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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.

  1. Is this formula correct or is there a more appropriate formula?
  2. 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.

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    You're missing the specific heat of air. Which I have as per pound, since you are in BTUs, and that leads to a conversion of cubic feet per pound of air. Perhaps that's baked into your figures in a non-obvious way, I don't recall ATM. Units are really good for keeping track and being sure it makes sense. I'll have to delve into my spreadsheets whence it lives. Nice hard engineering numbers, not RoT.
    – Ecnerwal
    Mar 2 at 23:16
  • It might be a good idea to edit your question and post where you got the formula from. Mar 2 at 23:44
  • @Triplefault - I don't know where I obtained that formula. My notes are a few years old. This is a project that has been delayed for two years. Mar 3 at 0:18
  • You might well have already factored this in, but thermal input includes the power for the fans. as well as all the computer equipment.
    – Ecnerwal
    Mar 3 at 2:37
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    In the spirit of making this more useful for posterity, can you include what "CFM" stands for/what it means?
    – Martha
    Mar 3 at 16:27

2 Answers 2

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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.

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  • Thank you. The number is 6 degrees (78 - 72). I understand what the variables and components are, I am both an engineer and architect in computers, I am trying to determine the formulas and not the explanations. Can you add to the bottom of your answer the formulas for my two questions (CFM and heat rise as CFM degrades or drops to zero)? The item that bugs me is the impact of mixing air at two temperatures. Although 1000 CFM is not that high, it does take a 10-inch duct fan to achieve that. Mistakes now will be expensive to correct later. Mar 3 at 0:16
  • Regarding your last edit. The temperature differential is only 6 degrees for the four walls and ceiling outside (72) to inside (78). I am not sure about the floor differential (insulated slab). Therefore the temperature differential is not large when functioning normally. When a threshold is reached (90 degrees) everything shuts down via sensors. In any event losses from inside to outside actually help my objective. Mar 3 at 0:24
  • I already answered for 0 CFM - I did misread your incoming as 74F for some reason. Actual rate of rise at 0 CFM will be less than 293°F/hour due to real walls/floor ceiling not being infinitely insulating. If airflow drops by 50%, temperature rise per hour will double to 12.2 °F/hr. It's all how many pounds are you dumping that heat into. But reality will intervene as the density lowers becasue the temperature is rising in ways that will alter the CFM/lb in a non-ideal manner. Or you'll shut it down ;-)
    – Ecnerwal
    Mar 3 at 0:24
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    @FreeMan - I actually mentioned that in my last comment which I then deleted. I agree with you. I find these details fascinating. I have a better appreciation for why putting a bunch of computers in my living room makes the room uncomfortable even with a good HVAC system. That is one of my reasons for a real computer room for my work which will be next to my home office. Air quality (dust), noise, and security are the other reasons. Mar 3 at 0:42
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    For your latest comment. I excluded the wall and ceiling R factor because the room is insulated to R-19 walls and ceiling (wine room specs). The reason is that the room can later be easily converted to a wine room if I want to sell the house. Computer rooms have a negative home value but wine rooms are a plus in the right areas. I have a computer room with majestic insulated redwood glass doors. Mar 3 at 0:50
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Excellent answer by Ecnerwal, who besides being one of the top experts on this site, knows more physics than I realized.

This question piqued my interest, so I worked through it myself. I get a slightly different answer because I used a different weight of air.

Note that the temperature is the exaust temperature. You will most likely have hot spots that are higher.

From the equation, if the air flow is zero, the temperature rise is infinite. Of course, this won't happen; but with an insulated room, it will get really hot until you get enough heat flow through the floor, walls, and ceiling to balance.

A bit of trivia: an average person generates about 100W of heat. If you have more than a few people in a room, this can be significant.

enter image description here

Formulas:

enter image description here

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  • Thank you. The spreadsheet helps my documentation objective. Is there a way you can make this downloadable? Mar 3 at 9:22
  • I am concerned with hot spots. There will be a 19-inch rack with its own fans pointing in the right direction - front-intake on the same side as room intake - exhaust high on the same side as room exhaust inlet. The exhaust inlet is high from the opposite side of the room from the low intake to help with airflow through the room (front bottom to top rear). Normally, no people in the room - designed for equipment only. Mar 3 at 9:28
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    @JohnHanley - Can't make it downloadable without posting it somewhere else. But, I can add the formulas here.
    – Mattman944
    Mar 3 at 12:54

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