一个热分析结果,有点疑问,请指点

boomchen

我现在遇到的情况也是这样，真是很矛盾，做不出来很急。我第一次作的截面很简单，用一个长方形来代替，做出来很好，温度也分层。后来我又作了一个，截面形状象大写的L，结果分析出来的结果和你现在一样，温度在厚度上没有区别，真是……，不知道为什么，希望能有高手致电我们。

boomchen

我现在遇到的情况也是这样，真是很矛盾，做不出来很急。我第一次作的截面很简单，用一个长方形来代替，做出来很好，温度也分层。后来我又作了一个，截面形状象大写的L，结果分析出来的结果和你现在一样，温度在厚度上没有区别，真是……，不知道为什么，希望能有高手致电我们。

tonnyw

逐渐减小对流系数,看情况如何? 我个人认为散热片的温度分布会逐渐明显. 另外,我们计算所得到的其实并不是温度,而是温差.

iomega

What's the thermal conductivity of your 散热片?  If it is Al with ~200 W/m-K, no wonder A and B will have close temperature rise.

To: tonnyw

>逐渐减小对流系数,看情况如何? 我个人认为散热片的温度分布会逐渐明显. 另外,我们计算所得到的其实并不是温度,而是温差.

Wrong, only increase the 对流系数 will result in a large temperature gradient on 散热片.

tonnyw

I think there is no problem about my assumptions.
The heat is conducted two ways in this case. First, it is conducted in the 散热片. Second, the air will circulated to carry away the heat which is convection. If we eliminate the convection, the heat is only conducted in the 散热片 and we can see more clearly the temperature gradient. Maybe we can compute a simple example to see what's going on.

iomega wrote:
What's the thermal conductivity of your 散热片?  If it is Al with ~200 W/m-K, no wonder A and B will have close temperature rise.

To: tonnyw

>逐渐减小对流系数,看情况如何? 我个人认为散热片的温度分布会逐渐明显. 另外,我们计算所得到的其实并不是温度,而是温差.

Wrong, only increase the 对流系数 will result in a large temperature gradient on 散热片.

iomega

>The heat is conducted two ways in this case.
Right, but the fact is that finally the heat has to dissipate into the air from 散热片 to air.

The way of your thinking is that heat can conduct in two ways:
(1) heat conduction through 散热片 to heat sink
(2) heat convection through air to the  heat sink.
If (1) and (2) are both hold, elimate the (2) might have effect.

But remember in this case, the heat sink is the air surrounding the 散热片. The conduction in 散热片 and convection in air are kind of in series. If you elimate the air convection. There will be no thermal ground for 散热片 and temperature in 散热片 will be uniform.

Hope this explaination works for you. :-)

tonnyw

I am not convinced at all. The attached picture is my explaination. I think the best way is to ask that guy to send us the input data and we can run the test to see what's going on. I would like to bet five bucks.

iomega wrote:
>The heat is conducted two ways in this case.
Right, but the fact is that finally the heat has to dissipate into the air from 散热片 to air.

The way of your thinking is that heat can conduct in two ways:
(1) heat conduction through 散热片 to heat sink
(2) heat convection through air to the  heat sink.
If (1) and (2) are both hold, elimate the (2) might have effect.

But remember in this case, the heat sink is the air surrounding the 散热片. The conduction in 散热片 and convection in air are kind of in series. If you elimate the air convection. There will be no thermal ground for 散热片 and temperature in 散热片 will be uniform.

Hope this explaination works for you. :-)

iomega

If you want , I can bet 100 bucks - all will be contributed to the forum though.

Let us put in this way: assume the air convection is elimated, then where the heat generated from the chip will go?  There will be no thermal ground to the system (if assume no heat radiation). Just like you touch a high voltage electrical cable, but you wear a insulating shoe, there will be no electrical current flow in your body to the ground. Therefore, everywhere in your body has the same voltage potential.  Same thing for the heat conduction.

Remember, temperature gradient means you have heat flow in the structure. if the air convection disappears, where the heat flow from  散热片 to?

Also your example prove what I said is right -> "There will force the heat to transfer along the cooling fin and there will be more obvious temperature gradient".

You assume the cooling fin is the 散热片, right? the whole 散热片 is in the air, since air convection is zero, 散热片 is no longer a coolin fin.

You only consider that the L_N boundary (air convection) adiabatic, but for this specific problem, if alf->0, both L_N and L_D boudary become adiabatic.

