Environmental Engineering Reference
In-Depth Information
Fig. 7.1
Physical problem and coordinate system
Finally, we obtain the PDS
⎧
⎨
q
k
+
=
−
, −
<
<
,
<
<
,
T
xx
T
yy
a
x
a
0
y
b
T
y
y
=
0
=
|
x
=
±
a
=
,
T
x
0
−
(7.53)
⎩
/
, |
| <
,
T
y
y
=
b
=
abq
kc
x
c
(
)=
f
x
0
,
|
x
|≥
c
.
(3) Solution of PDS (7.53)
: It is clear that the Poisson equation in PDS (7.53) has a
particular solution
qy
2
2
k
.
T
∗
=
−
T
∗
, the PDS (7.53)is transformed
By the function transformation
T
(
x
,
y
)=
v
(
x
,
y
)+
into
⎧
⎨
v
xx
+
v
yy
=
0
,−
a
<
x
<
a
,
0
<
y
<
b
v
y
y
=
0
=
v
x
|
x
=
±
a
=
0
(7.54)
⎩
v
y
y
=
b
=
g
(
x
)
.
where
bq
/
k
−
abq
/
kc
, |
x
| <
c
,
g
(
x
)=
bq
/
k
,
|
x
|≥
c
.
Note that the problem is symmetric with respect to the
Oy
-axis, so we can focus our
attention on the region: 0
≤
x
≤
a
,
0
≤
y
≤
b
by replacing the boundary conditions
0. Based on the given boundary conditions in
the
x
-direction, we use the eigenfunction set from Row 5 in Table 2.1 to expand
v
v
x
|
x
=
±
a
=
0 with
v
x
|
x
=
0
=
v
x
|
x
=
a
=
(
x
,
y
)
∞
m
=
0
Y
m
(
y
)
cos
m
π
x
v
(
x
,
y
)=
a
.
(7.55)
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