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1D Slab heated at a type 2 boundary.

The temperature in a 1D slab with a short-duration heat flux on one face and with a fixed temperature on the other face is discussed in this example. The initial temperature is zero and there is no internal heat generation. The boundary value problem for this case is given by:

$\displaystyle \frac{\partial^2 T}{\partial x^2}$ $\textstyle =$ $\displaystyle \frac{1}{\alpha} \frac{\partial T}{\partial t}; \; \; 0 < x < L$ (14)
$\displaystyle -k\frac{\partial T}{\partial x} \mid_{x^{\prime}=0}$ $\textstyle =$ $\displaystyle \left\{
\begin{array}{cc}
q_0; & 0 < t < t_1 \\
0; & t>t_1 \end{array}\right.$  
$\displaystyle T(x=L,t)$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle T(x,t=0)$ $\textstyle =$ $\displaystyle 0$  

This example is case X21B(t5)0T0. The method of GF is well suited to piecewise continuous heating. The temperature is found from the ``type 2'' boundary term of the GF solution equation with the heat flux substituted in the form $q(\tau)$:
\begin{displaymath}
T(x,t) = \left\{
\begin{array}{cc}
\frac{\alpha}{k} \int_{\...
... \int_{\tau=t_1}^{t} 0 \; d \tau; & t>t_1 \end{array} \right.
\end{displaymath} (15)

The integral for $t>t_1$ has been written in two pieces to emphasize the piecewise heating function. There are two forms of the GF for this geometry, and the one used here is best for large values of $(t-\tau)$. The large-time GF is given by
$\displaystyle G_{X21}(x,t \mid x^{\prime},\tau)$ $\textstyle =$ $\displaystyle \frac{2}{L}%
\sum_{m=1}^{\infty }\exp \left[ -\frac{\beta _{m}^{2...
...2}}%
\right] \cos (\beta _{m}\frac{x}{L})\cos (\beta _{m}\frac{x^{\prime }}{L})$ (16)
$\displaystyle \mbox{ where }\beta _{m}$ $\textstyle =$ $\displaystyle (2m-1)(\frac{\pi }{2})\mbox{, for }m=1,2,\ldots .$ (17)

The time integral in the GF solution may be distributed over the series term by term, and the integral falls only on the exponential portion of each term. When the GF is substituted into the temperature expression and evaluated, the result is:
$\displaystyle \mbox{For } 0<t<t_1:$      
$\displaystyle T(x,t)$ $\textstyle =$ $\displaystyle 2\frac{q_0L}{k} \sum_{m=1}^\infty \cos(\beta_m x/L)
\frac{1}{\beta_m^2}
[1- e^{-\beta_m^2 \alpha t/L^2}]$  
$\displaystyle \mbox{For } t>t_1:$      
$\displaystyle T(x,t)$ $\textstyle =$ $\displaystyle 2\frac{q_0L}{k} \sum_{m=1}^\infty \cos(\beta_m x/L)
\frac{1}{\beta_m^2}
[e^{-\beta_m^2 \alpha (t-t_1)/L^2}- e^{-\beta_m^2 \alpha t/L^2}]$  

Although the temperature is caused by a discontinous heating history, the temperature is continous and smoothly varying in space and time. At large time the introduced heat leaves the body and the steady-state temperature is zero.


next up previous
Next: Solid cylinder with internal Up: EXAMPLES, TEMPERATURE FROM GF Previous: Semi-infinite body heated at
2004-01-31