Seeking "Right" Formula for Heat Dissipation.

You've made an unrealistic set of conditions. Remember that heat transfer is the classic example of partial differential equation solving. The heat producers will always be warmer than the heat absorber.

Ok...so to lower the temperature as close as the ambient one

You still don't get it.

Compared to absolute zero it will start at close to ambient temperature.

Eventually the ambient temperature will rise to approach the heat source if this heat has nowhere else to go.

Then there is the absurdity of finding a singular, unique, univocal formula for heat transfer when it is already known that heat transfers from hot to cold via the different mechanisms of radiation, conduction and convection (conduction into a moving fluid media). Unless you want a non-linear, conditional equation, only suitable for computer programming your desire for one unique equation cannot be met.

Wait a sec, "Eventually the ambient temperature will rise to approach the heat source if this heat has nowhere else to go." my enclosure is outside the max outdoor temperature I get is 35°c

I just need to remove enough hot air inside the enclosure and replace it with the colder air from outside.

If the airflow is big enough I believe I can reduce the temperature inside and bring it close to the ambient temperature.

Is it not like that?

OK, let me try a different tact. You wish to cool the components inside this unknown volume of an enclosure by replacing the volume of fluid (air) inside the enclosure with cooler fluid. Does this sound correct?

On another note, did you account for the solar heating contribution?

Yes, and I know also the volume of this enclosure, the materials out of what is made and so on.

I calculated the internal temperature considering the heat load produced from the electric components, the heat dissipation through the enclosure wall and the solar gain.

If you calculated all that stuff, why are you having so much trouble now?

Moreover, you have furnished almost no meaningful data.

Great! Now do you understand how one of these work without any moving parts?

Now what do you think will happen if a fluid moves across this?

I can't use heat sinks outside the enclosure

I didn't say to use them on the outside of the enclosure. I asked if you understood how they work, apparently you don't.

It's impossible. Heat flows from hot to cold. There's got to be a temperature difference for heat to flow out of your enclosure. The only way to get the internal temperature equal to or less than the outside temperature is to use a heat pump such as a Peltier device.

https://en.wikipedia.org/wiki/Heat_pump

https://cr4.globalspec.com/thread/133648/How-Do-You-Correctly-Calculate-the-Heat-Transfer-Load

Thanks, I replied you there.

Wally, this enclosure...

Is this located in a wet area? Explosive or corrosive atmosphere? Subject to attack?

Are there vent slots in the enclosure and they simply aren't enough to thermosiphon off enough of the waste heat?

Is the true purpose of the exercise to shed some heat generated by the components located inside the enclosure, rather than every bit of the above ambient heat?

If you just tell us more about what you are really doing it would very likely result in better responses. We don't steal ideas around here, just tell us what you are up to.

You're right, so basically my enclosure is a digital totem. There are screens inside, transformers etc. that produce heat and and is located outside, exposed to the sun.

I've calculated that during a summer day where the ambient temperature reaches 35°c (at the totem location) the internal temperature [cosidering internal heat load, heat dissipation, solar gain] reaches 76°c.

I need to lower this temperature down to idyllically 35°c which is the recommended operating temperature of the electric components inside.

I believe that if I use an extractor that generates enough air flow through the totem I can obviously reduce the internal temperature.

My problem is I cannot find a formula which gives me this.

Please let me know if it's all clear.

The formula for convective heat transfer is called "Newtons law".

The rate of heat removed by forced convection is proportional to the surface area, the difference in temperature, and a factor, hc. The factor hc is determined by the rate of air flow.

"The equation for convection can be expressed as:

q = h_c A dT (1)

where

q = heat transferred per unit time (W, Btu/hr)

A = heat transfer area of the surface (m², ft²)

h_c = convective heat transfer coefficient of the process (W/(m^2oC, Btu/(ft² h ^oF))

dT = temperature difference between the surface and the bulk fluid (^oC, F)"

"

Convective Heat Transfer Coefficient for Air

The convective heat transfer coefficient for air flow can be approximated to

h_c = 10.45 - v + 10 v^1/2 (2)

where

h_c = heat transfer coefficient (kCal/m²h°C)

v = relative speed between object surface and air (m/s)"

https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html

From a practical standpoint, there are a few things to consider. Cover your "totem" with a reflective surface if it is in sunshine to minimize thermal heating. Forced convection cooling should have air entering the bottom and exiting from the top so that natural and forced convection are not fighting each other. If the forced air convection is not sufficient to keep the internal temperature low enough, some sort of heat pump can be used to refrigerate the incoming air.

