Archives: Solution Manual

Bonus A – Working Within the Legal Environment A-1 Working within the Legal Environment bonus c h a p t e r = what’s new in this edition A.3 brief chapter outline and learning goals A.3 lecture outline and lecture […]

Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 155 CHAPTER 19 PROGRAM MANAGEMENT, CONTROL, AND EVALUATION 1) In addressing the subject of goals and objectives for a systems engineering organization, one needs to deal with two […]

143 CHAPTER 18 SYSTEMS ENGINEERING PLANNING AND ORGANIZATION 1) Systems engineering requirements should be initiated at program inception: i.e., during the early stages of conceptual design when a “need” for a system is first identified. It is at this time […]

132 After determining the number of maintenance actions (estimated over the life cycle), the next step is to determine the expected resource consumption per maintenance action. Resources include both human and material resources. Human resources (in this instance) are measured […]

132 CHAPTER 17 DESIGN FOR AFFORDABILITY (LIFE – CYCLE COSTING) 1) Life–cycle cost (LCC) includes the total cost of a system over its entire life cycle. LCC includes all of the costs associated with the activities identified in Figure 17.1 […]

127 CHAPTER 16 DESIGN FOR PRODUCIBILITY, DISPOSABILITY, AND SUSTAINABILITY 1) To bring a system into being is to achieve a high degree of organization and order; to cease being is to return a system to a state of disorganization and […]

Systems Engineering and Analysis, Fifth Edition, by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the […]

109 CHAPTER 14 DESIGN FOR USABILITY (HUMAN FACTORS) 1) Human factors refer to those characteristics that are associated with the human being that must be considered in the design of systems where there are humans (i.e., operators and maintainers) involved […]

Systems Engineering and Analysis, Fifth Edition, by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the […]

85 CHAPTER 12 DESIGN FOR RELIABILITY 1) Reliability may be simply defined as the “probability that a system or product will perform its designated mission in a satisfactory manner for a given period of time when used under specified operating […]

79 CHAPTER 11 CONTROL CONCEPTS AND METHODS 1) Speed, the characteristic to be controlled, is measured by a well known sensory device called a speedometer and monitored by the automobile driver. Measured speed is compared to planned speed by the […]

73 CHAPTER 10 QUEUEING THEORY AND ANALYSIS 1) Monte Carlo analysis must be used in the study of a queuing system when the arrival and service time distributions, the queuing discipline, or other system characteristics cannot be represented mathematically. But, […]

61 CHAPTER 9 OPTIMIZATION IN DESIGN AND OPERATIONS 1) Let h = height; w = width; p = perimeter p = 2h + 2w A = hw = ph/2 – 2 h dA dh = p/2 – 2h = 0 […]

53 CHAPTER 8 MODELS FOR ECONOMIC EVALUATION P/F,6,8 1) P = $10,000 (0.6274) = $6,274 2) (a) F = $8,000 ( 2.004 ) = $16,032 F/P,8,5 (b) F = $52,500 ( 1.469 ) = $77,123 P/A,8,5 3) P = $6,000 […]

45 CHAPTER 7 ALTERNATIVES AND MODELS IN DECISION MAKING 1) The process of generating alternatives may begin with a hazy idea about the problem to be addressed but with good understanding about system requirements. A complete and all inclusive alternative, […]

38 CHAPTER 6 SYSTEM TEST, EVALUATION, AND VALIDATION 1) Specific test and evaluation requirements are first identified during the conceptual design phase when system operational requirements, the maintenance and support concept, and the prioritized TPMs are defined. As system requirements […]

31 CHAPTER 5 DETAIL DESIGN AND DEVELOPMENT 1) Conceptual Design primarily deals with the early definition of system requirements and the ultimate design of the overall system per se. The results are usually included in the System Specification (Type A […]

Systems Engineering and Analysis, Fifth Edition, by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the […]

