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Example 4: Two-dimensional convection and diffusion

This example demonstrates the effect of the spatial stabilization algorithm in 2D. A convection and diffusion of heat equation is analyzed on a 1 by 1 square. The two-dimensional mesh consists of distorted linear quadrilaterals

\begin{figure}
\centerline{\epsfig{file=ps/ex4mesh.ps,width=4cm}}\end{figure}

The convection velocity $\beta=1$ and the conductivity k=0.01. The boundary conditions for temperature are chosen such that the exact solution for a boundary layer in y-direction holds:

\begin{displaymath}
T(y) = T(y=0) +
(T(y=1)-T(y=0)) \frac{1- \exp (\beta\frac{y}{k})}{1-exp(\frac{\beta}{k})}
\end{displaymath}

where we choose T(y=0)=1 and T(y=1)=0. This is a severe test for the spatial stabilization algorithm. Many algorithms exist which solve this example exactly when using a one-dimensional domain, say with y-axis only, but few exist which do not show wiggles for irregular 2D grids. The node_dof records are initialized with temperature 1 as a first estimate for the solution field. The we check the results at x=0.7 and y=0.6. The exact solution is 1. The numerical solution with the 4-noded elements is 0.95. Splitting the elements in triangles (see control_mesh_split) would have given the solution 1.001. Triangles seem to behave better than distorted quads (in this example anyway). Both solutions are quite good however.


next up previous contents
Next: Example 5: Inverse modeling Up: Examples Previous: Example 3: Plasticity in   Contents
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1998-11-16