14-21
P14-7
Tubular Reactor
1st order, irreversible, pulse tracer test →
2
!
= 65
2
s
and
m
t
= 10 s
For a 1st order reaction, PFR: X =1– e-kτ = 0.98
We need τ and k. There being no data for diffusivity (Schmidt number) Da
and hence Per cannot
be obtained using tubular flow correlations.
τ = 0.75 m3 / (3*10-2 m3/s) = 25 s
This is greater than tm so channeling is occurring
k =
!
X“1
ln
= 3.91/25 s = 0.156 s-1
Therefore assume closed vessel dispersion model:
tm = τ = 10 s space-time
22
r
r
mPe
Pe
t
!=
“
(1-e-Per ) = 65/102 = 0.65
Iterating → Per = 1.5
X = 1–
( ) ( ) !
“
#
$
%
&‘
(
)
*
+
,–
––
!
“
#
$
%
&‘
(
)
*
+
,
+
‘
(
)
*
+
,
2
*
exp1
2
*
exp1
2
exp4
22 qPe
q
qPe
q
Pe
q
rr
r
Da = τ k = 25 * 0.156 = 3.9
q =
r
Pe
Da4
1+
=
5.1
9.3*4
1+
= 3.376
X = 1–
( ) ( ) !
“
#
$
%
&‘
(
)
*
+
,–
––
!
“
#
$
%
&‘
(
)
*
+
,
+
‘
(
*
+
2
376.3*5.1
exp376.31
2
376.3*5.1
exp376.31
2
exp376.3*4
22
= 0.88
Conversion for the real reactor assuming the closed dispersion (X=0.88) model is less than for the
ideal PFR (X=0.98)
τideal =
0
v
V
= 25 s
V = VD + VS
VS =
!
* V
Note,
XD
= 1-
( ) ( ) !
“
#
$
%
&‘
(
)
*
+
,–
––
!
“
#
$
%
&‘
(
)
*
+
,
+
‘
(
)
*
+
,
2
27.2*5.1
exp27.21
2
27.2*5.1
exp27.21
2
5.1
exp27.2*4
22
XD
= 1- (19.22/58.38) = 0.67
E(t):
Conversion T-I-S and Maximum Mixedness Model
XT-I-S=0.5
For a first order reaction Xmm=XT-I-S=
n
k
n!
“
#
$
%
&+
‘
(
1
1
1
( ) min225.0125.0)(
4
2
2
0
2
0
=!“+!=== ###
$
dtttdtttdttEtm
%
2
0
22
0
22 min
3
2
)())(( =!=!=““
##
$%
dtttEdttttE m
6
2
2
==
!
“
n
From the conversion it is possible to determine k at 300K:
1
min367.0
1
1
1
!
=
“
#
$
%
&
‘!
!
=
n
X
k
n
(
The conversion at T=310 K is given by:
1
300310 min422.1
310300
exp !
=
“
#
$
%
&
‘
(
!=KR
E
KR
E
kk
molK
cal
R9861.=
Xmm=XT-I-S=
903.0
1
1
1=
!
“
#
$
%
&+
‘n
k
n
(
14-24
( ) ( ) ( )
( )
!
!
!
F
E
CCCkCk
d
dC
AoAAA
A
“
“+“““=1
31
( ) ( ) ( )
( )
!
!
!
F
E
CCCkCk
d
dC
BoBBA
B
“
“+““=1
21
( ) ( ) ( )
( )
!
!
!
