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A long cylinder is initially at zero temperature and the boundary at
is maintained at zero temperature. Find the temperature in the cylinder
resulting from spatially uniform internal energy generation,
(W/m
).
The temperature satisfies the following equations:
This case is number R01B0T0G1. The GF solution equation for the temperature
contains only the internal heating term:
 |
(19) |
The large-time form of the GF for this case is given by
where
and
are Bessel functions and the eigenvalues given by
.
The two integrals in the above solution will be considered one at a time.
The spatial integral over
acts on one
term, and it may
be simplified by the substitution
and then
evaluated as shown below:
The time integral acts only on the exponential term of the series for the GF,
given by
![\begin{displaymath}
\int_{\tau=0}^t e^{-\beta _{m}^{2} \alpha (t-\tau )/b^2} \; ...
...= \frac{b^2}{\beta_m^2 \alpha}[1- e^{-\beta_m^2 \alpha t/b^2}]
\end{displaymath}](img137.png) |
(22) |
Then the above two integrals can be combined with the entire GF solution
to give:
![\begin{displaymath}
T(r,t)=2\frac{g_0b^2}{k} \sum_{m=1}^{\infty}
[1- e^{-\beta_m...
...t/b^2}] \frac{J_{0}( \beta_{m}r/b) }
{\beta_m^3 J_1(\beta_m) }
\end{displaymath}](img138.png) |
(23) |
For numerical evaluation, the steady term should be computed separately and
substituted for the slowly converging series term (the term without the
exponential).
When the steady-state portion of the solution is substituted (see below),
the temperature is given by:
![\begin{displaymath}
T(r,t)= \frac{g_0b^2}{4k} \left[ 1-(r/b)^2 \right] - 2\frac{...
... t/L^2} \frac{J_0 ( \beta _{m}r/b) }
{\beta_m^3 J_1(\beta_m) }
\end{displaymath}](img139.png) |
(24) |
This form of the solution converges rapidly for
.
Steady Solution. Next the steady solution for the solid sphere
with internal heating will be derived. The steady temperature satisfies
the following differential equation:
The general solution may be found by integrating the differential equation twice:
The constants of integration
and
may be determined by applying the
boundary condtions. For
to be bounded at
requires
. At
the boundary condition
then determines
. Then the steady temperature
is given by
Next: Solid sphere with convection.
Up: EXAMPLES, TEMPERATURE FROM GF
Previous: 1D Slab heated at
2004-01-31