Conjugate heat trasnfer analysis.

cysanghavi · July 7, 2016, 05:45

In conjugate heat transfer analysis, I have 3 air domains and one solid domain.

I have the following heat transfers --
1.Conduction along the length of the solid.
2.Convection between the fluid and the solid inside the cooling channels(2 channels)
3. And convection between the ouside air domain and the solid.
I do not understand how to couple the 3rd part with the first 2.
I have two questions:
A.I create an air domain for this. Should I just define the heat transfer coeffeicient of the fluid at the interface as a BC. ?
B. Which other BC should I apply, beacuse this domain is basically air at atm. pressure.
CFX.png

ghorrocks · July 7, 2016, 05:58

A: You do not define heat transfer coefficients for interfaces. The solver works it out for you. You only specify a HTC if you are not modelling the air.

B: First of all - do you need to model the outside air? Is a heat transfer coefficient boundary sufficiently accurate? If you decide you do need to model it then make an air domain big enough that the boundaries are far enough away to not affect results.

(I am assuming here that the surrounding air has no external motion and there are no bodies nearby which might affect the flow - this seems strange because a drill is hot when it is drilling and then it is inside a very constricted hole.)

cysanghavi · July 7, 2016, 06:13

Thanks a lot for your replies.
Part A: I did not know CFX assumes a HTC for air. I will read about it. Thanks.
PartB: I am trying to simulate the part after I finish drilling. How fast the tool cools down and the temperature profiles I get at corresponding time steps of the cutting tool.
As per my understanding, heat lost to the surroundings due to convection would be a high percentage.

ghorrocks · July 7, 2016, 06:17

A: CFX does not assume a HTC for air. If you define a convection boundary it uses the HTC you specify, if you define an interface CFX calculates the HTC from the local flow conditions.

B: OK, thanks for the clarification. I think you will find radiation losses may also be a significant proportion of the heat loss. But you can model radiation as a source term on the drill surface, so you don't need a radiation model (unless you want to be REALLY accurate and include the radiation transfer from one side of the flute to the other - you will need a radiation model to cover that, but this is going to be very small so don't do this unless you require super accuracy).

cysanghavi · July 7, 2016, 06:33

Part A: Perfect. Thanks
PArt B: So I model convection with the surroundings, and radiation as a source term. (Currently the temperature profile that I get after 5s. is really high compared to experimental results. the difference is more than 150K.But these results were without outside air domain.)

ghorrocks · July 7, 2016, 06:36

You can estimate the radiative heat losses with a simple hand calculation, providing you can assume your surroundings are a black body at a known temperature. Then you can see whether radiation is significant enough to bother including.

July 7, 2016, 05:45	Conjugate heat trasnfer analysis.	#1
cysanghavi Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 10	In conjugate heat transfer analysis, I have 3 air domains and one solid domain. I have the following heat transfers -- 1.Conduction along the length of the solid. 2.Convection between the fluid and the solid inside the cooling channels(2 channels) 3. And convection between the ouside air domain and the solid. I do not understand how to couple the 3rd part with the first 2. I have two questions: A.I create an air domain for this. Should I just define the heat transfer coeffeicient of the fluid at the interface as a BC. ? B. Which other BC should I apply, beacuse this domain is basically air at atm. pressure. CFX.png

July 7, 2016, 06:17		#4
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,703 Rep Power: 143	A: CFX does not assume a HTC for air. If you define a convection boundary it uses the HTC you specify, if you define an interface CFX calculates the HTC from the local flow conditions. B: OK, thanks for the clarification. I think you will find radiation losses may also be a significant proportion of the heat loss. But you can model radiation as a source term on the drill surface, so you don't need a radiation model (unless you want to be REALLY accurate and include the radiation transfer from one side of the flute to the other - you will need a radiation model to cover that, but this is going to be very small so don't do this unless you require super accuracy). cysanghavi likes this.

July 7, 2016, 06:36		#6
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,703 Rep Power: 143	You can estimate the radiative heat losses with a simple hand calculation, providing you can assume your surroundings are a black body at a known temperature. Then you can see whether radiation is significant enough to bother including. cysanghavi likes this.

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July 7, 2016, 05:58		#2
ghorrocks Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,703 Rep Power: 143	A: You do not define heat transfer coefficients for interfaces. The solver works it out for you. You only specify a HTC if you are not modelling the air. B: First of all - do you need to model the outside air? Is a heat transfer coefficient boundary sufficiently accurate? If you decide you do need to model it then make an air domain big enough that the boundaries are far enough away to not affect results. (I am assuming here that the surrounding air has no external motion and there are no bodies nearby which might affect the flow - this seems strange because a drill is hot when it is drilling and then it is inside a very constricted hole.)

July 7, 2016, 06:13		#3
cysanghavi Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 10	Thanks a lot for your replies. Part A: I did not know CFX assumes a HTC for air. I will read about it. Thanks. PartB: I am trying to simulate the part after I finish drilling. How fast the tool cools down and the temperature profiles I get at corresponding time steps of the cutting tool. As per my understanding, heat lost to the surroundings due to convection would be a high percentage.

July 7, 2016, 06:33		#5
cysanghavi Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 10	Part A: Perfect. Thanks PArt B: So I model convection with the surroundings, and radiation as a source term. (Currently the temperature profile that I get after 5s. is really high compared to experimental results. the difference is more than 150K.But these results were without outside air domain.)