F
E
CCCk
d
dC
CoCB
C
“
“+“=1
2
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 4 4
ca 1 0.6793055 1 0.6793055
cb 0 0 0.142158 0.142158
cc 0 0 0.0181892 0.0181892
cd 0 0 0.1603473 0.1603473
F 0.9999999 2.854E-06 0.9999999 2.854E-06
k2 0.1 0.1 0.1 0.1
k3 0.1 0.1 0.1 0.1
rc 0 0 0.0142158 0.0142158
rb 0.1 0.0537147 0.1 0.0537147
cao 1 1 1 1
cdo 0 0 0 0
t1 2 2 2 2
lam 4 0 4 0
E1 1 0 1 0
E2 0 0 1 1
E 0 0 0.4965476 0
EF 0 0 25.277605 0
Differential equations as entered by the user
[1] d(ca)/d(z) = –(-ra+(ca-cao)*EF)
[2] d(cb)/d(z) = –(-rb+(cb-cbo)*EF)
[3] d(cc)/d(z) = –(-rc+(cc-cco)*EF)
[4] d(cd)/d(z) = –(-rd+(cd-cdo)*EF)
[5] d(F)/d(z) = -E
14-25
Explicit equations as entered by the user
[1] k1 = 0.1
[2] k2 = 0.1
[3] k3 = 0.1
[4] rc = k2*cb
[5] ra = -k1*ca-k3*ca
[6] rb = k1*ca-k2*cb
[7] t2 = 4
[10] cbo = 0
[11] cco = 0
[12] cdo = 0
[13] t1 = 2
[14] lam = 4-z
[17] E = if (lam<t1) then (E1) else ( if (lam<=t2) then (E2) else (0))
[18] EF = E/(1-F)
P14-8 (c)
Use FEMLAB for the full solution
Complex reactions and Dispersion Model.
AA
ACkCk
d
dC
31 !!=
“
BA
BCkCk
d
dC
21 !=
“
B
CCk
d
dC
2
=
!
A
DCk
d
dC
3
=
!
14-26
P14-9 (a)
From P13-4:
!
“
2
==
m
t
min,
1590
2
1
2.==
!
“
min2 and k= 0.8 min-1
Tanks-in-series
For a first order reaction the conversion is given by:
From Example 14-1
P14-9 (b)
Closed-closed vessel dispersion model
For a first order reaction the conversion is given:
( )
( ) ( ) ( )
2/exp1)2/exp(1
2/exp4
122 qPeqqPeq
Peq
X
rr
r
!!!+
!=
( )
r
Pe
r
r
e
Pe
Pe
!
!!=1
22
250 2
.
Per=6.83
6380.== kDa
!
8851
836
63804
1
4
1.
.
.* =+=+=
r
Pe
Da
q
X=0.41
XT-I-S
XDispersion
0.447
0.41
Referring to P14.2 the approximate formula for the T-I-S is not good approximation.
Combination of ideal reactors
The cumulative distribution function F(t) is given:
14-27
The real reactor can be modelled as two parallel PFRs:
The relative
( ) ( )
!
“
#$+$=21 4
3
4
1
)(
%&%&
tttE
Model parameters
For two parallel PFRs, the parameters are τ1=10 min and τ2=30 min,
Fa01 =1/4Fa0 and Fa02=3/4Fa0 ,second order, liquid phase, irreversible reaction with k=0.1 dm3
/mol⋅min-1 and CAo=1.25 mol/dm3
3
1
dmmolC
Ck
CC Ao
Ao
!=
“
3
2
2
2/263.0
1
dmmoC
Ck
Ck
CC Ao
Ao
Ao
AoA =
+
!=
“
“
731.0
4
3
4
1
21
=
!!
=
Ao
AAAo
vC
vCvCvC
X
14-28
P14-11 (a)
From P13-6
min5
1=t
1
ttm=
and
2
2
2min167.4
6
25
6=== m
t
!
Tanks-in-series
6
2
2
==
!
“
n
(As in P14.8)
Second order reaction, liquid phase, kCAo=0.2min-1
!
=
“
2
Aout
AoutAi
kC
CC
Solving for CAout
!
!
k
Ck
CAin
Aout 2
411 ++“
=
Six-Reactor System: τ6=τ/6
The Damköhler number is given by Da0= kCAoτ6=0.167
2
411 1)( !
++!
=i
i
Da
Da
with i=1..6
In the following table the exit concentrations for each reactor are reported:
Da3
0.116
Da4
0.105
Da5
0.095
4730
0
60
0
60 .=
=
=Da
C
X
A
AA
( )
!
