Essentials of Chemical Reaction Engineering 2nd Edition

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

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 i, A, B, and , P2-10B, P3-5A, P3-8B, P3-11B, P3-13 A In-Class Problem: 3 - Bring i>clickers (tentative) - Test Run of System

10-1$.0Bs v B B B T BCCKP KPX KX$$$$$$$1.39BK=$ $1.038TK=$Therefore,$..TSBsCC=$f(X)$can$be$plotted.$$(ii)$KB$=$3$$(iii)$KB$=$1.2$$P10-1)(b))Example)10-2)(i)$ Increasing$the$pressure$will$increase$the$conversion$for$the$same$catalyst$weight.$At$40$atm,$we$had$68.2%$conversion$for$a$catalyst$weight$of$10000kg.$While,$at$80$atm$we$have$85.34%$conversion$for$the$same$weight$of$the$catalyst.$$$$At$1$atm,$we$get$0.95%$conversion$for$10000$kg$of$catalyst.$However,$its$not$practically$possible$to$operate$at$inlet$pressure$of$1$atm,$because$there$will$be$no$flow$of$feed$into$the$PFR,$due$to$absence$of$pressure$difference.$$(ii)$ If$the$flow$rate$is$decreased$the$conversion$will$increase$for$two$reasons:$(a)$Smaller$pressure$drop$(b)$Reactants$spend$more$time$in$the$reactor.$$(iii)-(iv)$Individualized$solution$$(v)$$$$$$From$figure$E10-3.1$we$see$that$when$X$=$0.6,$W$=$5800$kg.$$(vi)$$$$$k$has$to$be$varied$$P10-1)(c))Example)10-3)(1))With$the$new$data,$model$(a)$best$fits$the$data$(a)$POLYMATH)Results Nonlinear)regression)(L-M)) Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe) Variable Ini guess Value 95% confidence k 3 3.5798145 0.0026691 Kea 0.1 0.1176376 0.0014744 Ke 2 2.3630934 0.0024526 Precision R^2 = 0.9969101 R^2adj = 0.9960273 Rmsd = 0.0259656 Variance = 0.0096316 (b)$POLYMATH)Results Nonlinear)regression)(L-M))

10-21$ P10-10)(a))Continued$ ) ) ) ) ) ) ) ) $ $ $ 10-22$ P10-10)(b))$ ) ) ) ) )) )) ) ) Substituting$the$expressions$for$CV$and$CAS$into$the$equation$for$rA$ $ $ ) P10-10)(c))Individualized$solution$ ) 10-23$ P10-10)(d)) First$we$need$to$calculate$the$rate$constants$involved$in$the$equation$for$rA$in$part$ (a).$We$can$rearrange$the$equation$to$give$the$following$ $ $ $ $Thus$from$the$slope$and$intercept$data$$$ $Thus$from$the$slope$and$intercept$data$$$ $ Thus$from$the$slope$and$intercept$data$ $ $ $ $ $ $ See$Polymath$program$P10-10-d.pol.$ 10-24$ P10-10)(d)$continued$ POLYMATH)Results Calculated)values)of)the)DEQ)variables Variable initial value minimal value maximal value final value W 0 0 23 23 X 0 0 0.9991499 0.9991499 e 1 1 1 1 Pao 10 10 10 10 Pa 10 0.0042521 10 0.0042521 k1 560 560 560 560 k2 2.04 2.04 2.04

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$$ da d =$kaC$B$into$the$Polymath$program$ $ $ $ $ $ 10-42$ P10-17)(d)) $ $$ $$ $ $ $ 10-43$ P10-17)(e)) $ $ $$$$$$$$Not$a$good$solution.$Just$put$ da dW =kdCAa Us $into$Polymath$program.$$ $ $ 10-44$$$)P10-18)(a))Dda kdt =$SWUt=$SdW U dt=$DSkdt dadW dt U=$ $1DSkWaU=$If$01 DSkWU=$then$.5 2.5.2SDUWkgk===$( )( )( )22000 010AAAAA AakC XrdX a rdW F F F== =$( )20011ADSkC XkWdXdW U v=$( )02011ADSkC kWdX dWvUX=$ 10-44$$$)P10-18)(a))Dda kdt =$SWUt=$SdW U dt=$DSkdt dadW dt U=$ $1DSkWaU=$If$01 DSkWU=$then$.5 2.5.2SDUWkgk===$( )( )( )22000 010AAAAA AakC XrdX a rdW F

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.# (ii)#-3500#Cal/mol#(iii)#The#maximum#of#the#rate#increases#with#an#increase#in#T0#and#the#distance#down#the#reactor#where#this#maximum#is#achieved#decreases.#(iv)#Xe#reduces#and#X#increases#for#the#same#distance#down#the#reactor#with#an#increase#in#T0.#Since#the#reaction#is#exothermic,#according#to#Le#Chateliers#principle,#the#equilibrium#conversion#reduces.#However,#with#an#increase#in#inlet#temperature,#the#rate#of#the#reaction#increases.#Hence,#the#conversion#is#expected#to#increase.# (ii)#-3500#Cal/mol#(iii)#The#maximum#of#the#rate#increases#with#an#increase#in#T0#and#the#distance#down#the#reactor#where#this#maximum#is#achieved#decreases.#(iv)#Xe#reduces#and#X#increases#for#the#same#distance#down#the#reactor#with#an#increase#in#T0.#Since#the#reaction#is#exothermic,#according#to#Le#Chateliers#principle,#the#equilibrium#conversion#reduces.#However,#with#an#increase#in#inlet#temperature,#the#rate#of#the#reaction#increases.#Hence,#the#conversion#is#expected#to#increase.# (ii)#-3500#Cal/mol# (iii)#The#maximum#of#the#rate#increases#with#an#increase#in#T0#and#the#distance#down#the#reactor#where# this#maximum#is#achieved#decreases.# (iv)#Xe#reduces#and#X#increases#for#the#same#distance#down#the#reactor#with#an#increase#in#T0.#Since#the# reaction#is#exothermic,#according#to#Le#Chateliers#principle,#the#equilibrium#conversion#reduces.# However,#with#an#increase#in#inlet#temperature,#the#rate#of#the#reaction#increases.#Hence,#the# conversion#is#expected#to#increase.# (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.# (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.# (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 9.3 9.3 9.3 9.3 Fa0 146.7 146.7 146.7 146.7 T 330 330 341.27312 341.27312 Kc 3.099466 2.852278 3.099466 2.852278 k 4.2238337 4.2238337 9.3196276 9.3196276 Xe

11-15# P11-6)(b)) Mole#balance:# dX dV =rA FA0 # Rate#law:#### rA=kCA # Stoichiometry:### CA=CA01 1X 1+ X T0 T # =yA0 # =1+11=1 # yA0=FA0 F T0 =FA0 FA0+Fi0 =1 1+ i # =1 1+ i # T=XHRX +CPA + iCPi ( ) T0 CPA + iCPi # Enter#these#equations#into#Polymath# See#Polymath#program#P11-6-b.pol.# POLYMATH)Results Calculated)values)of)the)DEQ)variables Variable initial value minimal value maximal value final value V 0 0 500 500 X 0 0 0.417064 0.417064 Cao 0.0221729 0.0221729 0.0221729 0.0221729 Cio 0.0221729 0.0221729 0.0221729 0.0221729 theta 100 100 100 100 Fao 10 10 10 10 Cao1

12-1$ Solutions)for)Chapter)12))Steady-State)Nonisothermal) Reactor)Design$ ) P12-1)(a))) (i)-(vii)$Individualized$solution$ (viii)$They$separate$at$I$=$1.1.$At$I$=$1.2,$we$find$that$X$and$Xe$profiles$meet$at$low$$but$separate$at$ higher$.$(ix)$I$=$1.8$(x-xi)$I$$ higher$.$(ix)$I$=$1.8$(x-xi)$I$$ higher$.$ (ix)$I$=$1.8$ (x-xi)$I$$ (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 at W = 4500. We get Ta = 323.1302 K, for which Ta = Ta0 = 320 K at

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 <!!<16500 $(ii)$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$$(ii)$See$Polymath$program$P12-1-f-1.pol.$ 12-21$$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 <!!<16500 $(ii)$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$$(ii)$See$Polymath$program$P12-1-f-1.pol.$ 12-21$ $ Ta$(V=0)$$(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 <! ! <16500 $ (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$mol$ $ (ii)$See$Polymath$program$P12-1-f-1.pol.$ 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$ 12-22$ P12-1)(f))Continued$ POLYMATH Report Nonlinear Equations Calculated values of NLE variables Variable Value f(x) Initial Guess 1 T 563.7289 -9.319E-10 500. 2 X 0.3636087 3.864E-11 0.5 P12-1)(f))Continued$ Variable Value 1 A 1.696E+11 2 E 3.24E+04 3 k 0.0464898 4 R 1.987 5 tau 12.29

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$$Ta)$with$(Ta-T)$in$dTa/dV$equation$in$the$previous$Polymath$ program.$ program.$ program.$ See$Polymath$program$P12-7-d.pol.$ Calculated)values)of)the)DEQ)variables$ Variable$ initial$value$ minimal$value$ maximal$value$ final$value$ V$ 0$ 0$ 10$ 10$ X$ 0$ 0$ 0.3647241$ 0.3647241$ T$ 300$ 300$ 463.44558$ 450.37724$ Ta$ 440.71$ 440.71$ 457.98124$ 450.00189$ k$ 0.01$ 0.01$ 3.7132516$ 2.7077022$ Kc$ 286.49665$ 8.2274817$ 286.49665$ 9.9439517$ Fa0$ 0.2$ 0.2$ 0.2$ 0.2$ Ca0$ 0.1$ 0.1$ 0.1$ 0.1$ ra$ -1.0E-04$ -0.0256436$ -1.0E-04$ -9.963E-04$ Xe$ 0.8298116$ 0.3488462$ 0.8298116$ 0.381006$ DH$ -6000$ -6000$ -6000$ -6000$ Ua$ 20$ 20$ 20$ 20$ Fao$ 0.2$ 0.2$ 0.2$ 0.2$ sumcp$ 30$ 30$ 30$ 30$ mc$ 50$ 50$ 50$ 50$ Cpc$ 1$ 1$ 1$ 1$ 12-42$ P12-7)(d))continued$ ODE)Report)(RKF45)$ Differential$equations$as$entered$by$the$user$ $[1]$d(X)/d(V)$=$-ra$/$Fa0$ $[2]$d(T)/d(V)$=$(Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)$ $[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$$Explicit$equations$as$entered$by$the$user$$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$$[2]$Kc$=$10$*$exp(-6000$/$1.987$*$(1$/$450$-$1$/$T))$$[3]$Fa0$=$0.2$$[4]$Ca0$=$0.1$$[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$$[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$$[7]$DH$=$-6000$$[8]$Ua$=$20$$[9]$Fao$=$0.2$$[10]$sumcp$=$30$$[11]$mc$=$50$$[12]$Cpc$=$1$ $[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$$Explicit$equations$as$entered$by$the$user$$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$$[2]$Kc$=$10$*$exp(-6000$/$1.987$*$(1$/$450$-$1$/$T))$$[3]$Fa0$=$0.2$$[4]$Ca0$=$0.1$$[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$$[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$$[7]$DH$=$-6000$$[8]$Ua$=$20$$[9]$Fao$=$0.2$$[10]$sumcp$=$30$$[11]$mc$=$50$$[12]$Cpc$=$1$ $[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$ $ Explicit$equations$as$entered$by$the$user$ $[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$ $[2]$Kc$=$10$*$exp(-6000$/$1.987$*$(1$/$450$-$1$/$T))$ $[3]$Fa0$=$0.2$ $[4]$Ca0$=$0.1$ $[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$ $[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$ $[7]$DH$=$-6000$ $[8]$Ua$=$20$ $[9]$Fao$=$0.2$ $[10]$sumcp$=$30$ $[11]$mc$=$50$ $[12]$Cpc$=$1$ $ $ $ $ $ 12-43$ P12-7)(e))Adiabatic)$ See$Polymath$program$P12-7-e.pol.$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ Ca0$$ 0.1$$ 0.1$$ 0.1$$ 0.1$$ 2$$ Cpc$$ 1.$$ 1.$$ 1.$$ 1.$$ 3$$ DH$$ -6000.$$ -6000.$$ -6000.$$ -6000.$$ 4$$ Fa0$$ 0.2$$ 0.2$$ 0.2$$ 0.2$$ 5$$ k$$ 0.01$$ 0.01$$ 0.010587$$ 0.010587$$ 6$$ Kc$$ 286.4967$$ 276.8576$$ 286.4967$$ 276.8576$$ 7$$ mc$$ 50.$$ 50.$$ 50.$$ 50.$$ 8$$ Qg$$ 0.6$$ 0.6$$ 0.6286162$$ 0.6286162$$ 9$$ Qr$$ 0$$ 0$$ 0$$ 0$$ 10$$ ra$$ -0.0001$$ -0.0001048$$ -0.0001$$ -0.0001048$$ 11$$ sumCp$$ 30.$$ 30.$$ 30.$$ 30.$$ 12$$ T$$ 300.$$ 300.$$ 301.0235$$ 301.0235$$ 13$$ Ta$$ 450.$$ 450.$$ 450.$$ 450.$$ 14$$ Ua$$ 0$$ 0$$ 0$$ 0$$ 15$$ V$$ 0$$ 0$$ 10.$$ 10.$$ 16$$ X$$ 0$$ 0$$ 0.0051176$$ 0.0051176$$ 17$$ Xe$$ 0.8298116$$ 0.827152$$ 0.8298116$$ 0.827152$$ $ Differential)equations$$1$$d(X)/d(V)$=$-ra/Fa0$$ Differential)equations$$1$$d(X)/d(V)$=$-ra/Fa0$$ Differential)equations$$ 1$$ d(X)/d(V)$=$-ra/Fa0$$ 2$$ d(T)/d(V)$=$(Ua*(Ta-T)+(ra)*DH)/(Fa0*sumCp)$$ 3$$ d(Ta)/d(V)$=$Ua*(T-Ta)/mc/Cpc$$ $ Explicit)equations$$ 1$$ Kc$=$10*exp(-6000/1.987*(1/450-1/T))$$ 2$$ k$=$0.01*exp((10000/1.987)*(1/300-1/T))$$ 3$$ Fa0$=$0.2$$ 4$$ Ca0$=$0.1$$ 5$$ ra$=$-k*(Ca0^2)*((1-X)^2-X/Ca0/Kc)$$ 6$$ Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$$ 7$$ DH$=$-6000$$ 8$$ Ua$=$20*0$$ 9$$ sumCp$=$30$$ 10$$ mc$=$50$$ 11$$ Cpc$=$1$$ 12$$ Qg$=$ra*DH$$ 13$$ Qr$=$Ua*(T-Ta)$$ 12-44$ P12-7)(e))continued$ ) $$ $$ $ $ 12-45$ P12-7)(e))continued$ $ $$ P12-7)(f))$ $ We$see$that$it$is$better$to$use$a$counter-current$coolant$flow$as$in$this$case$we$achieve$the$maximum$ equilibrium$conversion$using$a$lesser$volume$of$the$PFR.$$ equilibrium$conversion$using$a$lesser$volume$of$the$PFR.$$ equilibrium$conversion$using$a$lesser$volume$of$the$PFR.$ $ P12-7)(g))$ )$ 12-46$ P12-7)(g))continued$ $ $ $ P12-8$ Refer)to)solution)P11-6$ Heat)Exchange$ $ $ $ $ $ $ $ $$$ (16B)$ $$$A.$Constant$Ta$(17B)$Ta$=$300$$$Additional$Parameters$(18B$$(20B):$$Ta,$ $,$Ua,$$$$B.$Variable$Ta$Co-Current$$$$(17C)$ $$$C.$Variable$Ta$Counter$Current$$$$(18C)$ $$$Guess$Ta$at$V$=$0$to$match$Ta$=$Tao$at$exit,$i.e.,$V$=$Vf$ $$$ (16B)$ $$$A.$Constant$Ta$(17B)$Ta$=$300$$$Additional$Parameters$(18B$$(20B):$$Ta,$ $,$Ua,$$$$B.$Variable$Ta$Co-Current$$$$(17C)$ $$$C.$Variable$Ta$Counter$Current$$$$(18C)$ $$$Guess$Ta$at$V$=$0$to$match$Ta$=$Tao$at$exit,$i.e.,$V$=$Vf$ $$$ (16B)$ $ $$A.$Constant$Ta$(17B)$Ta$=$300$ $$Additional$Parameters$(18B$$(20B):$$Ta,$ $,$Ua,$$ $$B.$Variable$Ta$Co-Current$ $$$ (17C) $ $ $$C.$Variable$Ta$Counter$Current$ $$$ (18C) $ $ $$Guess$Ta$at$V$=$0$to$match$Ta$=$Tao$at$exit,$i.e.,$V$=$Vf$ $ 12-47$ P12-8)(a))Variable)Ta)Co-Current$ $ $ $ $ 12-48$ P12-8)(b))Gas)Phase)Counter)Current)Heat)Exchange)Vf))=)20)dm3$ $ Note:$y$=$P$=$ P P 0 0 0 $in$5th$Edition$$ $ 12-49$ P12-8)(c)))Constant)Ta$ $ $ $ P12-8)(d))Refer$to$problem$P11-7$ $ P12-8)(e))Individualized$solution$ $ P12-8)(f))Refer$to$problem$P11-7$ $ $ 12-50$$P12-9)(a))Co-Current)Heat)Exchange$Gas$Phase$Co-current$Variable$Ta$$ 12-50$$P12-9)(a))Co-Current)Heat)Exchange$Gas$Phase$Co-current$Variable$Ta$$ 12-50$ $ P12-9)(a))Co-Current)Heat)Exchange$ Gas$Phase$Co-current$Variable$Ta$ $ P12-9$ Refer$to$solution$P11-7$$ )For)Heat)Exchanger $ A$ terms$$$$$$$$$$$$1$$15$$$$$$$$$$$$Same$as$adiabatic$ Energy$Balance$(20),$ $$ $ $ $Energy$Balance$ $ $ $ $ $Energy$Balance$ $ $ $ $ $ Energy$Balance$ $ $ $(21)$ $ A$ Constant$Ta$(22A)$ Ta$=$300$K$ B$ Co-Current$Exchange$(22B)$ dT a dW =mcCPCoolC$ Counter$Current$(22C)$ $,$$$where$$mc$=$018,$CPCool=18$ dW =mcCPCoolC$ Counter$Current$(22C)$ $,$$$where$$mc$=$018,$CPCool=18$ dW = mcCP Cool C$ Counter$Current$(22C)$

