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12-1
Solutions for Chapter 12 – Diffusion and Reaction in
Porous Catalysts
P12-1 Individualized solution
P12-2 (a)
(a)
"=5
12-2
1=A cosh Da
A=1
cosh Da , B =Tanh Da
cosh Da
"=cosh Da #
cosh Da $Tanh Da
cosh Da sinh Da #
(2) Monod Kinetics
De
d2CA
dz2"µmaxCACC
KS+CA
=0
dCC
dz2=µmaxCACC
KS+CA
=0
Use
Quasi
Steady
State
Analysis
No further solution to Monod Kinetics will be given.
(3) Variable Diffusion Coefficient
dFw
dt ="cWO 2 z=0Ac
WA="De
dCA
dz ="DeCA0
L
d#
d$$=0
dFw
dt =DeCA0
L
d#
d$$=0
%cAcV
Mole balance
d De
dCA
dz
"
#
$
%
&
'
dz (k=0
for hindered diffusion
De=DAB
1+"2Fw
21#Fw
( )
As a first approximation, assume no variation in De with λ
d2"
d#2$kL2
DeCA0
=0
as before
"0=kL2
2DeCA0
d2#
d$2%2"0=0
Solution the same as before Equation (E12-2.13)
"=#0$ $ % 2
( )
+1
WA=%De
dCA
dz z=0
=%DeCA0
L
d"
d$$=0
d"
d#=2$0# % 2$0
( )
#=0
=%2$0=%2kL2
2DeCA0
WO2=WA=%DeCA0
L
&
'
( )
*
+ %kL2
DeCA0
&
'
(
)
*
+ =kL
The flux of O2 in does not depend upon De which is not uprising since this reaction is
zero order.
For the build up of material that hinders diffusion
dFW
dt ="cAckL =AcLk =kV
FW=kVt
From a quasi steady state approximation as time goes on
FW
increases De decreases and
"0
increases.
However, the point at which the oxygen concentration is equal to zero has to be found.
We can parallel the analysis used in P12-10 switching the coordinates of λ = 0 and λ = 1
(see solution to P12-10(c) in which the solution manual) we will find
"c=1
12-4
P12-2 (c)
(1) For R1 η1 = 0.182 18.2% Surface reaction limited and
81.8% Diffusion limited
For R2 η = 0.856 85.6% Surface reaction limited and
14.6% diffusion limited
DeCAs
=0.95
( )
0.9
( )
2=0.77
which is less than 1 so there are no significant diffusion limitations.
P12-2 (d)
"=#
1+$ $
k
1
Sa%b
kcac
=1
1
#+$ $
k Sa%b
kcac
"=0.059
So 5.9% surface reaction resistance and 94/1% external and internal diffusion
limited
%R =0.941
1
"+# #
k Sa$b
kcac
=0.941
6.0 +10.96 == 0.941
16.96
% Internal diffusion reaction
0.941"100 "6.0
16.96 =33.3%
% External diffusion resistance =
0.941
( )
100
( )
"10.96
External Diffusion 60.8%
Internal Diffusion 33.3%
Surface Reactor 5.9%
100.0%
Increase temperature significantly. Surface reaction % resistance decreases. Increase gas
L2=L1ln 1
1"X2
ln 1
1"X2
ln 1
1"X1
=0.16 ln 10,000
( )
ln500
=0.16 #9.2
6.21 =0.24m
P12-2 (e)
(b) From Mears’s Criterion
"#HRx "$
r
A
( )
%bRE
hT2Rg
<0.15
1)
For large φ
"=3
R
De
k=3
R
De
# #
k $cSa
#
S
a=Sa
1+kDt
(1) Pore closure. Consider De As
t" #
pore throat closes
"=Area 2
Area 1 # $
,
"c#0
,
De"0
, and
"1# $
"#
r
A$0
(2) Loss of surface area Sa. As
t" #
then
"
S
a#0
then
" # 0
" #1
, but
"#
r
A=Sa$0
P12-2 (h)
The activation energy will be larger than that for diffusion control and hence the reaction is more
temperature sensitive. If the apparent reaction order is greater than one half, then the rate of reaction will be
less sensitive to concentration. If it is less than one half, the true order will be negative and the rate will
increase significantly at low concentration.
In example CDR12-1, the reactor is 5 m in diameter and 22 m high, whereas the reactor in CDR12-2 is
only 2 m3 in volume. The charge is much different. In CDR12-1 the charge is 100 kg/m3 and in CDR12-2 it
is only 3.9 kg/m3
P12-2 (k)
With the increase in temperature, the rate of reaction will increase. This will cause the slope of Ci/Ri vs.
