X
=41%.
Maximum Mixedness
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
Rate Law :
2
BAA CkCr =!
( )
XCC AoA !=1
( )
XCC BoB !=1
( )32 1XCkCr BoAoA !=
where k=175 dm6/mol2 min
( )32 1XkC
C
r
Bo
Ao
A!=
( )
zE
dz
dF !=
where z=14-λ
See Polymath program P13-9-b-2.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 14 14
X 0 0 0.3536026 0.3536026
F 0.9999 -0.1138842 0.9999 -0.1138842
Cbo 0.0313 0.0313 0.0313 0.0313
k 175 175 175 175
lam 14 0 14 0
Ca 0.0313 0.0202322 0.0313 0.0202322
Cb 0.0313 0.0202322 0.0313 0.0202322
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[2] k = 175
[3] lam = 14-z
[4] Cao = .0313
[5] Ca = Cao*(1-X)
[7] EF = E/(1-F)
[8] Cb = Cbo*(1-X)
[9] ra = -k*Ca*Cb^2
X=35.4%
( ) ( ) ( ) ( )
tE
dt
tdF
tFdttE
t
=!=
“0
where E(t) is obtained from the polynomial fit in Part (b).
( ) ( )
[ ]
!!
F
V
v
I“=1
Adiabatic reaction
Segregation model
( )32 1XkC
dt
dX
Bo !=
Where
( ) !
“
#
$
%
&‘
(
)
*
+
,–=
!
“
#
$
%
&‘
(
)
*
+
,–=TTToR
E
koTk 1
320
1
314.8
30000
exp175
11
exp
and
( ) XX
Cp
Hrx
ToKT
A
150320 +=
!
!
“
#
$
$
%
&‘(
+=
“
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 14 14
xbar 0 0 0.7585435 0.7585435
F 0 0 1.1137842 1.1137842
x 0 0 0.8973303 0.8973303
cbo 0.0313 0.0313 0.0313 0.0313
T 320 320 454.59954 454.59954
E 0 0 0.1565966 0.0199021
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(xbar)/d(t) = E*x
[2] d(F)/d(t) = E
[3] d(x)/d(t) = k*cbo^2*(1-x)^3
Explicit equations as entered by the user
[2] T = 320+150*x
[3] E = 0.0899237*t-0.0157181*t^2+0.000792*t^3-0.00000863*t^4
[4] k = 175*exp(30000/8.314*(1/320-1/T))
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
( )32 1XkC
C
r
Bo
Ao
A!!=
where
( ) !
“
$
%
(
*
+
!
“
$
%
(
*
+
koTk 1
320
314.8
exp175
exp
and
( ) XX
Cp
Hrx
ToKT
A
150320 +=
!
!
“
#
$
$
%
&‘(
+=
See Polymath program P13-9-d-2.pol
13-44
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 14 14
X 0 0 0.7189248 0.7189248
F 0.9999 -0.1138842 0.9999 -0.1138842
Cbo 0.0313 0.0313 0.0313 0.0313
T 320 320 427.83873 427.83873
lam 14 0 14 0
Ca 0.0313 0.0087977 0.0313 0.0087977
Cb 0.0313 0.0087977 0.0313 0.0087977
ra -0.0053663 -0.007411 -0.0020441 -0.0020441
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] Cbo = .0313
[2] T = 320+150*X
[3] lam = 14-z
[4] Cao = .0313
[5] Ca = Cao*(1-X)
[7] EF = E/(1-F)
[8] Cb = Cbo*(1-X)
[9] k = 175*exp(30000/8.314*(1/320-1/T))
[10] ra = -k*Ca*Cb^2
gives X=72%
13-45
P13-10
Irreversible, first order, long tubular reactor, constant volume, isothermal
For a PFR
A
Ar
dV
dF !=
( )
XkC
dV
dX
FAoAo !=1
! !
=
“
X V
o
dV
v
k
X
dX
0 0
1
kt
eX !
!=1
For
0.2865.0 =!=
“
kX
For laminar flow with negligible diffusion (LFR), the mean conversion is given by:
( ) ( ) ( ) ( )
!!
““
==
2/0
#
dttEtXdttEtXX
( ) kt
etX !
!=1
E(t) for laminar flow =
3
2
2t
!
where
2
!
