12-17
P12-10 (c)
Let’s try again with
“0
= 10
“=1+10
( )
20.12#1
[ ]
=1#1020.99
( )
“=1#99 =#99
not possible
ψ will be negative for any value of
“0
greater than one.
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
d“
d#2$2%0
2#+C1
“=%0
2#2+C1#+C2
0=“0
Subtracting
1=“0
2# “0
2$C
2+C11# $C
( )
Solving for C1
C1=
1” #0
21” $C
2
( )
1” $C
Solving for C2
C2=1” #0
2“
1” #0
21” $C
2
( )
( )
1” $C
“=#0
2$2+
1% #0
21% $C
2
( )
( )
1% $C
$+1% #0
2%
1% #0
21% $C
2
( )
( )
1% $C
At
“=“C
then
d“
d#=0=2$0
2#C%
1% $0
21% #C
2
( )
( )
1% #C
12-18
0=“0
2#C
2$2“0
2#C$1$ “0
2
( )
#C=2“0
2$
4“0
4$4“0
2
( )
$1$ “0
2
( )
( )
2“0
2=1$4“0
4+4“0
2$4“0
4
2“0
2=1$4“0
2
2“0
2
“C=1#1
$0
Sketch of concentration profile for different values of φ0
φ0 = 1 then λc = 0
φ0 = 2 then λc = 0.5
That is for φ0 = 2, the concentration of A is zero half way (λ = 0.5) through the slab.
1” #C=1
$0
“=#0
2$2+#0% #0
21+$C
( )
[ ]
+1% #0
2% #0% #0
21+$C
( )
[ ]
=#0
2$2+#0% #0
2% #0
2+#0
[ ]
$+1% #0
2%2#0+2#0
2
“=#0
2$2+2#01% #0
( )
$+1%2#0+#0
2
For λ > λc
P12-10 (e)
“=
#rAAcdz
0
zc
$+#rAAcdz
0
6 7 4 8 4
zc
L
$
#rAAcdz =#rAAczc+0
#rAAcdz
0
L
$=zc
L
“=zc
L=#c=1$1
%0
“rA=0 for zc<z<L
P12-10 (f)
P12-10 (g) No solution will be given
P12-10 (h) No solution will be given
P12-10 (i) Individualized solution
P12-11
12-20
12-21
P12-12 (a)
P12-13 (a)
( )( ) 0
12
2=!“!+
#
$
%
&
‘
(
ARxt rH
dr
dT
kr
dr
d
r
(1)
1
r2
d
dr
r2De
dCA
dr
“
#
$ %
&
‘ +rA=0
(2)
Dividing by ΔHRx and using Equation (2) to substitute for –rA
12-23
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.
CA+ktT
De“#HRx
( )
=C2
r = R
CA=CAs
and
T=T
S
( )( ) SAAs
t
Rxe
Rxe
St
As
Rxe
t
A
TCC
k
HD
T
Tk
C
Tk
C
+!
“!
=
+=
+
0at
max == A
CTT
Tmax =T
S+“#HRx
( )
DeCAs
kt
P12-13 (b) No solution will be given. Individualized solution and you need to use FEMLAB
(COMSOL multiphysics).
P12-14 No solution will be given.
P12-15 (a)
12-24
12-25
P12-15 (b)
12-26
12-27
12-28
P12-16 (c)
P12-16 (d)
P12-16 (e)
12-29
P12-17
d2y
d“2# $nyn=0
Multiply by 2 y
dy
d“
2y dy
d“
d
d“
dy
d“
#
$
% &
‘
( =)n
2yn2dy
d“
Manipulating the L.H.S.
d
d“
dy
d“
#
$
% &
‘
(
2
=2dy
d“
d2y
d“2
d
d“
dy
d“
#
$
% &
‘
(
2
=)nyn2dy
dy
#
2
yn+1
dy
d““=1
=2#n
n+1“=1
=2#n
n+1
The effectiveness factor is
“=
#R2DA
dCA
dr
$
%
& ‘
(
)
r=k
kCAs
n4
3#R3
In dimensionless form
“=
3d#
d$
%
&
‘ (
)
*
$=1
+n
2
“=#”
, differentiating gives
dy
d“=“d#
d“+#
at
“=1
dy
d““=1
=d#
d“+1
d#
d“=2$n
2
n+1%1
12-30
“=3
2#n
2
n+1$1
#n
2
%
&
‘
‘
‘
‘
(
)
*
*
*
*
=
32
n+1
#n
$3
#n
2
For larger
“n
“=32
n+1
1
#n
CDP12-A
CDP12-B (a) 3rd ed. 12-19 (a)
CDP12-B (b) 3rd ed. 12-19 (b)
CDP12-B (c) 3rd ed. 12-19 (c)
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-D (a) 3rd ed. 12-21 (a)
CDP12-D (a) 3rd ed. 12-21 (b)
CDP12-D (a) Individualized solution
CDP12-E 2nd ed. 11-18
CDP12-F 2nd ed. 11-19
CDP12-G 2nd ed. 11-20
CDP12-H 2nd ed. 11-21
12-31
CDP12-I 2nd ed. 11-22
CDP12-J (a) 2nd ed. 12-7 (a)
CDP12-J (b) 2nd ed. 12-7 (b)
CDP12-J (c) 2nd ed. 12-7 (c)
CDP12-J (d) 2nd ed. 12-7 (d)
CDP12-J (e) 2nd ed. 12-7 (e)
CDP12-K 2nd ed. 12-9
CDP12-L (a) 2nd ed. 12-8 (a)
CDP12-L (b) 2nd ed. 12-8 (b)
CDP12-L (c) 2nd ed. 12-8 (c)
CDP12-L (d) 2nd ed. 12-8 (d)
CDP12-M (a) 3rd ed. CDP12-L (a)
CDP12-M (b) 3rd ed. CDP12-L (b)
CDP12-M (c)
CDP12-M (d)
CDP12-N 3rd ed. CDP12-M
CDP12-O
CDP12-P
CDP12-Q
12-32
CDP12-R (a) 3rd ed. CDP12-Q (a)
CDP12-R (b) 3rd ed. CDP12-Q (b)
CDP12-S
CDP12-T
CDP12-U