According to the EPA's figures, 75% of residential energy use in the US in 2001 was for heating things up and cooling them down in fairly brute-force fashions --- of 9.86 quadrillion BTU, only 2.40 quadrillion were spent on anything else. [0] And essentially all of that is unnecessary.
Only about a fifth of total energy consumption in the US is residential [1], so this 75% is only about 16% of the total; but I think it is only slightly exaggerated from usage patterns in the US economy as a whole. I estimate that 50% of overall US energy consumption is spent on indoor climate control.
I don't have good numbers for the rest of the world. I assume they're roughly similar.
As global warming increases the amount of severe weather people must cope with, the importance of effective indoor climate control will increase; and fighting global warming will require the reduction of energy consumption. Additionally, sufficiently effective indoor climate control can render uninhabitable regions habitable and, if scalable to large buildings, enable the cultivation of a wider range of crops than the local climate would normally allow.
Here in Argentina, air conditioners are rated in "frigorías". One frigoría is one kilocalorie per hour, which is about four BTUs per hour. They recommend 50 frigorías per cubic meter. (This is a little strange, since you'd think you would pick your air conditioner capacity by the amount of heat that comes into your house per hour, rather than the amount of heat your house can hold. But I digress.)
Our apartment is 80 m², and about 2.75 meters in height, which means it contains about 220 m³, and therefore should need about 11000 kcal/h of cooling capacity. This is a bit of a problem for two reasons:
So I've been thinking some more about my 'desert cool and water economics' kragen-tol post from more than a year ago. Suppose we wanted to be able to sink 11000 kcal/h of electrical power for 40% of each day during the summer into some kind of heat storage tank (say, full of water) and then radiate it into space at night. How big of a tank would we need, and how much area? How much surface area would we need for heat-exchanger coils to suck 11000 kcal/h of heat out of the air during the day?
The first question is the easiest one. In the last few days, the highs have been around 28°C and the lows have been around 18°C. Suppose we want to lower the temperature of the air from 32°C to 22°C, which is 10 K of difference. So the water must start out cooler than 22°C and end up cooler than 32°C. Suppose we can start with water at 16°C and end with it at 30°C. Then we can get 14 kcal of heat out of the air per liter of water that flows from the cold tank through the heat exchanger into the hot tank. So 11000 kcal/h would be 790 liters per hour. I've hypothesized needing this cooling power for 9.6 hours (40%) out of each day, which means that each tank would need to hold 7580 liters of water, which is 7.58 metric tons and 7.58 cubic meters.
If we kept these 7.58 metric tons of water indoors, and didn't need any extra space for insulation, we would need 5.5 square meters of floor space for them, or about 2.3 meters square. That's about 7% of our total floor space, and that 7% should be no more and no less constant across dwellings than the figure of 50 frigorías per cubic meter.
The same amount of water can sink more heat, and sink it more quickly, when the difference between the air temperature and the cool-water temperature is greater. If the cool-water temperature stays at 16°C and I want to cool the air from 28°C, I can only sink, at most, 12 kcal per liter of water. But if the air temperature is 40°C, I can sink 24 kcal/L, twice as much heat for the same volume of water.
So while a conventional air conditioner needs to be sized for the amount of heat it needs to extract on the hottest days, and therefore cares a lot about how hot they are, the design of such a reservoir device should be relatively insensitive to the maximum temperature.
So, even if the 16°C number is correct, I might be able to get by with a much smaller amount of water, simply because the 11000-frigorías number is only relevant on the very hottest days --- if even then.
Cooling the water at night can happen in essentially three ways: evaporation, conduction, and radiation. I think evaporation is limited in its applicability (it won't help colonize the Grand Erg Occidental, and it will get expensive if water does) although being able to use brackish water might broaden its applicability.
This leaves conduction and radiation. Radiation has the major advantage that, on dry nights, the heat sink is the cosmic background radiation at a temperature of about 3 kelvins, so it might be possible to cool the water quite cold, perhaps even to freeze it. (This probably involves using an intermediate heat transfer fluid with a lower freezing point, but it's tractable.)
