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.
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:
Your calculations appear be within an order of magnitude of correct :-)
One other way to think of the number of shipping containers needed: actually the average american home uses 30 kwh/day. At our target energy density and efficiency we've been attempting to reach 30 kwh per m^3. 1 m^3 is approximately the internal volume of a refrigerator. So each home gets 1 fridge worth of storage. Not so bad ;-)
Outside diameter: 9.4in
Height: 52in
Weight (empty): 195lbs.
Air capacity at pressure: 510.5 ft^3
Internal volume: 2640 in^3
The 30kWh you are are aiming for would require about 360 m^3 of air (perm my prior calcs). This would require 25 of these tanks.
To double check, the internal volume of these tanks is given at 2640 in^3. 25 tanks come in at 66,000 in^3, which is just over 1 m^3.
What this highlights for me is just how large a vessel might be required to store such a volume of air at 6,000psi due to how strong it has to be. The external volume of these 25 tanks is approximately 1.5 m^3. Not too bad. We are taking about a 5 x 5 tank layout; about 4 ft x 4 ft and, say, 6 ft high with hoses, fittings and other hardware. They would weigh-in at about 5,000lbs, which might require some accommodations for a typical home garage.
Do these numbers describe what you are trying to accomplish to a reasonable approximation?
How noisy is the process of getting the energy back out of this storage system?
I had to look at a comparison with the energy density of current Lithium-Ion batteries:
If I assume 1,000 J/cm^3, that would require about 108,000 cm^3 in Lithium-Ion batteries or 0.108 m^3. Yikes! On first inspection, a 1 m^3 bank of Lithium-Ion batteries would allow you to run a house for ten days!
Not sure what that conclusion means, but Lithium-Ion, cost and other issues aside, looks very interesting.
How about gasoline? I know, horrible, but I have to ask.
Assuming 100% energy conversion we would need 0.0318 m^3 of gasoline to power a house for an entire day. Assuming a generator is 10% efficient that number becomes 0.3176 m^3 (317.6 liters or 83 US Gallons).
I won't do the numbers, but Liquid Propane looks very interesting.
Clearly your long term competition might very well be electrochemical battery or graphene supercapacitor technology.
I realize you are working on a method to be used in storing excess energy for later delivery (or smoothing out the spikes in infrastructure demands). If I was looking for emergency power backup today I think I might have a very serious look at Liquid Propane. I has none of the storage problems of gasoline (namely that it degrades if not attended to) and it is very easy to use for cooking as well as lighting, if required.
Would I want every house in my neighborhood to have LP tanks, gasoline tanks, compressed air tanks or huge banks of Lithium-Ion batteries? Probably not.
All of these options are scary in one way or another. Imagine Hurricane Sandy, Katrina or a good size earthquake here in CA in a scenario where every home has one of these technologies. Could get scary very fast.
Same issues as with electric cars. Very interesting until you have an incident involving several cars. Formula 1 teams had to make special accommodations to use their electrical KERS systems, some of which run at 375V.
Because of this I would think that your technology (or any other high-duration, high energy-density storage solution) might be best deployed at the substation or generation point rather than installed in every home. Most people are not really equipped to intelligently deal with electricity. Sometimes it is a good idea for power to go out.
We get a higher energy density per m^3 of air at higher pressures. We were aiming at 4500 psi eventually.
We aim for it not to be noisy -- any noise from high pressure air rushing out represents wasted energy. Sonic booms from exhaust have this problem in automobile engines, we avoid it.
Lithium ion is indeed much more dense :-)
Consider that every car has a gasoline tank, many houses have fuel oil, and we undergird our streets with natural gas pipes, which burned down San Francisco. I submit that air has its safety issues, but that most of these can be avoided, and in particular, chain reactions, which threaten flammable energy storage, can be made a non-issue.
> Consider that every car has a gasoline tank, many houses have fuel oil
Just guessing that this could be a gating issue once you have something to deploy. People can be irrational, even when faced with facts. I know people that will not go into the water at the beach for fear of being attacked by a shark. Yet, the same people don't think twice about getting into their cars in the morning and driving on Los Angeles freeways.
Here industrial design might be the key. If the unit looks, almost literally, like the typical freezer or refrigerator lots of people have in their garage it might mitigate irrational first impressions.
4500psi?! Haven't the ideal gas laws broken down by then?
There's a reason most (recreational) scuba tanks stop around 3000 - 3500 PSI max working pressure: you fairly quickly stop getting linear gains, at the expense of additional tank wall thickness and stronger valves required.
Edit: Nevermind, it appears that 4500 PSI/300 BAR is semi-standard in Compressed Air Powered cars, so I guess there is value in going to that pressure (storage density I guess).
You are of by a factor of ten: 108,000,000/34,000,000,000 = 108/34,000 = 0.003-ish m^3, or 3 liters of gas to produce those 30kWh. That is in line with http://en.wikipedia.org/wiki/Gasoline, which claims 9.7kWh/l. So, the volumetric difference between battery and gas is 30, not 3. If it were only a factor of 3, as in your calculations, I think all cars would be electric by now.
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.
Yes, of course, you wouldn't use compressed air storage for power 24/7. I just went for worst case to see where the numbers would lead.
> Conservation is the best possible starting point
Couldn't agree more. I happen to think that this is where we've failed miserably over the years. Homes are just not built to be efficient, despite what the propaganda might indicate.
I disagree with you on your point on efficiency. California has had a ton of success implementing efficiency programs, and these efforts are starting to be copied by other states.
My guess is that your perception of failure is largely driven by the McMansion trend of the 90's/early 00's. While these large, suburban homes definitely consume more energy than their smaller counterparts, if it wasn't for strict codes and standards and minimum efficiency requirements for appliances, our energy situation would look much worse.
I didn't read the article super-carefully, but my impression was that they were using the heat of vaporization of water for energy storage, not PdV work of the air. I haven't done a calculation, but my guess is that this is much more efficient.
You have to store both. Also, you can't vaporize much air at low temperatures, which we aim for. You have to hold onto water for its sensible (no evaporation) heat capacity.
1^m3 of 300 bar air = approx. 30 kwhr.
Approximately 1/4th of that in water storage will hold the heat.
We're aiming at a daily loss of 1%, but it caps out at 10% relative (~7% absolute) because the Energy Out/Energy In is proportional to T_exp/T_comp (in an absolute scale) -- if the temperature drops to ambient, T_exp is only slightly lower in an absolute scale.
The real problem with your analysis is you want far less energy storage than that. A reasonable goal for vary high levels of 'green' tech is ~1% or ~15 minutes of grid energy storage. Beyond that it's much more valuable to simply build some peaking power plants and have excess capacity.
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.
.
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%.