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5-1$
Solutions)for)Chapter)5)–)Isothermal)Reactor)Design-)Conversion)
)
P5-1)(a))Example)5-3$
For$50%$conversion,$X$=$0.5$and$k$=$3.07$sec-1$at$1100$K$(from$Example$5-3)$
Now,$we$have$from$the$example$
$
Now,$
n$=$31.44$ft3/0.0205$ft2$X40$ft$=$38.33$
$
P5-1)(b)$Example)5-4$
Individualized$Solution$
$
P5-1)(c))Example)5-5)
New$Dp$=$3D0/4$
Because$the$flow$is$turbulent$
$ $
$
Now,$
$
,$so$too$much$pressure$drop$P$=$0$and$the$flow$stops.$
P5-1)(d))Example)5-6$$
For$turbulent$flow$
$
5-2$
P5-1)(d))Example)5-6$Continued$
$
Therefore$there$is$no$change.$
$
P5-1)(e))Example)5-7$)
(i)$X$increases$and$f$decreases$with$an$increase$in$k’$for$the$same$W,$while$p$remains$unchanged,$and$
vice$versa.$This$is$an$expected$observation,$because$as$k’$increases,$the$rate$of$the$reaction$increases,$
$
P5-1)(f))Example)5-8$)
Individualized$solution$
$
P5-1)(g))Example)5-3,)Using)ASPEN,)we)get$(Refer$to$Aspen$Program$P5-2g$from$polymath$CD)$
(1) At$1000K,$for$the$same$PFR$volume$we$get$only$6.2%$conversion.$While$at$1200K,$we$get$a$
conversion$of$nearly$100%.$This$is$because$the$value$of$reaction$constant$‘k’$varies$rapidly$
(2) Earlier$for$an$activation$energy$of$82$kcal/mol$we$got$approx.$81%$conversion.$For$activation$
energy$of$74$kcal/mol$keeping$the$PFR$volume$the$same$we$get$a$conversion$of$71.1%.$
$
P5-1)(h)$Individualized$solution.$
$
P5-1)(i)$Individualized$solution$
)
$
P5-2)The$key$for$decoding$the$algorithm$to$arrive$at$a$numerical$score$for$the$Interaction$Computer$
Games$(ICGs)$is$given$at$the$front$of$this$Solutions$Manual.$
$
5-3$
P5-3)(c)$(1)$>0.234$$
$
−rA=kCA0
21−X
( )
2
p
−rA=∝kp2
p2<1
$
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$balance:$ $
$
$
Stoichiometry:$$$and$ $$
$
$
$ $
$
$
$
Solving$using$polymath$to$get$a$table$of$values$of$X$Vs$T.$
See$Polymath$program$P5-6.pol.$
$
5-5$
P5-6)Continued$$
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess$
X 0.4229453 3.638E-12 0.5 $
NLE Report (safenewt)
Nonlinear equations
[1] f(X) = (z/y)*X/((1-X)^2 - X^2/Kc) -V = 0$
$
Explicit equations $
[1] To = 300$
$
T(in$K)$
X$
300$
0.40$
301$
0.4075$
303$
0.4182$
304$
0.4213$
305$
0.4228$
305.5$
0.4229$
305.9$
0.4227$
307$
0.421$
310$
0.4072$
315$
0.3635$
$
We$get$maximum$X$=$0.4229$at$T$=$305.5$K.$
)
$
5-6$
P5-7)$
PFR$$
$
$
$ $ $
$ $ $
$CSTR$
$
$
$ $ $
$ $ $
)
$
5-7$
P5-8$
Base Case PFR above. Find k$
Mole Balance
$
Rate Law $
Stoichiometry$
$
Combine $
$
$
Now find X2$
5-8$
P5-8)Continued$
$
)
P5-9)(a))$
$
Using$the$Arrhenius$equation$at$the$CSTR$temperature$of$300$K$yields$the$new$specific$reaction$rate.$
$
So,$we$get$0.57$as$conversion$instead$of$0.5.$
