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What's the differences between the external emissivity and the internal emissivity

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Old   February 2, 2010, 07:17
Question What's the differences between the external emissivity and the internal emissivity
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Hello,I'm faced with a problem when simulating the radiation of heat transfer. I don't know how to define the emissivity coz there are two different emissivities in DO model in FLUENT. I also have looked to the FLUENT help documents but I found nothing.
Could anybody explain to me?
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Old   January 7, 2013, 23:49
Default Same situation
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Hi dude, I am facing similar situation. What is external emissivity and internal emissivity?
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Old   January 8, 2013, 01:08
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each wall or plate has two faces. internal means the face that is connected to the adjacent zone and external is the outer face.
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Old   January 8, 2013, 01:12
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Ok. emissivity of a surface depends on the material of the surface right? Or does it also depend on the environement to which the surface is exposed? If the wall is made of one material then shouldn't the internal and external emissivity be the same?

Thanks a lot for sharing your opinion.

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Old   August 16, 2017, 07:12
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External emissivity: When you have an external radiation source (outside the computational domain) the flux coming in to the fluid zone is calculated using the formula Q(ext)=Emissivity(external)*Boltzmann constant*Temperature^4. This is where you need the external emissivity (found under the thermal tab when radiation thermal condition is chosen). But, to make the radiation affect the fluid zone beside the external source, the wall (the source of radiation) should be opted as "semi-transparent" under the "radiation" tab.
Internal emissivity: You find this option in the "radiation" tab only if the "BC type" chosen is "opaque". This is the emission that comes from a wall on which the radiation falls and is absorbed and a fraction of it is emitted back. However, if you have a "semi-transparent" "BC condition", Fluent does not consider any emission from it unless a specified temperature boundary condition is specified.
Please find the link below for a better idea:
https://www.sharcnet.ca/Software/Flu...ug/node580.htm
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Old   November 15, 2017, 08:25
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Quote:
Originally Posted by Nimmy Kovoor View Post
External emissivity: When you have an external radiation source (outside the computational domain) the flux coming in to the fluid zone is calculated using the formula Q(ext)=Emissivity(external)*Boltzmann constant*Temperature^4. This is where you need the external emissivity (found under the thermal tab when radiation thermal condition is chosen). But, to make the radiation affect the fluid zone beside the external source, the wall (the source of radiation) should be opted as "semi-transparent" under the "radiation" tab.
Internal emissivity: You find this option in the "radiation" tab only if the "BC type" chosen is "opaque". This is the emission that comes from a wall on which the radiation falls and is absorbed and a fraction of it is emitted back. However, if you have a "semi-transparent" "BC condition", Fluent does not consider any emission from it unless a specified temperature boundary condition is specified.
Please find the link below for a better idea:
https://www.sharcnet.ca/Software/Flu...ug/node580.htm
Sorry, but the link above has a wrong formula under the section "Boundary Condition Treatment at Opaque Walls" The two formulae that are wrong are the specularly reflected energy and the absorped energy. So, for the specularly reflected energy, the formula in this page must be multiplied by the (1-emissivity) . Secondly the absorption at the surface wall must not be multiplied by the diffuse fraction. These are the formulae from the 18.2 theory guide.
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Old   November 22, 2017, 23:54
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I am simulating heat treatment furnace, in which non-premixed combustion is taking place. In real conditions the furnace max temperature is 1100K. When I simulate the same in fluent, max temp of 1900K was observed with P1 radiation model. I am heating a metal in the furnace. That i have modelled in fluent using wall BC of "mixed" (shell conduction, convection and radiation).

How can i get temperatures close to real conditions, any help.
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Old   February 25, 2019, 11:06
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Quote:
Originally Posted by Nimmy Kovoor View Post
External emissivity: When you have an external radiation source (outside the computational domain) the flux coming in to the fluid zone is calculated using the formula Q(ext)=Emissivity(external)*Boltzmann constant*Temperature^4. This is where you need the external emissivity (found under the thermal tab when radiation thermal condition is chosen). But, to make the radiation affect the fluid zone beside the external source, the wall (the source of radiation) should be opted as "semi-transparent" under the "radiation" tab.
Internal emissivity: You find this option in the "radiation" tab only if the "BC type" chosen is "opaque". This is the emission that comes from a wall on which the radiation falls and is absorbed and a fraction of it is emitted back. However, if you have a "semi-transparent" "BC condition", Fluent does not consider any emission from it unless a specified temperature boundary condition is specified.
Please find the link below for a better idea:
https://www.sharcnet.ca/Software/Flu...ug/node580.htm
Hi Nimmy, the walls in my model have a radiation thermal BC, it radiates to the inside of the fluid domain, so how should I set the internal and external emissivity?

