反应工程课件ppt

1、催催化反应工程化反应工程Catalytic Reaction Engineering李翔李翔 西部校区化工实验楼西部校区化工实验楼B325Xiang Li B-325, Chemical Engineering Laboratory, Western CampusTel： 84986124E-mail: 教材和讲义教材和讲义1. 王安杰，周裕之，赵蓓. 化学反应工程学化学反应工程学. 化学工业出版社2. O. Levenspiel Chemical Reaction Engineering. John Wiley & Sons, Inc. 1999.3. I. Chorkendorff, J.

2、W. Niemantsverdriet. Concepts of Modern Catalysis and Kinetics. WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim, 2003. Reaction rate2 . 1 Conversion and Molar Fractionn ConversionIn the Batch ReactorIn the Steady-State Flow Reactorn Molar Fraction2 . 1. 1 Conversion 1. In the Batch ReactorDadCacBabA00AAA

3、AnnnxAAAxnn10ABABxabnn/000/ABBnnSuppose that nA0 is the initial amount of A in the reactor at time t=0, and that nA is the amount of A at time t . Then the conversion of A in constant volume system is given by(2.1)(2.2)(2.3)ACACxacnn/000/ACCnnADADxadnn/000/ADDnn(2.4)(2.5)2 . 1. 1 Conversion Suppose

4、that I is an inert, then ：IAIInnn00nt0 is the initial amount of A in the reactor at time zero, nt is the amount of all components at time t :AAttxnadcbann00AAtAAAttxnxynn11000(2.6)(2.7)(2.8)adcbaA000/tAAnnyExpansion Factor：0AAAy(2.9)(2.10)Expansion Ratio：2 . 1. 1 Conversion The Steady-State Flow Rea

5、ctor00AAAAFFFx00AAfAAfFFFxSince the composition in steady-state reactor is uniform throughout, the conversion is given by：(2.11)(2.12)2 . 1. 2 Molar Fraction Batch Reactor：Steady-State Flow Reactor：tAAnny/tAAFFy/The expressions of molar fraction for he two systems are sameAA0AAAA0AAAA0AAAA0AAAA0AA11

6、)/(1)/(1)/(1)1 (xyyxxadyyxxacyyxxabyyxxyyIIDDCCBB(2.13)(2.14)(2.15)(2.16)(2.17)(2.18)(2.19)例例2.12A + B 2C 在一间歇反应器中进行CA0 = 2 kmolm-3，CB0 = 5 kmolm-3，CC0 =1 kmolm-3，CI0 = 10 kmolm-3。其中I为溶剂，不参与反应。假定反应液的密度不变，试求xA = 80 %时各组分的浓度和摩尔分率。a =2, b = 1, c = 2, B = 5/2 = 2.5, C = 1/2 = 0.5, I = 10/2 = 5例例2.12A +

7、B 2C 在一间歇反应器中进行CA0 = 2 kmolm-3，CB0 = 5 kmolm-3，CC0 =1 kmolm-3，CI0 = 10 kmolm-3。其中I为溶剂，不参与反应。假定反应液的密度不变，试求xA = 80 %时各组分的浓度和摩尔分率。a =2, b = 1, c = 2, B = 5/2 = 2.5, C = 1/2 = 0.5, I = 10/2 = 5例例2.12A + B 2C 在一间歇反应器中进行CA0 = 2 kmolm-3，CB0 = 5 kmolm-3，CC0 =1 kmolm-3，CI0 = 10 kmolm-3。其中I为溶剂，不参与反应。假定反应液的密度不

8、变，试求xA = 80 %时各组分的浓度和摩尔分率。a =2, b = 1, c = 2, B = 5/2 = 2.5, C = 1/2 = 0.5, I = 10/2 = 5例例2.12A + B 2Ca =2, b = 1, c = 2, B = 5/2 = 2.5, C = 1/2 = 0.5, I = 10/2 = 54 . 0)8 . 01)(2()1 ()1 (00AAAAAAxCVxnVnC CB = CA0 B - (b/a)xA = (2)2.5 (1/2)(0.8) = 4.2 Cc = CA0 c - (c/a)xA = (2)0.5 (2/2)(0.8) = 2.6 C