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$ 9.8927301$ $Fa0$ 0.2$ 0.2$ 0.2$ 0.2$$$ $Ca0$ 0.1$ 0.1$ 0.1$ 0.1$$$ $Ra$ -1.0E-04$ -0.0141209$ -1.0E-04$ -0.0019877$ $Xe$ 0.0333352$ 0.0333352$ 0.3801242$ 0.3801242$ $DH$ 6000$ 6000$ 6000$ 6000$$ $Ua$ 20$ 20$ 20$ 20$ $Fao$ 0.2$ 0.2$ 0.2$ 0.2$ $sumcp$ 30$ 30$ 30$ 30$ $mc$ 50$ 50$ 50$ 50$ $Cpc$ 1$ 1$ 1$ 1$ $ ODE)Report)(RKF45)$ Differential$equations$as$entered$by$the$user$ $[1]$d(X)/d(V)$=$-ra$/$Fa0$ $[2]$d(T)/d(V)$=$(Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)$$[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$$ $[2]$d(T)/d(V)$=$(Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)$$[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$$ $[2]$d(T)/d(V)$=$(Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)$ $[3]$d(Ta)/d(V)$=$Ua*(Ta-T)/mc/Cpc$ $ Explicit$equations$as$entered$by$the$user$ $[1]$k$=$.01$*$exp((10000$/$2)$*$(1$/$300$-$1$/$T))$ $[2]$Kc$=$10$*$exp(6000$/$2$*$(1$/$450$-$1$/$T))$ $[3]$Fa0$=$0.2$ $[4]$Ca0$=$0.1$$[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$$[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$$[7]$DH$=$6000$$[8]$Ua$=$20$$[9]$Fao$=$0.2$$[10]$sumcp$=$30$$[11]$mc$=$50$$[12]$Cpc$=$1$$ $[4]$Ca0$=$0.1$$[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$$[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$$[7]$DH$=$6000$$[8]$Ua$=$20$$[9]$Fao$=$0.2$$[10]$sumcp$=$30$$[11]$mc$=$50$$[12]$Cpc$=$1$$ $[4]$Ca0$=$0.1$ $[5]$ra$=$-k$*$(Ca0$^$2)$*$((1$-$X)$^$2$-$X$/Ca0/$Kc)$ $[6]$Xe$=$(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2$ $[7]$DH$=$6000$ $[8]$Ua$=$20$ $[9]$Fao$=$0.2$ $[10]$sumcp$=$30$ $[11]$mc$=$50$ $[12]$Cpc$=$1$ $ 12-62$ P12-10)(f))continued$ $ $ $ $ P12-11)(a)$ (1)$ Reaction:$+.$Equal$molar$feed,$irreversible$elementary$reaction.$ Mole$Balance:!" !" = ! ! !!! $ Rate$Law:$!=!!!$,$with$!=!!exp [!!(!!!!!)],$!(!) = 0.01,$E=10000$cal/mol,$ Rate$Law:$!=!!!$,$with$!=!!exp [!!(!!!!!)],$!(!) = 0.01,$E=10000$cal/mol,$ Rate$Law:$ !=! !!$,$with$!=! !exp [! !(! !! ! !)],$!( !) = 0.01,$E=10000$cal/mol,$ R=1.98$cal/mol/K,$ !=300$ Stoichiometry$(liquid):$$ !=!= !!1 ! $ Energy$balance:$$Reactor:$ =!+(!)(!" )!![!" +!" ]$!" =6000 /$!" =!" =15 /$!!= 0.1/$!!=0.2/$ Energy$balance:$$Reactor:$ =!+(!)(!" )!![!" +!" ]$!" =6000 /$!" =!" =15 /$!!= 0.1/$!!=0.2/$ Energy$balance:$$ Reactor:$ = !+( !)(!" ) !![!" +!" ]$ !" =6000 /$ !" =!" =15 /$ !!= 0.1/$ !!=0.2/$ 12-63$ P12-11)(a))continued$ !=323$ =0.08 .=0.0191 .$ =0.08 .=0.0191 .$ =0.08 . =0.0191 .$ $ $ $$$ $ $ 12-64$ P12-11)(a)$continued$ (2)$ when$!! !$increase$by$a$factor$of$3000:$$$ =240 .=57.4

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$$=$330$K$ $=$V/0$=$1.2$h$V$=$FA0X/-rA$=$CA00X/$-rA$$-rA$=$k(CA$$CB/KC)$=$kCA0(1-X$$X/KC)$k$=$0.001exp(30000/1.987(1/300-1/T)$X$=$k/(1+K+K/KC)$G(T) = (-HRx)X$G(T)$is$plotted$as$a$function$of$$T.$ $=$V/0$=$1.2$h$V$=$FA0X/-rA$=$CA00X/$-rA$$-rA$=$k(CA$$CB/KC)$=$kCA0(1-X$$X/KC)$k$=$0.001exp(30000/1.987(1/300-1/T)$X$=$k/(1+K+K/KC)$G(T) = (-HRx)X$G(T)$is$plotted$as$a$function$of$$T.$ $=$V/0$=$1.2$h$ V$=$FA0X/-rA$=$CA00X/$-rA$$ -rA$=$k(CA$$CB/KC)$=$kCA0(1-X$$X/KC)$ k$=$0.001exp(30000/1.987(1/300-1/T)$ X$=$k/(1+K+K/KC)$ G(T) = (-HRx)X$ G(T)$is$plotted$as$a$function$of$$T.$ $ See$Polymath$program$P12-18.pol.$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ Cpo$$ 250.$$ 250.$$ 250.$$ 250.$$ 2$$ DHrx$$ -4.2E+04$$ -4.2E+04$$ -4.2E+04$$ -4.2E+04$$ 3$$ E$$ 3.0E+04$$ 3.0E+04$$ 3.0E+04$$ 3.0E+04$$ 4$$ GT$$ 50.33959$$ 0.0001053$$ 3.61E+04$$ 0.0001053$$ 5$$ k$$ 0.001$$ 0.001$$ 8.48E+07$$ 8.48E+07$$ 6$$ Kc$$ 5.0E+06$$ 2.507E-09$$ 5.0E+06$$ 2.507E-09$$ 7$$ t$$ 0$$ 0$$ 300.$$ 300.$$ 8$$ T$$ 300.$$ 300.$$ 600.$$ 600.$$ 9$$ tau$$ 1.2$$ 1.2$$ 1.2$$ 1.2$$ 10$$ X$$ 0.0011986$$ 2.507E-09$$ 0.8596127$$ 2.507E-09$$ ) 12-82$ P12-17)continued$ Differential)equations$$ 1$$ d(T)/d(t)$=$1$$ $ Explicit)equations$$ 1$$ E$=$30000$$ 2$$ k$=$0.001*exp((E/1.987)*(1/300-1/T))$$ 3$$ DHrx$=$-42000$$ 4$$ Kc$=$5000000*exp((DHrx/1.987)*(1/300-1/T))$$ 5$$ Cpo$=$250$$ 6$$ tau$=$1.2$$ 7$$ X$=$tau*k/(1+tau*k$+$tau*k/Kc)$$ 8$$ GT$=$X*(-DHrx)$$ $ $ $ From)the)plot,)the)maximum)conversion)achieved,)Xmax)=)0.86$ At$Xmax,$T$=$367.6$K$and$G(T)$=$36100$cal/mol$ R(T)$=$G(T)$$$$=$2.84$Since$$ $$ R(T)$=$G(T)$$$$=$2.84$Since$$ $$ R(T)$=$G(T)$ $ $ $=$2.84$Since$$ $$ Therefore,)UA)=)(2.84)(10)mol/h)(250)cal/mol/K))=)7,100))cal/h/K$ $ 12-83$ P12-18) Note$p$=$y$ Adiabatic)$ Polymath)Report$ No$Title$$ Ordinary$Differential$Equations$ 08-May-2009$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ alpha$$ 0.019$$ 0.019$$ 0.019$$ 0.019$$ 2$$ Ca$$ 1.9$$ 0.7510709$$ 1.902453$$ 0.7510709$$ 3$$ Ca0$$ 1.9$$ 1.9$$ 1.9$$ 1.9$$ 4$$ Cc$$ 0$$ 0$$ 0.1974831$$ 0.1113793$$ 5$$ Cp0$$ 40.$$ 40.$$ 40.$$ 40.$$ 6$$ Cpc$$ 4200.$$ 4200.$$ 4200.$$ 4200.$$ 7$$ deltaCp$$ -30.$$ -30.$$ -30.$$ -30.$$ 8$$ DH$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ 9$$ epsilon$$ -1.$$ -1.$$ -1.$$ -1.$$ 10$$ Fa0$$ 5.$$ 5.$$ 5.$$ 5.$$ 11$$ k$$ 0.01$$ 0.01$$ 0.0261675$$ 0.0261675$$ 12$$ Kc$$ 10000.$$ 902.809$$ 10000.$$ 902.809$$ 13$$ m$$ 0.05$$ 0.05$$ 0.05$$ 0.05$$ 14$$ ra$$ -0.0361$$ -0.0627617$$ -0.014758$$ -0.014758$$ 15$$ T$$ 450.$$ 450.$$ 817.9727$$ 817.9727$$ 16$$ T0$$ 450.$$ 450.$$ 450.$$ 450.$$ 17$$ Ta$$ 500.$$ 500.$$ 500.$$ 500.$$ 18$$ Ua$$ 0$$ 0$$ 0$$ 0$$ 19$$ W$$ 0$$ 0$$ 50.$$ 50.$$ 20$$ X$$ 0$$ 0$$ 0.4949381$$ 0.4949381$$ 21$$ y$$ 1.$$ 0.2174708$$ 1.$$ 0.2174708$$ $ Differential)equations$$ 1$$ d(X)/d(W)$=$-ra/Fa0$$ 2$$ d(y)/d(W)$=$-alpha/2/y*(1+epsilon*X)*T/T0$$ 3$$ d(T)/d(W)$=$(ra*(DH+deltaCp*(T-298))$-$Ua*(T-Ta))/Fa0/Cp0$$ 4$$ d(Ta)/d(W)$=$Ua*(T-Ta)/m/Cpc$$ $ Explicit)equations$$ 1$$ Cp0$=$40$$ 2$$ DH$=$-20000$$ 3$$ epsilon$=$-1$$ 4$$ T0$=$450$$ 5$$ Ca0$=$1.9$$ 6$$ Fa0$=$5$$ 7$$ alpha$=$0.019$$ 8$$ Ca$=$Ca0*(1-X)/(1+epsilon*X)*T/T0*y$$ 9$$ Ua$=$0.8*0$$ $ 10$$ deltaCp$=$1/2*20-40$$ 11$$ Kc$=$10000*exp(DH/8.314*(1/450-1/T))$$ 12$$ Cc$=$Ca0*X/2/(1+epsilon*X)*T0/T*y$$ 13$$ m$=$0.05$$ 14$$ Cpc$=$4200$$ 15$$ k$=$0.01*exp(8000/8.314*(1/450-1/T))$$ 16$$ ra$=$-k*(Ca^2-Cc/Kc)$$ $ 12-84$ P12-18$(continued)$ $ Now,)constant)temperature)Ta)=)300K) $ Polymath)Report$ $ Ordinary$Differential$Equations$ 08-May-2009$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ alpha$$ 0.019$$ 0.019$$ 0.019$$ 0.019$$ 2$$ Ca$$ 1.9$$ 0.7615715$$ 1.906858$$ 0.7615715$$ 3$$ Ca0$$ 1.9$$ 1.9$$ 1.9$$ 1.9$$ 4$$ Cc$$ 0$$ 0$$ 0.1991675$$ 0.1180575$$ 5$$ Cp0$$ 40.$$ 40.$$ 40.$$ 40.$$ 6$$ Cpc$$ 4200.$$ 4200.$$ 4200.$$ 4200.$$ 7$$ deltaCp$$ -30.$$ -30.$$ -30.$$ -30.$$ 8$$ DH$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ 9$$ epsilon$$ -1.$$ -1.$$ -1.$$ -1.$$ 10$$ Fa0$$ 5.$$ 5.$$ 5.$$ 5.$$ 11$$ k$$ 0.01$$ 0.01$$ 0.0248669$$ 0.0248669$$ 12$$ Kc$$ 10000.$$ 1025.524$$ 10000.$$ 1025.524$$ 13$$ m$$ 0.05$$ 0.05$$ 0.05$$ 0.05$$ 14$$ ra$$ -0.0361$$ -0.0610755$$ -0.0144197$$ -0.0144197$$ 15$$ T$$ 450.$$ 450.$$ 783.9972$$ 783.9972$$ 16$$ T0$$ 450.$$ 450.$$ 450.$$ 450.$$ 17$$ Ta$$ 500.$$ 500.$$ 500.$$ 500.$$ 18$$ Ua$$ 0.8$$ 0.8$$ 0.8$$ 0.8$$ 19$$ W$$ 0$$ 0$$ 50.$$ 50.$$ 20$$ X$$ 0$$ 0$$ 0.4848172$$ 0.4848172$$ 21$$ y$$ 1.$$ 0.2300674$$ 1.$$ 0.2300674$$ ) Differential)equations$$ 1$$ d(X)/d(W)$=$-ra/Fa0$$ 2$$ d(y)/d(W)$=$-alpha/2/y*(1+epsilon*X)*T/T0$$ 3$$ d(T)/d(W)$=$(ra*(DH+deltaCp*(T-298))$-$Ua*(T-Ta))/Fa0/Cp0$$ 4$$ d(Ta)/d(W)$=$Ua*(T-Ta)/m/Cpc$*0$$ 12-85$ P12-18)continued$ Explicit)equations$$ 1$$ Cp0$=$40$$ 2$$ DH$=$-20000$$ 3$$ epsilon$=$-1$$ 4$$ T0$=$450$$ 5$$ Ca0$=$1.9$$ 6$$ Fa0$=$5$$ 7$$ alpha$=$0.019$$ 8$$ Ca$=$Ca0*(1-X)/(1+epsilon*X)*T/T0*y$$ 9$$ Ua$=$0.8$$ 10$$ deltaCp$=$1/2*20-40$$ 11$$ Kc$=$10000*exp(DH/8.314*(1/450-1/T))$$ 12$$ Cc$=$Ca0*X/2/(1+epsilon*X)*T0/T*y$$ 13$$ m$=$0.05$$ 14$$ Cpc$=$4200$$ 15$$ k$=$0.01*exp(8000/8.314*(1/450-1/T))$$ 16$$ ra$=$-k*(Ca^2-Cc/Kc)$$ $ Now,)Co-current)heat)exchanger) 12-86$ P12-18)continued$ Polymath)Report$ No$Title$$ Ordinary$Differential$Equations$ 08-May-2009$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ alpha$$ 0.019$$ 0.019$$ 0.019$$ 0.019$$ 2$$ Ca$$ 1.9$$ 0.7614153$$ 1.906781$$ 0.7614153$$ 3$$ Ca0$$ 1.9$$ 1.9$$ 1.9$$ 1.9$$ 4$$ Cc$$ 0$$ 0$$ 0.1990505$$ 0.1177523$$ 5$$ Cp0$$ 40.$$ 40.$$ 40.$$ 40.$$ 6$$ Cpc$$ 4200.$$ 4200.$$ 4200.$$ 4200.$$ 7$$ deltaCp$$ -30.$$ -30.$$ -30.$$ -30.$$ 8$$ DH$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ 9$$ epsilon$$ -1.$$ -1.$$ -1.$$ -1.$$ 10$$ Fa0$$ 5.$$ 5.$$ 5.$$ 5.$$ 11$$ k$$ 0.01$$ 0.01$$ 0.0249108$$ 0.0249108$$ 12$$ Kc$$ 10000.$$ 1021.01$$ 10000.$$ 1021.01$$ 13$$ m$$ 0.05$$ 0.05$$ 0.05$$ 0.05$$ 14$$ ra$$ -0.0361$$ -0.0610733$$ -0.0144393$$ -0.0144393$$ 15$$ T$$ 450.$$ 450.$$ 785.126$$ 785.126$$ 16$$ T0$$ 450.$$ 450.$$ 450.$$ 450.$$ 17$$ Ta$$ 500.$$ 499.0741$$ 521.1534$$ 521.1534$$ 18$$ Ua$$ 0.8$$ 0.8$$ 0.8$$ 0.8$$ 19$$ W$$ 0$$ 0$$ 50.$$ 50.$$ 20$$ X$$ 0$$ 0$$ 0.4849408$$ 0.4849408$$ 21$$ y$$ 1.$$ 0.2296895$$ 1.$$ 0.2296895$$ $ Differential)equations$$ 1$$ d(X)/d(W)$=$-ra/Fa0$$ 2$$ d(y)/d(W)$=$-alpha/2/y*(1+epsilon*X)*T/T0$$ 3$$ d(T)/d(W)$=$(ra*(DH+deltaCp*(T-298))$-$Ua*(T-Ta))/Fa0/Cp0$$ 4$$ d(Ta)/d(W)$=$Ua*(T-Ta)/m/Cpc$$ $ Explicit)equations$$ 1$$ Cp0$=$40$$ 2$$ DH$=$-20000$$ 3$$ epsilon$=$-1$$ 4$$ T0$=$450$$ 5$$ Ca0$=$1.9$$ 6$$ Fa0$=$5$$ 7$$ alpha$=$0.019$$ 8$$ Ca$=$Ca0*(1-X)/(1+epsilon*X)*T/T0*y$$ 9$$ Ua$=$0.8$$ 10$$ deltaCp$=$1/2*20-40$$ 11$$ Kc$=$10000*exp(DH/8.314*(1/450-1/T))$$ 12$$ Cc$=$Ca0*X/2/(1+epsilon*X)*T0/T*y$$ 13$$ m$=$0.05$$ 14$$ Cpc$=$4200$$ 15$$ k$=$0.01*exp(8000/8.314*(1/450-1/T))$$ 16$$ ra$=$-k*(Ca^2-Cc/Kc)$$ $ 12-87$ P12-18)continued$ $ $ Now,)counter))current)heat)exchanger$ $ 12-88$ P12-18)continued$ Polymath)Report$ No$Title$$ Ordinary$Differential$Equations$ 08-May-2009$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ alpha$$ 0.019$$ 0.019$$ 0.019$$ 0.019$$ 2$$ Ca$$ 1.9$$ 0.7671508$$ 1.910238$$ 0.7671508$$ 3$$ Ca0$$ 1.9$$ 1.9$$ 1.9$$ 1.9$$ 4$$ Cc$$ 0$$ 0$$ 0.2023329$$ 0.1203647$$ 5$$ Cp0$$ 40.$$ 40.$$ 40.$$ 40.$$ 6$$ Cpc$$ 4200.$$ 4200.$$ 4200.$$ 4200.$$ 7$$ deltaCp$$ -30.$$ -30.$$ -30.$$ -30.$$ 8$$ DH$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ -2.0E+04$$ 9$$ epsilon$$ -1.$$ -1.$$ -1.$$ -1.$$ 10$$ Fa0$$ 5.$$ 5.$$ 5.$$ 5.$$ 11$$ k$$ 0.01$$ 0.01$$ 0.025358$$ 0.025358$$ 12$$ Kc$$ 10000.$$ 976.5939$$ 10000.$$ 976.5939$$ 13$$ m$$ 0.05$$ 0.05$$ 0.05$$ 0.05$$ 14$$ ra$$ -0.0361$$ -0.062846$$ -0.0149206$$ -0.0149206$$ 15$$ T$$ 450.$$ 450.$$ 796.6909$$ 796.6909$$ 16$$ T0$$ 450.$$ 450.$$ 450.$$ 450.$$ 17$$ Ta$$ 520.$$ 500.0379$$ 521.8302$$ 500.0379$$ 18$$ Ua$$ 0.8$$ 0.8$$ 0.8$$ 0.8$$ 19$$ W$$ 0$$ 0$$ 50.$$ 50.$$ 20$$ X$$ 0$$ 0$$ 0.4958573$$ 0.4958573$$ 21$$ y$$ 1.$$ 0.2280604$$ 1.$$ 0.2280604$$ $ Differential)equations$$1$$d(X)/d(W)$=$-ra/Fa0$$ Differential)equations$$1$$d(X)/d(W)$=$-ra/Fa0$$ Differential)equations$$ 1$$ d(X)/d(W)$=$-ra/Fa0$$ 2$$ d(y)/d(W)$=$-alpha/2/y*(1+epsilon*X)*T/T0$$ 3$$ d(T)/d(W)$=$(ra*(DH+deltaCp*(T-298))$-$Ua*(T-Ta))/Fa0/Cp0$$ 4$$ d(Ta)/d(W)$=$-$Ua*(T-Ta)/m/Cpc$$ $ Explicit)equations$$ 1$$ Cp0$=$40$$ 2$$ DH$=$-20000$$ 3$$ epsilon$=$-1$$ 4$$ T0$=$450$$ 5$$ Ca0$=$1.9$$ 6$$ Fa0$=$5$$ 7$$ alpha$=$0.019$$ 8$$ Ca$=$Ca0*(1-X)/(1+epsilon*X)*T/T0*y$$ 9$$ Ua$=$0.8$$ 10$$ deltaCp$=$1/2*20-40$$ 11$$ Kc$=$10000*exp(DH/8.314*(1/450-1/T))$$ 12$$ Cc$=$Ca0*X/2/(1+epsilon*X)*T0/T*y$$ 13$$ m$=$0.05$$ 14$$ Cpc$=$4200$$ 15$$ k$=$0.01*exp(8000/8.314*(1/450-1/T))$$ 16$$ ra$=$-k*(Ca^2-Cc/Kc)$$ $ 12-89$ P12-18)continued$ $ $ $ $ P12-19) No$solution$will$be$given.$ $ ) P12-20)(a))Figure$1$matches$Figure$_C_$ ) P12-20)(b))Figure$2$matches$Figure$_A_$)P12-20)(c))Figure$3$matches$Figure$_D_$)P12-20)(d))Figure$4$matches$Figure$_B_$$) P12-20)(b))Figure$2$matches$Figure$_A_$)P12-20)(c))Figure$3$matches$Figure$_D_$)P12-20)(d))Figure$4$matches$Figure$_B_$$) P12-20)(b))Figure$2$matches$Figure$_A_$ ) P12-20)(c))Figure$3$matches$Figure$_D_$ ) P12-20)(d))Figure$4$matches$Figure$_B_$ $ ) 12-90$ P12-21$ First$note$that$CP$=$0$for$both$reactions.$This$means$that$HRx(T)$=$HRx$for$both$reactions.$ Now$start$with$the$differential$energy$balance$for$a$PFR:$ dT dV =Ua(TaT)+rij (HRxij ) FjCPj =Ua(TaT)+r 1A(H1A)+r 2B(HRx2B) FjCPj $ If$we$evaluate$this$differential$equation$at$its$maximum$we$get$ dTdV =0$and$therefore,$Ua(TaT)+r1A(HRx1C)+r2B(HRx2B)=0$$We$can$then$solve$for$r1A$from$this$information.$r1A=Ua(TaT)+r2B(HRx2B)(HRx1A)$ dTdV =0$and$therefore,$Ua(TaT)+r1A(HRx1C)+r2B(HRx2B)=0$$We$can$then$solve$for$r1A$from$this$information.$r1A=Ua(TaT)+r2B(HRx2B)(HRx1A)$ dT dV =0 $and$therefore,$ Ua(TaT)+r 1A(HRx1C)+r 2B(HRx2B)=0 $$ We$can$then$solve$for$r1A$from$this$information.$ r 1A= Ua(TaT)+r 2B(HRx2B) (HRx1A) $ r 1A= Ua(TaT)2k2DCBCC(HRx 2B) (HRx1A) $ 043.0 50000 )5000)(5.0)(2.0)(4.0(2)500325(10 1= = A r $ r 1A=0.043 =1 2 k1CCACB=1 2 k1C(0.1)(0.2) $ k1C=4.3 $ k1C(500) =k1C(400) exp E R 1 400 1 500 " # $% & ' ( ) * * + , - - $ 4.3 =0.043exp E 1.987 1 400 1 500 " # $% & ' ( ) * * + , - - $ E=18300 cal mol K $ $ Alternate)Solution:$ $ 12-91$ P12-21)continued$ $ $ $ ) P12-22! !$ $at$V$=$0,$T$=$T0$=$400$K$for$dTdV =0$"$" "$$ $at$V$=$0,$T$=$T0$=$400$K$for$dTdV =0$"$" "$$ $at$V$=$0,$T$=$T0$=$400$K$for$ dT dV =0 $ "$ " " $ $ 12-92$ $$ 12-92$ $$ 12-92$ $ $ P12-23)(a)" $ $ $ $ 12-93$ P12-23)(a))continued$ $ $ $ $ 12-94$ P12-23)(b)$ $ $ $ $ $ $ 12-95$ P12-23)(b)$continued$ $ $

