8-21$
P8-7)(c))Continued$
Therefore$the$reaction$should$be$run$at$a$low$temperature$to$maximize$SDU,$but$not$too$low$to$limit$the$
P8-7)(d)))))
)))))))
DA →
)and) –
AA CTKr )/12000exp(4280
1−=
)
)))))))
1
UD →
)and$$$$$$–
DD CTKr )/15000exp(100,10
2−=
)
)))))))
2
UA →
)and$$$$$$–
AA CTKr )/10800exp(26
3−=
)
AD
DA
UUD CC
CC
S1518
1814
21/ 1003.61094.1
1094.11018.1
−−
−−
×+×
×−×
=
1.18 1.96
.603
≈=
)
At$$T$=$1000K$
P8-7)(e))
)))))))
DBA →+
)and) –
BAA CCTKr )/10000exp(109
1−=
)
)))))))
BAD +→
)and$$$$$$–
DD CTKr )/2000exp(20
2−=
)
P8-7)(e))Continued$
The$desired$reaction$lies$very$far$to$the$left$and$CD$is$probably$present$at$very$low$concentrations$so$
that:$
BA
DBA
UD CC
CCC
S
045.0
025.01034.3 6
/
−×
=
−
0.000334 0.000074
4.5
≈=
$
/0
D U
S≈
$
At$T$=$1000K$
BA
DBA
UD CC
CCC
S
7.49
7.29.45399
/
−
=
45399 913
49.7
≈=
$
Here$we$need$a$high$temperature$for$a$lower$reverse$reaction$of$D$and$lower$formation$of$U$
Also$we$need$to$remove$D$as$soon$as$it$is$formed$so$as$to$avoid$the$decomposition.$
$
P8-7)(f)$$
)))))))
DBA →+
) ) and)
0.5
1800exp( 8000 / )
A A B
r K T C C−=−
)
)))))))
1
A B U+→
) ) and$$$$$$
210exp( 300 / )
B A B
r K T C C−=−
)
(1)$
( )
( )
( )
( )
1
0.5
/0.5
800exp 8000 / 80exp 8000 /
10exp 300 / exp 300 /
A B
D U
A B A
T C C T
S
T C C T C
− −
= =
− −
)
At$T$=$300$
1
10
/0.5
2.098*10
0.368
D U
A
S
C
−
=
)
At$T$=$1000$
8-23$
P8-7)(f))Continued$
(2)$$
( )
( ) ( )
( )
( ) ( )
0.5
/ 1 2 6
0.5
/ 1 2 6
800exp 8000 /
10exp 300 / 10 exp 8000 /
800 exp 8000 /
10exp 300 / 10 exp 8000 /
A B
D U U
A B D B
A
D U U
A D
T C C
S
T C C T C C
T C
S
T C T C
−
=−+−
−
=−+−
))
At$T$=$300$
9 0.5
/ 1 2 6
2.09*10 0
3.67 2.62*10
A
D U U
A D
C
S
C C
−
−
=≈
+
)
At$T$=$1000$and$very$low$concentrations$of$D$
0.5
/ 1 2 0.5
0.268 .03617
A
D U U
C
S
= =
)
)
P8-8)No$solution$will$be$given$
$
$
P8-9)(a)))
Species*A*:*
dCA
dt =rA
$
;$
-rA$$=$k1$*$Ca$
8-24$
P8–9)(a))continued$
See$polymath$program$P8-9-a.pol$
Calculated values of DEQ variables$$
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
1.6
6.797E–18
1.6
6.797E–18
2
Cb
0
0
1.455486
0.6036996
3
Cc
0
0
0.9963004
0.9963004
4
k1
0.4
0.4
0.4
0.4
5
k2
0.01
0.01
0.01
0.01
6
ra
–0.64
–0.64
–2.719E–18
–2.719E–18
7
rb
0.64
–0.0132415
0.64
–0.006037
8
rc
0
0
0.0145549
0.006037
9
t
0
0
100.
100.
