[Beowulf] Dutch Airco

[Lux, Jim (337C)]

On 12/1/11 11:10 PM, "Vincent Diepeveen" <diep at xs4all.nl> wrote:

>Hey all, I live in The Netherlands, so no surprise, solving the
>cooling in a barn for a small cluster i intend to solve by blowing
>some air to outside.
>
>Now the small cluster will be a few kilowatt only, so i wonder how
>much air i need to blow in and out of the room to outside.
>
>As this is a cluster for my chessprogram and it maybe 1 day a year
>reaches 30C outside, we don't have to worry about outside temperature
>too much,
>as i can switch off the cluster when necessary that single lucky day
>a year. If it's uptime 99% of the time this cluster i'm more than happy.
>
>Majority of the year it's underneath 18C. Maybe a day or 60 a year it
>might be above 18C and maybe 7 days it is above 25C outside.
>
>I wouldn't have the cash to buy a real airconditioning for the
>cluster anyway, as that would increase power usage too much, so
>intend to solve it Dutch style.
>
>Interesting is to have a function or table that plots outside
>temperature and number of kilowatts used, starting with 1, 1.5, 2,
>2.5, 3, 3.5, 4, 4.5 ,5, 5.5 , 6
>kilowatts. For sure cluster won't be above 6 kilowatt.
>
>First few weeks 1 kilowatt then it will be 2 kilowatt and i doubt
>it'll reach 4 kilowatt.
>
>Which CFM do i need to have to blow outside hot air and suck inside
>cold air, to get to what i want?


What you didn't say is what temperature you want your computers to be at
(or, more properly, what temperature rise you want in the air going
through).

It's all about the specific heat of the air, which is in units of
joules/(kg K)...  That is it tells you how many joules it takes to raise
one kilogram of air one degree.  For gases, there's two different numbers,
one for constant pressure and one for constant temperature, and for real
gases those vary with temperature, pressure, etc.

Q = cp * m * deltaT

Or rearranging

M = Q/(cp*deltaT)

But for now use Cp (constant pressure) which for air at typical room temp
is 1.012 J/(g*K)

You want to dump a kilowatt in (1000 Joules/sec), and lets assume a 10
degree rise (bring the air in at 10C, exhaust it at 20C)

M = 1000/(1.012E-3*10) = about 0.1 kg/sec

If the heat load is 5 times, then you need 5 times the air.
If you want half the temp rise, then twice the air, etc.


How many CFM is 0.1 kg/sec?   At 15 C, the density is 1.225 kg/m3, so you
need 0.08 m3/sec

(as a practical matter, when doing back of the envelopes, I figure air is
about 35 cubic feet/cubic meter... So 0.08*35...)


 About 170 cubic feet per minute per kilowatt for a 10 degree rise




Be aware that life is actually much more complicated and you need more
air.  For one thing the heat from your box is evenly transmitted to ALL
the air.. Some doesn't go through the box, so what happens is you have,
say, 200 cfm through the box with a 10degree rise and 200 cfm around the
box with zero rise, so the net rise is 5 degrees.


Also, the thermodynamics of gases is substantially more complex than my
simple "non-compressible constant density" approximation.  Since Tin and
Tout are close here (280K and 290K) the errors are small, but when you
start talking about rises of, say, 20-30C, it starts to make a difference.