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L15-1 L15: Nonisothermal Reactor Example Problems Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Multiple Steady States in CSTR 1 L15-2 XA,EB XA,MB 0.8 XA 0.6 0.4 0.2 0 0 100 200 300 400 500 600 T (K) • Plot of XA,EB vs T and XA,MB vs T • Intersections are the T and XA that satisfy both mass balance (MB) & energy balance (EB) equations • Each intersection is a steady state (temperature & conversion) • Multiple sets of conditions are possible for the same reaction in the same reactor with the same inlet conditions! Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-3 Review: Heat Removal Term R(T) & T0 Heat removed: R(T) Heat generated G(T) r V Cp0 1 T TC HRX A FA0 R(T) line has slope of CP0(1+) UA Cp0FA0 Tc Ta T0 1 R(T) =∞ =0 R(T) Increase Increase T0 T When T0 increases, slope stays same & line shifts to right For Ta < T0 Ta T0 T When increases from lowering FA0 or increasing heat exchange, slope and x-intercept moves Ta<T0: x-intercept shifts left as ↑ Ta>T0: x-intercept shifts right as ↑ =0, then TC=T0 =∞, then TC=Ta Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-4 Review: CSTR Stability G(T) Tc R(T) Ta T0 UA Cp0FA0 1 R(T) > G(T) → T falls to T=SS1 G(T) > R(T) → T rises to T=SS1 G(T) > R(T) → T rises to T=SS3 2 Heat removed: R(T) 1 3 R(T) > G(T) →T falls to T=SS3 Heat generated G(T) rA V FA0 Cp0 1 T TC HRX T • Magnitude of G(T) to R(T) curve determines if reactor T will rise or fall • G(T) = R(T) intersection, equal rate of heat generation & removal, no change in T • G(T) > R(T) (G(T) line above R(T) on graph): rate of heat generation > heat removal, so reactor heats up until a steady state is reached • R(T) > G(T) (R(T) line above G(T) on graph): rate of heat generation < heat removal, so reactor cools off until a steady state is reached Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. The elementary irreversible reaction A + B →C is carried out adiabatically in a flow L15-5 reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: liquid phase rxn CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K Strategy: 1. Mole balance 2. Rate Law 3. Stoichiometry 4. Combine rate law & stoichiometry 5. Energy balance 6. Solve 6a. Solve CSTR i. Use EB to find T as a function of XA ii. Calculate V using the CSTR design eq with k calculated at that T 6b. Solve PFR i. Use EB to construct table of T as a function of XA ii. Use k = Ae-E/RT to construct table of k as function of T & therefore XA iii. Calculate -rA as a function of T & XA iv. Calculate FA0/-rA for each T v. Use numeric technique to calculate V Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-6 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K CSTR Mole balance Rate law F X VCSTR A0 A rA PFR 0.85 F A0 VPFR dX A 0 rA rA kCACB dm3 1 10000cal mol 1 k 0.01 exp mol s 1.987cal mol K 300K T Stoichiometry Combine: CB CA CA0 1 X A 10000 1 1 2 2 rA 0.01exp C 1 X A0 A 1.987cal mol K 300K T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-7 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K Combine MB, rate law & stoichiometry for CSTR: CSTR VCSTR F X V A0 A rA CA0u0 X A dm3 1 10000cal mol 1 2 2 0.01 exp C 1 X A0 A mol s 1.987cal mol K 300K T VCSTR u0 XA dm3 1 2 1 0.01 exp 5032.7K CA0 1 X A mol s 300K T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-8 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K Energy balance n 0 Q Ws FA0 iCp,i T Ti0 HRX (T)FA0 X A Solve for T: i1 0 0 n FA0 iCp,i T Ti0 HRX (T)FA0 X A i1 HRX Substitute: H RX (TR ) CP T TR n iCp,i T Ti0 H RX (TR ) CP T TR X A i1 Multiply out quantities in brackets, bring T to 1 side of equation, factor out T, divide by quantity in bracket: n Evaluate Cp: iCp,iTi0 H RX (TR )X A CPTR X A C C C C P pC pA pB T i1 n cal CP 30 15 15 0 iCp,i CP X A mol K i1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-9 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K Energy balance n iCp,iTi0 H RX (TR )X A