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The problem with your article on isothermal air compression is that it deals only in academics, not the day-to-day use of compressed air systems. Perhaps the greatest advantage of pneumatics is their safety: moving cutters can be safely cleared no other way. Material moved through pneumatic (pressure-vacuum) systems clear debris from shops so that fall accidents are far less likely. Air tools can be safely used in wet environments and explosive atmospheres where electric tools are out of the question. Air-powered vacuums can easily remove even the most dangerous radioactive materials and asbestos without the dangers of contamination seen with electrical power. Air agitated paint tanks give the finest finish ever seen. And the list goes on. The physics of compressed air use are by definition, appealing: I can think of no less efficient way to deliver power. But we use air tools constantly, particularly indexing lifts, jacks, nailers, scalers, and air-motor driven tools. Because rotary screw compressors are now on-par with diesel powered equipment in deep-shaft mines, using the exhaust for breathing air and to cool the hot environment is a real advantage. Why didn't anyone else think to mention that? —Preceding unsigned comment added by 209.244.31.59 ( talk) 21:05, 17 April 2008 (UTC)
The Physics article seems completely wrong to me. By allowing the system to compress isothermally surely much of the energy is being lost? I'll check my thermo books tonight Greglocock 04:25, 21 March 2007 (UTC)
What about air-depression (you might call it fission as compression´s fusion)?
79.210.183.160 ( talk) 13:19, 23 June 2009 (UTC)
66kw (~90hp) to move a average car is quite high. While 90hp may be required to accelerate quickly, modern vehicles in the US can maintain 60mph with approx 20hp or ~15kw. The quick math to prove this is simple: Assuming a standard highway fuel economy of 30mpg & an almost universally accepted bsfc (brake specific fuel consumption) of 0.53 lbs/(hr*hp) for modern gasoline engines, an engine traveling 60mph will consume approx 2 gal /hr of gasoline. Gas in the US has a specific gravity of 0.73-0.75 and thus weighs approx 6 lbs, thus the engine is consuming 12 lbs of gasoline per hour. Using a BSFC of 0.53, the a vehicle traveling 60mph @ 30mpg has an average power of 22hp or 16.9kw. If an air car compressor is a heat pump,could it supplement a homes central heating?-- 79.67.199.164 ( talk) 17:41, 12 July 2008 (UTC)
The calculations are helpful, but impossible to understand with undefined variables. What are P_A and P_B? The article needs to be self-contained or contain references where terms are defined. —Preceding unsigned comment added by 68.7.105.15 ( talk • contribs) 03:12, 27 December 2007 (UTC)
This article talks mostly about the feasibility of compressed air cars. I'd love to see some discussion and math about using compressed air as a large-scale means to store energy. For example, storing the energy created by solar power during the day for use at night. -- Gadlen ( talk) 23:23, 18 January 2008 (UTC)
Hey Greglock, this is happytrombonist16, the alter-ego of whatever ip address my computer was using when I posted the facts tag. I'm kinda new, so it's very possible that wasn't the right tag to use for that situation. However, it looks like you've fixed most of the things I had a problem with, so thanks! Happytrombonist16 ( talk) 05:32, 27 January 2008 (UTC)
At the moment air engine redirects to this page. There exists an article with the much better name Compressed air engine. Why don't we move the air engine stuff there, and leave the physics and storage side here? Greg Locock ( talk) 09:45, 7 February 2008 (UTC)
The original focus of this article was on the use of compressed air energy storage for utility level power generation or grid energy storage. In my opinion the current version of this article focuses too much on the use of compressed air to power vehicles and not on power generation. Details like power output and storage time for the Mclntosh and Huntofr hybrid plants are missing. The section on types of compressed storage is inadequate and depends too much on the "German AACAES project information" which understates the efficiency of the hybrid CAES plants. The hybrid plants use the stored cooled compressed air to directly feed air into a gas turbine. This replaces the compressor that is powered directly by a normal gas turbine. The efficiency of a hybrid plant should be calculated by comparing the energy required to compress the stored air versus the energy required to run the compressor inline with the expansion turbine. See http://www.eere.energy.gov/de/compressed_air.html for details. —Preceding unsigned comment added by 68.2.181.124 ( talk) 21:26, 27 March 2008 (UTC)
The equations given in two sections of this article do not match. In one the specific energy is given per mole, in the other per N-m3. Probably an "n" missing somewhere. This needs to be checked and corrected.-- Theosch ( talk) 15:27, 4 April 2008 (UTC)
