How much air would you need to supply all the power required by all the homes in California? Here's my attempt to answer that question: The first stop is to g

How much air would you need to supply all the power required by all the homes in California? Here's my attempt to answer that question: The first stop is to get a sense of what the realistic energy density of these approaches might be. A quick search lands you here: http://en.wikipedia.org/wiki/Compressed_air_energy_storage#Energy_density_and_efficiency http://en.wikipedia.org/wiki/Compressed_air_energy_storage#E... My take-away: 1 m3 of air = about 300,000 J How much energy does a typical house in the US use per day? http://wiki.answers.com/Q/How_much_electricity_does_an_average_2-story_3-bedroom_house_use_per_day_per_month_or_per_year http://wiki.answers.com/Q/How_much_electricity_does_an_avera... I'll use 50KWh per day 1kWh = 1,000W x 3,600s = 3,600,000J This typical house, then, consumes 180,000,000J per day How much air do we need to compress to provide all of the energy needs of this one house (per day)? 180,000,000J / 300,000J = 600 m3 How many homes in California? http://quickfacts.census.gov/qfd/states/06/06037.html http://quickfacts.census.gov/qfd/states/06/06037.html Let's say it's about 14,000,000 homes How much air do we have to compress every day to service these homes: 600 m3 x 14,000,000 homes = 8,400,000,000 m3 OK, there's a number, whatever it means. Hmmm. How much of the available air are we using? What's the volume of air of the atmosphere? Tough question to answer. I think the number we'd want would be that of the Troposphere. http://en.wikipedia.org/wiki/Atmosphere_of_Earth http://en.wikipedia.org/wiki/Atmosphere_of_Earth I am going to use an over-simplification (you know, "assume a cow is a uniform sphere of milk" type stuff) to try to get a number. Sphere within a sphere to get the volume of the troposphere. Average Earth diameter: 12,742km http://www.universetoday.com/15055/diameter-of-earth/ http://www.universetoday.com/15055/diameter-of-earth/ Troposphere thickness: 17km http://en.wikipedia.org/wiki/Troposphere http://en.wikipedia.org/wiki/Troposphere Troposphere volume: 4,341,334,943,758,290,000 m3 That means that California would use 0.00000019% of the troposphere per day if every single home was powered using compressed air energy storage. Put a different way: It would take nearly 1.5 million years to process all of the air in the troposphere. I'm not sure if the above is complete nonsense or not. The problem is far more complex than these quickie calculations might suggest. On first inspection it sounds like we have plenty of air to go around. Would there be any environmental and/or air quality issues stemming from this approach? Do we end-up with cleaner air locally because of the process? Interesting stuff. . EDIT: A few more data points. How big of a container is required to store all of this air? The original assumption was that 1 m3 of air would compress into a 5L bottle, or 0.005 m3. Storage cube side length: 348m Storage sphere diameter: 431m How much would this much air weigh? 1 m3 of air at 20C = 1.204 kg 8,400,000,000 m3 = 10,113,600,000 kg The question, for me, begins to be about how realistic it might be to construct enough smaller storage vessels to capture this volume safely. The article mentions something about 40ft standard shipping containers. Assuming that the storage vessel has the internal dimensions of a standard 40ft container: http://en.wikipedia.org/wiki/Intermodal_container http://en.wikipedia.org/wiki/Intermodal_container Container volume: ~ 67 m3 Containers required to store enough compressed air to supply homes in California: ~627,000 units. That's a lot of containers, even if the calculations are off by 100%.