Urgent INFO needed
 

I only need an educated guess.

How much power does an airconditioner consume if it is to mantain a room 15C cooler than outside given that the room has a 1000W heatsource?

I ask this info to calculate how much a watercooling system will reduce costs given that the heat exchanging (evaporative) occurs outside. :)

Thanks :) :cool:

It depends on the quality of the insulation in the walls ;)

You'd have better luck googling for "how to size an air conditioner".

There's also a limit to air conditioners, in that they can only draw so much current from a normal wall outlet. I used to have an air conditioner that drew the maximum 15 amp @ 120 Vac. It would cool most of the appartment I used to live in, but the wiring was faulty, and the starting current kept tripping it up.

Let's say, walls found in a server room, and unlimited electricity.

Sorry for not giving enough info. I just need an average.

You can get 220 single phase out of two 110 lines, which will give you plenty of power. I wouldn't do this if you don't have any alot of experience. Or just get 220ac to that room.

Can't you just say the server room is underground? (I take it this is for another project).

Hehe, ok. The problem is that I don't know what BTUs stand for and all I wanted to proove is that watercooling would be gainful. But I want some numbers ;)

www.convert-me.com

There's your BTU converter ;)

1000Watts every hour is 1 Kilowatt/hour
Then convert it to BTUs about 3500

I just checked my 8000BTU AC unit and it uses about 800W of 120VAC. Keep in mind that is max output and it holds a good part of the house at 72F even if outside its well above 90F.

The figure you're looking for is the "coefficient of productivity" or COP. It's the ratio of heat extracted to power used by a heat pump. Most units have a COP of 2.5 to 3.5 - pulling out those 1000 W of heat will require 285 to 400 W of electricity.

If you search for COP and 'air condioning' you'll find the numbers posted by various manufacturers (they'll look good in your report). Some will list EER (energy efficiency ratio) instead of COP; COP=EER/3.412.