8-61$
P8–14)(d))continued$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.098
3.689E–08
0.098
3.689E–08
2
Cb
0.049
1.844E–08
0.049
1.844E–08
3
Cc
0
0
0.0150519
1.497E–12
4
Cd
0
0
0.0089776
1.018E–10
5
Ce
0
0
0.0496431
0.0496431
6
Cg
0
0
0.0477138
0.0477138
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0496431
0.0496431
9
Fa
10.
7.43E–06
10.
7.43E–06
10
Fb
5.
3.715E–06
5.
3.715E–06
11
Fc
0
0
1.854
3.016E–10
12
Fd
0
0
1.030757
2.05E–08
13
Fe
0
0
9.999993
9.999993
14
Fg
0
0
9.611354
9.611354
15
Ft
15.
14.91758
29.61135
29.61135
16
Fw
0
0
9.999993
9.999993
17
k1
0.014
0.014
30.64368
30.64368
18
k10
0.014
0.014
0.014
0.014
19
k2
0.007
0.007
7.341E+07
7.341E+07
20
k20
0.007
0.007
0.007
0.007
21
k3
0.014
0.014
6.707E+04
6.707E+04
22
k30
0.014
0.014
0.014
0.014
23
k4
0.45
0.45
984.9755
984.9755
24
k40
0.45
0.45
0.45
0.45
25
ra
–0.0003709
–0.0625305
–1.0E–07
–1.0E–07
26
rb
–0.0001855
–0.0312653
–5.002E–08
–5.002E–08
27
rc
0.0003037
–0.0238556
0.0236439
–5.93E–12
28
rd
6.723E–05
–0.0093009
0.0120515
–3.804E–10
29
re
0
0
0.0773459
1.004E–07
30
rg
0
0
0.0627128
1.003E–07
31
rw
0
0
0.0773459
1.004E–07
32
Sae
0
0
5.205E+05
7.43E–07
33
Scd
0
0
3.394658
0.0147083
34
Sdg
0
0
18.42613
2.133E–09
35
T
300.
300.
550.
550.
36
V
0
0
1000.
1000.
37
Yc
0
0
0.4483122
4.059E–05
8-62$
P8–14)(d))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(T)/d(V) = 0.25
Explicit equations
1
k20 = 0.007
2
k10 = 0.014
3
k1 = k10*exp((10000/1.97)*(1/300–1/T))
4
k2 = k20*exp((30000/1.97)*(1/300–1/T))
5
k30 = 0.014
6
k3 = k30*exp((20000/1.97)*(1/300–1/T))
7
k40 = 0.45
8
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
9
Cto = 0.147
10
Ca = Cto*(Fa/Ft)
11
Cb = Cto*(Fb/Ft)
12
Cc = Cto*(Fc/Ft)
13
Cd = Cto*(Fd/Ft)
14
Ce = Cto*(Fe/Ft)
15
Cg = Cto*(Fg/Ft)
16
ra = –(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
17
rb = ra/2
18
re = k3*Cc
19
k4 = k40*exp((10000/1.97)*(1/300–1/T))
20
Cw = Cto*(Fw/Ft)
21
rg = k4*Cd
22
rw = k3*Cc
23
rd = k2*(Ca)^2–k4*Cd
24
rc = k1*Ca*(Cb)^0.5 – k3*Cc+k4*Cd
25
Yc = Fc/Fa
26
Scd = if (V>0.0001)then (Fc/Fd) else (0)
27
Sae = if (V>0.0001)then (Fa/Fe) else (0)
28
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
P8–14)(d))continued
)
)
$
)
P8–15)
(1)$ C2H4$$+$$1/2O2$$→$$C2H4O$ $ (2)$ C2H4$$+$$3O2$$→$$2CO2$$+$$2H2O$
007626.082.0 == TOIO FF
$
)
P8–15)(a))
$Selectivity$of$D$over$CO2$$$
D
F
S=
)
8-64$
P8–15)(a))
POLYMATH)Results
Variable initial value minimal value maximal value final value
W 0 0 2 2
Fe 5.58E–04 1.752E–10 5.58E–04 1.752E–10
r1e –0.0024829 –0.0024829 –3.692E–10 –3.692E–10
r2e –0.0029803 –0.0029803 –8.136E–10 –8.136E–10
Differential equations as entered by the user
[1] d(Fe)/d(W) = r1e+r2e
Explicit equations as entered by the user
[1] Finert = 0.007626
[2] Ft = Fe+Fo+Fd+Fu1+Fu2+Finert
P8–15)(b))
Changes$in$equation$from$part$(a):$
$ $
×
OEE
ORrr
dF ++= 21 3
1
$and$$
( ) 0=oFO
$
$ $
skg
mol
W
RO.2
001116.00093.012.0 =
×
=
$
$ $ 𝐹
!0=0.0093 1−0.12 ∗0.06 =4.91𝐸−04 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒$
$
P8–15)(c))
Changes$in$equation$from$part$(a):$
$ $
EEE
ERrr
dF ++= 21
$and$$
( ) 0=oFE
$
8-66$
P8–16)(a))
See$Polymath$program$$P8-16.pol$
Calculated values of DEQ variables$$
Variable
Initial value
Minimal value
Maximal value
Final value
1
ConvC
0
0
0.8337198
0.8337198
2
ConvL
0
0
0.7379696
0.7379696
3
Ct0
0.0082791
0.0082791
0.0082791
0.0082791
4
Fc
0.0041152
0.0006843
0.0041152
0.0006843
5
Fc0
0.0041152
0.0041152
0.0041152
0.0041152
6
Fco
0
0
0.0342518
0.0342518
7
Fh
0
0
0.0383516
0.0383516
8
Fl
0.0018519
0.0004852
0.0018519
0.0004852
9
Fl0
0.0018519
0.0018519
0.0018519
0.0018519
10
Ft
0.0259671
0.0259671
0.0807757
0.0807757
11
Fw
0.02
0.0070028
0.02
0.0070028
12
k1c
3.0E+04
3.0E+04
3.0E+04
3.0E+04
13
k2l
1.4E+07
1.4E+07
1.4E+07
1.4E+07
14
Mc
0.6666673
0.1108536
0.6666673
0.1108536
15
Mco
0
0
0.9590499
0.9590499
16
Mh
0
0
0.0767032
0.0767032
17
Ml
0.333333
0.0873434
0.333333
0.0873434
18
Mw
0.36
0.1260502
0.36
0.1260502
19
P
1.
