I am designing a computer room inside my home. The room walls are inside the house. This question excludes floor, wall, and ceiling losses/gains.
I am looking for the formula to calculate the CFM required to keep the room interior temperature close to the exterior temperature.
I have been using this formula:
hs = 1.08 q dt
where:
- hs = sensible heat (Btu/hr)
- q = air volume flow (cfm, cubic feet per minute)
- dt = temperature difference (oF)
- 1.08 = the heating BTU multiplier at sea level, or .075-lbs. of air per CFM x .24 (the specific heat of air) x 60 (minutes in an hour.) This factor will vary at higher altitudes and temperatures.
Example:
6,800 BTUs of heat generated in a 12x14x8 (1344 cubic feet) room. The incoming air temperature is 72 degrees. The target room temperature is 78 degrees.
q = hs / (1.08 dt) = 6800 / (1.08 x 6) = 1079 CFM.
- Is this formula correct or is there a more appropriate formula?
- What is the formula for heat rise in case the ventilation system fails or the CFM degrades?
Note: I am not looking for rules of thumb. The room does have a mini split, but during the winter I want to ventilate the heat into my home and not condition the heat. During the winter the mini split is part of my ventilation failure cooling plan. The failure plan includes a temperature alarm that shuts down the UPS system which shuts down the computers.
[Update]
For those that are also thinking about a dedicated computer room at home:
- The room is 12 x 14 with 8-foot ceilings.
- The room has a dedicated subpanel with an outside generator hookup.
- 3,000 Watt APC Smart UPS
- Supports 2,000 watts of computer equipment.
- Dual gigabit Internet providers connected to a bonded router.
- 12,000 BTU mini split air conditioner with low-temperature cooling support (wine room).
- Temperature alarm shuts down UPS which shuts down the computers.
- Plumbed for water and drainage.
- In-wall dehumidifier.
- In-wall humidifier.
- Medify Air Purifier.
- Incoming air: MERV 13 4-inch filter.
- Two temperature-controlled variable speed 10-inch duct fans to exhaust heat during the heating season. Only one is required, two provide redundancy. Shutdown during the cooling season. Mini split ensures a temperature ceiling. If the mini split turns on the fans turn off (current sensing relay).
Google now recommends setting data center temperature to 80 degrees. This recommendation works with my goal of exhausting air into the house during the heating season. [link]
The purpose of this question is to calculate the CFM required to exchange the air in the computer room to maintain a reasonable temperature and take advantage of the heat generated during the heating season. During the cooling season the ducts are closed and the mini split provides for conditioned air. The MERV 13 filter ensures clean air in the computer room during heating season. A standalone air filter provides air quality during the cooling season.
There are additional construction details to support a low-cost conversion to a wine room to make a home sale easier in the future. My comments under the accepted answer provide some of those details such as R-19 insulation in the walls and ceiling, exterior grade insulated glass interior door, etc.
Best Answer
1 BTU is the amount of heat to raise 1 pound of water 1 °F (definition)
Specific heat of air is 0.24 (takes 0.24 BTU to raise 1 pound of air 1 °F)
A pound of air is about 13.9 cubic feet (at some specific temperature, probably 68 °F / 20 °C but that could be at a different temperature. The figure is in my spreadsheet, the notes regarding what temp/press that applied for are on decades old paper.) Also depends on the air pressure / altitude.
So you have 96.7 pounds of air in your room, approximately (subtract volume of room contents, quibble about temperature and pressure, etc...)
I'll presume you have (as people do, but really, getting the units helps to ensure correctness) shorthanded 6800 BTUs/hour as "6800 BTUs" - i.e. a 2 KW load, pretty much.
So without ventilation, the temperature rise per hour (discounting walls/floor/ceiling, or treating them as infinitely insulated) is 96.7 lbs air getting 6800 btus/hr dumped into it. The specific heat tells us that 96.7 pounds of air behaves like (96.7 * 0.24) 23.21 pounds of water.
A BTU is (pounds * °F) so BTU/Hr is (pounds * °F)/hour so to get °F per hour we divide by the pounds (of water, which gets the specific heat in there) leaving °F/hour. 6800BTU/hr / 23.21 bs (water) = 293 °F/hour which suggests that treating the walls/floor/ceiling as infinitely insulating might be a poor assumption for modeling actual temperature rise, as heat flow through them will become quite significant at large temperature differentials.
Thats about 2-3/4 gallons of water (8.35 lbs/gallon), and you can certainly bring that to a boil in less than an hour on a 2KW stove burner, so it sanity-checks.
You proposed to ventilate at 1079 CFM, or 77.6 lbs/min or 4657 lbs/hour (air) which is equivalent (in thermal mass) to 1118 lbs of water, and results in a 6.1 °F/hour rate of rise. That might be down to quibbles with constants, but suggests 1096 CFM to get a 6 degree rise on my figures.
With "R19" walls and ceiling, - you have 208 square feet of wall area, + 168 ft of ceiling, nominal differential of 6 °F, and resulting heat flow of 118 btu/hr from the walls (376 sq ft/(19 °F Hr Sqft)/BTU))* 6 °F) which will of course rise as the temperature differential increases. Assembeled stucture value is typically less than the insulation batts without framing, so likely a bit higher than that (figures vary with details, but R15 is more typical for the "whole wall" value - 150 btu/hr) Slab floor you'll get some conduction into adjacent rooms, typical 2 inch XPS is R10, but the temperature is "ground as affected by house" (I generally use 40F , since a slab is not deep enough to get to the 50-52F layer and I'm looking at heating season, mostly.)
Water cooling adds complexity and leak potential but makes it much easier to move heat around (or out of) the house (or into the water heater inlet) quietly.