“t
Therefore
( ) !!
“#
“
##=#=
2/
3
2
2/
3
2
2
1
2
1
$$
$$
dt
t
e
dt
t
eX
kt
kt
( )
( ) 782.0
4
44
5.0
5.0
=
++
!++
=
DaeDa
DaeDa
XDa
Da
85.0782.0 =<= PFR
XX
P13-11 (a)
First Moment about the mean: by definition is always equal to zero.
( ) ( ) ( ) ( )
! !!
” ““
=#=#=#=
0 00
10
$$$$
dttEdtttEdttEtm
0
111 === LFRPFRCSTR mmm
13-46
P13-11 (b)
Second-order liquid-phase reaction Da= τkCAo=1.0,τ=2min and kCAo=0.5min-1.
CSTR
VrFF AAAo !=!
XFFF AoAAo =!
( )exit
A
Ao
r
XF
V
!
=
Liquid-phase
( )
A
AAo
or
CC
v
V
!
!
==
“
Second-order
2
AA kCr =!
and
( )22 1XkC
X
kC
CC
Ao
A
AAo
!
=
!
=
“
Solved
PFR
A
Ar
dV
dF !=
!
AAo r
dV
dX
F!=
!“
=
X
A
Ao r
dX
FV
0
Second-order
!
=
X
A
Ao kC
dX
FV
0
2
where
( )
)1(
1
X
X
CC AoA
!
+
“
=
Liquid-phase ε = 0 and integrating
!
“
#
$
%
&
‘
=X
X
kC Ao 1
1
(
or
5.0
1=
+
=Da
Da
X
In the ring globule of radius r
A
Ar
dt
dC =
Where
2
AA kCr =!
Ao
Ao
ktC
ktC
X
+
=1
(2order batch)
13-47
!
“
#
$
<
=%min1min)2/(4
min10
)( 13 tfort
tfor
tE
(E(t) LFR )
( )
( )
( )
!
“
#
$
%
&‘
(
)
*
+
,+
–=
!
“
#
$
%
&‘
(
)
*
+
,+
+–=
=
!
!
“
#
$
$
%
&–
+
+=
+
==
.
...
///
2/
2/1
ln
2
1
1
ln
1
2
1
1
2
1
2
)(
2/
2
2/
2
2
2
2/
2
2
2
Da
DaDa
Da
t
ktC
kC
t
kC
dt
t
kC
ktC
kC
t
kC
tktC
dt
kCdttEXX
Ao
Ao
Ao
Ao
Ao
Ao
Ao
Ao
Ao
0
00
0
0
00
Evaluate for Da=1,
451.0=X
See Polymath program P13-11-b.pol
CSTR
PFR
LFR
0.382
0.5
0.451
P13-12
( )2
1AA
A
ACK
kC
r
+
=!
0
2
2
>
!
“!
A
A
C
r
Xseg>XMM
0
2
2
=
!
“!
A
A
C
r
Xseg=XMM
0
2
2
<
!
“!
A
A
C
r
Xseg<XMM
The following figure shows the reaction rate as function of the concentration.
.
0.01
-rA
CA
13-48
The second derivative is initially negative (Xseg<XMM), then positive (Xseg>XMM). The flex point
is for CA=8mol/dm3 (Xseg=XMM).
In the limit of low concentration
( )2
AAA CokCr +=!
(First order) and Xseg=XMM
!
!
#
$
$
&
+=‘22
1
AC
o
CK
k
r
(Reaction order=-1) and Xseg>XMM
P13-13 No solution will be given
P13-14 (a)
Liquid phase, Segregation Model, second order, non-ideal CSTR, adiabatic:
( ) ( ) 67.0,
0
== !
“
dttETtXX
( ) ( )
( )
!
“
=
0
dttC
tC
tE
and
( ) 3
0
min/1 dmmgdttC =
!
“
E(t)=IF (t<=1) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
For a batch globule:
AAo r
dt
dX
C!=
2
AA kCr =!
( )
XCC AoA !=1
( ) !