1/m and, therefore, the resistance to decrease.
P12-3 (a) Yes
P12-3 (b)
All temperatures, FT0 = 10 mol/hr. The rate of reaction changes with flow rate and increases linearly with
temperature
P12-3 (d)
T < 367 K, FT0 = 1000 mol/hr, 5000 mol/hr.
T < 362 K, FT0 = 100 mol/hr.
12-7
P12-3 (e) Yes
P12-3 (f)
T > 367 K, FT0 = 1000 mol/hr, 5000 mol/hr.
T > 362 K, FT0 = 100 mol/hr.
( )
( )
0
0
actual rate of reaction
ideal rate of reaction 362 , 5000 /
A T
A T
r at K F mol hr
"= = !=
0.26 0.37
0.70
!= =
P12-3 (h)
At FT0 = 5000 mol/hr, there is non external diffusion limitation, so the external effectiveness factor is 1.
( )
( )
T0
T0
actualrate of reaction at 362 K, F 5000 /
extrapolated rate of reaction at 362 K, F 5000 /
mol hr
mol hr
!
=
==
1.2 0.86
1.4
!
= =
P12-3 (i)
[ ]
2
3 cosh 1
0.86
! !
"!
#
= =
by iterative solution
1.60
!
=
( )
sinh
1
sinh
A
AS
C
C
!"
#" !
= =
P12-4 (a)
External mass transfer limited at 400 K and dP = 0.8 cm. Alos at all FT0 < 2000 mol/s
P12-4 (b)
Reaction rate limited at T = 300 K and dP = 0.3 cm. When T = 400 K: dP = 0.8, 0.1, and 0.03 cm.
P12-4 (c)
Internal diffusion limited at T = 400 K and 0.1 < dP < 0.8
P12-4 (d)
rate with 0.8 10 0.625
rate with 0.03 16
P
P
d
d
!
=
= = =
=
12-8
P12-5
Curve A: At low temperatures (high 1/T) the reaction is rate limited as evidenced by the high
activation energy. At high temperatures (low 1/T) the reaction is diffusion limited as evidenced
by the weak dependence on temperature,
Curve B: Weak dependence on temperature suggests diffusion limitations
12-9
P12-6 (c) Individualized solution
P12-7 (a)
P12-7 (c)
P12-7 (d)
P12-8
12-12
P12-8 (b)
P12-8 (c)
P12-8 (d) Individualized solution
12-13
P12-9 (a)
12-14
P12-9 (b)
A B
0
A A A
r r r
W rL W rL r r rL
! ! !
+"
#+"=
10
A A
drW r r
r dr + =
EMCD therefore,
A
A e
dC
W D
dr
=!
2
2
1 1
A
A A
e e
rdC
dd C dC
dr
D D
r dr dr r dr
! "
=#
$ %
& '
A A
r kC=!
10
A A
drW r r
r dr + =
2
2
10
A A
e A
d C dC
D kC
dr r dr
! "
# # =
$ %
& '
0
A
A
C
C
!
=
r
R
!
=
2 2
2
10
e
d d kR
d d D
! ! !
" " "
# # =
2
2
10
a
d d D
d d
! ! !
" " "
# # =
At λ = 1, ψ = 1 and at λ = 1,
0
d
d
!
"
=
Bessel Function Solution
P12-10 (a)
EMCD
WA="DdCA
dz , rA="k
In "Out +Gen =0
WAAcz"WAAcz+#z+rA#zAc=0
"dWA
dz +rA=0 , DA˜
d CA
dz2"k=0
$=CA
CAs
, %=Z
L , d2$
d%2
CA0
L2=d2CA
dz2
d2"
d#2$kL2
DACAs
=0
(1)
d"
d#=kL2
DACAs
#+C1
Using the symmetry B. C.
"=0 d#
d"=0
2DACAs
at
"=1 #=1
C2=1"kL2
2DACAs
#=1+kL2
2DaCAs
$2"1
( )
[ ]
Now let’s find what value of λ that ψ = 0 for different
"0
.
For
"0
2=1 : 0 =1+1#2$1
[ ]
=1+#2$1
"2=0
Therefore the concentration is zero (i.e., ψ = 0) at
"=0
For
"0
2=16 : 0 =1+#2$1
[ ]
=1+16#2$16
Seems okay, but let’s look further and calculate the concentration ratio ψ at λ = 0 for
"0
= 4.
"=1+16 0.2
( )
2#1
[ ]
=1+16 0.04 #1
[ ]
=#14.9
Negative concentration.
