Skip to content

2-D Transient Heat Conduction – Part 2

November 21, 2013

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).


From → GPU & GPGPU, Project

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: