P12-1)(c))Continued$
Case$4:$Countercurrent$conditions$
We$need$to$enter$Ta$(V$=0)$values$such$that$at$V=Vf,$Taf$=$1000K,$1175K$and$1350K$respectively$
12–22$
P12-1)(f))Continued$
POLYMATH Report
Nonlinear Equations
Calculated values of NLE variables
Variable
Value
Initial Guess
1
T
563.7289
500.
2
X
0.3636087
0.5
P12-1)(f))Continued$
Variable
Value
1
A
1.696E+11
2
E
3.24E+04
3
k
0.0464898
4
R
1.987
5
tau
12.29
Nonlinear equations
1
f(X) = X–(403.3*(T–535)+92.9*(T–545))/(36400+7*(T–528)) = 0
2
f(T) = X–tau*k/(1+tau*k) = 0
Explicit equations
1
tau = 12.29
2
A = 16.96*10^10
3
E = 32400
4
R = 1.987
5
k = A*exp(–E/(R*T))
$
(iii)$See$Polymath$program$P12-1-f-3.pol.$
$
POLYMATH Report
Nonlinear Equations
Calculated values of NLE variables
Variable
Value
Initial Guess
1
T
563.7289
564.
2
X
0.3636087
0.367
Variable
Value
1
A
1.696E+13
2
E
3.24E+04
3
k
4.648984
4
R
1.987
5
tau
0.1229
P12-1)(f))Continued$
Nonlinear equations
12–23$
1
f(X) = X–(403.3*(T–535)+92.9*(T–545))/(36400+7*(T–528)) = 0
2
f(T) = X–tau*k/(1+tau*k) = 0
Explicit equations
1
tau = 0.1229
2
A = 16.96*10^12
3
E = 32400
4
R = 1.987
5
k = A*exp(–E/(R*T))
$
(iv)$See$Polymath$program$P12-1-f-4.pol.$
Change$CP$=$29$and$–ΔH$=$38700$
Polymath)Results$
NLES)Solution)$
Variable$
Value$
f(x)$
Ini$Guess$
$X$
0.7109354$
2.444E–11$
0.367$
$T$
593.6885$
1.2E–09$
564$
$Tau$
0.1229$
$
$
$A$
1.696E+13$
$
$
$E$
3.24E+04$$
$
$
$R$
1.987$
$
$
$k$
20.01167$
$
$
$
NLES)Report)(safenewt)$
Nonlinear$equations$$
$[1]$f(X)$=$X–(397.3*(T–535)+92.9*(T–545))/(38700+7*(T–528))$=$0$
(v)$Modify$the$polymath$code$to$incorporate$area$as$shown$in$polymath$program$P12-1-5.pol$
f(X)=X–(403.3*(T–535)+(U*area*(T–545)/Fao))/(36400+7*(T–528))
$
Change$the$area$in$the$above$program$to$get$the$values$of$conversion$corresponding$to$the$value$of$the$
12–24$
P12-1)(f))Continued$
$
$
P12–1)(g)
(i)$0.178$mol/L$
(ii)$Ua$=$10,460$J/m3s$C$
(iii)$See$Polymath$program$P12-1-g.pol.$
Calculated)values)of)DEQ)variables$$
$$
Variable$
Initial$value$
Minimal$value$
Maximal$value$
Final$value$
1$$
alpha$$
1.05$$
1.05$$
