Environmental Engineering Reference
In-Depth Information
Solution.
It is straightforward to show that the Poisson equation in PDS (7.52) has
a particular solution
1
6
(
u
∗
=
x
2
y
2
z
2
+
+
)
.
Consider the function transformation
u
∗
.
u
(
x
,
y
,
z
)=
v
(
x
,
y
,
z
)+
The PDS (7.52) is thus transformed into
⎧
⎨
⎩
x
2
y
2
z
2
v
xx
+
v
yy
+
v
zz
=
0
,
+
+
<
1
,
53
6
.
v
|
x
2
+
y
2
+
z
2
=
1
=
Its solution is, by the extremum principle,
53
6
,
x
2
y
2
z
2
v
=
+
+
≤
1
.
Thus the solution of PDS (7.52) is
x
2
y
2
z
2
53
6
+
+
+
u
=
.
6
7.3.3 Four Examples of Applications
Example 1.
Consider two-dimensional heat conduction in a rectangular plate. Inside
the plate, there exists a uniform heat source of strength
q
. The heat can flow out
through the central part of one side of the plate boundary. All the other parts of plate
boundary are well insulated. Find the steady temperature distribution in the plate.
Solution. (1) Coordinate system
: Consider the Cartesian coordinate system in
Fig. 7.1. Let 2
a
and
b
be the length and the width of plate. 2
c
is the length of
the part of the plate side where heat can be conducted to the outside.
(2) PDS
:Let
k
be the thermal conductivity of the plate material. The steady temper-
ature
T
(
x
,
y
,
t
)
at point
(
x
,
y
)
satisfies
T
xx
+
T
yy
=
−
q
/
k
.
The heat generated by the heat source inside the plate per unit of time is 2
abq
by the
definition of
q
. The heat conducted to the outside via the boundary is
y
=
b
2
kc
∂
T
−
∂
y
per unit of time. At a steady state, we have
y
=
b
=
y
=
b
=
−
2
kc
∂
T
∂
T
∂
abq
kc
, |
−
2
abq
or
x
| <
c
∂
y
y
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