 0 = F A 0 C v = Need relationships: X T V Stoichiometry: A = The energy balance provides this relationship 0 (1 X ) C C A A0 A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

2. L12-2 Review: Terms in Energy Balance dE n n s sy ys s dt = + Q W F E F E i i i i in out = i 1 = i 1 Rate of accum of energy in system n W = Rate of work done by syst energy added to syst by mass flow in energy leaving syst by mass flow out Heat in - = - + n P : pressure iV specific volume WS: shaft work = + + FPV FPV W i i i i s in out i 1 = i 1 Flow work Internal energy is major contributor to energy term dE Q- F( dt i H E U i i n n s sy ys s = + + + W U PV ) - F( U P V ) i s i i i U n i n + i i i out = i 1 = 1 i = PV i n = + 0 Q W i0 i0 F H i i FH Steady state: s i 1 = Energy & work added by flow in i 1 = Heat in shaft work Energy & work removed by flow out Accum of energy in system - =0= - + Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

3. L12-3 Review: Relate T to Conversion n s i 1 n = + 0 Q W i0 i0 F H F H Steady state: i i = = i 1 F F ( ) i0 = + i A0 F = + i = F F X F F X where i i0 A i A0 i A i A0 dE n n s sy ys s dt ( ) = + + If XA0=0, then: Q W H i A0 F H F X s i0 i A0 i i A = i 1 = i 1 dE n n Multiply out: sys dt ( ) = + + Q W H F H F i A0 i HF X s i0 i A0 i A0 i A = i 1 = i 1 dE Total energy balance (TEB) n n s sy ys s dt ( ) = + i i i H F Q W F H H X s A0 i 0 i A0 A = i 1 = i 1 i ( ) T H heat of reaction 0 at steady state RX n n ( ) ( ) T F = + 0 Q W A0 F H H H X s i0 i i RX A0 A i 1 = i 1 = Energy & work added by flow in Energy & work removed by flow out Heat in shaft work Accum of energy in system - = - + Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

4. L12-4 Review: Q in a CSTR CSTR with a heat exchanger, perfectly mixed inside and outside of reactor FA0 = Q UA T ( T) Ta a Ta T, X T, X The heat flowto the reactor is in terms of: Overall heat-transfer coefficient, U Heat-exchange area, A Difference between the ambient temperature in the heat jacket, Ta, and rxn temperature, T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

5. L12-5 Review: Tubular Reactors (PFR/PBR): Integrate the heat flux equation along the length of the reactor to obtain the total heat added to the reactor : A V = = Q U(T T)dA Ua(T T)dV a a a: heat-exchange area per unit volume of reactor A a V = Heat transfer to a perfectly mixed PFR in a jacket For a tubular reactor of diameter D, a = 4 / D dQ dV= Ua(T -T) a For a jacketed PBR (perfectly mixed in jacket): 1 dQ dV dQ dW Ua(T = = Heat transfer to a PBR T) a b b Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

6. L12-6 L12: Nonisothermal Reactor Design Goal: Use TEB to design nonisothermal steady-state reactors Steady state total energy balance (TEB): n s A0 i 1 Needs to be simplified before we can apply it to reactor design dE n s sy ys d t s ( ) ( ) T = = + 0 Q W F H H H F X i0 i i RX A0 A = = i 1 ( ) T TR = + At a particular temperature: H H (T ) C dT no phase change i i R pi T Ti0C dT T T T T = i 0 = + + H H H (T ) C dT H (T ) C dT pi i i 0 i R pi i R pi R R (Hi Hi0) = - (Hi Hi0) dE T n s sy ys s dt ( ) T F = Substitute Q W F i p,i C dT H X s A0 RX A0 A = i 1Ti0 T n For a SS nonisotherm flow reactor: Constant (average) heat capacities : ( ) T F = i p,i C dT 0 Q W F H X s A0 RX A0 A = i 1Ti0 n A0 i 1 = = i p,i C 0 Q W F T T H (T)F X s i0 RX A0 A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

7. Relating HRX(T) to HRX(TR) and Overall Change in Heat Capacity T n s A0 i p,i i 1Ti 0 L12-7 ( ) T = + 0 Q W F C dT H A0 F X RX A = n n ( ) T T T ( ) T T TR = + H H (T ) C dT = i i H (T ) + i pi C H d T i R X RX R P RX R R = i 1 = 1 n = i pi C overall heat capaci t y: C P = i 1 H n ( ) ( ) = overall heat of reaction at reference t emp: T i i H T RX R R = i 1 ( ) T n T TR = + + 0 Q W F i p,i C dT H ( T ) C dT F X s A0 R X R P A0 A = i 1 Ti0 Only considering constant (average) heat capacities: = i 1 T = reaction temp Ti0 = initial (feed) temp n ( ) C = + 0 Q W F i p C T T H (T ) T T F X s A0 ,i i0 R X R P R A0 A TR= reference temp Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

