978-0134663890 Chapter 2 Part 1

Solutions)for)Chapter)2)–)Conversion)and)Reactor)Sizing)

)

P2–1)(a))Example)2–1)through)2-3$

If$flow$rate$FAO$is$cut$in$half.$

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$X=0.2,$

From$previous$example;$V1$(volume$of$first$CSTR)$$=$0.188$m3$

Also$the$next$reactor$is$PFR,$Its$volume$is$calculated$as$follows$

$

(2)$

Now)the)sequence)of)the)reactors)remain)

unchanged.$

But$all$reactors$have$same$volume.$

Now$

For$PFR:$

$$$$$$

X2$=$.183$

For$CSTR,$

VCSTR2$=)

$

2-3$

P2–1)(e))Example)2–5)Continued)

Conversion$

Original$Reactor$Volumes$

Worst$Arrangement$

X1$=$0.20$

V1$=$0.188$(CSTR)$

V1$=$0.23$(PFR)$

X2$=$0.60$

V2$=$0.38$(PFR)$

V2$=$0.53$(CSTR)$

X3$=$0.65$

V3$=$0.10$(CSTR)$

$

For$PFR,$

$$X1$=$0.2$

A

P2–1)(f)$No$solution$will$be$given.$

)

$

P2-2$The$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.$

)

P2-3)

$

X$

0$

0.1$

0.2$

0.4$

0.6$

0.7$

0.8$

FAO/-rA$(m3)$

0.89$

1.08$

1.33$

2.05$

3.54$

5.06$

8.0$

$

V$=$1$m3$$$$$$$$$$$$$$$$$$$$$$$$

2-4$

P2–3)(a)$Two$CSTRs$in$series$$$$$$

For$first$CSTR,$

$

P2–3)(b)$

Two$PFRs$in$series$

$$$$$$ $

P2–3)(c)$$

Two$CSTRs$in$parallel$with$the$feed,$FAO,$divided$equally$between$two$reactors.$FANEW/-rAX1$=$0.5FAO/-rAX1$$

P2–3)(d)$

Two$PFRs$in$parallel$with$the$feed$equally$divided$between$the$two$reactors.$

FANEW/-rAX1$=$0.5FAO/-rAX1$

P2–3)(e)$$

A$CSTR$and$a$PFR$are$in$parallel$with$flow$equally$divided$

Since$the$flow$is$divided$equally$between$the$two$reactors,$the$overall$conversion$is$the$average$of$the$

P2–3)(f)$$

A$PFR$followed$by$a$CSTR,$

XPFR$$$=$0.565$(using$part$(b))$

P2–3)(g)$$

A$CSTR$followed$by$a$PFR,$

$$$$$XCSTR$=$0.435$(using$part(a))$

2-5$

P2–3)(h)$$

A$1$m3$PFR$followed$by$two$0.5$m3$CSTRs,$

)

P2–4))$

)Exothermic$reaction:$$$A$$→$$B$$+$$C$

X$

r(mol/dm3.min)$

1/–r(dm3.min/mol)$

0$

1$

0.20$

1.67$

0.6$

0.40$

5$

0.2$

0.45$

5$

0.2$

0.50$

5$

0.2$

0.60$

5$

0.2$

0.80$

1.25$

0.8$

0.90$

0.91$

1.1$

$

P2–4)(a))$

To$solve$this$problem,$first$plot$1/–rA$vs.$X$from$the$chart$above.$Second,$use$mole$balance$as$given$

below.$

CSTR:$

P2–4)(b))$

For$a$feed$stream$that$enters$the$reaction$with$a$previous$conversion$of$0.40$and$leaves$at$any$

conversion$up$to$0.60,$the$volumes$of$the$PFR$and$CSTR$will$be$identical$because$of$the$rate$is$constant$

over$this$conversion$range.$

$

2-6$

P2–4)(b))Continued)$

$

P2–4)(c)$

VCSTR$=$105$dm3$

P2–4)(d)))$

P2–4)(e)$

From$part$(a),$we$know$that$X1$=$0.40.$$Use$trial$and$error$to$find$X2.$

Mole$balance:$

$

2-7$

P2–4)(f)$$

See$Polymath$program$P2-4-f.pol.$

$

)

P2-5$

We$must$first$find$a$CSTR$up$to$X$=$0.2$to$minimize$the$volume,$and$hence$the$cost$of$reactor.$

We$can$either$use$a$CSTR$or$a$PFR$for$X$=$0.2$to$X$=$0.6$since$the$rate$is$independent$of$X$in$that$range.$

