James.P.Lux at jpl.nasa.gov
Thu Jun 19 16:14:59 EDT 2003
Simple.. all the electrical power is converted to heat. 1 unit @338W and 10
units at 295W gives you, to a first order, 2950+338 = 3288 Watts... So, in
energy terms, that's 3.3 kJ/second.
You can convert kJ to calories, BTU, tons of ice, etc.
A comment: 295 and 338 appear to be overly precise numbers. I would venture
to guess that the instantaneous power consumption of a PC varies somewhat,
depending on instruction stream (fine scale) and environmental effects
(line voltage, air temperature, etc.) You might just assume 350W per node
and be done with it...
Nit picky note: indeed, the information coming out of the cluster is
somewhat more organized (i.e. lower entropy) than the information going in,
and there is some small, but non-zero, amount of energy required to reduce
the information entropy. Then again, one can also use the energy to
randomize the data.
At 09:16 AM 6/19/2003 -0700, you wrote:
>We need to know the heat generated (that is, how much
>cooling is needed) by a beowulf cluster of 11 units.
>Each unit has 2 Pentium 4 processors. The master node
>has a power consumption of 338 W and each slave node
>Where can I get this information?
James Lux, P.E.
Spacecraft Telecommunications Section
Jet Propulsion Laboratory, Mail Stop 161-213
4800 Oak Grove Drive
Pasadena CA 91109
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