Resistance on heat sink fins???

[Arun81]

Dear all,

I am working on a liquid cooling system and have to deal with resistance on base,fins and fluid flow. I found the traditionally used equation for fin resistance doesnt have any dependance on fin materials thermal conductivity. Isit that if i use a plastic fin which has a very low thermal conductivity will not affect the overall heat transfer? I see only the huge dependance of materials thermal conductivity on base resistance. i tried to make a analytical tool with excel and found out that the resistance of fin is not changing much with change in the fin material. But in a broader look we can say that if a fin matreial is plastic there wont be a good conduction of heat to coolant. Please provide your guidance on this..

thanks in advance

Arun

 

[IRstuff]

I think you are misinterpreting the situation.

The "standard" equations are assuming that the thermal conductivity of the material used to make the fin is small compared to the convective transfer coefficient.

TTFN

 

[PZas]

I think you meant to say that the conductivity is large (not small) compared to the convective coefficient.

However in a forced flow liquid cooling situation this may be totally untrue.

In this situation the resistance of the fin material and the resistance of the convection area are both going to have a large impact.

How accurate are you looking for here. You could always use a FMEA tool to give you an answer but the biggest problem is going to be getting the fluid velocity over the fins and therefore the convective heat rate (which is dependent on the conduction because of the surface temp
).

If you are looking for a quick dirty way to get an order of magnitude estimate look at the system and try and evaluate the thermal resistance of each side individually and then balance them for the final qdot.

 

[gwolf]

Try "Heat Transfer" by Holman, if you already haven't looked at his work - it tells you how to do tapered fins. I'm not sure if it will be enough for you but it is the best and most comprehensive heat transfer book that I have ever found, now in about it's 8th edition. Good worked examples too for a book of it's age.

I take my hat off to the man if he's still alive.

gwolf.

 

[Arun81]

Thank you to all who have replied, So i could see from all your answers the general resistance on fins equation assume that thermal conductivity of the material used to make the fin is small compared to the convective transfer coefficient. So i could understand that it is just an assumption. But could it work out the same way if i use plastic as a fin which has a thermal conductivity of say 1W/mK? In my analytical solution it shows that the resistance of a copper fin (thermal conductivity=385W/mK) and the resistance on plastic fin has only a difference. PZas told that if the velocity of liquid (i.e., the heat transfer coeff.) is high enough we can use any material for fin. In my case i am using a heat transfer coeff=10000-15000 W/m2K. So do you think i can use plastic fins??? thank u IRstuff you crisp answer blasted my puzzle in one second. gwolf i will have a look on to the Holman's book. i think we have it in our library.

 

[IRstuff]

I don't even begin to see how you came up with that conclusion. Particularly given the somewhat incredible HTC, why would a 400x increase in thermal resistance NOT make a difference?

For a 3 in wide, 0.1 in thick, and 0.5 in long fin, I get 65 K/W using your plastic, and 0.17 K/W for your copper value. How is that only a little difference?

TTFN

 

[arto]

Also check out chapter 508.2 of the GE Heat Transfer Book:

http://www.genium.com/drafting/ht.shtml

 

[PZas]

you understood me incorrectly. I was saying that if the convection coeff is LOW enough and the conduction coeff is proportionally HIGH it doesn't matter how high it is.

This is to say if you have a convection coeff of 1w/m^2K and a conduction coeff of 500W/m^2K then changing the conduction coeff by even as much as 10% will make little difference.

Think in terms of thermal resistance. If the conduction makes up only 1% of the total then doubling it is only a 1% change in the overall system. This is what the basic calcs are based on.

Now you have forced flow of a liquid over the fins. With this setup your convection coeff could be as high or higher than your conduction coeff. Because of this you need to figure out what your convection coeff is as it is largely independent of fin material.

Start by figuring out the resistance on 1 rectangular channel (between 2 fins) on the fluid flow. Then use this to find the total flow with your given pump. once you have that you can extimate the average flow rate within the channnel (averaged by surface area of the fins not area of the fluid flow) and use this to calculate the convection coeff. I assume you will not be hot enough to take radiation into account (but keep it in mind).

