P14-17
The presence of a minimum (run2) imply the presence of a radial temperature profile that
effects the reaction rate and determines the deviation from the concentration profile given
by Eq. (14.51).
The enthalpy balance allows the identification of the dimensionless thermal parameters:
Defining
00 U
U
U
C
C
T
T
L
z
R
r
A
A
isot
=====
!“#$
( ) 0
1
0
2
2
2
2=
!”
#
$
$
#
%
%
%
%
&
‘
(
(
(
(
)
*
$
$
+
,
–
.
/
0
+
$
+
+
,
–
.
.
/
0
$
$
$
isotp
reac
nn
A
p
k
TC
CkU
L
R
CR
123
4
3
4
3
5
5
3
5
5
14
( )
!
“
!
“
“
!
“
#=
$
$
#==
$
$
=1100 BiM
where
k
hR
Bi
!
=
ratio convection-radial conduction
p
k
CR2
!“
ratio conduction-enthalpy flux
coolantp
reac
TC
!”
dimensionless adiabatic heat
( ) ( )
!““
n
A
CkDa 0
=
A parametric study of the numerical solutions of Eq.(14.51) and the enthalpy balance would give
the “exact” conditions for a minimum in the concentration.
A decrease in the thermal conductivity of the reaction mixtures determines an increase of the bulk
fluid temperature and can determine a minimum in concentration.
Overall heat transfer coefficient increases:
For a given flux this determines less deviation from the isothermal case: no minimum
Overall heat transfer coefficient decreases:
For a given flux this determines more deviation from the isothermal case
and the possibility of a minimum.
The external heat transfer coefficient increases, this implies a lower wall temperature and a
resulting lower temperature in the reacting mixture and less deviation from the isothermal case:
no minimum
The external heat transfer coefficient increases, this implies an increase in the wall temperature
and a resulting higher temperature in the reacting mixture that can determine a minimum in the
concentration.
(2 ) Constants and scalar expressions
– Constants
14-43
– Scalar expressions
(3 ) Subdomain settings
– Physics
(Mass balance)
14-44
(Energy balance)
– Initial Values
(Mass balance) cA(t0) = cA0
(Energy balance) T(t0) = T0
– Boundary Conditions
@ r = 0, Axial symmetry
@ inlet, cA = cA0 (for mass balance)
(for energy balance)
14-45
@ outlet, Convective flux
@ wall, Insulation/Symmetry (for mass balance)
(for energy balance)
P14-18
Vary the Peclet number and the reaction order in laminar flow (Example 14–3(c))
1
0
0
1
0
!! == n
A
n
AkC
U
L
kCDa
“
AB
DLUPe /
0
=
mL 36.6=
mR 05.0=
min./25.0 3moldmk =
min/24.1
0mU =
3
0/5.0 dmmolC A=
min/106.7 25 mDAB
!
“=
n
Pe
Conversion
Parameter
1.04×103
/
100* DAB
1.04×104
/
10 * DAB
1.04×105
/
DAB
1.04×106
/
0.1*DAB
0.1
Attempt to evaluate
non-integral power of
negative number!
1.04×103
/
100* DAB
1.04×105
/
DAB
1.04×106
/
0.1*DAB
0.5
Attempt to evaluate
non-integral power of
negative number!
1.04×104
0.818
10 * DAB
1.04×105
0.783
DAB
14-46
1.04×107
/
0.01*DAB
1.04×103
0.721
100* DAB
1.04×104
0.717
10 * DAB
1.04×106
0.653
0.1*DAB
1
1.04×107
0.641
0.01*DAB
1.04×104
0.522
10 * DAB
1.04×105
0.503
DAB
1.04×106
0.481
0.1*DAB
1.04×103
0.390
100* DAB
1.04×104
0.388
10 * DAB
1.04×105
0.374
DAB
2
1.04×106
0.270
0.1*DAB
1.04×107
0.265
0.01*DAB
1.04×104
0.218
10 * DAB
1.04×105
0.211
DAB
1.04×106
0.203
0.1*DAB
1.04×103
0.164
100* DAB
1.04×104
0.163
10 * DAB
1.04×105
0.159
DAB
3.5
1.04×107
0.150
0.01*DAB
1.04×104
0.122
10 * DAB
1.04×105
0.119
DAB
1.04×106
0.115
0.1*DAB
1.04×107
0.113
0.01*DAB
14-47
103104105106107108
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Conversion
Pe
n
0.8
1.0
1.5
2.0
2.5
3.0
3.5
4.0
(1) With the increase of the reaction order n, the conversion will decrease. The conversion
will also decrease when the Peclet number increases.
