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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 […]

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)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) […]

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$ $ […]

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$ $ […]

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 […]

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+ […]

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 […]

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$mol$ […]

12–41$ 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$ […]

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$ […]

12–81$ 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$$ –4.2E+04$$ […]

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)$ […]

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 […]

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)) ) ) ) ) […]

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$ […]

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.$It$ […]

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 […]

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$ […]

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.$ ) $$$ $ $$$$ $ $ $ $ […]

5-41$ 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 $ $ […]

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 […]

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 […]

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 […]

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 […]

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$P8–12–f.pol.$ $$ $ […]

8-61$ 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 […]

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

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$case.$ $ $ […]