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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$
0$
10$
10$
$X$
0$
0$
0.3458817$
0.3458817$
$T$
300$
300$
449.27319$
449.27319$
$Ta$
423.8$
423.8$
450.01394$
450.01394$
$K$
0.01$
0.01$
2.5406259$
2.5406259$
$Kc$
0.3567399$
0.3567399$
9.8927301$
9.8927301$
$Fa0$
0.2$
0.2$
0.2$
0.2$$$
$Ca0$
0.1$
0.1$
0.1$
0.1$$$
$Ra$
-1.0E-04$
-0.0141209$
-1.0E-04$
-0.0019877$
$Xe$
0.0333352$
0.0333352$
0.3801242$
0.3801242$
$DH$
6000$
6000$
6000$
6000$$
$Ua$
20$
20$
20$
20$
$Fao$
0.2$
0.2$
0.2$
0.2$
$sumcp$
30$
30$
30$
30$
$mc$
50$
50$
50$
50$
$Cpc$
1$
1$
1$
1$
$
ODE)Report)(RKF45)$
Differential$equations$as$entered$by$the$user$
$[1]$d(X)/d(V)$=$-ra$/$Fa0$
Explicit$equations$as$entered$by$the$user$
$[1]$k$=$.01$*$exp((10000$/$2)$*$(1$/$300$-$1$/$T))$
$[2]$Kc$=$10$*$exp(6000$/$2$*$(1$/$450$-$1$/$T))$
$[3]$Fa0$=$0.2$
12-62$
P12-10)(f))continued$
$
$
$
$
P12-11)(a)$
(1)$
Reaction:$𝐴+𝐵→𝐶.$Equal$molar$feed,$irreversible$elementary$reaction.$
Mole$Balance:!"
!"
= −!
!
!!!
$
R=1.98$cal/mol/K,$𝑇
!=300𝐾$
Stoichiometry$(liquid):$$
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$$
0$$
0.0498859$$
0.0498859$$
3$$
Cc$$
0$$
0$$
0.0273679$$
0.0060305$$
4$$
Cpa$$
40.$$
40.$$
40.$$
40.$$
5$$
Cpb$$
25.$$
25.$$
25.$$
25.$$
6$$
Cpc$$
15.$$
15.$$
15.$$
15.$$
7$$
Ct0$$
0.2710027$$
0.2710027$$
0.2710027$$
0.2710027$$
8$$
dHrx$$
-2.0E+04$$
-2.0E+04$$
-2.0E+04$$
-2.0E+04$$
9$$
Fa$$
5.42$$
3.215735$$
5.42$$
3.215735$$
10$$
Fa0$$
5.420054$$
5.420054$$
5.420054$$
5.420054$$
11$$
Fb$$
0$$
0$$
2.204265$$
2.204265$$
12$$
Fc$$
0$$
0$$
0.9446979$$
0.2664623$$
13$$
Ft$$
5.42$$
5.42$$
6.364698$$
5.686462$$
14$$
Hc$$
-3.735E+04$$
-3.735E+04$$
-2.988E+04$$
-2.988E+04$$
15$$
k$$
0.133$$
0.133$$
10.91169$$
10.91169$$
16$$
kc$$
1.5$$
1.5$$
1.5$$
1.5$$
17$$
Kc$$
0.0690539$$
0.0041692$$
0.0690539$$
0.0041692$$
18$$
ra$$
-0.0360434$$
-0.1033995$$
-0.0067748$$
-0.0067748$$
19$$
Rc$$
0$$
0$$
0.0410519$$
0.0090457$$
20$$
T$$
450.$$
450.$$
947.6106$$
947.6106$$
21$$
To$$
450.$$
450.$$
450.$$
450.$$
22$$
V$$
0$$
0$$
100.$$
100.$$
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$
0$
1000$
1000$$
$T$
350$
350$
450$
450$$$
$RT$
0$
0$
1.5E+04$
1.5E+04$$$
$EoR$
2.013E+04$
2.013E+04$
2.013E+04$
2.013E+04$
$k$
0.0066$
0.0066$
2346.7972$
2346.7972$
$tau$
100$
100$
100$
100$$$
$X$
0.3975904$
0.3975904$
0.9999957$
0.9999957$
$GT$
2981.9277$
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$ $