$
Differential equations$$
1
d(Cc)/d(t) = rc
2
d(Cb)/d(t) = rb
3
d(Ca)/d(t) = ra
$
Explicit equations$$
1
k2 = 0.01
2
rc = k2*Cb
3
k1 = 0.4
4
ra = – k1*Ca
5
rb = k1*Ca – k2*Cb
)
)
P8-9)(b))
Now$$the$$rate$laws$will$change$-$
ra$=$k1-1Cb$–$k1*Ca$
8-25$
P8–9)(b))continued$
See$polymath$program$P8-9-b.pol$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
1.6
0.3949584
1.6
0.3949584
2
Cb
0
0
0.8736951
0.5191343
3
Cc
0
0
0.6859073
0.6859073
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
ra
–0.64
–0.64
–0.0022431
–0.0022431
8
rb
0.64
–0.0047669
0.64
–0.0029483
9
rc
0
0
0.008737
0.0051913
10
t
0
0
100.
100.
Differential equations
1
d(Ca)/d(t) = ra
2
d(Cb)/d(t) = rb
3
d(Cc)/d(t) = rc
Explicit equations
1
k1 = 0.4
2
k2 = 0.01
3
k1b = 0.3
4
ra = k1b*Cb – k1*Ca
5
rb = k1*Ca – k1b*Cb – k2*Cb
6
rc = k2*Cb
$
8-26$
P8-9)(c))
rC$=$k2*CB$–$k-2*Cc$
See$polymath$program$P8-9-c.pol$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
1.6
0.4500092
1.6
0.4500092
2
Cb
0
0
0.8742432
0.5953978
3
Cc
0
0
0.554593
0.554593
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
k2c
0.005
0.005
0.005
0.005
8
ra
–0.64
–0.64
–0.0013843
–0.0013843
9
rb
0.64
–0.0044688
0.64
–0.0017967
10
rc
0
0
0.0085186
0.003181
11
t
0
0
100.
100.
Differential equations
1
d(Ca)/d(t) = ra
2
d(Cb)/d(t) = rb
3
d(Cc)/d(t) = rc
Explicit equations
1
k1 = 0.4
2
k2 = 0.01
3
k1b = 0.3
4
k2c = 0.005
5
ra = k1b*Cb – k1*Ca
6
rb = k1*Ca – k1b*Cb – k2*Cb + k2c*Cc
7
rc = k2*Cb – k2c*Cc
$
P8-9)(d))Individualized$Solution$
)
P8-9)(e))
When$k1$>$100$and$k2$<$0.1$the$concentration$of$B$immediately$shoots$up$to$1.5$and$then$slowly$comes$
back$down$,$while$CA$drops$off$immediately$and$falls$to$zero$.$This$is$because$the$first$reaction$is$very$$
fast$and$the$second$reaction$is$slower$with$no$reverse$reactions.$
$
P8–10)(a))
Intermediates$(primary$K–phthalates)$are$formed$from$the$dissociation$of$K–benzoate$with$a$CdCl2$
catalyst$reacted$with$K–terephthalate$in$an$autocatalytic$reaction$step:$
) )
1 2
k k
A R S⎯⎯→ ⎯⎯→
))))))))))))))))))Series$
3
3
110 0.02 /
AO
P kPa
C mol dm
= = =
)
POLYMATH)Results
Variable initial value minimal value maximal value final value
t 0 0 1500 1500
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = –k1*A
$
$
P8–10)(b))
(1))T$=$703$K$
)))))))))CAO$=$0.019$mol/dm3$
$$$$$
1
(42600 / ) 1 1
(1.08 10 ) exp 0.42 10
(1.987 / . ) 683 663
cal mol
k s s
cal mol K K K
⎛ ⎞
=× − =×
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
‘1
1 1 ‘
1 1
exp E
k k
⎛ ⎞
⎛ ⎞
=−
⎜ ⎟
⎜ ⎟
$
‘ 3 1 3 1
1
(42600 / ) 1 1
(1.08 10 ) exp 2.64 10
(1.987 / . ) 683 703
cal mol
k s s
cal mol K K K
− − − −
⎛ ⎞
⎛ ⎞
=× − =×
