978-0134663890 Chapter 12 Part 4

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$

10$

$X$

0$

0.3458817$

$T$

300$

449.27319$

$Ta$

423.8$

450.01394$

$K$

0.01$

2.5406259$

$Kc$

0.3567399$

9.8927301$

$Fa0$

0.2$

0.2$$$

$Ca0$

0.1$

0.1$$$

$Ra$

–1.0E–04$

–0.0141209$

–1.0E–04$

–0.0019877$

$Xe$

0.0333352$

0.3801242$

$DH$

6000$

6000$$

$Ua$

20$

$Fao$

0.2$

$sumcp$

30$

$mc$

50$

$Cpc$

1$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(X)/d(V)$=$–ra$/$Fa0$

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$=$0.2$

12–62$

P12–10)(f))continued$

$

P12–11)(a)$

(1)$

Reaction:$𝐴+𝐵→𝐶.$Equal$molar$feed,$irreversible$elementary$reaction.$

Mole$Balance:!”

!”

= −!

!

!!!

$

R=1.98$cal/mol/K,$𝑇

!=300𝐾$

Stoichiometry$(liquid):$$

𝐶

!=𝐶!=𝐶

!!1−𝑋

𝑇

!

𝑇$

12–63$

P12-11)(a))continued$

𝑇

!=323𝐾$

$

$$$ $

$

12–64$

P12–11)(a)$continued$

(2)$

when$!!

!$increase$by$a$factor$of$3000:$$$

$

12–65$

P12–11)(a))continued$

$

(3)$$

Taking$the$pressure$drop$into$account:$$

𝑑𝑦

𝛼

1+𝜖𝑋

𝑇

$

In$addition,$the$expression$for$concentration$of$A$changed$into:$$

𝐶

!=𝐶

!!1−𝑋

𝑇

!

𝑇𝑦$

Solving$the$system$using$Polymath:$$

$

12–66$

P12–11)(a))continued$

$

P12–11)(b)$

(1)$

Co–current$Coolant:$$

𝑑𝑇

𝑈𝑎

𝜌(𝑇−𝑇

!)

$

12–67$

P12–11)(b))continued$

$

12–68$

P12–11)(b))continued$

(2)$

Counter–current$Coolant:$$

𝑑𝑇

𝑈𝑎

𝜌(𝑇

!−𝑇)

$

$ $

12–69$

P12–11)(b))continued$

$

(3)$

For$an$adiabatic$process,$no$heat$transfer$occurred,$the$reactor$energy$balance$became:$$

𝑑𝑇

!)(−Δ𝐻!” )

$

12–70$

P12–11)(c)$

For$a$CSTR$reactor:$$

Mole$balance:$$

𝑊=

𝐹

!!𝑋

−𝑟

$

Rate$Law:$$

−𝑟

!=𝑘𝐶

!

!$

Combining$the$mole$balance$and$the$rate$law:$$

𝑊=

𝐹

!!𝑋

!

$

𝑋!” =

𝑈𝐴

𝐹

!!

(𝑇−𝑇

!)

−Δ𝐻𝑟𝑥𝑛(𝑇

119

0.2

6000 (𝑇−300)$

$

Make$plot$of$𝑋!” $and$𝑋!”$on$the$same$plot:$$

$

T=306K$

0″

0.1″

0.2″

0.3″

0.4″

0.5″

0.6″

0.7″

0.8″

0.9″

1″

298″ 300″ 302″ 304″ 306″ 308″ 310″

XEB” XMB”

12–71$

P12–11)(d))$

Now$W=80kg$

Rate$Law:$−𝑟

!=𝑘!𝐶

!𝐶!−𝑘!𝐶!$

$

P12–11)(e))Individualized$solution$

$

P12–12)(a)$

Start$with$the$complete$energy$balance:$

ˆ

S i i in i i out

dE Q W E F E F

dt =− − Σ − Σ



$

The$following$simplifications$can$be$made:$

• It$is$steady$state.$

• In$part$(a),$there$is$no$heat$taken$away$or$added$

• There$is$no$shaft$work$

That$leaves$us$with$

0i i in i i out

E F E F=−Σ − Σ

$

Evaluating$energy$terms:$

In:$

0 0 0 0 0 0A A B B C C

H F H F H F+ +

$

Out:$

( ) ( ) ( )

