[TriLUG] OT: thermodynamics of A/C question

[Jeremy Portzer]

On 7/5/2012 1:28 AM, Joseph Mack NA3T wrote:
> On Wed, 4 Jul 2012, Steve Litt wrote:
>
>>> The assumption was that the A/C is a Carnot machine and in
>>> the early morning the condenser would be operating at a
>>> lower temperature. From the Carnot cycle efficiency formula,
>>> I expected I'd get 50% more efficiency.
>>>
>>> No-one had an answer to this
>
>> I saw lots of answers to this.
>
> I must have missed them. What's the increase in efficiency (or cost 
> savings) from moving a fixed number of BTUs in the early morning when 
> the air is cool compared to late afternoon when the air is hot?
>

Um, the answer was that it's unanswerable without knowing how much loss 
of the cool sink would occur during the day, which depends on many 
factors such as the house construction, insulation, ventilation, and 
shade factors, and whether this significant loss would offset the 
savings from cooling the house in the morning at lower outside air 
temps.  It was mentioned that practical systems as currently installed 
require a water reservoir, or significant differences in construction 
materials from a standard wooden house, implying that your theory is not 
commonly thought to be cost effective without additional capital 
investment.  I would really suggest you re-read if you truly missed this 
all!

But instead of harping on about it, why don't you conduct a simple 
experiment and answer for yourself?   It will be slightly inaccurate as 
the weather will not be exactly the same day to day, but if you can 
reasonably predict two days with the same high temp and humidity, do this:

Day 1: control:  Thermostat at your normal setting for morning, day, and 
evening setting (presuming that most A/C usage will be in late afternoon 
as the evening setpoint is attained).  Note the electric meter reading 
at 6am and 6pm for that day's usage.  Don't use appliances that you 
won't also use on Day 2:ideally go away to work to avoid any inadvertent 
usage, keep fridge closed, etc.
Day 2: hypothesis: At 6am, note meter reading.  Program thermostat to 72 
(or whatever) and supercool the house, with higher set point at 9am.  
Normal setpoint for 6pm.  Note meter reading and ensure appliance usage 
is the same as Day 1.

Compare the usage on Day 1 versus Day 2.  That should answer the 
question for YOUR house and YOUR A/C unit, right?

Obviously there is the important wildcard of differing weather from day 
to day, and possible differences in appliance power usage, but if you 
get lucky on that, you could have a pretty good test.  For better 
results, record the temperature and dewpoint every hour with a weather 
station (or use NWS data), repeat the experiment multiple times over 
many weeks, and build in some controls/adjustments for the weather.  
Also, figure out how to measure the time your compressor runs or 
something else more precise than overall power usage, since the other 
appliances may skew this if you're at home using them.  Maybe you could 
hook up a clamp-on current ammeter to the power feed to the compressor.

Hope this helps,
Jeremy

P.S. I personally think that the Day 2 method will use MORE power, and 
end up being a worse idea than letting the house heat up during the day 
and cooling again at night, assuming power costs the same at all times. 
    Maybe if you have really good insulation, shade, attic fan, etc, it 
will be a break-even supposition.  Would actually be quite curious to 
see the results... and the availability of demand-based power pricing 
really changes the equation.