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Electrical/Electronic Heat Dissipation

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Twoballcane

Mechanical
Jan 17, 2006
951
Hi guys and gals,

I have done alot of analysis at the CCA level and have always used the "power" that was being used by the device as the heat dissipated. Well these days Im doing work with racks of equipment in a room. My logic is that the power the equipment is using is also the heat dissipated. If the deivce is using 500W it will dissipate 500W to the air thru convetion and conduction (thru the chassis to the rack to the air at steady state). At a meeting another Engineer said that I was being way over conservitive by four times.

Part of my goal is to size the AC for the room. If I used all of the power being used I will have a good size AC, but if my coworker is correct, the AC will be way over kill.

I know for mechanical equipment there is a power factor that you have to consider to get the heat dissipated, but is that the same for electronic equipment? I know there are fans, but I would think that they would be neglectable.

I guess my question is "if the router is using 100W, is it dissipating 100W to the air both thru conduction and convection at steady state'?

Thanks in advance for your time and knowledge

Tobalcane
"If you avoid failure, you also avoid success."
 
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How do you know what the actually power being draw is? Is this being measured or based on specs of the product? A computer power supply (or other power supply) will draw a certain VA but that is not the same as the watts due to the power factor of the power supply itself. Also, a computer will have a range of power draw based on load, so it might have a 600W power supply, but it could only be necessary at startup or maximum load, average watt usage could be half of that. Of course you would need to size based on maximum load, which is also different than startup (power draw at startup is higher for hard drives, etc... so their spec sheet rating depends on operating conditions (idle, load, spin up)).

You might wish to chart the spec'd power requirements versus some measurements of the equipment. You'd want to see the VA and watts.
 
I design HVAC for server rooms and high-density telecom equipment based on power consumption numbers provided by the client, or use nameplate + a healthy margin. I would not count any diversity or make any judgements regarding the nameplate-vs-actual power values; the risk is all yours and historically the power density increases constantly. Never go low! Equipment gets upgraded frequently and no one comes back to ask if any HVAC changes are appropriate.

Your friend is correct that nameplate can be 4x steady-state loads, but you have no control over how the equipment is used.

 
Thanks guys for responding.

Keeping it cool,
At this point I have nothing to measure, I just have spec sheets. I checked a few and found that they only report the power in watts. Nothing on power factors or apparent power (V-Amp). However, real power (watts) is what I am assuming to be the heat dissipated and not the apparnet power (V-Amps). Apparent power would be a bigger number. In the past I have always associated the power factor with big pumps and big fans, but not with electrical equipment. Is this typical that electronic equpment like routers and computers have a power factor?

MintJulep,
I agree with you ... conservation of energy right? Energy in and then the energy has to go some where. If there are no big gears or mechanical parts to move then it most like dissipate as heat. That is my rational electronic equipment. Except for the small muffin fan, everyting else is CCAs.

RossABQ,
Oh Yeah, I agree with you. However, the first AC just marginaly will accomdate the heat load, I'm recomending to go to the next higher size which is twice as big. I need some good evidence to convince my self and then them.

Thanks

Tobalcane
"If you avoid failure, you also avoid success."
 
With some electronic equipment (I'm thinking of power supplies), the rated output can be far lower than the actual input power.

For example, a 500W computer power supply could have as much as 1000W of input power to generate the 500W output. Even the most efficient single-phase power supplies are very inefficient. In this case, there would be an extra 500W of "ghost" heat to remove.
 
Stephenw22,

I agree with you there, however, the other 500w that goes to the electronic equipment, in my opinion since there are no moving parts, a small portion of it will do its job in the circuit and the rest (maybe 499w) will dissipate as heat. That is my rational.

Thanks =)

Tobalcane
"If you avoid failure, you also avoid success."
 
UPS manufacturers are pretty good about listing the heat dissipated to the room (this would be the 499W consumed in the UPS to produce 500W for the servers).
 
RossABQ,
So is that 500w of dissipated heat to the air?

Tobalcane
"If you avoid failure, you also avoid success."
 
Yes it is dissipated to the room air. So is the 499W, so total heat to the room is 999; but you shouldn't double-count! The 500W is dissipated by the servers, not the UPS.

In other words, in this case:

999 W goes into the UPS; it dissipates 499W. 500W goes to the servers and is didipated by them. Total cooling load is 999W, not 1499.
 
RossABQ,
I agree with you on that. If you ever had to size an AC would you use the 999W (3,409 btu/hr, AC use these units)?
I would, but other engineers are telling me that this is not the case. These same people do not know why. I'm thinking of thermal mass, but that is in transient cases.

Thanks

Tobalcane
"If you avoid failure, you also avoid success."
 
I have no idea why it would be different from the 999 in this example. Thermal mass is not involved. I think you have answered the question; "...These same people do not know why."
 
I would go with your original analysis in any case. Electronic equipment last longer with cooler temperatures.

It's also a bitch to debug electronics in a hot room.

TTFN



 
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