[Nadaaaa22]

I’m working in a heat sink similar to the attached one, I applied a heat source at the bottom and I added the fluid domain for the heat sink as attached. It is a natural convection flow. The issue here that I don’t know the way of getting the heat transfer coefficient. If I use the expression (Q/deltaT) which Q can I use and from where I can get it. Can someone please help me with this?.

[Nadaaaa22]

There is no one?.

[ashokac7]

Originally Posted by Nadaaaa22

I’m working in a heat sink similar to the attached one, I applied a heat source at the bottom and I added the fluid domain for the heat sink as attached. It is a natural convection flow. The issue here that I don’t know the way of getting the heat transfer coefficient. If I use the expression (Q/deltaT) which Q can I use and from where I can get it. Can someone please help me with this?.

There are few things first.

Is all the heat produced by heat source is dissipated to an atmosphere by natural convection?
Do you know the equilibrium surface temperature at which it is dissipating heat to environment.
( HTC is constant for pair of surfaces and fluid flow(constant) over it, So temperature of surface will increase at first when heat starts to dissipate and then it will achieve a equilibrium deltaT value such that whatever heat is being produced will be dissipated because of that temperature difference )



Then you can use Q/deltaT for htc. Q is heat flux here.


Please do keep in mind that you are considering radiation heat transfer into convection heat transfer here. So you will be predicting higher HTC in this case. Which is okay because you are dissipating whatever heat is produce anyway.



If you consider radiation too as a heat transfer media, you will get lower HTC value. This HTC is often termed as convective HTC. If you don't want to go into radiation, Convection is sufficient and will take care of heat transfer by radiation as well.


Let me know if you have some doubt in this. Do refer Wikipedia first for natural convection and radiation article.


Hope this helps.

[Nadaaaa22]

Originally Posted by ashokac7

There are few things first.

Is all the heat produced by heat source is dissipated to an atmosphere by natural convection?
Do you know the equilibrium surface temperature at which it is dissipating heat to environment.
( HTC is constant for pair of surfaces and fluid flow(constant) over it, So temperature of surface will increase at first when heat starts to dissipate and then it will achieve a equilibrium deltaT value such that whatever heat is being produced will be dissipated because of that temperature difference )



Then you can use Q/deltaT for htc. Q is heat flux here.


Please do keep in mind that you are considering radiation heat transfer into convection heat transfer here. So you will be predicting higher HTC in this case. Which is okay because you are dissipating whatever heat is produce anyway.



If you consider radiation too as a heat transfer media, you will get lower HTC value. This HTC is often termed as convective HTC. If you don't want to go into radiation, Convection is sufficient and will take care of heat transfer by radiation as well.


Let me know if you have some doubt in this. Do refer Wikipedia first for natural convection and radiation article.


Hope this helps.

Thank you, for the good explanation, the heat source applied it is dissipated by natural convection and I have radiation as well+ I considered the solid so there is a conduction as well. The issue I didn’t get what do you mean by if i’m not considering the radiation the natural convection will take care of it. The heat flux do I consider it for the surface + the bottom of the heat sink(where heat source is applied). Or just for the surface without the bottom of the heat sink since I don’t have natural convection at the bottom of the heat sink.

[ashokac7]

Originally Posted by Nadaaaa22

Thank you, for the good explanation, the heat source applied it is dissipated by natural convection and I have radiation as well+ I considered the solid so there is a conduction as well. The issue I didn’t get what do you mean by if i’m not considering the radiation the natural convection will take care of it. The heat flux do I consider it for the surface + the bottom of the heat sink(where heat source is applied). Or just for the surface without the bottom of the heat sink since I don’t have natural convection at the bottom of the heat sink.

Generally we don't know how much heat is transferred because of convection alone. So normal method to calculate HTC is from Nusselt's number. For Nusselt's number calculation, we need to calculate Rayleigh's and Grashoff's number. Refer some document on internet for it.
But here, as you are considering all the heat is dissipate by convection itself, HTC will be more than convective HTC from Nusselt's number.

Depending on what and where you want to use these you can choose how you want to move forward. If you are interested in Heat transfer by radiation as well, then you are better with calculation of HTC from Nusselt's number.
Just for a rough idea, to see the ratio of convection vs radiation, calculate heat flux by radiation as well as heat flux by convection. You can consider HTC around 10-20 for this. As for natural convection, you will be getting htc value near to 10-20.


Hope this helps.

[Nadaaaa22]

Originally Posted by ashokac7

Generally we don't know how much heat is transferred because of convection alone. So normal method to calculate HTC is from Nusselt's number. For Nusselt's number calculation, we need to calculate Rayleigh's and Grashoff's number. Refer some document on internet for it.
But here, as you are considering all the heat is dissipate by convection itself, HTC will be more than convective HTC from Nusselt's number.