Also remember that as a rule of thumb you will not need fins that are more than 10 times as long as they are wide.
In this situation you may only need 3 times....

If you find your flow velocities to be slow you may need to take natural convection into account.

If you give me a better description of the problem we can help out a little more.

 

[sailoday28]

For information only
With the exception of the fin tip area, the major assumption in fin analysis is no temperature gradient in the direction that is normal or perpendicular to the convective surface.

 

[drkillroy]

Pzas I'm confused here. There is a heat source that needs to be cooled. A fluid is going to be the medium used to transfer the heat from the source to elsewhere. Wasn't teh origional question asking wheather plastic heat sinks could conduct the heat from the source to the fluid almost as fast as a copper heat sink? If 2 setups were active, one with a plastic fins and the other copper, wouldn't there be a much greater temperature differential between the source temperature and fluid temperature using plastic material over copper while transfering the same btu rate?
Thanks drkillroy

 

[PZas]

drkillroy, not necessarily. Think of the system in terms of a resistance circuit. This is assuming that you have had some training in heat transfer and therefore have seen heat transfer resistance circuits. If you have not seen these let us know and we can explain a little more in detail.

You will have primarily 3 resistances to heat transfer in this system. The first is the contact resistance from the source to the heat sink. The second is the conductive resistance through the heat sink. The third is the convection coefficient from the heat sink to the fluid medium.

Now if you make the assumption that the resistance resulting from the convection is >> the resistance from the conduction then making changes to the conduction resistance will have little effect.

That said to directly answer your question yes there will be a higher dT for conduction as the conduction coefficient drops. However if originally you had a dT of 0.5 degrees through the copper and 50 degrees across the convective boundary then changing the dT of conduction to say 5 degrees (factor of 10) will only result in an increase of the overall dT of less than 10%. How much room you have in picking materials for the heat sink is 100% based on the whole system.

If you do not have a super clear picture of the system and can afford to use copper or aluminum do it. It is rarely a bad idea to over design.

 

[ko99]

PZas explanation is very good, but his example is perhaps over simplified. Please remember fin conduction effects convection.

You should calculate the fin efficiency. Let us know if you need the formulas.

The effect is small for a typical aluminum vs copper analysis with relatively thick fins. Expect a dramatically worse fin efficiency with plastic unless you have unusually low, thick fins -- the fin tips will be approx the same as the air temperature and very little convective heat transfer will occur.



ko (

http://www.ecooling.biz)

 

[Arun81]

Hi all,

sorry Pzas, i was out of that topic for long time. I have calculated the resistance in the channel heat sink with the following dimensions:
Fin thickness=1.07mm
Length of heatsink=30mm
width of heat sink=15mm
Base thickness=0.8mm
fin heigth =1mm
number of fins=12 with Aspect ratio of 5
here i considered a spreading heat source of width=5mm

And then calculating the total thermal resistance by adding resistance in base, resistance in fin and channel resistance.

For Copper(k=385W/mK)the total thermal resistance i found is Rth= 0.26 K/W. (Rfin=0.15,Rbase=0.06 and channel resistance is negligible.)

For Plastic(k=1W/mK)the total thermal resistance i found is Rth= 24.76K/W.(Rfin=0.47,Rbase=24.24.channel resistance is negligible.)

Okay let me consider a material with thermal resistance of 20W/mK.the total thermal resistance i found is Rth= 1.49K/W.Rfin=0.23,Rbase=1.21.(channel resistance is negligible.)

So can you see there is a huge difference of heat conduction/convection effect in copper and normal plastic.But there is only a small difference of 1K/W in material having 20W/mK and copper.

So let me say the material i stated having 20W/mK is a plastic(there are some thermally conductive plastics with huge thermal conductivity). So my question is in a right application why not plastic(k=20W/mK)can be used for heat transfer application. One more thing is as Ko99 said, there is an impact of thermal conductivity of fin material. I agree with his statement But the difference is very small when compared to the base resistance, which u can see in the above example. So in this case of small difference in fin resistance which agrees with Pzas statement of conduction(thermal conductivity) is overtaken by convection.

Awaiting your critics and views.

