Watts to CFM

I have a unit that draws say 100 watts.

It's going into a outer box; I'm supplying it with fan-supplied air.

How do I SWAG the CFM I need to keep the temperature rise reasonable in the box?

Yes, this likely oversimplifies the whole issue, but I need to start somewhere.

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Reply to
David Lesher
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Yeah, very oversimplified. I did this once, but don't have the computations handy. But, you need to figure out what temperature rise is acceptable, and figure out how much heat it takes to heat the air that much. Then, you need to exchange the air such that an amount of hot air is pushed out when it reaches that temperature.

When I did this, I assumed there was no heat transfer to the outside, it was all carried in the air. Well, in fact, when pushing a lot of air into the box, you create turbulence, and that DRAMATICALLY aids heat transfer from the interior gadget to the box. So, I ended up overdoing the cooling by at least a factor of 4.

You should be able to find charts giving cubic measure of air and heat capacity. And, then, you can work it out from there.

Jon

Reply to
Jon Elson

1 CFM of air flow has an equivalent theta of about 1.8 K/w.

You can apply that to an insulated black box. The temperature rise of internal bits is more complicated.

What size fans are candidates? Something in the 20 to maybe 50 CFM range, I'd wild guess.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

1 CFM of air flow has an equivalent theta of about 1.8 K/w.

You can apply that to an insulated black box. The temperature rise of internal bits is more complicated.

What size fans are candidates? Something in the 20 to maybe 50 CFM range, I'd wild guess.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

The specific heat of air is close to 1 kJ/kg/K = 1 J/g/K. Assuming maximum temperature increase of 10 K and P= 100 J/s, the mass flow required is

100 J/s / (10 K x 1 k / (g x K)) = 10 g/s
Reply to
upsidedown

It is nice that 1 l/s is close to 1 K/W.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Not that I'm disagreeing... but where did you get that number from? I wonder if I could hand wave some heat capacity argument?

George H.

Reply to
George Herold

The calc is in my thermal cheat-sheet.

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Corrections and additions are welcome. It does need an update.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Great! Thanks. I have part of that saved somewhere on my 'puter, now I've got it all.

My only quick addition is that the volumetric heat capacity of every solid near room temp is about (factor of 2) 3 J/(cc* watt).

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George H.

Reply to
George Herold

oops... 3 J/ (cc* K)

GH (always check your units)

Reply to
George Herold

There are empirical convection heat transfer coefficients involved, theoreticals are useful only as guide to developing them.

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Reply to
bloggs.fredbloggs.fred

Free Convection - air, gases and dry vapors : 0.5 - 1000 (W/(m2K))

That doesn't look very useful.

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Reply to
John Larkin

They are all 1000:1 ratios like that. That's where the coefficients come in.

Reply to
bloggs.fredbloggs.fred

The IEEE floating point standard has a value for NAN. It needs another for "whatever."

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

That's an understatement...

Let's look at the limit case. If the 100W is concentrated in one junction, what's the maximum junction temperature, the maximum input air temperature and the elephant in the room...What's the coupling from that junction to the air? This applies in any subset of the contents.

The other end limit is if you can couple all of the heat to the air. It's then the heat capacity of air, temperature rise and the volume you pass thru. I don't remember the constant, but it should be readily available in HVAC texts.

Unless your heat is generated uniformly across a surface, any gross estimates will be hugely inadequate.

More air won't help much if it's not tightly coupled to the heat source. If you're using commercial heat sinks, the vendor should have curves as a function of velocity, which can be converted to CFM by detailed analysis of how air flows inside your box.

There is no such thing as a SWAG for cooling.

Reply to
Mike

50 CFM maybe.
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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

Maybe? That sums it up nicely.

Why do computers have a fan sitting on the CPU? That fan probably affects the total airflow thru the box not at all. Might as well save a few bux and remove it.

Reply to
Mike

To cool the CPU chip!

We do that too.

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The FPGA would heat itself up a lot without that fan, but doesn't dissipate enough power to warm up the box much. And that fan *does* stir up the air in the box a useful amount.

There's a pin-fin heat sink glued to the top of the FPGA, but it would be practically useless without that fan blasting directly into the pins.

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John Larkin         Highland Technology, Inc 

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Reply to
John Larkin

I found useful tools at and am looking through them.

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Reply to
David Lesher

To some extent, that's right. A well-designed enclosure would have one fan and duct the airflow so it streams over all the items to be cooled, most sensitive first, then vent from the box.

Unducted fan-on-heatsink-on-chip typically has zero airflow in the center (the fan hub) and maximum airflow at the periphery (blade velocity highest) despite the sensitive item being.. in the center.

Remember: air flow can serve DEVICES IN SERIES. The airflow is the SAME for all such DEVICES IN SERIES, and it matters not whether the fan is near the device.

Reply to
whit3rd

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