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Rating of equipment (watts) and total contribution to heat load

Rating of equipment (watts) and total contribution to heat load

(OP)
I'll start by saying most of my experience is not with CFD but more with structural FEA, vibration etc.

Anyway, I'm currently working on a project where we have a "heating problem". We take off the shelf electronics and integrate them into solutions for the military. In this particular instance we have various electronic components in EDAK and  Zargus transportable cases. Cases are approximately 27"W x 27"D x 36"H. In evaluating the forced cooling required I have used some simple approximations for calculating BTU/hour. Example, a 800Watt piece of equipment would yield... 800W x 3.41 = 8385 BTU/H. This seems to be a standard.

My “problem” is that the case with the lower combined wattage of equipment is supposedly producing more heat. I will have the chance to verify. We are currently building a test setup to verify effects of a cooling fan. My question is as follows. Is using the simple formula above accurate at all? If not how does one even progress to inputing this into a CFD analysis or do any upfront work? Wouldn’t you have to test each component to know what its “real” heat output is?

Thanks

RE: Rating of equipment (watts) and total contribution to heat load

winpop123,

I think you last statement is probably correct, anyways that would explain your problem.  Many times the nameplate rating on equipment is the max power draw and not the continuous rating.  Put an ammeter on each piece of equipment and get the real power draw.  Your formula is correct assuming that all the power goes into heat.

Timelord

RE: Rating of equipment (watts) and total contribution to heat load

winpop123-
Yes, to find out how much real power is being dissipated the most practical thing to do is to measure the current draw as the equipment is working under the worst-case duty cycle then P(Watts)=V(volts)I(amps).  Many times the current draw is the same no matter what the duty cycle.  Consult your EE.

In the best case (no airstream constrictions or expansions between the intake and exhaust) the heat dissipated Q for a desired temperature rise dT above ambient (20C is pretty standard but it depends on the environment and temperature sensitivity of the eqpt) Q=(dm/dt)(Cp)(dT)=p(dV/dt)(Cp)(dT) where p=density, Cp=specific heat of air, dV/dt=flow rate (cfm).  This tells you how much air (dV/dt) you need to push (or pull) through the enclosure because you know Q (by measurement) and you have chosen a dT.  The hard part is determining how much air you are really moving through the enclosure.  You need to find the intersection of the fan's static pressure curve with the enclosure's static pressure curve.  The former is published by the fan mfgr.  The latter you can get by sending the stuffed enclosure to a test lab so they can determine the curve by measurement.

Of course you can always do what most people do and that is to simply design something and test it.  Determine what components are most temperature-critical and put thermocouples on them.  Compare the measurements with the maximum case temperatures published by the component mfgrs.

Tunalover

RE: Rating of equipment (watts) and total contribution to heat load

Winpop123,

Then adding onto Tunalover's perfect statement.  Will this equipment be used at altitude....ie in a pressurized cabin, or on top of Mt. Whitney.  You need to factor in the density of air to gage your fan performance.  Also, don't forget the take into count any resistance to air flow like card cages, 90 deg turns, and hard drives.

The cooling zone has some good data that is free

http://www.coolingzone.com/

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2005 SP 5.0 & Pro/E 2001
Dell Precision 370
P4 3.6 GHz, 1GB RAM
XP Pro SP2.0
o
_\(,_
(_)/ (_)

Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

RE: Rating of equipment (watts) and total contribution to heat load

winpop123-
The effect of the turns and obstructions mentioned by Heckler are quantified when you test the stuffed enclosure.  The effect of altitude is easy to figure:  D=Palt/Psl=the derating factor for altitude (also =the ratio of the densities at altitude and sea level).  Multiply this times the heat dissipated at sea level to come up with the heat dissipation at altitude (will always be less).  The terms Palt and Psl are the standard air pressures at altitude and sea level.  You can get these from just about any thermo book and any text dealing with practical heat transfer analysis.  Of course Psl=14.7psia as we recall from high school physics or chemistry.

Tunalover

RE: Rating of equipment (watts) and total contribution to heat load

But you should do some analysis using the first order method that way you have a good idea how the design will perform.  I'm a firm believer of doing the appropriate analysis on the front end thus saving you headaches once you spent  on hardware that doesn't perform within your design specifications.

Tunalover - The density of the air is a very important parameter in any air cooled enclosure design and especially important for high altitude air cooled enclosures.  "High altitude air cooling has always been somewhat of a mystery to the uninformed."

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2005 SP 5.0 & Pro/E 2001
Dell Precision 370
P4 3.6 GHz, 1GB RAM
XP Pro SP2.0
o
_\(,_
(_)/ (_)

Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

RE: Rating of equipment (watts) and total contribution to heat load

Heckler-
There are many ways to skin a cat and both of our ways are acceptable approaches.  In my experience struggling through first-order approximations of pressure losses (e.g. Steinberg) the test data has proven the first-order predictions off by a large margin in three separate (and unrelated) tests.

BTW I don't recall saying that density loss at altitude was not important.

Tunalover

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