1.
1.
1.
20
R
0.082
0.082
0.082
0.082
21
r1c
–0.2509955
–0.2509955
–0.0015102
–0.0015102
22
r2l
–0.3361048
–0.3361048
–0.0003587
–0.0003587
23
T
1473.
1473.
1473.
1473.
24
V
0
0
0.417
0.417
$
Differential equations$$
1
d(Fc)/d(V) = r1c
2
d(Fl)/d(V) = r2l
3
d(Fw)/d(V) = r1c+7*r2l
4
d(Fh)/d(V) = –6*r1c–13*r2l
5
d(Fco)/d(V) = –6*r1c–10*r2l
1
T = 1473
2
R = 0.082
3
P = 1
4
k2l = 14000000
5
k1c = 30000
6
Ft = Fc+Fl+Fw+Fh+Fco
7
Ct0 = P/(R*T)
8
Fc0 = 0.00411523
9
Fl0 = 0.00185185
8-67$
P8–16)(a)$continued$
10
Mc = Fc*162
11
Ml = Fl*180
12
Mh = Fh*2
13
Mco = Fco*28
14
Mw = Fw*18
15
r1c = –k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft))
16
r2l = –k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
17
ConvC = (Fc0–Fc)/Fc0
18
ConvL = (Fl0–Fl)/Fl0
)
$
Plot))of))Fl))vs))V)
8-68$
P8–16)(a)$continued$
)
Plot)of)Fw))vs))V)
$
Plot)of)Fh)vs)V)
8-69$
P8–16)(a)$continued$
)
Plot)of)Fco)vs)V)
)
P8–16)(b)$
W$$is$$the$$key$reactant$
YC$=$(–dCC/-dCW)$$=$–r1C/(–r1C$-7*r2L)$$
$
8-70$
P8–16)(b))continued$
Now$$YL$=$$-dL/(-dW)$$=$$–r2L/(–r1C$–$7*r2L)$$$
The$$plot$is$as$follows$:$
$$
Overall$selectivity$$
2
/CO H
S
$=$$
2
CO
H
N
N
$$=$$ exit$molar$flow$rate$of$CO/exit$molar$flow$rate$of$H2$=$$$
CCO$|exit$/CH2$|$exit$$$$at$any$point$in$the$reactor.$
$
)
P8–16)(c)$Individualized$solution$
$
8-71$
P8–17)(a))
The$reactions$are$$$
(i)$$L$$+$$$3$W$$!$$$3$H2$+$$3$CO$$+$char$
-r2CH$=$$$$$k2ChCchCw
2$$$$$$$$$$$$$$$$$$$$$$;$$$k2CH$=$$1000$($dm3/mol)2/sec$
$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
ConL
0
0
0.9899505
0.9899505
2
Ct0
0.2
0.2
0.2
0.2
3
Fch
0
0
0.0075887
0.0047166
4
Fco
0
0
0.0887476
0.0887476
5
Fh
0
0
0.1111269
0.1111269
6
Fl
0.0123
0.0001236
0.0123
0.0001236
7
Fl0
0.0123
0.0123
0.0123
0.0123
8
Ft
0.1233
0.1233
0.2493464
0.2493464
9
Fw
0.111
0.0446317
0.111
0.0446317
10
k1l
3721.
3721.
3721.
3721.
11
k2ch
1000.
1000.
1000.
1000.
