14-21
P14-7
Tubular Reactor
1st order, irreversible, pulse tracer test →
2
!
= 65
2
s
and
m
t
= 10 s
Calculate, k
k =
!
X“1
1
ln
= 3.91/25 s = 0.156 s-1
Therefore assume closed vessel dispersion model:
tm = τ = 10 s space-time
Calculate, Pe
Calculate, X using the measured V and
0
v
giving τ = 25 s
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
14-22
τideal =
0
v
V
= 25 s
Dead Volume = (1- α) V= 0.6 V, Vs = 0.4V
Use tm = 10s, Da = tm * k = 10*0.156 = 1.56
q =
5.1
56.1*4
1+
= 2.27
P14-8 (a)
E(t):
14-23
For a first order reaction Xmm=XT-I-S=
n
k
n!
#
$
&+
‘
(
1
1
1
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:
P14-8 (b)
Complex Reactions
AA
ACkCk
d
dC
31 !!=
“
14-24
( ) ( ) ( )
( )
!
!
!
F
E
CCCkCk
d
dC
AoAAA
A
“
“+“““=1
31
POLYMATH Results
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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(z) = –(-ra+(ca-cao)*EF)
14-25
Explicit equations as entered by the user
[1] k1 = 0.1
[2] k2 = 0.1
[3] k3 = 0.1
P14-8 (c)
Use FEMLAB for the full solution
Complex reactions and Dispersion Model.
AA
ACkCk
d
dC
31 !!=
“
14-26
P14-9 (a)
From P13-4:
!
“
2
==
m
t
min,
1590
2
1
2.==
!
“
min2 and k= 0.8 min-1
P14-9 (b)
Closed-closed vessel dispersion model
For a first order reaction the conversion is given:
XT-I-S
XDispersion
0.447
0.41
Referring to P14.2 the approximate formula for the T-I-S is not good approximation.
P14-10 (a)
14-27
The real reactor can be modelled as two parallel PFRs:
P14-10 (b)
Model parameters
For two parallel PFRs, the parameters are τ1=10 min and τ2=30 min,
P14-10 (c)
Conversion
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
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
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:
Da1
0.145
!
$
<
ttif
t
t
1
2
1
14-29
P14-11 (b)
Peclet number
The Peclet number
P14-11 (c)
Conversion
Linearizing the 2st order according 14.2 (g):
AA
A
AA CkC
C
kkCr ‘
2
0
2=!=“
it is possible to obtain an approximated solution;
P14-12 (a)
Remembering the physical meaning of the Peclet number:
Per=rateoftransportbyconvection
rateoftransportbydiffusionordispersion =Ul
Da
14-30
q=1+4Da
Per
Da =
“
k
For Per>>1 (or for small vessel dispersion number)
Introducing (2) into (1) and neglecting terms
2
1
!
!
“
#
$
$
%
&
r
Pe
o
P14-12 (b)
!
k
PFR eX “
“=1
and
( )
r
Pe
k
k
DisoPe eX
2
1
1
!
!
+“
>> “=
In order to achieve the same conversion:
P14-12 (c)
Defining:
1.0==
a
PFR
rD
Ul
Pe
14-31
P14-12 (d)
Starting from 14.12.1 and subtracting the conversion for a PFR:
P14-12 (e)
According P12.3 dispersion doesn’t affect zero order reaction.
P14-13
From P13.19:
P14-13 (a)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Segregation Model
Segregation Model
14-32
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 70 70
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
P14-13 (b)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Maximum Mixedness Model
( )
( )X
F
E
C
r
d
dX
Ao
A
!
!
!
“
+= 1
14-33
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
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
(Check F(t))
P14-13 (c)
Tanks in series and 1st order reaction with k=0.1min-1
P14-13 (d)
Dispersion models and 1st order reaction with k=0.1min-1
Peclet number open system
14-34
P14-13 (e)
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):
P14-13 (f)
Two parameters model (backflow)
14-35
P14-13 (g)
Two parameters model models and 2nd order reaction with k=0.1dm3/mol
⋅
s and CA0=1mol/dm3
P14-13 (h)
Table of the conversions
XT_I_S
XMM
Xseg
X Dispersion
Xtwoparameters
P14-13 (i)
Adiabatic reaction with Segregation Model
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Segregation Model
Segregation Model
14-36
P14-14 (a)
Product distribution for the CSTR and PFR in series
T
k2/k1
τk1CA0
T1
5.0
0.2
Considering Arrhenius equation and applying the following notation:
rT
r
r
rT E
RT
E
A
k!=!=ln
We can write this linear system of 5 equations for the 5 unknowns (
114131211 ,,,, AEEEE
):
P14-14 (b-d) No solution will be given at this time.
P14-15 (a)
Two parameters model
( )
)1.0exp(110)(min10min0
ttCt
!!“=<#
.
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===
!“#
P14-15 (b)
Conversion
2nd order, vo=1 dm3/min, k=0.1dm3/mol⋅min, CA0=1.25mol/dm3
vPFR=βvo
V1=αV
t(min)
14-38
Where
25.1== AoPFR CkDa
!
“
#
3
/556.0 dmmolCPFR =
For the CSTR:
P14-15 (c)
Two parameters model
0)(min10min0
1
=<#
tFt
α= Fractional volume
β= Fraction of flow
For the PFR:
Second-order
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:
vb=βvo
V1=αV
node1
V2
P14-16