CPTR XA T i1 n iCp,i CP X A i1 Evaluate SiCp: FB0 1 A 1 B FA0 CP 0 T Ti0 H RX (TR )X A n iCp,i i1 n cal cal 30 iCpi CpA CpB 15 15 mol K mol K i1 Evaluate H°RX(TR): cal cal H RX TR HC HA HB 41000 20000 15000 6000 mol mol Simplify EB: T 300K 6000 cal mol X A T 300K 200K X A 30 cal mol K Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-10 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Extra info: CpA=CpB=15 cal/mol•K CpC= 30 cal/mol E = 10,000 cal/mol HA°(273K)= -20 kcal/mol HB°(273K)= -15 kcal/mol HC°(273K)= -41 kcal/mol k 0.01 dm3 mol s at 300K Solve for CSTR volume: use EB to find T when XA=0.85 T 300K 200K X A VCSTR T 300K 200K 0.85 T 470K u0 XA dm3 1 2 1 0.01 exp 5032.7K CA0 1 X A mol s 300K T dm3 2 0.85 s VCSTR dm3 1 mol 2 1 0.01 exp 5032.7K 0.1 1 0.85 mol s 300K 470K dm3 VCSTR 175L Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-11 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA (We’re interested in range where XA = 0 to XA = 0.85) T 300K 200K X A XA T (K) 0 300 T 300K 200K 0 300K k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3) 0.2 0.4 0.6 0.8 0.85 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-12 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 T 300K 200K X A ii. XA T (K) 0 300 0.2 340 0.4 380 0.6 420 0.8 460 0.85 470 k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3) Calculate k(T) for each T in the table dm3 1 1 k 0.01 exp 5032.7K mol s 300K T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-13 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 T 300K 200K X A ii. k (dm3/mol·s) -rA (mol/dm3·s) XA T (K) 0 300 0.01 0.2 340 0.072 0.4 380 0.34 0.6 420 1.21 0.8 460 3.42 0.85 470 4.31 FA0/-rA (dm3) Calculate k(T) for each T in the table dm3 1 1 k 0.01 exp 5032.7K mol s 300K T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-14 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 ii. Calculate k(T) iii. k (dm3/mol·s) -rA (mol/dm3·s) XA T (K) 0 300 0.01 0.2 340 0.072 0.4 380 0.34 0.6 420 1.21 0.8 460 3.42 0.85 470 4.31 FA0/-rA (dm3) Calculate –rA each XA and k in the table rA kCA02 1 XA 2 2 dm3 mol mol 2 rA 0.01 0.1 1 0 0.0001 3 mol s dm dm3 s Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-15 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 ii. Calculate k(T) iii. k (dm3/mol·s) -rA (mol/dm3·s) XA T (K) 0 300 0.01 0.0001 0.2 340 0.072 0.00046 0.4 380 0.34 0.00122 0.6 420 1.21 0.00194 0.8 460 3.42 0.0014 0.85 470 4.31 0.00097 FA0/-rA (dm3) Calculate –rA each XA and k in the table rA kCA02 1 XA 2 2 dm3 mol mol 2 rA 0.01 0.1 1 0 0.0001 3 mol s dm dm3 s Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-16 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 ii. Calculate k(T) iii. Calculate –rA each XA and k in the table iii. k (dm3/mol·s) -rA (mol/dm3·s) XA T (K) 0 300 0.01 0.0001 0.2 340 0.072 0.00046 0.4 380 0.34 0.00122 0.6 420 1.21 0.00194 0.8 460 3.42 0.0014 0.85 470 4.31 0.00097 FA0/-rA (dm3) Calculate FA0/–rA for each XA in the table mol dm3 mol FA0 CA0u0 0.1 3 2 0.2 s s dm Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-17 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: i. Use EB to construct table of T as a function of XA -Temperature range should cover XA = 0 to XA = 0.85 ii. Calculate k(T) iii. Calculate –rA each XA and k in the table iii. k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3) XA T (K) 0 300 0.01 0.0001 2000 0.2 340 0.072 0.00046 434.8 0.4 380 0.34 0.00122 163.9 0.6 420 1.21 0.00194 103.1 0.8 460 3.42 0.0014 142.9 0.85 470 4.31 0.00097 206.2 Calculate FA0/–rA for each XA in the table mol dm3 mol FA0 CA0u0 0.1 3 2 0.2 s s dm Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-18 The elementary irreversible reaction A + B →C is carried out adiabatically in a flow reactor with ẆS=0. An equal molar feed of A & B enters at 300K with u0 = 2 dm3/s and CA0 = 0.1 mol/dm3. What is the PFR & CSTR volume required to achieve XA=0.85? Solve for PFR: k (dm3/mol·s) -rA (mol/dm3·s) FA0/-rA (dm3) XA