Chemistry Reference
In-Depth Information
with boundary conditions
Δ
T
|
x
=0
= 0
,
(1.15)
Δ
T
|
x
→
∞
= 0
.
(1.16)
Equation (1.14) can be transformed into a linear one by using inequality Eq. (1.13)
as follows:
d
2
T
dx
2
Δ
+
M
δ
d
T
dx
−
Δ
Bi
δ
T
=
M
M
Δ
M
(
T
S
−
2
Δ
T
∞
)
δ
2
1 +
1 + 4Bi
/
M
2
M
2
1 +
1 + 4Bi
/
M
2
M
2
⎛
⎞
x
δ
⎝
−
⎠
×
exp
M
M
1 +
1 + 4Bi
/
M
2
M
2
⎡
⎤
exp
E
R
T
S
exp
Qk
0
ac
E
(
T
S
−
T
∞
)
x
δ
⎣
−
⎦
±
M
M
R
T
S
(1.17)
with boundary conditions Eqs. (1.15) and (1.16). After some rearrangement, the
solution to Eq. (1.17) is
1 +
1 + 4Bi
/
M
2
2
⎛
⎞
T
∞
)
exp
x
δ
⎝
−
⎠
Δ
T
=(
T
S
−
M
1 +
1 + 4Bi
/
M
2
M
2
⎛
⎞
x
δ
⎝
−
⎠
−
exp
M
M
Qk
0
exp(
−
E
/
RT
S
)
±
2
2
]
ac
[
β
−
(M
/
δ
)
β
−
Bi
/
δ
exp
x
)
,
M
1 +
1 + 4Bi
/
M
2
2
x
δ
×
−
−
exp(
−
β
(1.18)
where
1 +
1 + 4Bi
/
M
2
M
2
E
(
T
S
−
T
∞
)
M
M
δ
β
=
−
.
R
T
S
It can be shown that the following relationships are always valid:
M
δβ
<<
1
(1.19)
