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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$PFR:$
$$$$$$
X2$=$.183$
For$CSTR,$
VCSTR2$=)
$
$
2-3$
P2-1)(e))Example)2-5)Continued)
Conversion$
Original$Reactor$Volumes$
Worst$Arrangement$
X1$=$0.20$
V1$=$0.188$(CSTR)$
V1$=$0.23$(PFR)$
X2$=$0.60$
V2$=$0.38$(PFR)$
V2$=$0.53$(CSTR)$
X3$=$0.65$
V3$=$0.10$(CSTR)$
V3$=$0.10$(CSTR)$
$
For$PFR,$
$$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:$$$A$$→$$B$$+$$C$
X$
r(mol/dm3.min)$
1/-r(dm3.min/mol)$
0$
1$
1$
0.20$
1.67$
0.6$
0.40$
5$
0.2$
0.45$
5$
0.2$
0.50$
5$
0.2$
0.60$
5$
0.2$
0.80$
1.25$
0.8$
0.90$
0.91$
1.1$
$
P2-4)(a))$
To$solve$this$problem,$first$plot$1/–rA$vs.$X$from$the$chart$above.$Second,$use$mole$balance$as$given$
below.$
CSTR:$
P2-4)(b))$
For$a$feed$stream$that$enters$the$reaction$with$a$previous$conversion$of$0.40$and$leaves$at$any$
conversion$up$to$0.60,$the$volumes$of$the$PFR$and$CSTR$will$be$identical$because$of$the$rate$is$constant$
over$this$conversion$range.$
$
2-6$
P2-4)(b))Continued)$
$
P2-4)(c)$
VCSTR$=$105$dm3$
P2-4)(d)))$
P2-4)(e)$
From$part$(a),$we$know$that$X1$=$0.40.$$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$below.$
