Need ball park power and cooling requirements
muno at aem.umn.edu
Wed Oct 2 10:22:34 EDT 2002
On Wed, Oct 02, 2002 at 09:19:31AM -0400, Robert G. Brown wrote:
> > Rough BTU/hr cooling = (((Amps * Volts) * 0.8) / 1000) * 3413
> > The 0.8 is something called "power factor" I think, and largely
> > relates to how efficiently the power is converted and used. I
> > believe the 3413 is just a conversion factor; it takes 3413 BTUs/hr
> > to cool a constant load of 1250VA at 0.8 power factor. I think. =]
> > Rough AC Tonnage = BTUs per hour/12000
> It is a lot easier to combine these two computations, especially when
> the power factor is irrelevant to the cooling requirements (it increases
> the peak current required to produce a given AVERAGE power consumption,
> but the AC has to remove the AVERAGE power being consumed, not the peak
> VA). If you do a long and tedious computation that converts a "ton" of
> AC (air conditioning that can remove enough heat to melt a ton if ice at
> 32F in 24 hours, so help me -- God help the evil fiends who came up with
> this non-SI-unit abomination:-) into something approximating sane units
> (SI) one concludes that a one ton air conditioner removes a teen bit
> more than 3500 Joules of energy per second (3500 watts) continuously.
> HOWEVER, when computing the wattage that needs to be removed, one has to
> remember that heat bleeds into any cool space from any contiguous hotter
> place, AND that you need to room to be considerably cooler than ambient
> air in most situations. Building ambient might be (e.g.) 72F but you'd
> like your server room to be (e.g.) 60F. You'd need some AC capacity to
> maintain this temperature differential even for an empty room with no
> nodes at all.
> The >>baseline<< estimate for cooling capacity is therefore based on the
> number of nodes.
> d) T = N*W/3500
> where T is the number of tons of AC, N is the number of nodes to be run,
> W is their average power, and 3500 converts Watts in to Tons. You can
> then apply a fudge factor to come up with your actual requirement:
> d') T' = T*1.2 or T*1.5 or whatever.
> The correction should be sufficient to deal with human bodies in the
> space (100W each), electric lights in the space (add their wattage),
> heat coming in through walls, windows(! -- a window can add LOTS of
> wattage in the form of sunlight and conduction), and any other non-node
> electrical equipment you might have handy -- monitors, switches, video
> games to play while waiting for jobs to run.
> Warning/disclaimer: Although I've been gradually building up the
> understanding explicitly represented above for some years now, it could
> still have some errors in it (although it almost certainly is, if
> anything, too conservative and not too liberal). I invite comment and
> correction from those on the list that are wiser than I. I'm very
> shortly going to go ahead and add the above to my online book so that
> this FAQ can be referred to the book in the future.
In addition, code will require some fresh air addition to the room, in most
cases, to accomodate the average number of people that will occupy the room.
In our case, "fresh air" means humidity laden warm air drawn from an intake
on the roof. You need additional cooling capacity to handle that additional
load as well.
Ray Muno http://www.aem.umn.edu/people/staff/muno
University of Minnesota e-mail: muno at aem.umn.edu
Aerospace Engineering and Mechanics Phone: (612) 625-9531
110 Union St. S.E. FAX: (612) 626-1558
Minneapolis, Mn 55455
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