12-96$ 12-96$ 12-96$ P12-23)(c)$ $ $$) $$) $ $ ) $ $ $ 12-97$ P12-23)(c))continued$ $$$)P12-24$ $$$)P12-24$ $ $ $ ) 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)$ Mole$balance:$ $ Rate$Laws:$ $ Stoichiometry:$ $ Energy$balance:$ $ $$$$$$See$Polymath$program$P12-25.pol.$ $$$$$$See$Polymath$program$P12-25.pol.$ $ $ $ $ $ $ See$Polymath$program$P12-25.pol.$ 12-100$ P12-25)(a))Continued$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ w$$ 0$$ 0$$ 100.$$ 100.$$ 2$$ fb$$ 1.$$ 0.9261241$$ 1.368476$$ 0.9261241$$ 3$$ fa$$ 1.$$ 0.6296429$$ 1.$$ 0.8694625$$ 4$$ fc$$ 0$$ 0$$ 0.2044134$$ 0.2044134$$ 5$$ T$$ 330.$$ 330.$$ 416.3069$$ 416.3069$$ 6$$ Ua$$ 16.$$ 16.$$ 16.$$ 16.$$ 7$$ Ta$$ 500.$$ 500.$$ 500.$$ 500.$$ 8$$ Dhr1a$$ -1800.$$ -1800.$$ -1800.$$ -1800.$$ 9$$ Dhr3a$$ -1100.$$ -1100.$$ -1100.$$ -1100.$$ 10$$ cpa$$ 100.$$ 100.$$ 100.$$ 100.$$ 11$$ cpb$$ 100.$$ 100.$$ 100.$$ 100.$$ 12$$ cpc$$ 100.$$ 100.$$ 100.$$ 100.$$ 13$$ k1$$ 0.5312401$$ 0.5312401$$ 0.7941566$$ 0.7941566$$ 14$$ k3$$ 0.0008165$$ 0.0008165$$ 0.0030853$$ 0.0030853$$ 15$$ ct$$ 2.$$ 2.$$ 2.$$ 2.$$ 16$$ ft$$ 2.$$ 2.$$ 2.$$ 2.$$ 17$$ To$$ 330.$$ 330.$$ 330.$$ 330.$$ 18$$ Kc$$ 3.885029$$ 1.062332$$ 3.885029$$ 1.062332$$ 19$$ k2$$ 0.1367403$$ 0.1367403$$ 0.74756$$ 0.74756$$ 20$$ ca$$ 1.$$ 0.5682599$$ 1.$$ 0.6892094$$ 21$$ cb$$ 1.$$ 0.7341242$$ 1.253213$$ 0.7341242$$ 22$$ r1a$$ -0.5312401$$ -0.5748799$$ -0.362406$$ -0.5473402$$ 23$$ r3a$$ -0.0008165$$ -0.002138$$ -0.0007594$$ -0.0021264$$ 24$$ rc$$ 0.0008165$$ 0.0007594$$ 0.002138$$ 0.0021264$$ 25$$ r2b$$ -0.1367403$$ -0.5770243$$ -0.1367403$$ -0.5488019$$ 26$$ rb$$ 0.3944998$$ -0.0522707$$ 0.3944998$$ -0.0014617$$ 27$$ ra$$ -0.3953164$$ -0.3953164$$ 0.0510521$$ -0.0006647$$ $ Differential)equations$$ 1$$ d(fb)/d(w)$=$rb$$ 2$$ d(fa)/d(w)$=$ra$$ 3$$ d(fc)/d(w)$=$rc$$ 4$$ d(T)/d(w)$=$(Ua$*$(Ta$-$T)$+$(-r1a)$*$(-Dhr1a)$+$(-r2b)$*$(Dhr1a)$*$(-r3a)$*$(-Dhr3a))$/$(fa$*$cpa$+$fb$*$cpb$+$fc$*$cpc)$$ $ Explicit)equations$$ 1$$ Ua$=$16$$ 2$$ Ta$=$500$$ 3$$ Dhr1a$=$-1800$$ 4$$ Dhr3a$=$-1100$$ 5$$ cpa$=$100$$ 6$$ cpb$=$100$$ 7$$ cpc$=$100$$ 8$$ k1$=$.5$*$exp(2$*$(1$-$320$/$T))$$ 9$$ k3$=$.005$*$exp(4.6$*$(1$-$(460$/$T)))$$ 10$$ ct$=$2$$ 11$$ ft$=$2$$ 12$$ To$=$330$$ 13$$ Kc$=$10$*$exp(4.8$*$(430$/$T$-$1.5))$$ 14$$ k2$=$k1$/$Kc$$ 15$$ ca$=$ct$*$fa$/$ft$*$To$/$T$$ 16$$ cb$=$ct$*$fb$/$ft$*$To$/$T$$ 17$$ r1a$=$-k1$*$ca$$ 18$$ r3a$=$-k3$*$ca$$ 19$$ rc$=$-r3a$$ 20$$ r2b$=$-k2$*$cb$$ 21$$ rb$=$-r1a$+$r2b$$ 22$$ ra$=$-r2b$+$r1a$+$r3a$$ 12-101$ P12-25)(b)) As$seen$in$the$above$table,$the$lowest$concentration$of$o-xylene$(A)$=$.568$mol/dm3$ $ $ $ P12-25$(c)$ The$maximum$concentration$of$o-xylene$=$1$mol/dm3$ $ $ $ P12-25$(d)) The$same$equations$are$used$except$that$FB0$=$0.$$ The$lowest$concentration$of$o-xylene$=$0.638$mol/dm3.$The$highest$concentration$of$$m-xylene$=$1.09$mol/dm3.$The$maximum$concentration$of$o-xylene$=$2$mol/dm3.$$ The$lowest$concentration$of$o-xylene$=$0.638$mol/dm3.$The$highest$concentration$of$$m-xylene$=$1.09$mol/dm3.$The$maximum$concentration$of$o-xylene$=$2$mol/dm3.$$ The$lowest$concentration$of$o-xylene$=$0.638$mol/dm3.$The$highest$concentration$of$$ m-xylene$=$1.09$mol/dm3.$The$maximum$concentration$of$o-xylene$=$2$mol/dm3.$ $ P12-25$(e)$ Decreasing$the$heat$of$reaction$of$reaction$1$slightly$decreases$the$amount$of$E$formed.$Decreasing$the$ heat$of$reaction$of$reaction$3$causes$more$of$C$to$be$formed.$Increasing$the$feed$temperature$causes$ less$of$A$to$react$and$increases$formation$of$C.$Increasing$the$ambient$temperature$causes$a$lot$of$C$to$be$formed.$$ less$of$A$to$react$and$increases$formation$of$C.$Increasing$the$ambient$temperature$causes$a$lot$of$C$to$be$formed.$$ less$of$A$to$react$and$increases$formation$of$C.$Increasing$the$ambient$temperature$causes$a$lot$of$C$to$ be$formed.$ $ P12-25$(f)$Individualized$solution$ $ ) P12-26)(a)$ $ We$want$the$exiting$flow$rates$B,$D$and$F$ Start$with$the$mole$balance$in$PFR:$ $ $Rate$Laws:$ $Rate$Laws:$ $ Rate$Laws:$ $ $ $ $$$$$$ $$$$$$ $ $ $ $ $ $ 12-102$ P12-26)(a)$continued$ Stoichiometry:$ $ $$$ $$$ $ $ $ $ Energy$Balance:$ $ $ See$Polymath$program$P12-29.pol.$ For$T0$=$800K$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ v$$ 0$$ 0$$ 10.$$ 10.$$ 2$$ fa$$ 0.00344$$ 0.002496$$ 0.00344$$ 0.002496$$ 3$$ fb$$ 0$$ 0$$ 0.0008974$$ 0.0008974$$ 4$$ fc$$ 0$$ 0$$ 0.0008615$$ 0.0008615$$ 5$$ fd$$ 0$$ 0$$ 1.078E-05$$ 1.078E-05$$ 6$$ fe$$ 0$$ 0$$ 1.078E-05$$ 1.078E-05$$ 7$$ ff$$ 0$$ 0$$ 3.588E-05$$ 3.588E-05$$ 8$$ fg$$ 0$$ 0$$ 3.588E-05$$ 3.588E-05$$ 9$$ T$$ 800.$$ 765.237$$ 800.$$ 765.237$$ 10$$ Hla$$ 1.18E+05$$ 1.18E+05$$ 1.18E+05$$ 1.18E+05$$ 11$$ H2a$$ 1.052E+05$$ 1.052E+05$$ 1.052E+05$$ 1.052E+05$$ 12$$ H3a$$ -5.39E+04$$ -5.39E+04$$ -5.39E+04$$ -5.39E+04$$ 13$$ p$$ 2137.$$ 2137.$$ 2137.$$ 2137.$$ 14$$ phi$$ 0.4$$ 0.4$$ 0.4$$ 0.4$$ 15$$ Kl$$ 0.0459123$$ 0.0196554$$ 0.0459123$$ 0.0196554$$ 16$$ sr$$ 14.5$$ 14.5$$ 14.5$$ 14.5$$ 17$$ fi$$ 0.04988$$ 0.04988$$ 0.04988$$ 0.04988$$ 18$$ ft$$ 0.05332$$ 0.05332$$ 0.0542282$$ 0.0542282$$ 19$$ Pa$$ 0.1548387$$ 0.1104652$$ 0.1548387$$ 0.1104652$$ 20$$ Pb$$ 0$$ 0$$ 0.0397155$$ 0.0397155$$ 21$$ Pc$$ 0$$ 0$$ 0.0381277$$ 0.0381277$$ 22$$ r2b$$ 2.991E-06$$ 5.16E-07$$ 2.991E-06$$ 5.16E-07$$ 23$$ rd$$ 2.991E-06$$ 5.16E-07$$ 2.991E-06$$ 5.16E-07$$ 24$$ re$$ 2.991E-06$$ 5.16E-07$$ 2.991E-06$$ 5.16E-07$$ 25$$ r3t$$ 0$$ 0$$ 4.196E-06$$ 4.151E-06$$ 26$$ rf$$ 0$$ 0$$ 4.196E-06$$ 4.151E-06$$ 27$$ rg$$ 0$$ 0$$ 4.196E-06$$ 4.151E-06$$ 28$$ rls$$ 0.0002138$$ 2.481E-05$$ 0.0002138$$ 2.481E-05$$ 29$$ rb$$ 0.0002138$$ 2.481E-05$$ 0.0002138$$ 2.481E-05$$ 30$$ rc$$ 0.0002138$$ 2.066E-05$$ 0.0002138$$ 2.066E-05$$ 31$$ ra$$ -0.0002167$$ -0.0002167$$ -2.948E-05$$ -2.948E-05$$ 12-103$ P12-26)(a)$continued$ Differential)equations$$ 1$$ d(fa)/d(v)$=$ra$$ 2$$ d(fb)/d(v)$=$rb$$ 3$$ d(fc)/d(v)$=$rc$$ 4$$ d(fd)/d(v)$=$rd$$ 5$$ d(fe)/d(v)$=$re$$ 6$$ d(ff)/d(v)$=$rf$$ 7$$ d(fg)/d(v)$=$rg$$ 8$$ d(T)/d(v)$=$-(rls$*$Hla$+$r2b$*$H2a$+$r3t$*$H3a)$/$(fa$*$299$+$fb$*$273$+$fc$*$30$+$fd$*$201$+$fe$*$90$+$ff$*$68$+$fi$*$40)$$ $ Explicit)equations$$ 1$$ Hla$=$118000$$ 2$$ H2a$=$105200$$ 3$$ H3a$=$-53900$$ 4$$ p$=$2137$$ 5$$ phi$=$.4$$ 6$$ Kl$=$exp(-17.34$-$1.302e4$/$T$+$5.051$*$ln(T)$+$((-2.314e-10$*$T$+$1.302e-6)$*$T$+$-0.004931)$*$T)$$ 7$$ sr$=$14.5$$ 8$$ fi$=$sr$*$.00344$$ 9$$ ft$=$fa$+$fb$+$fc$+$fd$+$fe$+$ff$+$fg$+$fi$$ 10$$ Pa$=$fa$/$ft$*$2.4$$ 11$$ Pb$=$fb$/$ft$*$2.4$$ 12$$ Pc$=$fc$/$ft$*$2.4$$ 13$$ r2b$=$p$*$(1$-$phi)$*$exp(13.2392$-$25000$/$T)$*$Pa$$ 14$$ rd$=$r2b$$ 15$$ re$=$r2b$$ 16$$ r3t$=$p$*$(1$-$phi)$*$exp(.2961$-$11000$/$T)$*$Pa$*$Pc$$ 17$$ rf$=$r3t$$ 18$$ rg$=$r3t$$ 19$$ rls$=$p$*$(1$-$phi)$*$exp(-0.08539$-$10925$/$T)$*$(Pa$-$Pb$*$Pc$/$Kl)$$ 20$$ rb$=$rls$$ 21$$ rc$=$rls$-$r3t$$ 22$$ ra$=$-rls$-$r2b$-$r3t$$ $ P12-26)(a)$ Fstyrene$=$0.0008974$ Fbenzene$=$1.078E-05$ Ftoluene$=$3.588E-05$SS/BT$=$19.2$) Ftoluene$=$3.588E-05$SS/BT$=$19.2$) Ftoluene$=$3.588E-05$ SS/BT$=$19.2$ ) P12-26)(b)$ T0$=$930K$ Fstyrene$=$0.0019349$ Fbenzene$=$0.0002164$Ftoluene$=$0.0002034$SS/BT$=$4.6$$ Fbenzene$=$0.0002164$Ftoluene$=$0.0002034$SS/BT$=$4.6$$ Fbenzene$=$0.0002164$ Ftoluene$=$0.0002034$ SS/BT$=$4.6$ $ P12-26)(c)$ T0$=$1100$K$ Fstyrene$=$0.0016543$ Fbenzene$=$0.0016067$Ftoluene$=$0.0001275$SS/BT$=$0.95$$ Fbenzene$=$0.0016067$Ftoluene$=$0.0001275$SS/BT$=$0.95$$ Fbenzene$=$0.0016067$ Ftoluene$=$0.0001275$ SS/BT$=$0.95$ $ 12-104$ P12-26)(d)$ Plotting$the$production$of$styrene$as$a$function$of$To$gives$the$following$graph.$The$temperature$that$is$ ideal$is$995K$ ideal$is$995K$ ideal$is$995K$ $ $ P12-26)(e)$ Plotting$the$production$of$styrene$as$a$function$of$the$steam$gives$the$following$graph$and$the$ratio$that$ is$the$ideal$is$25:1$$$P12-26)(f)$ is$the$ideal$is$25:1$$$P12-26)(f)$ is$the$ideal$is$25:1$ $ $ P12-26)(f)$ See$Polymath$program$P12-29-f.pol.$ When$we$add$a$heat$exchanger$to$the$reactor,$the$energy$balance$becomes:$ $With$Ta$=$1000$K$Ua$=$100$kJ/min/K$=$1.67$kJ/s/K$The$recommended$entering$temperature$would$be$T0$=$440$K.$This$gives$the$highest$outlet$flow$rate$of$styrene.$ $With$Ta$=$1000$K$Ua$=$100$kJ/min/K$=$1.67$kJ/s/K$The$recommended$entering$temperature$would$be$T0$=$440$K.$This$gives$the$highest$outlet$flow$rate$of$styrene.$ $ With$Ta$=$1000$K$ Ua$=$100$kJ/min/K$=$1.67$kJ/s/K$ The$recommended$entering$temperature$would$be$T0$=$440$K.$This$gives$the$highest$outlet$flow$rate$of$ styrene.$ $ $ 12-105$ P12-26)(g))Individualized$solution$ $ P12-26)(h))Individualized$solution$ $ ) P12-27$ Let$$$ a$=$A$ b$=$A2$ c$=$A4$ Polymath)Report$ $ Ordinary$Differential$Equations$ 21-Jun-2010$ Calculated)values)of)DEQ)variables$ $ $ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ Ca$$ 2.$$ 0.155599$$ 2.$$ 0.155599$$ 2$$ Cb$$ 0$$ 0$$ 0.4963834$$ 0.1995088$$ 3$$ Cpa$$ 25.$$ 25.$$ 25.$$ 25.$$ 4$$ Cpb$$ 50.$$ 50.$$ 50.$$ 50.$$ 5$$ Cpc$$ 100.$$ 100.$$ 100.$$ 100.$$ 6$$ DH1a$$ -3.25E+04$$ -3.25E+04$$ -3.25E+04$$ -3.25E+04$$ 7$$ DH2b$$ -2.75E+04$$ -2.75E+04$$ -2.75E+04$$ -2.75E+04$$ 8$$ E1$$ 4000.$$ 4000.$$ 4000.$$ 4000.$$ 9$$ E2$$ 5000.$$ 5000.$$ 5000.$$ 5000.$$ 10$$ Fa$$ 100.$$ 7.779951$$ 100.$$ 7.779951$$ 11$$ Fb$$ 0$$ 0$$ 24.81917$$ 9.975439$$ 12$$ Fc$$ 0$$ 0$$ 18.06729$$ 18.06729$$ 13$$ k1$$ 0.6$$ 0.6$$ 0.6$$ 0.6$$ 14$$ k1a$$ 0.6$$ 0.6$$ 76.97387$$ 6.133163$$ 15$$ k2$$ 0.35$$ 0.35$$ 0.35$$ 0.35$$ 16$$ k2b$$ 0.2072016$$ 0.2072016$$ 89.46089$$ 3.787125$$ 17$$ Qg$$ 7.8E+04$$ 8971.335$$ 1.569E+06$$ 8971.335$$ 18$$ Qr$$ -1.5E+04$$ -1.5E+04$$ 7.696E+05$$ 1.44E+05$$ 19$$ r1a$$ -2.4$$ -42.01417$$ -0.1484903$$ -0.1484903$$ 20$$ r1b$$ 1.2$$ 0.0742452$$ 21.00708$$ 0.0742452$$ 21$$ r2b$$ 0$$ -18.03538$$ 0$$ -0.1507418$$ 22$$ ra$$ -2.4$$ -42.01417$$ -0.1484903$$ -0.1484903$$ 23$$ rb$$ 1.2$$ -8.261933$$ 17.21274$$ -0.0764966$$ 24$$ rc$$ 0$$ 0$$ 9.017691$$ 0.0753709$$ 25$$ sumCp$$ 2500.$$ 2500.$$ 2500.$$ 2500.$$ 26$$ T$$ 300.$$ 300.$$ 1084.641$$ 459.0062$$ 27$$ T1$$ 300.$$ 300.$$ 300.$$ 300.$$ 28$$ T2$$ 320.$$ 320.$$ 320.$$ 320.$$ 29$$ Ta$$ 315.$$ 315.$$ 315.$$ 315.$$ 30$$ Ua$$ 1000.$$ 1000.$$ 1000.$$ 1000.$$ 31$$ V$$ 0$$ 0$$ 10.$$ 10.$$ 32$$ vo$$ 50.$$ 50.$$ 50.$$ 50.$$ $ Differential)equations$$ 1$$ d(T)/d(V)$=$(Qg-Qr)/sumCp$$ 2$$ d(Fc)/d(V)$=$rc$$ 3$$ d(Fb)/d(V)$=$rb$$ 4$$ d(Fa)/d(V)$=$ra$$ 12-106$ P12-27$continued$ Explicit)equations$$ 1$$ Cpc$=$100$$ 2$$ Cpb$=$50$$ 3$$ Cpa$=$25$$ 4$$ k2$=$.35$$ 5$$ Ta$=$315$$ 6$$ Ua$=$1000$$ 7$$ DH2b$=$-27500$$ 8$$ DH1a$=$-$32500$$ 9$$ T1$=$300$$ 10$$ E1$=$4000$$ 11$$ k1$=$0.6$$ 12$$ T2$=$320$$ 13$$ E2$=$5000$$ 14$$ k2b$=$k2*exp((E2/1.987)*(1/T2-1/T))$$ 15$$ vo$=$50$$ 16$$ Cb$=$Fb/vo$$ 17$$ Ca$=$Fa/vo$$ 18$$ sumCp$=$(Fa*Cpa+Fb*Cpb+Fc*Cpc)$$ 19$$ k1a$=$k1*exp((E1/1.987)*(1/T1-1/T))$$ 20$$ r2b$=$-k2b*Cb^2$$ 21$$ rc$=$-r2b/2$$ 22$$ r1a$=$-k1a*Ca^2$$ 23$$ r1b$=$-r1a/2$$ 24$$ rb$=$r1b$+$r2b$$ 25$$ ra$=$r1a$$ 26$$ Qg$=$r1a*DH1a+r2b*DH2b$$ 27$$ Qr$=$Ua*(T-Ta)$$ $ 12-107$ P12-27)Continued$ $ $ $ $ P12-27)(b)$ The$required$reactor$volume$=$3.5$dm3$ $ $ $ 12-108$ P12-27)(c)) $ $ $ $ $ $ $ 12-109$ P12-27)(c)) "$ $ ) $

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)$$ (1)$Runaway$occurs$at$Ta1$=$292$K$ (2)$As$Ta1$increases,$the$maxima$of$the$Qr$versus$t,$and$Qg$versus$T$curves$occur$at$shorter$times.$This$is$ because$the$inlet$temperature$of$the$coolant$is$increased,$due$to$which$the$driving$force$for$heat$transfer$becomes$lesser.$As$NA0$increases,$Qr,max$and$Qg,max$increase.$This$is$because$we$are$starting$with$a$larger$number$of$reactant$molecules,$which$gives$rise$to$higher$generated$heat.$As$mc$is$increased,$the$heat$generated$decreases.$$This$is$because$a$larger$amount$of$heat$is$removed$with$an$increase$in$flow$rate$of$the$coolant.$(3)$Individualized$solution$$$ because$the$inlet$temperature$of$the$coolant$is$increased,$due$to$which$the$driving$force$for$heat$transfer$becomes$lesser.$As$NA0$increases,$Qr,max$and$Qg,max$increase.$This$is$because$we$are$starting$with$a$larger$number$of$reactant$molecules,$which$gives$rise$to$higher$generated$heat.$As$mc$is$increased,$the$heat$generated$decreases.$$This$is$because$a$larger$amount$of$heat$is$removed$with$an$increase$in$flow$rate$of$the$coolant.$(3)$Individualized$solution$$$ because$the$inlet$temperature$of$the$coolant$is$increased,$due$to$which$the$driving$force$for$heat$ transfer$becomes$lesser.$As$NA0$increases,$Qr,max$and$Qg,max$increase.$This$is$because$we$are$starting$with$ a$larger$number$of$reactant$molecules,$which$gives$rise$to$higher$generated$heat.$As$mc$is$increased,$the$ heat$generated$decreases.$$This$is$because$a$larger$amount$of$heat$is$removed$with$an$increase$in$flow$ rate$of$the$coolant.$ (3)$Individualized$solution$$$ $ (iii)$Coolant$flow$rate$does$not$have$a$significant$effect$on$the$conversion.$The$temperature$of$the$reactor$decreases$with$an$increase$in$coolant$flow$rate.$This$is$an$intuitive$observation.$$$ (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.$$$ (iii)$Coolant$flow$rate$does$not$have$a$significant$effect$on$the$conversion.$The$temperature$of$the$ reactor$decreases$with$an$increase$in$coolant$flow$rate.$This$is$an$intuitive$observation.$$ $ (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$new$HRn$and$T.$With$T=497+89.8X$the$polymath$program$of$ example$13-1$gives$t=$8920$s$for$90$%$conversion.$$ example$13-1$gives$t=$8920$s$for$90$%$conversion.$$ example$13-1$gives$t=$8920$s$for$90$%$conversion.$ $ 13-2$ P13-1)(a)$Continued$ (vi)$ Calculated values of DEQ variables$$ Variable Initial value Minimal value Maximal value Final value 1 t 0 0 9000. 9000. 2 X 0 0 1. 1. 3 T 480. 480. 894.193 894.193 4 k 8.307E-06 8.307E-06 56.65781 56.65781 5 Ta 896.4 896.4 896.4 896.4 6 Ua 0.22 0.22 0.22 0.22

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$$ $See$Polymath$program$P13-3-b.pol$$ $ 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$P13-3-c.pol$ See$Polymath$program$P13-3-c.pol$ See$Polymath$program$P13-3-c.pol$ 13-23$ P13-3)(c)) $ $ $ $ $ $ 13-24$ P13-4)(a)) ) ) )))) )))) ) ) ) ) $ $ P13-4)(b)) ) )) )) ) ) 13-25$ P13-4)(b)$Continued$ ) $ $ $ $ P13-5) ) ))$$$ ))$$$ ) ) $ $ $ 13-26$ P13-5)Continued$ $ $See$Polymath$program$P13-5.pol$ $See$Polymath$program$P13-5.pol$ $ See$Polymath$program$P13-5.pol$ $ POLYMATH)Results Calculated)values)of)the)DEQ)variables Variable initial value minimal value maximal value final value t 0 0 10 10 X 0 0 0.2504829 0.2504829 T 373 373 562.91803 562.91802 k1 0.002 0.002 106.13627 106.13612 Ca 0.1 0.0749517 0.1 0.0749517 Cb 0.125 0.0999517 0.125 0.0999517 k2 3.0E-05 3.0E-05 366.75159

