Re: Heatsink cooling

1kW is nothing really. My amps can generate about that much ( in 2u of 19" rackmount ) and I get rid of it with a max heatsink temp of 95C with 2 80mm 'boxer' style fans.

Graham

That's very nice to know PB. 1 kW is a bit of an overestimation. 800 watts would be the maximum (200 amps x 2 volts voltage drop x 2 sides of the bridge), and welding would not be continuous, as well.

i

Back in the office I have a neat little equation for temp rise in airflow vs energy absorbed. You can do it from first principles but it may come in handy. I'll post it later. It shows importantly that for certain temp rises vs watts you need a given mimimum airflow.

Graham

energy absorbed. You

Thanks! Yes, I would like indeed to do some calculations. Now that I am packaging stuff for real installation inside the welder, I want to avoid making stupid avoidable mistakes.

I made some pictures last night, of the heatsink assembly with some comments.

i

Found this nice article:

Will check it out. Turns out that my heatsink with glued in fins is the highest performance heatsink.

i

energy absorbed. You

I used this calculator:

Volumetric Flow Rate m^3/s 0.06 Number of Fins 13 Fin Width m 0.001 Fin Length m 0.3 Fin Height m 0.13 Sink Width m 0.12 Fin Thermal Conductivity W/mK 237

Result: Thermal Resistance C/W 7.46266e-02

So, for a continuous 800 watt load, the temperature rise would be

56C. That's probably acceptable. In reality, my load would be less than 800 watt (current lower than maximum, non-100% duty cycle in a welder). So, I am not worried about adequacy of my heat sink assembly. i

Here's the info I had in mind. It conveniently mixes units so as to use the ones that are hopefully most accessible.

Graham

Heat transfer equation from Sunon ( fan manufacturer )

Q = Cp . W . T

Q = Amount of heat transferred Cp = specific heat of air T = temperature rise W = mass flow

putting in the relevant values I got to this..................

Air flow required ( CFM ) = 1.76 * power / temp rise ( degrees C )

Worked example e.g. 200W and delta T = 50C

gives 1.76 * ( 200 / 50 ) = 7 CFM

ones that are hopefully

Thanks. I also tried to find some place with a calculator and found something nice. Below is a copy of ,y text file with a little detail of these calculations. It looks like my setup is just about adequate for 100% duty, which it will obviously not see due to application (welder).

Heatsink calculations:

Size 7x12 width, 5" depth, 10 fins.

According to

Volumetric Flow Rate m^3/s 0.06 Number of Fins 13 Fin Width m 0.001 Fin Length m 0.3 Fin Height m 0.13 Sink Width m 0.12 Fin Thermal Conductivity W/mK 237

Result: Thermal Resistance C/W 7.46266e-02

ones that are hopefully

That looks about right. But that assumes perfect coupling from the heatsink to the air stream, which doesn't happen.

This gets complex, and I don't really understand it, but it's something like...

If the fins on a heatsink are few and far between, the 7cfm will zip through easily and not pick up much heat. So if the sink is dissipating the 200 watts, the air exit temperature will certainly be

50K above intake (by conservation of energy) but the heatsink may be a lot hotter. Imagine half the air being in good contact with fins, half zipping through unaffected, and the halves mixing at exit, averaging +50. The heatsink only contacts half the air, so it rises +100.

If the fins are very dense, coupling from sink to air will be good, but you'll develop a lot of back pressure, and a "7 cfm" fan won't move 7 cfm, so again the sink temp rise will be above 50.

My working theory is that you get the best heat transfer if the heatsink restricts the fan to delivering about half its rated cfm. And the fins should be nearly isothermal (ie, not long and skinny) because cool fins restrict flow without coupling heat very well.

Anybody know more about this?

John

I wish more people would do what you do - post updates of the progrees in your project. You do a very good job of keeping interested people informed. Thanks!

Ed