The basic radiation law is Stefan's Law or the Stefan-Boltzmann law:
Q/t = e{sigma}A(T_hot^4 - T_cold^4)
e is the emissivity, a number between 0 and 1 indicating how close
to an ideal black-body radiator the object in question is (at the
relevant wavelengths). I think it's (1 - albedo) at the relevant
wavelengths.
Sigma is the Stefan-Boltzmann constant: 5.67e-8 W/m²/K^4
Q is the heat transferred.
t is time.
T_hot and T_cold are the temperatures of the hot and cold bodies,
measured in kelvins.
So if our e is, say, 0.75; T_hot is at least 16°C (289 K); T_cold is insignificantly small; t is, say, 8 hours; and we want the Q to be 100Mcal (11000 kcal/h * 9.6 h); how much emitting area do we need?
100Mcal is 420 MJ, so the required heat dumping rate is 14.5kW; T_hot^4 is 7.0e9 K^4; so we have
14.5kW / (0.75 * (5.67e-8 W/m²/K^4) * 7.0e9 K^4) = A
m² * 14.5e3 / (0.75 * 5.67e-8 * 7.0e9) = A
m² * 14.5e3 / (0.75 * 5.67 * 70) = A = 49 m²
So I'd need about 7 meters square of roof space to beam the heat from my little apartment back out into space at night, plus enough pipes and aluminum to keep it all warm. And on cloudy nights, it wouldn't work. Note that 49m² is more than half the floor area of the apartment.
Conduction, on the other hand, requires little extra equipment (just a fan and some flaps to direct air from the outdoors through the same heat exchanger used to cool the indoors) and is known to be feasible.
We can estimate the "cost" of losing the space at 7% of our rent: for us, $150 per month, or $1850 per year. So it costs us $40 per month during the life of the thing, and as long as the electrical rates are so deeply subsidized, it would actually cost us almost all of the $150 amount. We're not in a particularly expensive area of Buenos Aires, but areas outside the Capital Federal are cheaper still. So for people in areas that cost less per month per square meter, it might save them money each month. Even for us, it might cost less up front than a $7000 refrigerative cooler, at least if it doesn't break the floor and can be adequately insulated without building anything really expensive.
I don't really have a clue what size of heat exchanger I would need. Presumably it would be a few times bigger than the heat exchanger that an air conditioner would use, because the temperature difference between the coolant and the air is smaller.
The hot water tank must retain its heat until it can be exhausted into the great outdoors at night, if it is inside; and the cold water tank must retain its cool until it can be exhausted into the indoors during the hot day, if it is outside. In the winter, their roles would be reversed. (Beatrice suggested reversing their roles during the winter. I need to think more about this.) Insulating materials like fiberglass typically have thermal conductivities around 0.04 W/m/K. If a cubical tank contains 5.28 cubic meters, it's 1.74 meters on a side, and therefore has 18.2 m² of surface area, so that's around 0.72 m W/K, and I've been hypothesizing about a 20 K temperature difference, so that's 14.4 m W. We need to divide that by enough centimeters of insulation that the resulting uncontrolled heat flux is small compared to the average 4400 kcal/h (=5100W) heat flux that the thing is designed to control. If "small" means 5%, that's 510W, which means we need 2.8 cm of insulation.
3 cm of insulation around cubical tanks seems like it might be a much more reasonable proposition than welding up some giant dewar flasks for the thing, which would presumably need to be round instead of some more convenient shape. (Right?) That's roughly a cubic meter of fiberglass (or whatever: cork, cotton, felt, "mineral wool", styrofoam, are all in the same ballpark; silica aerogel could cut the space by half, and straw would double it) to insulate all ten cubic meters of reservoir.
Water has a very high specific heat, so using most other materials for the hot and cool reservoirs in place of water (air, say, or iron, or concrete) would increase the weight required, rather than decrease it. (And solid materials have the additional problem that they're kind of inconvenient to move between the warm reservoir and the cool reservoir.)