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$4$)Reactor)1) ) ) ) Reactor)2$V1$=$0.82m3$$$$V2$=$3.2$m3$V$=$(FAO/-rA)X$0.82 =FA0rA!"##$%&&X1X1( )$ $ $3.2 =FA0rA"#$$%&''X2X2( )$By$trial$and$error$we$get:$X1$=$0.546$ $ $ $ and$$$$$$$$X2$=$0.8$Overall$conversion$XOverall$=$(1/2)X1$+$(1/2)X2$=$(0.546+0.8)/2$=$0.673$$ 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$4$)Reactor)1) ) ) ) Reactor)2$V1$=$0.82m3$$$$V2$=$3.2$m3$V$=$(FAO/-rA)X$0.82 =FA0rA!"##$%&&X1X1( )$ $ $3.2 =FA0rA"#$$%&''X2X2( )$By$trial$and$error$we$get:$X1$=$0.546$ $ $ $ and$$$$$$$$X2$=$0.8$Overall$conversion$XOverall$=$(1/2)X1$+$(1/2)X2$=$(0.546+0.8)/2$=$0.673$$ 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$ ) Reactor)1) ) ) ) Reactor)2$ V1$=$0.82m3$$$$V2$=$3.2$m3$ V$=$(FAO/-rA)X$ 0.82 = FA0 rA ! " # # $ % & & X1 X1 ( ) $ $ $ 3.2 = FA0 rA " # $ $ % & ' ' X2 X2 ( ) $ By$trial$and$error$we$get:$ X1$=$0.546$ $ $ $ and$$$$$$$$X2$=$0.8$ Overall$conversion$XOverall$=$(1/2)X1$+$(1/2)X2$=$(0.546+0.8)/2$=$0.673$ $ Solutions)for)Chapter)2))Conversion)and)Reactor)Sizing) ) P2-1)(a))Example)2-1)through)2-3$ If$flow$rate$FAO$is$cut$in$half.$ $v1$=$v/2$,$$F1=$FAO/2$$and$CAO$will$remain$same.$Therefore,$volume$of$CSTR$in$example$2-1,$$If$the$flow$rate$is$doubled,$$$$$$$$$$$$$$$$F2$=$2FAO$$$$and$CAO$will$remain$same,$Volume$of$CSTR$in$example$2-1,$V2$=$F2X/-rA$=$12.8$m3$$ $v1$=$v/2$,$$F1=$FAO/2$$and$CAO$will$remain$same.$Therefore,$volume$of$CSTR$in$example$2-1,$$If$the$flow$rate$is$doubled,$$$$$$$$$$$$$$$$F2$=$2FAO$$$$and$CAO$will$remain$same,$Volume$of$CSTR$in$example$2-1,$V2$=$F2X/-rA$=$12.8$m3$$ $v1$=$v/2$,$$F1=$FAO/2$$and$CAO$will$remain$same.$ Therefore,$volume$of$CSTR$in$example$2-1,$ $ If$the$flow$rate$is$doubled,$ $$$$$$$$$$$$$$$F2$=$2FAO$$$$and$CAO$will$remain$same,$ Volume$of$CSTR$in$example$2-1,$ V2$=$F2X/-rA$=$12.8$m3$ $ P2-1)(b))No$solution$will$be$given$ $ P2-1)(c))No$solution$will$be$given$ $ P2-1)(d))Example)2-4$$$ $ 2-2$ P2-1)(e))Example)2-5)$ (1))For$$first$CSTR,$$ At$X=0.2,$ 30m 94.0=AArF$;$V1=0.94&m3( )0.2( )=0.188&m3$ 30m 94.0=AArF$;$V1=0.94&m3( )0.2( )=0.188&m3$ 3 0m 94.0= A A r F $;$ V 1=0.94&m3 (

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.$ $ The$necessary$catalyst$weight$is$approximately$13$kg.$$ The$necessary$catalyst$weight$is$approximately$13$kg.$$ The$necessary$catalyst$weight$is$approximately$13$kg.$ $ P2-10)(f)$ $ $ $ $ ) $ 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.$ $ (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.$It$does$not$mean$that$they$stop$occurring.$This$can$be$seen$from$the$fact$that$there$is$always$a$small$fluctuation$about$the$equilibrium$concentration.$$ (ii)$The$trajectories$remain$similar,$but$the$time$taken$to$attain$equilibrium$changes,$as$the$rate$constants$are$lower.$$(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.$$$(vi)$No,$equilibrium$only$means$that$the$forward$reaction$rate$is$equal$to$the$reverse$reaction$rate.$It$does$not$mean$that$they$stop$occurring.$This$can$be$seen$from$the$fact$that$there$is$always$a$small$fluctuation$about$the$equilibrium$concentration.$$ (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.$$ $ (vi)$No,$equilibrium$only$means$that$the$forward$reaction$rate$is$equal$to$the$reverse$reaction$rate.$It$ does$not$mean$that$they$stop$occurring.$This$can$be$seen$from$the$fact$that$there$is$always$a$small$ fluctuation$about$the$equilibrium$concentration.$$ $ P3-1)(c)$ $ 3-2$ P3-1)(c))Continued$ ln =ln 1 $ From$the$graph$of$ln$k$vs$1/T$above,$we$get:$ ln =27.577, =10889 $ Thus,$ =9.474 10!! !!, =90.531 /$ =9.474 10!! exp 10889 !!$ =9.474 10!! exp 10889 !!$ =9.474 10!! exp 10889 !!$ $ P3-1)(d))Example)3-1$ For,$E$=$60kJ/mol$ $ $ $ $$$$$$$$$$$$For,$E$=$240kJ/mol$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ T)(K)$ k)(1/sec)$ 1/T$ ln(k)$ $ T)(K)$ k)(1/sec)$ 1/T$ ln(k)$ 310$ 1023100$ 0.003226$ 13.83918$ $ 310$ 4.78E-25$ 0.003226$ -56.0003$ 315$ 1480488$ 0.003175$ 14.2087$ $ 315$ 2.1E-24$ 0.003175$ -54.5222$ 320$ 2117757$ 0.003125$ 14.56667$ $ 320$ 8.77E-24$ 0.003125$ -53.0903$ 325$ 2996152$ 0.003077$ 14.91363$ $ 325$ 3.51E-23$ 0.003077$ -51.7025$ 330$ 4194548$ 0.00303$ 15.25008$ $ 330$ 1.35E-22$ 0.00303$ -50.3567$ 335$ 5813595$ 0.002985$ 15.57648$ $ 335$ 4.98E-22$ 0.002985$ -49.0511$ $ $ $ 3-3$ P3-1)(e))No$solution$will$be$given$ P3-1)(f)$ A+1 2 B1 2 C $ Rate$law:$ $$and$ $ $ $ $ $ ==12.5$ ==12.5$ ==12.5 $ $ ) P3-2)(a)$ Refer$to$Fig$3-4$ The$fraction$of$molecular$collisions$having$energies$less$than$or$equal$to$35$Kcal$is$given$by$the$area$ under$the$curve,$f(E,T)dE$$from$EA$=$0$to$35$Kcal.$ under$the$curve,$f(E,T)dE$$from$EA$=$0$to$35$Kcal.$ under$the$curve,$f(E,T)dE$$from$EA$=$0$to$35$Kcal.$ $ P3-2(b))$ The$fraction$of$molecular$collisions$having$energies$between$10$and$20$Kcal$is$given$by$the$area$under$ the$curve$f(E,T)$from$EA$=$10$to$20$Kcal.$ the$curve$f(E,T)$from$EA$=$10$to$20$Kcal.$ the$curve$f(E,T)$from$EA$=$10$to$20$Kcal.$ $ P3-2)(c)$ The$fraction$of$molecular$collisions$having$energies$greater$than$the$activation$energy$EA=$25$Kcal$is$ given$by$the$area$under$the$curve$f(E,T)$from$$EA$=25$to$50$Kcal.$ given$by$the$area$under$the$curve$f(E,T)$from$$EA$=25$to$50$Kcal.$ given$by$the$area$under$the$curve$f(E,T)$from$$EA$=25$to$50$Kcal.$ $ ) P3-3)(a)$ $ $ (a)$ (b)$ $ ) P3-4)No$solution$will$be$given$ $ 3-4$ P3-5)(a)$ The$fraction$of$collisions$having$energy$between$E$=$3$and$E$=$5$is$the$area$under$the$graph$between$ those$two$boundaries$=$2*0.5*(0.1875$+$0.25)$=$0.4375$ those$two$boundaries$=$2*0.5*(0.1875$+$0.25)$=$0.4375$ those$two$boundaries$=$2*0.5*(0.1875$+$0.25)$=$0.4375$ P3-5)(b)$ $ P3-5)(c)$ $ P3-5)(d)$ Since$f(E,T)$=$0$for$E$>$8$kcal,$the$fraction$with$energies$greater$than$8$kcal$=$0$ $ $ $ ) P3-6)(a)$$ Note:$This$problem$can$have$many$solutions$as$data$fitting$can$be$done$in$many$ways.$ Using$Arrhenius$Equation$ For$Fire$flies:$ T(in$K)$ 1/T$ Flashes/min$ ln(flashes/min)$ 294$ 0.003401$ 9$ 2.197$ 298$ 0.003356$ 12.16$ 2.498$ 303$ 0.003300$ 16.2$ 2.785$ $ Plotting$ln$(flashes/min)$vs.$1/T,$$ We$get$a$straight$line.$ $ For$Crickets:$$ T(in$K)$ 1/T$x103$ chirps/min$ ln(chirps/min)$ 287.2$ 3.482$ 80$ 4.382$ 293.3$ 3.409$ 126$ 4.836$ 300$ 3.333$ 200$ 5.298$ $ Plotting$ln$(chirps/min)$Vs$1/T,$$ We$get$a$straight$line.$ Both,$Fireflies$and$Crickets$data$$ follow$the$Arrhenius$Model.$ln$y$$=$A$+$B/T$,$and$have$the$similar$activation$energy.$ follow$the$Arrhenius$Model.$ln$y$$=$A$+$B/T$,$and$have$the$similar$activation$energy.$ follow$the$Arrhenius$Model.$ ln$y$$=$A$+$B/T$,$and$have$the$similar$activation$energy.$ ) ) ) ) 3-5$ P3-6)(b)$$ For$Honeybee:$$ T(in$K)$ 1/T$x103$ V(cm/s)$ ln(V)$ 298$ 3.356$ 0.7$ -0.357$ 303$ 3.300$ 1.8$ 0.588$ 308$ 3.247$ 3$ 1.098$ $ Plotting$ln$(V)$vs.$1/T,$almost$straight$line.$ ln$(V)$=$44.6$$1.33E4/T$$At$T$=$40oC$(313K)$V$=$6.4cm/s$At$T$=$-5oC$(268K)$V$=$0.005cm/s$(But$bee$would$not$be$alive$at$this$temperature)$ ln$(V)$=$44.6$$1.33E4/T$$At$T$=$40oC$(313K)$V$=$6.4cm/s$At$T$=$-5oC$(268K)$V$=$0.005cm/s$(But$bee$would$not$be$alive$at$this$temperature)$ ln$(V)$=$44.6$$1.33E4/T$ $ At$T$=$40oC$(313K)$V$=$6.4cm/s$ At$T$=$-5oC$(268K)$V$=$0.005cm/s$(But$bee$would$ not$be$alive$at$this$temperature)$ $ P3-6)(c)$$ For$Ants:$$ T(in$K)$ 1/T$x103$ V(cm/s)$ ln(V)$ 283$ 3.53$ 0.5$ -0.69$ 293$ 3.41$ 2$ 0.69$ 303$ 3.30$ 3.4$ 1.22$ 311$ 3.21$ 6.5$ 1.87$ $ Plotting$ln$(V)$$vs.$$1/T,$$ We$get$almost$a$straight$line.$$ $ So$activity$of$bees,$ants,$crickets$and$fireflies$follow$Arrhenius$model.$So$activity$increases$with$an$ increase$in$temperature.$Activation$energies$for$fireflies$and$crickets$are$almost$the$same.$ $ Insect$ Activation$Energy$ Cricket$ 52150$ Firefly$ 54800$ Ant$ 95570$ Honeybee$ 141800$ $ P3-6)(d)$$ There$is$a$limit$to$temperature$for$which$data$for$any$one$of$the$insect$can$be$extrapolate.$Data$which$ would$be$helpful$is$the$maximum$and$the$minimum$temperature$that$these$insects$can$endure$before$ death.$Therefore,$even$if$extrapolation$gives$us$a$value$that$looks$reasonable,$at$certain$temperature$it$could$be$useless.$ death.$Therefore,$even$if$extrapolation$gives$us$a$value$that$looks$reasonable,$at$certain$temperature$it$could$be$useless.$ death.$Therefore,$even$if$extrapolation$gives$us$a$value$that$looks$reasonable,$at$certain$temperature$it$ could$be$useless.$ $ P3-6)(e)$ 1)$The$rate$at$which$the$beetle$can$push$a$ball$of$dung$is$directly$proportional$to$its$rate$constant,$ therefore$ -rA$=c*k,$where$c$is$a$constant$related$to$the$mass$of$the$beetle$and$the$dung$and$k$is$the$rate$constant$ =!!!!" $ =!!!!" $ = !!! !" $ $ 3-6$ P3-6)(e))Continued$ From$the$data$given$ $$$$$$$$$-rA$ $$$$$$T(K)$ $ 1/T$ ln$k$ 6.5$ 300$ $ 0.003333$ 1.871802$ 13$ 310$ $ 0.003226$ 2.564949$ 18$ 313$ $ 0.003195$ 2.890372$ $ Refer$to$P3-8$(similar$procedure)$ Therefore,$A$=$1.299X1011$$$ $$$$$$$$$$$$$$$$$$$$$E$=$59195.68$J/mol$k$=$1.299X1011$exp(-7120/T)$Now$at$T$=$41.5$C$=$314.5$K$k$=$19.12$cm/s$Therefore,$beetle$can$push$dung$at$19.12$cm/s$at$41.5$C$ $$$$$$$$$$$$$$$$$$$$$E$=$59195.68$J/mol$k$=$1.299X1011$exp(-7120/T)$Now$at$T$=$41.5$C$=$314.5$K$k$=$19.12$cm/s$Therefore,$beetle$can$push$dung$at$19.12$cm/s$at$41.5$C$ $$$$$$$$$$$$$$$$$$$$$E$=$59195.68$J/mol$ k$=$1.299X1011$exp(-7120/T)$ Now$at$T$=$41.5$C$=$314.5$K$ k$=$19.12$cm/s$ Therefore,$beetle$can$push$dung$at$19.12$cm/s$at$41.5$C$ $ P3-6)(e))$ 2))Individualized$solution$ $ ) P3-7$ There$are$two$competing$effects$that$bring$about$the$maximum$in$the$corrosion$rate:$Temperature$and$ HCN-H2SO4$concentration.$The$corrosion$rate$increases$with$increasing$temperature$and$increasing$ concentration$of$HCN-H2SO4$complex.$The$temperature$increases$as$we$go$from$top$to$bottom$of$the$column$and$consequently$the$rate$of$corrosion$should$increase.$However,$the$HCN$concentrations$(and$the$HCN-H2SO4$complex)$decrease$as$we$go$from$top$to$bottom$of$the$column.$There$is$virtually$no$HCN$in$the$bottom$of$the$column.$These$two$opposing$factors$results$in$the$maximum$of$the$corrosion$rate$somewhere$around$the$middle$of$the$column.$ concentration$of$HCN-H2SO4$complex.$The$temperature$increases$as$we$go$from$top$to$bottom$of$the$column$and$consequently$the$rate$of$corrosion$should$increase.$However,$the$HCN$concentrations$(and$the$HCN-H2SO4$complex)$decrease$as$we$go$from$top$to$bottom$of$the$column.$There$is$virtually$no$HCN$in$the$bottom$of$the$column.$These$two$opposing$factors$results$in$the$maximum$of$the$corrosion$rate$somewhere$around$the$middle$of$the$column.$ concentration$of$HCN-H2SO4$complex.$The$temperature$increases$as$we$go$from$top$to$bottom$of$the$ column$and$consequently$the$rate$of$corrosion$should$increase.$However,$the$HCN$concentrations$(and$ the$HCN-H2SO4$complex)$decrease$as$we$go$from$top$to$bottom$of$the$column.$There$is$virtually$no$ HCN$in$the$bottom$of$the$column.$These$two$opposing$factors$results$in$the$maximum$of$the$corrosion$ rate$somewhere$around$the$middle$of$the$column.$ $ ) y$=$-7120.4x$+$25.593$ R$=$0.9893$ 0) 0.5) 1) 1.5) 2) 2.5) 3) 3.5) 0.00318) 0.00321) 0.00324) 0.00327) 0.0033) 0.00333) 0.00336) ln(k)) 1/T))(K-1)$ 3-7$Since, = !"!"$$$therefore$ln$A$=$41.99$and$E/R$=$14999$ 3-7$Since, = !"!"$$$therefore$ln$A$=$41.99$and$E/R$=$14999$ 3-7$ Since, = !" !"$$$therefore$ln$A$=$41.99$and$E/R$=$14999$ P3-8)Antidote$did$not$dissolve$from$glass$at$low$temperatures.$ $ ) P3-9)(a)$ If$a$reaction$rate$doubles$for$an$increase$in$10C,$at$T$=$T1$let$k$=$k1$and$at$T$=$T2$=$T1+10,$let$k$=$k2$=$2k1.$$ Then$with$k$=$Ae-E/RT$in$general,$ $and$ $,$or$ $$$ $$$or$$$$ $ $ Therefore:$ $ $ $ $ which$can$be$approximated$by$ .$Consequently,$for$this$doubling$rate$fule$of$thumb$to$ be$valid,$the$temperature$at$which$the$doubling$will$take$place$must$be$related$to$the$activation$energy$ by$this$relationship.$$ by$this$relationship.$$ by$this$relationship.$$ $ P3-9)(b))Individualized$solution$ $ ) P3-10)$ From$the$$given$data$ -rA(dm3/mol.s)$ T(K)$ k=$-rA/(4*1.5)$ 1/T$(!!)$ ln$(k)$ 0.002$ 300$ 0.00033333$ 0.003333$ -8.00637$ 0.046$ 320$ 0.00766667$ 0.003125$ -4.87087$ 0.72$ 340$ 0.12$ 0.002941$ -2.12026$ 8.33$ 360$ 1.38833333$ 0.002778$ 0.328104$ Plotting$ln(k)$vs$(1/T),$we$have$a$straight$line:$$ $ 3-8$ P3-10)(a)$ Activation)energy)(E),)) $$ $E$=$14999*8.314$=124700$J/mol$=$124.7)kJ/mol$) $E$=$14999*8.314$=124700$J/mol$=$124.7)kJ/mol$) $E$=$14999*8.314$=124700$J/mol$=$124.7)kJ/mol$ ) P3-10)(b)$ Frequency)Factor)(A),)ln)A)=)41.99)))))$ $=1.72 10!" !"! !"#!.! !)$ ) ) ) P3-10)(c)$ $=1.72

