chemicalreactionengineering3ededition作者octaveLevenspiel课后习题问题详解

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1、wordCorresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING1CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS3CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA7CHAPTER 4 INTRODUCTION TO REACTOR DESIGN19CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR22CHAPTER

2、6 DESIGN FOR SINGLE REACTIONS26CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR32CHAPTER 11 BASICS OF NON-IDEAL FLOW34CHAPTER 18 SOLID CATALYZED REACTIONS43Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a smal

3、l munity (Fig.P1.1). Waste water, 32000 m3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O2CO2 + H2O A typical entering feed has a BOD (biological oxyge

4、n demand) of 200 mg O2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Waste water32,000 m3/dayWaste waterTreatment plant Clean water 32,000 m3/day200 mg O2needed/literMean residencetime =8 hr Zero O2 neededSolution:1.2 Coal burnin

5、g electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed bustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsew

6、here in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within the beds, based on the oxygen used.Solution:Chapter 2 Kinetics of Homogeneous Reactions A reaction has the stoichiome

7、tric equation A + B =2R . What is the order of reaction?Solution:Because we dont know whether it is an elementary reaction or not, we cant tell the index of the reaction.2.2 Given the reaction 2NO2 + 1/2 O2 = N2O5 , what is the relation between the rates of formation and disappearance of the three r

8、eaction ponents?Solution:2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rate expression-rA = 2 C0.5 ACB What is the rate expression for this reaction if the stoichiometric equation is written as A + 2B = 2R + SSolution: No change. The stoichiometric equation cant

9、effect the rate equation, so it doesnt change. For the enzyme-substrate reaction of Example 2, the rate of disappearance of substrate is given by-rA = , mol/m3s What are the units of the two constants?Solution:2.5 For the plex reaction with stoichiometry A + 3B 2R + S and with second-order rate expr

10、ession -rA = k1AB are the reaction rates related as follows: rA= rB= rR? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: A certain reaction has a rate given by -rA = 0.005 C2 A , mol/cm3min If the concentration is to be expressed in mol/li

11、ter and time in hours, what would be the value and units of the rate constant?Solution: For a gas reaction at 400 K the rate is reported as - = 3.66 p2 A, atm/hr (a) What are the units of the rate constant? (b) What is the value of the rate constant for this reaction if the rate equation is expresse

12、d as -rA = - = k C2 A , mol/m3sSolution:(a) The unit of the rate constant is (b)Because its a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to So we can get that the value of The pyrolysis of ethane proceeds with an activation energy of

13、 about 300 kJ/mol. How much faster the deposition at 650 than at 500?Solution:In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ant

14、s, I find Running speed, m/hr150160230295370Temperature, 1316222428 What activation energy represents this change in bustliness?Solution:Suppose , so intercept1501602302953701316222428-Also , intercept = 15.686 ,Chapter 3 Interpretation of Batch Reactor Data If -rA = - (dCA/dt) =0.2 mol/litersec whe

15、n CA = 1 mol/liter, what is the rate of reaction when CA = 10 mol/liter?Note: the order of reaction is not known.Solution: Information is not enough, so we cant answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A is converted in a 5-minute ru

16、n. How much longer would it take to reach 75% conversion?Solution:Because the deposition of A is a 1st-order reaction, so we can express the rate equation as:We know that for 1st-order reaction, , , So equ(1) equ(2)So 3.3 Repeat the previous problem for second-order kinetics.Solution:We know that fo

17、r 2nd-order reaction, ,So we have two equations as follow:, equ(1), equ(2)So , A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a -order rate. What would be the fraction converted in a half-hour run?Solution:In a order reaction: ,After integration, we can get

18、: ,So we have two equations as follow:, equ(1), equ(2)bining these two equations, we can get:, but this means , which is impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted .In a hmogeneous isothermal liquid polymerization, 20% of the

19、monomer disappears in34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution:The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,

20、And After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction.Solution:In 1st order reaction, , dissatisfied.In 2nd order reaction, , satisfied.According to the information, the reaction is

21、a 2nd-order reaction.nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alikeinto the joint with his weeks salary of 180, steady gambling at “2-up for two hours, then home to his family leaving 45 behind. Snake Eyess betting pattern is predictable. He always bets in amounts

22、 proportional to his cash at hand, and his losses are also predictableat a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with 135. How much was his raise?Solution:, , , , So we obtain , , The first-order r

23、eversible liquid reaction A R , CA0 = 0.5 mol/liter, CR0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibriumconversion is 66.7%. Find the equation for the this reaction.Solution: Liquid reaction, which belongs to constant volume system,1st order reversible re

