P8–12)(e)$continued$
[18] rb = –2*rd1–rf3
[19] rc = rd1+re2–2*rf3
P8–12)(f)))
The$only$change$from$part$(e)$is:$
D
D cD D
dF r k C
$$
$
P8–12)(g)))
The$only$change$from$part$(e)$is:$
0B
B
B
F
dF r
=−
where$VT$=$500$dm3$and$FB0$=$20$mol/min$
8-42$
P8–13)(a))
Isothermal$gas$phase$reaction$in$a$membrane$reactor$packed$with$catalyst.$
ABC
⎯⎯→ +
←⎯⎯
$$$$$$$$$$$$$$$$$$$$
‘
1 1
1
B C
C C A
C
C C
r k C
K
⎡ ⎤
=−
⎢ ⎥
⎣ ⎦
$
A$!$D$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
‘
2 2D D A
r k C=
$
2C$+$D$!$$2E$$$$$$$$$$$$$$$$$$
‘ 2
3 3E E C D
r k C C=
$
( )
3
3
24.6 0.6 /
0.082 / . (500 )
AO
P atm
C mol dm
RT dm atm mol K K
= = =
$
Fa(0)$=$10$mol/min$
Fb(0)$=$Fc(0)$=$Fd(0)$=$Fe(0)$=$0$
See$Polymath$program$P8–13–a.pol$
Calculated)values)of)DEQ)variables$$
))
Variable)
Initial)value)
Minimal)value)
Maximal)value)
Final)value)
1$$
alfa$$
0.008$$
0.008$$
0.008$$
0.008$$
2$$
Ca$$
0.6$$
0.0029024$$
0.6$$
0.0029024$$
3$$
Cb$$
0$$
0$$
0.1525032$$
0.0075281$$
4$$
Cc$$
0$$
0$$
0.1425644$$
0.0541764$$
5$$
Cd$$
0$$
0$$
0.0254033$$
0.0036956$$
6$$
Ce$$
0$$
0$$
0.274725$$
0.235079$$
7$$
Cto$$
0.6$$
0.6$$
0.6$$
0.6$$
8$$
Fa$$
10.$$
0.0702103$$
10.$$
0.0702103$$
9$$
Fb$$
0$$
0$$
3.391481$$
0.1821054$$
10$
Fc$$
0$$
0$$
3.16936$$
1.310531$$
11$
Fd$$
0$$
0$$
0.5564071$$
0.0893965$$
12$
Fe$$
0$$
0$$
5.686575$$
5.686575$$
13$
Ft$$
10.$$
7.338818$$
12.8179$$
7.338818$$
14$
Fto$$
10.$$
10.$$
10.$$
10.$$
15$
k1c$$
2.$$
2.$$
2.$$
2.$$
16$
K1c$$
0.2$$
0.2$$
0.2$$
0.2$$
17$
k2d$$
0.4$$
0.4$$
0.4$$
0.4$$
18$
k3e$$
400.$$
400.$$
400.$$
400.$$
19$
kb$$
1.$$
1.$$
1.$$
1.$$
20$
r1c$$
1.2$$
0.0017264$$
1.2$$
0.0017264$$
21$
r2d$$
0.24$$
0.001161$$
0.24$$
0.001161$$
22$
r3e$$
0$$
0$$
0.200243$$
0.0043387$$
23$
ra$$
–1.44$$
–1.44$$
–0.0028874$$
–0.0028874$
24$
rb$$
1.2$$
0.0017264$$
1.2$$
0.0017264$$
25$
rc$$
1.2$$
–0.0492057$$
1.2$$
–0.0026123$
26$
rd$$
0.24$$
–0.0164816$$
0.24$$
–0.0010084$
27$
re$$
