[aspect-devel] Question regarding heat flux output

[John Naliboff]

Hi Timo,

Thanks for the quick reply! From heat_flux_stastics.cc (lines 96-103) there is

local_normal_flux
+=
  -thermal_conductivity *
  (temperature_gradients[q] *
  fe_face_values.normal_vector(q)) *
  fe_face_values.JxW(q);

Is this the equation you are referencing below?  If it is, I'm not sure what normal_vector(q) and JxW refer to.  Is normal_vector(q) a directional multiplier?

If the only dimensional values in the equation above are thermal conductivity (W/(K*m)) and temperature (K), we would get the correct units (W/m).  The listed output value of Watts, though, would have to be the integrated heat flux over each boundary.

John


On Jun 22, 2012, at 7:09 PM, Timo Heister wrote:

> Hi John,
> 
>> Above, the heat flux through the boundary parts is listed in terms of Watts
>> (I assume "W" is for watts).  Are these values in fact the heat flux values
>> integrated over the length of each boundary?
> 
> Yes, for each boundary indicator we compute the heat flux through that
> part of the surface and W stands for Watts.
> 
> Let us check if the units are correct:
> The formula used is  \int thermal_conductivity * (\nabla T) * normal * dS
> Thermal conductivity has units W / (K*m), the derivative has K/s, [dS]
> = m^(dim-1).
> 
> @Wolfgang: why is the answer here in Watt? I get W/s or W * m/s with
> the formula above.
> 
> -- 
> Timo Heister
> http://www.math.tamu.edu/~heister/
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> Aspect-devel at geodynamics.org
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