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Name: Persia
Status: student
Age: N/A
Location: CA
Country: N/A
Date: N/A


Question:
What exactly is "electrical density"? How does it relate to the design of a computer facility? I am in JETS, and my mission statement is to design a computer lab; one of the givens is the electrical density of the computer room is 200 w/ft. What exactly does this mean? I know it has something to do with electrons and surface area (D=Q/A), but that is about it.



Replies:
Hi Persia

Electrical density in the design of IT infrastructure has to do with considerations for available power in watts. The figure you quoted should be 200 watts per square foot. This number is as high as it is because IT facilities are generally populated by vertical racks of equipment that pack a lot of power consumers in a small footprint. Electrical density is tied into heat dissipation, so your air handling and cooling equipment should also be sized accordingly. Hope this helps.

Bob Froehlich


Perhaps this is an engineering and regulatory term rather than a physics term.

The words "electrical density" by themselves are a bit general, could start to refer to a bunch of things:
electric current density, (amps/mm2 in a wire)
electric charge surface-density (what you guessed),
electric outlet density (sort of what I think it means).
All significantly differing. You would have to decide which from the context.

For architects, architectural. engineers, builders, electricians, electrical safety people, and even air-conditioning engineers who have to provide cooling to remove the generated heat, it is important to know how many 120vac loads will be plugged in, within a given workspace. How many plugs, how many watts, how many wires and circuit-breakers...

Thanks for providing the specific context info:
---one of the givens is the electrical density of the computer room is 200 w/ft---
So we know it means power density, in watts per [whatever]. My "whatever' could refer to volume, area, or length, such as cubic feet of room volume, square feet of floor, or linear feet of wall-face. Sounds like they are thinking of watts available per linear foot of wall. If you know the perimeter of the room is 50 ft, then your design needs to provide 50 feet x 200 watts/foot = 10,000 watts of electrical power to that room, complete with the appropriate number and size of sockets, wires, conduits, and circuit-breakers, and a matching amount of air-conditioning to remove the dissipated heat. Sounds like quite a lot of power, I guess. Certainly it is more than the people in the room emit by their body-heat. Compare it to the sunlight falling on the roof over that room (1kw/meter2 or less)...

You also might compare this to the electrical density of one of your rooms at home. The kitchen would be much higher than the living-room or a bedroom.

It might help if you can borrow a copy of the NEC (National Electrical Code) book.

Hope you find your design exercise creative...

Jim Swenson



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