!
!
“
!
!
!
#
$
%%&&
<
=
otherwise
tttiftt
t
ttif
t
t
tE
0
22
1
)( 111
2
1
1
2
1
P14-11 (b)
Peclet number
The Peclet number
a
rD
Ul
Pe =
For Closed-Closed System (no dispersion at the entrance and at the exit of the reactor P13.6)
( )
r
Pe
e
!
!!== 1
22
1
2
#
14-30
q=1+4Da
Per
Da =
“
k
For Per>>1 (or for small vessel dispersion number)
q=1+4Da
Per
=1+2Da
Per
“2Da
Per
#
$
%
&
‘
(
2
+oDa
Per
#
$
%
&
‘
(
3
(2)
Introducing (2) into (1) and neglecting terms
2
1
!
!
“
#
$
$
%
&
r
Pe
o
!
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&‘+
!
“
#
$
%
&
!
“
#
$
%
&+(
!
“
#
$
%
&+(
‘=
rr
r
r
Pe
Da
Pe
Da
Pe
Pe
e
Pe
k
e
Pe
k
X
2
22
1
2
2
2
14
2
14
1
)
)
( )
!
!
“
#
$
$
%
&+‘
‘=r
Pe
k
k
eX
2
1
(
(
P14-12 (b)
!
k
PFR eX “
“=1
and
( )
r
Pe
k
k
DisoPe eX
2
1
1
!
!
+“
>> “=
In order to achieve the same conversion:
( )
r
Disp
DispPFR Pe
k
kk
2
!
!!
“=
( )
2
11
U
kD
Pe
k
V
Va
r
Disp
P
Disp
PFR !=!==
“
“
“
P14-12 (c)
Defining:
1.0==
a
PFR
rD
Ul
Pe
( )
( ) ( ) !
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&
‘‘‘
!
!
“
#
$
$
%
&
+
!
!
“
#
$
$
%
&
!
!
“
#
$
$
%
&
‘=‘‘
2/exp1)2/exp(1
2/exp4
1199.0
22 q
l
l
Peqq
l
l
Peq
l
l
Peq
e
PFR
r
PFR
r
PFR
r
k
(
r
PFR
Pe
k
q
!
4
1+=
The following figure shows the solution as function of kτPFR for different values of l/lPFR.
P14-12 (d)
Starting from 14.12.1 and subtracting the conversion for a PFR:
Pe
k
C
C
plug
A
A
!
=ln
For small deviations from the plug flow:
P14-12 (e)
According P12.3 dispersion doesn’t affect zero order reaction.
P14-13
From P13.19:
τ=10 min and σ2=74min2
$
%&%+%&=‘‘
&&&
107492.4101355.1101675.1)(30
223243
1
ttttEtfor
14-32
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 70 70
Xbar 0 0 0.4224876 0.4224876
kCao 0.1 0.1 0.1 0.1
E1 0 -2.436E+04 0.0836855 -2.436E+04
E2 0.092343 -21.261016 0.092343 -21.261016
E4 0 0 0 0
E 0 0 0.0836855 0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = E*X
Explicit equations as entered by the user
[1] kCao = 0.1
[2] E1 = -0.0011675*t^4+0.011355*t^3-0.047492*t^2+0.0995005*t
[3] E2 = -1.8950*10^(-6)*t^4+8.7202*10^(-5)*t^3-1.1739*10^(-3)*t^2-1.7979*10^(-4)*t+0.092343
[4] E3 = 1.2618*10^(-8)*t^4-2.4995*10^(-6)*t^3+1.8715*10^(-4)*t^2-6.3512*10^(-3)*t+0.083717
[5] E4 = 0
[7] X = kCao*t/(1+kCao*t)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Maximum Mixedness Model
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
Rate Law :
2
AA kCr =!
