Looking to create a temperature rate of rise table for air

[Sola]

How did you find PF?: Google

l was asked how hot would our IT space get if we lost all air conditioning. The space is 20' x 15' x 10', We keep set point at 70 F and have 80 kW of load. Assuming this exists in a well insulate container with 20 racks of equipment. Doesn't have to be perfect but a general idea as we compare to various equipment failure thresholds

 

[berkeman]

Welcome to PF.

How much airflow is there in and out of the room under those conditions? Are you assuming no forced air at all, so only convection flow? What is the ambient temperature outside of the room and outside of the building? How big are the vents into the bottom of the room and out of the top of the room?

Also, how hot does it need to get in there before some of that load automatically sheds? I assume there are temperature sensors that have some shutdown capability when the racks start overheating?

 

[Sola]

I am making broad analysis without many specifics,
No air flow,
minor leakage (10%)
There are split systems in the space.
Ambient I choose is 50 F.
The various IT equip will shut off at various temps above 105 F.
My question is "what is the heat RoR" with the 80 continuing to heat the space. How much time in an emergency before we do a graceful shutdown vs hard stop. I am trying to graph this so we can show operators the data

 

[Sola]

BTW as a new user, where would I find posting procedures? TY for the help

 

[berkeman]

BTW as a new user, where would I find posting procedures? TY for the help

Under INFO at the top of the page you'll find the Rules and About Us pages. But for general posting guidelines, there are only a few:

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That's about it, I think. Feel free to send me a Personal Message (PM, click my avatar and "Start a conversation") if you have any more questions about PF. Enjoy!

 

[russ_watters]

l was asked how hot would our IT space get if we lost all air conditioning...Doesn't have to be perfect but a general idea as we compare to various equipment failure thresholds

HVAC engineer here. The short answer is infinity. The longer answer is extremely difficult to calculate but suffice to say the room would be uninhabitable and the IT equipment would fail before it got to an equilibrium - in a matter of minutes.

For the room air, you can calculate the rate of rise from mass, heat capacity and heat input. The mass of equipment won't matter immediately so that should get you through the first 5-10 min.

[Edit] Nope, just ran the numbers and it's seconds, not minutes.

 

[Sola]

TY Russ, Is there a formula or calculator to input the room data. Per above without getting very specific on building envelope for a telco hut. 20 x 15 x 10' CMU exterior, r-19 wall insulation, no windows, metal truss roof with 4" roof rigid roof insulation on a metal corrugated deck. We use split hvac systems. I am learning Newton's laws and looking to build an excel sheet that I can do high level comparisons at different loads. TY for comments.

 

[russ_watters]

TY Russ, Is there a formula or calculator to input the room data. Per above without getting very specific on building envelope for a telco hut. 20 x 15 x 10' CMU exterior, r-19 wall insulation, no windows, metal truss roof with 4" roof rigid roof insulation on a metal corrugated deck. We use split hvac systems. I am learning Newton's laws and looking to build an excel sheet that I can do high level comparisons at different loads. TY for comments.

m * Cp * ΔT = Q (heat input)

And pick a rate (E.G. heat rate per minute, ΔT per minute).

You have a fixed mass of air in the room, it has a certain heat capacity and you apply heat at a certain rate, resulting in a rate of temperature change. That enough to run the calculation yourself?

The rate of temperature rise is way too fast for the heat loss through the walls to matter.

 

[Dullard]

I just did a similar analysis for a customer who wanted me to build him a bitcoin mining container. 80KW in that space (without ventilation or refrigeration) will get really hot really fast (like: RUN!). 80KW is approx 23 Tons of heat load.

 

[Sola]

m * Cp * ΔT = Q (heat input)

And pick a rate (E.G. heat rate per minute, ΔT per minute).

You have a fixed mass of air in the room, it has a certain heat capacity and you apply heat at a certain rate, resulting in a rate of temperature change. That enough to run the calculation yourself?

The rate of temperature rise is way too fast for the heat loss through the walls to matter.