1.05$$
1.05$$
2$$
Ca$$
0.1$$
1.306E–07$$
0.1$$
1.306E–07$$
3$$
Cb$$
0$$
0$$
0.0208092$$
8.93E–07$$
4$$
Cc$$
0$$
0$$
0.0038445$$
1.739E–07$$
5$$
Cto$$
0.1$$
0.1$$
0.1$$
0.1$$
6$$
Fa$$
100.$$
9.521981$$
100.$$
9.521981$$
7$$
Fb$$
0$$
0$$
65.11707$$
65.11707$$
8$$
Fc$$
0$$
0$$
12.68047$$
12.68047$$
9$$
Ft$$
100.$$
87.31953$$
100.$$
87.31953$$
10$$
Fto$$
100.$$
100.$$
100.$$
100.$$
11$$
k1a$$
482.8247$$
482.8247$$
1.753E+04$$
1.706E+04$$
12$$
k2a$$
553.0557$$
553.0557$$
1.79E+06$$
1.683E+06$$
13$$
r1a$$
–48.28247$$
–136.5345$$
–0.0022276$$
–0.0022276$$
14$$
r2a$$
–5.530557$$
–90.93151$$
–2.87E–08$$
–2.87E–08$$
15$$
T$$
423.$$
423.$$
682.1122$$
678.9481$$
16$$
To$$
423.$$
423.$$
423.$$
423.$$
17$$
V$$
0$$
0$$
0.8$$
0.8$$
18$$
y$$
1.$$
1.922E–05$$
1.$$
1.922E–05$$
$
Differential)equations$$
1$$
d(Fa)/d(V)$=$r1a+r2a$$
2$$
d(Fb)/d(V)$=$–r1a$$
3$$
d(Fc)/d(V)$=$–r2a/2$$
4$$
d(T)/d(V)$=$(4000*(373–T)+(–r1a)*20000+(–r2a)*60000)/(90*Fa+90*Fb+180*Fc)$$
5$$
d(y)/d(V)$=$–alpha/(2*y)*(Ft/Fto)*(T/To)$$
12–25$
P12-1)(g))Continued$
Explicit)equations$$
1$$
k1a$=$10*exp(4000*(1/300–1/T))$$
2$$
k2a$=$0.09*exp(9000*(1/300–1/T))$$
3$$
Cto$=$0.1$$
4$$
Ft$=$Fa+Fb+Fc$$
5$$
To$=$423$$
6$$
Ca$=$Cto*(Fa/Ft)*(To/T)*y$$
7$$
Cb$=$Cto*(Fb/Ft)*(To/T)*y$$
8$$
Cc$=$Cto*(Fc/Ft)*(To/T)*y$$
9$$
r1a$=$–k1a*Ca$$
10$$
r2a$=$–k2a*Ca^2$$
11$$
Fto$=$100$$
12$$
alpha$=$1.05$$
$
$
$
(iv)$In$the$Polymath$program$of$part$(f)$(iii),$rate$equation$will$be$changed$as$–$$
$$$$$$$r1a$=$–k1a*(Ca–Cb/Kc)$
12–26$
P12-1)(g))Continued$
$
$
$
(v)$The$selectivity$would$increase$with$an$increase$in$Ua,$and$vice$versa.$
$
P12-1)(h)$
(i)$The$maximum$temperature$that$can$be$kept$is$440$K.$
(ii)$235$K$<$T0$<$355$K$
P12-1)(i)$
(i)$Flow$rate$will$be$close$to$zero.$
(ii)$E2$
12–27$
P12-1)(i))Continued$
Calculated)values)of)DEQ)variables$$
)$
Variable$
Initial)value$
Minimal)value$
Maximal)value$
Final)value$
1$$
V$$
0$$
0$$
10.$$
10.$$
2$$
Fa$$
5.$$
0.3455356$$
5.$$
0.3455356$$
3$$
Fb$$
10.$$
0.7169561$$
10.$$
0.7169561$$
4$$
Fc$$
0$$
0$$
4.62858$$
4.62858$$
5$$
Fd$$
0$$
0$$
0.0258848$$