“
#
$
%
&‘
(
)
*
+
,–=T
Ea
Tk 1
300
1
314.8
exp5.0
CpXC
XH
ToT
ps
rx
!+
!“
+=
Where
!“=+== KmolJCCC papiips /500
#
0502/1002/1 =!=!=“papb CCCp
T=300+150X
Iterate with Ea, using the ODE solver for values of X(t,T) and substitute these into the polynomial
regression to evaluate the integral
( ) ( ) 67.0,
0
== !
“
dttETtXX
An activation energy Ea of 10000 J/mol gives approximately the correct mean conversion,
X
See Polymath program P13-14-a.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 3 3
xbar 0 0 0.6673438 0.6673438
x 0 0 0.9198721 0.9198721
cao 2 2 2 2
T 300 300 437.98082 437.98082
E 0 0 0.9933851 0
Ea 1.25E+04 1.25E+04 1.25E+04 1.25E+04
k 0.5 0.5 2.4247011 2.4247011
Differential equations as entered by the user
[1] d(xbar)/d(t) = E*x
[2] d(x)/d(t) = k*cao*((1-x)^2)
Explicit equations as entered by the user
[2] T = 300+150*x
[3] E1 = t
[4] E2 = 2-t
[5] E = if (t<=1) then (E1) else (if (t>=2) then (0) else (E2))
[6] Ea = 12500
[7] k = 0.5*exp(Ea/8.314*((1/300)-(1/T)))
Parallel reactions, isothermal, segregation model
Batch globules
BACAAAAA
ACCkCkrrr
dt
dC
2
2
121 !!=+==
BACAAAAB
BCCkCkrrr
dt
dC
2
2
121 !=+!==
13-50
BACAC
cCCkrr
dt
dC
22 =!==
Exit concentrations
( )
tEC
dt
dC
A
Abar =
( )
tEC
dt
dC
B
Bbar =
( )
tEC
dt
dC
C
Cbar =
E(t)=IF (t<=1) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
Selectivity
38.2==
Cbar
Bbar
C
C
S
See Polymath program P13-14-b.pol
P13-15
Reactor: fluidised CSTR (V=1m3; F=10dm3/s, CC
0=2 Kmol/m3)
τ = 1000dm3s/10dm3 = 100s
AP
CCkCk
d
dC
21 !!=
“
PAC
PCkCkCk
d
dC
341 !+=
“
AAPC
ACkCkCkCk
d
dC
5432 !!+=
“
A
OCk
d
dC
5
=
!
P13-15 (a)
Segregation Model
See Polymath program P13-15-a.pol
13-51
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 20 20
ca 0 0 900.42918 900.42918
cp 0 0 689.61265 689.61265
cc 2000 1383.5708 2000 1383.5708
cabar 0 0 137.10239 137.10239
co 0 0 466.78696 466.78696
cobar 0 0 9.1391124 9.1391124
k1 0.012 0.012 0.012 0.012
k2 0.046 0.046 0.046 0.046
k3 0.02 0.02 0.02 0.02
k4 0.034 0.034 0.034 0.034
k5 0.04 0.04 0.04 0.04
ra 92 10.804749 92 10.804749
tau 1.667 1.667 1.667 1.667
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cp)/d(t) = rp
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cpbar)/d(t) = cp*E
[6] d(ccbar)/d(t) = cc*E
[7] d(co)/d(t) = ro
[8] d(cobar)/d(t) = co*E
Explicit equations as entered by the user
[1] k1 = 0.012
[2] k2 = 0.046
[3] k3 = 0.020
[4] k4 = 0.034
[5] k5 = 0.04
[6] rc = -k1*cp-k2*ca
[7] ra = k2*cc+k3*cp-k4*ca-k5*ca
[8] rp = k1*cc+k4*ca-k3*cp
[9] tau = 1.667
The exiting selectivity is 0.928
P13-15 (b)
Maximum Mixedness model
See Polymath program P13-15-b.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 20 20
Ca 0 0 875.11273 875.11273
Cp 0 0 631.80972 631.80972
Cc 2000 1463.3111 2000 1463.3111
Co 0 0 408.43732 408.43732
k2 0.046 0.046 0.046 0.046
k3 0.02 0.02 0.02 0.02
k4 0.034 0.034 0.034 0.034
k5 0.04 0.04 0.04 0.04
ra 92 15.190162 92 15.190162
tau 1.667 1.667 1.667 1.667
EF 5.999E+07 3.695E-06 5.999E+07 3.695E-06
sigma 3 3 3 3
Cao 0 0 0 0
Cpo 0 0 0 0
Cco 2000 2000 2000 2000
Coo 0 0 0 0
[1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF)
[2] d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF)
[3] d(Cc)/d(z) = -(-rc+(Cc-Cco)*EF)
[4] d(Co)/d(z) = -(-ro+(Co-Coo)*EF)
[5] d(F)/d(z) = -E
[2] k2 = 0.046
[3] k3 = 0.020
[4] k4 = 0.034
[5] k5 = 0.04
[6] rc = -k1*Cp-k2*Ca
[7] ra = k2*Cc+k3*Cp-k4*Ca-k5*Ca
[8] rp = k1*Cc+k4*Ca-k3*Cp
[9] tau = 1.667
[13] EF = E/(1-F)
[14] sigma = 3
[15] Cao = 0
[16] Cpo = 0
[17] Cco = 2000
[18] Coo = 0
13-53
P13-15 (c)
The selectivities are reported in the following table:
SMMPFR
XMMCSTR
XsmPFR
XsmCSTR
4.83
0.99
4.174
0.92
P13-15 (d)
( ) ( ) ( ) !