13 CHAPTER 3 CONCEPTUAL SYSTEM DESIGN 1) The first step is to thoroughly define the problem, or the current deficiency, which then leads to the identification of a need for a system that will ultimately provide a solution. What functions […]

Systems Engineering and Analysis, Fifth Edition, by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the […]

INSTRUCTOR’S GUIDE TO PROBLEM SOLUTIONS For SYSTEMS ENGINEERING AND ANALYSIS Fifth Edition By Benjamin S. Blanchard Wolter J. Fabrycky Systems Engineering and Analysis, Fifth Edition, by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle […]

13–20$ P13-3)(a)$Continued$ 13–21$ P13-3)(a)$Continued$ $ $ P13-3)(b)) This$is$the$same$as$part$(a)$except$the$energy$balance.$ Energy$balance:$ $ $ See$Polymath$program$P13-3-b.pol$ $ 13–22$ P13-3)(b)$Continued$ $ $ $ P13-3)(c)) This$is$the$same$as$part$(b)$except$the$reaction$is$now$reversible.$ $ $ See$Polymath$program$P13-3-c.pol$ 13–23$ P13–3)(c)) $ $ $ $ 13–24$ P13-4)(a)) ) ) $ $ P13-4)(b)) ) ) ) ) […]

13-1$ Solutions)for)Chapter)13)–)Unsteady)State)Non–isothermal) Reactor)Design) ) P13-1)(a))Example)13-1) (i)$The$reaction$runs$away$at$T0$=$282.2$K$ $ (ii)$$ (1)$Runaway$occurs$at$Ta1$=$292$K$ (2)$As$Ta1$increases,$the$maxima$of$the$Qr$versus$t,$and$Qg$versus$T$curves$occur$at$shorter$times.$This$is$ $ (iii)$Coolant$flow$rate$does$not$have$a$significant$effect$on$the$conversion.$The$temperature$of$the$ reactor$decreases$with$an$increase$in$coolant$flow$rate.$This$is$an$intuitive$observation.$$ $ (iv)$$ If$the$heat$of$mixing$had$been$neglected,$the$shape$of$the$graphs$would$have$been$as$follows:$ ) (v)$$ The$new$T0$of$20$˚F$(497$˚R)$gives$a$new$δHRn$and$T.$With$T=497+89.8X$the$polymath$program$of$ example$13–1$gives$t=$8920$s$for$90$%$conversion.$ $ because$the$inlet$temperature$of$the$coolant$is$increased,$due$to$which$the$driving$force$for$heat$ transfer$becomes$lesser.$As$NA0$increases,$Qr,max$and$Qg,max$increase.$This$is$because$we$are$starting$with$ a$larger$number$of$reactant$molecules,$which$gives$rise$to$higher$generated$heat.$As$mc$is$increased,$the$ heat$generated$decreases.$$This$is$because$a$larger$amount$of$heat$is$removed$with$an$increase$in$flow$ rate$of$the$coolant.$ (3)$Individualized$solution$$$ 13-2$ P13-1)(a)$Continued$ (vi)$ Calculated values of DEQ variables$$ Variable Initial value Minimal value […]

P12–23)(c)$ 12–96$ $ $ $ ) $ $ $ 12–97$ P12–23)(c))continued$ $ $ $ ) P12-24$ T(K)$ $ 800$ 0.92$ 700$ 1.06$ 600$ 1.0755$ 500$ 0.78$ 650$ 1.1025$ 625$ 1.099$ 675$ 1.088$ 12–98$ P12–24)Continued$ $ $ $ ) 12–99$ P12-25)(a)$ […]