0.0258848$$
6$$
T$$
300.$$
300.$$
914.7896$$
518.2818$$
7$$
Ta$$
325.$$
322.7616$$
506.8902$$
506.8902$$
8$$
E2$$
1.2E+04$$
1.2E+04$$
1.2E+04$$
1.2E+04$$
9$$
y$$
1.$$
1.$$
1.$$
1.$$
10$$
R$$
1.987$$
1.987$$
1.987$$
1.987$$
11$$
Ft$$
15.$$
5.716956$$
15.$$
5.716956$$
12$$
To$$
310.$$
310.$$
310.$$
310.$$
13$$
k2c$$
2.$$
2.$$
1.511E+06$$
9619.448$$
14$$
E1$$
8000.$$
8000.$$
8000.$$
8000.$$
15$$
Cto$$
0.2$$
0.2$$
0.2$$
0.2$$
16$$
Ca$$
0.0688889$$
0.0072303$$
0.0688889$$
0.0072303$$
17$$
Cc$$
0$$
0$$
0.096852$$
0.096852$$
18$$
r2c$$
0$$
–0.0081056$$
0$$
–0.0004569$$
19$$
Cpco$$
10.$$
10.$$
10.$$
10.$$
20$$
m$$
50.$$
50.$$
50.$$
50.$$
21$$
Cb$$
0.1377778$$
0.0150021$$
0.1377778$$
0.0150021$$
22$$
k1a$$
40.$$
40.$$
3.318E+05$$
1.14E+04$$
23$$
r1a$$
–0.0523079$$
–3.727241$$
–0.0185467$$
–0.0185467$$
24$$
r1b$$
–0.1046159$$
–7.454481$$
–0.0370935$$
–0.0370935$$
25$$
rb$$
–0.1046159$$
–7.454481$$
–0.0370935$$
–0.0370935$$
26$$
r2a$$
0$$
–0.0081056$$
0$$
–0.0004569$$
27$$
DH1b$$
–1.5E+04$$
–1.5E+04$$
–1.5E+04$$
–1.5E+04$$
28$$
DH2a$$
–10000.$$
–10000.$$
–10000.$$
–10000.$$
29$$
r1c$$
0.0523079$$
0.0185467$$
3.727241$$
0.0185467$$
30$$
Ta55$$
325.$$
325.$$
325.$$
325.$$
31$$
Cpd$$
16.$$
16.$$
16.$$
16.$$
32$$
Cpa$$
10.$$
10.$$
10.$$
10.$$
33$$
Cpb$$
12.$$
12.$$
12.$$
12.$$
34$$
Cpc$$
14.$$
14.$$
14.$$
14.$$
35$$
sumFiCpi$$
170.$$
77.2731$$
170.$$
77.2731$$
36$$
rc$$
0.0523079$$
0.0180899$$
3.725968$$
0.0180899$$
37$$
Ua$$
80.$$
80.$$
80.$$
80.$$
38$$
r2d$$
0$$
0$$
0.0162112$$
0.0009137$$
39$$
ra$$
–0.0523079$$
–3.728513$$
–0.0190036$$
–0.0190036$$
40$$
rd$$
0$$
0$$
0.0162112$$
0.0009137$$
41$$
Qg$$
1569.238$$
560.9706$$
1.118E+05$$
560.9706$$
42$$
Qr$$
–2000.$$
–2000.$$
4.139E+04$$
911.3246$$
$
Differential$equations$$
1$$
d(Fa)/d(V)$=$ra$$
2$$
d(Fb)/d(V)$=$rb$$
3$$
d(Fc)/d(V)$=$rc$$
4$$
d(Fd)/d(V)$=$rd$$
5$$
d(T)/d(V)$=$(Qg$-$Qr)$/$sumFiCpi$$
6$$
d(Ta)/d(V)$=$Ua$*$(T$-$Ta)$/$m$/$Cpco$$
$
12–28$
P12-1)(i))Continued$
Explicit$equations$$
1$$
E2$=$12000$$
2$$
y$=$1$$
3$$
R$=$1.987$$
4$$
Ft$=$Fa$+$Fb$+$Fc$+$Fd$$
5$$
To$=$310$$
6$$
