Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
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
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$
$
$
