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Steady-State Thermal Analysis

313 bytes added, 19:04, 16 June 2018
where q is the heat transfer rate or heat flux density with units of W/m<sup>2</sup>, T is the temperature expressed in &deg;C or &deg;K, &nabla; is the gradient operator and k is the thermal conductivity with units of W/(m.K). It can be shown that the distribution of temperature is governed by the heat diffusion equation subject to the appropriate boundary conditions:
<math> \frac{\partial T}{\partial t} - \alpha \nabla^2 T(\mathbf{r}) = - \frac{1} {\alpha}\frac{\partial T}{\partial t} - \alpha = \left( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} \right) - \frac{1} {\alpha}\frac{\partial T}{\partial t} = 0 -w(\mathbf{r}) / k </math>
where &alpha; is the thermal diffusivity with units of W/mand w(r) is the volume heat source with units of W/m<sup>3</sup>.
In the steady-state regime, the time derivative vanishes and the diffusion equation reduces to the Laplace Poisson equation:
<math> \nabla^2 T(\mathbf{r}) = 0 -w(\mathbf{r}) / k </math>
Under the static assumptions, Maxwell's equations reduce to elliptic partial differential equations known as the Poisson and Laplace equations. These equations can be solved analytically only for a few canonical geometries with very simple boundary conditions. For most practical and realistic problems, you need to utilize a numerical technique and seek a computer solution. The Poisson and Laplace equations can be solved numerically using the finite difference (FD) method.
 
== Thermal Boundary Conditions ==
 
The simplest thermal boundary condition at the walls of the computational domain is the Dirichlet boundary condition:
 
<math> T(\mathbf{r}) = T_0 </math>
 
 
 
== Electrostatics Analysis==
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