Computer Heat Gain 3

[sunsentinel]

My boss argues that ALL the power drawn by a computer will end up as heat gain in a room. I know that energy needs to be conserved (1st law of thermo.)
My questions:
1-How does ALL the power drawn to rotate a hard disk or an inserted CD (Rotational power=Torque X rotational speed) end up as heat gain to the room? (This is not the only operation a computer performs, I know)
2-ASHRAE Fundamentals 2005 section 30.8: "Actual power consumption of office equipment is assumed to equal total (radiant plus convective) heat gain,.."(is this line proving what my boss is talking about?) How do we argue that ALL the radiant heat gain will end up as heat gain to the room? (Convective is understandable)
3-What are you guys using as heat gain from a computer nowadays? (ASHRAE handbook talks about a conservative 155W together with a monitor.)My boss wants me to use a 500W! Checking Dell site regular office computers have 300-400W NAMEPLATE (which is the max?) inputs.

Thanks in advance for answers.

 

[IRstuff]

1-The HD spinning is produced by a motor that continually overcomes the friction of the bearings.

2-Why do you think that radiant energy is not conserved? If the equipment is hotter than the room, then it will have a net radiance into the room.


TTFN

FAQ731-376

 

[MintJulep]

If the power doesn't go to the room, where else would it go?

 

[TADiep]

In commercial cooling load estimation and HVAC design, we view computers, printers, scanners and most other office equip as black boxes. We dont care what goes on inside these devices. We are mainly concerned with how the overall operation of these equip will affect the room or space conditions.

Your Dell computer has a max nameplate power consumption of 400 Watts. Due to safety reasons, most equipment never draw more than 80% of the rated power even during peak demand. With this in mind, lets assume 320W power consumption.

1st law of thermo states...engery in = engery out. The pc draws 320W therefore 320W must be dissipated in the form of heat, sound and friction.

So if you were to design a cooling system for this computer, design it for 320W. NOT so fast....

But why does ashrae say 155W?

This is because both you and I both know the computer will not be operating at 100% CPU speed all the time. (Even high density data centers do not operate at 100% nameplate capacity all the time.) The actual time your CPU is running at 100% is about 1-10% of the time. ASHRAE takes this diversity factor into account and recommends a heat dissipation for computers at 155W. This implies that during normal operation, the computer will draw 155W from your power outlet.

To comment on your boss's rule of thumb of 500W per computer....this will result in oversizing the HVAC system. Oversized system equals wasted money.

Imagine a room with 50 computers. According to your boss, you will need an HVAC system capable of cooling 25,000 W.

According to ASHRAE, you will need to cool 7,750 W.

The difference between the two is 17,250W and can amount to an extra $15,000 of operating expenses each year. (assuming 24 hour operation). You might want to spring this on your boss and enlighten him.

Hope I didnt lose you with the long explanation.



----
A green thought..."We don't inherit the earth from our ancestors, we borrow it from our children." (unknown)

 

[mike7941]

Everything TADiep said is true.

But you do have to look at the application and environment. Yes, I use heavy computer loads for computers in general but then I'm using that to cover miscellaneous office equipment that no one can tell me about when I'm running loads (like copiers fax machines and scanners etc.). The other item to take into account is the type of computers they plan to install. Higher end Cad machines and workstation for fairly intense engineering or the low end terminals that often show up in call centers.

Its a judgement call. If you pick the low end be careful to account for all the equipment. If you pick the high end make sure your equipment can do part load efficiently.

Mike

 

[sunsentinel]

IRStuff: HD accelerates, right? Doesn't this mean power applied is larger than the power lost to friction?(ieart of power to the disk's mass to give it a rotation, how did this power end up as heat gain is my curious question) Also, I do believe radiant component of heat gain is also conserved but how can we guarantee it will end up in the same room as the computer? May be it radiates to the room next door or even outside?

TADiep: Thanks for your explanation. "Black box" meaning I will simply need to consider a computer nothing different than a dummy resistance from now on I think!