8. L12-8 Solving TEB for Conversion Always start with this TEB: n ( ) = + 0 Q W F i p,i C T T H (T ) C T T F X s A0 i0 RX R P R A0 A = i 1 Rearrange to isolate terms with XA on one side of eq: n ( ) + = + F i p,i C T T W Q H (T ) C T T F X A0 i0 s RX R P R A0 A = i 1 n Solve for XA: i p,i C + F T T W Q A0 i0 s i 1 = = X A ( ) + H (T ) C T T F RX R P R A0 Plug in Q for the specific type of reactor, and solve this eq simultaneously with design equation Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

9. L12-9 Solving TEB for XA for an Adiabatic Rxn s A0 i p,i i0 i 1 = Rearrange: n A0 i p,i i0 s i 1 n ( ) = + 0 Q W F C T T H (T ) C T T F X RX R P R A0 A ( ) = + F C T T Q W H (T ) C T T F X RX R P R A0 A = Which term in this equation is zero because we re solving for an adiabatic reaction? a) dEsys/dt b) c) When the reaction is adiabatic (Q=0): d) e) FA0 None of the above Q n ( ) = + F i p,i C T T Q W H (T ) C T T F X A0 i0 s RX R P R A0 A = i 1 n ( ) = + F i p,i C T T 0 W H (T ) C T T F X A0 i0 s RX R P R A0 A = i 1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

10. L12-10 Solving TEB for XA for an Adiabatic Rxn s A0 i p,i i0 i 1 = Rearrange: n A0 i p,i i0 i 1 n ( ) = + 0 Q W F C T T H (T ) C T T F X RX R P R A0 A ( ) = + F C T T Q W H (T ) C T T F X s RX R P R A0 A = When shaft work can be neglected ( =0) and the reaction is adiabatic (Q=0): = i 1 n A0 i p,i i 1 H (T ) + i p,i i0 i 1 H (T ) C T T + T = reaction temp Ti0 = initial (feed) temperature n ( ) = + F i p,i C T T 0 0 H (T ) C T T F X A0 i0 RX R P R A0 A F C T T Solve for XA: i0 = = X A ( ) C T T F RX R P R A0 Solve this eq simultaneously with design equation Design eqs do not change, except k will be a function of T n C T T = = X A ( ) RX R P R TR= reference temp Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

11. L12-11 Nonisothermal Adiabatic Operation Constant or mean heat capacities n Q W F C ( T T )F X H ( T ) C( T T ) + = 0 s A i pi i A RX R p R 0 0 0 i = 1 Q W = = 0 0 For a system with no shaft work ( ) & adiabatic operation ( ): s n Q = 0 i pi C (T T ) i0 i 1 = = X W = 0 Xenergy balance + s H (T ) C (T T ) RX R p R CSTR, PFR, PBR, Batch C( T T ) H ( T ) Usually, Temperature p R RX R Adiabatic exothermic reactions Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

12. L12-12 Nonisothermal CSTR F X A0 r Design equation (From mass balance) : = V A Coupled Energy balance: n + = Q W F i pi C (T T ) F X H (T ) C (T T ) 0 s A0 i0 A0 RX R p R = i 1 With the exception of processes involving highly viscous materials, the work done by the stirrer can be neglected (i.e. ) = a Q UA(T With heat exchanger: A0 i pi i0 i 1 n a i pi i0 i 1 = W = 0 s T) n + = UA(T T) 0 F C (T T ) F X H (T ) C (T T ) 0 a A0 RX R p R = = + + UA(T T) F C (T T ) F X H (T ) C (T T ) A0 A0 RX R p R n UA(T T) a = + + i pi C (T T ) X H (T ) C (T T ) i0 RX R p R F = i 1 A0 n UA(T T) = a + X H (T ) C ( T T ) i pi C (T T ) RX R p R i0 F = i 1 A0 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

13. L12-13 Application to CSTR Case 1: Given FA0, CA0, A, E, Cpi, H I, and XA, calculate T & V a) b) c) Solve TEB for T at the exit (Texit = Tinsidereactor) Calculate k = Ae-E/RT where T was calculated in step a Plug the k calculated in step b into the design equation to calculate VCSTR Case 2: Given FA0, CA0, A, E, Cpi, H I, and V, calculate T & XA a) b) c) Solve TEB for T as a function of XA Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT ) Plot XA,EB vs T & XA,MB vs T on the same graph. The intersection of these 2 lines is the conditions (T and XA) that satisfies the energy & mass balance XA,EB = conversion determined from the TEB equation XA,MB = conversion determined using the design equation XA,exit Intersection is T and XA that satisfies both equations XA,MB XA XA,EB T Texit Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