k2c$=$2$*$exp((E2$/$R)$*$(1$/$300$-$1$/$T))$$
7$$
E1$=$8000$$
8$$
Cto$=$0.2$$
9$$
Ca$=$Cto$*$(Fa$/$Ft)$*$(To$/$T)$*$y$$
10$$
Cc$=$Cto$*$(Fc$/$Ft)$*$(To$/$T)$*$y$$
11$$
r2c$=$–k2c$*$Ca$^$2$*$Cc$^$3$$
12$$
Cpco$=$10$$
13$$
m$=$50$$
14$$
Cb$=$Cto$*$(Fb$/$Ft)$*$(To$/$T)$*$y$$
15$$
k1a$=$40$*$exp((E1$/$R)$*$(1$/$300$-$1$/$T))$$
16$$
r1a$=$–k1a$*$Ca$*$Cb$^$2$$
17$$
r1b$=$2$*$r1a$$
18$$
rb$=$r1b$$
19$$
r2a$=$r2c$$
20$$
DH1b$=$–15000$$
21$$
DH2a$=$–10000$$
22$$
r1c$=$–r1a$$
23$$
Ta55$=$325$$
24$$
Cpd$=$16$$
25$$
Cpa$=$10$$
26$$
Cpb$=$12$$
27$$
Cpc$=$14$$
28$$
sumFiCpi$=$Cpa$*$Fa$+$Cpb$*$Fb$+$Cpc$*$Fc$+$Cpd$*$Fd$$
29$$
rc$=$r1c$+$r2c$$
30$$
Ua$=$80$$
31$$
r2d$=$–2$*$r2c$$
32$$
ra$=$r1a$+$r2a$$
33$$
rd$=$r2d$$
34$$
Qg$=$r1b$*$DH1b$+$r2a$*$DH2a$$
35$$
Qr$=$Ua$*$(T$-$Ta)$$
$
$
Addition)of)Inerts:$
The$adding$of$inerts$will$decrease$the$peak$in$the$reactor$temperature.$By$trial$and$error,$an$inert$flow$
rate$of$6.2$mol/min$is$seen$to$be$sufficient$to$keep$the$reactor$temperature$below$700K.$
12–29$
P12-1)(i))Continued$
Calculated$values$of$DEQ$variables$$
$$
Variable$
Initial$value$
Minimal$value$
Maximal$value$
Final$value$
1$$
V$$
0$$
0$$
10.$$
10.$$
2$$
Fa$$
5.$$
0.4822244$$
5.$$
0.4822244$$
3$$
Fb$$
10.$$
0.9811024$$
10.$$
0.9811024$$
4$$
Fc$$
0$$
0$$
4.501122$$
4.501122$$
5$$
Fd$$
0$$
0$$
0.0166535$$
0.0166535$$
6$$
T$$
300.$$
300.$$
697.1199$$
570.593$$
7$$
Ta$$
325.$$
322.0417$$
477.6044$$
477.6044$$
8$$
E2$$
1.2E+04$$
1.2E+04$$
1.2E+04$$
1.2E+04$$
9$$
y$$
1.$$
1.$$
1.$$
1.$$
10$$
R$$
1.987$$
1.987$$
1.987$$
1.987$$
11$$
Ft$$
15.$$
5.981102$$
15.$$
5.981102$$
12$$
To$$
310.$$
310.$$
310.$$
310.$$
13$$
k2c$$
2.$$
2.$$
1.914E+05$$
2.8E+04$$
14$$
E1$$
8000.$$
8000.$$
8000.$$
8000.$$
15$$
Cto$$
0.2$$
0.2$$
0.2$$
0.2$$
16$$
Ca$$
0.0688889$$
0.0087606$$
0.0688889$$
0.0087606$$
17$$
Cc$$
0$$
0$$
0.081772$$
0.081772$$
18$$
r2c$$
0$$
–0.0040086$$
0$$
–0.0011748$$
19$$
Cpco$$
10.$$
10.$$
10.$$
10.$$
20$$
m$$
50.$$
50.$$
50.$$
50.$$
21$$
Cb$$
0.1377778$$
0.0178237$$
0.1377778$$
0.0178237$$
22$$
k1a$$
40.$$
40.$$
8.369E+04$$
2.323E+04$$
23$$
