Review: Nonideal Flow in a CSTR

L3- Review: Nonideal Flow in a CSTR Ideal CSTR: uniform reactant concentration throughout the vessel Real stirred tank Relatively high reactant concentration at the feed entrance Relatively low concentration in the stagnant regions, called dead zones (usually corners and behind baffles) Short Circuiting Dead Zone Dead Zone

Review: Nonideal Flow in a PBR Ideal plug flow reactor: all reactant and product molecules at any given axial position move at same rate in the direction of the bulk fluid flow Real plug flow reactor: fluid velocity profiles, turbulent mixing, & molecular diffusion cause molecules to move with changing speeds and in different directions channeling Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois Dead at zones Urbana-Champaign.

Review: Residence Time Distribution RTD E(t) residence time distribution function RTD describes the amount of time molecules have spent in the reactor RTD is experimentally determined by injecting an inert tracer at t= and measuring the tracer concentration C(t) at exit as a function of time Measurement of RTD Pulse injection Reactor X C(t) Detection tracer conc at exit between t & t+ Δt RTD = E( t) = = C t sum of tracer conc at exit for infinite time Et= C t Fraction of material leaving reactor that has been inside reactor for a time between t & t C curve E(t)= for t< since no tracer can exit before it enters E(t) for t> since mass fractions are always positive t = t Et t

pulse of tracer was injected into a reactor, and the effluent concentration as a function of time is in the graph below. Construct a figure of C(t) & E(t) and calculate the fraction of material that spent between 3 & 6 min in the reactor t min 3 4 5 6 7 8 9 4 C g/ m 3 C(t) (g/m 3 ) 8 6 4 5 8 8 6 4 3..5.6 Plot C vs time: 4 6 8 4 t (min) To tabulate E(t): divide C(t) by the total area under the C(t) curve, which must be numerically evaluated as shown below: = C t C( t) + 4 C( t) X N Δt f x dx f 4f f 4f 3 f 4... 4f N X f 3 N = ( + + + + + + ) X Δt f( x) dx = ( f + 4f + f ) X 3 gmin gmin C( t) 47. 4.6 3 = 5 3 m m m 3 = + gmin

pulse of tracer was injected into a reactor, and the effluent concentration as a function of time is in the graph below. Construct a figure of C(t) & E(t) and calculate the fraction of material that spent between 3 & 6 min in the reactor t min 3 3 4 4 5 5 6 6 7 7 88 9 4 C g/ m 3 55 8 8 8 8 6 6 4 4 33..5.6 E(t)...6..6..8.6.44.3. gmin Ct = 5 m 3 Tabulate E(t): divide C(t) by the total area under the C(t) curve: Et 5 = = 5 Et ( ) = =. 5 E t = C t Et = =. 5 C t 8 E( t3 ) = =.6 5 Plot E(t): E(t) (min - ).5..5..5 4 6 8 4 t (min)

Review: RTD Profiles & Cum RTD Function F(t) E(t) E(t) E(t) E(t) τ Nearly ideal PFR t F(t) = E t F t = when t< F t when t F = F t = E t t t Nearly ideal CSTR F(t) t 4 τ t PBR with channeling & dead zones t (min) t CSTR with dead zones F(t)=fraction of effluent in the reactor less for than time t.8 8% of the molecules spend 4 min or less in the reactor

L3-7 Review: Relationship between E & F F(t) = fraction of effluent that has been in the reactor for less than time t t F(t) = E t E t = C t C t E(t)= Fraction of material leaving reactor that was inside for a time between t & t

L3-8 Review: Mean Residence Time, t m For an ideal reactor, the space time τ is defined as V/υ The mean residence time t m is equal toτ in either ideal or nonideal reactors t te t m E t = = te t = τ V τ t υ = = By calculating t m, the reactor V can be determined from a tracer experiment m The spread of the distribution (variance): σ ( t t m ) E ( t ) = Space time τ and mean residence time t m would be equal if the following two conditions are satisfied: No density change No backmixing In practical reactors the above two may not be valid, hence there will be a difference between them

L3-9 Significance of Mixing RTD provides information on how long material has been in the reactor RTD does not provide information about the exchange of matter within the reactor (i.e., mixing)! For a st order reaction: k X = ( ) Concentration does not affect the rate of conversion, so RTD is sufficient to predict conversion But concentration does affect conversion in higher order reactions, so we need to know the degree of mixing in the reactor Macromixing: produces a distribution of residence times without specifying how molecules of different age encounter each other and are distributed inside of the reactor Micromixing: describes how molecules of different residence time encounter each other in the reactor