14. L12-14 Application to a Steady-State PFR FA0 FA XA T distance Negligible shaft work ( S=0) and adiabatic (Q=0) a) b) c) d) Use TEB to construct a table of T as a function of XA Use k = Ae-E/RT to obtain k as a function of XA Use stoichiometry to obtain rA as a function of XA Calculate: XA A0 XA0 dX A = V F ( ) r X ,T A A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

15. A first order reaction A(l) B(l) is to be carried out adiabatically in a CSTR. Given A, E, T0, 0, CA0, and FA0, find the reactor volume that produces a conversion XA. The heat capacities of A & B are approximately equal, & S=0. a) Solve TEB for T: s A0 i p,i i0 i 1 = 0 0 L12-15 n ( ) = + 0 Q W F C T T H (T ) C T T F X RX R P R A0 A n ( ) i p,i C = + A0 F T T H (T ) C T T A0 F X i0 RX R P R A i 1 = n ( ) i p,i C = + T T H (T ) C T T X Multiply out i0 RX R P R A = i C 1 n i p,i = + C T T i p,i i0 C T H (T ) X C T X X Isolate T RX R A P A P R A = i 1 n n i p,i C + = P R C T X + + i p,i i C T C T X H (T )X T P A RX R A A 0 i 1 = i 1 = n i p,i C + = P R C T X + + T C X H (T )X p,A A0 C T Factor out T P A RX R A A i 1 = Temp when specified XA is reached Plug in values ( Cp, H RX(TR), Cp,i) given in problem statement (look them up if necessary) & solve n + P R C T + i p,i i0 C T H (T ) X X RX R A A i 1 = = T n i p,i C + C X P A = i 1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

16. A first order reaction A(l) B(l) is to be carried out adiabatically in a CSTR. Given A, E, T0, 0, CA0, and FA0, find the reactor volume that produces a conversion XA. The heat capacities of A & B are approximately equal, & S=0. a) Solve TEB for T of reaction when the specified XA is reached: L12-16 n P R C T X + + i p,i i0 C T H (T )X RX R A A i 1 = = T n i p,i C + C X P A i 1 = b) Calculate k = Ae-E/RT where T was calculated in step (a) Look up E in a thermo book c) Plug the k calculated for the reaction s temperature when the specified XA is reached (in step b) into the design equation to calculate VCSTR A0 0 F X r F X F X 1 X C X A0 A A0 kC A A0 A A = = = = V V V V ( ) ( ) kC kC 1 X A A A0 A A0 A X 0 A = V ( ) k 1 X A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

17. L12-17 Now, the first order reaction A(l) B(l) is carried out adiabatically with and inlet temp of 300 K, CPA = 50 cal/mol K, and the heat of reaction = -20,000 cal/mol. Assume S=0. The energy balance is: A0 i p,i i0 s i A A0 R H (T ) C T T F + 0 0 n + F C T T W Q ( ) T ( ) = + H H (T ) C T T = 1 = RX RX R P R X ( ) RX R P n ( ) n i pi C T T ( ) C T T 0 i pi C 1 C = P 0 p i 1 = A = = X X A i 1 = EB EB ( ) T ( ) T H H RX RX From thermodynamics ( 20000 ) XEB 50 T 300 = X EB From energy balance T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

18. L12-18 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K Start with SS EB & solve for T: s 0 Q W = n A0 i p,i i 1 = A0 i p,i i0 R i 1 = i p,i i0 RX R P i 1 = n i p,i P A RX i 1 = H (T )X T C = CpI = 15 cal/mol K n i p,i C F T T H (T)F X A0 i0 RX A0 A i 1 = = 0 0 0 F C T T H (T)F X i0 RX A0 A n ) ( ) ( ) ( = + F C T T H T C T T F X X R P R A 0 A n Multiply out brackets & bring terms containing T to 1 side ( ) = + C T T H (T ) C T T X R A n + = P R C T X + + i p,i i0 C T C T C TX H (T )X R A A i 1 = n P R C T X + + i p,i i0 C T RX R A A i 1 = = n i p,i + C X P A i 1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

19. L12-19 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K = CpI = 15 cal/mol K n + + i p,i C H (T ) X C T X T R X R A P R A i0 i 1 = T Start with SS EB & solve for T: n i p,i C + C X P A i 1 = 1 2 cal cal bC a = C 0 = = Cp 30 15 Cp C p p p mol K mol K ( T B B A n cal cal cal o b a ( ) ( ) = = = i p,i C = 1 0 1 = 1 15 + 1 15 30m ) T R A B I mo l K ) mo ) l K l K = i 1 d a c a ( ) ( ) ( ( = + H T H T H T H H RX R D R C R R A cal mol 1 2 cal mol cal m ( ) ( ) = 5000 = 50,000 H T HR T 20,000 RX R X R o l Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