24、action, according to page56 eq. 53b, we obtain , , so we obtain eq(1) eq(1), , so we obtain eq(2), eq(2)bining eq(1) and eq(2), we obtain So the rate equation is Aqueous A reacts to form R (AR) and in the first minute in a batch reactor its concentration drops from CA0 = 2.03 mol/liter to CAf = 1.97

25、 mol/liter. Find the rate equation from thereaction if the kinetics are second-order with respect to A.Solution: Its a irreversible second-order reaction system, according to page44 eq 12, we obtain ,so so the rate equation is At room temperature sucrose is hydrolyzed by the catalytic action of the

26、enzyme sucrase as follows: Aucrose products Starting with a sucrose concentration CA0 = 1.0 millimol/liter and an enzyme concentration CE0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):CA, millimol/lite

27、r t,hr1234567891011Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or -rA = where CM = Michaelis constant If the fit is reasonable, evaluate the constants k3 and CM. Solve by the integral method. Solution: Solve the question by the integral m

28、ethod:, ,mmol/L1234567891011Suppose y=, x=, thus we obtain such straight line graph, intercept=So , Enzyme E catalyzes the transformation of reactant A to product R as follows: A R, -rA = If we introduce enzyme (CE0 = 0.001 mol/liter) and reactant (CA0 = 10 mol/liter) into a batch rector and let the

29、 reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.Solution: Rearranging and integrating, we obtain: M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present t

30、he data inTable P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at 22.9: H2SO4 + (C2H5)2SO4 2C2H5SO4HInitial concentrations of H2SO4 and (C2H5)2SO4 are each 5.5 mol/liter. Find a rate equation for this reaction.t, minC2H5SO4H, mol/litert, minC2H5SO4H, mol/liter0018041

31、19448212552677531896368127379146410162(5.80)Solution: Its a constant-volume system, so we can use XA solving the problem:i) We postulate it is a 2nd order reversible reaction system The rate equation is: , , , When , So ,After integrating, we obtain eq (1)The calculating result is presented in follo

32、wing Table., min000004148557596127146162180194212267318368379410Draw t plot, we obtain a straight line:,When approach to equilibrium, ,so So the rate equation isii) We postulate it is a 1st order reversible reaction system, so the rate equation is After rearranging and integrating, we obtain eq (2)D

33、raw t plot, we obtain another straight line:,So So the rate equation isWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by paring eq.(1) and eq.(2), especially when XAe =0.5 , the two equations are identical. This means these two equations would have alm

34、ost the same fitness of data when the experiment data of the reaction show that XAe =0.5.(The data that we use just have XAe =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the follo

35、wing rates, and CA alone determines this rate:CA,mol/liter12467912-rA, mol/literhrWe plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from CA0 = 10 mol/liter to CAf = 2 mol/liter.So

36、lution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002468101214Ca-1/Ra The thermal deposition of hydrogen iodide 2HI H2 + I2 is reported by M.Bodenstein Z.phys.chem.,29,295(1899) as follows:T,508427393356283k,cm3/mols80.910-60.94210-6Find the plete rate eq

37、uation for this reaction. Use units of joules, moles, cm3, and seconds.According to Arrhenius Law,k = k0e-E/R Ttransform it,- In(k) = E/R(1/T) In(k0)Drawing the figure of the relationship between k and T as follows:From the figure, we get slope = E/R = 7319.1 intercept = - In(k0E = 60851 J/mol k0 =

38、105556 cm3/molsFrom the unit k we obtain the thermal deposition is second-order reaction, so the rate expression is- rA = 105556e-60851/R TCA2Chapter 4 Introduction to Reactor Design Given a gaseous feed, CA0 = 100, CB0 = 200, A +B R + S, XA = 0.8. Find XB,CA,CB.Solution: Given a gaseous feed, , , ,

39、 find , , , Given a dilute aqueous feed, CA0 = CB0 =100, A +2B R + S, CA = 20. Find XA, XB, CB.Solution: Given a dilute aqueous feed, , , , find , , Aqueous reaction system, so When , When , So , , which is impossible.So , Given a gaseous feed, CA0 =200, CB0 =100, A +B R, CA = 50. Find XA, XB, CB.So

40、lution: Given a gaseous feed, , , .find , , , which is impossible.So Given a gaseous feed, CA0 = CB0 =100, A +2B R, CB = 20. Find XA, XB, CA.Solution: Given a gaseous feed, ，, ,Find , , , , , Given a gaseous feed, T0 =1000 K, 05atm, CA0=100, CB0=200, A +B5R,T =400 K, =4atm, CA =20. Find XA, XB, CB.S