L3- Quality of Mixing RTDs alone are not sufficient to determine reactor performance Quality of mixing is also required Goal: use RTD and micromixing models to predict conversion in real reactors Maximum mixedness: molecules are free to move anywhere, like a microfluid. This is the extreme case of early mixing Extremes of Fluid Mixing

L3- Quality of Mixing RTDs alone are not sufficient to determine reactor performance Quality of mixing is also required Goal: use RTD and micromixing models to predict conversion in real reactors Extremes of Fluid Mixing Maximum mixedness: molecules are free to move anywhere, like a microfluid. This is the extreme case of early mixing Complete segregation: molecules of a given age do not mix with other globules. This is the extreme case of late mixing

L3- Complete Segregation Model Flow is visualized in the form of globules Each globule consists of molecules of the same residence time Different globules have different residence times No interaction/mixing between different globules j j j Mixing of different age groups at the last possible moment The mean conversion is the average conversion of the various globules in the exit stream: X = X t E t Δt Conversion achieved after spending time t j in the reactor Δ t X = X t E t Fraction of globules that spend between t j and t j + Δt in the reactor X (t) is from the batch reactor design equation

L3-3 Complete Segregation Example First order reaction, Products Batch reactor design equation: N = r V N = kcv N = kc ( X ) V N = kn ( X ) = k ( X ) X t = e kt To compute conversion for a reaction with a st order rxn and complete segregation, insert E(t) from tracer experiment and X (t) from batch reactor design equation into: X = X t E t & integrate

L3-4 Maximum Mixedness Model In a PFR: as soon as the fluid enters the reactor, it is completely mixed radially with the other fluid already in the reactor. Like a PFR with side entrances, where each entrance port creates a new residence time: λ υ E( λ) Δλ λ υ υ λ +Δ λ υ λ V = λ +Δλ λ V = V λ: time it takes for fluid to move from a particular point to end of the reactor υ(λ): volumetric flow rate at λ, = flow that entered at λ+δλ plus what entered through the sides υ E(λ)Δλ: Volumetric flow rate of fluid fed into side ports of reactor in interval between λ + Δλ & λ Volumetric flow rate of fluid fed to reactor at λ: υ( λ) = υ λ E( λ) dλ = υ F( λ) fraction of effluent in reactor for less than time t Volume of fluid with life expectancy between λ + Δλ & λ: V F Δ = υ λ Δλ

L3-5 Maximum Mixedness & Polymath Mole balance on gives: residence time distribution function r E( λ) = + X λ ( λ) fraction of effluent in reactor for less than time t d C F E(t) must be specified Often it is an expression that fits the experimental data curves, one on the increasing side, and a second for the decreasing side Use the IF function to specify which E is used when See section 3.8 in book E E E t lso need to replace λ because Polymath cannot calculate as λ gets smaller z = T λ where T is the longest time measured ( z) r ET dz C F T z Note that the sign on = X each term changes

Review: Nonideal Flow & Reactor Real CSTRs Relatively high reactant conc at entrance Relatively low conc in stagnant regions, called dead zones (corners & behind baffles) Short Circuiting Design Real PBRs fluid velocity profiles, turbulent mixing, & molecular diffusion cause molecules to move at varying speeds & directions channeling Dead Zone Dead zones Dead Zone Goal: mathematically describe non-ideal flow and solve design problems for reactors with nonideal flow

Residence Time Distribution (RTD) RTD E(t) residence time distribution function RTD describes the amount of time molecules have spent in the reactor RTD is experimentally determined by injecting an inert tracer at t= and measuring the tracer concentration C(t) at exit as a function of time Measurement of RTD Pulse injection Et Et= Ct = = C t Reactor X C(t) Detection tracer conc at exit between t & t+ Δt sum of tracer conc at exit for infinite time Fraction of material leaving reactor that has been inside reactor for a time between t & t The C curve E(t)= for t< since no fluid can exit before it enters E(t) for t> since mass fractions are always positive t = t Et t

