L13-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Nonisothermal Reactor Design Steady–state total energy balance (TEB): For a SS nonisotherm flow reactor: “Simplified” TEB: Constant (average) heat capacities : Can rearrange this equation to solve for T T = reaction tempT i0 = initial (feed) temperatureT R = reference temp

L13-2 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Solve TEB for Conversion Solve for X A : Plug in Q for the specific type of reactor T = reaction tempT i0 = initial (feed) temperatureT R = reference temp For an adiabatic reaction (Q=0) and shaft work can be neglected ( Ẇ S =0)

L13-3 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Application to CSTR a)Solve TEB for T at the exit (T exit = T inside reactor ) b)Calculate k = Ae -E/RT where T was calculated in step a c)Plug the k calculated in step b into the design equation to calculate V CSTR Case 1: Given F A0, C A0, A, E, C pi, H° I, and X A, calculate T & V a)Solve TEB for T as a function of X A (make a table of T vs X A using EB) b)Solve CSTR design equation for X A as a function of T (plug in k = Ae -E/RT ) (use design eq to make a table of X A vs T) c)Plot X A,EB vs T & X A,MB vs T on the same graph. The intersection of these 2 lines is the conditions (T and X A ) that satisfies the energy & mass balance Case 2: Given F A0, C A0, A, E, C pi, H° I, and V, calculate T & X A X A,EB = conversion determined from the TEB equation X A,MB = conversion determined using the design equation XAXA T X A,EB X A,MB X A,exit T exit Intersection is T and X A that satisfies both equations

L13-4 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review: Application to a SS PFR PFR F A0 FAFA distance T XAXA Negligible shaft work ( Ẇ S =0) and adiabatic (Q=0) a)Use TEB to construct a table of T as a function of X A b)Use k = Ae -E/RT to obtain k as a function of X A c)Use stoichiometry to obtain –r A as a function of X A d)Calculate: May use numerical methods

L13-5 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. L13: Equilibrium Conversion in Nonisothermal Reactor Design The highest conversion that can be achieved in reversible reactions is the equilibrium conversion –For reversible reactions, the equilibrium conversion is usually calculated first The equilibrium conversion increases with increasing temperature for endothermic reactions The equilibrium conversion decreases with increasing temperature for exothermic reactions

L13-6 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Review of Equilibrium Kinetics K C : equilibrium constant (capital K): Gas-phase reaction: If K C is given at a single temperature T 2, &  C P can be neglected then: K P : equilibrium constant in terms of partial pressures P i : For ideal gases, K P = K C (RT) Δn where Temp dependence of K P is given by van’t Hoff’s equation: If  C P can be neglected then:

L13-7 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Equilibrium Conversion X Ae T X A,e 0 1 exothermic endothermic Example) A ⇌B C A0 =1 C B0 =0 Rearrange to solve in terms of X Ae This equation enables us to express X ae as a function of T

L13-8 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. X Ae and Temperature Substitute for K C : Divide numerator & denominator by K C Changed sign

L13-9 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. X Ae and Temperature Makes sense from Le Chatelier’s principle Exothermic rxn produces heat → increasing temp adds heat (product) & pushes rxn to left (lower conversion) Makes sense from Le Chatelier’s principle Heat is a reactant in an endothermic rxn → increasing temp adds reactant (heat) & pushes rxn to right (higher conversion)

L13-10 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. For the elementary solid-catalyzed liquid-phase reaction 1.Make a plot of equilibrium conversion as a function of temperature. 2.Determine the adiabatic equilibrium temperature and conversion when pure A is fed to the reactor at a temperature of 298 K. Adiabatic Equilibrium T Example

L13-11 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Rate law: equilibrium -r A = 0 The Van’t Hoff equation: X e = f (T) X e only depends on thermodynamics! Nothing to do with the energy balance!

L13-12 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Reaction is carried out adiabatically with an inlet temp of 298 K, C PA = 50 cal/mol∙K, & the heat of reaction = 20,000 cal/mol. The energy balance is: From thermodynamics X T From energy balance 0 0 How to increase the conversion?