 

[IRstuff]

The work done to accelerate the HD gets dissipated when the HD is powered and spun down, either directly through bearing friction or through the motor back EMF that's dissipated against the motor driver.

Don't understand your last comment. "Radiation" refers to blackbody radiation, which is photonic and will be absorbed by walls, etc. What little there is that's RF in nature is miniscule.



TTFN

FAQ731-376

 

[marcoh]

All electrical input will be converted to heat at some stage. (Though a colleague of mine did point out that the photons leaving a computer monitor that escape through a window to outside might be lost!)

Typically we use 15 or 20W/m2 for small power gains in a typical offices based on 10m2/person, ie 150 to 200 W/person to allow for PC's, photocopiers etc.

I have seen some papers that say the small power usage is often overestimated in AC assumptions (but not always!)

 

[IRstuff]

Well, just for giggles, assume your 17" display has an average luminance of 50*pi*cd/ft^2 (50 ft-L), then, there's no more than 625 mW of light power leaving the monitor.

TTFN

FAQ731-376

 

[atlas06]

Expansion of systems in computer rooms is almost a certainty these days.
We've installed CRAC units for a computer room that became jam-packed with additional racks with 2 years, the two 5-ton units we had originally were no longer sufficient, while at the project commissioning, one unit was sufficient.

Sizing for 500W/SF as your boss says makes sense for computer room, I am told that as computers gain in speed, they also use more power (to be verified though-I'd think more efficient).

Use multiple units, one unit sized for the ASHRAE recommneded 155W/SF so no one says you're not within industry standards. Let the owner know of your rational, and write a report about your rational in your calculations file, should another engineer take over for you while you're not around for whatever reason.

 

[ActiveSolar05]

Look on the SEAGATE hard disk mfg site; they give case temps for all of their drives >> DElta T, A, >> Watts heat...

Second, look at the ANTEC site; some power supplies coming out now are E-80 rated; at least 80% efficient at load values between 20% and 95%.

The Intel Core 2 Duo E6700 draws 21 watts at idle......

 

[kenvlach]

re 80% efficient power supplies: Perhaps 80% efficient at converting 120 VAC to 12, 5, 3.3, & 2.5 V DC. Thereafter circuits, semiconductors, fan & drive motors. So, consider the computer a heat-creating box. Of course, only rarely used at full power but the difference maybe approximates the power used (& given off as heat) by displays, printers, etc.

Computer users generate heat, too. An average person just sitting burns maybe 550 calories over 8 hours. And, 1 food calorie = 1000 thermodynamic calories = 4184 Joules.
Since 1 Watt = 1 joule/sec, (550 x 4184 J)/28,800 seconds = 80 Watts.

But, maybe users absorb photons of knowledge from their computers, package them and transmit as product to customers, maybe over the bundle of tubes known as the Internet?

TGIF.

 

[RossABQ]

I have asked a couple of computer OEM reps's, and never gotten a really straight answer, whether a 300W power supply delivers 300W of power to the motherboard and peripherals, or consumes 300W. Anyone know for sure?

 

[kenvlach]

Looking at one of the lamest computer P/S's ever, a P/S DPS-160GB B (removed from an HP Pavilion mfd. in 2000). Its label says
INPUT: 100-127V @ ~6.0A (or 200-240V ~3.0A,
OUTPUT: 5 V @ 25A*, 3.3V @ 16A*, 12V @ 4A.
* (5V + 3.3V combined total cannot exceed 25A).

So total max output is 173 W. Divided by total input 720 W, gives efficiency 24%. It was called a 160 Watt unit, and still is:

http://www.impactcomputers.com/0950-3751.html

So, a P/S is rated by the delivered power.

 

[lzh007]

I believe the power supply must consider the power consume of other devices attached to the computer later, such as external hard drivers ( depends on how many usb ports ), it makes sense the total power supply is much larger than the data from ashrae. If in the real situlation there is not much future power share, just trust ASHRAE publication.