12–81$ P12–16)(i))continued$ See$Polymath$program$P12–16–i.pol.$ $ $ P12–16)(j)$ Lowing$T0$or$Ta$or$increasing$UA$will$help$keep$the$reaction$running$at$the$lower$steady$state.$ $ ) P12-17$ TC$=$Ta$=$T0$$=$330$K$ $ See$Polymath$program$P12–18.pol.$ Calculated)values)of)DEQ)variables$$ τ$=$V/υ0$=$1.2$h$ V$=$FA0X/-rA$=$CA0υ0X/$-rA$$ -rA$=$k(CA$–$CB/KC)$=$kCA0(1–X$–$X/KC)$ k$=$0.001exp(30000/1.987(1/300–1/T)$ X$=$τk/(1+τK+τK/KC)$ G(T) = (–∆HRx)X$ G(T)$is$plotted$as$a$function$of$$T.$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ Cpo$$ 250.$$ 250.$$ 250.$$ 250.$$ 2$$ DHrx$$ –4.2E+04$$ –4.2E+04$$ –4.2E+04$$ […]

12–61$ P12–10)(f))continued$ $ For$counter–current$flow,$ See$Polymath$program$P12–10-f-counter.pol.$ Calculated)values)of)the)DEQ)variables$ Variable$ initial$value$ minimal$value$ maximal$value$ final$value$ $V$ 0$ 0$ 10$ 10$ $X$ 0$ 0$ 0.3458817$ 0.3458817$ $T$ 300$ 300$ 449.27319$ 449.27319$ $Ta$ 423.8$ 423.8$ 450.01394$ 450.01394$ $K$ 0.01$ 0.01$ 2.5406259$ 2.5406259$ $Kc$ 0.3567399$ 0.3567399$ 9.8927301$ […]

12–41$ P12–7)(c))continued$ $ $ Next$increase$the$coolant$flow$rate$and$run$the$same$program$to$compare$results.$ $ P12–7)(d)$ For$counter–current$flow,$swap$(T$–$Ta)$with$(Ta–T)$in$dTa/dV$equation$in$the$previous$Polymath$ See$Polymath$program$P12-7-d.pol.$ Calculated)values)of)the)DEQ)variables$ program.$ Variable$ initial$value$ minimal$value$ maximal$value$ final$value$ V$ 0$ 0$ 10$ 10$ X$ 0$ 0$ 0.3647241$ 0.3647241$ T$ 300$ 300$ 463.44558$ 450.37724$ Ta$ 440.71$ 440.71$ 457.98124$ 450.00189$ k$ 0.01$ 0.01$ 3.7132516$ […]

P12-1)(c))Continued$ Case$4:$Countercurrent$conditions$ We$need$to$enter$Ta$(V$=0)$values$such$that$at$V=Vf,$Taf$=$1000K,$1175K$and$1350K$respectively$ 12–21$ $ Ta$(V=0)$$(K)$ Ta(V=Vf)$(K)$ 983.75$ 1000$ 992$ 1175$ 999$ 1350$ $ $ $ P12-1$(d))Aspen$problem$ No$solution$will$be$provided$ $ P12-1$(e))$ (i)$There$are$at$least$two$solutions$for$16415 𝑅<! ! <16500 𝑅$ (ii)$The$conversion$is$0.8$at$T$=$605$R$ $ (iii)$525$K$<$T0$<$542$K$ $ If$the$flow$rate$of$methanol$were$increased$by$a$factor$of$4,$the$new$operating$range$is:$ 527$K$<$T0$<$558$K$ $ P12-1$(f))$ (i)$Increase$in$activation$energy$decreases$the$conversion$obtained$from$mole$balance,$but$has$no$effect$ on$the$conversion$obtained$from$energy$balance.$Increase$in$enthalpy$of$the$reaction$decreases$the$ conversion$obtained$from$energy$balance,$but$has$no$effect$on$the$conversion$obtained$from$mass$ balance.$$ $ A$set$of$values$where$80%$conversion$is$achieved$while$maintaining$the$temperature$below$125$F$is:$ E$=$30,000$Btu/lb$mol$R;$Enthalpy$of$reaction$=$–20,000$Btu/lb$mol$ […]