Another way to approach it is to realize that the amount of electrical power into your totem is converted to heat. If this heat is removed by forced convection, the air is heated by delta T.

Delta T (deg K) can be calculated as the mass flow of air (g/sec) times Power (watts) divided by the specific heat of air. (1.012 J/(g K). For a given delta T, you should be able to calculate the required air flow.

https://www.google.com/search?q=air+heat+capacity&oq=air+heat+capacity&aqs=chrome..69i57j0l7.7898j0j15&sourceid=chrome&ie=UTF-8

I'm uncertain why this fixation with being provided a formula.

If you hear thundering hooves, think horses - not zebras.

Shade the enclosure, perhaps a kiosk (pay phone type shield) facing away from the sun. Solar gain is (a guess at this point) likely a very large contributor to your trouble. As well, vent the enclosure to promote thermosiphon. No formulae required, just some action.

It is well known that I very much enjoy complicated, wonderful and expensive solutions. Simple solutions do have their places, this seems to be applicable here.

As has been stated, you still have not told us very much. Is this installation covert?

Wally, I just went through your other thread - you tell us that shading the enclosure is not possible. 'Not Possible' and 'Not Practical' are two different obstacles entirely, but we can go with 'Not Possible'. This being the case you will be unable to match the inside enclosure temp with ambient using air transfer, there will always be a diurnal solar gain.

Previously suggested was a peltier device. Several TEC's exist for use with NEMA/UL Type 4, 4X or 12 enclosures that are rated at 300BTU/Hr or better, With an enclosure volume that is being withheld, a geographic latitude that is being withheld, and the funding available withheld, this is as good of a guess as I am able to make.

I don't know about anyone else, but right now the original post is asking us to help the poster to (effectively, it seems) steal a motorcycle.

???

Must be some wires crossed at CR4???

Original post: https://cr4.globalspec.com/thread/136953/Hot-Wired-but-No-Spark got stuck on this thread.

That's what happens when you violate the laws of Thermodynamics!!!

Steal a motorcycle?

This is a cheap Chinese recreational toy that would cost more to fix and maintain than it does to buy a new one.

The place I almost ran it over outside my garage is where I park my truck and it is on my property.

I have generous neighbors that will give me their old bikes and equipment etc.

If I don't have a use for it I will get it to the next best place even if its just the recyclery.

Given that it was in at least one accident and not functioning I'm not surprised it ended up here.

WHAT THE FUCK...SOMEONE HACKED MY ACCOUNT...

The shit about the motorcycle is not mine!!!

Why would you assume someone hacked your account?

And why so verbally abusive about an obvious glitch?

...dissapte your heat a little less dramatically... please

The two different threads including this one have had their original content restored. Your account was not hacked or compromised in any way. Sorry for the mix up, we're looking into what happened. Thanks for your patience!

ADMIN ALERT. The title and the text are not related. Some wires are crossed. See post 19.

Something has happened here...I've never asked about a motorcycle

HELP ADMIN

Two big ass fans 3 times the size of your enclosure with insane CFM ratings ought to do it.

Now, back to stealing that bike.....

Not that it's anybody's business, but I've brought more bikes than anyone I know. to the community bike collective that puts tools, bikes and knowledge in the hands of those who need them most.

I have another pile of 10 or so (that have been dropped off here) to go soon.

If they will take the motor bike?

I will bring it there as well.

We have all the bikes we need.

assume nothing.

Easy. The answer is infinite.

• If the temperature is intended to be <...35^oC...> and the outside temperature is the same, then just turn the thing off and open the door.

Your approach Temperature is impossible. You need move of a gap. Between the ambient temperature and heat producing temperature, you will never get to ambient temperature.

The amount of <...air flow...> needed is infinitely large, which is why it is difficult to calculate.

It would be easier to leave the door/lid of the <...enclosure...> open.