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
$
Similarly,$$
‘ 3 1
23.3 10k s
− −
=×
$
And,$
‘ 3 3
33.1 10 / .k dm mol s
−
=×
$
8-29$
P8–10)(b))continued$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
t 0 0 10000 10000
ODE)Report)(RKF45)$
Differential$equations$as$entered$by$the$user$$$$$
[1] d(A)/d(t) = –k1*A
[2] d(R)/d(t) = (k1*A)–(k2*R)–(k3*R*S)
[3] d(S)/d(t) = (k2*R)–(k3*R*S)
Explicit$equations$as$entered$by$the$user$
[1] k1 = 0.42e-3
[2] k2 = 0.4e-3
[3] k3 = 0.78e-3
$Independent$variable$$
variable name : t
initial value : 0
final value : 10000
Maxima$in$R$occurs$around$t$=$2500$sec.$
$
P8–10)(c))
Use$the$Polymath$program$from$part$(a)$and$change$the$limits$of$integration$to$0$to$1200.$We$get:$
)))))))CAexit$=$0.0055$mol/dm3$
)
)
P8–11)(a)))))))))))P8–11)(b))
$$$$$$$$$ $
P8–11)(c))))))))))P8–11)(d)) )
$$$$ $
P8–11)(e))
$
P8–11)(f))
$
)
P8–11)(g))
)
)
$
P8–11)(h))
Mole$balance:$
( )
τ
AAAO rCC −=−
$
$$$
( )
τ
CC rC =
$
Solving$in$polymath:$$
MCA0069.0=
$$
MCB96.0=
$$
MCC51.0=
$$
MCD004.0=
$
$$$SB/D$=$rB/rD$=$247$ $ $$$SB/C$=$1.88$ $ $
P8–11)(h))continued$
POLYMATH)Results
NLES)Solution)
Variable Value f(x) Ini Guess
Ca 0.0068715 –2.904E–10 3
NLES)Report)(safenewt)
Nonlinear equations
[1] f(Ca) = Cao–Ca+ra*tau = 0
P8–11)(i))
For$PFR$and$gas$phase:$
Mole$balance:$
dFA
dV =rA
$
dF
B
dV =r
B
$
dF
C
dV =r
C
$
dF
D
dV =r
D
$
dF
E
dV =r
E
$
Rate$law:$
rA=−k1ACA+1
3
k2DCACC
2
⎡
⎣
⎢⎤
⎦
⎥
$
8-32$
P8–11)(i))continued$
Stoichiometry:$
CA=CTO
FA
p
$$$$$$$$
CC=CTO
FC
p
$$$$$$$$
CD=CTO
FD
p
$
Plot$of$CB$and$CC$are$overlapping.$
See$Polymath$program$P8–11–i.pol.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
ka 7 7 7 7
rb 0.4666667 3.191E–05 0.4666667 3.191E–05
Cd 0 0 3.0E–04 2.56E–04
rd 0 –7.012E–05 8.653E–04 –6.742E–05
X 0 0 0.9999543 0.9999543
ODE)Report)(RKF45)$
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3] d(Fc)/d(V) = rc
[4] d(Fd)/d(V) = rd
[5] d(Fe)/d(V) = re
[6] d(y)/d(V) = –alfa*Ft/(2*y*Fto)
8-33$
P8–11)(i))continued$
Explicit equations as entered by the user
[1] Ft = Fa+Fb+Fc+Fd+Fe
[2] Cto = 0.2
[3] Cc = Cto*Fc/Ft*y
[4] ka = 7
[5] kd = 3
[6] ke = 2
[7] Ca = Cto*Fa/Ft*y
[8] rb = ka*Ca/3
[9] ra = –(ka*Ca+kd/3*Ca*Cc^2)
[10] Cd = Cto*Fd/Ft*y
[11] Fto = 0.2*100
[12] rc = ka*Ca/3 – 2/3*kd*Ca*Cc^2 – ke*Cd*Cc
[13] rd = kd*Ca*Cc^2 – 4/3*ke*Cd*Cc
[14] re = ke*Cd*Cc
[15] alfa = 0.0001
[16] X = 1–Fa/20
$$$$ $
)
P8–11)(j)$Changes$in$equation$from$part$(i):$ $
CC
CRr
dF −=
$
CdiffuseCCkR =
$$$$$$
1
min2−
=
diffuse
k
$
)
8-34$
P8–12)(a))
)
)
)
See$Polymath$program$P8-12–a.pol.$
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca
1.5
0.3100061
1.5
0.3100061
2
Cb
2.
