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Commit 56a69ce1 authored by Tom Fischer's avatar Tom Fischer Committed by Dmitri Naumov
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[web] Corrections in heat conduction neumann docu.

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...@@ -25,32 +25,28 @@ $$ ...@@ -25,32 +25,28 @@ $$
\eqalign{ \eqalign{
T(x, t=0) = T_0,\cr T(x, t=0) = T_0,\cr
T(x) = g_D(x) &\quad \text{on }\Gamma_D,\cr T(x) = g_D(x) &\quad \text{on }\Gamma_D,\cr
k{\partial T(x) \over \partial n} = g_N(x) &\quad \text{on }\Gamma_N, \lambda {\partial T(x) \over \partial n} = g_N(x) &\quad \text{on }\Gamma_N,
} }
$$ $$
where $T$ could be temperature, the subscripts $D$ and $N$ denote the Dirichlet- and Neumann-type boundary conditions, $n$ is the normal vector pointing outside of $\Omega$, and $\Gamma = \Gamma_D \cup \Gamma_N$ and $\Gamma_D \cap \Gamma_N = \emptyset$. where $T$ could be temperature, the subscripts $D$ and $N$ denote the Dirichlet- and Neumann-type boundary conditions, $n$ is the normal vector pointing outside of $\Omega$, and $\Gamma = \Gamma_D \cup \Gamma_N$ and $\Gamma_D \cap \Gamma_N = \emptyset$.
## Problem specification and analytical solution ## Problem specification and analytical solution
We solve the Parabolic equation on a line domain $[60\times 1]$ with $k = 3.2$ and $\rho C_\textrm{p} = 2.5e10^6$ w.r.t. the specific boundary conditions: We solve the parabolic equation on a line domain $[60\times 1]$ with $\lambda = 3.2$ and $\rho C_\textrm{p} = 2.5 \times 10^6$ w.r.t. the specific boundary conditions:
$$ $$
\eqalign{ \eqalign{
k {\partial T(x) \over \partial n} = 2 &\quad \text{on } (x=0) \subset \Gamma_N.\cr \lambda {\partial T(x) \over \partial n} = 2 &\quad \text{on } (x=0) \subset \Gamma_N.\cr
k {\partial T(x) \over \partial n} = 0 &\quad \text{on } (x=60) \subset \Gamma_N. \lambda {\partial T(x) \over \partial n} = 0 &\quad \text{on } (x=60) \subset \Gamma_N.
} }
$$ $$
The solution of this problem is The solution of this problem is
TODO: is not rendered correct
$$ $$
\begin{equation} \begin{equation}
T(x,t) = \frac{2q}{\lambda}\left((\frac{\alpha t}{\pi})^{\frac{1}{2}} e^{-x^2/4\alpha t} - \frac{x}{2}\textrm{erfc}( \frac{x}{2\sqrt{\alpha t}})\right), T(x,t) = \frac{2q}{\lambda}\left(\left(\frac{\alpha t}{\pi}\right)^{\frac{1}{2}} e^{-x^2/4\alpha t} - \frac{x}{2}\textrm{erfc}\left( \frac{x}{2\sqrt{\alpha t}}\right)\right),
\end{equation} \end{equation}
$$ $$
where $T_\textrm{b}$ is the boundary temperature, $\textrm{erfc}$ is the complementary error function and $\alpha = \lambda/(C_p \rho)$ is the thermal diffusivity. where $T_\textrm{b}$ is the boundary temperature, $\textrm{erfc}$ is the complementary error function and $\alpha = \lambda/(C_p \rho)$ is the thermal diffusivity.
## Input files ## Input files
......
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