r1a$$
–0.0523079$$
–1.876998$$
–0.0523079$$
–0.0646601$$
24$$
r1b$$
–0.1046159$$
–3.753997$$
–0.1046159$$
–0.1293201$$
25$$
rb$$
–0.1046159$$
–3.753997$$
–0.1046159$$
–0.1293201$$
26$$
r2a$$
0$$
–0.0040086$$
0$$
–0.0011748$$
27$$
DH1b$$
–1.5E+04$$
–1.5E+04$$
–1.5E+04$$
–1.5E+04$$
28$$
DH2a$$
–10000.$$
–10000.$$
–10000.$$
–10000.$$
29$$
r1c$$
0.0523079$$
0.0523079$$
1.876998$$
0.0646601$$
30$$
Ta55$$
325.$$
325.$$
325.$$
325.$$
31$$
Fi$$
6.2$$
6.2$$
6.2$$
6.2$$
32$$
Cpa$$
10.$$
10.$$
10.$$
10.$$
33$$
Cpb$$
12.$$
12.$$
12.$$
12.$$
34$$
Cpc$$
14.$$
14.$$
14.$$
14.$$
35$$
Cpd$$
16.$$
16.$$
16.$$
16.$$
36$$
rc$$
0.0523079$$
0.0523079$$
1.876239$$
0.0634852$$
37$$
Ua$$
80.$$
80.$$
80.$$
80.$$
38$$
r2d$$
0$$
0$$
0.0080172$$
0.0023497$$
39$$
ra$$
–0.0523079$$
–1.877758$$
–0.0523079$$
–0.0658349$$
40$$
rd$$
0$$
0$$
0.0080172$$
0.0023497$$
41$$
Qg$$
1569.238$$
1569.238$$
5.632E+04$$
1951.55$$
42$$
Qr$$
–2000.$$
–2000.$$
2.358E+04$$
7439.087$$
43$$
Cpi$$
10.$$
10.$$
10.$$
10.$$
44$$
sumFiCpi$$
232.$$
141.8776$$
232.$$
141.8776$$
$
Differential$equations$$
1$$
d(Fa)/d(V)$=$ra$$
2$$
d(Fb)/d(V)$=$rb$$
3$$
d(Fc)/d(V)$=$rc$$
4$$
d(Fd)/d(V)$=$rd$$
5$$
d(T)/d(V)$=$(Qg$-$Qr)$/$sumFiCpi$$
6$$
d(Ta)/d(V)$=$Ua$*$(T$-$Ta)$/$m$/$Cpco$$
12–30$
P12-1)(i))Continued$
Explicit$equations$$
1$$
E2$=$12000$$
2$$
y$=$1$$
3$$
R$=$1.987$$
4$$
Ft$=$Fa$+$Fb$+$Fc$+$Fd$$
5$$
To$=$310$$
6$$
k2c$=$2$*$exp((E2$/$R)$*$(1$/$300$-$1$/$T))$$
7$$
E1$=$8000$$
8$$
Cto$=$0.2$$
9$$
Ca$=$Cto$*$(Fa$/$Ft)$*$(To$/$T)$*$y$$
10$$
Cc$=$Cto$*$(Fc$/$Ft)$*$(To$/$T)$*$y$$
11$$
r2c$=$–k2c$*$Ca$^$2$*$Cc$^$3$$
12$$
Cpco$=$10$$
13$$
m$=$50$$
14$$
Cb$=$Cto$*$(Fb$/$Ft)$*$(To$/$T)$*$y$$
15$$
k1a$=$40$*$exp((E1$/$R)$*$(1$/$300$-$1$/$T))$$
16$$
r1a$=$–k1a$*$Ca$*$Cb$^$2$$
17$$
r1b$=$2$*$r1a$$
18$$
rb$=$r1b$$
19$$
r2a$=$r2c$$
20$$
DH1b$=$–15000$$
21$$
DH2a$=$–10000$$
22$$
r1c$=$–r1a$$
23$$
Ta55$=$325$$
24$$
Fi$=$6.2$$
25$$
Cpa$=$10$$
26$$
Cpb$=$12$$
27$$
Cpc$=$14$$
28$$
Cpd$=$16$$
29$$
rc$=$r1c$+$r2c$$
30$$
Ua$=$80$$
31$$
r2d$=$–2$*$r2c$$
32$$
ra$=$r1a$+$r2a$$
33$$
rd$=$r2d$$
34$$