0$$
0$$
0.200243$$
0.0043387$$
28$
W$$
0$$
0$$
100.$$
100.$$
29$
y$$
1.$$
0.5056359$$
1.$$
0.5056359$$
8-43$
P8–13)(a))continued$
Differential)equations$$
1$
d(Fa)/d(W)$=$ra$$
2$
d(Fb)/d(W)$=$rb–(kb*Cb)$$
3$
d(Fc)/d(W)$=$rc$$
4$
d(Fd)/d(W)$=$rd$$
5$
d(Fe)/d(W)$=$re$$
6$
d(y)/d(W)$=$–alfa*Ft/(2*Fto*y)$
$
Explicit)equations$$
1$$
k2d$=$0.4$$
2$$
K1c$=$0.2$$
3$$
Ft$=$Fa+Fb+Fc+Fd+Fe$
4$$
Cto$=$0.6$$
5$$
Cb$=$Cto*(Fb/Ft)*y$$
6$$
Ca$=$Cto*(Fa/Ft)*y$$
7$$
Cd$=$Cto*(Fd/Ft)*y$$
8$$
Cc$=$Cto*(Fc/Ft)*y$$
9$$
kb$=$1$$
10$
k1c$=$2$$
11$
r2d$=$k2d*Ca$$
12$
k3e$=$400$$
13$
r1c$=$k1c*(Ca–(Cb*Cc/K1c))$
14$
ra$=$–r1c–r2d$$
15$
r3e$=$k3e*(Cc^2)*Cd$$
16$
rd$=$r2d–(r3e/2)$$
17$
rb$=$r1c$$
18$
rc$=$r1c–r3e$$
19$
re$=$r3e$$
20$
Ce$=$Cto*(Fe/Ft)*y$$
21$
alfa$=$0.008$$
22$
Fto$=$10$$
$
$
P8–13)(b)))
Species$ B,$ C,$ D,$ and$ E$ all$ go$ though$ a$ maximum.$ $ The$ concentration$ of$ species$ are$ affected$by$ two$
factors:$reaction$and$pressure$drop.$$$We$can$look$at$species$B$for$example:$
$
Species$B:$$
8-44$
P8-13)(b))Continued$
Species$E:$
The$reasoning$for$species$E$is$similar$to$species$B.$
P8–13)(c)))
Calculated)values)of)DEQ)variables$)
))
Variable)
Initial)value)
Minimal)value)
Maximal)value)
Final)value)
1$$
alfa$$
0.008$$
0.008$$
0.008$$
0.008$$
2$$
Ca$$
0.6$$
0.0024106$$
0.6$$
0.0024106$$
3$$
Cb$$
0$$
0$$
0.2447833$$
0.0593839$$
4$$
Cc$$
0$$
0$$
0.1181101$$
0.0028718$$
5$$
Cd$$
0$$
0$$
0.0408517$$
0.0119119$$
6$$
Ce$$
0$$
0$$
0.1034882$$
0.0252849$$
7$$
Cto$$
0.6$$
0.6$$
0.6$$
0.6$$
8$$
Fa$$
10.$$
0.2791706$$
10.$$
0.2791706$$
9$$
Fb$$
0$$
0$$
6.877205$$
6.877205$$
10$
Fc$$
0$$
0$$
2.623093$$
0.3325842$$
11$
Fd$$
0$$
0$$
1.379513$$
1.379513$$
12$
Fe$$
0$$
0$$
2.928221$$
2.928221$$
13$
Ft$$
10.$$
10.$$
13.00145$$
11.79669$$
14$
Fto$$
10.$$
10.$$
10.$$
10.$$
15$
k1c$$
2.$$
2.$$
2.$$
2.$$
16$
K1c$$
0.2$$
0.2$$
0.2$$
0.2$$
17$
k2d$$
0.4$$
0.4$$
0.4$$
0.4$$
18$
k3e$$
400.$$
400.$$