A A A B B B C C C

H F R V H F R V H F R V+ + + + +

$

Now$we$evaluate$Fi$

0 0A A A

F F F X=−

0 0B B A

F F F X= +

0 0C C A C

F F F X R V= + −

FB0$=$FC0$=$0$

$

Inserting$these$into$our$equation$gives:$

$

Combining$and$substituting$terms$gives:$

$

Differentiating$with$respect$to$V$with$ΔCP$=$0$

$

12–72$

P12–12)(a))continued$

Combine$that$with$the$mole$balance$and$rate$law:$

$

See$Polymath$program$P12–12.pol.$

Calculated)values)of)DEQ)variables$$

$$

Variable$

Initial$value$

Minimal$value$

Maximal$value$

Final$value$

1$$

Ca$$

0.2710027$$

0.072777$$

0.2710027$$

0.072777$$

2$$

Cb$$

0$$

0.0498859$$

3$$

Cc$$

0$$

0.0273679$$

0.0060305$$

4$$

Cpa$$

40.$$

5$$

Cpb$$

25.$$

6$$

Cpc$$

15.$$

7$$

Ct0$$

0.2710027$$

8$$

dHrx$$

–2.0E+04$$

9$$

Fa$$

5.42$$

3.215735$$

5.42$$

3.215735$$

10$$

Fa0$$

5.420054$$

11$$

Fb$$

0$$

2.204265$$

12$$

Fc$$

0$$

0.9446979$$

0.2664623$$

13$$

Ft$$

5.42$$

6.364698$$

5.686462$$

14$$

Hc$$

–3.735E+04$$

–2.988E+04$$

15$$

k$$

0.133$$

10.91169$$

16$$

kc$$

1.5$$

17$$

Kc$$

0.0690539$$

0.0041692$$

0.0690539$$

0.0041692$$

18$$

ra$$

–0.0360434$$

–0.1033995$$

–0.0067748$$

19$$

Rc$$

0$$

0.0410519$$

0.0090457$$

20$$

T$$

450.$$

947.6106$$

21$$

To$$

450.$$

22$$

V$$

0$$

100.$$

23$$

Xe$$

0.4506284$$

0.1771437$$

0.4506284$$

0.1771437$$

$

Differential)equations$$

1$$

d(Fc)/d(V)$=$–ra–Rc$$

2$$

d(Fb)/d(V)$=$–ra$$

3$$

d(Fa)/d(V)$=$ra$$

4$$

d(T)/d(V)$=$(ra*dHrx$-$Rc*Hc)/(Fa0*Cpa)$$

$

Explicit)equations$$

1$$

k$=$.133*exp((31400/8.314)*(1/450–1/T))$$

2$$

To$=$450$$

3$$

Ct0$=$10/0.082/450$$

4$$

Ft$=$Fa$+$Fb$+$Fc$$

5$$

Cc$=$Ct0*Fc/Ft*To/T$$

6$$

Fa0$=$Ct0*20$$

7$$

Ca$=$Ct0*Fa/Ft*To/T$$

8$$

dHrx$=$–20000$$

9$$

Cb$=$Ct0*Fb/Ft*To/T$$

10$$

Kc$=$1*exp((dHrx/8.314)*(1/300–1/T))$$

11$$

Cpc$=$15$$

12$$

ra$=$–k*(Ca–Cb*Cc/Kc)$$

13$$

kc$=$1.5$$

12–73$

14$$

Rc$=$kc*Cc$$

15$$

Xe$=$((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5$$

16$$

Cpa$=$40$$

17$$

Cpb$=$25$$

18$$

Hc$=$–40000+Cpc*(T–273)$$

$

Concentration$profile:$

$

As$the$equilibrium$constant$Kc$increases,$more$product$will$form$and$be$removed$through$the$

membrane.$

$

P12–12)(b)$

Now,$the$heat$balance$equation$needs$to$be$modified.$

)

$

12–74$

P12–12)(b))continued$

$

The$difference$between$three$cases$are$not$obvious,$because$the$reaction$rate$is$very$low$due$to$the$

$

P12–13)(a))

$A)$Increase$conversion.$

$ $

12–75$

P12-13)(a))continued$

$

If$it$is$an$adiabatic$system,$then$it$has$to$be$endothermic,$because$

the$temperature$decreases$and$the$heat$of$reaction$is$positive.$

P12–13)(b))

Answer:$C$and$D$

)

P12–14$

Assume$Adiabatic$

$

P12–14)(a)$

$ –12,000$cal/mol$

)

P12–15)(a)$

G(T)=ΔHRX

X=

τ

k

1+

τ

k,k=6.6×10−3exp E

R

1

350 −1

T

#

$

%&

‘

(

)

*

+

,

–

.