4-1$ Solutions)for)Chapter)4))Stoichiometry) ) P4-1)(a))Example)4-4)) (i)$The$critical$value$is$around$0.5$atm$ $(ii)$Kp$has$the$least$effect,$and$KSO3$has$the$greatest$effect$$(iii)$0.25$$ $(ii)$Kp$has$the$least$effect,$and$KSO3$has$the$greatest$effect$$(iii)$0.25$$ $ (ii)$Kp$has$the$least$effect,$and$KSO3$has$the$greatest$effect$ $ (iii)$0.25$ $ (iv)$Individualized$Solution$ $ (v)$See$Polymath$program$P4-1-av.pol.$$ (v)$See$Polymath$program$P4-1-av.pol.$$ (v)$See$Polymath$program$P4-1-av.pol.$ $ 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 Fao 3. 3. 3. 3. 3 k 9.7 9.7 9.7 9.7 4 KO2 38.5 38.5 38.5 38.5 5 KP 930. 930. 930.

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$5-3)$ FB$=$200X106$/$(365$X$24$X$3600$X$28)$lbmol/sec$=$0.226$lbmol/sec$FAO$=$$$$$$$Also,$ $$ FB$=$200X106$/$(365$X$24$X$3600$X$28)$lbmol/sec$=$0.226$lbmol/sec$FAO$=$$$$$$$Also,$ $$ FB$=$200X106$/$(365$X$24$X$3600$X$28)$lbmol/sec$=$0.226$lbmol/sec$ FAO$=$$ $ $$$$Also,$ $$ Now,$we$have$from$the$example$ $ $ $=$35.47$X$0.886$ft3$=$31.44$ft3$ $ $=$35.47$X$0.886$ft3$=$31.44$ft3$ $ $ =$35.47$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$reduced.$ So,$we$see$that$for$lower$conversion$and$required$flow$rate$the$volume$of$the$reactor$is$reduced.$ 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$ $ $ $ $$p$ $ $$p$ $ $ $p $ $ Now,$ $ ,$so$too$much$pressure$drop$P$=$0$and$the$flow$stops.$ P5-1)(d))Example)5-6$$ For$turbulent$flow$ $ 5-2$ P5-1)(d))Example)5-6$Continued$ $ Therefore$there$is$no$change.$ $ P5-1)(e))Example)5-7$) (i)$X$increases$and$f$decreases$with$an$increase$in$k$for$the$same$W,$while$p$remains$unchanged,$and$ vice$versa.$This$is$an$expected$observation,$because$as$k$increases,$the$rate$of$the$reaction$increases,$ and$naturally,$X$increases.$Since$the$conversion$down$the$reactor$increases,$more$product$is$formed,$lesser$amount$of$reactants$are$present.$Since$$<$0,$the$volumetric$flow$rate$decreases$with$an$increase$in$conversion.$$$(ii)$As$$and$FA0$increase,$X$and$p$decrease$for$the$same$W,$while$f$increases.$$$(iii)$Individualized$solution$ and$naturally,$X$increases.$Since$the$conversion$down$the$reactor$increases,$more$product$is$formed,$lesser$amount$of$reactants$are$present.$Since$$<$0,$the$volumetric$flow$rate$decreases$with$an$increase$in$conversion.$$$(ii)$As$$and$FA0$increase,$X$and$p$decrease$for$the$same$W,$while$f$increases.$$$(iii)$Individualized$solution$ and$naturally,$X$increases.$Since$the$conversion$down$the$reactor$increases,$more$product$is$formed,$ lesser$amount$of$reactants$are$present.$Since$$<$0,$the$volumetric$flow$rate$decreases$with$an$increase$ in$conversion.$$ $ (ii)$As$$and$FA0$increase,$X$and$p$decrease$for$the$same$W,$while$f$increases.$$ $ (iii)$Individualized$solution$ $ P5-1)(f))Example)5-8$) Individualized$solution$ $ P5-1)(g))Example)5-3,)Using)ASPEN,)we)get$(Refer$to$Aspen$Program$P5-2g$from$polymath$CD)$ (1) At$1000K,$for$the$same$PFR$volume$we$get$only$6.2%$conversion.$While$at$1200K,$we$get$a$ conversion$of$nearly$100%.$This$is$because$the$value$of$reaction$constant$k$varies$rapidly$ with$reaction$temperature.$ with$reaction$temperature.$ with$reaction$temperature.$ (2) Earlier$for$an$activation$energy$of$82$kcal/mol$we$got$approx.$81%$conversion.$For$activation$ energy$of$74$kcal/mol$keeping$the$PFR$volume$the$same$we$get$a$conversion$of$71.1%.$ While$for$an$activation$energy$of$90$kcal/mol$we$get$a$conversion$of$89.93%.$(3) On$doubling$both$flow$rate$and$pressure$we$find$that$the$conversion$remains$the$same.$ While$for$an$activation$energy$of$90$kcal/mol$we$get$a$conversion$of$89.93%.$(3) On$doubling$both$flow$rate$and$pressure$we$find$that$the$conversion$remains$the$same.$ While$for$an$activation$energy$of$90$kcal/mol$we$get$a$conversion$of$89.93%.$ (3) On$doubling$both$flow$rate$and$pressure$we$find$that$the$conversion$remains$the$same.$ $ P5-1)(h)$Individualized$solution.$ $ P5-1)(i)$Individualized$solution$ ) $ P5-2)The$key$for$decoding$the$algorithm$to$arrive$at$a$numerical$score$for$the$Interaction$Computer$ Games$(ICGs)$is$given$at$the$front$of$this$Solutions$Manual.$ ) ) ) $ P5-3)(a))(3)$50%$$$(Liquid$phase$reactions$do$not$depend$on$pressure)$$P5-3)(b)$(2)$<50%$$$(Low$concentration$if$we$lower$the$pressure)$),)low)rate$ P5-3)(a))(3)$50%$$$(Liquid$phase$reactions$do$not$depend$on$pressure)$$P5-3)(b)$(2)$<50%$$$(Low$concentration$if$we$lower$the$pressure)$),)low)rate$ P5-3)(a))(3)$50%$$$(Liquid$phase$reactions$do$not$depend$on$pressure)$ $ P5-3)(b)$(2)$<50%$$$(Low$concentration$if$we$lower$the$pressure)$ ),) low)rate $ 5-3$ P5-3)(c)$(1)$>0.234$$ $ rA=kCA0 21X ( ) 2 p rA=kp2 p2<1 ktruep2=k=0.234ktrue =0.234p2>0.234)$P5-4)(a)$Ans.$(1)$X$>$0.5$$ $~1D$$Increase$D,$decrease,$increase$X.$$P5-4)(b)$Ans.$(3)$Xe$=$0.75$$ $ $P5-4)(c)$Ans.$(3)$remain$the$same$$ $Kc=CACAe=CA01Xe( )pCA0Xep=1XXe$$ Xe$is$not$a$function$of$$$P5-4)(d)$Ans.$(3)$remain$the$same$$$Xe$is$not$a$function$of$$$P5-4)(e)$Ans.$(4)$insufficient$information$to$tell$$ $ $ $)$P5-5$r$=$-$k$CA2$$ ktruep2=k=0.234ktrue =0.234p2>0.234)$P5-4)(a)$Ans.$(1)$X$>$0.5$$ $~1D$$Increase$D,$decrease,$increase$X.$$P5-4)(b)$Ans.$(3)$Xe$=$0.75$$ $ $P5-4)(c)$Ans.$(3)$remain$the$same$$ $Kc=CACAe=CA01Xe( )pCA0Xep=1XXe$$ Xe$is$not$a$function$of$$$P5-4)(d)$Ans.$(3)$remain$the$same$$$Xe$is$not$a$function$of$$$P5-4)(e)$Ans.$(4)$insufficient$information$to$tell$$ $ $ $)$P5-5$r$=$-$k$CA2$$ ktruep2=k=0.234 ktrue =0.234 p2>0.234 ) $ P5-4)(a)$Ans.$(1)$X$>$0.5$ $ $ ~1 D $ $Increase$D,$decrease,$increase$X.$$ P5-4)(b)$Ans.$(3)$Xe$=$0.75$ $ $ $ P5-4)(c)$Ans.$(3)$remain$the$same$ $ $ Kc=CA CAe = CA01Xe ( ) p CA0Xep=1X Xe $ $ Xe$is$not$a$function$of$$ $ P5-4)(d)$Ans.$(3)$remain$the$same$ $$Xe$is$not$a$function$of$$ $ P5-4)(e)$Ans.$(4)$insufficient$information$to$tell$ $ $ $ $ ) $ P5-5$ r$=$-$k$CA 2$ $ $ $$ $$ $ $ 5-4$ P5-5)Continued$$ Thus,$k$=$5$ For$a$CSTR,$ $ $ $ ) $ P5-6$ $$$$$$$$$$$$$$$$$$$$$ $ To$=$300K$$ KCO$(300K)=$3.0$V$=$1000gal$=$3785.4$dm3$ Mole$balance:$ $ $Rate$law:$

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$ $ $$$ $$$$$$ $ $$ $For$a$plug$flow$reactor:$ $$$ $$$$$$ $ $$ $For$a$plug$flow$reactor:$ $$$ $ $$$$ $ $ $ $ $ For$a$plug$flow$reactor:$ $ $ $ $ $ $ $ $ At$T2$=$260F$=$720R,$with$k1$=$0.0015$min-1$at$T1$=$80F$=$540R,$ $ $$$$$$Therefore$14$pipes$are$necessary.$) $$$$$$Therefore$14$pipes$are$necessary.$) $ $ $ $$ $ Therefore$14$pipes$are$necessary.$ ) 5-23$ P5-16)(a)$ )$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$A$$$$B/2$ $ $ Combining$ $ (for$the$integration,$refer$to$Appendix$A)$ from$the$Ideal$Gas$assumption.$ $ Substituting$Eqn.$(5),$X$=$0.8$and$$=$1/4$to$Eqn.$(4)$yields,$.(6)$ Substituting$Eqn.$(5),$X$=$0.8$and$$=$1/4$to$Eqn.$(4)$yields,$.(6)$ Substituting$Eqn.$(5),$X$=$0.8$and$$=$1/4$to$Eqn.$(4)$yields,$ .(6)$ $ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ $ 5-24$ P5-16)(b))Individualized$solution.$ ) $ P5-17)(a))$ Given:$The$metal$catalyzed$isomerization$$$liquid$phase$reaction$ with$Keq$=$5.8 $$For$a$plug$flow$reactor$with$yA$=$1.0,$X1$=$0.55$ $$For$a$plug$flow$reactor$with$yA$=$1.0,$X1$=$0.55$ $$ For$a$plug$flow$reactor$with$yA$=$1.0,$X1$=$0.55$ Case$1:$an$identical$plug$flow$reactor$connected$in$series$with$the$original$reactor.$ $ Since$yA$=$1.0,$B$=$0.$For$a$liquid$phase$reaction$ $$and$ $$ $ For$the$first$reactor,$ $ or$$ or$$ or$ $ $ Take$advantage$of$the$fact$that$two$PFRs$in$series$is$the$same$as$one$PFR$with$the$volume$of$the$two$ combined.$$ VF$=$V1$+$V2$=$2V1$and$at$VF,$$$X$=$X2$ $ $ $ $ 5-25$ P5-17)(a))Continued$ $ $X2$=$0.74$ $X2$=$0.74$ $ X2$=$0.74$ $ P5-17)(b))$ Case$2:$Products$from$1st$reactor$are$separated$and$pure$A$is$fed$to$the$second$reactor,$ $ The$analysis$for$the$first$reactor$is$the$same$as$for$case$1.$ $ By$performing$a$material$balance$on$the$separator,$FA0,2$=$FA0(1-X1)$ Since$pure$A$enters$both$the$first$and$second$reactor$CA0,2$=$CA0,$CB0,2$=$0,$B$=$0$$ $$$ $for$the$second$reactor.$ $$and$since$V1$=$V2$$or$$ $$and$since$V1$=$V2$$or$$ $ $ and$since$V1$=$V2$ $ or$$ 5-26$ P5-17)(b))continued$ $ $$ $$ $ $ Overall$conversion$for$this$scheme:$ $ $ $ $ ) P5-17)(c))Individualized$Solution) ) $ P5-18$ Given:$Meta-$to$ortho-$and$para-$isomerization$of$xylene.$ $ $$Pressure$=$300$psig$T$=$750F$V$=$1000$ft3$cat.$$Assume$that$the$reactions$are$irreversible$and$first$order.$Then:$$$$ $$Pressure$=$300$psig$T$=$750F$V$=$1000$ft3$cat.$$Assume$that$the$reactions$are$irreversible$and$first$order.$Then:$$$$ $ $ Pressure$=$300$psig$ T$=$750F$ V$=$1000$ft3$cat.$ $ Assume$that$the$reactions$are$irreversible$and$first$order.$ Then:$ $ $ $ Check$to$see$what$type$of$reactor$is$being$used.$ Case$1:$ $ 5-27$ P5-18)Continued$ Case$2:$ $ $ $ $ Assume$plug$flow$reactor$conditions:$ $$$or$ $$$ $$$ $ $ $ CM0,$k,$and$V$should$be$the$same$for$Case$1$and$Case$2.$ Therefore,$ $ $$ $$ $ $ The$reactor$appears$to$be$plug$flow$since$(kV)Case$1$=$(kV)Case$2$ As$a$check,$assume$the$reactor$is$a$CSTR.$ $ )))or$$$ $ Again$kV$should$be$the$same$for$both$Case$1$and$Case$2.$ $ $ kV$is$not$the$same$for$Case$1$and$Case$2$using$the$CSTR$assumption,$therefore$the$reactor$must$be$ modeled$as$a$plug$flow$reactor.$ $ $ $ $ 5-28$ P5-18)Continued$ For$the$new$plant,$with$v0$=$5500$gal$/$hr,$XF$=$0.46,$the$required$catalyst$volume$is:$ $ This$assumes$that$the$same$hydrodynamic$conditions$are$present$in$the$new$reactor$as$in$the$old.$ This$assumes$that$the$same$hydrodynamic$conditions$are$present$in$the$new$reactor$as$in$the$old.$ This$assumes$that$the$same$hydrodynamic$conditions$are$present$in$the$new$reactor$as$in$the$old.$ ) $ P5-19$$ A$B$in$a$tubular$reactor$ $ Tube$dimensions:$L$=$40$ft,$D$=$0.75$in.$ nt$=$50$ $ $$$$with$ $ $$$$with$ $ $ $ $ $ with$ $ $or$$ $ Assume$Arrhenius$equation$applies$to$the$rate$constant.$ At$T1$=$600R,$k1$=$0.00152$ $ At$T2$=$760R,$k2$=$0.0740$ $ At$T2$=$760R,$k2$=$0.0740$ $ At$T2$=$760R,$k2$=$0.0740$ $ 5-29$ P5-19)continued$ $ $$$so$$ $$$so$$ $ $ $ so$ $ $ From$above$we$have$ $ so$$ so$$ so$ $ $ Dividing$both$sides$by$T$gives:$ $ $ Evaluating$and$simplifying$gives:$ $ Solving$for$T$gives:$T$=$738R$=$278F$ Solving$for$T$gives:$T$=$738R$=$278F$ Solving$for$T$gives:$ T$=$738R$=$278F$ ) $ P5-20) $ 5-30$ P5-20)Continued$ $ $ $ 5-31$ P5-20)(b)$ $ $ $ $ P5-20)(c)$ $ $) $) $ ) $ P5-21$ Production$of$phosgene$in$a$micro$reactor$ $$$$$$$$$$$CO$$+$$Cl2$$ $$COCl2$$$$(Gas$phase$reaction)$ $$$$$$$$$$$$$$A$$+$$B$$$$ $$$C$ $$$$$$$$$$$$$$A$$+$$B$$$$ $$$C$ $$$$$$$$$$$$$$A$$+$$B$$$$ $$$C$ The$equations$that$need$to$be$solved$are$as$follows:$ d(X)/d(W)$=$-rA/FA0$ d(y)/d(W)$=$-*(1+*X)/(2*y)$$$$$$($from$4.30)$ d(y)/d(W)$=$-*(1+*X)/(2*y)$$$$$$($from$4.30)$ d(y)/d(W)$=$-*(1+*X)/(2*y)$$$$$$($from$4.30)$ P5-21)Continued$ FB0$=$FA0;$ Fb$=$FB0-FA0*X;$ Fc$=$FA0*X$See$Polymath$program$P5-22.pol.$$ Fc$=$FA0*X$See$Polymath$program$P5-22.pol.$$ Fc$=$FA0*X$ See$Polymath$program$P5-22.pol.$ $ 5-32$ P5-21)Continued$ POLYMATH)Results$ Calculated)values)of)the)DEQ)variables$ Variable initial value minimal value maximal value final value$ W 0 0

5-41$ P5-24)Continued$ $ $ ) $ P5-25))$ =0","=0 p=1 W ( ) 1 2 $ Mole$Balance/Design$Equation$ $ 5-42$ P5-25)Continued$ Rate$Law$ $ Stoichiometry$CA=CA0 1X( )p$Combining$FA0dXdW =kCA021X( )2p2$$$ Stoichiometry$CA=CA0 1X( )p$Combining$FA0dXdW =kCA021X( )2p2$$$ Stoichiometry$ CA=CA0 1X ( ) p $ Combining$ F A0 dX dW =kCA0 21X ( ) 2p2 $ $ $ ) $ P5-26)Individualized$solution$ ) $ 6-1$ Solutions)for)Chapter)6))Isothermal)Reactor)Design-)Molar) Flow)rates) ) P6-1)(a))Example)6-1)) (i)$The$conversion$increases$with$an$increase$in$both$E$and$T.$This$is$because$the$rate$constant$is$higher$ for$higher$E$and$T.$ $(ii)$For$pressure$doubled$and$temperature$decrease$by$20$C,$CTO$=$2*Po/RT$and$T$=$678K$See$Polymath$program$P6-1-aii.pol.$$ $(ii)$For$pressure$doubled$and$temperature$decrease$by$20$C,$CTO$=$2*Po/RT$and$T$=$678K$See$Polymath$program$P6-1-aii.pol.$$ $ (ii)$For$pressure$doubled$and$temperature$decrease$by$20$C,$ CTO$=$2*Po/RT$and$T$=$678K$ See$Polymath$program$P6-1-aii.pol.$ $ POLYMATH Report Ordinary Differential Equations Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value 1 Ca 0.5822357 0.0292158 0.5822357 0.0292158 2 Cto 0.5822357 0.5822357 0.5822357 0.5822357 3 E

6-15$ P6-6$(a))Continued$ Rename$ Transport$out$the$sides$of$the$reactor:$ RA$=$kACA$= $$RB$=$kBCB$=$$$Stoichiometery:$ $$RB$=$kBCB$=$$$Stoichiometery:$ $$ RB$=$kBCB$=$ $ $ Stoichiometery:$ $ 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 61.916043 61.916043$ Kc 0.01 0.01 0.01 0.01 $ Ft 100 100 122.2435 121.06199$ Co 1 1 1 1 $ K 10

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 = $ 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 4 ) See$Polymath$program$P7-1-a.pol.$ POLYMATH)Results Nonlinear)regression)(L-M)) Model: r = k*(PCO^alfa)*(PH2^beta) Variable Ini guess Value 95% confidence k 0.1 0.0060979 6.449E-04 alfa

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$phase),$ $2321 AAACkCkkddC =$ $ $1kddC X=$$ABCkddC2=$$$$$23AYCkddC =$$In$PFR$with$V$=$1566$dm3$we$get$$=$V/v0$=$783$$X$=$0.958$,$SB/XY$(instantaneous)$=$0.244,$and$SB/XY$(overall)$=$0.624$ $2321 AAACkCkkddC =$ $ $1kddC X=$$ABCkddC2=$$$$$23AYCkddC =$$In$PFR$with$V$=$1566$dm3$we$get$$=$V/v0$=$783$$X$=$0.958$,$SB/XY$(instantaneous)$=$0.244,$and$SB/XY$(overall)$=$0.624$ $ 2 321 AA ACkCkk d dC = $ $ $ 1 k d dC X= $ $ A BCk d dC 2 = $$$$$ 2 3A YCk d dC = $ $In$PFR$with$V$=$1566$dm3$we$get$$=$V/v0$=$783$ $X$=$0.958$,$SB/XY$(instantaneous)$=$0.244,$and$SB/XY$(overall)$=$0.624$ also$at$$=$350,$SB/XY$(instantaneous)$$is$at$its$maximum$value$of$0.84$ See$polymath$problem$P8-1-a.pol$ $ Calculated)values)of)the)DEQ)variables Variable initial value minimal value maximal value final value tau 0 0 783 783 Ca 0.4 0.0166165 0.4 0.0166165 Cx 1.0E-07 1.0E-07 0.0783001 0.0783001 Cb 0 0 0.1472919

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$the$ 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.$$ production$of$desired$product.$The$reaction$should$also$take$place$in$high$concentration$of$A$and$the$concentration$of$D$should$be$limited$by$removing$through$a$membrane$or$reactive$distillation.$$ 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$distillation.$ $ P8-7)(d))))) ))))))) DA )and) - AA CTKr )/12000exp(4280 1= ) ))))))) 1 UD )and$$$$$$- DD CTKr )/15000exp(100,10 2= ) ))))))) 2 UA )and$$$$$$- AA CTKr )/10800exp(26 3= ) ADDAUUDUUD CTKCTKCTKCTKrrrS)/10800exp(26)/15000exp(100,10)/15000exp(100,10)/12000exp(42802121/ +=+=)At$T$=$300K$k1$=$1.18$x$10-14$$&$ k2$=$1.94$X$10-18$&$k3$=$6.03$x$10-15$If$we$keep$CA$>$1000CD$ ADDAUUDUUD CTKCTKCTKCTKrrrS)/10800exp(26)/15000exp(100,10)/15000exp(100,10)/12000exp(42802121/ +=+=)At$T$=$300K$k1$=$1.18$x$10-14$$&$ k2$=$1.94$X$10-18$&$k3$=$6.03$x$10-15$If$we$keep$CA$>$1000CD$ AD DA UU D UUD CTKCTK CTKCTK rr r S )/10800exp(26)/15000exp(100,10 )/15000exp(100,10)/12000exp(4280 21 21/ + = + = ) At$T$=$300K$ k1$=$1.18$x$10-14$$&$ k2$=$1.94$X$10-18$&$k3$=$6.03$x$10-15$ If$we$keep$CA$>$1000CD$ AD DA UUD CC CC S1518 1814 21/ 1003.61094.1 1094.11018.1 + = 1.18 1.96 .603 = ) At$$T$=$1000K$ k1$=$0.026$&$ k2$=$3.1$X$10-3$&$k3$=$5.3$x$10-4$If$we$keep$CA$>$1000CD$ADDAUUD CCCCS43321/ 103.5101.3101.3026.0+=.026 49.00053=$Here,$in$order$to$lower$U1$use$low$temperature$and$high$concentration$of$A$But$low$temperature$and$high$concentration$of$A$favors$U2$$So,$its$a$optimization$problem$with$the$temperature$and$concentration$of$A$as$the$variables$.$Membrane$reactor$in$which$D$is$diffusing$out$can$be$used.$$ k1$=$0.026$&$ k2$=$3.1$X$10-3$&$k3$=$5.3$x$10-4$If$we$keep$CA$>$1000CD$ADDAUUD CCCCS43321/ 103.5101.3101.3026.0+=.026 49.00053=$Here,$in$order$to$lower$U1$use$low$temperature$and$high$concentration$of$A$But$low$temperature$and$high$concentration$of$A$favors$U2$$So,$its$a$optimization$problem$with$the$temperature$and$concentration$of$A$as$the$variables$.$Membrane$reactor$in$which$D$is$diffusing$out$can$be$used.$$ k1$=$0.026$&$ k2$=$3.1$X$10-3$&$k3$=$5.3$x$10-4$ If$we$keep$CA$>$1000CD$ AD DA UUD CC CC S43 3 21/ 103.5101.3 101.3026.0 + = .026 49 .00053 = $ Here,$in$order$to$lower$U1$use$low$temperature$and$high$concentration$of$A$ But$low$temperature$and$high$concentration$of$A$favors$U2$$ So,$its$a$optimization$problem$with$the$temperature$and$concentration$of$A$as$the$variables$.$ Membrane$reactor$in$which$D$is$diffusing$out$can$be$used.$ $ P8-7)(e)) ))))))) DBA + )and) - BAA CCTKr )/10000exp(109 1= ) ))))))) BAD + )and$$$$$$- DD CTKr )/2000exp(20 2= ) )))))))UBA +)and$$$$$$-BAA CCTKr )/3000exp(1033=)BADBAUDUD CCTKCTKCCTKrrS)/3000exp(10)/2000exp(20)/10000exp(1039/==$At$T$=$300K$k1$=$3.34$x$10-6$$&$ k2$=$0.025$$&$$k3$=$0.045$ )))))))UBA +)and$$$$$$-BAA CCTKr )/3000exp(1033=)BADBAUDUD

8-41$TdV VSee$Polymath$program$P8-12-g.pol.$$$ $ 8-41$TdV VSee$Polymath$program$P8-12-g.pol.$$$ $ 8-41$ T dV V See$Polymath$program$P8-12-g.pol.$ $ $ $ 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) [20] Scd = rc/(rd+.0000000001) [21] Sef = re/(rf+.00000000001) [20] Scd = rc/(rd+.0000000001) [21] Sef = re/(rf+.00000000001) P8-12)(f))) The$only$change$from$part$(e)$is:$ D D cD D dF r k

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 5 Ce 0 0 0.0496431 0.0496431 6 Cg 0 0 0.0477138

9-8$ P9-3)Continued$ $ $ $ P9-4)(a)$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $Active$intermediates:$ $$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $Active$intermediates:$ $$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ $ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ Active$intermediates:$ $$ $ $$$$$$$$$$$$$$$ $ 9-9$$$) 9-9$$$) 9-9$ $ $ ) P9-4)(a))Continued$ $$$$$$$$$$$$$$$ $ $$$$$$$$$$$$$$$ $ $ $$$$$$$$$$$ $ Plugging$in$expression$for$ $:$$$$$$$$$$ $$$$$$$$$$$$ $Now,$substitute$expressions$for$ $into$equation$for$ :$$$$$$$$$$$ $$$$$$$$$$ $$ Plugging$in$expression$for$ $:$$$$$$$$$$ $$$$$$$$$$$$ $Now,$substitute$expressions$for$ $into$equation$for$ :$$$$$$$$$$$ $$$$$$$$$$ $$ Plugging$in$expression$for$ $:$ $$$$$$$$$ $ $$$$$$$$$$$ $ Now,$substitute$expressions$for$ $into$equation$for$ :$ $ $$$$$$$$$ $ $$$$$$$$$ $ $ P9-4)(b)$ When$ $$$ $ P9-4)(c)$ $ $$$$$$$$$$$$$CH3$CHO$CO$CH4$H2$kkkkC H $$$$$$$$$$$$$CH3$CHO$CO$CH4$H2$kkkkC H $ $ $ $ $ $ $ $ $ $ $ $ $ CH3$ CHO$ CO$ CH4$ H2$ k k k k C H CH3CHO$ 9-10$ P9-5)(a)$ $ $$$$$ $$$$$ $ $ $ $ $ P9-5)(b)$ $ $ $ $ $ $ 9-11$ 9-11$ 9-11$ P9-5)(b))Continued$ $ $$ $$ $ $ add$rCOCl$to$rCl$ $ $ $ $$$$ $$$$ $ $ $ $ P9-5)(c))$ The$proposed$mechanism$for$this$reaction$is:$ $ Through$this$mechanism,$it$may$be$deduced$that$the$net$rate$of$formation$of$HBr$(product)$is$given$by:$ $ where$the$concentrations$of$the$intermediates$may$be$determined$by$invoking$the$steady$state$ approximation,$i.e.:$ $ 9-12$ P9-5)(c))Continued$ and:$ $ i.e.$from$(ii)$+$(i)$we$obtain:$ i.e.$from$(ii)$+$(i)$we$obtain:$ i.e.$from$(ii)$+$(i)$we$obtain:$ $ i.e.:$ $ which$substituted$back$in$(i)$gives:$$i.e.:$$i.e.:$$or:$dividing$'top'$and$'bottom'$by$k`b$:$$ which$substituted$back$in$(i)$gives:$$i.e.:$$i.e.:$$or:$dividing$'top'$and$'bottom'$by$k`b$:$$ which$substituted$back$in$(i)$gives:$ $ i.e.:$ $ i.e.:$ $ or:$dividing$'top'$and$'bottom'$by$k`b$:$ $ Thus,$the$net$rate$of$formation$of$HBr$may$be$written$solely$in$terms$of$reactants,$i.e.,:$ 9-13$ P9-5)(c))Continued$ $ which$simplifies$to:$ $as$predicted$by$the$empirical$rate$law:$$where:$ $as$predicted$by$the$empirical$rate$law:$$where:$ $ as$predicted$by$the$empirical$rate$law:$ $ where:$ $ $ ) 9-14$ P9-6)(a))$ $ $$$$$ $$$$$ $ $ $ $ $ $ $ $$$$ $$$$ $ $ $ $ $ $ 9-15$ P9-6)(a))Continued$ $ $ P9-6)(b))$ Low$temperatures$with$anti-oxidant$$ $$$$$$$$$$$ $$$$$$$$$$$ $ $ $ $ $ $ $ $ $ $ $ $ 9-16$ P9-6)(b))Continued$ $ $ $ P9-6(c))$ If$the$radicals$are$formed$at$a$constant$rate,$then$the$differential$equation$for$the$concentration$of$the$ radicals$becomes:$ $and$$$The$substitution$in$the$differential$equation$for$R$also$changes.$Now$the$equation$is:$$and$solving$and$substituting$gives:$$$Now$we$have$to$look$at$the$balance$for$RO2.$$and$if$we$substitute$in$our$expression$for$[R]$we$get$$which$we$can$solve$for$[RO2].$$Now$we$are$ready$to$look$at$the$equation$for$the$motor$oil.$$and$making$the$necessary$substitutions,$the$rate$law$for$the$degradation$of$the$motor$oil$is:$$$ $and$$$The$substitution$in$the$differential$equation$for$R$also$changes.$Now$the$equation$is:$$and$solving$and$substituting$gives:$$$Now$we$have$to$look$at$the$balance$for$RO2.$$and$if$we$substitute$in$our$expression$for$[R]$we$get$$which$we$can$solve$for$[RO2].$$Now$we$are$ready$to$look$at$the$equation$for$the$motor$oil.$$and$making$the$necessary$substitutions,$the$rate$law$for$the$degradation$of$the$motor$oil$is:$$$ $ and$ $ $ The$substitution$in$the$differential$equation$for$R$also$changes.$Now$the$equation$is:$ $ and$solving$and$substituting$gives:$ $ $ Now$we$have$to$look$at$the$balance$for$RO2.$ $ and$if$we$substitute$in$our$expression$for$[R]$we$get$ $which$we$can$solve$for$[RO2].$ $ Now$we$are$ready$to$look$at$the$equation$for$the$motor$oil.$ $ and$making$the$necessary$substitutions,$the$rate$law$for$the$degradation$of$the$motor$oil$is:$ $ $ 9-17$ P9-6)(d)$ $Without$antioxidants$$With$antioxidants$$ $ $ $ $ P9-6)(e))Individualized$solution$ $ ) P9-7)(a)$ $ $ $$ $ $$ $ $ $ 9-18$ P9-7)(b)$ $ $ P9-7)(c)$ $ $$$ $$$ $ $ $ P9-7)(d)$ See$Polymath$program$P9-7-d.pol.$ 9-19$ P9-7)(d))Continued$ )$ Everyone$becomes$ill$rather$quickly,$and$the$rate$at$which$an$ill$person$recovers$to$a$healthy$person$is$ much$slower$than$the$rate$at$which$a$healthy$person$becomes$ill.$Eventually$everyone$is$ill$and$people$start$dying.$ much$slower$than$the$rate$at$which$a$healthy$person$becomes$ill.$Eventually$everyone$is$ill$and$people$start$dying.$ much$slower$than$the$rate$at$which$a$healthy$person$becomes$ill.$Eventually$everyone$is$ill$and$people$ start$dying.$ $ P9-7)(e))Individualized$solution$ $ $ $ ) P9-8)(a)$ $ $ $ $ $ $ By$applying$PSSH$for$the$complex$[E.S],$we$have$ $ $$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$