400.$$
400.$$
19$
kb$$
1.$$
1.$$
1.$$
1.$$
20$
r1c$$
1.2$$
0.0031158$$
1.2$$
0.0031158$$
21$
r2d$$
0.24$$
0.0009642$$
0.24$$
0.0009642$$
22$
r3e$$
0$$
0$$
0.1548711$$
3.93E–05$$
23$
ra$$
–1.44$$
–1.44$$
–0.0040801$$
–0.0040801$
24$
rb$$
1.2$$
0.0031158$$
1.2$$
0.0031158$$
25$
rc$$
1.2$$
0.001223$$
1.2$$
0.0030765$$
26$
rd$$
0.24$$
0.0009446$$
0.24$$
0.0009446$$
27$
re$$
0$$
0$$
0.1548711$$
3.93E–05$$
28$
W$$
0$$
0$$
100.$$
100.$$
29$
y$$
1.$$
0.169772$$
1.$$
0.169772$$
)
8-45$
P8–13)(c))Continued$
Differential)equations$$
1$
d(Fa)/d(W)$=$ra$$
2$
d(Fb)/d(W)$=$rb$$
3$
d(Fc)/d(W)$=$rc–(kb*Cc)$$
4$
d(Fd)/d(W)$=$rd$$
5$
d(Fe)/d(W)$=$re$$
6$
d(y)/d(W)$=$–alfa*Ft/(2*Fto*y)$
$
Explicit)equations$$
1$$
k2d$=$0.4$$
2$$
K1c$=$0.2$$
3$$
Ft$=$Fa+Fb+Fc+Fd+Fe$$
4$$
Cto$=$0.6$$
5$$
Cb$=$Cto*(Fb/Ft)*y$$
6$$
Ca$=$Cto*(Fa/Ft)*y$$
7$$
Cd$=$Cto*(Fd/Ft)*y$$
8$$
Cc$=$Cto*(Fc/Ft)*y$$
9$$
kb$=$1$$
10$
k1c$=$2$$
11$
r2d$=$k2d*Ca$$
12$
k3e$=$400$$
13$
r1c$=$k1c*(Ca–(Cb*Cc/K1c))$
14$
ra$=$–r1c–r2d$$
15$
r3e$=$k3e*(Cc^2)*Cd$$
16$
rd$=$r2d–(r3e/2)$$
17$
rb$=$r1c$$
18$
rc$=$r1c–r3e$$
19$
re$=$r3e$$
20$
Ce$=$Cto*(Fe/Ft)*y$$
21$
alfa$=$0.008$$
22$
Fto$=$10$$
$
(3) The$concentration$of$E$is$lower$because$of$the$same$reasoning$as$(2).$
$
P8–13)(d)$Individualized$Solution$
$
)
8-46$
P8–14)(a)))
Equations:)
$
Rate)law:)
Mole)Balance:)
$
8-47$
P8–14)(a))continued)
Maximum$values$can$be$found$out$from$this$table:$$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.098
0.0064314
0.098
0.0064314
2
Cb
0.049
0.0032157
0.049
0.0032157
3
Cc
0
0
0.0138769
0.0004265
4
Cd
0
0
0.0001148
6.459E–07
5
Ce
0
0
0.0642751
0.0642751
6
Cg
0
0
0.0083756
0.0083756
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0642751
0.0642751
9
Fa
10.
0.9041349
10.
0.9041349
10
Fb
5.
0.4520674
5.
0.4520674
11
Fc
0
0
1.463177
0.0599575
12
Fd
0
0
0.0116183
9.08E–05
13
Fe
0
0
9.035817
9.035817
14
Fg
0
0
1.177438
1.177438
15
Ft
15.