“
#
$
%
&
‘
(
(=
!
“
#
$
%
&(
(=2
2
2
2
32
5
exp
.23
1
.2
exp
2
1tt
tE
)
*
+
)*
Segregation Model
See Polymath program P13-15-d-1.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 2 2
ca 0 0 171.75027 171.75027
cp 0 0 52.879376 52.879376
cc 2000 1991.2989 2000 1991.2989
cabar 0 0 11.116671 11.116671
cobar 0 0 0.3200147 0.3200147
k1 0.012 0.012 0.012 0.012
k2 0.046 0.046 0.046 0.046
k3 0.02 0.02 0.02 0.02
k4 0.034 0.034 0.034 0.034
rc 0 -8.5350648 0 -8.5350648
rp 24 24 28.677508 28.677508
sigma 3 3 3 3
ro 0 0 6.8700107 6.8700107
tau 5 5 5 5
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cp)/d(t) = rp
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cpbar)/d(t) = cp*E
[6] d(ccbar)/d(t) = cc*E
[7] d(co)/d(t) = ro
Explicit equations as entered by the user
[1] k1 = 0.012
[2] k2 = 0.046
[3] k3 = 0.020
[4] k4 = 0.034
[5] k5 = 0.04
[6] rc = -k1*cp-k2*ca
[9] sigma = 3
[10] ro = k5*ca
[11] tau = 5
[12] Spo = rp/(ro+0.0000001)
[13] E1 = 1/(tau*2*(1-0.99))
[14] E = exp(-(t-tau)^2/(2*sigma^2))/(sigma*(2*3.14)^0.5)
POLYMATH Results
POLYMATH Report 08-25-2005, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 2 2
Ca 0 0 120.77791 120.77791
Cp 0 0 35.906087 35.906087
Cc 2000 1995.2809 2000 1995.2809
Co 0 0 3.8182818 3.8182818
k1 0.012 0.012 0.012 0.012
k2 0.046 0.046 0.046 0.046
k3 0.02 0.02 0.02 0.02
k4 0.034 0.034 0.034 0.034
rc 0 -5.986657 0 -5.986657
rp 24 24 27.331698 27.331698
lam 2 0 2 0
ro 0 0 4.8311165 4.8311165
tau 5 5 5 5
Coo 0 0 0 0
EF 2.968E+06 0.089981 2.968E+06 0.089981
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF)
[2] d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF)
[3] d(Cc)/d(z) = -(-rc+(Cc-Cco)*EF)
[4] d(Co)/d(z) = -(-ro+(Co-Coo)*EF)
[5] d(F)/d(z) = -E
13-55
Explicit equations as entered by the user
[1] k1 = 0.012
[2] k2 = 0.046
[3] k3 = 0.020
[4] k4 = 0.034
[5] k5 = 0.04
[9] lam = 2-z
[10] ro = k5*Ca
[11] tau = 5
[12] Spo = rp/(ro+0.0000001)
[13] sigma = 3
[16] Cpo = 0
[17] Cco = 2000
[18] Coo = 0
[19] EF = E/(1-F)
Multiple parallel reactions, isothermal
E(t)=0.0279693 – 0.0008527t + 1.2778e-5 t2 – 1.0661e-7 t3 + 4.5747e-10 t4 – 7.73108e-13 t5
