Load calculation of computer system

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kevinware

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Can someone help me to calculate the load for this branch circuit.

I have 15 servers in one rack.

13 of them are pulling 8 amps at 208V if fully loaded.

The other two are pulling 24 amps at 208V if fully loaded.

The calculation I have are as follows:
8 x 208 = 1664VA x 13 = 21632VA
12 x 208 = 2496VA x 2 = 4992VA
Total VA = 26624VA

Article 645-5(a) Tells me I SHALL have an ampacity of not less then 125% of the total connected load.

26624VA x 1.25 = 33280VA

Is there a section of Article 210 that allows me to derate this load, because all servers will never be 100 % loaded?

Thanks for your help
Kevin
 
Re: Load calculation of computer system

If all the equipment is on one branch-circuit, then no demand factor can be made. The branch circuit must be able to handle the total possible load that CAN be applied. Even if under normal operation they will not all be on together, the branch0circuit must be sized for that condition.

Now, if these were fed from a feeder, there is a possibilty that a small derating could be made, however there is not a specific factor specified by the code.

Good Luck! :)
 
Re: Load calculation of computer system

Kevin
645.5A says that the branch circuit shall have the ampacity of 125% of the "connected load".
Not much discussion there. This load is sure to have the dreaded 3rd harmonic. You need to verify
how much this will be. You will need to consider increasing the neutral size.
 
Re: Load calculation of computer system

Kevin, I agree with the other two responses. As a data center power engineer I suggest you never skimp on brnach circuits. It is far easier to explain over engineering, rather than a failure.

Use the name plate rating, add them up, multiply by 125%, and size conductors for 3% voltage drop. You will sleep better.

The servers will never be 100%, Hmmm, reboot.

[ June 04, 2003, 07:10 PM: Message edited by: dereckbc ]
 
Re: Load calculation of computer system

Just a note, U might want to use tvss circuit breakers for the servers, does anybody else have information on TVSS? I have to design one for about 16 computers.
 
Re: Load calculation of computer system

Bonding Jumper, ANSI 62.41.1992 is the best application guide for TVSS, but it is not cheap. Most TVSS applications are junk if installed as an after thought of the original electrical installation. The add on models are shunt type devices and you won't be able to keep the lead lengths short enough to be effective.

The best divices are Kelvin Clamps or series TVSS. These types have to be part of the original design or retrofitted. For example buying service entrance switch gear with a built-in Kelven clamp device

It starts at the service entrance by installing a class "C" device, then adding class "B" devices on all sub-panels and step-down/isolation transformers. The class "A" device (you know the ones you buy for your PC) installed on branch circuits are pretty much useless unless backed up by upstream class "C" and "B" devices.

Here are a few web sites with info, but none as as in depth as 62.41.1991. Try google and search for TVSS APPLICATION.

http://www.atis.org/pub/peg/peg2000/reed.pdf

http://alds.stts.edu/APPNOTE/PowerControl/SurgeApplication.pdf
 
Re: Load calculation of computer system

Originally posted by kevinware:
Can someone help me to calculate the load for this branch circuit.

I have 15 servers in one rack.

13 of them are pulling 8 amps at 208V if fully loaded.

The other two are pulling 24 amps at 208V if fully loaded.

The calculation I have are as follows:
8 x 208 = 1664VA x 13 = 21632VA
12 x 208 = 2496VA x 2 = 4992VA
Total VA = 26624VA

Article 645-5(a) Tells me I SHALL have an ampacity of not less then 125% of the total connected load.

26624VA x 1.25 = 33280VA

Is there a section of Article 210 that allows me to derate this load, because all servers will never be 100 % loaded?

Thanks for your help
Kevin
Click to expand...
I find this really hard to believe. I have never seen a rack mounted Server that has a 1600 watt power supply. The largest one i found on the Dell web-site was 900watts. Some of these servers have redundant power supplies, but they are just that, redundant. They do not draw 1800 watts, but 900. Each power supply has to be able to take the entire load. But just because the power supply is rated at 900 watts, does not mean that is the connected load for that computer. The computer would seldom if ever run at 100%. That would mean you have all the expansion slots full, all the hard drive bays full, etc. And all that is running at 100% inside that computer at the same time. If did your UPS sizing calculation for this computer room based on these loads, you would have a very hard time finding a ups to feed a computer room larger than a couple racks. We have done large computer rooms with a 80kva ups that had many more then the three racks that you would be able to install. the computer room has been up and running for almost two years with no problem. The UPS is about 43% loaded also. We used a 40% diversity factor. The UPS companies use about 25-30% for their sizing purposes. These percentages are based on the name-plate rating. The 43% it is loaded right now is after they added computers after our initial design.
That rack is probaly about 2' x 4' at the most, so that is 8SF. leave about 3' in front and back and you have about 20SF. If you take your 26624va/20SF and you end up with 1331 va/sf. That is very high. Never seen a computer room designed for that kind of density in my life. Nothing even close to that. I think you need to check your numbers again.
 
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