12-1$ Solutions)for)Chapter)12)–)Steady–State)Nonisothermal) Reactor)Design$ ) P12-1)(a))) (i)–(vii)$Individualized$solution$ (viii)$They$separate$at$θI$=$1.1.$At$θI$=$1.2,$we$find$that$X$and$Xe$profiles$meet$at$low$α$but$separate$at$ (xii–xiv)$Individualized$solution$ (xv)$For$countercurrent$flow,$the$only$equation$which$changes$is:$ d(Ta)/d(W) = –Uarho*(T–Ta)/(mc*Cpcool) Note that the right hand side of the equation has been multiplied by –1. Now, we have to guess a value of Ta such that it matches Ta0 […]

11-15# P11-6)(b)) Mole#balance:# dX dV =−rA FA0 # Rate#law:#### −rA=kCA # Stoichiometry:### CA=CA01 1−X 1+ ε X T0 T # ε =yA0 δ # δ =1+1−1=1 # yA0=FA0 F T0 =FA0 FA0+Fi0 =1 1+ θ i # ε =1 1+ […]

11-1# Solutions)for)Chapter)11)–)Non-isothermal)Reactor)Design-The) Steady-State)Energy)Balance)and)Adiabatic)PFR)Applications) ) P11-1)(a))Example)Table)11-2) (i)#T0#and#CPA#have#the#greatest#effect,#FA0#and#k0#have#the#least#effect#on#the#temperature#profiles.# (v)#Individualized#answer# P11-1)(b))Example)11-3) (i)#No#solution#will#be#provided# (ii)#A#minimum#feed#temperature#of#320#K#must#be#maintained.# (iii)#The#critical#value#is#0.6.# (iv)#At#low#values#of#heat#of#reaction,#the#heat#that#is#released#is#negligible.#As#a#result,#the#reaction# mixture#does#not#get#heated#so#much.#Hence,#the#temperature#along#the#reactor#remains#virtually#a# constant.# (v)#T0#affects#the#rate#the#most.#The#maximum#of#the#rate#increases#with#an#increase#in#T0#and#the# distance#down#the#reactor#where#it#is#achieved#reduces.# (vi)#Set#Vfinal#=#0.8#m3# See#Polymath#program#P11-1-b.pol.# POLYMATH)Results Calculated)values)of)the)DEQ)variables Variable initial value minimal value maximal value final value V 0 0 0.8 0.8 X 0 0 0.2603491 0.2603491 Ca0 […]

10–40$ P10–17)(b)) ) $assume$CA$changes$very$slowly$w.r.t.$“a”$changing$ ) $ $ $ 10–41$ P10–17)(c)$ Find$the$new$equation$for$a:$again$assuming$CB$changes$slowly$w.r.t.$“a”.$A$better$solutions$is$to$put$$ da d τ =$–kaC$B$into$the$Polymath$program$ $ $ $ 10–42$ P10–17)(d)) $ $ $ $ 10–43$ P10–17)(e)) $ $ $$$$$$$$Not$a$good$solution.$Just$put$ da dW =kdCAa Us $into$Polymath$program.$$ $ $ P10–17)(e))Continued$ $ 10–44$ $ […]

10–21$ P10–10)(a))Continued$ ) ) ) ) ) $ $ $ ) 10–22$ P10–10)(b))$ ) ) ) ) Substituting$the$expressions$for$CV$and$CA·S$into$the$equation$for$–r’A$ $ $ ) P10–10)(c))Individualized$solution$ ) ) ) 10–23$ P10–10)(d)) First$we$need$to$calculate$the$rate$constants$involved$in$the$equation$for$–r’A$in$part$ (a).$We$can$rearrange$the$equation$to$give$the$following$ $ $ $ $ $ $ $ See$Polymath$program$P10–10–d.pol.$ $ Thus$from$the$slope$and$intercept$data$ $ […]

