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14-1
Solutions for Chapter 14 – Models for Non-ideal
Reactors
P14-1 Individualized solution
P14-2 (a)
Approximated formula for Segregation Model (1st reaction)
!
k
Error
0.1
-0.0001667
P14-2 (b)
Parameters Dispersion Model
Closed-Closed dispersion model
14-2
U
l
kkDa ==
!
Damköhler number
a
rD
Ul
Pe =
Peclet number
dt
Re
ReSc
L/dt
D/(U*dt)
D (cm2/s)
1 cm
10
104
30.9
0.18 From
Fig14.10
0.018
Is there a diameter that would maximize or minimize conversion in this range ?
P14-2 (c)
(1) Vary the Damköhler number for a second-order reaction (Example 14-3(b))
14-3
Damköhler number/ Da
Conversion
Parameter
0.1603
0.138
8*U0
(2) Vary the Peclet and Damköhler numbers for a second–order reaction in laminar flow
(Example 14-3(c))
For a second order reaction
0
0
0AA kC
U
L
kCDa ==
!
;
AB
DLUPe /
0
=
mR 05.0=
Da
Pe
Conversion
Parameter
0.1603
0.132
8*U0
0.7
0.8
0.9
1.0
When Peclet Number decreases less than 2 ×105, the conversion is influenced significantly.
14-4
Below is a FEMLAB anaylsis of the problem.
(1) Vary the Damköhler number for a second-order reaction (Example 14-3(b))
For a second order reaction
0
0
0AA kC
U
L
kCDa ==
!
mR 05.0=
Damköhler number/ Da
Conversion
Parameter
0.1603
0.138
8*U0
- Femlab Screen Shots
[1] Domain
[2] Constants and scalar expressions
- Constants
14-5
- Scalar expressions
[3] Subdomain Settings
- Physics
(Mass Balance)
- Boundary Conditions
@ r = 0, Axial symmetry
@ inlet,
[4] Results
(Concentration, cA)
(2) Vary the Peclet and Damköhler numbers for a second–order reaction in laminar flow
(Example 14-3(c))
For a second order reaction
0
0AA kC
U
L
kCDa ==
!
;
AB
DLUPe /
0
=
Da
Pe
Conversion
Parameter
0.1603
8.32×105
0.132
8*U0
14-7
0.8
0.9
1.0
When Peclet Number decreases less than 2 ×105, the conversion is influenced significantly.
- Femlab Screen Shots
[1] Domain
[2] Constants and scalar expressions
- Constants
- Scalar expressions
[3] Subdomain Settings
- Physics
(Mass Balance)
- Initial Values
(Mass balance) cA(t0) = cA0
- Boundary Conditions
@ r = 0, Axial symmetry
14-9
@ outlet, Convective flux
@ wall, Insulation/Symmetry
[4] Results
(Concentration, cA)
P14-2 (d)
Two parameters model
min10==
o
v
V
!
14-10
min.250==
s
s
sv
V
!
Ideal CSTR
CSTR with Dead Space and
Bypass
CSTR with Dead Space and
Bypass
P14-2 (e)
Two CSTR with interchange (1st order reaction)
v1
v1
V2
CA0 v0
14-11
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 200 200
CT1 2000 31.814045 2000 31.814045
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(CT1)/d(t) = (beta*CT2-(1+beta)*CT1)/alpha/tau
[2] d(CT2)/d(t) = (beta*CT1-beta*CT2)/(1-alpha)/tau
Explicit equations as entered by the user
[1] beta = 0.15
Comparison experimental and predicted (β=0.15 α=0.75) concetration.
14-12
α
β
X
0.75
0.15
0.51
14-13
P14-2 (f)
Tubular Reactor Design
The correlations between Re and Da show what flow conditions (characterised by Re) give the
P14-2 (g)
Linearizing non-first-order reactions
May be a good approximation if CA does not change very much with time, i.e. A is in excess, in
P14-2 (h)
Figures 14.3 and 14.1b
The curves in Fig.14.3 represent the residence time distributions for the Tanks in Series model as
function of the number of reactors.
Given a CSTR of volume 1 (V=V1), we divide the CSTR in two CSTRs (V2=1/2). The mean
14-14
V
V1
=
!
v
v1
=
!
Stoichiometry: liquid phase
CA=CAo 1"X
( )
2nd order PFR :
V
v=1
kCAo
22
"
1+
"
( )
ln 1#X
( )
+
"
2X+
1+
"
2
( )
X
1#X
$
%
&
&
'
(
)
)
"
=yAo
#
where
"
=c
a+b
a#1=1
2+1
2#1=0
"
=0
Determining
!
and
!
:
Gives
25.0
10
5.2
1
== v
vi
!
and
50
52
5
5
52
1.
./
./ ===
!
"
V
V
Now
min5
2
//
2
221121 =
+
=
+
=vVvV
!!
!
Reactor 1:
!
"
#
$
%
&
'
=
1
1
21210
1
52
X
X
*.
.
Reactor 2:
!
"
#
$
%
&
'
=
2
2
21
2*1.0
1
5.7
X
X
P14-3 (a)
Money for buying reactors
Using the tank in series model:
Assume that
t
!!
=
and that in reactors medelled as more than one tank, that
n
t
!
!
=
. Number of
tanks
2
2
!
"
=n
rounded to the nearest integer.
Reactor
Σ(min)
τ(min)
n
X
Maze & blue
2
2
1
0.50
14-16
Where
Scarlet & grey: X1 = 0.5, CA1=0.5 → X2 = 0.38, CA2=0.31→ X = 0.69
P14-3 (b)
More money for buying reactors
P14-3 (c)
Ann Arbor, MI
East Lansing, MI
P14-4
Packed bed reactor with dispersion
1st order, k1=0.0167/s, ε=0.5, dp=0.1 cm,
14-17
P14-5 (a)
Number of tanks in series
Assuming the Peclet-Bodenstein relation:
1
2+= Bo
n
Where
a
D
UL
Bo =
2
P14-5 (b)
Conversion
Using individual reactor material balances:
Reactor 1:
Reactor 2:
( ) 507.0
1
2
2
21
2=!
"
=X
v
XkC
XA
( ) 3
212 /001607.0)507.01(00326.01 dmmolXCC AA =!=!=
Reactor 3:
( ) 387.0
1
3
2
32
3=!
"
=X
v
XkC
XA
P14-5 (c)
Change of the fluid velocity
Let U=0.1cm/s
Re=50 and Sc=2
From gas phase dispersion correlation chart,
5.0=
t
a
ud
D
Let U=100cm/s
Re=50000 and Sc=2
From gas phase dispersion correlation chart,
21.0=
a
ud
D
14-19
P14-5 (d)
Change of the superficial velocity
80
01.0
4*2.0
Re === v
udt
From packed bed dispersion correlation chart,
55.0=
p
a
ud
D
!
P14-5 (e)
Liquid instead of gas
Part (a)
2.0
1.0
001.0*4*5
Re === v
udt
14-20
P14-6 (a)
Peclet numbers
From Example 13.2 σ2=6.19min2 and tm=5.15min
Closed:
P14-6 (b)
Space–time and dead volume
P14-6 (c)
Conversions for 1st order isomerization
Dispersion model
Da=kτ=0.927
1221
4
1=+=
r
Pe
Da
q
P14-6 (d)
Conversions PFR and CSTR
PFR:
!
k
eX "
"=1
X=0.604
