5-41$
P5–24)Continued$
$
$
)
$
P5–25))$
δ=0“,”ε=0
∴p=1− αW
( )
1 2
$
Mole$Balance/Design$Equation$
$
5-42$
P5-25)Continued$
Rate$Law$
$
)
$
P5–26)Individualized$solution$
)
$
6-1$
Solutions)for)Chapter)6)–)Isothermal)Reactor)Design-)Molar)
Flow)rates)
)
P6–1)(a))Example)6–1))
(i)$The$conversion$increases$with$an$increase$in$both$E$and$T.$This$is$because$the$rate$constant$is$higher$
for$higher$E$and$T.$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.5822357
0.0292158
0.5822357
0.0292158
2
Cto
0.5822357
0.5822357
0.5822357
0.5822357
3
E
2.4E+04
2.4E+04
2.4E+04
2.4E+04
4
Fa
2.26E–05
1.659E–06
2.26E–05
1.659E–06
5
Fao
2.26E–05
2.26E–05
2.26E–05
2.26E–05
6
Fb
0
0
2.094E–05
2.094E–05
7
Fc
0
0
1.047E–05
1.047E–05
8
Ft
2.26E–05
2.26E–05
3.307E–05
3.307E–05
9
k
164.7179
164.7179
164.7179
164.7179
10
ra
–55.83911
–55.83911
–0.1405976
–0.1405976
11
rateA
55.83911
0.1405976
55.83911
0.1405976
12
rb
55.83911
0.1405976
55.83911
0.1405976
13
rc
27.91956
0.0702988
27.91956
0.0702988
14
T
678.
678.
678.
678.
15
Tau
0
0
0.2576264
0.2576264
16
V
0
0
1.0E–05
1.0E–05
17
vo
3.882E–05
3.882E–05
3.882E–05
3.882E–05
18
X
0
0
0.9265741
0.9265741
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = rb
3
d(Fc)/d(V) = rc
6-2$
Explicit equations
1
T = 678
2
Cto = 2*1641/8.314/T
3
E = 24000
4
Ft = Fa+Fb+Fc
5
Ca = Cto*Fa/Ft
6
k = 0.29*exp(E/1.987*(1/500–1/T))
7
Fao = 0.0000226
8
vo = Fao/Cto
9
Tau = V/vo
10
ra = –k*Ca^2
11
X = 1–Fa/Fao
12
rb = –ra
13
rc = –ra/2
14
rateA = –ra
General
Total number of equations
17
Number of differential equations
3
Number of explicit equations
14
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
)
(iii)$See$Polymath$program$P6-1-aiii.pol.$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.2827764
0.1426944
0.2827764
0.1426944
2
Cb
0
0
0.093388
0.093388
3
Cc
0
0
0.046694
0.046694
4
Cto
0.2827764
0.2827764
0.2827764
0.2827764
5
E
2.4E+04
2.4E+04
2.4E+04
2.4E+04
6
Fa
2.26E–05
1.366E–05
2.26E–05
1.366E–05
7
Fao
2.26E–05
2.26E–05
2.26E–05
2.26E–05
8
Fb
0
0
8.94E–06
8.94E–06
9
Fc
0
0
4.47E–06
4.47E–06
10
Ft
2.26E–05
2.26E–05
2.707E–05
2.707E–05
11
k
274.4284
274.4284
274.4284
274.4284
12
Kc
0.02
0.02
0.02
0.02
13
ra
–21.94397
–21.94397
–1.497E–10
–1.497E–10
14
rateA
21.94397
1.497E–10
21.94397
1.497E–10
15
rb
21.94397
1.497E–10
21.94397
1.497E–10
6-3$
16
rc
10.97199
7.487E–11
10.97199
7.487E–11
17
T
698.
698.
698.
698.
18
Tau
0
0
0.1251223
0.1251223
19
V
0
0
1.0E–05
1.0E–05
20
vo
7.992E–05
7.992E–05
7.992E–05
7.992E–05
21
X
0
0
0.3955739
0.3955739
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = rb
3
d(Fc)/d(V) = rc
1
T = 698
2
Cto = 1641/8.314/T
3
E = 24000
4
Kc = 0.02
5
Ft = Fa+Fb+Fc
6
Ca = Cto*Fa/Ft
7
Fao = 0.0000226
8
vo = Fao/Cto
9
Tau = V/vo
10
Cb = Cto*Fb/Ft
11
X = 1–Fa/Fao
12
k = 0.29*exp(E/1.987*(1/500–1/T))
13
Cc = Cto*Fc/Ft
14
ra = –k*(Ca^2 – Cb^2*Cc/Kc)
15
rb = –ra
16
rc = –ra/2
17
rateA = –ra
General
Total number of equations
20
Number of differential equations
3
Number of explicit equations
17
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
6-4$
$
(iv)$
𝑝=1− ∝!𝑉
!
!$
$
See$Polymath$program$P6-1-aiv.pol$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.2827764
0.005108
0.2827764
0.005108
2
Cto
0.2827764
0.2827764
0.2827764
0.2827764
3
E
2.4E+04
2.4E+04
2.4E+04
2.4E+04
4
Fa
2.26E–05
5.616E–06
2.26E–05
5.616E–06
5
Fao
2.26E–05
2.26E–05
2.26E–05
2.26E–05
6
Fb
0
0
1.698E–05
1.698E–05
7
Fc
0
0
8.492E–06
8.492E–06
8
Ft
2.26E–05
2.26E–05
3.109E–05
3.109E–05
9
k
274.4284
274.4284
274.4284
274.4284
10
p
1.
0.1
1.
0.1
11
ra
–21.94397
–21.94397
–0.0071603
–0.0071603
12
rateA
21.94397
0.0071603
21.94397
0.0071603
13
rb
21.94397
0.0071603
21.94397
0.0071603
14
rc
10.97199
0.0035801
10.97199
0.0035801
15
T
698.
698.
698.
698.
6-5$
16
Tau
0
0
0.1251223
0.1251223
17
V
0
0
1.0E–05
1.0E–05
18
vo
7.992E–05
7.992E–05
7.992E–05
7.992E–05
19
X
0
0
0.7514895
0.7514895
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = rb
3
d(Fc)/d(V) = rc
Explicit equations
1
T = 698
2
Cto = 1641/8.314/T
3
E = 24000
4
p = (1 – 99000*V)^0.5
5
k = 0.29*exp(E/1.987*(1/500–1/T))
6
Ft = Fa+Fb+Fc
7
Fao = 0.0000226
8
vo = Fao/Cto
9
Tau = V/vo
10
Ca = p*Cto*Fa/Ft
11
X = 1–Fa/Fao
12
ra = –k*Ca^2
13
rb = –ra
14
rc = –ra/2
15
rateA = –ra
General
Total number of equations
18
Number of differential equations
3
Number of explicit equations
15
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
$
6-6$
$
$
P6–1)(b))Example)6–2)$
(i)$The$conversion$increases$with$increase$in$KC,$k,$CT0,$and$kc.$This$is$because$in$each$of$the$cases,$the$
(iii)$Individualized$solution$
(iv)$RB$is$maximum$when$KC,$k,$CT0,$and$kc$are$at$their$maximum$values$
(v)$Individualized$solution$
(vi)$See$Polymath$program$P6-1-b.pol.$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Cto
0.2
0.2
0.2
0.2
2
Fa
10.
3.338718
10.
3.338718
3
Fao
10.
10.
10.
10.
4
Fb
0
0
3.687651
3.421463
5
Fc
0
0
6.661282
6.661282
6
Ft
10.
10.
13.68765
13.42146
7
k
0.7
0.7
0.7
0.7
8
kc
0.2
0.2
0.2
0.2
9
Kc
0.05
0.05
0.05
0.05
10
p
1.
0
1.
0
11
ra
–0.14
–0.14
0
0
12
Rate
0.14
0
0.14
0
13
Rb
0
0
0.0107766
0.010197
14
V
0
0
500.
500.
15
X
0
0
0.6661282
0.6661282
6-7$
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = –ra–kc*Cto*(Fb/Ft)*p
3
d(Fc)/d(V) = –ra
Explicit equations
1
p = (1 – 0.002*V)^0.5
2
Ft = Fa+Fb+Fc
3
k = 0.7
4
Cto = 0.2
5
Kc = 0.05
6
kc = 0.2
7
ra = –k*Cto*((Fa/Ft)*p – Cto/Kc*(Fb/Ft)*(Fc/Ft)*p*p)
8
Rate = –ra
9
Rb = kc*Cto*(Fb/Ft)
10
Fao = 10
11
X = (Fao–Fa)/Fao
General
Total number of equations
14
Number of differential equations
3
Number of explicit equations
11
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
$
The$conversion$goes$up$from$60%$to$66.6%$
(vii)$Individualized$solution$
$
(viii)$Individualized$solution$
$
6-8$
P6–1)(c))Example)6–2)$
See$Polymath$program$P6-1-c.pol$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Cto
0.2
0.2
0.2
0.2
2
Fa
10.
0.4432696
10.
0.4432696
3
Fao
10.
10.
10.
10.
4
Fb
0
0
6.058466
5.126694
5
Fc
0
0
9.55673
9.55673
6
Ft
10.
10.
16.05847
15.12669
7
k
0.7
0.7
0.7
0.7
8
kc
0.2
0.2
0.2
0.2
9
Kc
0.05
0.05
0.05
0.05
10
p
1.