However, materials that change phase in the relevant temperature range --- say, solid to liquid, or liquid to gas --- have a much higher effective specific heat than any substance. If we were trying to cool the apartment from by heating water to 1°C from -1°C, for example, we could dump half a calorie per kilogram of ice going from -1°C to 0°C, then 80 calories per kilogram as it melted, then a calorie per kilogram of water going from 0°C to 1°C. This allows you to reduce the weight of water you need by more than a factor of 40. If previously you would have needed five metric tons, now you only need 123 kg.
(Air conditioners in the US are often rated in "tons", which are tons of ice melted per day. Before refrigerative air conditioners were invented, you would cut ice out of frozen lakes in the winter and store it in an "icehouse", insulated by straw, all year round. So this idea of using phase-change materials as heat reservoirs is nothing new.)
(Note that this works even if you want to cool the air from 25°C to 20°C.)
But this only works if the relevant reservoir can be induced to keep its temperature at the melting or boiling point of the reservoir fluid, or oscillate back and forth across it. If you can dependably get access to a reservoir of below-freezing cold, you can use this approach with water; otherwise, you might have to use a material whose phase change temperature is somewhere more convenient. (If it can be actually inside the range where you want to maintain the air temperature, you don't need tanks and stuff. You can just put the stuff out where it can conduct heat to the air, like in the wall or on the coffee table or whatever.)
There are all kinds of research going on on phase-change materials as lower-mass heat reservoirs. See, for example, J.R.Gates's phase-change material home page:
http://freespace.virgin.net/m.eckert/index.htm
They also have this nice thermostatic property, which has been used to calibrate thermometers for centuries, that they tend to heat up things that are colder than their phase-change temperature, and cool down things that are warmer.
Obviously, indoor climate control with insulated heat reservoirs, passive solar design, phase-change materials for reservoirs, and so on, is nothing new. Adobe houses, stone castles, dugout houses, soddies, caves, iceboxes using ice stored since the winter in the icehouse, walls shared between houses to prevent heat loss, and so on, all go back generations if not millennia or longer. But they were all abandoned in the developed world over the last few decades. Why were they abandoned, and what makes me think they're worth recovering?
They were abandoned for several reasons:
I think there are a number of new developments that make these approaches relevant again:
Traditional adobe construction. Traditional dugout construction. Earth berms. Trombe walls. The German Passivhaus program. Seasonal thermal stores. Earth-berm construction. Straw-bale construction. Dewar flasks. Earthships. Thermal energy storage in general, in particular, "full storage systems" that run the air conditioner chillers only at night. John Hait's Passive Annual Heat Storage. Water walls. Isolated solar gain. Annualized geo-solar. Superinsulation. Heat recovery ventilation. Ground source heat pumps. The SHPEGS Solar Heat Pump Electrical Generation System.
[0] "Energy Consumption and Expenditures RECS 2001", from the US Department of Energy Residential Energy Consumption Survey:
http://www.eia.doe.gov/emeu/recs/recs2001/detailcetbls.html#total In particular, see page 3 of the total consumption PDF: ftp://ftp.eia.doe.gov/pub/consumption/residential/2001ce_tables/enduse_consump2001.pdf
Of the 9.86 quadrillion BTU, 2.21 quadrillion are spent on air conditioners and refrigerators, which someone might argue aren't really part of "climate control".
[1] US Department of Energy Energy Information Administration (DOE EIA) Monthly Energy Review, December, 2007
http://www.eia.doe.gov/emeu/mer/consump.html In particular, the third page of section 2, "Energy Consumption by Sector", p.27: http://www.eia.doe.gov/emeu/mer/pdf/pages/sec2.pdf
Currently this shows a 9-month total for 2007 of 16.43 quadrillion BTU for the residential sector, out of 76.16 quadrillion overall; there's another 13.84 quadrillion attributed to the "commercial" sector, which has roughly similar seasonal and source energy usage patterns, suggesting that a large part of its energy consumption is also devoted to indoor climate control.
However, that total also includes 30.87 quadrillion BTU of energy used by the "electric power sector" --- 40% of the total --- and presumably that usage is roughly proportional to total usage of electricity.
So my best estimate is that (0.75 * (16.43 + 13.84) / (76.16 - 30.87)) = 50% of US energy consumption is devoted to climate control.