You might want to factor in any live bodies in the room also -
esprcially if someone'd going to be there for an extended period.
RP

On 16/01/07, Justin Richards <justin.richards@gmail.com> wrote:
> Hi Harold,
>
> thanks for confirming/reinforcing my understanding of the problem.
>
> Cheers
>
> Justin
>
> On 1/16/07, Harold Hallikainen <harold@hallikainen.org> wrote:
> >
> > > A discussion came up about sizing an airconditioner for a
> > > computer/equipment room.
> > >
> > > I suggested simply measure the current and voltage applied to the room
> > > to determine power consumed (and average over some time) and this will
> > > be the heat/energy that needs to be removed.  (Unless there are cables
> > > that transfer the power out of the room)
> > >
> > > My collegues indicated that not all the power consumed by the
> > > processors and other hardware is converted into heat.  The energy is
> > > converted into work done by the processor.
> > >
> > > My understanding is that this is not the case and all the work done by
> > > a processor and other hardware must be converted into heat (or light
> > > which is still heat, or movement but when anything that finally comes
> > > to rest must have converted its stored energy to heat via friction)
> > > otherwise it is kind of 'disappearing'.
> > >
> > > Am I off the mark here.
> > >
> > > Regards
> > > Justin
> >
> > I'd say you're on the mark with one exception. Since the power factor is
> > unlikely to be 1.000, the power in watts is not going to be quite as high
> > as the apparent power in volt-amps. It's probably close enough, though. If
> > you wanted to be more accurate, use a wattmeter instead of a volt meter
> > and ammeter. Overall, if it's a sealed room with AC power coming in and
> > maybe netowrk cables going out, power in is converted to a relatively
> > small amount of electrical power going out on the network cable (possibly
> > the same as that coming in on the network cable). The rest of the power
> > stays in the room and is converted to heat, which leaks out to outside
> > through the room insulation, is exhausted by fans, or is pumped out with
> > an air conditioner.
> > So, I think your original statement is almost precisely correct.
> >
> > Harold
> >
> >
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