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Figure 19.4.
Heat conduction along a rod.
flux q 0 at one end, a constant temperature at the other, and no heat loss in between.
An example of this is an insulated wire. We wish to determine the temperature T(x)
along the rod. The well-known differential equation that describes this situation is
2
K dT
dx
-
=
Q
for
0
<
x
<
L
,
(19.6)
2
K dT
dx
-
=
q
for
x
=
0,
and
(19.7)
0
T
=
T
for
x
=
L
,
(19.8)
L
where K is the material's thermal conductivity, Q is the heat generation in the rod per
unit volume, and
K dT
dx
() =-
qx
is the heat flux. We shall assume that K and Q are constant. For reasons we shall see
shortly, equation (19.6), along with the boundary conditions (19.7) and (19.8), is called
the strong form of the one-dimensional heat flow equation . It is easy to show that the
exact solution to this equation is
q
K
Q
K
(
)
0
() =+
(
) +
2
2
Tx
T
Lx
-
Lx
-
.
(19.9)
L
2
In general of course, one would not have an exact solution, so that having the solu-
tion (19.9) is not important. However, one is always interested in the accuracy of
approximations and so it is worthwhile comparing the solution we get using the FEM
to the one in (19.9).
Applying the Galerkin weighted residual method to (19.6) we get
2
K dT
dx
Ê
Á
ˆ
˜
L
Ú 0
() -
wx
-
Qdx
=
0.
(19.10)
2
The only problem now is the high degree of differentiability that using (19.10) would
require of any approximation. For example, dividing the domain [0, L] into elements
and looking for a solution of the form (19.3) with linear approximations over each
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