Qg$=$r1b$*$DH1b$+$r2a$*$DH2a$$
35$$
Qr$=$Ua$*$(T$-$Ta)$$
36$$
Cpi$=$10$$
37$$
sumFiCpi$=$Cpa$*$Fa$+$Cpb$*$Fb$+$Cpc$*$Fc$+$Cpd$*$Fd$+$Cpi$*$Fi$$
$
$
$
12–31$
P12-1)(i))Continued$
(viii)$The$only$way$to$make$more$of$species$D$is$to$increase$E2.$
$
(ix)$No$solution$will$be$provided$
$
(x)$The$molar$flow$rate$of$C$does$not$go$through$a$maximum$because$the$activation$energy$of$the$
second$reaction$is$higher$than$that$of$the$first$reaction,$and$the$specific$rate$constant$is$higher$for$the$
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = rb
3
d(Fc)/d(V) = rc
4
d(Fd)/d(V) = rd
5
d(T)/d(V) = (Qg–Qr)/sumFiCpi
6
d(Ta)/d(V) = Ua*(T–Ta)/m/Cpco
Explicit equations
1
E2 = 12000
2
alpha_rho = 0
3
R = 1.987
4
k2c = 2*exp((E2/R)*(1/300–1/T))
5
Cto = 0.2
6
To = 300
7
E1 = 8000
8
Ft = Fa+Fb+Fc+Fd
9
p = (1 – alpha_rho*V)^0.5
10
Ca = Cto*(Fa/Ft)*(To/T)*p
11
k1a = 40*exp ((E1/R)*(1/300–1/T))
12
Cpco = 10
13
m = 50
14
Cb = Cto*(Fb/Ft)*(To/T)*p
15
Cc = Cto*(Fc/Ft)*(To/T)*p
16
r1a = –k1a*Ca*Cb^2
17
r2c = –k2c*Ca^2*Cc^3
18
r1b = 2*r1a
19
r2a = 2/3*r2c
20
DH1b = –15000
21
DH2a = –10000
12–32$
22
Cpd = 16
23
Cpa = 10
24
Cpb = 12
25
Cpc = 14
26
sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd
27
r1c = –r1a
28
Ua = 80
29
rb = r1b
30
r2d = –1/3*r2c
31
ra = r1a+r2a
32
Qg = r1b*DH1b+r2a*DH2a
33
Qr = Ua*(T–Ta)
34
rc = r1c+r2c
35
rd = r2d
$$
P12-1)(j))No$solution$will$be$given$
P12-1)(k))No$solution$will$be$given$
P12-1)(l))No$solution$will$be$given$
$
$
P12–2)(a)))
(i)$The$axial$temperature$profile$is$virtually$unaffected$by$change$in$heat$transfer$coefficient.$The$
maximum$of$the$radial$temperature$profile$decreases$with$an$increase$in$heat$transfer$coefficient.$This$
is$an$expected$result$as$more$heat$is$exchanged$with$an$increase$in$heat$transfer$coefficient.$$
$
$
P12–3))
(a)$FA0$
(b)$CT0$
$
12–33$
P12–4))
$
P12–5))
NH 4NO3
( )
→2H2O g
( )
+N2O g
( )
A
( )
→2W g
( )
+B g
( )
$
From$Rate$Data$
ln k2
k1
=E
R
T2−T
1
T2T
1
#
$
%
&
‘
( =ln 2.912
0.307
#
$
% &
‘
( =E
R
50
( )
970
( )
1020
( )
$
E
R=44518R
$
k=0.307exp E
R
1
970 −1
T
#
$
% &
‘
(
)
*
+
,
–
.
