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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
P12-2 (b)
(1) First Order Reaction Kinetics
De
d2CA
dz2"kCA=0 , #=CA
CA0
, $=z
b
12-2
1=A cosh Da
A=1
cosh Da , B =Tanh Da
cosh Da
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
As a first approximation, assume no variation in De with λ
d2"
d#2$kL2
DeCA0
=0
12-3
Solution the same as before Equation (E12-2.13)
"=#0$ $ % 2
( )
+1
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
12-4
P12-2 (c)
(1) For R1 η1 = 0.182 18.2% Surface reaction limited and
81.8% Diffusion limited
P12-2 (d)
"=#
1+$ $
k
1
Sa%b
kcac
Summary of Resistances
External Diffusion 60.8%
Internal Diffusion 33.3%
Surface Reactor 5.9%
12-5
P12-2 (e)
(b) From Mears’s Criterion
"#HRx "$
r
A
( )
%bRE
hT2Rg
<0.15
1)
The value from the question
"HRx =#25kcal /mol =#104.6kJ /mol
"b=(1# $)"c=(1#0.5)(2.8 *106)=1.4 *106g /m3
R=3*10"3m
T=1173K
At the inlet of the reactor the fraction of NO =0.02
From ideal gas law
P12-2 (f)
For γ = 30 use Figure 12-7.
P12-2 (g)
a=1
1+k0t , "1=R# #
k $cSa
De
(1) Pore closure. Consider De As
t" #
pore throat closes
"=Area 2
Area 1 # $
,
"c#0
,
De"0
, and
"1# $
"#
r
A$0
P12-2 (h)
The activation energy will be larger than that for diffusion control and hence the reaction is more
P12-2 (i)
P12-2 (j) No solution will be given at this time
P12-2 (k)
P12-2 (l) No solution will be given at this time.
P12-3 (a) Yes
P12-3 (b)
P12-3 (c) Yes
P12-3 (d)
P12-3 (e) Yes
P12-3 (f)
P12-3 (g)
P12-3 (h)
At FT0 = 5000 mol/hr, there is non external diffusion limitation, so the external effectiveness factor is 1.
P12-3 (i)
[ ]
2
3 cosh 1
0.86
! !
"!
#
= =
P12-4 (a)
External mass transfer limited at 400 K and dP = 0.8 cm. Alos at all FT0 < 2000 mol/s
P12-4 (c)
Internal diffusion limited at T = 400 K and 0.1 < dP < 0.8
P12-4 (d)
P
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
P12-6 (a)
P12-6 (b)
P12-6 (c) Individualized solution
P12-7 (a)
12-10
P12-7 (b)
P12-7 (c)
P12-7 (d)
P12-8
12-12
P12-8 (a)
P12-8 (b)
P12-8 (c)
P12-8 (d) Individualized solution
12-13
P12-9 (a)
12-15
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
10
A A
e A
d C dC
D kC
dr r dr
! "
# # =
$ %
& '
0
A
A
C
C
!
=
r
R
!
=
P12-10 (a)
EMCD
WA="DdCA
dz , rA="k
In "Out +Gen =0
WAAcz"WAAcz+#z+rA#zAc=0
12-16
d"
d#=kL2
DACAs
#+C1
Using the symmetry B. C.
"=0 d#
d"=0
C1=0
Integrating equation (1)
"=kL2#2
2DACAs
+C2
P12-10 (b)
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
