反应工程基础(程易)chpt14-fixedbedrea.ppt

《反应工程基础(程易)chpt14-fixedbedrea.ppt》由会员分享，可在线阅读，更多相关《反应工程基础(程易)chpt14-fixedbedrea.ppt（67页珍藏版）》请在三一办公上搜索。

1、Summary of Chapter 13,The concentration profile for a first-order reaction occurring in a spherical catalyst pellet is where 1 is the Thiele modulus.For a first-order reactionThe internal effectiveness factor and overall effectiveness factor.,1,Summary(contd),For large values of the Thiele modulus f

2、or an n order reaction,For internal diffusion control,the me reaction order is related to the measured reaction order by The true and apparent activation energies are related by,2,Summary(contd),The Weisz-Prater Parameter The Weisz-Prater criterion dictates that:If CWP1,internal diffusion limitation

3、s presentMears Criteria for Neglecting External Diffusion and Heat Transfer,3,4,Chapter 14Fixed-Bed Reactor,Department of Chemical EngineeringTiefeng Wang,Contents,Introduction of fixed-bed reactorAdiabatic fixed bed reactorFixed-bed reactor with heating or coolingModel of the fixed-bed reactor,5,14

4、.1 Types of the fixed bed reactor,6,7,Fixed-bed reactors for methanol synthesis,8,多段冷激式反应器,列管式合成反应器,径向反应器,9,Fluidized bed reactor,10,Selection of the reactor type,11,Reactor type preferred as a function of on stream time of the catalyst between two regenerations(van Swaaij et al.,Chem.Eng.J.,2002,90

5、:2545),Characters of the fixed-bed reactor,In passing through fixed beds,gases approximate plug flow.If efficient contacting in a reactor is of primary importance,then the fixed bed is favored.Effective temperature control of large fixed beds can be difficult because such systems are characterized b

6、y a low heat conductivity.If operations are to be restricted within a narrow temperature range,then the fluidized bed is favored.Fixed beds cannot use very small sizes of catalyst because of plugging and high-pressure drop,whereas fluidized beds are well able to use small-size particles.If the catal

7、yst has to be treated(regenerated)frequently because it deactivates rapidly,then the liquid-like fluidized state allows it to be pumped easily from unit to unit.There is no catalyst attrition in the fixed-bed reactor.,12,Temperature profile of the fixed bed reactor,13,Exothermic reaction,Hot spot,14

8、.2 Pressure Drop in the Packed Bed Reactor,In liquid phase,effect of pressure drop can be ignored.In gas phase,for an ideal gas,the concentration of reacting species i isFor isothermal operation:,14,where,See Section 4.5,Analyze the following second order gas phase reaction that occurs isothermally

9、in a PBR2A B+C Mole balance(must use differential form),Rate law,Stoichiometry,Combine,Need to find(P/P0)as a function of W(or V if you have a PFR).,15,Pressure drop in a packed bed,P/P0=f(Volume,V)or f(catalyst weight,W)Majority gas phase reactions happened in a packed bed of catalyst particles.Equ

10、ation used to calculate pressure drop is the Ergun Eqn.:,16,Ergun Equation,17,b=bulk densityc=solid catalyst density=porosity(void fraction),Pressure drop in reactor,For single reaction:For isothermal operation,T=T0 with=0;,18,This Eqn.is for multiple reactions and reaction in a membrane reactor.,Re

11、action with pressure drop,19,1.0,(a),(b),(d),(c),(e),Effect of pressure drop on P(a),CA(b),-rA(c),X(d),and(e),Reaction with pressure drop,20,AB,2nd order isothermal reaction,Rate law:,Stoichiometry:Gas phase isothermal reaction with=0,Mole balance:,Combining:,Separating variables:,Integrating with l

12、imits X=0 when W=0 and substituting for FA0=CA00 yields:,21,Solving for conversion gives:,Solving for catalyst weight,we have:,22,14.3 Adiabatic fixed-bed reactor,23,Dependence of-rA on T,In a general form,For irreversible reaction:T-rA,For first-order reactions:,For reversible exothermic reactions:

13、-rA has a maximum,and E-E+,With fixed conversion,24,14.3.1 Equilibrium temperature and optimum temperature,when-rA=0,the temperature is the equilibrium temperature Tewhen-rA gets its maximum,the temperature is the optimum,Topt,For each conversion x:,Te is obtained by setting,Topt is obtained by sett

14、ing,25,Curves of Te and Topt for reversible exothermic reactions,The relationship between the equilibrium and optimum temperatures:,For a specific conversion,the same reaction rate corresponds to two temperatures,one smaller than Topt,and the other larger than Topt,26,14.3.2 Adiabatic tubular reacto

15、r,Energy balance for adiabatic operation of PFR,Differential mole balance:,Almost always we will use an ODE solver,Femlab?Matlab?,Eq.(8-30)Fogler,TR:reference T,Adiabatic temperature and equilibrium conversion,28,Exothermic reactions,T0,T01,Xe,Energy balance,Adiabatic temp.,Equilibrium,T01T0,For 1st

16、-order reaction,Make XeT curve,Mean heat capacity:,The elementary reversible gas-phase reaction,Pressure drop neglectedPure A enters,Mole balance,Rate law,with,Stoichiometry,Combine,Energy balance,If pure A enters and if Cp=0,then,Optimum operating temperature for exothermic reversible reaction,Why

17、is there a maximum in the rate of reaction with respect to conversion(hence with respect to temperature and reactor volume)for an adiabatic reactor?,Rate Law:,Curve A:Reaction rate slow,conversion dictated by rate of reaction and reactor volume.As temperature increases rate increases and therefore c

18、onversion increases.Curve B:Reaction rate very rapid.Virtual equilibrium reached in reaction conversion dictated by equilibrium conversion.,For fixed volume of catalyst,there is an Topt,in to give maximum XFor fixed conversion,there is an Topt,in to use minimum volume of catalyst If Vcat,1Vcat,2,Top

19、t,in,1Topt,in,2,Fixed Volume Exothermic Reactor,Optimum feed temperature for exothermic reversible reaction,in,14.3.3 Staged adiabatic packed bed reactors,33,Optimization of the operations of staged packed bed,34,T0,T1,x1,x1,T1,x3,To get the minimum,the following conditions should be fulfilled:,T2,x

20、2,x2,T2,优化变量,Ti为i-1级入口温度,xi为 i 级出口转化率,以上条件的物理含义：第 i 级出口反应速率与第 i+1级反应速率相等第 i 级入口温度是第 i 级出口转换率的最优温度,35,Two-stage adiabatic packed bed reactors,36,Optimum two-stage packed bed reactor.,The optimization of operations reduces to minimizing the total amount of catalyst needed to achieve a given conversion

21、.,Three variables:(1)the incoming temperature Ta(2)the amount of catalyst used in the first stage(3)the amount of inter-coolingProceduresGuess Ta.Move along the adiabatic line until the following condition is satisfied:Cool to point c which has the same rate of reaction as point b;thusMove along the

22、 adiabatic from point c until the criterion of Eq.1 is satisfied,giving point d.Repeat until point d is at the desired final conversion.,37,Eq.1,Cold shot cooling multistage fixed-bed reactor,38,Optimum two-stage mixed flow reactor set up,39,Infinite recycle for staged packed beds,Optimum two-stage

23、packed bed reactor with recycle,40,The conversions shown represent a recycle ratio R=1 in both stages.,14.4 Fixed-bed reactor with heating or cooling,Heat transfer in the fixed-bed reactorHeat transfer between the particle and fluidEffective heat conductance of the bedHeat transfer between the bed a

24、nd wall,41,42,14.4.1 Heat transfer between the particle and fluid,Heat flux:,For forced convection,the heat transfer coefficient is normally correlated in terms of three dimensionless groups:,am is the surface area per unit weight of catalyst is a factor related to the catalyst shape,14.4.2 Effectiv

25、e thermal conductivity of packed bed,The principle mechanisms of heat transfer in packed beds are as follows:Conduction through the gas and solid phasesRadiation between particles combined with conductionConvective flow of fluid elements,43,ktd is the turbulent diffusion thermal conductivitykt,f is

26、the thermal conductivity of the fluidkte is the effective thermal conductivity of the bedkr is the radiation conductivity,(1)Conduction through the gas and solid particles,44,45,(2)Radiant heat transfer,The radial flux between two surfaces is:,hr is the radial heat transfer coefficientb is the bed v

27、oidage,Radiation between particles and conduction though the solid are accounted for in the model of Schotte:,For 3-mm catalyst particles in air,kr is about 1.2kt,f at 300oC and 2.0kt,f at 500oC.,(3)Convective flow of fluid elements,46,Taking the Peclet number as 10,then,14.4.3 Heat transfer between

28、 bed and wall,47,Tm:The average bed temperatureTR:The temperature that would be reached by extending the parabola to the wall.Tw:The wall temperatureTc:The temperature in the center.,Which temperature difference is used to calculate the heat transfer rate?,14.4.3 Heat transfer between bed and wall,4

29、8,一维模型：,二维模型：,The average bed temperature(Tm)is used to calculate the temperature difference,and the heat transfer coefficient is h0,The bed temperature at the boundary layer(TR)is used to calculate the temperature difference,and the heat transfer coefficient used is hw.,Using the overall heat trans

30、fer coefficient(U),where,To relate kte to hb,the bed geometry and shape of the temperature profile are needed.If q,the rate of heat generation per unit volume of bed,is assumed to be independent of the bed radius,a heat balance for a unit length of cylindrical bed is:,49,The total heat generated is

31、equated to the heat conducted from the bed,using the driving force(Tm-TR),hb的定义式,热量平衡,50,Integration leads to:,As a result:,According to the definition of Tm,there is:,51,In summary,the heat transfer rate can be calculated based on different driving force(T),but the corresponding heat transfer coeff

32、icient must be properly used(hb,h0,hw,hj,U).,14.5 Design equation for the packed bed reactor,52,Assumptions:(1)Plug flow(2)Overall heat transfer coefficient is U,Molar balance:,Energy balance:,Boundary conditions:,自换热固定床反应器,53,Iso-thermal reactor:,The reaction rate:,Tm is the temperature of the cool

33、ing/heating fluid,54,Adiabatic packed bed:,Adiabatic temperature increase is:,55,Non-isothermal and non-adiabatic fixed bed:,The numerical schemes:,（1）,（2）,Iteration:2121.until convergence,56,Typical results:,14.6 2D model of the packed bed reactor,57,In Out+Generation=Accumulation,2D model of the p

34、acked bed reactor(contd),58,Molar balance:,Energy balance:,kte,r is the bed effective conductivity in the radial directionkte,z is the bed effective conductivity in the axial direction,Er is the bed effective diffusion coefficient in the radial directionEz is the bed effective diffusion coefficient

35、in the axial direction,59,If Ez,Er,cp,u,kte,z and kte,r are constant:,Boundary conditions:,60,If EZ and kez are further neglected:,Boundary conditions:,61,偏微分方程的差分离散及求解,一、差分原理,y=f(x)在xi+1和xi1Taylor展开：,这里的增量取负值。,从而得精度为0(x)的两个一阶差商：,一个精度为0(x2)的一阶差商：,及一个精度为0(x2)的二阶差商：,62,二、显式差分格式,坐标离散：,最终得到迭代公式：,变量离散：,利

36、用LHopitol法则修正轴心：,优点是可以顺序迭代，不需联立求解；缺点是收敛性差。,离散边界条件：,63,三、隐式差分格式,Crank-Nicholson法（在轴向格点中间对变量离散）：,最终得到迭代公式：,优点是收敛性好；缺点是需要联立求解代数方程组。,邻二甲苯氧化制邻苯二甲酸酐,64,A:o-xylene(邻二甲苯)B:phthalic anhydride(邻苯二甲酸酐)C:CO+CO2,65,Design equations:,Boundary conditions:,66,Modeling Using Femlab,Start FEMLAB and select modelsDefine constants and expressionsGeometryPhysics settingsBoundary conditionsSubdomain settings(coefficients,reaction rate,velocity)Mesh generationComputing the solution,67,