🚀 go-pugleaf

In article <17660134@hpfcdj.HP.COM> myers@hpfcdj.HP.COM (Bob Myers) writes:
>someone (must be named 'Tom') wrote:
>>Don't forget, you can have machines that you supply with, say, 1kW of
>>electricity and which put >1kW of heat into your room. The _useful_
>>efficiency is thus >100%.
>
>OK, Tom, taking as implied that the 1kW of electricity is the only source
>of power for the machine in question, exactly how is this done?
>
Not difficult, really, note the '_useful_' qualifier (I generally use
_effective_ in the same context).  Here's an example (with which I am *very*
familiar :-):

	Ground-coupled heat-pump -- mine has a coefficient of performance (COP)
	of 3.20; this means that 3.2 watts of heat energy is delivered to my 
	house for every watt of electric power supplied.  Hence:  320%
	EFFECTIVE Efficiency.

You will quickly note that I weasel-worded my reply, you don't get 'something
for nothing'; the heatpump extracts heat from the ground via a normal
refrigeration cycle and the electric power is *NOT* the only energy input.  But
as far as the dollars are concerned, it's 320% efficient.

Walt K.