L13-13 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Does increasing the entering temperature increase X A ? X Ae XAXA T (K) X A,EB for adiabatic operation (slants up for exothermic rxn) Equilibrium curve for exothermic rxn Higher T moves X A,EB curve to the right Adiabatic op, X A,EB at T i0,2 X A,e (T adiabatic ) decreases for an exothermic reaction X A,e at Ti0,1 X A,e at T i0,2

L13-14 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Optimum Feed Temperature For a reversible and exothermic reaction, the feed temperature should be optimized to maximize the conversion From thermodynamics X EB T XAXA W T 0 = 600 T 0 = 500 T 0 = 350 There is an optimum inlet temperature! High T 0 : reaction reaches equilibrium fast, but X A is low Low T 0 would give high X A,e but the specific reaction rate k is so small that most of the reactant passes through the reactor without reacting (never reach X A,e )

L13-15 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. How does one increase X A for adiabatic operation of an exothermic reaction? with interstage cooling! X EB T final conversion X A,EB1 cooling process X A,EB2 X A,EB3 X A,EB4 T0T0 Reactor 4Reactor 1 Cooling, C1C2C3 Reactor 3Reactor 2 Each reactor operates adiabatically Endothermic reactions are similar, but with heating instead of cooling

L13-16 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. The equilibrium conversion increases with increasing temperature, so use interstage heating to increase the conversion Endothermic Reactions X EB T heating process final conversion Red lines are from the energy balance, slant backwards because  H° RX >0 for endothermic reaction

L13-17 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Suppose pure A enters a reactor at 298K. What is the maximum X A achievable in an adiabatic reactor? Assume Ẇ S =0, and  C P = 0, C P.A =60 J/molK,  H° RX (T R )= 20,000 J/mol, K C =10 exp[2405T-7.2] Solve TEB for X A : 00 Plot X EB vs T and X A,e vs T to compute the maximum X A graphically

L13-18 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Suppose pure A enters a reactor at 298K. What is the maximum X A achievable in an adiabatic reactor? Assume Ẇ S =0, and  C P = 0, C P.A =60 J/molK,  H° RX (T R )= 20,000 J/mol, K C =10 exp[2405T-7.2] TX A,EB TX Ae

L13-19 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Suppose pure A enters a reactor at 298K. What is the maximum X A achievable in an adiabatic reactor? Assume Ẇ S =0, and  C P = 0, C P.A =60 J/molK,  H° RX (T R )= 20,000 J/mol (endothermic), K C =10 exp[2504T-7.2] X Ae EB for adiabatic operation, (slants down for endothermic rxn) T adiabatic Nearly 0 conversion, not good! T adiabatic : Outlet T if reactor had an infinite volume X A,e at T adiabatic is max achievable X A in adiabatic reactor

L13-20 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Does increasing the inlet temperature to 600K improve the conversion of this reaction? Ẇ S =0, and  C P = 0, C P.A =60 J/molK,  H° RX (T R )= 20,000 J/mol (endothermic), & K C =10 exp[2504T-7.2] X Ae T adiabatic : Outlet T if reactor had an infinite volume EB for adiabatic operation (endothermic) T 0 =298K T adiabatic X A,e at T adiabatic is max achievable X A in adiabatic reactor Nearly 0 conversion

L13-21 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign. Does increasing the inlet temperature to 600K improve the conversion of this reaction? Ẇ S =0, and  C P = 0, C P.A =60 J/molK,  H° RX (T R )= 20,000 J/mol (endothermic), & K C =10 exp[2504T-7.2] X Ae ≈ 0 at T 0 = 298K EB for adiabatic operation (endothermic) T 0 =298K T adiabatic EB for adiabatic operation T 0 =600K X Ae ≈ 0.2 when T 0 = 600K T adiabatic (T 0 =600K) Yes, higher conversion is achieved