( )
XCC AoA !=1
( )2
1XkC
C
r
Ao
Ao
A!=
See Polymath program P14-13-b.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
z 0 0 60 60
X 0 0 0.4773052 0.4047103
F 0.99 -0.010344 0.99 -0.010344
Cao 1 1 1 1
lam 60 0 60 0
Ca 1 0.52273 1 0.5952897
k 0.1 0.1 0.1 0.1
E2 -9.8680524 -9.8680524 0.092343 0.092343
E3 2.228E-05 1.806E-05 0.083717 0.083717
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] Cao = 1
[2] lam = 60-z
[3] Ca = Cao*(1-X)
[4] k = .1
[5] ra = -k*Ca^2
[7] E1 = -0.0011675*lam^4+0.011355*lam^3-0.047492*lam^2+0.0995005*lam
[8] E2 = -1.8950*10^(-6)*lam^4+8.7202*10^(-5)*lam^3-1.1739*10^(-3)*lam^2-1.7979*10^(-
4)*lam+0.092343
[9] E3 = 1.2618*10^(-8)*lam^4-2.4995*10^(-6)*lam^3+1.8715*10^(-4)*lam^2-6.3512*10^(-
[10] E = if(lam<=3)then(E1)else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)))
(Check F(t))
P14-13 (c)
Tanks in series and 1st order reaction with k=0.1min-1
35.1
2
2
==
!
“
n
For 1st order reaction is acceptable a non integer value of n for calculating the conversion
527.0
1
1
1=
!
“
#
$
%
&+
‘=n
k
n
X
(
P14-13 (d)
Dispersion models and 1st order reaction with k=0.1min-1
Peclet number open system
14-34
906.4
82
22
2
=!+= Pe
Pe
Pe
tm
“
FEMLAB application
( ) 98.01
22
22
2
=!““=“Pee
Pe
Pe
t
Pe
m
#
( )
( ) ( ) ( )
2/exp1)2/exp(1
2/exp4
122 qPeqqPeq
Peq
X
rr
r
!!!+
!=
r
Pe
Da
q4
1+=
kDa
!
=
X=4.59
Dispersion models and 2nd order reaction with k=0.1dm3/mol
⋅
s and CA0=1mol/dm3
Conversion
Linearizing the 2st order reaction rate according 14.2 (g):
AA
A
AA CkC
C
kkCr ‘
2
0
2=!=“
it is possible to obtain an approximated solution;
1
005.0
2
‘!
== s
C
kk A
30‘ == kDa
!
Per=0.98 (from Part(b))
111.11
98.0
30*4
1
4
1=+=+=
r
Pe
Da
q
( )
( ) ( ) ( )
2/exp1)2/exp(1
2/exp4
122 qPeqqPeq
Peq
X
rr
r
!!!+
!=
X=0.998
Full solution can be obtained with FEMLAB.
P14-13 (f)
Two parameters model (backflow)
For tracer pulse input
1121
1
1TT CvCv
dt
dC
V!=
14-35
2011
2
2TT CvCv
dt
dC
V!=
Defining:
00
11 ,v
V
and
v
v
V
V===
!“#
We arrive at two coupled differential equations:
For tracer pulse input
( )
12
1
TT CC
dt
dC !=
“#$
( ) ( )
21
2
1TT CC
dt
dC !=!
“#
See Polymath program P14-13-f.pol
P14-13 (g)
Two parameters model models and 2nd order reaction with k=0.1dm3/mol
⋅
s and CA0=1mol/dm3
2
1120 AAAA CkCCC
!”##
$=$+
( )
!“##
$$=$$ 1
2
2221 AAAA kCCCC
0
20
A
AA
C
CC
X!
=
P14-13 (h)
Table of the conversions
XT_I_S
XMM
Xseg
X Dispersion
Xtwoparameters
0.527
0.405
0.422
0.5
?