T (K) 0 300 0.01 0.0001 2000 0.2 340 0.072 0.00046 434.8 0.4 380 0.34 0.00122 163.9 0.6 420 1.21 0.00194 103.1 0.8 460 3.42 0.00137 146 0.85 470 4.31 0.00097 206.2 Numeric evaluation by parts: 5-point rule for XA interval 0 to 0.8, 2-point rule for XA interval from 0.8 to 0.85: X5 X4 h1 h2 h1 X4 X0 4 f X dx f X dx f 4f 2f 4f f f f 0 1 2 3 4 4 5 3 2 X X h X X 0 VPFR 4 2 5 4 0.2 0.05 2000 4 434.8 2 163.9 4 103.1 146 146 206.2 3 2 VPFR 317L Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-19 For these conditions, what is the max T0 that would keep T≤ 550K at complete conversion? T Ti0 H RX (TR )X A n iCp,i i1 n cal iCpi 30 mol K i1 H RX TR 6000 cal mol cal 6000 XA mol T Ti0 200K X A T Ti0 cal 30 mol K T≤ 550K at complete conversion, XA=1: 550K Ti0 200K 1 350K Ti0 Max T0 is 350K Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-20 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K Need to find where G(T)=R(T) for T0 = 450K 1. Put R(T) & G(T) in terms of constants in the problem statement 2. Plot R(T) vs T and G(T) vs T on the same graph & find where they intersect R T Cp0 1 T TC UA Cp0FA0 Tc Ta T0 1 r V G T HRX A FA0 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-21 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K I = inert A I B I Put R(T) in terms of constants in the problem statement, starting with CP0: T T0 R T Cp0 1 T TC UA Cp0FA 0 Tc a 1 n cal Cp0 iCpi Cp0 A CpA ICpI A I 1 Cp0 20 30 50 mol K i1 Put in terms of constants from the problem statement: 8000 cal min K UA 2 cal mol Cp0FA 0 50 80 mol K min Put TC in terms of constants in the problem statement: 2 300K 450K T T0 Tc a Tc 1 1 2 Tc 350K Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-22 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K A I B I I = inert Find steady state temp [G(T)=R(T) ] for T0 = 450K cal R T Cp0 1 T TC Tc 350K Cp0 50 2 mol K Plug and Tc into R(T): cal R T Cp0 1 T TC R T 50 1 2 T 350K mol K cal cal R T 150 T 52500 mol K mol Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-23 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K A I B I I = inert Find steady state temp [G(T)=R(T) ] for T0 = 450K cal cal R T 150 T 52500 mol K mol Now put G(T) in terms of constants from the problem statement: r V G T HRX A FA0 E 1 1 k k1 exp R T1 T Combine: Rate law: rA kC A Stoichiometry: C A CA0 1 X A 40000 cal 6.6 10 3 1 1 mol k exp min 1.987 cal 350 T mol K 6.6 10 3 1 20130.85 1 rA exp C A0 1 X A min K 350 T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-24 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K A I B I I = inert Find steady state temp [G(T)=R(T) ] for T0 = 450K cal cal rA V R T 150 T 52500 G T HRX mol K mol F A0 Plug rate law into G(T) & simplify: G T HRX 6.6 10 3 1 20130.85 1 exp C A0 1 X A V min K 350 T C A0u0 V t u0 6.6 103 1 20130.85 1 G T HRX exp 1 X A t min K 350 T Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-25 The first order irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains A and an inert liquid in equimolar amounts, where FA0 = 80 mol/min. What is the reactor temp when the inlet temp T0 is 450K? UA= 8000 cal/min·K Ta= 300K HRX=-7500 cal/mol CpA = CpB =20 cal/mol·K Cpi =30 cal/mol·K t=100 min E=40,000 cal/mol k=6.6 x 10-3 min-1 at 350K A I B I I = inert Steady state temp [G(T)=R(T) ] for T0 = 450K cal cal 150 T 52500 HRX mol K mol 6.6 103 1 20130.85 1 exp 1 X A t min K 350 T Rearrange so can be solved with Polymath nonlinear equation solver: cal cal f T 0 52500 150 T HRX mol mol K 6.6 10 3 1 20130.85 1 exp 1 X A t min K 350 T Use design equation to get XA as an explicit equation: VkC A0 1 X A V rA FA0 X A X A t k 1 X A X A X V A C A0u0 FA0 rA t k t kXA XA t k XA t kXA t k X A 1 t k t k 1 t k X A Explicit equation for Polymath Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-26 Enter in Polymath: Select graph If you want to see this graphically, click the Graph button above Calculated values of NLE variables 1 Variable Value T 399.9425 Initial