[aspect-devel] Question regarding heat flux output

[John Naliboff]

Hi Wolfgang,

Thank you for the reply and yes, I agree with everything below in terms of the units.

In order to prevent any confusion from future users reading the output, I would do one of two things:
1) Change the output value to the actual heat flux W/m (2D) or W/m2 (3D)
or 
2) Change the description of the output values from "heat flux" to "total heat output across boundary" (or something along those lines).

I think option 1) would be preferable as I imagine most users would want the heat flux.

Does anyone else have an opinion or preference on this?

Cheers,
John




On Jun 26, 2012, at 2:25 AM, Wolfgang Bangerth wrote:

> 
>> 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?
> 
> This formula is the numerical approximation to the integral
> 
>   \int_{\Gamma_i}   - k  nabla T . n  dS
> 
> that Timo referenced. Here,  n=normal_vector(q)  at quadrature point q, 
> has no physical units, and  JxW(q)=area element dS has units meters in 
> 2d and meters^2 in 3d.  Gamma_i is the part of the boundary of the 
> domain with boundary indicator i.
> 
> k, the thermal conductivity, is documented as having units W/m/K, so in 
> 3d I indeed get units  W  for the entire expression. (Timo: The 
> derivative has units K/m, not K/s.) That is exactly the (integrated) 
> heat flux through the surface we were looking for.
> 
> John: Do you agree? This may of course not be what you need from a 
> simulation -- if so, let us know what it is that you want and we can see 
> how to implement it.
> 
> Cheers
>  W.
> 
> PS: The units of the conductivity are
>   heat flux per unit area perpendicular to the flux direction
>   - per -
>   unit gradient in temperature
> In 3d, that is  (W/m^2) / (K/m)  =  W/(K m) as documented. In 2d, on the 
> other hand, it is  (W/m) / (K/m)  =  W/K. In other words, in 2d, the 
> units of the above formula would be
>   W/K  *  K/m  *  1  *  m
> which is again just Watts. I've fixed the description of the units in 
> the documentation to this end.
> 
> ------------------------------------------------------------------------
> Wolfgang Bangerth               email:            bangerth at math.tamu.edu
>                                 www: http://www.math.tamu.edu/~bangerth/
> 
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