Solutions)for)Chapter)10)–)Catalysis)and)Catalytic)Reactors$ ) P10–1)(a))Example)10-1) (i)$$ 10-1$ .0 Bs v B B B T B CCKP KPX KX $$$$$$$ 1.39 B K= $ $ 1.038 T K= $ Therefore,$ . . TS Bs C C =$f(X)$can$be$plotted.$ $ (ii)$KB$=$3$ $ (iii)$KB$=$1.2$ $ P10–1)(b))Example)10-2) […]

9-28$ P9–11)(a))continued$ $ $ $ P9–11)(b)$ $ $ $ $ 9-29$ P9–11)(b))continued$ $ $ P9–11)(c))Individualized$solution$ P9–11)(d)$Individualized$solution$ $ ) P9–12)$ For$No$Inhibition,$using$regression,$ $Equation$model:$ $ $$$$a0$=$0.008$$a1$=$0.0266$ For$Maltose,$ $Equation$model:$ $ $ $$$a0$=$0.0098$a1$=$0.33$ For$α–dextran,$ $Equation$model:$ $ $ $$$a0$=$0.008$a1$=$0.0377$ ⇒$Maltose$show$non–competitive$inhibition$as$slope$and$intercept,$both$changing$compared$to$no$ inhibition$case.$ $ ⇒$α-dextran$show$competitive$inhibition$as$intercept$same$but$slope$increases$compared$to$no$ inhibition$case.$ $ $ […]

9-8$ P9–3)Continued$ $ $ $ P9-4)(a)$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ Active$intermediates:$ $$ $ $$$$$$$$$$$$$$$ $ P9–4)(a))Continued$ $$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$ $ $ $$$$$$$$$$$ $ Plugging$in$expression$for$ $:$ $$$$$$$$$ $ $$$$$$$$$$$ $ Now,$substitute$expressions$for$ $into$equation$for$ :$ […]

8-61$ P8–14)(d))continued$ Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca 0.098 3.689E–08 0.098 3.689E–08 2 Cb 0.049 1.844E–08 0.049 1.844E–08 3 Cc 0 0 0.0150519 1.497E–12 4 Cd 0 0 0.0089776 1.018E–10 […]

P8–12)(e)$continued$ [18] rb = –2*rd1–rf3 [19] rc = rd1+re2–2*rf3 [20] Scd = rc/(rd+.0000000001) [21] Sef = re/(rf+.00000000001) P8–12)(f))) The$only$change$from$part$(e)$is:$ 8-41$ T dV V See$Polymath$program$P8–12–g.pol.$ $ $ $ D D cD D dF r k C dV =− See$Polymath$program$P8–12–f.pol.$ $$ $ […]

8-21$ P8-7)(c))Continued$ Therefore$the$reaction$should$be$run$at$a$low$temperature$to$maximize$SDU,$but$not$too$low$to$limit$the$ P8-7)(d))))) ))))))) production$of$desired$product.$The$reaction$should$also$take$place$in$high$concentration$of$A$and$the$ concentration$of$D$should$be$limited$by$removing$through$a$membrane$or$reactive$distillation.$ $ DA → )and) – AA CTKr )/12000exp(4280 1−= ) ))))))) 1 UD → )and$$$$$$– DD CTKr )/15000exp(100,10 2−= ) ))))))) 2 UA → )and$$$$$$– AA CTKr )/10800exp(26 3−= ) AD DA UU […]

8-1$ Solutions)for)Chapter)8)–)Multiple)Reactions) ) P8-1)(a))Example)8-1) (i)$k1$affects$the$selectivity$and$conversion$the$most.$ (ii)$No$solution$will$be$given.$ (iii)$For$PFR$(gas$phase$with$no$pressure$drop$or$liquid$phase),$ $ 2 321 AA ACkCkk d dC −−−= τ $ $ $ 1 k d dC X= τ $ $ A BCk d dC 2 = τ $$$$$ 2 3A YCk d dC […]