Guess 400. ( 300. -4.547E-12 < T < 500. ) f(x) Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-27 Format the graph Create a table, save it as a text file, import into Excel, & make a graph with correct labels Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-28 Plot reactor temperature as a function of feed temperature, with T0 between 350 and 450K Need to re-run the Polymath program using various T0 between 350 and 450K to find the new steady state reactor temperatures Do we need to change G(T) or R(T) when T0 changes? rA V G T HRX G(T) does not depend on T0 FA0 T T0 R T Cp0 1 T TC UA Cp0FA0 Tc a 1 CP0 and do not depend on T0, but TC does Enter equations into Polymath so that R(T) varies according to T0, & run program with varied values of T0 In Excel, create a table of T vs T0, and make a graph Run over and over again, varying T0 from 350 K to 450 K Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-29 Steady state temperature as a function of T0 T0 (K) T1 (K) T2 (K) T3 (K) 350 316.7 360 320.15 370 323.6 357.16 370.3 380 327.3 353.4 375.1 390 331.2 350.1 379.1 400 336.3 346 382.8 410 386 420 389.8 430 393.2 440 396.6 450 399.9 420 400 380 T 360 340 320 300 350 370 390 410 430 450 T0 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-30 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat generated term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) rA V G(T)= HRX T F A0 HRX T H RX TR CP T TR Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-31 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat generated term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) rA V G(T)= HRX T F A0 HRX T H RX TR CP T TR b cal CPB CPA Cp 15 15 0 mol K a cal H RX H RX TR 0 10,000 mol CP t V u0 tu0 V Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-32 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat generated term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) rA V G(T)= HRX T F A0 HRX T H RX TR CP T TR b cal CPB CPA Cp 15 15 0 mol K a cal H RX H RX TR 0 10,000 mol CP t V u0 L tu0 V V 120 min 300 36,000L min mol 0.0015 36000 L cal cal L min G(T)= 10,000 G(T) 9000 mol mol mol 60 min Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-33 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat removal term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) R T Cp0 1 T TC n cal cal cal Cp0 iCP Cp0 1 15 0 15 C 15 p0 mol K mo l K mol K i i 1 UA Cp0FA0 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-34 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat removal term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) R T Cp0 1 T TC n cal cal cal Cp0 iCP Cp0 1 15 0 15 C 15 p0 mol K mo l K mol K i i 1 cal 3200 UA min K 3.56 Cp0FA0 cal mol 15 60 mol K min Ta T0 Tc 1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-35 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 What is the value of the heat removal term (with units) when a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K? (The reactor is NOT at the steady state.) R T Cp0 1 T TC n cal cal cal Cp0 iCP Cp0 1 15 0 15 C 15 p0 mol K mo l K mol K i i 1 cal 3200 UA min K 3.56 Cp0FA0 cal mol 15 60 mol K min Ta T0 3.56 340K 310K Tc Tc 333.4K Tc 1 1 3.56 R T 15 cal cal 1 3.56 360K 333.4K R T 1819 mol m ol K Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L15-36 The elementary, irreversible reaction A(l) → B(l) is carried out in a jacketed CSTR. The feed contains pure A, & it enters the reactor at 310 K, where FA0 = 60 mol/min and u0 = 300 L/min. Extra info: UA= 3200 cal/min•K Ta= 340 K ∆HRX(TR) = -10,000 cal/mol CpA=15 cal/mol·K CpB=15 cal/mol·K t =120 min E =20,000 cal/mol k(400K) = 1 min-1 When a disturbance causes the temperature in the reactor to drift to 360 K if –rA = 0.0015 mol/L•min at 360K, cal cal G(T) 9000 and R T 1819 mol mol Will the reactor temperature heat up, cool down, or stay at 360 K? G(T) > R(T) so the reactor will heat up Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.