So ?
m= mass (volume of air in the space )3000 cu ft or 229 lbs ?
Cp=total heat capacity of air would I use 1 or .718 (Cv)
Delta T
I think so.. Let me take a shot.
My initial approach was to convert the 80 kW to Btuh (276,300) / 60 (minutes), div by 229 (lbs) = 20 deg F per minute. That seemed really high so here I am

 

[Sola]

Additionally I would have a start point of 70 F and average ambient of 50 F, as we have locations across the country

 

[gmax137]

To figure the equilibrium temperature you need the surface area (I got 1700 ft2) and an estimate of the heat transfer coefficient from interior to ambient. With no forced convection (fans) the free convection coefficient would be limiting (unless the wall insulation is really good).

A typical free convection coefficient would be 1 or 2 Btu/hr-ft2-F. The the temperature drop across the wall would then be 80 to 160F (to transfer your 273,000 Btu/hr). So for 50F ambient, the interior would be 130 to 210F. That's about as rough an estimate as you're likely to get. No leakage, no equip shutoff, ...

 

[russ_watters]

So ?
m= mass (volume of air in the space )3000 cu ft or 229 lbs ?
Cp=total heat capacity of air would I use 1 or .718 (Cv)
Delta T
I think so.. Let me take a shot.
My initial approach was to convert the 80 kW to Btuh (276,300) / 60 (minutes), div by 229 (lbs) = 20 deg F per minute. That seemed really high so here I am

Close, but I think you mixed units there. In English units, the heat capacity is about 0.24 (but you didn't use it?). So the answer is about 84F / min.

So, even worse but you were in the right OOM.

 

[russ_watters]

To figure the equilibrium temperature you need the surface area (I got 1700 ft2) and an estimate of the heat transfer coefficient from interior to ambient. With no forced convection (fans) the free convection coefficient would be limiting (unless the wall insulation is really good).

A typical free convection coefficient would be 1 or 2 Btu/hr-ft2-F.

Wall insulation was provided and it's really good: R-19 = U of 0.053 BTU/hr/ft2-F. That's why convective heat transfer is usually ignored for walls for building envelope loads.

 

[gmax137]

Thanks @russ_watters -- I didn't know or didn't remember that the R "value" is the inverse of the U (with U in "imperial" units: Btu/hr ft2 F).

The short answer is infinity.

Yes, using that 0.05 Btu/hr ft2 F the equilibrium temperature works out to >3000 F. Showing it is silly to ignore the leakage etc. Rather, best to focus on the rate of rise and the given temperature limits, as you said in your original reply.

 

[Sola]

Close, but I think you mixed units there. In English units, the heat capacity is about 0.24 (but you didn't use it?). So the answer is about 84F / min.

So, even worse but you were in the right OOM.

Russ,

Thank you, This is a great learning experience.

 

[Sola]

Russ,
Ok now the formula makes sense but I am stunned how much heat is generated and how quick. Still thinking I am missing some variable

 

[russ_watters]

Ok now the formula makes sense but I am stunned how much heat is generated and how quick. Still thinking I am missing some variable

I think you're just expecting another variable to be there that isn't. Watts is watts and 80 kW is a lot of watts. Period. Every watt that the data center uses is converted to heat.

Maybe if you think about it in terms of a unit of heat you recognize. Like a hairdryer for example (on full). This is 55 hairdryers.

Or, I live in a 1500 sq ft townhouse in the northeast of the US and my central heating is generously sized. This data center is 4x the heat output of my house's heater in 1/5 the space. It's enough to run all of the air in the room through the heater, with it's 60F temperature rise, in 30 seconds.

 

[Lnewqban]

Russ,
Ok now the formula makes sense but I am stunned how much heat is generated and how quick. Still thinking I am missing some variable

I would say that you can’t have so much generated heat within that space.
In my experience designing cooling for server rooms, it is safer and more practical to measure the actual amount of amps used by all the electronic equipment in the room in normal operation.
When the approach was adding up all the amps shown in the nameplate of each piece of equipment, the AC units resulted grossly oversized.

Another way to approach your problem could be to calculate how much heat your AC equipment is normally moving out of the room while keeping set temperature, based on its performance and capacity at normal working temperatures.