$
5-9$
P5-9)(b))
$
$
P5-9)(d)$
1)$CSTR$and$PFR$are$connected$in$series:$
$ $
Solving$the$quadratic$equation,$XCSTR$=$0.44$
For$PFR,$
$ $
$
2)$when$CSTR$and$PFR$are$connected$in$parallel,$
$ $ $
5-10$
P5-9)(d))Continued$
XPFR$=$0.92$
$
P5-9)(e)$
To$process$the$same$amount$of$species$A,$the$batch$reactor$must$handle$
$
If$the$reactants$are$in$the$same$concentrations$as$in$the$flow$reactors,$then$
$
So$the$batch$reactor$must$be$able$to$process$14400$dm3$every$24$hours.$
Now$we$find$the$time$required$to$reach$90%$conversion.$Assume$the$reaction$temperature$is$300K.$
$
$
Assume$that$it$takes$three$hours$to$fill,$empty,$and$heat$to$the$reaction$temperature.$
tf$=$3$hours$
Therefore,$we$can$run$4$batches$in$a$day$and$the$necessary$reactor$volume$is$
$
Referring$to$Table$1-1$and$noting$that$3600$dm3$is$about$1000$gallons,$we$see$that$the$price$would$be$
approximately$$85,000$for$the$reactor.$
$
P5-9)(f)$
The$points$of$the$problem$are:$
1)$To$note$the$significant$differences$in$processing$times$at$different$temperature$$(i.e.$compare$
part$(b)$and$(c)).$
5-11$
P5-9)(g)$
Individualized$solution$
)
$
P5-10)(a))$
The$blades$makes$two$equal$volumes$zones$of$500gal$each$rather$than$one$‘big’$mixing$zone$of$1000gal.$
So,$we$get$0.57$as$conversion$instead$of$0.5.$
$ $ $
5-12$
P5-10)(b))$
CAO$=$2$mol/dm3$
A$→$B$
Assuming$1st$order$reaction,$
For$CSTR,$
$
Now$assuming$2nd$order$reaction,$
For$CSTR,$$Now,$assuming$2nd$order$reaction,$
For$CSTR,$
$
$
So,$while$calculating$PFR$conversion$they$considered$reaction$to$be$1st$order.$But$actually$it$is$a$second$
order$reaction.$
$
P5-10)(c))$
A$graph$between$conversion$and$particle$size$is$as$follows:$Originally$we$are$at$point$A$in$graph,$when$
particle$size$is$decreased$by$15%,$we$move$to$point$C,)which$have$same$conversion$as$particle$size$at$A.$
But$when$we$decrease$the$particle$size$by$20%,$we$reach$at$point)D,$so$a$decrease$in$conversion$is$
5-13$
P5-11)
Reaction:$
A$+$B$→$C$+$D$
Mole$Balance:$$
F
A0
dX
dW =−"
r
A
$
Rate$Law:$
−"
r
A=KCACB
$
Stoichiometry:$
CA=F
A
υ
F
A0 1−X
( )
υ=CA0 1−X
( )
p((;(( P
P
0
1− αW
( )
1 2
CA=CA0 1−X
( )
1− αW
( )
1 2
$
Equimolar$flow$rate:$
CB=CA=CA0 1−X
( )
1− αW
( )
1 2
$
Combine:$
−"
r
A=kCA0
21−X
( )
2
1− αW
( )
dX
dW =
kCA0
21−X
( )
2
1− αW
( )
F
A0
$
Evaluate:$
$
#
&
Solution:$
First$we$must$solve$for$KW:$
$
5-14$
$
$
)
$
5-15$
P5-12)$
)
P5-12)(a)$
$
$
P5-12)(b)$
$
5-16$
P5-12)(c)$
$
$
P5-12)(d)$
The$conversion$will$remain$unchanged$since$the$reactor$is$well-mixed$ideal$CSTR.$
$
P5-13)(a)$
V$=$5$dm3$
Mass$of$one$capsule$=$18$g$
$
P5-13)(b))
For$a$batch$reactor,$
Solving$the$differential$equation,$we$get:$
ln
𝐶
!
𝐶
= 𝑘!𝑡$
P5-13)(c)))
𝑘!