Thanks in advance for your help
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Old   November 5, 2019, 13:04
Default modeling radiattion without using model in Fluent
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hi everyone, i'm modelling Species transport and Gaseous combustion in fluent. i have finished simulating with the boundary condition of outer-wall is: "temperature: 300K" and no use any condition of radiation, and the max temperature of the result is 2309.21K
after that, I was changing the boundary condition of outer-wall is :"radiation" with external emissivity=1 and temp=300K but no using radiation model. The contour of this time result seems right, however, the max temperature of this time is 2312.13K. this temp larger than the temp with no use radiation boundary condition. i think it is unphysical
So i want to know am i did right and what and where is the problem?
i really appreciate with your help!!!

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Old   November 5, 2019, 14:13
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Quote:
Originally Posted by maiminh0110 View Post
hi everyone, i'm modelling Species transport and Gaseous combustion in fluent. i have finished simulating with the boundary condition of outer-wall is: "temperature: 300K" and no use any condition of radiation, and the max temperature of the result is 2309.21K
after that, I was changing the boundary condition of outer-wall is :"radiation" with external emissivity=1 and temp=300K but no using radiation model. The contour of this time result seems right, however, the max temperature of this time is 2312.13K. this temp larger than the temp with no use radiation boundary condition. i think it is unphysical
So i want to know am i did right and what and where is the problem?

It completely makes sense physically. You had a wall temperature of 300K. Now you changed it to a radiation condition, also with a temperature reservoir of 300 K but now the radiation is a thermal resistance, so less heat will leave the domain than before. The max temperature will increase.
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Old   November 5, 2019, 14:26
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Originally Posted by LuckyTran View Post
It completely makes sense physically. You had a wall temperature of 300K. Now you changed it to a radiation condition, also with a temperature reservoir of 300 K but now the radiation is a thermal resistance, so less heat will leave the domain than before. The max temperature will increase.
but in the boundary condition of outer-wall i chose radiation at thermal tab and the external emissivity = 1 (default) so i think Qloss was maximum and the peak of temp has to decrease because it is external emissivity , not internal emissivity
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Old   November 5, 2019, 15:05
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Originally Posted by maiminh0110 View Post
but in the boundary condition of outer-wall i chose radiation at thermal tab and the external emissivity = 1 (default) so i think Qloss was maximum and the peak of temp has to decrease because it is external emissivity , not internal emissivity

Why would that be the maximum heat loss? You have a certain mind set, recalibrate it. Your thinking is emissivity is 1, that should be a blackbody, and that is the maximum heat transfer as far as radiation is concerned. But boundary conditions is something else.

The cool thing about doing CFD is you can actually check to see if you are right. Check your outer wall temps and compare the two. And check the heat flux (i.e. the heat loss) and compare the two.

Think about what it means for a wall to be a fixed temperature. What needs to take place for the wall to be 300K no matter what?
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Old   November 5, 2019, 23:43
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Why would that be the maximum heat loss? You have a certain mind set, recalibrate it. Your thinking is emissivity is 1, that should be a blackbody, and that is the maximum heat transfer as far as radiation is concerned. But boundary conditions is something else.

The cool thing about doing CFD is you can actually check to see if you are right. Check your outer wall temps and compare the two. And check the heat flux (i.e. the heat loss) and compare the two.

Think about what it means for a wall to be a fixed temperature. What needs to take place for the wall to be 300K no matter what?
I have a little bit confused about emissivity and blackbody like you said, and are u vietnamese person? can u explain to me by vietnamese or if you have any page, link, or document about radiation (blackbody), can u send and post link for me? i really appriciate that
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Old   November 6, 2019, 11:44
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Simple, there's no reason that radiation to to a 300 K reservoir with an emissivity of 1 is the maximum heat loss. Actually, you can use a heat flux boundary condition and just set it to a bigger number and get more heat loss.


We only use English on these forums because there's no practical way to moderate other languages.
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Old   November 6, 2019, 12:21
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Originally Posted by LuckyTran View Post
Simple, there's no reason that radiation to to a 300 K reservoir with an emissivity of 1 is the maximum heat loss. Actually, you can use a heat flux boundary condition and just set it to a bigger number and get more heat loss.


We only use English on these forums because there's no practical way to moderate other languages.
Thank you so much, i'll modeling with other boundary condition to have more better result
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