Environmental Engineering Reference
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so
π
2
a
n
4
T
0
a
n
n
3
n
b
n
=
T
0
θ
(
π
−
θ
)
sin
n
θ
d
θ
=
π
[
1
−
(
−
1
)
]
.
π
0
Thus
∞
n
=
1
4
T
0
a
n
n
3
n
r
n
sin
n
u
(
r
,
θ
)=
π
[
1
−
(
−
1
)
]
θ
.
Example 2
. Consider heat conduction in a sphere
Ω
of center
O
and radius
a
.Its
boundary is denoted by
. There is no heat source/sink inside the sphere. The
temperature on
S
1
and
S
2
is
T
0
(constant) and 0, respectively. Here
S
1
is the part of
∂Ω
∂Ω
inside a cone of vertex O and vertex angle 2
α
,and
S
2
is the part of
∂Ω
outside
the cone. Find the steady temperature distribution in
Ω
.
Solution.
Take the center of the sphere (also the cone vertex) as the origin of the
coordinate system. Let
u
be the temperature; it must satisfy
⎧
⎨
x
2
y
2
z
2
a
2
Δ
u
=
0
,
+
+
<
,
T
0
,
(
x
,
y
,
z
)
∈
S
1
,
(7.20)
⎩
u
=
0
,
(
x
,
y
,
z
)
∈
S
2
.
Without loss of generality, take the positive
Oz
-axis as the axis of the cone. In a
spherical coordinate system
(
r
,
θ
,
ϕ
)
,
u
is independent of
ϕ
basedonasymmetryof
the boundary conditions. Therefore,
u
satisfies
⎧
⎨
r
2
∂
sin
=
∂
∂
u
1
sin
∂
∂θ
θ
∂
u
∂θ
Δ
u
+
=
0
,
0
<
r
<
a
,
0
<
θ
<
π
,
r
∂
r
θ
T
0
,
⎩
0
≤
θ
≤
α
,
u
(
a
,
θ
)=
0
,
α
<
θ
≤
π
.
(7.21)
Assume
u
=
R
(
r
)
Θ
(
θ
)
. Substituting it into the equation in PDS (7.21) yields
r
2
R
(
)
θ
)
(
Θ
(
θ
)
r
1
sin
sin
=
−
=
λ
,
R
(
r
)
θ
Θ
(
θ
)
where
λ
is the separation constant. Thus we have
⎧
⎨
1
sin
θ
)
+
λΘ
(
θ
)=
θ
(
Θ
(
θ
)
sin
0
,
0
<
θ
<
π
,
(7.22)
⎩
|
Θ
(
0
)
| <
∞
,
θ
∈
(
0
,
π
)
.
and
r
2
R
(
2
rR
(
r
)+
r
)
−
λ
R
(
r
)=
0
,
|
R
(
0
)
| <
∞
.
(7.23)
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