$
Mole$Balance$
V=FA0X
−r
A
$
XMB =−r
AV
FA0
=
kM
V⋅V
FA0
=kM
FA0
$
Energy$Balance$
FA0HA0+FW0HW0−FAHAg
( )
−FWHWg
( )
−FBHBg
( )
=0
FA0HA0+FA0ΘWHW0−FA01−X
( )
HAg
( )
−FA0ΘW+2FA0X
( )
HWg
( )
−FA0XHBg
( )
=0
HAg,T
( )
=HA,T
( )
+ΔHVap
ΔHRx =2HWg
( )
+HBg
( )
−HA
( )
HA0−HAg
( )
+ΘWHW0−HWg
( )
( )
−2HWg
( )
+HBg
( )
−HA
( )
ΔHRx
− ΔHVap
%
&
‘
‘
(
)
*
*
X=0
HA,T
( )
−HA0
( )
CPAT−660
( )
+1−X
( )
ΔHVap +ΘWHS500°F
( )
−HW200°F
( )
+CPST−500
( )
[ ]
=−ΔHRx X
XE=CPT−660
( )
+ΘWHS500°F
( )
−HW200°F
( )
+CPST−500
( )
[ ]
−ΔHRx
$
P12–5)continued$
ΘW=F
W
FA
=0.17
( )
18
( )
0.83
( )
80
( )
=0.9103
CPA=0.38 BTU
lb°R×80 lb
mol =30.4 BTU
lbmol°R
CPS=0.47 BTU
lb°R×18 lb
mol =8.46 BTU
lbmol°R
ΔHRx =−336 BTU
lb ×80 lb
mol =−26,880 BTU
lbmol
H200°F
( )
=168 BTU
lb =2,916 BTU
lbmol
HW500°F
( )
=1,202 BTU
lb =21,636 BTU
lbmol
$
$
$
12–35$
P12–5)continued$
$
$
$
P12–6))
A+B→2C
A$
B$
C$
Fio
lb mole
hr
−
” #
$ %
& ‘
10
10
0.0
Tio(F)
80
80
–
~
Pio
Btu
C
lb mole F
! “
# $
°
% &
51
44
47.5
,lb
MW
lb mol
! “
# $
% &
128
94
222
,3
i
lb
ft
ρ
” #
$ %
& ‘
63
67.2
65
20, 000
R
Btu
H
lb mol A
Δ=
,
Energy$balance$with$work$term$included$is:$
[ ]
0
0
0
10
1, 1, 1
10
( )
S
A R i Pi o
A
B
A B AF
A
Q W X H C T T
F
FX
F
Q UA Ts T
θ
θ θ
−− Δ =∑ −
= = = = =
=−
$
12–36$
P12–6)continued$
Substituting$into$energy$balance,$
[ ]
{ }
[ ]
0 0 0
0 0 0
0
0
0
( )
( )
( )
63525
199
S S A R AF A pA pB
S S A R A pA pB
S S A R
A pA pB
s
UA T T W F H X F C C T T
UA T T W F H F C C UA T T
UA T T W F H
T T F C C UA
Btu
Whr
T F
! “
− − − Δ = + −
% &
! “
⇒ − − − Δ = + + −
% &
− − − Δ
= + ! “
+ +
% &
−=
∴= °
$
$
$
P12-7)(a))
$
Since$the$feed$is$equimolar,$CA0$=$CB0$=$0$.1$mol/dm3$
CA$=$CA0(1-X)$
T=T0+X[−ΔHR(T0)]
φ
i
CP
i
∑+XΔ
CP
ΔCP=CpC −CpB −CpA =30 −15−15 =0
$
ΔHR(T) =HC−HB−HA
$=$-$41000$-$(-15000$)$-$(–20000)$=$–6000$cal/mol$A$
15 15 30
i i pA B pB
cal
C C C mol K
θ θ
= + = + =
∑
$
6000
300 300 200
30
X
T X= + = +
2 2 2
0(1 ) .01 (1 )
A A
r k C X k X−=−=−
0
0
PFR A
A
A
CSTR
A
dX
V F
r
F X
V
r
=−
=−
∫
FA0$=$CA0v0$=$(.1)(2)$=$0.2$mols/dm3$
k$=$.01*exp((10000$/$2)$*$(1$/$300$-$1$/$T))$
See$Polymath$program$P12-7-a.pol.$
Calculated)values)of)NLE)variables)$
$$$
Variable$$
Value$$
f(x)$$
Initial$Guess$$
