[Beowulf] cooling question: cfm per rack?

[David Mathog]

In designing a computer room two key factors are:

1.  Power in   (electricity)
2.  Power out  (A/C)

The second term really has two parts: 

  A.  the amount of air moved
  B.  the reduction in temperature of that air across the A/C unit

The latter part is specified in tons.  The A/C guys I've spoken
with recently utilize some more or less standard relationship
between cubic feet per minute (cfm) and A/C tons for the units they
maintain.  These run off the campus cold water supply, so
it makes sense that heat out is proportional to flow across, assuming
that the cold water has a very large heat capacity.

However, in terms of cooling the units themselves, the amount of
air flow through the racks is also important.  That flow is
also in cfm.  Ideally cfm through the racks would be equal to cfm
through the A/C, ie, all air goes once through the racks and then
directly through the A/C.  Even more ideally cfm through _each_ rack
could be modulated somehow, since some racks move much more 
air than others and putting a low flow rack next to a high flow rack
might drive the air the wrong way through the low flow unit.

How does one calculate an optimal cfm through a rack? 

For a specific example with round numbers, let's say it's a
25U rack, dissipates 10kW, and has a single 50 cfm per minute output
fan per 1U node.  (Ie, all air out must go through that path.)

There seem to be a bunch of variables that are hard to deal with. 
For instance, adding the exhaust fans would be 50*25 = 1250 cfm.
Is that all there is to it? But that type of fan only runs at
the stated flow rate if the pressures are exactly as specified. 
Without incredibly careful balancing of the pressure across the
rack it won't generally run at 50 cfm.

Is cfm the key unit here or should one think in terms of pressure
at various points in the room?

Thanks,

David Mathog
mathog at caltech.edu
Manager, Sequence Analysis Facility, Biology Division, Caltech