14.87618
20.66532
20.66532
16
Fw
0
0
9.035817
9.035817
17
k1
0.014
0.014
0.014
0.014
18
k2
0.007
0.007
0.007
0.007
19
k3
0.014
0.014
0.014
0.014
20
k4
0.45
0.45
0.45
0.45
21
ra
–0.0003709
–0.0003709
–5.395E–06
–5.395E–06
22
rb
–0.0001855
–0.0001855
–2.698E–06
–2.698E–06
23
rc
0.0003037
–3.328E–05
0.0003037
–5.744E–07
24
rd
6.723E–05
–8.355E–07
6.723E–05
–1.1E–09
25
re
0
0
0.0001943
5.971E–06
26
rg
0
0
5.166E–05
2.906E–07
27
rw
0
0
0.0001943
5.971E–06
28
Sae
0
0
25.89133
0.1000612
29
Scd
0
0
660.3424
660.3424
30
Sdg
0
0
0.0501992
7.711E–05
31
V
0
0
2.0E+05
2.0E+05
32
Yc
0
0
0.2299208
0.0663148
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(Fe)/d(V) = re
6
d(Fw)/d(V) = rw
7
d(Fg)/d(V) = rg
8-48$
P8–14)(a))continued)
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10
ra = –(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11
rb = ra/2
12
re = k3*Cc
13
Cd = Cto*(Fd/Ft)
14
Ce = Cto*(Fe/Ft)
15
Cg = Cto*(Fg/Ft)
16
Cw = Cto*(Fw/Ft)
17
rg = k4*Cd
18
rw = k3*Cc
19
rd = k2*(Ca)^2–k4*Cd
20
rc = k1*Ca*(Cb)^0.5 – k3*Cc+k4*Cd
21
Yc = Fc/Fa
22
Scd = if (V>0.0001)then (Fc/Fd) else (0)
23
Sae = if (V>0.0001)then (Fa/Fe) else (0)
24
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
)
)
8-49$
P8–14)(a))continued)
)
8-50$
P8–14)(a))continued)
)
8-51$
P8–14)(a))continued)
)
8-52$
P8–14)(b)))
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.098
0.0951293
0.098
0.0951293
2
Cb
0.049
0.0475647
0.049
0.0475647
3
Cc
0
0
0.0032152
0.0032152
4
Cd
0
0
0.0001412
0.0001411
5
Ce
0
0
0.0002229
0.0002229
6
Cg
0
0
0.0005038
0.0005038
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0002229
0.0002229
9
Fa
10.
9.63739
10.
9.63739
10
Fb
5.
4.818695
5.
4.818695
11
Fc
0
0
0.3257277
0.3257277
12
Fd
0
0
0.0143026
0.014299
13
Fe
0
0
0.0225834
0.0225834
14
Fg
0
0
0.0510394
0.0510394
15
Ft
15.
14.89232
15.
14.89232
16
Fw
0
0
0.0225834
0.0225834
17
k1
0.014
0.014
0.014
0.014
18
k2
0.007
0.007
0.007
0.007
19
k3
0.014
0.014
0.014
0.014
20
k4
0.45
0.45
0.45
0.45
21
ra
–0.0003709
–0.0003709
–0.0003538
–0.0003538
22
rb
–0.0001855
–0.0001855
–0.0001769
-0.0001769
23
rc
0.0003037
0.0003037
0.000336
0.000309
24
rd
6.723E–05
–1.674E–07
6.723E–05
–1.674E–07
25
re
0
0
4.501E–05
4.501E–05
26
rg
0
0
6.352E–05
6.351E–05
27
rw
0
0
4.501E–05
4.501E–05
28
Sae
0
0
7.875E+05
426.7466
29
Scd
0
0
22.77975
22.77975
30
Sce
0
0
594.3536
14.42332
31
Sdg
0
0
18.10358
0.2801563
32
V
0
0
1000.
1000.
33
Yc
0
0
0.0337983
0.0337983
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(Fe)/d(V) = re
6
d(Fw)/d(V) = rw
7
d(Fg)/d(V) = rg
P8–14)(b))continued)
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10
ra = –(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11
rb = ra/2
12
re = k3*Cc
13
Cd = Cto*(Fd/Ft)
14
Ce = Cto*(Fe/Ft)
15
Cg = Cto*(Fg/Ft)
16
Cw = Cto*(Fw/Ft)
17
rg = k4*Cd
18
rw = k3*Cc
19
rd = k2*(Ca)^2–k4*Cd
20
rc = k1*Ca*(Cb)^0.5 – k3*Cc+k4*Cd
21
Yc = Fc/Fa
22
Scd = if (V>0.0001)then (Fc/Fd) else (0)
23
Sae = if (V>0.0001)then (Fa/Fe) else (0)
24
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
25
Sce = if (V>0.0001)then (Fc/Fe) else (0)
P8–14)(b))continued)
$
$
Yc)
ΘO2)
SD/G$
ΘO2$
8-55$
P8–14)(c)$)
Now$we$have$a$pressure$drop$parameter$α$=$0.002,$so$we$modify$our$polymath$program$as$follows:$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
alpha
0.002
0.002
0.002
0.002
2
Ca
0.098
0.0041647
0.098
0.0041647
3
Cb
0.049
0.0020823
0.049
0.0020823
4
Cc
0
0
0.0004581
3.917E–05
5
Cd
0
0
5.56E–05
3.278E–06
6
Ce
0
0
8.655E–06
9.533E–07
7
Cg
0
0
3.75E–05
3.755E–06
8
Cto
0.147
0.147
0.147
0.147
9
Cw
0
0
8.655E–06
9.533E–07
10
Fa
10.
