Calculating Energy Required to Cool a Computer

The best way is to get your handy dandy clamp on amp meter and see what you have. Use 3400 BTUH per KVA for a rough sizing of the A/C but don't forget to add the load of the lights and the latent heat load from the users. Usually that is between 200-400 BTUH depending on how hard they are working

Yes, it's a completely linear relationship. Every watt that the computer consumes winds up as heat. In fact, every watt anything consumes inside that room will wind up as heat and this includes the lights.

No, don't bother trying to figure out the energy of the light that escapes under the door or through the windows. It's not enough to count and it's usually more than balanced by the light coming in.

Anthony

The computer produces 3.413 BTUs per watt/hour it takes in. Your AC as an installed system has an efficiency - probably around 10 BTUs per watt/hour. It's going to be something a bit less than the SEER rating. So, barring complicated calculations, for every three watts you spend on computing you will have to spend about a watt on extracting the heat produced by that computer. This doesn't account for outside temperature, but if the room was going to be conditioned anyway that expense doesn't really count into the cost of the computing.

In a computer room I deal with, that bears out: we spend around 3000 watts on computing and about 1000 on air conditioning. Snazzy!

Notable: in the last couple of years, power supplies have gotten significantly more efficient, if you're willing to spend for it. A common power supply from a few years ago would have an efficiency somewhere around 65% and a power factor of about 0.70. You can now buy a whole selection of commodity power supplies that have efficiencies over 80% through their entire operating curve, and power factors of

Quote from

The 80 PLUS performance specification requires power supplies in computers and servers to be 80% or greater energy efficient at 20%, 50% and 100% of rated load with a true power factor of 0.9 or greater.

This makes an 80 PLUS certified power supply substantially more efficient than current power supplies and creates a unique market differentiation opportunity for power supply and computer manufacturers.

Anyway, I'm running power supplies rated as such in all my servers and computers. As a side effect, they're quieter, too, since there's less waste heat to get rid of.

-Keith

One watt of AC can remove about 3 watts of heat.

One kW is 3412 Btu per hour.

You could measure the power consumption of the devices in the room and subtract the power dissipated outside the room, if they supply that power. An old PBX or a large power transformer might dissipate most of its power outside the room.

If you remove all the heat power that's dissipated in the room, it would be the same temperature as the rooms around it. If you remove less, it would be warmer. The size of the room and the wall insulation would determine how much warmer, using Ohm's law for heatflow. For instance, if you dissipate 10K Btu/h and remove 8K Btu/h from an 8' cube with R2 walls and a thermal conductance of 6x8'x8'/R2 = 192 Btu/h-F and the rooms around it are 70 F, the cube temp will be 70+(10K-8K)/192 = 80.4 F.

You might cool it by adding high and low vents in the walls to surrounding rooms, esp if they are large, with vent area A = P/(16.6sqrt(H)dT^1.5) ft^2, where P is heat power in Btu/h, H (ft) is the height difference between vents, and dT (F) is the air temp difference between vents.

For instance, to move P = 10K Btu/h from an 83 F room to a 70 F room with H = 16'and dT = 83-70 = 13 F, A = 10K/(16.6sqrt(16)13^1.5) = 3.2 ft^2, so

Nick

It's a lot harder to measure heated air than heated liquid. I suggest one of the many PC liquid cooling kits. Then you can measure the temperature of the coolant and the volume, to come up with a better idea of the waste heat generated.

As a practical matter virtually all of the electrical energy that gets put into the room ends up as heat. The small amount that escapes as light or some mechanical energy product not eventually converted to heat will be within your margin of error in the calculations. You are measuring something with a micrometer that will be marked with chalk and cut with an ax anyway. Since computers come and go in a room a well designed system will be a staged affair where smaller units start the process and larger units come on line as the load increases. That prevents overloading the room with cool air and losing your humidity control. Compared to the old "big iron" days these systems are miniscule loads anyway.