3
$\begingroup$

Many of us may remember the problems from thermodynamics where we had to determine if a pretend air conditioning unit supplier's claims violated the 2nd law. I want to pose a similar question regarding a new "green" Air Conditioning (AC) technology.

An ice energy storage unit for AC works by making ice during the night and using the latent heat (mostly) to cool the building during the day. The major motive for doing this is to move the energy demand from the peak energy use times to the low-demand times at night when it is cheaper. The cost of electricity is not relevant to this question. The claims of a vendor of this technology, Ice Bear Energy Storage System, include the following:

Bottom line: The Ice Bear system reduces total net energy consumption for most buildings under virtually all operating conditions and installations.

Question: Is this claim physically plausible, given reasonable assumptions about climate? There are some example cities below, I imagine that considering an average location or just the extremes would be sufficient.

Here are a few things to note, but these are certainly not comprehensive. Creativity may be important for identifying all the factors at play.

  • The temperature of ice is $0^{\circ} C$ ($32^{\circ} F$)
  • Room temperature is $22.2^{\circ} C$ ($72^{\circ} F$)
  • The outside ambient temperature is lower at night than during the day. Exactly how much depends on the location and humidity particularly. Locations mentioned on their website include Glendale, CA; Howell, NJ; Las Vegas, NV; Fort Collins, CO; Orange County, CA; and Redding, CA.
  • The ice storage duration is 6 hours and since these are multiple MW products, I'll say it's fair to assume almost perfect insulation unless someone wants to challenge this.
  • You can obtain useful work while exchanging the heat from the office building to the ice. I believe this would be important considering the wording of "total net energy consumption".

Temperatures

I have looked up the temperatures for most of the mentioned cities. For some I couldn't get data for I used a nearby city instead.

  • (City/State) (night-day difference) (seasonal difference) (hottest month high)
  • Redding CA, 27.4, 44, 99
  • Las Vegas NV, 26.5, 48, 104
  • Denver CO, 27.8, 45, 88
  • Lakehurts NJ, 27.8, 45, 88
  • Santa Anna CA, 20.5, 18, 83

These seem to be fairly consistent for the night-day difference in particular. I hope that simplifies the problem. It seems clear that running at night "buys" you about 27.5 degrees F. Now, the absolute temperatures as well as the seasonal variation will affect the efficiency, but to a much smaller degree.

My conflict: Can a $27.5$ degree difference really make up for a $72-32=40$ degree penalty on the thermal cycle? Really? It's not a trivial problem, but I am skeptical. I will certainly send them an email letting them know they violated the 2nd law depending on the outcome here.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The technology you describe can actually save some energy using the fact that the night temperature can be much lower than the day temperature. When you cool the ice using a refrigerator-like unit it's very important how cold the air you use to cool the refrigerator radiator is.

Since you can place the radiator in such way that it is intensely cooled with outside air and use much colder air available during the night you can cool the ice more efficiently than you would cool the inside air during the day with a usual AC system. So not only you shift to cheaper electricity but you could also use less electricity because with colder outside air cooling can be more efficient during the night.

One point to illustrate this: my own electricity usage pattern is mostly constant during the year. Now the electricity bill is about 30% larger in the hottest summer months because of the refrigerator that constitutes the most of my electricity consumption. Of course the refrigerator also heats up faster with warmer room air, but it also cools less efficiently and these two factors noticeably boost energy consumption.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Sometimes it's nice to see how an equation relates to physics. In this case it would be to derive the actual efficiency from entropy and the differences in temperature. $\endgroup$
    – Carl Brannen
    Jul 12, 2011 at 7:34
  • $\begingroup$ Point taken, but you haven't answered the question. When you exchange between the cold(er) outside air and ice, both the hot and cold reservoirs are colder than the "conventional" scenario. The ice, however, is colder by a larger margin. btw, I should volunteer that the ice/water mix has better heat transfer so "conventional" operation may require cooling to lower than 72 degrees F, like maybe 65 instead. It still doesn't look good for this company IMO. $\endgroup$
    – Alan Rominger
    Jul 12, 2011 at 13:29
  • 1
    $\begingroup$ Zass, is asking the right question. I think he is assuming the wrong numbers however, the cold heat exchangers on an AC unit run a lot colder than the room temperature air. In fact I've had portable units that could freeze up(accumulate ice that blocks the fins, even during a hot day). So we don't have the numbers to work with. Also the efficiency of the heat engine is not even close to the theoretical Carnot efficiency, so we would need real engineering data to make a valid comparison. $\endgroup$
    – Omega Centauri
    Jul 12, 2011 at 14:21
  • $\begingroup$ @Omega Since this is only a relative comparison, bad assumptions may still lead to the right conclusion. But you're right that AC systems have to cool to lower than air temperature, and it will still be present for the water/ice mix. For instance, the $\Delta T$ due to an imperfect heat exchanger (HX) might be 15-20 F for the conventional HX and 5 F for the water/ice HX. But when you transfer the heat from the ice to the building during the day, if you extract energy from that cycle you are penalized by the same 15-20 F (same air circ system). Hope this makes sense. $\endgroup$
    – Alan Rominger
    Jul 12, 2011 at 15:06
  • 2
    $\begingroup$ I'm guessing it is a bit of a "marketing" claim. They probably compare against some well known, but older (and probably not the most efficient in class) compressor/AC. I worked in go fast in computers, competitors always compared their still on the drawing board systems, against your older models running the application that was most favorable to them. Its very likely their system is more efficient then a twenty-year old system it is replacing. $\endgroup$
    – Omega Centauri
    Jul 12, 2011 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.