9.896873
10.
9.896873
11
Fb
5.
4.948437
5.
4.948437
12
Fc
0
0
0.0930717
0.0930717
13
Fd
0
0
0.0082669
0.0077896
14
Fe
0
0
0.0022655
0.0022655
15
Fg
0
0
0.0089238
0.0089238
16
Ft
15.
14.95962
15.
14.95963
17
Ft0
15.
15.
15.
15.
18
Fw
0
0
0.0022655
0.0022655
19
k1
0.014
0.014
0.014
0.014
20
k2
0.007
0.007
0.007
0.007
21
k3
0.014
0.014
0.014
0.014
22
k4
0.45
0.45
0.45
0.45
23
ra
–0.0003709
–0.0003709
–2.782E–06
–2.782E–06
24
rb
–0.0001855
–0.0001855
–1.391E–06
–1.391E–06
25
rc
0.0003037
3.587E–06
0.0003037
3.587E–06
26
rd
6.723E–05
–4.638E–06
6.723E–05
–1.354E–06
27
re
0
0
6.413E–06
5.483E–07
28
rg
0
0
2.502E–05
1.475E–06
29
rw
0
0
6.413E–06
5.483E–07
30
Sae
0
0
3.344E+06
4368.521
31
Scd
0
0
11.94813
11.94813
32
Sce
0
0
1218.255
41.08224
33
Sdg
0
0
37.46977
0.8729093
34
V
0
0
500.
500.
35
y
1.
0.0428242
1.
0.0428242
36
Yc
0
0
0.0094042
0.0094042
8-56$
P8–14)(c))continued)
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(Fe)/d(V) = re
6
d(Fw)/d(V) = rw
7
d(Fg)/d(V) = rg
8
d(y)/d(V) = –alpha/2/y*(Ft/Ft0)
Explicit equations
1
alpha = 0.002
2
Ft0 = 15
3
Cto = 0.147
4
k1 = 0.014
5
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
6
k2 = 0.007
7
Cb = Cto*(Fb/Ft)*y
8
k3 = 0.014
9
k4 = 0.45
10
Ca = Cto*(Fa/Ft)*y
11
Cc = Cto*(Fc/Ft)*y
12
ra = –(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
13
rb = ra/2
14
re = k3*Cc
15
Cd = Cto*(Fd/Ft)*y
16
Ce = Cto*(Fe/Ft)*y
17
Cg = Cto*(Fg/Ft)*y
18
Cw = Cto*(Fw/Ft)*y
19
rg = k4*Cd
20
rw = k3*Cc
21
rd = k2*(Ca)^2–k4*Cd
22
rc = k1*Ca*(Cb)^0.5 – k3*Cc+k4*Cd
23
Yc = Fc/Fa
24
Scd = if (V>0.0001)then (Fc/Fd) else (0)
25
Sae = if (V>0.0001)then (Fa/Fe) else (0)
26
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
27
Sce = if (V>0.0001)then (Fc/Fe) else (0)
8-57$
P8–14)(c))continued)
$
8-58$
P8–14)(c))continued)
$
8-59$
P8–14)(c))continued)
$
8-60$
P8–14)(c))continued)
$
P8–14)(d)))
Since$C(Formic$acid)$is$our$desired$product,$so$temperature$corresponding$to$the$maximum$yield$of$C$