> 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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