K1=5.0 m6/kmol2min, k2=2.0 m3/kmol.min, k3=10 m6/kmol2min, k4=5.0 m2/kmol2/3min
P13-16 (a)
3/2
42
2
13
2
ACBABA
ACCkCCkCCk
dt
dC !!!=
2
32
2
175.025.1 CBBABA
BCCkCCkCCk
dt
dC !!!=
3/2
4
2
3
2
12ACBCBA
CCCkCCkCCk
dt
dC !!=
3/2
42
2
15.15.1 ACBABA
DCCkCCkCCk
dt
dC ++=
3/2
42 6
5
5.0 ACBA
ECCkCCk
dt
dC +=
BC
FCCk
dt
dC 2
3
2=
( )
tEC
dt
Cd
A
A=
( )
tEC
dt
Cd
B
B=
( )
tEC
dt
Cd
C
C=
13-56
( )
tEC
dt
Cd
D
D=
( )
tEC
dt
Cd
E
E=
( )
tEC
dt
Cd
F
F=
D
C
CD C
C
S=
E
D
DE C
C
S=
F
E
EF C
C
S=
See Polymath program P13-16-a.pol
Exit concentrations: Selectivities:
3
/026.0 dmmolC A=
00196.0=
CD
S
3
/008.0 dmmolC B=
009.3=
DE
S
3
/5955.6 dmmoleCC!=
28433=
EF
S
3
/036.0 dmmolCD=
3
/012.0 dmmolCE=
3
/7156.4 dmmoleCF!=
P13-16 (b)
Maximum Mixedness model
E(t)=0.0279693-0.0008527λ+1.2778e-5λ2-1.0661e-7 λ3+4.5747e-10 λ4-7.73108e-13 λ5
And λ=tf-z where z=t, tf=200min (extent of E(t)).
( ) ( )
( )!
!
“
#
$
$
%
&
‘
‘‘=
(
(
F
E
CCr
dz
dC
AoAA
A
1
and
!
“
#
$
%
&‘‘‘=3/2
42
2
13
2
ACBABAA CCkCCkCCkr
and so on for other species.
( ) ( )
!
!
E
dz
dF “=
gives F(λ)
See Polymath program P13-16-b.pol
Exit concentrations: Selectivities:
3
/028.0 dmmolC A=
0009.0=
CD
S
3
/010.0 dmmolCB=
004.3=
DE
S
3
/5927.2 dmmoleCC!=
1343110=
EF
S
3
/033.0 dmmolCD=
3
/011.0 dmmolCE=
3
/9985.7 dmmoleCF!=
13-57
P13-16 (c)
Ideal CSTR
tm=τ and
( )
!
!
/t
e
tE
“
=
Mol balances:
( )
AAo
ArF
dt
dF !!=
where
3/2
42
2
13
2
ACBABAA CCkCCkCCkr !!!=
BooAoAo
FmolvCF ==!== min/5.01005.0
and so on for the other species.
….etc
F
F
CC
T
A
ToA !=
1.005.005.0 =+=+= BoAoTo CCC
FEDCBAT FFFFFFF +++++=
See Polymath program P13-16-c.pol
Exit concentrations: Selectivities:
3
/050.0 dmmolC A=
068.0=
CD
S
3
/049.0 dmmolCB=
342.3=
DE
S
3
/5639.5 dmmoleCC!=
815844=
EF
S
3
/0008.0 dmmolCD=
3
/0002.0 dmmolCE=
3
/1003.3 dmmoleCF!=
Ideal PFR
tm=τ and RTD function
( ) ( )
!“
#=ttE
( )
A
Ar
dV
dF !=
Where
3/2
42
2
13
2
ACBABAA CCkCCkCCkr !!!=
and so on for other species.
….etc
F
F
CC
T
A
ToA !=
1.005.005.0 =+=+= BoAoTo CCC
FEDCBAT FFFFFFF +++++=
Exit concentrations: Selectivities: