化学反应工程Chapter 6.ppt

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1、化化 学学 反反 应应 工工 程程Chapter 6 Design for Single Reactions The reactor system selected will influence the economics of the process by dictating the size of the units needed and by fixing the ratio of products formed.The first factor,reactor size,may well vary a hundredfold(百倍)among competing designs whi

2、le the second factor,product distribution,is usually of prime(主要的)consideration where it can be varied and controlled.化化 学学 反反 应应 工工 程程In this chapter we deal with single reactions.These are reactions whose progress can be described and followed adequately by using one and only one rate expression c

3、oupled with the necessary stoichiometric and equilibrium expressions.For single reactions product distribution is fixed;hence,the important factor in comparing designs is the reactor size.Design for multiple reactions,for which the primary consideration is product distribution,is treated in the next

4、 two chapters.化化 学学 反反 应应 工工 程程6.1 SIZE COMPARISON OF SINGLE REACTORS Batch ReactorGenerally,the batch reactor is well suited to produce small amounts of material and to produce many different products from one piece of equipment.On the other hand,for the chemical treatment of materials in large amo

5、unts the continuous process is nearly always found to be more economical.Regarding reactor size,a comparison of Eqs.5.4 and 5.19 for a given duty and for =0 shows that an element of fluid reacts for the same length of time in the batch and in the plug flow reactor.Thus,the same volume of these react

6、ors is needed to do a given job.化化 学学 反反 应应 工工 程程Mixed Versus Plug Flow Reactors,First-and Second-Order Reactions For a given duty the ratio of sizes of mixed and plug flow reactors will depend on the extent of reaction,the stoichiometry,and the form of the rate equation.For the general case,a compa

7、rison of Eqs 5.11 and 5.17 will give this size ratio.Let us make this comparison for the large class of reactions approximated by the simple nth-order rate law 化化 学学 反反 应应 工工 程程where n varies anywhere form zero to three.For mixed flow Eq.5.11 giveswhereas for plug flow Eq.5.17 gives 化化 学学 反反 应应 工工 程

8、程Dividing we find that(1)With constant density,or=0,this expression integrates to化化 学学 反反 应应 工工 程程or(2)Equation 1 and 2 are displayed in graphical form in Fig.6.1 to provide a quick comparison of the performance of plug flow with mixed flow reactors.化化 学学 反反 应应 工工 程程Figure6.1 comparison of performan

9、ce of single mixed flow and plug flow reactors for the nth-order reactions 化化 学学 反反 应应 工工 程程For identical feed composition CA0 and flow rate FA0 the ordinate(纵座标)of this figure gives directly the volume ratio required for any specified conversion.Figure 6.1 shows the following.1.For any particular d

10、uty and for all positive reaction orders the mixed reactor is always larger than the plug flow reactor.The ratio of volumes increases with reaction order.2.When conversion is small,the reactor performance is only slightly affected by flow type.The performance ratio increases very rapidly at high con

11、version;consequently,a proper representation of the flow becomes very important in this range of conversion.化化 学学 反反 应应 工工 程程3.Density variation during reaction affects design;however,it is normally of secondary importance compared to the difference in flow type.Figure 6.5 and 6.6 show the same firs

12、t-and second-order curves for=0 but also include dashed lines which represent fixed values of the dimensionless reaction rate group,defined as for second-order reactionfor first-order reaction化化 学学 反反 应应 工工 程程Variation of Reactant Ratio for Second-Order ReactionsSecond-order reactions of two compone

13、nts and of the type (3.13)behave as second-order reactions of one component when the reactant ratio is unity.Thus when M=1 (3)化化 学学 反反 应应 工工 程程On the other hand,when a large excess of reactant B is used then its concentration does not change appreciably()and the reaction approaches first-order behav

14、ior with respect to the limiting component A,orwhen M1 (4)Thus in Fig.6.1,and in terms of the limiting component A,the size ratio of mixed to plug flow reactors is represented by the region between the first-order and the second-order curves.化化 学学 反反 应应 工工 程程General Graphical Comparison For reaction

15、s with arbitrary but known rate performance capabilities of mixed and plug flow reactors are best illustrated in Fig.6.2.The ratio of shaded(阴影的)and of hatched(阴影线的)areas gives the ratio of space-times needed in these two reactors.The rate curve drawn in Fig.6.2 is typical of the large class of reac

16、tions whose rate decreases continually on approach to equilibrium(this includes all nth-order reactions,n0).For such reactions it can be seen that mixed flow always needs a larger volume than does plug flow for any given duty.化化 学学 反反 应应 工工 程程Figure 6.2 Comparison of performance of mixed flow and pl

17、ug flow reactors for any reaction kinetics 化化 学学 反反 应应 工工 程程6.2 MULTIPLE-REACTOR SYSTEMSPlug Flow Reactors connected in SeriesConsider N plug flow reactors connected in series,and let XA1,XA2,.,XAN be the fractional conversion of component A leaving reactor 1,2,N.Basing the material balance on the f

