[Helium]

Hi guys,

I have to measure the temperature in a point between two semi-transparent walls (glasses). The problem is that air (which is heated in a previous part of the model) passes through this point too, so I cannot say who impacts most at the vertex average value of T in this point.
For example, let's say that in this point I measure 40 °C and let's say that 35°C are due to convection and 5°C are due to external radiation coming from the wall, so in this case I know that radiation contributed in "raising" only 5°C in this point.
Does Fluent have a monitor like this? Can Fluent evaluate contributions separately? Or can I create anything like this?

Gracias!

[vinerm]

You can monitor incident radiation and then calculate temperature due to that using .

[Helium]

Originally Posted by vinerm

You can monitor incident radiation and then calculate temperature due to that using .

Hi vinerm!
In that point I have an incident radiation of 3904.5 w/m2.
So, if I understood correctly, I should try to:

3904.5 = 1 * 5.674 * 10^-4 * T^4

where emissivity = 1 because I am between 2 glasses and nothing emits.
So the value for T that I get (the positive one) is due only to radiation? Is this correct?

[vinerm]

Emissivity of 1 implies blackbody. For gray or real bodies, value is always less than 1. is 5.67e-8 and not 5.67e-4. And this will return absolute temperature.

[Helium]

Originally Posted by vinerm

Emissivity of 1 implies blackbody. For gray or real bodies, value is always less than 1. is 5.67e-8 and not 5.67e-4. And this will return absolute temperature.

Regarding the emissivity: the point does not lie on a surface, it's just placed in between two surfaces. I try to explain better:

wall 1 ------ point ------ wall 2

And the space between the walls is occupied by air. The point is not something real, I just place it in order to see the value of T in order to confront it with the value of T in points on wall 1 and wall 2.

Plus, with this method, I do not seem to have a reliable number because the value of T is much higher if compared to the value of T over the surfaces of both walls.

Thanks

[vinerm]

Doesn't matter where the point is, the emissivity is always between 0 and 1. And in reality, while determining temperature from incident radiation, it is the absorptivity that is to be used and not emissivity, however, for most of the materials, both are assumed equal. If it is dry air, the absorptivity is very low. But 3.9 kW/sq.m. is a high flux and is supposed to give a high temperature.