Re: Capacity and Efficiency of Small A/C Units

[Peter Lawton]

In <bd4dcl$lvk@acadia.ee.vill.edu> Nick Pine wrote:
> Tim Keating <NotForJunkEmail@directinternet11.com1> wrote:
>
>>A device which outputs 17.1 BTU's for every watt hour of input would
>>have a COP of ~5
>>
>> BTU/ (Wh*3.412) == COP.
>
> Right...
>
> Nick


The formula is right, but the description is a little off, I think.
Nick, correct me if I'm wrong, but I think the COP is based on the heat
moved (numerator), not counting energy in (work req'd to make system run,
denominator).

I.e., I don't think the numerator is the sum of the heat removed from
the cold side and the work of compression (and motor efficiency factor
in...).


OT: Usually I just lurk here, but since I'm posting, ... Thanks, Nick,
for all your diligent comments when folks mis-use units and terminology.
I am seriously interested in the field of energy, and I realize how
important it is to communicate clearly and correctly, especially when
it's me that's learning.


Peter


(when replying directly, remove the obvious in my posted address)

[daestrom]

"Peter Lawton" <peter@IHateSpam.alwaysthinking.com> wrote in message
news:20030623010659990-0400@news.earthlink.net...
> In <bd4dcl$lvk@acadia.ee.vill.edu> Nick Pine wrote:
> > Tim Keating <NotForJunkEmail@directinternet11.com1> wrote:
> >
> >>A device which outputs 17.1 BTU's for every watt hour of input would
> >>have a COP of ~5
> >>
> >> BTU/ (Wh*3.412) == COP.
> >
> > Right...
> >
> > Nick
>
>
> The formula is right, but the description is a little off, I think.
> Nick, correct me if I'm wrong, but I think the COP is based on the heat
> moved (numerator), not counting energy in (work req'd to make system run,
> denominator).
>
> I.e., I don't think the numerator is the sum of the heat removed from
> the cold side and the work of compression (and motor efficiency factor
> in...).

Actually, the thermo texts that I have say there are two different COPs.
One for refrigeration and one for heat pumps. The COP for refrigeration is
(Heat removed from cold box)/(Energy needed to remove said heat). And the
COP for heat-pumps is (Heat delivered to hot box)/(Energy needed to deliver
that heat).

The difference being that COP for heat-pumps considers that all the energy
used by the heat pump is also delivered to the condenser, whereas the energy
used by a refrigerator is not energy removed from the evaporator. Given the
COP for a heat pump, the COP for refrigeration is simply:

COP refrig = COP heatpump - 1

So for A/C, I would advise the COP for refrigeration <heat removed from
cold> / <energy needed to operate device>

daestrom