7-1$ Solutions)for)Chapter)7)–)Collection)and)Analysis)of)Rate)Data) ) P7-1)(a))Example)7-4) rate$law:$ 4 2 CH CO H r kP P α β = $ Regressing$the$data$ r’(gmolCH4/gcat.min) PCO (atm) PH2 (atm) 5.2e-3 1 1 13.2e-3 1.8 1 30e–3 4.08 1 4.95e-3 1 0.1 7.42e-3 1 0.5 5.25e-3 1 […]

6-15$ P6-6$(a))Continued$ Rename$ Transport$out$the$sides$of$the$reactor:$ RA$=$kACA$= $ –rA$=$rB$=1/2$rC$ Combine$and$solve$in$Polymath$code:$ See$Polymath$program$P6-6-a.pol.$ POLYMATH)Results$ Calculated)values)of)the)DEQ)variables$ Variable initial value minimal value maximal value final value$ v 0 0 20 20 $ Fa 100 57.210025 100 57.210025$ Fb 0 0 9.0599877 1.935926 $ Fc 0 0 […]

5-41$ P5–24)Continued$ $ $ ) $ P5–25))$ δ=0“,”ε=0 ∴p=1− αW ( ) 1 2 $ Mole$Balance/Design$Equation$ $ 5-42$ P5-25)Continued$ Rate$Law$ $ Stoichiometry$ CA=CA0 1−X ( ) p $ Combining$ F A0 dX dW =kCA0 21−X ( ) 2p2 $ $ […]

5-21$ P5–14)(c))continued$ $ So,$we$see$the$maximum$rate$in$case$with$pressure$drop$is$at$catalyst$weight$equal$to$around$600$Kg.$ $$ To$achieve$70%$conversion,$catalyst$weight$required$is$932.3$kg$.$ $ In$case$of$(a),$915.5$kg$of$catalyst$is$required$to$achieve$70%$conversion.$ 5-22$ P5–14)(d))Individualized$solution$ P5–14)(e))Individualized$solution$ ) $ P5–15)$ Gaseous$reactant$in$a$tubular$reactor:$A$→$B$ $ $ $ $ $ $ $ $ $ At$T2$=$260°F$=$720°R,$with$k1$=$0.0015$min-1$at$T1$=$80°F$=$540°R,$ $ $ $ $ $$ $ Therefore$14$pipes$are$necessary.$ ) $$$ $ $$$$ $ $ $ $ […]

5-1$ Solutions)for)Chapter)5)–)Isothermal)Reactor)Design-)Conversion) ) P5–1)(a))Example)5-3$ For$50%$conversion,$X$=$0.5$and$k$=$3.07$sec–1$at$1100$K$(from$Example$5–3)$ Now,$we$have$from$the$example$ $ $ $ =$35.47$X$0.886$ft3$ =$31.44$ft3$ Now,$ n$=$31.44$ft3/0.0205$ft2$X40$ft$=$38.33$ So,$we$see$that$for$lower$conversion$and$required$flow$rate$the$volume$of$the$reactor$is$reduced.$ $ P5–1)(b)$Example)5-4$ Individualized$Solution$ $ P5–1)(c))Example)5-5) New$Dp$=$3D0/4$ Because$the$flow$is$turbulent$ $ $ $ $ $p $ $ Now,$ $ ,$so$too$much$pressure$drop$P$=$0$and$the$flow$stops.$ P5–1)(d))Example)5–6$$ For$turbulent$flow$ $ FB$=$200X106$/$(365$X$24$X$3600$X$28)$lbmol/sec$=$0.226$lbmol/sec$ FAO$=$$ $ $$$$Also,$ $$ 5-2$ […]

4-1$ Solutions)for)Chapter)4)–)Stoichiometry) ) P4-1)(a))Example)4-4)) (i)$The$critical$value$is$around$0.5$atm$ (iv)$Individualized$Solution$ $ (v)$See$Polymath$program$P4-1-av.pol.$ $ $ (ii)$Kp$has$the$least$effect,$and$KSO3$has$the$greatest$effect$ $ (iii)$0.25$ $ POLYMATH Report Ordinary Differential Equations Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 epsilon –0.14 –0.14 –0.14 –0.14 2 […]

