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Semi-infinite body heated at a type 1 boundary.

Consider the temperature in semi-infinite body with a specified temperature applied to the $x=0$ boundary. The temperature satisfies the following equations:

$\displaystyle \frac{\partial^2 T}{\partial x^2}$ $\textstyle =$ $\displaystyle \frac{1}{\alpha} \frac{\partial T}{\partial t}; \; \; 0 < x < \infty$ (8)
$\displaystyle T(x=0,t)$ $\textstyle =$ $\displaystyle T_1$  
$\displaystyle T(x,t=0)$ $\textstyle =$ $\displaystyle T_0$  

The above problem has two non-homogeneous terms, however one may be eliminated to simplify the problem by normalizing the temperature. Let $T^+(x,t) = (T(x,t) - T_0)/(T_1 - T_0)$. If you replace $T^+$ into the above problem, the initial condition is zero and the boundary temperature is unity. Then this example is described by number X10B1T0. The temperature is given by the type 1 boundary term of the GF solution equation, as follows:
\begin{displaymath}
\frac{T(x,t)-T_0}{T_1 - T_0} = \alpha \int_{\tau=0}^t
\left...
...al G_{X10}}{\partial x^{\prime}} \right]_{x^{\prime}=0} d \tau
\end{displaymath} (9)

Note that here the sign of the derivative has been chosen by the outward normal according to $-\partial / \partial n^{\prime} = + \partial / \partial x^{\prime} $. The GF is given by
\begin{displaymath}
G_{X10}(x,t \mid x^{\prime},\tau ) =[4 \pi \alpha (t-\tau)]^...
...-\frac{(x+x^{\prime })^{2}}{4 \alpha (t-\tau)}\right] \right\}
\end{displaymath} (10)

The required derivative of the GF with respect to $x^{\prime}$ evaluated at $x^{\prime}=0 $ is given by


\begin{displaymath}
\left[ \frac{\partial G_{X10}}{\partial x^{\prime}} \right]_...
...]^{3/2} }
\exp \left[ \frac{-x^{2}}{4 \alpha (t-\tau)}\right]
\end{displaymath} (11)

Replace the above GF derivative into the GF solution equation to find the temperature; the integral may be evaluated as an error function.


\begin{displaymath}
\frac{T(x,t)-T_0}{T_1 - T_0} = \left[ 1- erf \left( \frac{x}{(4 \alpha t)^{1/2}} \right) \right ]
\end{displaymath} (12)

or as the complementary error function
\begin{displaymath}
\frac{T(x,t)-T_0}{T_1 - T_0} = erfc \left( \frac{x}{(4 \alpha t)^{1/2}} \right)
\end{displaymath} (13)

Alternately, this problem could have been solved as an initial condition problem if the temperature were normalized as $(T(x,t) - T_1)/(T_0 - T_1)$.


next up previous
Next: 1D Slab heated at Up: EXAMPLES, TEMPERATURE FROM GF Previous: 1D infinite body with
2004-01-31