Depending on what and where you want to use these you can choose how you want to move forward. If you are interested in Heat transfer by radiation as well, then you are better with calculation of HTC from Nusselt's number.
Just for a rough idea, to see the ratio of convection vs radiation, calculate heat flux by radiation as well as heat flux by convection. You can consider HTC around 10-20 for this. As for natural convection, you will be getting htc value near to 10-20.


Hope this helps.

Thank you very much I have a question as well if I remove the radiation from the model, the heat will be transferred will be only by convection so can I use the heat flux in this case?.because in the Nusselt number equation I will need the characteristic length which will make it harder to get.

[ashokac7]

Let's take it one by one.


1) If you don't consider radiation


Then you will get higher surface temperature in calculation than in actual practical. Because HTC don't depend on temperatures, but as heat by radiation is dissipated by convection now, deltaT needs to be higher. So calculation (or simulation) temps are higher than practical.


2) If you consider radiation as well.


Now your practical and calculation (or simulation) temps will be near to each. This will be more accurate approach if you can calculate (or simulate) the case with radiation as well.

[ashokac7]

Empirical relation for pin-fin are difficult but are already formulated.

Search for Yoav Peles naturak convection on internet.

Refer to below Links for Natural convection empirical relations.

Yoav Peles has lots of lectures and papers on this.
https://slideplayer.com/slide/9286917/

This is one of the lecture on natural convection.
These are some documentation form his lectures.

yoav peles 1.jpg
yoav peles 2.jpg

Do refer this before going foreword.

Hope this helps.

[Nadaaaa22]

Originally Posted by ashokac7

Empirical relation for pin-fin are difficult but are already formulated.

Search for Yoav Peles naturak convection on internet.

Refer to below Links for Natural convection empirical relations.

Yoav Peles has lots of lectures and papers on this.
https://slideplayer.com/slide/9286917/

This is one of the lecture on natural convection.
These are some documentation form his lectures.

Attachment 71456
Attachment 71457

Do refer this before going foreword.

Hope this helps.

Thank you, I went through the theory of natural convection but I think i’m not able to relate anything I got confused now I’m working in the model of star-ccm+ I did two models one with the radiation and the other one without the radiation and as you said I got lower temperature when I added the radiation, then I used the surface average to get the heat transfer coefficient and here where I got stuck, I found out that there is a radiation from the bottom of the heat sink where I applied the heat source (in the model of no radiation) even if there is no radiation in the model. I found out that I have a heat transfer as well although I don’t have a fluid domain at the bottom of the heat sink I just added the fluid domain above the bottom of the heat sink to consider it insulted, I got lost to be honest. Sorry for bothering and. Thank you again

[ashokac7]

Yes you look very confused. First thing is take your own time for asking question. Read it twice before posting. Ask every question differently. Don't mix them up.


Explain this to me first. It is not making any sense to me.

I found out that there is a radiation from the bottom of the heat sink where I applied the heat source (in the model of no radiation) even if there is no radiation in the mode

Is it like below heat sink you have atmospheric conditions, so it is radiating heat out from it (How is this possible without radiation modelling??).
Your wall must be adiabatic as you have insulation there. Am I right?



Second thing is what you wish to get from this simulation. Do you want to match the results of some experiment or something else?

[Nadaaaa22]

Originally Posted by ashokac7

Yes you look very confused. First thing is take your own time for asking question. Read it twice before posting. Ask every question differently. Don't mix them up.


Explain this to me first. It is not making any sense to me.



Is it like below heat sink you have atmospheric conditions, so it is radiating heat out from it (How is this possible without radiation modelling??).
Your wall must be adiabatic as you have insulation there. Am I right?



Second thing is what you wish to get from this simulation. Do you want to match the results of some experiment or something else?

I want to match it with an article, I don’t know why i’m getting this there is no fluid in the region (bottom of the heat sink) and there is no atmospheric conditions it is just pure solid.thank you very much

[Nadaaaa22]

Originally Posted by ashokac7

Let's take it one by one.


1) If you don't consider radiation


Then you will get higher surface temperature in calculation than in actual practical. Because HTC don't depend on temperatures, but as heat by radiation is dissipated by convection now, deltaT needs to be higher. So calculation (or simulation) temps are higher than practical.


2) If you consider radiation as well.


Now your practical and calculation (or simulation) temps will be near to each. This will be more accurate approach if you can calculate (or simulate) the case with radiation as well.

If it doesn’t bother you can you explain this to me again? I worked in two models one without radiation (I got higher surface temperature+ heat transfer Q is equal to heat input). Another model with radiation (I got lower surface temperature and Q is less as well it shouldn’t be more?.