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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$$
fast$and$the$second$reaction$is$slower$with$no$reverse$reactions.$
$
P8-10)(a))
Intermediates$(primary$K-phthalates)$are$formed$from$the$dissociation$of$K-benzoate$with$a$CdCl2$
catalyst$reacted$with$K-terephthalate$in$an$autocatalytic$reaction$step:$
) )
1 2
k k
A R S⎯⎯→ ⎯⎯→
))))))))))))))))))Series$
3
3
110 0.02 /
AO
P kPa
C mol dm
= = =
)
POLYMATH)Results
Variable initial value minimal value maximal value final value
t 0 0 1500 1500
ODE)Report)(RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1*A
$
$
P8-10)(b))
(1))T$=$703$K$
)))))))))CAO$=$0.019$mol/dm3$
$$$$$
'1
1 1 '
1 1
exp E
k k
⎛ ⎞
⎛ ⎞
=−
⎜ ⎟
⎜ ⎟
$
' 3 1 3 1
1
(42600 / ) 1 1
(1.08 10 ) exp 2.64 10
(1.987 / . ) 683 703
cal mol
k s s
cal mol K K K
− − − −
⎛ ⎞
⎛ ⎞
=× − =×
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
$
Similarly,$$
' 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