18、eed rate of A to the first reactor,we find for the ith reactor from Eq.5.18 or for the N reactors in series化化 学学 反反 应应 工工 程程Hence,N plug flow reactors in series with a total volume V gives the same conversion as a single plug flow reactor of volume V.化化 学学 反反 应应 工工 程程For the optimum hook up(连接)of pl

19、ug flow reactors connected in parallel or in any parallel-series combination,we can treat the whole system as a single plug flow reactor of volume equal to the total volume of the individual units if the feed is distributed in such a manner that fluid streams that meet have the same composition.Plug

20、 Flow Reactors connected in Parallel 化化 学学 反反 应应 工工 程程Where fi is the volumetric(or molar)fraction which the feed flow into i vessel and ,Let XA1=XA2=XAf,We get 化化 学学 反反 应应 工工 程程Thus,for reactors in parallel V/F or must be the same for each parallel line.Any other way of feeding is less efficient.化化

21、 学学 反反 应应 工工 程程EXAMPLE 6.1 OPERATING A NUMBER OF PLUG FLOW REACTORSThe reactor setup shown in Fig.E6.1 consists of three plug flow reactors in two parallel branches.Branch D has a reactor of volume 50 liters followed by a reactor of volume 30 liters.Branch E has a reactor of volume 40 liters.What fr

22、action of the feed should go to branch D?化化 学学 反反 应应 工工 程程SOLUTIONBranch D consists of two reactors in series;hence,it may be considered to be a single reactor of volumeNow for reactors in parallel V/F must be identical if the conversion is to be the same in each branch.Therefore,orTherefore,化化 学学 反

23、反 应应 工工 程程Equal-size Mixed Flow Reactors in SeriesIn plug flow,the concentration of reactant decreases progressively through the system;in mixed flow,the concentration drops immediately to a low value.Because of this fact,a plug flow reactor is more efficient than a mixed flow reactor for reactions

24、whose rates increase with reactant concentration,such as nth-order irreversible reactions,n 0.Consider a system of N mixed flow reactors connected in series.Though the concentration is uniform in each reactor,there is,nevertheless,a change in concentration as fluid moves from reactor to reactor.化化 学

25、学 反反 应应 工工 程程Figure 6.3 Concentration profile through an N-stage mixed flow reactor system compared with single flow reactors.化化 学学 反反 应应 工工 程程This stepwise(阶梯式的)drop in concentration,illustrated in Fig.6.3,suggests that the larger the number of units in series,the closer should the behavior of the

26、system approach plug flow.Let us now quantitatively(定量地)evaluate the behavior of a series of N equal-size mixed flow reactors.Density changes will be assumed to be negligible;hence =0 and .As a rule,with mixed flow reactors it is more convenient to develop the necessary equations in terms of concent

27、rations rather than fractional conversions;therefore,we use this approach.The nomenclature(命名法,术语)used is shown in Fig.6.4 with subscript i referring to the i th vessel.化化 学学 反反 应应 工工 程程Figure 6.4 Notation for a system of N equal-size mixed reactors in series.化化 学学 反反 应应 工工 程程First-order Reactions.F

28、rom Eq.5.12 a material balance for component A about vessel i gives Because =0 this may be written in terms of concentrations.Hence or(5)化化 学学 反反 应应 工工 程程Now the space-time (or mean residence time )is the same in all the equal-size reactors of volume Vi ,Therefore,(6a)Rearranging,we find for the sys

29、tem as a whole(6b)化化 学学 反反 应应 工工 程程With Eqs.6b and 7 we can compare performance of N reactors in series with a plug flow reactor or with a single mixed flow reactor.This comparison is shown in Fig.6.5 for first-order reactions in which density variations are negligible(=0).In the limit,for N ,this e

30、quation reduces to the plug flow equation(7)化化 学学 反反 应应 工工 程程Figure 6.5 Comparison of performance of a series of N equal-size mixed flow reactors with a plug flow reactor for the first-order reaction A R,(=0)1 XA=CA/CA0化化 学学 反反 应应 工工 程程Second-Order Reactions.We may evaluate the performance of a seri

31、es of mixed flow reactors for a second-order,bimolecular-type reaction,no excess of either reactant,by a procedure similar to that of a first-order reaction.Thus,for N reactors in series we find(8a)whereas for plug flow(8b)化化 学学 反反 应应 工工 程程A comparison of the performance of these reactors is shown i

32、n Fig.6.6.Figures 6.5 and 6.6 support our intuition(直觉)by showing that the volume of system required for a given conversion decreases to plug flow volume as the number of reactors in series is increased,the greatest change taking place with the addition of a second vessel to a one-vessel system.化化 学

33、学 反反 应应 工工 程程Figure 6.6 Comparison of performance of a series of N equal-size mixed flow reactors with a plug flow reactor for elementary second-order reactions 2 A productsA+B products,CA0=CB0，with negligible expansion.化化 学学 反反 应应 工工 程程EXAMPLE 6.2 MIXED FLOW REACTORS IN SERIES At present 90%of reac