41、olution:Given a gaseous feed, , , , , , , , find , , ., , According to eq page 87, A mercial Popcorn Popping Popcorn Popper. We are constructing a 1-liter popcorn to be operated in steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/min of mixed exit

42、 stream. Independent tests show that when raw corn pops its volume goes from 1 to 31. With this information determine what fraction of raw corn is popped in the unit.Solution:, , Chapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A R + 2S with unknown kinetics. If a space

43、 velocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution:,Varying volume system, so cant be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is co

44、nverted in 13 min. What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor?Solution:Liquid reaction system, so According to eq.4 on page 92, Eq.13, , cant be certain. Eq.17, , so We plan to replace our present mixed flow reactor with

45、 one having double the bolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented by A R, -rA = kC1.5 ASolution:Liquid reaction system, so , Now we know: , , , So we obtain An aqueous feed of A and B (400liter/min, 100 mmol A/liter

46、, 200 mmol B/liter) is to be converted to product in a plug flow reactor. The kinetics of the reaction is represented by A +B R, -rA = 200CACB Find the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so According to page 102 eq.19, , ,5.9 A specific e

47、nzyme acts as catalyst in the fermentation of reactant A. At a given enzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (CA0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given

48、 by A R , -rA = Solution:P.F.R, according to page 102 eq.18, aqueous reaction, Enzyme E catalyses the fermentation of substrate A (the reactant) to product R. Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzy

49、me. The kinetics of the fermentation at this enzyme concentration are given by A R , -rA = Solution:, , , Constant volume system, M.F.R., so we obtain,5.14 A stream of pure gaseous reactant A (CA0 = 660 mmol/liter) enters a plug flow reactor at a flow rate of FA0 = 540 mmol/min and polymerizes the a

50、s follows3A R, -rA = 54How large a reactor is needed to lower the concentration of A in the exit stream to CAf = 330 mmol/liter?Solution:, So we obtain Gaseous reactant A deposes as follows:A 3 R, -rA-1)CAFind the conversion of A in a 50% A 50% inert feed (0 = 180 liter/min, CA0 =300 mmol/liter) to

51、a 1 m3 mixed flow reactor.Solution:, M.F.R. According to page 91 eq.11, So we obtain Chapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in a series. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the

52、concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V2/V1 =2.Solution:V2/V1 = 2, 1 = = , = = CA0=1mol/l , CA1=0.5mol/l , , -rA1=kC2 A1 ,-rA2=kC2 A2 (2nd-order) , 2= So we obtain 2A2)/(kCA22)CA2= 0.25 mol/l Water containing a short-lived radioact

53、ive species flows continuously through a well-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but wed like to lower it still more. One of our office secretari

54、es suggests that we insert a baffle down the middle of the tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream pared to the entering stream.Solution:1st-order reaction, constant

55、volume system. From the information offered about the first reaction, we obtain=If a baffle is added,=6/k =3/k= bining equation and we obtain:CA21A0 ; CA22A21=So it will help, and the expected activity of the exit stream is 1/16 of the feed. An aqueous reactant stream (4 mol A/liter) passes through

56、a mixed flow reactor followed by a plug flow reactor. Find the concentration at the exit of the plug flow reactor if in the mixed flow reactor CA = 1 mol/liter. The reaction is second-order with respect to A, and the volume of the plug flow unit is three times that of the mixed flow unit.Solution:Co

57、nstant volume system and 2nd-order reaction:=3/k =9/k= -= -= bining equation. and we obtain: CAf= 0.1 mol/liter6.4 Reactant A (A R,CA0=26 mol/m3) passes in steady flow through four equal-size mixed flow reactors in series (total =2 min). When steady state is achieved the concentration of A is found

58、to be 11, 5, 2, 1 mol/m3 in the four units. For this reaction, what must be plug so as to reduce CA from CA0 = 26 to CAf = 1 mol/m3?Solution:=CA0=26mol/liter, CA1=11 mol/liter, CA2=5 mol/liter, CA3= 2mol/liter, CA4=1mol/literSo we abtain: 15/(-rA1) = 6/(-rA2) = 3/(-rA3) = 1/(-rA4) We postalate the r

59、eaction rate is 1 unit when CA4=1 mol/literSo we obtain CA, mol11521-rA3012621/(-rA)1/301/121/61/2=So we obtain 2.63 min. At 100 pure gaseous A reacts away with stoichiometry 2A R + S in a constant volume batch reactor as follows:t, sec020406080100120140160pA, atmWhat size of plug flow reactor operating at 100 and 1 atm can treat 100 moles A/hr in a feed consisting of 20% inserts to obtain 95