(2) When n is less than 1.5 and Pe is 105, the diffusion in the system becomes more
Below show FEMLAB screenshots useful for this problem.
– Constants
14-48
– Scalar expressions
(5 ) Subdomain settings
– Physics
14-49
– Initial Values
(Mass balance) cA(t0) = cA0
– Boundary Conditions
@ r = 0, Axial symmetry
@ inlet,
@ outlet, Convective flux
@ wall, Insulation/Symmetry
P14-19 (a)
First order reaction
Input Parameters:
1=n
mR 05.0=
mL 36.6=
min/25.0=k
min/24.1
0mU =
3
0/5.0 dmmolC A=
min/106.7 25 mDAB
!
“=
Conversion:
xA= 0.687 @ Open-vessel Boundary: –Ni·n =2 *U0*(1–(r/Ra)^2)*CA0
xA= 0.726 @ Close-vessel Boundary: –Ni·n =U0 *CA0
Conversion:
xA= 0.755 @ Open-vessel Boundary: –Ni·n =2 *U0*(1–(r/Ra)^2)*CA0
xA= 0.778 @ Close-vessel Boundary: –Ni·n =U0 *CA0
I. Variation of Da number
Conversion
Damköhler number/ Da
Closed-vessel
Open-vessel
Parameter
0.449
0.404
0.287
8*U0
0.898
0.527
0.439
4*U0
1.795
0.658
0.606
2*U0
3.59
0.778
0.755
U0
7.18
0.870
0.862
U0/2
14.36
0.930
0.928
U0/4
28.72
0.964
0.964
U0/8
14-50
II. Variation of Pe number
Conversion
Peclet number/Pe
Closed-vessel
Open-vessel
Parameter
1.04e3
0.781
0.781
100* DAB
1.04e4
0.781
0.776
10 * DAB
1.04e5
0.778
0.755
DAB
1.04e6
0.772
0.740
0.1*DAB
1.04e7
0.770
0.737
0.01*DAB
III. Femlab Screen Shots
(1) Domain
(2) Constants and scalar expressions
– Constants
– Scalar expressions
14-51
(3) Subdomain settings
– Physics
– Initial Values
(Mass balance) cA(t0) = cA0
– Boundary Conditions
@ r = 0, Axial symmetry
@ inlet, Flux N0 = 2*u0*(1-(r/Ra)^2)*cA0 : Open-Vessel Boundary
Flux N0 = u0*cA0 : Close-Vessel Boundary
@ outlet, Convective flux
14-52
(4) Results
(Open-vessel Boundary)
(Close-vessel Boundary)
14-53
P14-19 (b)
Third order reaction with k *C2
A0= 0.7 min–1
1
0
0
1
0
!! == n
A
n
AkC
U
L
kCDa
“
AB
DLUPe /
0
=
Input Parameters:
3=n
mR 05.0=
mL 36.6=
min./8.2 3moldmk =
min/24.1
0mU =
3
0/5.0 dmmolC A=
min/106.7 25 mDAB
!
“=
I. Variation of Da number
Conversion
Damköhler number/ Da
Closed-vessel
Open-vessel
Parameter
0.255
0.381
0.253
8*U0
0.51
0.466
0.375
4*U0
1.02
0.560
0.506
2*U0
4.06
0.741
0.730
U0/2
8.12
0.813
0.809
U0/4
16.24
0.867
0.865
U0/8
II. Variation of Pe number
Conversion
Peclet number/Pe
Closed-vessel
Open-vessel
Parameter
1.04e3
0.650
0.649
100* DAB
1.04e4
0.654
0.645
10* DAB
1.04e5
0.655
0.628
DAB
1.04e6
0.652
0.617
0.1*DAB
1.04e7
0.650
0.615
0.01*DAB
(c) Half order reaction with k= 0.495 (mol/dm3)1/2min–1
0
0
0
A
AkC
U
L
kCDa
“
AB
DLUPe /
0
=
Input Parameters:
2/1=n
mR 05.0=
mL 36.6=
12/13 min)/(495.0 !
=dmmolk
min/24.1
0mU =
3
0/5.0 dmmolC A=
min/106.7 25 mDAB
!
“=