P14-13 (i)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Segregation Model
Segregation Model
( )2
01XkC
t
X
A!=
“
“
XT 150320 +=
))/1320/1(*314.8/45000exp(*1.0 Tk !=
“
14-36
P14-14 (a)
Product distribution for the CSTR and PFR in series
T
k2/k1
τk1CA0
T1
5.0
0.2
T2
2.0
2
T3
0.5
20
T4
0.1
200
Considering Arrhenius equation and applying the following notation:
rT
r
r
rT E
RT
A
We can write this linear system of 5 equations for the 5 unknowns (
114131211 ,,,, AEEEE
):
=!=1112
12
11
ln EE
k
k
=!=1113
13
11
ln EE
k
k
=!=1114
14
11
ln EE
k
k
1121
2
1
21
11 lnln EE
A
A
k
k!+=
1222
2
1
22
12 lnln EE
A
A
k
k!+=
1323
2
1
23
13 lnln EE
A
A
k
k!+=
P14-14 (b-d) No solution will be given at this time.
P14-15 (a)
Two parameters model
( )
( )
)1.0exp(1105)(min10
)1.0exp(110)(min10min0
ttCt
ttCt
!!“+=#
!!“=<#
( )
( )
)1.0exp(1
3
2
3
1
)(min10
)1.0exp(1
3
)(min10min0
2
1
ttFt
ttFt
!!“+=#
!!“=<#
14-37
.
0 5 10 15 20
0
0.5
1
The F (t) can be representative of the ideal PFR and ideal CSTR in parallel model:
α= Fractional volume
β= Fraction of flow
min10,
3
1
,
3
1===
!“#
( )
( ) min10
1
1
min10
1=
!
!
====
“
#
$
““
#
$
“
CSTR
PFR
PFR v
V
P14-15 (b)
Conversion
2nd order, vo=1 dm3/min, k=0.1dm3/mol⋅min, CA0=1.25mol/dm3
Balance around node 1
For the PFR:
Second-order
556.0
1
0
0=
+
=
!
=
PFR
PFR
A
PFRA
PFR Da
Da
C
CC
X
vPFR=βvo
V2
V1=αV
node1
β=1/3
t(min)
F(t)
Where
25.1== AoPFR CkDa
!
“
#
3
/556.0 dmmolCPFR =
For the CSTR:
( ) 42.0
2
4121
0
0=
+!+
=
!
=
CSTR
CSTRCSTR
A
CSTRA
CSTR Da
DaDa
C
CC
X
Where
25.1
1
1=
!
!
=AoCSTR CkDa
“
#
$
3
/725.0 dmmolCCSTR =
465.0
0
=
A
C
X
P14-15 (c)
Two parameters model
[ ]
{ }
)10(2.0exp1
3
2
3
1
)(min10
0)(min10min0
2
1
!!!“+=#
=<#
ttFt
tFt
.
0 5 10 15 20
0
0.5
1
t
F(t)
14-39
α= Fractional volume
β= Fraction of flow
min
3
40
,
4
3
,
3
1===
!“#
( )
( ) min5
1
1
min10
1=
!
!
====
“
#
$
“$”“
CSTR
o
PFR v
V
Conversion
2nd order, vo=1 dm3/min, k=0.1dm3/mol⋅min, CA0=1.25mol/dm3
Balance around node 1
( ) ( ) ACSTRPFRoCSTRbo CvCvCvCvCv 000 11 =!+=!+
““““
For the PFR:
556.0
1
0
0=
+
=
!
=
PFR
PFR
A
PFRA
PFR Da
Da
C
CC
X
Where
25.1== AoPFR CkDa
!”
3
/556.0 dmmolCPFR =
For the CSTR:
( ) 185.0
2
4121
=
+!+
=
!
=
CSTR
CSTRCSTR
PFR
CSTRPFR
CSTR Da
DaDa
C
CC
X
Where
278.0
1
1=
!
!
=PFRCSTR CkDa
“
#
$
3
/453.0 dmmolCCSTR =
0
A
C
Where
( ) 3
/487.01 dmmolCCC CSTRPFRA =!+=
““
vb=βvo
V1=αV
node1
V2
14-40
P14-16