9、I = CA0 I = 25 =10 例例2.12A + B 2Ca =2, b = 1, c = 2, B = 5/2 = 2.5, C = 1/2 = 0.5, I = 10/2 = 5A = ( -2 1 + 2 )/2 = -0.5 yA0 = 2/(2+5+1+10)= 0.111 A = AyA0 = (-0.5)(0.111) = -0.0551 + AxA = 1 + (-0.0555)(0.8) = 0.956例例2.12A + B 2CA = -0.5 yA0= 0.111 A = -0.0551 + AxA = 0.956yA = (0.111)(1-0.8)/0.956

10、 = 0.0232yB = (0.111)2.5 (1/2)(0.8)/0.956 = 0.244yC = (0.111)0.5 + (2/2)(0.8) /0.956 = 0.151yI = (0.111)(5)/0.956 = 0.581nSingle ReactionnMultiple ReactionsnHeterogenous Reactions2 . 2 Rate Equation2 . 2 . 1 Single ReactiondDcCbBAa The most useful measure of reaction rate for reactant A is thenVdtdN

11、rAA-The rates of reaction of all components are related bydcbaDCBA)r ()r ()-r()-r(r(2.20)(2.21)(amount of A disappearing)(volume) (time)，smmol3Note that this is an intensive measureThe minus sign means disappearance2 . 2 . 2 Multiple Reactions riA is the rate of reaction of material i.miiAmAiAAAArrr

12、rrr121(2.22)2 . 3 Reaction orderPower Rate Law2 . 3 Reaction order2.3.1 Steady-state approximationcP- tcA- tcB- tct2.4 Steady-state approximationThe rates of the elementary steps are： 2A A* + A r1 = k1CA2 A* + A 2A r2 = k 2CA*CA A* C + D r3 = k3CA*From equation (2.21)： r1A* = (1)r1 = k1CA2 r2A* = (-

13、1)r2 = - k2CA*CA r3A* = (-1)r3 = - k3CA*According to equation(2.22)，the rate equation of A* is ： rA* = r1A*+ r2A* + r3A*= k1CA2 - k2CA*CA - k3CA*To give the SSA concentration of A* ： k1CA2 - k2CA*CA - k3CA* = 02.3.1 Steady-state approximationwith the result： 3221*kCkCkCAAAFrom equation (2.21) ： r =(

14、- rA)/1 = rC/1 = rD/1 then rC = (1)r3 = k3CA*32213kCkCkkrrAACConsistent with equations (a) and (b)：At high pressure, k2CA k3，the rate equation becomes： Order with respect to A is 1. r = (k3k1/k2)CA (a)(b) At low pressure, k2CA k3，the rate equation becomes： r = k1CA2 Order with respect to A is 2.2 .

15、3 Reaction orderCatalytic Reactions:2 . 3 Reaction orderCatalytic Reactions:Steady state approximation2 . 3 Reaction orderCatalytic Reactions:2 . 3 Reaction orderCatalytic Reactions:If the conversions from adsorbed intermediate into either product or reactant are fast, then catalyst surface will be

16、mostly empty. In this case the rate expression becomes equal to that of the uncatalyzed reaction, and the order in the reactant becomes +1.When the denominator becomes large, the surface is heavily occupied because either the reaction from intermediate to product or the reverse reaction to reactant

17、is slow. the rate has become zero order in R.2 . 3 Reaction orderThe special case of a pseudo-zero order reaction arises if a reactant is present in large excess, and the reaction does not noticeably change the concentration of the reactant.Hence, the orders of overall reactions should certainly not

18、 be treated as universal constants but rather as a convenient parameterization that is valid for a specific set of reaction conditions.2.3.2 The Quasi-equilibrium ApproximationLangmuirHinshelwood Kinetics:2.3.2 The Quasi-equilibrium approximation2.3.2 The Quasi-equilibrium approximation2.3.2 The Quasi-equilibrium approximation2.3.2 The Quasi-equilibrium approximation2 . 3 Reaction orderFor overall reactions the rate often cannot be written as a simple power law. In this case orders will generally assume non-integral v