Peter Thiel, Bill Gates, Khosla fund LightSail Energy in $37M Deal

Below is what my understanding: Current energy storage mechanisms (batteries) are inefficient. In theory one way to store energy that would be more efficient (why?) is to use the energy to compress air (with a conventional air compressor?). The air is stored in a container until it is converted to electrical energy through some inverse process (powering rotary screws?). Unfortunately compressed air is very hot and difficult to store. Instead water droplets can be injected into the compressed air container. These droplets will absorb most of the energy of the compressed air. The vapour is separated from the compressed air and stored in other containers (still as vapour?). How is the vapour then converted back to electrical energy? Isn't the vapour just as hot as the compressed air? Is it easier to deal with because it can be stored in a larger volume?

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#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_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

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

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/

Troposphere thickness: 17km

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.

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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

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%.

This is akin to a battery, so there is a 'charge' and a 'discharge' cycle, you'd be using the charge cycle when there is an excess and the discharge when you need more than is available or when the price of your stored energy is lower than what you'd be buying from the grid. So likely while you're charging (I'm assuming that's the better part of a day) you're not consuming from the device.

So your 50KWh initial value is more likely only half of that or even less, the portion that you'd be consuming that was previously stored. I've lived off a 48KWh lead/acid battery and it would - in a very energy efficient home - power the house for up to 5 days before needing a top-up absent sufficient sun and wind. This still holds when the storage capacity is centralized, only the flow would be slightly different and the houses would be in 'sink' mode all the time.

Another point regarding consumption:

Conservation is the best possible starting point for any renewable installation, large scale or small scale does not matter. It is easier to save a KWh than it is to generate or store one, up to a point, so that low hanging fruit is where you start.