20. L12-20 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be the temperature inside of a steady-state CSTR that achieved XA= 0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K = CpI = 15 cal/mol K n + + i p,i C H (T ) X C T X T R X R A P R A i0 i 1 = T Start with SS EB & solve for T: n i p,i C + C X P A i 1 = n cal cal mol ( ) = C 0 i p,i C = = 30mol K H T 5000 p RX R i 1 = cal mol cal cal mol cal mol 5000 + + X 0 30mol 294 K + 5000 X 8820 A A K = = T T ca l cal + 30mol K 0 30mol K ( ) = T 427.3K = + T 166.67K 0.8 294K = X 0.8 = X 0.8 A A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

21. L12-21 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. What would be volume of the steady-state CSTR that achieves XA= 0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K CpI = 15 cal/mol K F X = CSTR V A0 -r A Solve the CSTR design eq for V at XA = 0.8 & T = 427.3K: A C X ( ) = = 2 A0 2 0 A X Combine : V Ar kC = 1 X Stoichiometry :C C ( ) C STR A 2 A A0 A k C 1 A0 A 3 dm mol s 10,000cal mol 1.987cal mol K 350K 1 1 = Need at 427.3K: k k 0.02 exp 427.3 3 3 dm dm mol s ( ) = k 0.2696mol s = k 0.02 exp 2.60124 3 dm s ( ) 5 0.8 = 3 = CSTR V 370.9dm CSTR V 3 dm mol s mol dm ( ) 2 1 0.8 0.2696 1 3 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

22. L12-22 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. Use the 2-point rule to numerically calculate the PFR volume required to achieve XA=0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K Use the energy balance to construct table of T as a function of XA For each XA , calculate k, -rA and FA0/-rA Use numeric evaluation to calculate VPFR XA T(K) k(dm3/mol s) 0 294* 0.8 427.3* 0.2696* *Calculated in CSTR portion of this problem 3 dm 1 k 0.02 exp 5032.7126K mol s 350K dm r 0.00129 mol s 3 dm mol r 0.2696 1 1 0.8 mol s dm CpI = 15 cal/mol K -rA(mol/dm3 s) 0.00129 0.010784 FA0/-rA(dm3) 0.00129 3 1 dm = = k 0.00129mol s 294 3 2 mol dm mol dm s ( ) ( ) 2 2 = 1 0 = = 1 X 2 1 r 0.00129 r kC A A A A0 A 6 3 = = X 0 X 0 A A 2 mol dm s ( ) 2 = = r 0.010784 A A 6 3 = = X 0.8 X 0.8 A A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

23. L12-23 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. Use the 2-point rule to numerically calculate the PFR volume required to achieve XA=0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K Use the energy balance to construct table of T as a function of XA For each XA , calculate k, -rA and FA0/-rA Use numeric evaluation to calculate VPFR XA T(K) k(dm3/mol s) 0 294 0.00129 0.8 427.3 0.2696 CpI = 15 cal/mol K -rA(mol/dm3 s) 0.00129 0.010784 FA0/-rA(dm3) 3876 463.6 3 mol dm dm s mol s = = F 1 5 5 = F C A0 3 A0 A0 0 mol s mol s 5 5 F F = = = = 3 3 3876 dm 463.6 dm A0 r A0 r mol mol 0.00129dm s 0.010784dm s A A = = X 0 X 0.8 A A 3 3 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.

24. L12-24 The irreversible, elementary liquid-phase reaction 2A B is carried out adiabatically in a flow reactor with S=0 and without a pressure drop. The feed contains equal molar amounts of A and an inert liquid (I). The feed enters the reactor at 294 K with 0 = 5 dm3/s and CA0= 1 mol/dm3. Use the 2-point rule to numerically calculate the PFR volume required to achieve XA=0.8? Extra info: E = 10,000 cal/mol CpA= 15 cal/mol K CpB= 30 cal/mol K HA (TR) = -20 kcal/mol HB (TR) = -50 kcal/mol HI (TR) = -15 kcal/mol k = 0.02 dm3/mol s at 350 K Use the energy balance to construct table of T as a function of XA For each XA , calculate k, -rA and FA0/-rA Use numeric evaluation to calculate VPFR XA T(K) k(dm3/mol s) 0 294 0.00129 0.8 427.3 0.2696 CpI = 15 cal/mol K -rA(mol/dm3 s) 0.00129 0.010784 FA0/-rA(dm3) 3876 463.6 X 1 h ( ) ( ) ( ) 2-point rule: f x dx f X f X where h X X h 0 8 . h 0 8 . = + = = = 0 0 1 1 0 2 X 0 . 0 83876 2 3 3 V dm . dm = + 463 6 PFR 3 V dm = 1736 PFR Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.