0
1.
0
11
ra
–0.14
–0.14
0
0
12
Rate
0.14
0
0.14
0
13
Rb
0
0
0.015091
0.0135567
14
V
0
0
500.
500.
15
X
0
0
0.955673
0.955673
Differential equations
1
d(Fa)/d(V) = ra
2
d(Fb)/d(V) = –ra–kc*Cto*(Fb/Ft)*p
3
d(Fc)/d(V) = –ra
Explicit equations
1
p = (1 – 0.002*V)^0.5
2
Ft = Fa+Fb+Fc
3
k = 0.7
4
Cto = 0.2
5
Kc = 0.05
6
kc = 0.2
7
ra = –k*Cto*(Fa/Ft)*p
8
Rate = –ra
9
Rb = kc*Cto*(Fb/Ft)
10
Fao = 10
11
X = (Fao–Fa)/Fao
6-9$
General
Total number of equations
14
Number of differential equations
3
Number of explicit equations
11
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
$
$
P6–1)(d))Example)6–3))
See$Polymath$program$P6-1-d.pol$
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
0.05
0.0052864
0.05
0.0052864
2
Cao
0.05
0.05
0.05
0.05
3
Cb
0
0
0.0177864
0.0177864
4
Cbo
0.025
0.025
0.025
0.025
5
Cc
0
0
0.0041087
0.0030469
6
Cd
0
0
0.0041087
0.0030469
7
k
2.2
2.2
2.2
2.2
8
Kc
0.1
0.1
0.1
0.1
9
ra
0
–0.0001295
0
–2.62E–06
10
rate
0
0
0.0001295
2.62E–06
11
t
0
0
500.
500.
12
V
5.
5.
30.
30.
13
Vo
5.
5.
5.
5.
14
vo
0.05
0.05
0.05
0.05
15
X
0
0
0.3656277
0.3656277
Differential equations
1
d(Ca)/d(t) = ra– vo*Ca/V
2
d(Cb)/d(t) = ra+ (Cbo–Cb)*vo/V
3
d(Cc)/d(t) = –ra–vo*Cc/V
4
d(Cd)/d(t) = –ra–vo*Cd/V
Explicit equations
1
vo = 0.05
2
Vo = 5
3
V = Vo+vo*t
4
k = 2.2
5
Cbo = 0.025
6-10$
6
Kc = 0.1
7
ra = –k*(Ca*Cb – Cc*Cd/Kc)
8
Cao = 0.05
9
rate = –ra
10
X = (Cao*Vo–Ca*V)/(Cao*Vo)
General
Total number of equations
14
Number of differential equations
4
Number of explicit equations
10
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps
0.000001
$
On$comparison$with$the$solution$obtained$from$the$irreversible$case,$we$find$that$the$conversion$after$t$
=$500$s$drops$down$from$99.9%$to$about$36%.$This$is$an$expected$result$as$the$reverse$reaction$also$
takes$place.$
$
P6–1)(e))Example)6-4$
P6–1)(f))Example)6-5)
Individualized$solution$
$
)
P6–2)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.$
)
P6-3$
Polymath$
d(Cb) / d(t) = rb
Cb(0) = 2
d(Ca) / d(t) = ra
Ca(0) = 2
6-11$
P6-3)Continued$
At$50$mins:$X$=$0.82$
At$100$mins:$X$=$0.9$
For$reactor$1:$
$
$
$
For$reactor$2:$
$
Therefore$the$design$equation$and$concentration$profiles$are$the$same$as$in$part$a)$for$the$two$reactors.$
As$they$have$the$same$concentration$profile$at$any$time,$the$conversions$are$the$same$for$the$two$
reactors$as$in$part$a).$$
At$10$mins:$X$=$0.47,$Ca$=$Cb$=$1.06$mol/dm3,$Cc$=$0.94$mol/dm3.$
At$50$mins:$X$=$0.82$
$
6-12$
P6–4)(a)$
$
P6–4)(b)$
$$$$ $
)
P6-5)
$
$
6-13$
P6-5)Continued$
$
Using$Polymath$to$solve$the$differential$equation$gives$a$volume$of$290$dm3$
See$Polymath$program$P6-5-c.pol.$
PFR$with$pressure$drop:$Alter$the$Polymath$equations$from$part$(c).$
See$Polymath$program$P6-5-c-pressure.pol.$
$
$
$
P6–5)continued$
Calculated)values)of)the)DEQ)variables$
Variable initial value minimal value maximal value final value$
v 0 0 500 500 $
ODE)Report)(STIFF)$
Differential equations as entered by the user$
[1] d(x)/d(v) = –r/fo$
Explicit equations as entered by the user$
[1] Kc = .025$
[2] alfa = 0.001$
[3] Cao = 0.3$