1$$
T$$
483.8314$$
–1.421E–13$$
700.$$
2$$
X$$
0.919157$$
–1.516E–10$$
0.99$$
$
12–37$
P12–7)(a))continued$
$$$
Variable$$
Value$$
1$$
Ca0$$
0.1$$
2$$
Fa0$$
0.2$$
3$$
k$$
5.625546$$
4$$
ra$$
–0.0003677$$
5$$
V$$
500.$$
$
Nonlinear)equations$$
1$$
f(T)$=$T-$300$-$200$*$X$=$0$$
2$$
f(X)$=$X$+$V*ra/Fa0$=$0$$
$
Explicit)equations$$
1$$
V$=$500$$
2$$
k$=$0.01*exp((10000$/$2)$*$(1$/$300$-$1$/$T))$$
3$$
Fa0$=$0.2$$
4$$
Ca0$=$0.1$$
5$$
ra$=$–k$*$(Ca0$^$2)$*$((1$-$X)$^$2)$$
$
For)500)dm3)CSTR,)X)=)0.92$
P12–7)(b))Constant)heat)exchanger)temperature)Ta$
When$heat$exchanger$is$added,$the$energy$balance$can$be$written$as$
$
So$with$ =$0$,$$$$,$$$$=$–6000$cal/mol$
$
Where$Ua$=$20$cal/m3/s/K,$Ta$=$450$K$
See$Polymath$program$P12-7-b.pol.$
Calculated)values)of)the)DEQ)variables$
Variable$
initial$value$
minimal$value$
maximal$value$
final$value$
V$
0$
0$
10$
10$
X$
0$
0$
0.3634806$
0.3634806$
T$
300$
300$
455.47973$
450.35437$
K$
0.01$
0.01$
3.068312$
2.7061663$
Kc$
286.49665$
9.2252861$
286.49665$
9.9473377$
Fa0$
0.2$
0.2$
0.2$
0.2$
Ca0$
0.1$
0.1$
0.1$
0.1$
Ra$
–1.0E–04$
–0.0221893$
–1.0E–04$
–0.0010758$
Xe$
0.8298116$
0.3682217$
0.8298116$
0.3810642$
DH$
–6000$
–6000$
–6000$
–6000$
Ua$
20$
20$
20$
20$
Ta$
450$
450$
450$
450$
Fao$
0.2$
0.2$
0.2$
0.2$
sumcp$
30$
30$
30$
30$
$
12–38$
P12–7)(b))continued$
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$/$1.987)$*$(1$/$300$-$1$/$T))$
$[2]$Kc$=$10$*$exp(–6000$/$1.987$*$(1$/$450$-$1$/$T))$
$
$
12–39$
P12–7)(b))continued$
$
$
$
P12–7)(c)$
For$a$co–current$heat$exchanger,$
CpC$=$1cal/g/K,$Ta1=450$K,$
m=50 g
$
12–40$
P12–7)(c))continued$
Calculated)values)of)the)DEQ)variables$
Variable$
initial$value$
minimal$value$
maximal$value$
final$value$
V$
0$
0$
10$
10$
X$
0$
0$
0.3611538$
0.3611538$
T$
300$
300$
442.15965$
442.15965$
Ta$
450$
434.90618$
450$
441.60853$
k$
0.01$
0.01$
2.1999223$
2.1999223$
Kc$
286.49665$
11.263546$
286.49665$
11.263546$
Fa0$
0.2$
0.2$
0.2$
0.2$
Ca0$
0.1$
0.1$
0.1$
0.1$
ra$
–1.0E–04$
–0.0160802$
–1.0E–04$
–0.0019246$
Xe$
0.8298116$
0.4023362$
0.8298116$
0.4023362$
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$
$[1]$k$=$.01$*$exp((10000$/$1.987)$*$(1$/$300$-$1$/$T))$
$[12]$Cpc$=$1$
$
$