E(t) E(t) E(t) E(t) Nice multiple choice question τ Nearly ideal PFR t Nearly ideal CSTR t τ t PBR with dead zones t CSTR with dead zones The fraction of the exit stream that has resided in the reactor for a period of time shorter than a given value t: t t = F( t) E t = E t F t F t = when t< F t when t F =.8 F(t) is a cumulative distribution function 4 8% of the molecules spend 4 min or less in the reactor

L-9 Review: Mean Residence Time, t m For an ideal reactor, the space time τ is defined as V/υ The mean residence time t m is equal to τ in either ideal or nonideal reactors t te t m E t = = te t = τ V τ t υ = = m By calculating t m, the reactor V can be determined from a tracer experiment The spread of the distribution (variance): σ ( t t m ) E ( t ) = Space time τ and mean residence time t m would be equal if the following two conditions are satisfied: No density change No backmixing In practical reactors the above two may not be valid, hence there will be a difference between them

Review: Complete Segregation Model Flow is visualized in the form of globules Each globule consists of molecules of the same residence time Different globules have different residence times No interaction/mixing between different globules j j j Mixing of different age groups at the last possible moment The mean conversion is the average conversion of the various globules in the exit stream: X = X t E t Δt Conversion achieved after spending time t j in the reactor Δ t X = X t E t Fraction of globules that spend between t j and t j + Δt in the reactor X (t) is from the batch reactor design equation

L- Review: Maximum Mixedness Model In a PFR: as soon as the fluid enters the reactor, it is completely mixed radially with the other fluid already in the reactor. Like a PFR with side entrances, where each entrance port creates a new residence time: λ υ E( λ) Δλ λ υ υ λ +Δ λ υ λ V = λ +Δλ λ V = V λ: time it takes for fluid to move from a particular point to end of the reactor υ(λ): volumetric flow rate at λ, = flow that entered at λ+δλ plus what entered through the sides υ E(λ)Δλ: Volumetric flow rate of fluid fed into side ports of reactor in interval between λ + Δλ & λ Volumetric flow rate of fluid fed to reactor at λ: υ( λ) = υ E( λ) dλ = υ F( λ) λ fraction of effluent that in reactor for less than time t Volume of fluid with life expectancy between λ + Δλ & λ: V F Δ = υ λ Δλ

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for () an ideal PFR and () for the complete segregation model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Start with PFR design eq & see how far can we get: dv r = F kccb dv C υ = C = C ( X ) C = C ( X ) 3 kc C X = dv C υ B Get like terms together & integrate B B X V 3 ( ) kc = X υ B dv ( ) X kc B = X υ V = kc ( X ) B τ X = B kc τ +

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for () an ideal PFR and () for the complete segregation model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. t*e(t)...48.8.8.7.56.48.396.3.44 4 X = B kc τ + Use numerical method to determine t m : How do we determine τ? For an ideal reactor, τ = t m t m = te t 4 t = te t = te t + te t m + 4. +. + 4.48 +.8 + 4.8 te( t) = = 4.57 3 +.7 + 4.56 +.48 + 4.396 +.3 te( t) =.3 + 4(.44) + =.584 3 tm = 4.57 +.584 = 5.5min

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. X = X t E t X (t) is from batch reactor design eq Numerical method. Solve batch reactor design equation to determine eq for X. Determine X for each time 3. Use numerical methods to determine X Segregation model: Polymath Method. Use batch reactor design equation to find eq for X. Use Polymath polynomial curve fitting to find equation for E(t) 3. Use Polymath to determine X

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. X = X t E t X (t) is from batch reactor design eq Batch design eq: N = CV N 3 = rv N = kccb ( X ) V = kcb X Segregation model: Stoichiometry: kccb r = C = C X C = C X B B X t 3 ( X ) = = kc t ( X ) kc B B X = ( ) X kc B = X + B kc t 3 t

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. X.37.3.98.35.39.48.458.483.55.55.558.585 Segregation model: Numerical method X = X( t) E( t) X = + kc t B = +.349min t t X = X t E t = X t E t d + X t E t + 4.37. +.3. + 4.98.6 X ( t) E( t) = +.35. + 4.39.6 +.48. + 4.458.8 3 + + +.483.6 4.55.44.55.3 = X t E t.35 4

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. X.37.3.98.35.39.48.458.483.55.55.558.585 Segregation model: 4 X = X( t) E( t) X = + kc t Numerical method B = t 3 +.349min t X = X t E t d =. 5 + X t E t X ( t) E( t) = (.55)(.3) + 4(.558)(.) + (.585) =.45 3 X = X t E t =. 35 +.4 X =.39 4