R(T)=Cp01+

κ

( )

T−Tc

( )

Cp0=ΘiCpi =50

∑

12–76$

P12-15)(a))continued$

κ

=UA

Cp0FA0

=8000

50 ×80 =2

Tc =

κ

Ta+T0

1+

κ

=350K

To$find$the$steady$state$T,$we$must$set$G(T)$=$R(T).$This$can$be$done$either$graphically$or$solving$the$

equations.$We$find$that$for$T0$=$450$K,$steady$state$temperature$399.94$K.$

Graphical$method$

$

Polymath)Results$02–22–2006,$$$Rev5.1.233$$

Variable$

initial$value$

Minimal$value$

maximal$value$

final$value$

$t$

0$

1000$

1000$$

$T$

350$

450$

450$$$

$RT$

0$

1.5E+04$

1.5E+04$$$

$EoR$

2.013E+04$

$k$

0.0066$

2346.7972$

$tau$

100$

100$$$

$X$

0.3975904$

0.9999957$

$GT$

2981.9277$

7499.968$

7499.968$$

$at$

2981.9277$

–7500.032$

4336.6841$

–7500.032$$

$

ODE)Report)(RKF45)$

Differential$equations$as$entered$by$the$user$

$[1]$d(T)/d(t)$=$0.1$

Explicit$equations$as$entered$by$the$user$

$[1]$RT$=$150*(T–350)$

$[2]$EoR$=$40000/1.987$

12–77$

P12-15)(a))continued$

Polymath)Results$02–22–2006,$$$Rev5.1.233$$

NLE)Solution)$

Variable$

Value$

f(x)$

Ini$Guess$

$T$

399.9425$

3.181E–09$

300$$$

$RT$

7491.375$$

$

$EoR$

2.013E+04$

$

$K$

8.6856154$

$

$tau$

100$$$

$

$X$

0.99885$$$

$

$GT$

7491.375$$

$

NLE)Report)(safenewt)$

Nonlinear$equations$$

$[1]$f(T)$=$RT–GT$=$0$

$Explicit$equations$$

$[1]$RT$=$150*(T–350)$

First,$we$must$plot$G(T)$and$R(T)$for$many$different$T0’s$on$the$same$plot.$From$this$we$must$generate$

data$that$we$use$to$plot$Ts$vs$To.$

$

P12–15)(c)$

For$high$conversion,$the$feed$stream$must$be$pre–heated$to$at$least$404$K.$At$this$temperature,$X$=$.991$

P12–15)(d)$

For$a$temperature$of$369.2$K,$the$conversion$is$0.935$$

P12–15)(e)$

The$extinction$temperature$is$360$K$(87oC).$

)

12–78$

P12–16)(a)$

$

The$following$plot$gives$us$the$steady$state$temperatures$of$310,$377.5$and$418.5$K$

See$Polymath$program$P12–16–b.pol.$

$

12–79$

P12–16)(c)$

$

P12–16)(d)$

$

P12–16)(e)$

The$plot$below$shows$Ta$varied.$

$

The$next$plot$shows$how$to$find$the$ignition$and$extinction$temperatures.$The$ignition$temperature$is$

358$K$and$the$extinction$temperature$is$208$K.$

$

12–80$

P12–16)(f)$

$

P12–16)(g)$

At$the$maximum$conversion$G(t)$will$also$be$at$its$maximal$value.$This$occurs$at$approximately$T$=$404$

K.$G(404$K)$=$73520$cal.$For$there$top$be$a$steady$state$at$this$temperature,$R(T)$=$G(T).$$See$Polymath$

program$P12–16–g.pol.$

$

P12–16)(h))Individualized$solution$

$

P12–16)(i))$

The$adiabatic$blowout$flow$rate$occurs$at$ $