0.3384816
2.
0.3384816
3
Cc
0
0
0.1970387
0.105508
4
Cd
0
0
0.6751281
0.6516263
5
Ce
0
0
0.1801018
0.1801018
6
Cf
0
0
0.3621412
0.3621412
7
kd1
0.25
0.25
0.25
0.25
8
ke2
0.1
0.1
0.1
0.1
9
kf3
5.
5.
5.
5.
10
ra
–1.5
–1.5
–0.0694818
–0.0694818
11
rb
–3.
–3.
–0.0365984
–0.0365984
12
rc
1.5
–0.0490138
1.5
–0.0085994
13
rd
1.5
–0.0128609
1.5
–0.0126825
14
rd1
1.5
0.0088793
1.5
0.0088793
15
re
0
0
0.0523329
0.0202008
16
re2
0
0
0.0523329
0.0202008
17
rf
0
0
0.2571468
0.0188398
18
rf3
0
0
0.2571468
0.0188398
19
V
0
0
50.
50.
20
vo
10.
10.
10.
10.
8-35$
P8–12)(a))
Differential equations
1
d(Ca)/d(V) = ra/vo
2
d(Cb)/d(V) = rb/vo
3
d(Cc)/d(V) = rc/vo
4
d(Cd)/d(V) = rd/vo
5
d(Ce)/d(V) = re/vo
6
d(Cf)/d(V) = rf/vo
Explicit equations
1
vo = 10
2
kf3 = 5
3
ke2 = .1
4
kd1 = 0.25
5
rf3 = kf3*Cb*Cc^2
6
rd1 = kd1*Ca*Cb^2
7
re2 = ke2*Ca*Cd
8
rf = rf3
9
re = re2
10
rd = rd1–2*re2+rf3
11
ra = –rd1–3*re2
12
rb = –2*rd1–rf3
13
rc = rd1+re2–2*rf3
$
8-36$
P8–12)(a))Continued$
$
P8–12)(b))
Determine$the$effluent$concentration$and$conversion$from$a$50$dm3$CSTR.$
Mole$Balance:$
$
$ $
0$
0.1$
0.2$
0.3$
0.4$
0.5$
0.6$
0.7$
0.8$
0.9$
0$ 10$ 20$ 30$ 40$ 50$ 60$
X
“
Time)
Conversion$
8-37$
P8–12)(b))Continued$
$
)
P8–12)(c))
V0$=$40$dm3$Semi–Batch$reactor.$(1)$A$is$fed$to$B,$(2)$B$is$fed$to$A$(Case$1)$A$is$fed$to$B,$
$
8-38$
P8–12)(c))
$
$
8-39$
P8–12)(c))
$$$$ $
$
$
P8–12)(d)$$
As$θB$increases$the$outlet$concentration$of$species$D$and$F$increase,$while$the$outlet$concentrations$of$
$
8-40$
P8–12)(e)$$
When$the$appropriate$changes$to$the$Polymath$code$from$part$(a)$are$made$we$get$the$following.$
See$Polymath$program$P8–12–e.pol.$
POLYMATH)Results
Calculated)values)of)the)DEQ)variables
Variable initial value minimal value maximal value final value
V 0 0 500 500
Fa 20 18.946536 20 18.946536
Ft 40 38.931546 40 38.931546
Cto 0.4 0.4 0.4 0.4
kd1 0.25 0.25 0.25 0.25
kf3 5 5 5 5
Ca 0.2 0.1946651 0.2 0.1946651
re2 0 0 1.691E–04 1.691E–04
Sef 0 0 83.266916 1.9686327
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
Explicit equations as entered by the user
[1] vo = 100
[2] Ft = Fa+Fb+Fc+Fd+Fe+Ff
[3] Cto = .4
[4] kd1 = 0.25