On the other hand, I really like to discuss the fundmentals of the heat transfer with you  -> I think that is what this forum lacks...

tonnyw

I don't feel confused.
Let's make clear one thing. Is this transient or steady-state problem?

My understanding is that at the beginning it is transient and as time goes on it will becomes steady-state which means at each point the temperature is not changing any more and different points may have different temperatures. If in this case, air is the only way to carry away the heat, I think when the convection coefficient goes small, the temperature will keep rising. We should have no doubt about this point. Since heat has no where to go and it will just accumulate. My original thought was that we have two types of boundary condition in this case, one is convection which is Neumann and one is temperature which is Dirichlet, for instance, the fin touches something containing coolant. If convection is gone, heat can only dissipate toward the Dirichlet boundary. Do you agree with this?
By the way you could say L_N boundary become adiabatic as alpha->0. But you cannot say L_D boundary become adiabatic because I just specify the temperature and I don't specify any flux. This is agaist intuition. But it is correct.
Luckily I passed the qualifying exam and was not failed by this problem.

Anyway it is such a pleasure to have a discussion with you.

iomega wrote:
If you want , I can bet 100 bucks.

Let us put in this way: assume the air convection is elimated, then where the heat generated from the chip will go?  There will be no thermal ground to the system (if assume no heat radiation). Just like you touch a high voltage electrical cable, but you wear a insulating shoe, there will be no electrical current flow in your body to the ground. Therefore, everywhere in your body has the same voltage potential.  Same thing for the heat conduction.

Remember, temperature gradient means you have heat flow in the structure. if the air convection disappears, where the heat flow from  散热片 to?

Also your example prove what I said is right -> "There will force the heat to transfer along the cooling fin and there will be more obvious temperature gradient".

You assume the cooling fin is the 散热片, right? the whole 散热片 is in the air, since air convection is zero, 散热片 is no longer a coolin fin.

You only consider that the L_N boundary (air convection) adiabatic, but for this specific problem, if alf->0, both L_N and L_D boudary become adiabatic.

My dear friend, this could be a typical qualify exam problem and I see many people were confused before.

If you look at the Ph.D thesis from MIT, Stanford, berkeley students, at least five people's research structures were based this type of so-called fin problem.

iomega

"一组发光芯片(0.16mm*0.8mm)贴在散热片上(14mm*55mm),热从芯片中出发,在散热片被23W/m-K的自然对流带走,空气温度为25摄氏度", from the description, your agrument of "for instance, the fin touches something containing coolant. If convection is gone, heat can only dissipate toward the Dirichlet boundary" is not correct. There is no such L_D boundary condition but only L_N condition.

Did you notice the temperature in 散热片 is around 33C, while air temperature is 25C. The difference is exactly caused from the thermal resistance of the air convection to the heat sink (air). If you you have Dirichlet boundary condition, you have to apply a T0 on the surface of the 散热片, so the surface of the 散热片 will be kept at T0.

By looking at the description of the problem, can you find such T0 applied on the 散热片?

tonnyw

If air is the only medium to which the heat has to dissipate, I agree when alpha->0 the temperature will keep rising as I mentioned before. That also means there is no Dirichlet boundary. I thought the cooling fin is such stuff that the air will circulate around some of its surfaces to bring away the heat and in the meantime other surfaces will touch something to conduct the heat.

I enjoy this discussion. I also realize you gave me one point. Thanks for that.

Hope to discuss some other subjects too.

iomega wrote:
"一组发光芯片(0.16mm*0.8mm)贴在散热片上(14mm*55mm),热从芯片中出发,在散热片被23W/m-K的自然对流带走,空气温度为25摄氏度", from the description, your agrument of "for instance, the fin touches something containing coolant. If convection is gone, heat can only dissipate toward the Dirichlet boundary" is not correct. There is no such L_D boundary condition but only L_N condition.

Did you notice the temperature in 散热片 is around 33C, while air temperature is 25C. The difference is exactly caused from the thermal resistance of the air convection to the heat sink (air). If you you have Dirichlet boundary condition, you have to apply a T0 on the surface of the 散热片, so the surface of the 散热片 will be kept at T0.

By looking at the description of the problem, can you find such T0 applied on the 散热片?

iomega

Sure, really nice to find a case that can brings such warm discussion...

bluewaiter

有可能是导热系数的单位不对，如果在按ansys中填入23W/m-K的自然对流
导热系数就应该在国际单位的基础上加大1000倍