Thank you,

Arun

 

[PZas]

As for Ko99's statement. It is generally not the case that you will have fins that are wider than they are tall. It is quite common to have a fin height 10x the width. As the height of the fin increases or the width decreases you will see more effect on the fin efficiency with a reduction of the conduction coefficient of the base material (fin material assumption is they are the same).

I have been a little sick so I may just be missing something here...however it appears you did not consider the convection to the air. As stated before this is very important when determining what materials you can use. For example some base line estimates allow your thermal resistance from convection to be as high as 200 K/W in still air. dependent on orientation and all that of course. I can't really solve this with the information at hand. However lets just assume that you have a convective resistance of 100K/W from your given heat sink.

If this is the case than a change from a material with K=300+ to one with k=20 results in a change in the overall thermal resistance of less than 1%! As you can see depending on the situation thermally conductive plastics may very well be functional and less expensive.

Now with forced conduction of a fluid as you stated in your original post you will likely have a convective resistance << 100K/W. Keep that in mind. If say your convective resistance was 1K/W then a change from copper to plastic could double your source's temperature.

Really the moral here is to keep your mind open when approaching the problem. You do not have to assume that copper/aluminum are your only options when creating heat sinks. It may very well be that a slightly larger sink made of a material that is inert to the environment is the better choice. Gather your data and do some good engineering.

As to why you don't see plastic heat sinks out there ....you do. Why do you think the box that holds the power supply for your lab top has fins on it . Plastic heat sinks are not very common because much of the design work today is done by individuals who do not understand the heat transfer of the entire system. One of the problems that arises when companies try to cut costs by having people without an education do their designs because their salary is lower...but lets not discuss that in this thread.

 

[IRstuff]

For most real applications, the junction temperatures are already max'd out and even a 3 degree change can substantially reduce the reliability of the component.

TTFN

 

[ko99]

Arun, is there perhaps a typo in your example heat sink? You didn't mention airflow or ducting, but Rth looks quite optimistic for 1mm tall fins. Otherwise, I agree that your short fins look like good candidates for a ~20W/mK material.

Also, the base conduction loss looks too optimistic for a 5mm source on a 15 x 30 mm plastic base only 0.8mm thick. The heat will not get to the outer fins. If possible, consider making the base thicker or making the source larger (such as adding a cu spreader between the source and the plastic sink).



ko (

http://www.ecooling.biz)

 

[PZas]

IRstuff (Aerospace)

I believe that you are looking at the small picture here. To make the assumption that the junction temperature is already maxxed out in most "real" applications is just a leap of faith. I personally work in R&D. When I look at a junction temperature it is never defined already. When I finish a project is the junction temperature maxxed out? Hardly ever. Remember the OEM has to design for the system to be used in the hottest environment likely.

Just take a look at the computers in your office. Do you think the heat sinks on even the same processors are all the same?

Ko99
I agree with most of your statement. However without knowing the flow it is probably a bit early to choose a material.

 

[ko99]

There are far too many design tradeoffs for me to presume to choose a material. That's Arun's job. Sorry if I implied otherwise.

ko (

http://www.ecooling.biz)

 

[IRstuff]

I'm talking about spec limits, not RT operation.

In any case, for many semiconductor failure modes, a 3ºC increase in temperature results in about a 40% reduction in life.

TTFN

 

[Arun81]

Sorry, i forgot to mention that I am using a forced liquid coolant in the duct. And i can go upto flow velocity of 1m/s. The results i gave in my last posting are taking for the flow velicity of 0.5m/s.

PZas,

Thank you for that motivating suggestion...i will go ahead. Takecare of your health...

Ko99,

Your suggestion of using Cu heat spreaders are already in my mind. I am also thinking to use the heat sink upside down to have the base on top to get rid of huge base resistance. In that case i can use a highly conductive think layer of heat spreader between source and the channel heat sink to reduce the overall thermall resistance.

IRstuff,

I know even a small increase in jn temp will cause a huge damage on semiconductor chip. Thats why i have specified in my previous posting that "In right application why shouldnt we use the plastic heat sinks". But if i get rid of base resistance i can get the thermal resitance of plastic heatsink(k=20W/mK) close to that of copper or aluminium.

Thank you all,

Arun