34、tant A is converted into product by a second-order reaction in a single mixed flow reactor.We plan to place a second reactor similar to the one being used in series with it.(a)For the same treatment rate as that used at present,how will this addition affect the conversion of reactant?(b)For the same

35、 90%conversion,by how much can the treatment rate be increased?化化 学学 反反 应应 工工 程程SOLUTIONThe sketch of Fig.E6.2 shows how the performance chart of Fig.6.6 can be used to help solve this problem.Figure E6.2化化 学学 反反 应应 工工 程程(a)Find the conversion for the same treatment rate.(b)For the single reactor at

36、 90%conversion we have from Fig.6.6 For the two reactors the space-time or holding time is doubled;hence,the operation will be represented by the dashed line of Fig.6.6 whereThis line cuts the N=2 line at a conversion ,point a.化化 学学 反反 应应 工工 程程(b)find the treatment rate for the same conversion.Stayi

37、ng on the 90%conversion line,we find for N=2 that,point b Comparing the value of the reaction rate group for N=1 and N=2,we find Since the ratio of flow rates becomes 化化 学学 反反 应应 工工 程程Thus,the treatment rate(0)can be raised to 6.6 times of the original(0).Note.If the second reactor had been operated

38、 in parallel with the original unit then the treatment rate could only be doubled.Thus,there is a definite(明确的)advantage in operating these two units in series.This advantage becomes more pronounced(显著的)at higher conversions.化化 学学 反反 应应 工工 程程For arbitrary kinetics in mixed flow reactors of different

39、 size,two types of questions may be asked:how to find the outlet conversion from a given reactor system,and the inverse question,how to find the best setup to achieve a given conversion.Mixed Flow Reactors of Different Sizes in SeriesDifferent procedures are used for these two problems.We treat them

40、 in turn.化化 学学 反反 应应 工工 程程Finding the Conversion in a Given System A graphical procedure for finding the outlet composition from a series of mixed flow reactors of various sizes for reactions with negligible density change has been presented by Jones(1951).All that is needed is an r versus C curve f

41、or component A to represent the reaction rate at various concentrations.Let us illustrate the use of this method by considering three mixed flow reactors in series with volumes,feed rates,concentrations,space-times(equal to residence times because =0),and volumetric flow rates as shown in Fig.6.7.No

42、w from Eq.5.11,noting that =0,we may write for component A in the first reactor 化化 学学 反反 应应 工工 程程Figure 6.7 Notation for a series of unequal-size mixed flow reactors.化化 学学 反反 应应 工工 程程or(9)Similarly,from Eq.5.12 for the i th reactor we may write(10)化化 学学 反反 应应 工工 程程Plot the C versus r cure for compon

43、ent A and suppose that it is as shown in Fig.6.8.To find the conditions in the first reactor note that the inlet concentration CA0 is known(point L),that CA1 and(-rA)1 correspond to a point on the curve to be found(point M),and that the slope of the line from Eq.6.9.Hence,from CA0 draw a line of slo

44、pe until it cuts the rate curve;this gives CA1.Similarly,we find from Eq.6.10 that a line of slope(1/2)from point N cuts the curve at P,giving the concentration CA2 of material leaving the second reactor.This procedure is then repeated as many times as needed.化化 学学 反反 应应 工工 程程Figure 6.8 Graphical pr

45、ocedure for finding compositions in a series of mixed flow reactors.化化 学学 反反 应应 工工 程程Determining the Best System for a Given Conversion.Suppose we want to find the minimum size of two mixed flow reactors in series to achieve a specified conversion of feed which reacts with arbitrary but known kineti

46、cs.The basic performance expressions,Eqs.5.11 and 5.12,then give,in turn,for the first reactor(11)and for the second reactor(12)化化 学学 反反 应应 工工 程程These relationships are displayed in Fig.6.9 for two alternative reactor arrangements,both giving the same final conversion XA2.Note,as the intermediate co

47、nversion XA1 change,so does the size ratio of the units(represented by the two shaded areas)as well as the total volume of the two vessels required(the total area shaded).Figure 6.9 shows that the total reactor volume is as small as possible(total shaded area is minimized)when the rectangle KLMN is

48、as large as possible.This brings us to the problem of choosing XA1(or point M on the curve)so as to maximize the area of this rectangle.Consider this general problem.化化 学学 反反 应应 工工 程程Figure 6.9 Graphical representation of the variables for two mixed flow reactors in series.化化 学学 反反 应应 工工 程程Maximizat

49、ion of Rectangles.In Fig.6.10,construct a rectangle between the x-y axes(轴)and touching the arbitrary curve at point M(x,y).The area of the rectangle is then(13)This area is maximized whenor when(14)化化 学学 反反 应应 工工 程程Figure 6.10 Graphical procedure for maximizing the area of a rectangle.化化 学学 反反 应应 工

50、工 程程In words,this condition means that the area is maximized when M is at that point where the slope of the curve equals the slope of the diagonal(对角线)NL of the rectangle.Depending on the shape of the curve,there may be more than one or there may be no“best”point.However,for nth-order kinetics,n 0,t