I removed the "m³-N" units and values like "110 kJ per m³" (instead of 100), because it wasn't clarified why they were used at all, they were used inconsistently, and the differences aren't that significant. W=pV ln(p1/p2) doesn't depend on temperature, so the "110 at 24°C" value has no sense, unless we measure the gas' volume at 0°C and use it (decompress isothermally) at 24°C. Maybe this is required when talking about energy density (I don't see why the decompressed volume is used anyway), so please correct me if I'm wrong. Tokenzero ( talk) 12:27, 8 November 2008 (UTC)
100kJ * ln(20MPa/100kPa)=230 kJ and not 530 kJ as written in the article. please check., Jan 25 2012 — Preceding
unsigned comment added by
Kommentier (
talk •
contribs)
14:16, 25 January 2012 (UTC)
No you have to use log e but I accidentally pushed the log base ten button instead of the log base e (the buttons are right next to each other on my calculator) !! That is the author of the calculation rightly wrote: "Nevertheless it is useful to describe the maximum energy storable using the isothermal case, which works out to about 100 kJ/m3 [ ln(PA/PB)]. Thus if 1.0 m3 of ambient air is very slowly compressed into a 5 L bottle at 20 MPa, the potential energy stored is 530 kJ." The formula is from above where you take the integral over 1/V dV, which is ln V, i.e. the log over e, so I think the formula is right. now the typical air pressure is about 100kPa so 20MPa/100kPa=200 and now finally my calculator gives ln 200= 5.3. — Preceding unsigned comment added by Kommentier ( talk • contribs) 10:24, 27 January 2012 (UTC)
At first sight this looks like an intelligent suggestion. But, I have never seen a reference to it, and certainly lake storage is rather odd. This is energetically equivalent to pumping water into the lake (to raise the water level) and then using it to generate power later on. But, it is more efficiewnt and cost effective to do this directly with water than with air. So, find some cites or I'll blow it away. Greg Locock ( talk) 01:00, 29 April 2008 (UTC)
I added citations, including one relating to a patent (expired or about to expire). The patent covers the same material: underwater storage bags and their advantages. As for pumping water into a lake, that's not the correct analogy for understanding the energies: highly pressurized air is useful stored energy, water sitting in a high dam is also useful, but water in a lake can't do anything. The criticism, however, may point to other problems: for example, at extreme depth, effects relating to hydrostatic equilibrium and pressure gradient force could create problematic heating and cooling and potentially catastrophic "hammer" effects (like "water hammer" in domestic plumbing). (I'll try to add some math on this - into this talk section - if I can remember the appropriate integral calculus equation to use or have the patience to rough it out with a sums table.)
Clearly, the deep water storage system has thermal efficiency disadvantages compared to adiabatic storage. (Though ... I suppose the air bags could be insulated with "wet suits" - good for perhaps 12 hours of storage.) Carnot efficiency can only be improved by complex staged pumping, possibly with large pressurized equalization chambers (or columns) containing the necessary heat exchangers. On the other hand, energy recovery is greatly simplified by the constant pressure of the source (the turbines can be optimized for a particular intake pressure and rate of flow). Also, the huge expense of mining is eliminated. Anthony717 ( talk) 04:58, 30 April 2008 (UTC)
As an outsider, I found these formula's to be surprising and completely non-intuitive to the point that it would be great if someone with a good handle on them could add some text on the significance to the apparent fact (based on the ln function in the equation) that as you increase the pressure, the energy storage becomes less and less dramatic.
Intuitively, if I have a handpump that pumps in 1 liter of low pressure gas per stroke into a 1 cubic meter tank it takes me 1000 strokes to double the pressure, and then 2000 strokes to double the pressure again (both times I pump slowly and all heat conducts away). Not only did the second doubling of pressure take more strokes with my handpump, each stroke was significantly harder that during the first doubling of pressure, and yet, the formula for energy content claims the second doubling of pressure only added as much energy as the first doubling.
Instead I would expect the second doubling to provide ~4X the amount of energy since it was 2X the number of strokes at an average of 2X the force.
If the ln function is correct, I could also do some surprising things like take a 1 cubic meter tank that is at 3 atmospheres connected via a closed valve to a tank that at one atmosphere and calculate the energy. Then I could open the valve and hear a hiss and even drive a tiny turbine and power a flashlight while the tanks adjusted to be 2 tanks of 2 atmospheres each. Then if I recalculate the stored energy I would find the energy stored went up, not down because ln(3) < 2*ln(2). Are the formula's correct or am I wildly off the rails. If the formulas are correct, then some text about the implications of this would be a great addition.