12
r1l
–2.406642
–2.406642
–0.0004728
–0.0004728
13
r2ch
0
–0.0997755
0
–0.0048484
14
V
0
0
0.417
0.417
Differential equations
1
d(Fl)/d(V) = r1l
2
d(Fch)/d(V) = r2ch–r1l
3
d(Fw)/d(V) = 3*r1l+4*r2ch
4
d(Fh)/d(V) = –3*r1l–10*r2ch
5
d(Fco)/d(V) = –3*r1l–7*r2ch
Explicit equations
1
k2ch = 1000
2
k1l = 3721
3
Ft = Fch+Fl+Fw+Fh+Fco
4
Ct0 = 0.2
5
r1l = –k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
6
r2ch = –k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2)
7
Fl0 = 0.0123
8
ConL = (Fl0–Fl)/(Fl0)
8-72$
P8–17)(a))continued$
$
P8–17)(b))
$
8-73$
P8–17)(c)$SCO/Ch$
Let$$$S1$=$S~
CO/Ch$$$=$Fco/Fch$
The$plot$is$as$follows$
Yw$=$1$
and$Yl$=$(rl)/(rw)$=$$(r1l)/(3*r1l$+$4*r2ch)$
the$plot$is$as$follows:$
$
)
P8–17)(d)))
)
P8–18)(a)$Individualized$Solution)
P8–18)(b)$Individualized$Solution$
$
)
9-1$
Solutions)for)Chapter)9)–)Reaction)Mechanisms,)Pathways,)
Bioreactions)and)Bioreactors)
)
P9-1)(a))Example)9–1)$$
(i)$The$graph$of$Io/I$will$remain$same$if$CS2$concentration$changes.$If$concentration$of$M$increases$the$
slope$of$line$will$decrease.$
$
P9-1)(b))Example)9–2)$
(i)$It$is$a$constant$at$very$low$KM$
(ii)$Vmax$=$2$kmol/m3s,$$and$KM$=$0.04$kmol/m3$
$
$
P9-1)(c))Example)9-3$$
(i)$Infinity$
(ii)$Now$Curea$=$1$mol/dm3$and$t$=$15$min$=$900$sec.$
$ $ $$
9-2$
P9-1)(d))Example)9–4)$
$
P9-1)(e))Example)9–5))
(i)$YP/C$
(ii)$107$g/dm3$
See$Polymath$program$P9-1-e.pol.$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value
t 0 0 24 24
Cc 2.0E–04 1.276E–04 0.002742 0.002742
Cs 5.0E–04 5.0E–04 4.5794959 4.5794959
mp 0 0 0.20716 0.20716
$
ODE)Report)(RKF45)$
Differential$equations$as$entered$by$the$user$
$[1]$d(Cc)/d(t)$=$rg–rd–vo*Cc/V$
9-3$
P9–1)(e))Example)9–5)Continued$
Explicit$equations$as$entered$by$the$user$
$[1]$rd$=$Cc*.01$
$[2]$Ysc$=$1/.08$
$[14]$mp$=$Cp*V$
$
(vi)$$
Change$the$cell$growth$rate$law:$
$
9-4$
P9–1)(e))Example)9–5)Continued$
$$$ $
$
Notice$that$uncompetitive$inhibition$by$the$substrate$causes$the$concentration$of$cells$to$decrease$with$
time,$ whereas$ without$ uncompetitive$ inhibition,$ the$ concentration$ of$ cells$ increased$ with$ time.$$
Consequently,$ the$ concentration$ of$ product$ is$ very$ low$ compared$ to$ the$ case$ without$ uncompetitive$
inhibition.$
$
(vii)$
Change$the$observed$reaction$rate$constant:$
$
$
All$other$equations$are$the$same$as$in$part$(v).$
Since$CP$is$so$small,$the$factor$
$
whether$CP
*$=$93$g/dm3$or$CP
*$=$10,000$g/dm3.$$Thus$the$
plots$for$this$part$are$approximately$the$same$as$the$plots$in$part$(v).$
$
P9-1)(f))Example)on)the)Professional)Reference)Shelf)R9.1$
For$t$=$0$to$t$=$0.35$sec,$PSSH$is$not$valid$as$steady$state$not$reached.$
9-5$
P9–1)(f))Continued$
POLYMATH)Results$
Calculated)values)of)the)DEQ)variables$
Variable$$initial$value$$minimal$value$$maximal$value$$final$value$
t 0 0 12 12
C1 0.1 2.109E–04 0.1 2.109E–04
C2 0 0 1.311E–09 1.311E–09
k4 9.53E+08 9.53E+08 9.53E+08 9.53E+08
k3 5.71E+04 5.71E+04 5.71E+04 5.71E+04
$
ODE)Report)(STIFF)$
$Differential$equations$as$entered$by$the$user$
$[1]$d(C1)/d(t)$=$–k1*C1–k2*C1*C2–k4*C1*C6$
$
P9-1)(g))Individualized$solution))
$
$
P9-2)(a))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.$
$
P9-2)(b))Individualized$Solution$
$
P9-2)(c))
Applying$the$pseudo–steady$state$hypothesis,$
$
9-6$
P9-3$
Burning:$
$
9-7$
P9–3)Continued$
$
$