978-0134663890 Chapter 5 Part 1

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 $

2000. $

2 $

x $

0 $

0.9978192 $

3 $

FBuO $

50. $

4 $

PBuO $

10. $

5 $

PBu $

10. $

0.0109159 $

10. $

0.0109159 $

6 $

k $

0.054 $

7 $

K $

0.32 $

8 $

rBu $

–0.0306122 $

–0.0421872 $

–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 $

1100. $

2 $

x $

0 $

0.7750365 $

3 $

y $

1. $

0.2689385 $

1. $

0.2689385 $

4 $

FBuO $

50. $

5 $

f $

1. $

6.600157 $

6 $

PBuO $

10. $

7 $

PBu $

10. $

0.3408456 $

10. $

0.3408456 $

8 $

k $

0.054 $

9 $

K $

0.32 $

10 $

rBu $

–0.0306122 $

–0.0421871 $

–0.0149635 $

11 $

alpha $

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 $

$