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,$$Equation$model:$$$$$$a0$=$0.0098$a1$=$0.33$For$-dextran,$$Equation$model:$$$$$$a0$=$0.008$a1$=$0.0377$$Maltose$show$non-competitive$inhibition$as$slope$and$intercept,$both$changing$compared$to$no$inhibition$case.$$$-dextran$show$competitive$inhibition$as$intercept$same$but$slope$increases$compared$to$no$inhibition$case.$$ $$$$$a0$=$0.008$$a1$=$0.0266$For$Maltose,$$Equation$model:$$$$$$a0$=$0.0098$a1$=$0.33$For$-dextran,$$Equation$model:$$$$$$a0$=$0.008$a1$=$0.0377$$Maltose$show$non-competitive$inhibition$as$slope$and$intercept,$both$changing$compared$to$no$inhibition$case.$$$-dextran$show$competitive$inhibition$as$intercept$same$but$slope$increases$compared$to$no$inhibition$case.$$ $ $$$$a0$=$0.008$$a1$=$0.0266$ For$Maltose,$ $Equation$model:$ $ $ $$$a0$=$0.0098$a1$=$0.33$ For$-dextran,$ $Equation$model:$ $ $ $$$a0$=$0.008$a1$=$0.0377$ $Maltose$show$non-competitive$inhibition$as$slope$and$intercept,$both$changing$compared$to$no$ inhibition$case.$ $ $-dextran$show$competitive$inhibition$as$intercept$same$but$slope$increases$compared$to$no$ inhibition$case.$ $ $ 9-30$ P9-13)(a)$ $ $ $ $$$$$$$Now$plug$the$value$of$(EH)$into$rP$$$At$very$low$concentrations$of$H+$(high$pH)$rS$approaches$0$and$at$very$high$concentrations$of$H+$(low$pH)$rS$also$approaches$0.$Only$at$moderate$concentrations$of$H+$(and$therefore$pH)$is$the$rate$much$greater$than$zero.$This$explains$the$shape$of$the$figure.$ $$$$$$$Now$plug$the$value$of$(EH)$into$rP$$$At$very$low$concentrations$of$H+$(high$pH)$rS$approaches$0$and$at$very$high$concentrations$of$H+$(low$pH)$rS$also$approaches$0.$Only$at$moderate$concentrations$of$H+$(and$therefore$pH)$is$the$rate$much$greater$than$zero.$This$explains$the$shape$of$the$figure.$ $ $ $ $ $ $ $ Now$plug$the$value$of$(EH)$into$rP$ $ $ At$very$low$concentrations$of$H+$(high$pH)$rS$approaches$0$and$at$very$high$concentrations$of$H+$(low$ pH)$rS$also$approaches$0.$Only$at$moderate$concentrations$of$H+$(and$therefore$pH)$is$the$rate$much$ greater$than$zero.$This$explains$the$shape$of$the$figure.$ $ P9-13)(b))Individualized$solution$ P9-13)(c)$Individualized$solution$ $ ) P9-14$No$solution$will$be$given$ $ ) P9-15$No$solution$will$be$given$ $ ) 9-31$ P9-16$ rs$=$(max*Cs*Cc)/(Km+Cs)$ max=$1hr-1$ Km$=$0.25$gm/dm3$Yc/s$=$0.5g/g$ Km$=$0.25$gm/dm3$Yc/s$=$0.5g/g$ Km$=$0.25$gm/dm3$ Yc/s$=$0.5g/g$ $ P9-16)(a)$ Cc0$=$0.1g/dm3$ Cs0$=$20$g/dm3$Cc=Cc0$+$Yc/s(Cs0-Cs)$ Cs0$=$20$g/dm3$Cc=Cc0$+$Yc/s(Cs0-Cs)$ Cs0$=$20$g/dm3$ Cc=Cc0$+$Yc/s(Cs0-Cs)$ See$polymath$problem$P9-16-a.pol$ Calculated)values)of)DEQ)variables$$ $$ Variable$ Initial$value$ Minimal$value$ Maximal$value$ Final$value$ 1$$ Cc$$ 0.1$$ 0.1$$ 10.1$$ 10.1$$ 2$$ Cc0$$ 0.1$$ 0.1$$ 0.1$$ 0.1$$ 3$$ Cs$$ 20.$$ 6.301E-11$$ 20.$$ 6.301E-11$$ 4$$ Cs0$$ 20.$$ 20.$$ 20.$$ 20.$$ 5$$ Km$$ 0.25$$ 0.25$$ 0.25$$ 0.25$$ 6$$ rc$$ 0.0493827$$ 1.273E-09$$ 4.043487$$ 1.273E-09$$ 7$$ rs$$ -0.0987654$$ -8.086974$$ -2.546E-09$$ -2.546E-09$$ 8$$ t$$ 0$$ 0$$ 10.$$ 10.$$ 9$$ umax$$ 1.$$ 1.$$ 1.$$ 1.$$ 10$$ Ycs$$ 0.5$$ 0.5$$ 0.5$$ 0.5$$ ) Differential)equations$$ 1$$ d(Cs)/d(t)$=$rs$$ ) Explicit)equations$$1$$Cc0$=$0.1$$ Explicit)equations$$1$$Cc0$=$0.1$$ Explicit)equations$$ 1$$ Cc0$=$0.1$$ 2$$ Ycs$=$0.5$$ 3$$ Cs0$=$20$$ 4$$ umax$=$1$$ 5$$ Cc$=$Cc0+Ycs*(Cs0-Cs)$$ 6$$ Km$=$0.25$$ 7$$ rs$=$-umax*Cs*Cc/(Km+Cs)$$ 8$$ rc$=$-rs*Ycs$$ 9-32$ P9-16)(a))continued$ $ Plot$$of$$Cc$and$$Cs$$versus$time$ $ Plot$of$$rs$and$rc$with$time$ $ P9-16)(b)$Change$the$polymath$code$to$include$ rg$=$max*(1-Cc/C)*Cc$ C=$1$g/dm3$ C=$1$g/dm3$ C=$1$g/dm3$ $ 9-33$ P9-16)(b))continued$ Calculated)values)of)DEQ)variables$$ )$ Variable$ Initial)value$ Minimal)value$ Maximal)value$ Final)value$ 1$$ Cc$$ 0.1$$ 0.1$$ 0.5751209$$ 0.5751209$$ 2$$ Cc0$$ 0.1$$ 0.1$$ 0.1$$ 0.1$$ 3$$ Cinf$$ 1.$$ 1.$$ 1.$$ 1.$$ 4$$ Cs$$ 20.$$ 19.04976$$ 20.$$ 19.04976$$ 5$$ Cs0$$ 20.$$ 20.$$ 20.$$ 20.$$ 6$$ Km$$ 0.25$$ 0.25$$ 0.25$$ 0.25$$ 7$$ rc$$ 0.0225$$ 0.0225$$ 0.0624999$$ 0.0610892$$ 8$$ rg$$ 0.09$$ 0.09$$ 0.2499995$$ 0.2443569$$ 9$$ rs$$ -0.045$$ -0.1249997$$ -0.045$$ -0.1221784$$ 10$$ t$$ 0$$ 0$$ 10.$$ 10.$$ 11$$ umax$$ 1.$$ 1.$$ 1.$$ 1.$$ 12$$ Ycs$$ 0.5$$ 0.5$$ 0.5$$ 0.5$$ $ Differential)equations$$1$$d(Cs)/d(t)$=$rs$$ Differential)equations$$1$$d(Cs)/d(t)$=$rs$$ Differential)equations$$ 1$$ d(Cs)/d(t)$=$rs$$ $ Explicit)equations$$ 1$$ Cc0$=$0.1$$ 2$$ Ycs$=$0.5$$ 3$$ Cs0$=$20$$ 4$$ umax$=$1$$ 5$$ Cinf$=$1$$ 6$$ Cc$=$Cc0+Ycs*(Cs0-Cs)$$ 7$$ Km$=$0.25$$ 8$$ rg$=$umax$*$(1-Cc/Cinf)$*$Cc$$ 9$$ rs$=$-Ycs*rg$$ 10$$ rc$=$-rs*Ycs$$ 9-34$ P9-16)(b))continued$ $Plot$of$Cc$with$time$ $Plot$of$Cc$with$time$ $ Plot$of$Cc$with$time$ $ P9-16)(c)) Cs0$=$20g/dm3$ Cc0$=$0$ The$dilution$rate$at$which$wash-out$occurs$will$be$by$setting$Cc=0$in$equation;$Cc$=$Ycs*(Cs0$$(DKs)/(umax$$D))$Dmax$=$$$ Dmax$=$$$=$0.987$hr-1$Thus$dilution$rate$at$which$washout$occurs$is$0.987hr-1.$$ The$dilution$rate$at$which$wash-out$occurs$will$be$by$setting$Cc=0$in$equation;$Cc$=$Ycs*(Cs0$$(DKs)/(umax$$D))$Dmax$=$$$ Dmax$=$$$=$0.987$hr-1$Thus$dilution$rate$at$which$washout$occurs$is$0.987hr-1.$$ The$dilution$rate$at$which$wash-out$occurs$will$be$by$setting$Cc=0$in$equation;$ Cc$=$Ycs*(Cs0$$(DKs)/(umax$$D))$ Dmax$=$ $ $ Dmax$=$ $$ =$0.987$hr-1$ Thus$dilution$rate$at$which$washout$occurs$is$0.987hr-1.$ $ P9-16)(d)$ $$,$$ $$,$$ $$,$$$$ $ Divide$by$CCV,$ $$Now$$ $ $$$DCc$$=$D$YC/S$(CS0$$ )$ $$Now$$ $ $$$DCc$$=$D$YC/S$(CS0$$ )$ $ $ Now$$ $ $ $ $DCc$$=$D$YC/S$(CS0$$ )$ $ 9-35$ P9-16)(d))continued$ Now,$for$ $,$ $ $ $ $Dmax,prod$$=$0.88$hr-1$ Using$this$value$of$D$we$can$find$the$value$of$Cs$Cs$=$ $$$$$$$$$$$=$1.83$g/dm3$$$$$$$$$$=$9.085g/dm3$rs$=$D(Cs0$$Cs)$=$15.98$g/dm3/hr$ Using$this$value$of$D$we$can$find$the$value$of$Cs$Cs$=$ $$$$$$$$$$$=$1.83$g/dm3$$$$$$$$$$=$9.085g/dm3$rs$=$D(Cs0$$Cs)$=$15.98$g/dm3/hr$ Using$this$value$of$D$we$can$find$the$value$of$Cs$ Cs$=$ $$ $$$$$$$$$=$1.83$g/dm3$ $ $$$$$$$$=$9.085g/dm3$ rs$=$D(Cs0$$Cs)$=$15.98$g/dm3/hr$ $ P9-16)(e)$Cell$death$cannot$be$neglected.$ Kd$=0.02$hr-1$ DCc$=$rg$rd$$$ And$D(CS0$Cs)=$rS$$ And$D(CS0$Cs)=$rS$$ And$D(CS0$Cs)=$rS$$ For$steady$state$operation$to$obtain$mass$flow$rate$of$cells$out$of$the$system,$Fc$ FC$=CCv0$=$(rg-rd)V=$(-kd)$CCV$ After$dividing$by$CcV;$D=-kd$$Now$since$maintenance$is$neglected.$Substituting$for$$in$terms$of$substrate$concentration;$Cs$=$$$The$stoichiometry$equation$can$be$written$as$:$-rs$=$rg$YS/C$$CC$=$Yc/s$$$Now$the$dilution$rate$at$can$be$found$by$substituting$Cc$=0;$Thus$$$Dmax$$=$$=$0.96$hr-1$Similarly$the$expression$for$dilution$rate$for$maximum$production$is$given$by$solving$the$equation$;$(D+kd)$Cc$$=$(D+kd$)$YC/S$(CS0$-$ $)$Now$for$ $,$$$Thus$we$obtain$Dmax,prod$=$0.86$hr-1$ After$dividing$by$CcV;$D=-kd$$Now$since$maintenance$is$neglected.$Substituting$for$$in$terms$of$substrate$concentration;$Cs$=$$$The$stoichiometry$equation$can$be$written$as$:$-rs$=$rg$YS/C$$CC$=$Yc/s$$$Now$the$dilution$rate$at$can$be$found$by$substituting$Cc$=0;$Thus$$$Dmax$$=$$=$0.96$hr-1$Similarly$the$expression$for$dilution$rate$for$maximum$production$is$given$by$solving$the$equation$;$(D+kd)$Cc$$=$(D+kd$)$YC/S$(CS0$-$ $)$Now$for$ $,$$$Thus$we$obtain$Dmax,prod$=$0.86$hr-1$ After$dividing$by$CcV;$ D=-kd$$ Now$since$maintenance$is$neglected.$ Substituting$for$$in$terms$of$substrate$concentration;$ Cs$=$ $ $ The$stoichiometry$equation$can$be$written$as$:$ -rs$=$rg$YS/C$$ CC$=$Yc/s$ $ $ Now$the$dilution$rate$at$can$be$found$by$substituting$Cc$=0;$ Thus$$$Dmax$$=$ $ =$0.96$hr-1$ Similarly$the$expression$for$dilution$rate$for$maximum$production$is$given$by$solving$the$equation$;$ (D+kd)$Cc$$=$(D+kd$)$YC/S$(CS0$-$ $)$ Now$for$ $,$ $ $ Thus$we$obtain$Dmax,prod$=$0.86$hr-1$ 9-36$ P9-16)(f)$ In$this$case$the$maintenance$cannot$be$neglected.$ $m$=$0.2$g/hr/dm3$ The$correlation$for$steady$substrate$concentration$will$remain$the$same.$ Cs$=$ $But$the$cell$maintenance$cannot$be$neglected.$Thus$the$stoichiometry$equation$will$be$changed.$The$equation$will$be$-$$-rs$=$Ys/Crg$+$mCc$=>$-rs$=$rg/YS/C$+$mCc$=>$(Cs0$$Cs)$=$Cc/YC/S$+$$$$Also$we$know$that$by$mass$balance$-$$ $Using$this$relation,$the$stoichiometric$equation$for$substrate$consumption$changes$to$-$$ Cc$=$$Now$for$finding$the$dilution$rate$at$which$wash$out$occurs,$Cc$=0;$So$Dwashout$=$0.98$hr-1$Similarly$to$calculate$$,$$$Thus$we$obtain$Dmax,prod$=$0.74$hr-1$ $But$the$cell$maintenance$cannot$be$neglected.$Thus$the$stoichiometry$equation$will$be$changed.$The$equation$will$be$-$$-rs$=$Ys/Crg$+$mCc$=>$-rs$=$rg/YS/C$+$mCc$=>$(Cs0$$Cs)$=$Cc/YC/S$+$$$$Also$we$know$that$by$mass$balance$-$$ $Using$this$relation,$the$stoichiometric$equation$for$substrate$consumption$changes$to$-$$ Cc$=$$Now$for$finding$the$dilution$rate$at$which$wash$out$occurs,$Cc$=0;$So$Dwashout$=$0.98$hr-1$Similarly$to$calculate$$,$$$Thus$we$obtain$Dmax,prod$=$0.74$hr-1$ $ But$the$cell$maintenance$cannot$be$neglected.$Thus$the$stoichiometry$equation$will$be$changed.$The$ equation$will$be$-$$ -rs$=$Ys/Crg$+$mCc$ =>$-rs$=$rg/YS/C$+$mCc$ =>$(Cs0$$Cs)$=$Cc/YC/S$+$$ $ $ Also$we$know$that$by$mass$balance$-$ $ $ Using$this$relation,$the$stoichiometric$equation$for$substrate$consumption$changes$to$-$ $ Cc$= $ $ Now$for$finding$the$dilution$rate$at$which$wash$out$occurs,$ Cc$=0;$ So$Dwashout$=$0.98$hr-1$ Similarly$to$calculate$$, $$ $ Thus$we$obtain$Dmax,prod$=$0.74$hr-1$ $ ) P9-16)(g)$Individualized$solution$ ) P9-16)(h)$Individualized$solution$ $ ) P9-17)$ Tessier$Equation,$ $ $ $ $ Redoing)P9-16)part)(a))$For$batch$reaction,$$ $ ,$ ,$ $ $) ) $See$Polymath$program$P9-16-a.pol.$ Redoing)P9-16)part)(a))$For$batch$reaction,$$ $ ,$ ,$ $ $) ) $See$Polymath$program$P9-16-a.pol.$ Redoing)P9-16)part)(a))$ For$batch$reaction,$ $ $