[學(xué)習(xí)]反應(yīng)工程基礎(chǔ)程易cha

1、,回 顧,1－ 基本概念，解題思路,－ 解題思路,－ 基本環(huán)節(jié)：設(shè)計(jì)方程,CSTR,PFR,Batch,1級(jí)和2級(jí)反應(yīng),恒容變?nèi)?,,,熟記、熟知、熟查,,,,2－,膨脹因子,膨脹率,,3－理想反應(yīng)器：極端行為,CSTR,PFR,,實(shí)際反應(yīng)器,4－教學(xué)與學(xué)習(xí),A. 科學(xué)概念、知識(shí)庫(kù),B. 技能：解決問(wèn)題的能力（思路＋細(xì)節(jié)）,實(shí)驗(yàn)解析數(shù)值,(工具)i,,,,C. 相關(guān)的數(shù)學(xué),0維： CSTR 代數(shù)方程（組）,1維：

2、 PFR 常微分方程（組）,2維： 2D-PFR 偏微分方程（組）,直接求解迭代（試差）,直接解析數(shù)值解（如Matlab）,直接解析（如拉氏變換）數(shù)值解（如COMSOL）,D. 收獲：知識(shí)，方法論，信心,,,物理－建模－求解－應(yīng)用,Chapter 8 Steady-State Nonisothermal Reactor Design,Overview,Heat effects in chemical reac

3、torsMole balancesRate lawsStoichiometryEnergy balance,Objectives,Describe the algorithm for CSTRs, PFRs, and PBRs that are not operated isothermally. Size adiabatic and nonadiabatic CSTRs, PFRs, and PBRs. Use react

4、or staging to obtain high conversions for highly exothermic reversible reactions. Carry out an analysis to determine the Multiple Steady States (MSS) in a CSTR along with the ignition and extinction temperatures. Analy

5、ze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables. Use CO

6、MSOL to solve for both axial and radial temperature and concentration profiles.,8.1 Rationale: info necessary to design nonisothermal reactors,Example: highly exothermic reaction, adiabatic, plug-flow reactor,To calcul

7、ate the reactor volume necessary for 70% conversion,A ? B,1. Mole balance (design equation),,2. Rate law,Arrhenius equation,,,3. Stoichiometry (liquid phase),,4. Combining,,,T varies along the length of the reactor, k wi

8、ll also vary,,To solve this, we need another relationship between X and T, or T and V,5. Energy balance,,T0: entering temperature,?HRx: heat of the reaction,CPA: heat capacity,,8.2 The energy balance,8.2.1 First law of t

9、hermodynamics,For a closed system: no mass crosses the system boundaries,,Total energyof the system,Heat flowto the system,Work done by the system on the surroundings,? : not exact differentials of a state function,,,

10、,Open system: a continuous-flow reactor,,Energy balance on an open system,Rate of accumulation of energy within the system,=,Rate of flow of heat to the system from the surroundings,-,Rate of work done by the system on

11、 the surroundings,+,Rate of energy added to the system by mass flow into the system,-,Rate of energy leaving the system by mass flow out of the system,Unit: Joule/s,,Energy balance on a well-mixed open system: schematic,

12、,,,,,,The starting point,,,,How to express these terms?,,8.2.2 Evaluating the work term,,Flow work and Shaft work,P: pressure (Pa),,Flow work unit:,,,Flow work: combined with those terms in the energy balance that repre

13、sent the energy exchange by mass flow across the system boundaries,,In almost all chemical reactor situations, the kinetic, potential, and “other” energy terms are negligible in comparison with the enthalpy, heat transfe

14、r, and work terms, and hence:,,,Enthalpy:,J/mol,,,Enthalpy:,,,0: inlet conditions,8.2.4 Dissecting the steady-state molar flow rates to obtain the heat of reaction,Steady-state energy balance,,,,In:,,Out:,Expressing mol

15、ar flow rates in terms of conversion,Reaction:,,,,Heat of reaction at temperature T,,Steady-stateenergy balance,,,,8.2.5 Dissecting the enthalpies,Calculate enthalpy when phase changes are involved,Enthalpy of species

16、 i inthe gas at T,=,Enthalpy of formation of species i in the solid phase at TR,?HQ in heating solid from TR to Tm,+,+,Heat of melting at Tm,+,?HQ in heating gas from Tb to T,+,Heat of vaporation at Tb,+,?HQ in

17、heating liquid from Tm to Tb,+,Example:,,If no phase change:,,,,,,Mean heat capacity,Phase change considered.,,When reacting fluid is heated without phase change from entrance temperature, Ti0, to a temperature T:,,,,,,A

18、ssumption: CPi=const. (or mean),,,The next ...,,,Already known!,8.2.6 Relating ?HRx(T), ?H?Rx(TR) and ?CP,,,,,,,,Heat of reaction at temperature T,,Energy balance in terms of mean or constant heat capacities,,,,Last page

19、,See example: P338-339,i.e.,8.3 Adiabatic operation,8.3.1 Adiabatic energy balance,Assume,,T,XEB,,CSTRPFRPBRBatch,,X~T: Linear relationship,XEB: conversion from Energy Balance,,small,Example: Effect of inerts,The foll

20、owing conversion temperature relationship is for an adiabatic reaction, A ? B, containing 50% inerts:,Sketch X vs. T forwhen the inerts are increased to 75% when the inerts are decreased to 25%.,Solution:,,The followi

21、ng conversion temperature relationship is also for an adiabatic reaction, A ? B, containing 50% inerts:,Example (cont’d):,Solution:,Note: The inerts NEVER enter into these calculations of,or,.,Sketch X vs. T forwhen the

22、 inerts are increased to 75% when the inerts are decreased to 25%.,As you increase inerts, there is more sensible heat to supply the reaction and the temperature does not drop as much when the inerts are increased.,8.3.

23、2 Adiabatic tubular reactor,,Energy balance for adiabatic operation of PFR,,Differential mole balance:,,Adiabatic PFR/PBR algorithm,The elementary reversible gas-phase reaction,Pressure drop neglectedPure A enters,Mole

24、balance,,,Rate law,With,,,Stoichiometry,Gas,,,Combine,,,Energy balance,,If pure A enters and iff ?CP=0, then,,,Example: p353-357,VCSTR vs. VPFR,Algorithm Adiabatic Reactions,1. Choose X,Calculate T,Calculate k,Calculat

25、e T/To,Calculate CA,Calculate CB,Calculate KC,Calculate -rA,2. Increment X and then repeat calculations.,3. When finished, plot,vs. X or use numerical technique to find V.,Levenspiel Plot for anexothermic, adiabatic rea

26、ction,Consider:,PFR: Shaded area is the volume,For an exit conversion of 40%,For an exit conversion of 70%,CSTR,For an exit conversion of 40%,For an exit conversion of 70%,CSTR+PFR,For an intermediate conversion of 40% a

27、nd exit conversion of 70%,The best arrangement is a CSTR with a 40% conversion followed by a PFR up to 70% conversion.,8.4 Steady-state tubular reactor with heat exchange,,,?V,,,FA0T0,FAeTe,,,T,,8.4.1 Deriving the ener

28、gy balance for a PFR,,,Expanding,,,where,,,,PFR energy balance,,PBR energy balance,,,,,(A),(B),(C),,Coupled differential equations,If coolant temperature varies down the reactor,,,The next ...,8.4.2 Balance on the coolan

29、t heat transfer fluid,,,,FA0, T0,,Tao,Ta,,,,,Tao,,,,Heat transfer fluidR2,,,ReactantsR1,Case A: Co-current flow,Rate of energyin at V,-,Rate of energyout at V+?V,+,Rate of heat addedby conduction through The inner

30、wall,= 0,,,,,Variation of coolant temperature Ta down the length of reactor,,,,exothermic,endothermic,,V,Ta,Ta,V,and,,Case B: Counter current flow,,Solution to this counter current flow problem to find the exit conversio

31、n and temperature requires a trial-and-error procedure,,(1) Ta0=300K,(2) Ta2=340K,(3) T’a0=310K,?,(4) Ta2=330K,(5) Ta0=300K,Trial and Error procedure for counter current flow problems,4. Now guess a coolant temperature a

32、t V = 0 and X = 0 of 330 K. We see that the exit coolant temperature of Ta2 = 330 K will give a coolant temperature at V = V1 of 300 K.,1. Consider an exothermic reaction where the coolant stream enters at the end of the

33、 reactor at a temperature Ta0, say 300 K.,2. Assume a coolant temperature at the entrance (X = 0, V = 0) to the reactor Ta2 =340 K.,3. Calculate X, T, and Ta as a function of V. We can see that our guess of 340 K for Ta2

34、 at the feed entrance (X = 0) gives a coolant temperature of 310 K, which does not match the actual entering coolant temperature of 300 K.,Example P360-365,8.5 Equilibrium conversion,For reversible reactions, the equilib

