12-17
P12-10 (c)
Let’s try again with
“0
= 10
P12-10 (d)
We now need to resolve the problem with the fact that there is a critical value of λ, λc, for which both
ψ = 0 and
d“
d#=0
d2“
d#2$2%0
2=0
Subtracting
1=“0
2# “0
2$C
2+C11# $C
( )
Solving for C1
“C=1#1
$0
Sketch of concentration profile for different values of φ0
That is for φ0 = 2, the concentration of A is zero half way (λ = 0.5) through the slab.
1” #C=1
$0
12-19
P12-10 (e)
P12-10 (f)
P12-10 (g) No solution will be given
P12-10 (h) No solution will be given
P12-11
12-20
12-21
P12-12 (a)
12-22
P12-12 (b) No solution will be given.
P12-13 (a)
( )( ) 0
12
2=!“!+
#
$
%
&
‘
(
ARxt rH
dr
dT
kr
dr
d
r
(1)
Integrating
( )
( ) 2
1
1
20
C
r
C
HD
Tk
C
CC
HD
Tk
dr
d
r
Rxe
t
A
A
Rxe
t
+=
!“
+
=+
#
$
%
&
‘
(+
!“
C1=0
because T & C must be finite at r = 0.
P12-13 (b) No solution will be given. Individualized solution and you need to use FEMLAB
(COMSOL multiphysics).
P12-15 (a)
12-24
12-25
P12-15 (b)
12-26
P12-16
12-28
P12-16 (b)
P12-16 (c)
P12-16 (d)
To make the catalyst more effective, we should use a smaller diameter.
P12-16 (e)
12-29
P12-17
d2y
d“2# $nyn=0
Multiply by 2 y
dy
dy
d“
#
$
% &
‘
(
2
=2)n
yn+1
n+1+C1
y=“# =CA
CA0
, #=0 y =0 d“
d#=0
therefore C1 = 0
Taking the derivative of y and evaluating at λ =1
In dimensionless form
“=
3d#
d$
%
&
‘ (
)
*
$=1
+n
2
“=#”
, differentiating gives
12-30
For larger
“n
“=32
n+1
1
#n
CDP12-A
CDP12-B (a) 3rd ed. 12-19 (a)
CDP12-C (a) 3rd ed. 12-20 (a)
CDP12-C (b) 3rd ed. 12-20 (b)
CDP12-C (c) 3rd ed. 12-20 (c)
CDP12-E 2nd ed. 11-18
CDP12-F 2nd ed. 11-19
12-31
CDP12-I 2nd ed. 11-22
CDP12-J (a) 2nd ed. 12-7 (a)
CDP12-K 2nd ed. 12-9
CDP12-L (a) 2nd ed. 12-8 (a)
CDP12-M (a) 3rd ed. CDP12-L (a)
CDP12-M (b) 3rd ed. CDP12-L (b)
CDP12-N 3rd ed. CDP12-M
CDP12-O
CDP12-S
CDP12-T
CDP12-U