Significance of Boundary condition based Power source

[ee13b125]

In 2-D z normal simulation, if the heat power source is defined as a part of the boundary conditions, how is the power distributed? Is the power given in Watts divided by the Norm length of the solver or is the power divided by the length of the source defined in the z-direction?

Also, how to find the temperature at a particular point or find the integrated power in a region (minus the losses). Is this possible in Devices at all??

[aalam]

In 2-D z normal simulation, if the heat power source is defined as a part of the boundary conditions, how is the power distributed? Is the power given in Watts divided by the Norm length of the solver or is the power divided by the length of the source defined in the z-direction?

For a 2D simulation the solver will assume that the solid has a length equal to the “norm length” in the third dimension and calculate the power flux to be applied based on that.

Also, how to find the temperature at a particular point or find the integrated power in a region (minus the losses). Is this possible in Devices at all??

The simplest solution would be to use a line monitor to get the temperature profile along a line. Unlike 2D or 3D data when you plot this 1D data, you can bring the mouse over a data point and it will show you the value at that data point.