35、rium conversion, Xe, is usually calculated first.For endothermic reactions, Xe increases with increasing temperatureFor exothermic reactions, Xe decreases with increasing temperature,Example:,Rate law:,Concentration

36、equilibrium constant,Van't Hoff Equation,For the special case of,Integrating the Van't Hoff Equation gives:,8.5.1 Adiabatic temperature and equilibrium conversion,Exothermic reactions,,,,T0,T01,Xe,,Energy balance

37、,,,Adiabatic temp.,Equilibrium,,T01>T0,,For 1st-order reaction,Make Xe~T curve,,Reactor staging with interstage cooling or heating,Higher conversions can be achieved for adiabatic operation by connecting reactors in s

38、eries with interstaging cooling,,,T,Equilibrium,X,,,,,,Endothermic reactions,Interstage heating,,,X,T,,,,,,8.5.2 Optimum feed temperature,Why is there a maximum in the rate of reaction with respect to conversion (hence w

39、ith respect to temperature and reactor volume) for an adiabatic reactor?,Exothermic, Reversible Reaction,Rate Law:,,,,T,Equilibrium,X,,,,,,,T03,T02,T01,Xe3,Xe2,Xe1,Fixed reactor size or catalyst weight,Exothermic, Revers

40、ible Reaction,,X,V,,,,,X3 @ T03,X2 @T02,,X1 @T01,T0?, -rA ?,,,Reactor exit,,,,,,T01,T02,T03,X,T,,Curve A: Reaction rate slow, conversion dictated by rate of reaction and reactor volume. As temperature increases rate incr

41、eases and therefore conversion increases.Curve B: Reaction rate very rapid. Virtual equilibrium reached in reaction conversion dictated by equilibrium conversion.,Optimum Inlet Temperature,Fixed Volume Exothermic React

42、or,8.6 CSTR with heat effects,,,CSTR mole balance:,,Note: CSTR - well mixed with uniform temperature, but reaction can be carried out non-isothermally. Isothermal operation: feed temperature = temperature in CSTR,term

43、in the CSTR,8.6.1 Heat added to the reactor,,For exothermicreactions,,For endothermicreactions,X, T,,,,,,Ta1,Ta2,T0, FA0,T, X,,Heat exchanger:,Rate of energyin by flow,-,Rate of energyout by flow,-,Rate of heat trans

44、fer from exchanger to reactor,= 0,,,,,,Heat transfer to a CSTR,,,,At a quasi-steady state:,For large values of the coolant flow rate,,,where,,,,,For large coolant flow rate,,,Design equation,,8.6 CSTR with heat effects

45、,8.7 Multiple steady states,X, T,,,,,,Ta1,Ta2,T0, FA0,T, X,,,For large coolant flow rate,,CSTR,,,,,,Forms of the energy balance for a CSTR with heat exchange,Example: Adiabatic Liquid Phase in a CSTR,Second Order React

46、ion Carried Out Adiabatically in a CSTR. The acid-catalyzed irreversible liquid-phase reaction,The reaction is second order in A. The feed, which is equimolar in a solvent (which contains the catalyst) and A, enters the

47、 reactor at a total volumetric flow rate of 10 dm3 /min with the concentration of A being 4M. The entering temperature is 300 K. (a) What CSTR reactor volume is necessary to achieve 80% conversion? (b) What conversion

48、can be achieved in a 1000 dm3 CSTR? What is the new exit temperature? (c) How would your answers to part (b) change, if the entering temperature of the feed were 280 K?,Additional Information:,Solution,1. CSTR Design Eq

49、uation:,2. Rate Law:,3. Stoichiometry:,(liquid phase),4. Combine:,5. Determine T:,,,=380 K,(a) What CSTR reactor volume is necessary to achieve 80% conversion?,6. Solve for the Rate Constant (k) at T = 380 K:,7. Calculat

50、e the CSTR Reactor Volume (V):,Solution,(b) What conversion can be achieved in a 1000 dm3 CSTR? What is the new exit temperature?,1. CSTR Design Equation:,2. Rate Law:,3. Stoichiometry:,(liquid phase),4. Combine:,NOTE: W

51、e will find it more convenient to work with this equation in terms of space time, rather than volume:,Given reactor volume (V), you must solve the energy balance and the mole balance simultaneously for conversion (X), si

52、nce it is a function of temperature (T).,Analysis:,5. Solve the Energy Balance for XEB as a function of T,From the adiabatic energy balance (as applied to CSTRs),,,,6. Solve the Mole Balance for XMB as a function of T,,,

53、7. Plot XEB and XMB,Plot XEB and XMB on the same graph (as functions of T) to see where they intersect. This will tell you where your steady-state point is,X = 0.87 and T = 387 K,(c) How would your answers to part (b) ch

54、ange, if the entering temperature of the feed were 280 K?,Solution:,1. CSTR Design Equation:,2. Rate Law:,3. Stoichiometry:,(liquid phase),4. Combine:,Given reactor volume (V), you must solve the energy balance and the m

55、ole balance simultaneously for conversion (X), since it is a function of temperature (T).,5. Solve the Energy Balance for XEB as a function of T,6. Solve the Mole Balance for XMB as a function of T,7. Plot XEB and XMB,X

56、= 0.062 at T = 286 K (stable)X = 0.38 at T = 318 K (unstable)X = 0.75 at T = 355 K (stable),X = 0.062 at T = 286 K (stable)X = 0.38 at T = 318 K (unstable)X = 0.75 at T = 355 K (stable),1. Solve the Energy Balance fo

57、r XEB as a function of T,2. Solve the Mole Balance for XMB as a function of T,Start-up of a CSTR where there are three steady state solutions,8.7 Multiple steady states,Steady state operation of a CSTR in which a first-o

58、rder reaction is taking place,then,where,,If,,,Mole balance,,Heat generated term,,Heat removed term,,Vary entering temperature,,R(T),T,,,,,,Increase T0,8.7.1 Heat removed term, R(T),Vary non-adiabatic parameter, ?,,R(T),

59、T,,,,,Increase ?,T0,Ta,,? = ?,? = 0,,8.7.2 Heat of generation, G(T),,,,For 1st-order liquid phase reaction,,Low temperature,High temperature,Low T,High T,,G(T),T,T,,,Low E,High E,,G(T),,,,,,Increase ?,Heat generated curv

60、es, G(T),For 2nd-order liquid phase reaction,,8.7.3 Ignition-extinction curve,Intersection of R(T) and G(T): the temperature at which the reactor canoperate at steady state,,R(T)G(T),,,,,,TC1,TC2,TS2,TS3,,,R(T)G(T),,,

61、,,,,,,,,,,Ignition-extinction curve,TS1,,,,,,,Ts,T0,1,2,3,4,5,6,7,8,9,10,11,12,Upper steady state,Lower steady state,,,,Unstable steady states,,R(T)G(T),,,,,,R>G,G>R,R>G,G>R,,,,?,?,?,?,Ignition-extinction cu

62、rve,T01,T02,T03,T04,T05,T06,,Ignition temp.,Extinction temp.,,8.7.4 Runaway reactions in a CSTR,,R(T)G(T),,,,,R(T),G(T),,Runaway,Slope=,,TC,T*,,T,,,,,,,At T*:,,,,where,,Difference between the reactor temperature and TC,

63、,If,is exceeded, transition to the,upper steady state will occur.,For many industrial reactions, E/RT is typically between 16 and 24, and the reaction temperatures may be between 300 to 500K. Consequently, this critical

64、temperature difference will be somewhere around 15 to 30?C.,Stability diagram,,,To show regions of stable operation and unstable operation.,(2),(1),(3),(4),(5),G(T*),(6),1st-order reaction,,,Tc,S,,S*,,CP0(1+?),Runaway,Un

65、stable,Stable,,,,Tc1,Tc2,,,,Runaway will occur.,No runaway, won’t move to upper steady states,Please refer to CD: Professional Reference Shelf,8.8 Nonisothermal multiple chemical reactions,Tie together all the previous

66、chapters to analyze multiple reactions that do not take place isothermally.,8.8.1 Energy balance for multiple reactions in plug-flow reactors,,,For a single reaction in a PFR,For q multiple reactions in the PFR,,,,Reacti

67、on 1:,Reaction 2:,,PFR energy balance:,q independent reactions,8.8.2 Energy balance for multiple reactions in CSTR,For q reactions and m species:,,,,,,Two reactions,,8.9 Radial and axial variations in a tubular re

68、actor,,,,,Overview of energy balances,1. Adiabatic,CSTR, PFR, Batch or PBR.,The relationship between conversion and temperature for,,,For an exothermic reaction,,T,XEB,,T0,,2. CSTR with heat exchanger,3. PFR/PBR with hea