In this previous post, I have demonstrated the mathematical concept behind a Gauss-Seidel iterative solution to the Laplace Equation, i.e. the equation below:

$\frac{\partial T}{\partial t} = \alpha\nabla^2T.$

For a numerical computation, this differential equation has been discretized by the Finite Difference Scheme to obtain an approximate representation:

$\frac{\Delta T}{\Delta t} = \alpha(\frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{\Delta x^2} + \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{\Delta y^2}).$

Now coming to HOW we obtain a value of   $\alpha_{factor}.$ = 0.2.

Consider Aluminium as your 2-D sheet material. As we know, $\alpha_{factor} = \frac{\alpha.\Delta t}{\Delta x^2}$. For Aluminium, the Thermal Diffusivity is roughly 84.18 mm2/s. Next, we chose total number of time steps as 30,000. So taking one time step as 0.01 seconds, we are allowing 300 seconds to reach steady state (sounds comfortable). Our 2-D square metal sheet has 512 grid nodes in each direction. Since  $\Delta x = \Delta y$, we know that $511\times\Delta x = l$ (where $l$ is the side length of the square). Since we chose   $\alpha_{factor} = 0.2$ , we can now calculate that   $\Delta x = 2.052 mm$, i.e. roughly 2 mm.

That means we are approximating a grid length of $511\times 2 = 1022 mm$ , which seems perfectly fine (a grid size of 1×1 m2).