# How to define a report to evaluate different "impacts" on temperature in a point?

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 May 15, 2020, 11:43 How to define a report to evaluate different "impacts" on temperature in a point? #1 New Member   Josh Join Date: May 2019 Posts: 16 Rep Power: 5 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!

 May 15, 2020, 12:54 Incident Radiation #2 Senior Member     Vinerm Join Date: Jun 2009 Location: Nederland Posts: 2,946 Blog Entries: 1 Rep Power: 34 You can monitor incident radiation and then calculate temperature due to that using . berkmm likes this. __________________ Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority.

May 18, 2020, 02:58
#3
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Josh
Join Date: May 2019
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Quote:
 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?

 May 18, 2020, 04:23 Equation #4 Senior Member     Vinerm Join Date: Jun 2009 Location: Nederland Posts: 2,946 Blog Entries: 1 Rep Power: 34 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. berkmm likes this. __________________ Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority.

May 18, 2020, 04:35
#5
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Josh
Join Date: May 2019
Posts: 16
Rep Power: 5
Quote:
 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

 May 18, 2020, 04:44 Temperature #6 Senior Member     Vinerm Join Date: Jun 2009 Location: Nederland Posts: 2,946 Blog Entries: 1 Rep Power: 34 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. berkmm likes this. __________________ Regards, Vinerm PM to be used if and only if you do not want something to be shared publicly. PM is considered to be of the least priority.