$

)

P2–6)(a))Individualized$Solution)$

2-8$

P2–6)(b))

1)$In$order$to$find$the$age$of$the$baby$hippo,$we$need$to$know$the$volume$of$the$stomach.$The$

metabolic$rate,$-rA,$is$the$same$for$mother$and$baby,$so$if$the$baby$hippo$eats$one$half$of$what$the$

mother$eats$then$Fao$(baby)$=$½$Fao$(mother).$The$Levenspiel$Plot$is$shown:$

$

))))))))))))))))))))Autocatalytic)Reaction)

$

Since$the$volume$of$the$stomach$is$proportional$to$the$age$of$the$baby$hippo,$and$the$volume$of$the$

baby’s$stomach$is$half$of$an$adult,$then$the$baby$hippo$is$half$the$age$of$a$full$grown$hippo.$

$

P2–6)(b))

2)$$If$Vmax$and$mao$are$both$one$half$of$the$mother’s$then$

$

will$be$identical$for$both$the$baby$and$mother.$$$

Assuming$that$like$the$stomach$the$intestine$volume$is$proportional$to$age$then$the$volume$of$the$

intestine$would$be$0.75$m3$and$the$final$conversion$would$be$0.40$

$

2-9$

P2–6)(c))$

Vstomach$=$0.2$m3$

From$the$web$module$we$see$that$if$a$polynomial$is$fit$to$the$autocatalytic$reaction$we$get:$

$

=$127X4$-$172.36X3$+$100.18X2$-$28.354X$+$4.499$

P2–6)(d))$

PFR→$CSTR$

PFR:$

Outlet$conversion$of$PFR$=$0.111$

$

We$must$solve$$

V$=$0.46$=$(X–0.111)(127X4$-$172.36X3$+$100.18X2$-$28.354X$+$4.499)$

P2-7$

Irreversible$gas$phase$reaction$See$Polymath$program$P2–7.pol.$

P2–7)(a)$

PFR$volume$necessary$to$achieve$50%$conversion$

Mole$Balance$

$ $ $

Volume$=$Geometric$area$under$the$curve$of$

(FA0/-rA)$vs$X)$

$

V$=)150000$m3$

$

P2–7)(b))$

CSTR$Volume$to$achieve$50%$conversion$

Mole$Balance$

$

V$=$50000m3$

$

P2–7)(c))$

Volume$of$second$CSTR$added$in$series$to$achieve$80%$conversion$

$

V2$=$150000m3$

$

P2–7)(d)$

Volume$of$PFR$added$in$series$to$first$CSTR$to$achieve$

80%$conversion$

$

VPFR$=$90000m3$

$

2-11$

P2-7)(e)$$

For$CSTR,$$

V$=$60000$m3$(CSTR)$

Mole$Balance$

$ $

X$=$0.463$

For$PFR,$$

V$=$60000$m3$(PFR)$

Mole$balance$

$ $

X$=$0.134$

$

P2–7)(f))$

Real$rates$would$not$give$that$shape.$$The)reactor)volumes)are)absurdly)large.$

$

)

P2-8$

Problem$2–8$involves$estimating$the$volume$of$three$reactors$from$a$picture.$The$door$on$the$side$of$the$

building$was$used$as$a$reference.$It$was$assumed$to$be$8$ft$high.$

The$following$estimates$were$made:$

CSTR$

2-12$

P2–10)(a)$

The$smallest$amount$of$catalyst$necessary$to$achieve$80$%$conversion$in$a$CSTR$and$PBR$connected$in$

series$and$containing$equal$amounts$of$catalyst$can$be$calculated$from$the$figure$below.$$

$

P2–10)(b)$

Calculate$the$necessary$amount$of$catalyst$to$reach$80$%$conversion$using$a$single$CSTR$by$determining$

the$area$of$the$shaded$region$in$the$figure$below.$

$

2-13$

P2–10)(c)$

The$CSTR$catalyst$weight$necessary$to$achieve$40$%$conversion$can$be$obtained$by$calculating$the$area$

of$the$shaded$rectangle$shown$in$the$figure$below.$

$

The$area$of$the$rectangle$is$approximately$7.6$kg$of$catalyst.$

$

P2–10)(d)$

The$catalyst$weight$necessary$to$achieve$80$%$conversion$in$a$PBR$is$found$by$calculating$the$area$of$the$

shaded$region$in$the$figure$below.$