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. X.37.3.98.35.39.48.458.483.55.55.558.585 lternative approach: segregation model by Polymath: X = X( t) E( t) = X X = ( t) E( t) + kc t B Need an equation for E(t) k=76 C B =.33 Use Polymath to fit the E(t) vs t data in the table to a polynomial

time E(t) For the irreversible, liquidphase, nonelementary rxn +B C+D, -r =kc C B Calculate the X using the complete segregation model using Polymath Gave best fit E(t) = at t= Model: C= a*c + a*c^ + a3*c^3 + a4*c^4 a=.88937 a= -.578 a3=.796 a4= -8.63E-6 Final Equation: E=.88937*t -.578*t +.796*t 3 8.63E-6*t 4

Complete segregation model by Polymath +B C+D -r =kc C B Segregation model by Polymath: X =.36

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, nonelementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Maximum mixedness model: r E( λ) dλ C F( λ) 3 B = + X λ=time r = kc C X C = CB =.33mol L Polymath cannot solve because λ (needs to increase) df E dλ = F(λ) is a cumulative distribution function k L = 76 mol min Substitute λ for z, where z=t -λ where T =longest time interval (4 min) r ET ( z) E must be in terms of T -z. df = + X ET dz C F( T z) ( z) dz = Since T -z=λ & λ=t, simply substitute λ for t E(λ)=.88937*λ-.578*λ +.796*λ 3 8.63E-6*λ 4

Maximum Mixedness Model, nonelementary reaction +B C+D -r =kc C B df ET dz = ( z) ( z) ( ) ET r = + dz C F T z X Denominator cannot = z = T λ λ = T z Eq for E describes RTD function only on interval t= to 4 minutes, otherwise E= X, maximum mixedness =.347

For a pulse tracer expt, C(t) & E(t) are given in the table below. The irreversible, liquidphase, nonelementary rxn +B C+D, -r =kc C B will be carried out in this reactor. Calculate the conversion for the complete segregation model under adiabatic conditions with T = 88K, C =C B =.33 mol/l, k=76 L /mol min at 3K, ΔH RX =-4 cal/ mol, E/R =36K, C P =C PB =cal/mol K & C PC =C PD =3 cal/mol K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Polymath eqs for segregation model: = X ( t) E( t) E(t)=.88937*t -.578*t +.796*t 3 8.63E-6*t Express 4 k as function of T: k T 76 L exp 36K = mol min 3K T Need equations from energy balance. For adiabatic operation: T 3 n o RX ( R ) i pi P R i= n Θ icp + X i ΔCP i= Δ H T X + Θ C T + X ΔC T = = kc X B

For a pulse tracer expt, C(t) & E(t) are given in the table below. The irreversible, liquidphase, nonelementary rxn +B C+D, -r =kc C B will be carried out in this reactor. Calculate the conversion for the complete segregation model under adiabatic conditions with T = 88K, C =C B =.33 mol/l, k=76 L /mol min at 3K, ΔH RX =-4 cal/ mol, E/R =36K, C P =C PB =cal/mol K & C PC =C PD =3 cal/mol K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. n o RX ( R ) i pi i= P R n Δ H T X + Θ C T + X ΔC T Energy balance for T = adiabatic operation: Θ icp + X i ΔCP Not zero! n i= cal Θ i C p = cal cal i C p + C P = B 4 = mol K Δ Cp = ( 3+ 3 ) = mol K mol K i T = cal cal 7 X + 576 mol mol cal cal + X mol K mol K ( t) E( t) E(t)=.88937*t -.578*t +.796*t 3 8.63E-6*t 4 = X B 3 = kc X k ( T L ) 76 exp 36K = 3K T mol min

Segregation model, adiabatic operation, nonelementary reaction kinetics +B C+D -r =kc C B X =.93

The following slides show how the same problem would be solved and the solutions would differ if the reaction rate was still -r =kc C B but the reaction was instead elementary: +B C+D These slides may be provided as an extra example problem that the students may study on there own if time does not permit doing it in class.