2-14$ P2–10)(e)$ The$amount$of$catalyst$necessary$to$achieve$40$%$conversion$in$a$single$PBR$can$be$found$from$ calculating$the$area$of$the$shaded$region$in$the$graph$below.$ $ P2–10)(f)$ $ $ ) $ The$necessary$catalyst$weight$is$approximately$13$kg.$ $ 3-1$ Solutions)for)Chapter)3)–)Rate)Laws) ) P3–1)(a))$ (i)$Individualized$solution$ (ii)$2550$K$ (iii)$Individualized$solution$ ) P3–1)(b)$ (i)$The$equilibrium$concentration$changes,$but$the$equilibrium$conversion$remains$the$same$in$all$the$ three$cases$(50%).$The$time$taken$to$attain$equilibrium$remains$the$same$in$all$the$three$cases.$ $ $ P3–1)(c)$ $ (ii)$The$trajectories$remain$similar,$but$the$time$taken$to$attain$equilibrium$changes,$as$the$rate$ constants$are$lower.$ $ (iii)$The$forward/reverse$reaction$goes$to$completion.$ $ (iv)$This$is$the$nature$of$a$stochastic$simulation.$The$number$of$molecules$in$the$simulation$are$very$less,$ as$compared$to$the$large$number$in$the$deterministic$model$(number$of$molecules$are$of$the$order$of$ the$Avogadro$number).$The$fluctuations$reduce$as$you$increase$the$number$of$molecules,$and$the$ stochastic$model$is$identical$to$a$deterministic$model$when$there$are$infinite$molecules.$ $ (v)$The$fluctuations$in$the$trajectories$reduce.$$ $ (vi)$No,$equilibrium$only$means$that$the$forward$reaction$rate$is$equal$to$the$reverse$reaction$rate.$It$ […]

Solutions)for)Chapter)2)–)Conversion)and)Reactor)Sizing) ) P2–1)(a))Example)2–1)through)2-3$ If$flow$rate$FAO$is$cut$in$half.$ $v1$=$v/2$,$$F1=$FAO/2$$and$CAO$will$remain$same.$ Therefore,$volume$of$CSTR$in$example$2-1,$ $ If$the$flow$rate$is$doubled,$ $$$$$$$$$$$$$$$F2$=$2FAO$$$$and$CAO$will$remain$same,$ Volume$of$CSTR$in$example$2-1,$ V2$=$F2X/-rA$=$12.8$m3$ $ P2–1)(b))No$solution$will$be$given$ $ P2–1)(c))No$solution$will$be$given$ $ P2–1)(d))Example)2-4$$$ $ 2-1$ $ Now,$FAO$=$0.4/2$=$0.2$mol/s,$ Table:$Divide$each$term$ $ in$Table$2–3$by$2.$ X$ 0$ 0.1$ 0.2$ 0.4$ 0.6$ 0.7$ 0.8$ [FAO/-rA](m3)$ 0.445$ 0.545$ 0.665$ 1.025$ 1.77$ 2.53$ 4$ […]

xi Study Problems: P2-7A 4) Monday, September 21 Topic: Lecture 4 – Chapter 4, Stoichiometry Batch Systems Read: Chapter 4 Section 4.1 Hand In: Problem Set 3: Define θi, θA, θB, and δ, P2-10B, P3-5A, P3-8B, P3-11B, P3-13 A In-Class […]

Solutions Manual for Essentials of Chemical Reaction Engineering Second Edition H. Scott Fogler Ame and Catherine Vennema Professor of Chemical Engineering and The Arthur F. Thurnau Professor at the University of Michigan, Ann Arbor, Michigan Boston • Columbus • Indianapolis […]