=exp −
𝐸
1
−
1
$
$
5-17$
P5-13)(c))Continued$
Substituting$values$of$kA0$=$0.00944$hr-1,$E$$=$20500$cal/mol,$R$=$1.986$cal/mol$K,$T$=$311.7$K$and$T0$=$310$
K,$we$get:$
$
$
P5-14)(a)$
POLYMATH Report$
Ordinary Differential Equations$
Calculated values of DEQ variables $
$
Variable$
Initial value$
Minimal value$
Maximal value$
Final value$
1 $
w $
0 $
0 $
2000. $
2000. $
2 $
x $
0 $
0 $
0.9978192 $
0.9978192 $
3 $
FBuO $
50. $
50. $
50. $
50. $
4 $
PBuO $
10. $
10. $
10. $
10. $
5 $
PBu $
10. $
0.0109159 $
10. $
0.0109159 $
6 $
k $
0.054 $
0.054 $
0.054 $
0.054 $
7 $
K $
0.32 $
0.32 $
0.32 $
0.32 $
8 $
rBu $
-0.0306122 $
-0.0421872 $
-0.0005854 $
-0.0005854 $
$
Differential equations $
1 $
d(x)/d(w) = -rBu / FBuO $
$
Kgcat-1 $
Explicit equations $
1 $
FBuO = 50 $
$
Kmol.hr-1 $
2 $
PBuO = 10 $
$
atm $
3 $
PBu = PBuO * (1 - x) / (1 + x) $
$
atm $
4 $
k = 0.054 $
$
kmol.kgcat-1.hr-1.atm-1 $
5 $
K = 0.32 $
$
atm-1 $
6 $
rBu = -k * PBu / (1 + K * PBu) ^ 2 $
$
Kmol.kgcat-1.hr-1 $
5-18$
P5-14)(a)$continued$
$
$
Hence,$weight$of$catalyst$required$for$80%$conversion$is$1054.1$kg.$$
$
P5-14)(b)$
Differential)Mole)Balance:)
F
A0 =dX
dW =−"
r
Bu
$
PBu$=$PBuO$
$Here,$$ $$
5-19$
P5-14)(b)$continued$
For)Fluidized)CSTR:$ $ $$
$
Given,$$
X$=$0.8$
$FBuO$=$50$Kmol.hr-1 $
Therefore,$putting$the$values$in$the$above$equation$
w$=$50*0.8
$=$1225)Kg$
P5-14)(c))With$pressure$drop$parameter$alpha$=$0.0006$kg-1$
POLYMATH Report$
Ordinary Differential Equations$
Calculated values of DEQ variables $
$
Variable$
Initial value$
Minimal value$
Maximal value$
Final value$
1 $
w $
0 $
0 $
1100. $
1100. $
2 $
x $
0 $
0 $
0.7750365 $
0.7750365 $
3 $
y $
1. $
0.2689385 $
1. $
0.2689385 $
4 $
FBuO $
50. $
50. $
50. $
50. $
5 $
f $
1. $
1. $
6.600157 $
6.600157 $
6 $
PBuO $
10. $
10. $
10. $
10. $
7 $
PBu $
10. $
0.3408456 $
10. $
0.3408456 $
8 $
k $
0.054 $
0.054 $
0.054 $
0.054 $
9 $
K $
0.32 $
0.32 $
0.32 $
0.32 $
10 $
rBu $
-0.0306122 $
-0.0421871 $
-0.0149635 $
-0.0149635 $
11 $
alpha $
0.0006 $
0.0006 $
0.0006 $
0.0006 $
Differential equations $
1 $
d(x)/d(w) = -rBu / FBuO $
$
Kgcat-1 $
2 $
d(y)/d(w) = -alpha * (1 + x) / 2 / y $
$
Kgcat-1 $
5-20$
P5-14)(c))continued$
Explicit equations $
1 $
FBuO = 50 $
$
Kmol.hr-1 $
2 $
f = (1 + x) / y $
3 $
PBuO = 10 $
$
atm $
4 $
PBu = PBuO * (1 - x) / (1 + x) * y $
$
atm $
5 $
k = 0.054 $
$
kmol.kgcat-1.hr-1.atm-1 $
6 $
K = 0.32 $
$
atm-1 $
7 $
rBu = -k * PBu / (1 + K * PBu) ^ 2 $
$
Kmol.kgcat-1.hr-1 $
8 $
alpha = 0.0006 $
$
Kg-1 $
$
$
