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Watts to CFM

My whole point here is to point out the absurdity of the question and elicit parameters useful in making a recommendation.

It's relatively easy to calculate the air flow to maintain a given temperature rise due to 100W. But that information is almost irrelevant in the current (lack of) context.

The internal heat sources will reach an equilibrium temperature to make that happen.

That may not be acceptable to those devices. The most precise info we have is the name of this newsgroup. I suggest that is insufficient specificity.

As an engineering manager, my most useful contribution was to shine a light on issues that the engineers ignored or swept under the rug that might result in the failure of their endeavor.

Often, the GUESS answer to a VAGUE question is not the most direct path to success.

My intent is not to give anybody a fish. My intent is to teach them HOW to fish... and what to do when they catch one... or suggest that chasing a chicken might be more productive in this context.

Reply to
Mike
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The array processor I worked on used ducted fans to blow on a number of dev ices in parallel. The PCB used ECL gate arrays in LCC packages. Each one had an aluminum heat sink glued on top with radial fins sticking up into a tube from the cooling plenum. A single fan blew air into the plenum.

We couldn't run the machine with the board out of the plenum so probing the chips directly was not possible. The PCB had via holes the same diameter as the probe tips on our scopes so we could plug them in without the top ha t. Once in a while we would bump the probe and break the tip. They cost $

75 which was quite the dollar back then so management gave us some crap pro be that wouldn't work. We should have found a way to jury rig something.

Rick C.

Reply to
gnuarm.deletethisbit

It's prudent to aim the fan at the hot bits. If it's an exhaust fan, aiming doesn't work, so adapt accordingly.

It brackets a ballpark. That is what he asked for.

Be a good manager and buy them $60K worth of thermal simulation software, and expect a month or two of learning curve.

What I usually do is make a box out of cardboard and duct tape and try it.

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com
Reply to
John Larkin

I think we're in technical agreement, but far apart on the PROJECT issues.

Yankee Stadium is a ballpark. And that information is just about as useful as 50 CFM.

Yep. That's a good way to do it. Now, take the problem statement and come up with a cardboard model.

We have 100W. Next we model the...well...we have 100W...and a box that's...well, we don't know about the box...but the temperature rise is...well, wonder what that is...and the particulars of the contents can be modeled by...well, we have 100W...

When you answer a question, it's good to present an answer that will be helpful to the questioner in the context. Based on the question, the OP likely needs way more than a random number. It's extremely difficult to achieve a goal that you can't/won't even define.

I've pissed off a lot of people, some not even in my division, by crashing their design review meeting and asking questions that they should have asked months ago. But once, they did put my name on the patent for the solution I suggested to the problem they didn't even know they had.

Wrangling engineers is a dirty job, but somebody's gotta do it.

Reply to
Mike

We don't know what spot heat sources are inside his box. I figured that 10 CFM would be plenty to keep the overall box temp rise low, so

50 might be good to cool off things inside.

Not 3, not 500.

I just did a box that does use a 3 CFM fan. It's tiny. And good enough.

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The overall box theta (in/out air temp differential) was only 0.7 K/W, and the heat is fairly well distributed across the PCB.

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The giant power opamps are on the bottom of the board.

Air enters slots on the bottom, blows on the amps, sneaks around the edges of the board, and get exhausted by the fan on top.

It's fun and friendly if you know how.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

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I'm not sure why people are making such a big deal about this. First, the question wasn't for anyone to solve his problem. He was asking for tools o r an approach to analyze the problem. Second, he isn't looking for details about the thing he is trying to cool he is just asking how much air needs to be moved to dissipate the heat from the unit.

Sure seems people are trying to complicate the issues. A fan is bringing a ir into the box. A unit is dissipating 100 watts of heat inside the box. What is the relation between air flow and heat removal? I think it is a re asonable assumption that the air is fairly well mixed so the air leaving th e box is the average/typical air temperature in the box. Then all that is needed is to define how much heat raises air by a degree, then do the math to get the relationship between air flow and temperature rise.

It is irrelevant how well the unit creating the heat couples to the air flo w because that isn't the question. The guy clearly knows something about h ow the unit dissipates in open air and just wants to know the ambient tempe rature in the outer box for a given air flow.

I see someone said this information is irrelevant in the "current (lack of) context". But the question asked is the only context. Everything else is irrelevant including all the noise people have been making about this simp le issue.

Rick C.

Reply to
gnuarm.deletethisbit

The OP did not say anything, is it some germanium transistors or perhaps a 100 W incandescent lamp.

enough.

In this case, the output air temperature would be 50 C warmer than input air. With 30 C input air temperature, you may have to avoid causing burns to people touching the exhaust air :-) .

Reply to
upsidedown

+1
Reply to
John S

Where: Watts = Internal Heat Load in watts

CFM = Required airflow in ft^3/min.

Credit is hereby given to Pentair app note "Thermal managemenT Heat Dissipation in electrical enclosures"

Reply to
John S

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