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, elementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Start with PFR design eq & see how far can we get: dv C = C X r = F kccb = dv C υ dτ = b ν = = C = C X a b B B kccb C = dτ kc C X X B C = kc X X dτ B C B =.33 k =.33 Could solve with Polymath if we knew the value of τ

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, elementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. t*e(t)...48.8.8.7.56.48.396.3.44 kc X X dτ = B How do we determine τ? For an ideal reactor, τ = t m t m = te t Use numerical method to determine t m : 4 t = te t = te t + te t m + 4. +. + 4.48 +.8 + 4.8 te( t) = = 4.57 3 +.7 + 4.56 +.48 + 4.396 +.3 te( t) =.3 + 4(.44) + =.584 3 tm = 4.57 +.584 = 5.5min 4

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, elementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Segregation model with Polymath: Batch reactor design eq: Stoichiometry: kccb r = N C = C X C = C X B B X X t E t = = X ( t) E( t) X (t) is from batch reactor design eq = rv N = k C X X N = CV = kc B X X k=76 C B C B =.33 Best-fit polynomial line for E(t) vs t calculated by Polymath (slide 9) E(t)=.88937*t -.578*t +.796*t 3 8.63E-6*t 4 V

Segregation model, isothermal operation, elementary rxn: +B C+D X,seg =.7

For a pulse tracer expt, the effluent concentration C(t) & RTD function E(t) are given in the table below. The irreversible, liquid-phase, elementary rxn +B C+D, -r =kc C B will be carried out isothermally at 3K in this reactor. Calculate the conversion for an ideal PFR, the complete segregation model and maximum mixedness model. C =C B =.33 mol/l & k=76 L /mol min at 3K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Maximum mixedness model: Polymath cannot solve because λ (must increase) ( λ) ( λ) r E = + X λ=time dλ C F r = kc C X X B C = C =.33mol L B df E dλ = k L = 76 mol min Substitute λ for z, where z=t -λ where T =longest time interval (4 min) r ET ( z) E must be in terms of T -z. df = + X ET dz C F( T z) ( z) dz = Since T -z=λ & λ=t, simply substitute λ for t E(λ)=.88937*λ-.578*λ +.796*λ 3 8.63E-6*λ 4

Maximum Mixedness Model, elementary reaction +B C+D, -r =kc C B df ET dz = ( z) ( z) ( ) ET r = + dz C F T z X Denominator cannot = z = T λ λ = T z Eq for E describes RTD function only on interval t= to 4 minutes, otherwise E= X, maximum mixedness =.5

For a pulse tracer expt, C(t) & E(t) are given in the table below. The irreversible, liquidphase, elementary rxn +B C+D, -r =kc C B will be carried out in this reactor. Calculate the conversion for the complete segregation model under adiabatic conditions with T = 88K, C =C B =.33 mol/l, k=76 L /mol min at 3K, ΔH RX =-4 cal/ mol, E/R =36K, C P =C PB =cal/mol K & C PC =C PD =3 cal/mol K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. Polymath eqs for segregation model: = X ( t) E( t) E(t)=.88937*t -.578*t +.796*t 3 8.63E-6*t Express 4 k as function of T: k T 76 L exp 36K = mol min 3K T Need equations from energy balance. For adiabatic operation: n o Δ H RX ( T ) R X + Θ icp T + XΔCPT i R T = i= n Θ icp + X i ΔCP i= = kc X X B

For a pulse tracer expt, C(t) & E(t) are given in the table below. The irreversible, liquidphase, elementary rxn +B C+D, -r =kc C B will be carried out in this reactor. Calculate the conversion for the complete segregation model under adiabatic conditions with T = 88K, C =C B =.33 mol/l, k=76 L /mol min at 3K, ΔH RX =-4 cal/ mol, E/R =36K, C P =C PB =cal/mol K & C PC =C PD =3 cal/mol K t min 3 4 5 6 7 8 9 4 C g/ m 3 5 8 8 6 4 3..5.6 E(t)...6..6..8.6.44.3. diabatic EB: n o Δ H ( T ) X + Θ C T + X ΔC T i T = i= n Θ icp + X i ΔCP i= RX R i p P R T = 88K + X = X ( t) E( t) ( ) Δ C = 3+ 3 = n i= E(t)=.88937*t -.578*t +.796*t 3 8.63E-6*t 4 p cal Θ i C p = i C p + C P = B 4 mol K k ( T L ) 76 exp 36K = 3K T mol min = kc X X B

Segregation model, adiabatic operation, elementary reaction kinetics +B C+D -r =kc C B Because B is completely consumed by X =.5 X =.5 Why so much lower than before?