Johnkoger (
talk)
04:08, 1 August 2009 (UTC)
While tidying the "Types" section I removed the following unreferenced text: "see CHARLES HODGES AND THE PORTER COMPOUND AIR LOCOMOTIVES". If anyone knows what this relates to and can add a ref please do so. Cheers -- Timberframe ( talk) 10:52, 21 April 2009 (UTC)
A similar proposal to the caisson method is being funded by EON - [1] The reference quotes the energy content at a depth of about 1,970 feet,as some 6,945 MWh of energy for every cubic meter, however this is clearly wrong, since 1 cubic meter of air at that depth (600 meters) would contain 600m*gravity*1000kg/m^3 = 600*9.8*1000/(3600*1000000) = 0.0016 MWh
bear in mind, it is not the energy content of the air you are interested in. It is the energy contect of the 1 m3 of water than you have raised by 1 meter - this contains about 10 times the energy of the compressed air per se.
Are you suggesting that in an adiabatic process there would be any difference in the energy stored, whichever way you calculate it? Greglocock ( talk) 02:40, 2 May 2009 (UTC)
References
Hi, I don't understand the formulas. The article says Energy W is a volume integral from V_A to V_B. But the volume of the vessel is constant all the time, may V means the amount of air at atmospheric pressure? 91.8.113.69 ( talk) 21:56, 28 November 2010 (UTC)
Compressed Air Energy Storage is a very specific term, applying to a particular approach to large scale energy storage within the utility industry. It's relatively simple. But the article as written dives right into all the talk about adiabatic, isothermal, etc. without first explaining the basis of how CAES projects (and there are currently two specific CAES projects operating) work. These more obscure engineering references might be discussed much further down, in the context of some of the R&D going on for CAES. — Preceding unsigned comment added by Onefinity ( talk • contribs) 08:13, 5 January 2011 (UTC)
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I am looking to different energy storage process, and I think a french version of you article will be welcome, how can I proceed ahead ? — Preceding unsigned comment added by Mbariou ( talk • contribs) 10:33, 7 November 2011 (UTC)
I added some material to the history section with cites in response to the expand section tag placed there. After making the additions, I removed the tag. If anyone feels further expansion is needed, feel free to restore it. There certainly have been many more announced projects and these were not mentioned. J JMesserly ( talk) 17:45, 14 April 2012 (UTC)
Note on a retracted bit of info regarding salt dome capacities. In the Times article I cited, Wald wrote that the salt cavern could handle 5 times the 1100 psi pressure that McIntosh now uses. Others may see this and want to mention it as I did. I wound up retracting it because of an authoritative report in Gas Turbine World, Sept-Oct 2009, editor de Biasing states that the max allowable pressure is calculated at .9 of depth. So the ceiling of the McIntosh cavern at 1500 foot depth allows a max of only 1350psi. a reprint may be found at www.espcinc.com/library/GTW%20 ACAES%20Article.pdf. I suspect Wald is in error, but I don't know. Since it is contested, I pulled it. - J JMesserly ( talk) 10:30, 17 April 2012 (UTC)
I wonder why there is no mention of a trombe as a means of storing compressed air. — Preceding unsigned comment added by 83.240.200.90 ( talk) 18:15, 3 December 2012 (UTC)
Please see http://www.lightsail.com/tech.html and their CEO's Google Solve for X talk from a few days ago for the details, but I'm not sure which category (e.g. hybrid etc.) this fits into. so could someone please add it to the article? Apparently using water mist as a heat exchanger is a completely new idea bringing efficiency from 30% to 70%, and LightSail has patents on it. I'd feel more comfortable if an expert took a look at it before it was added. Neo Poz ( talk) 23:15, 12 February 2013 (UTC)
Just came upon that article, interesting. I'm just very surprised to hear about Dresden, which is my hometown - I never heard of a pneumatic storage system installed there anywhere and I'm very tech interested. Can't find anything with Google as well. So is it possible to have a source for that??
Rheinisch-Westfälisches Elektrizitätswerk AG is developing 2 new systems; see ALACAES and ADELE. Ref 2: http://content.lib.utah.edu/utils/getfile/collection/etd2/id/86/filename/1430.pdf Mention in article KVDP ( talk) 17:36, 24 July 2013 (UTC)
Maybe its just late at night and I'm being stupid, but isn't the formula for the stored energy ignoring the work done by the ambient pressure, which should not be included in the calculation of the "stored" energy, as it is not recoverable?
Let's look at the situation of moving from state A: volume , pressure , to state B: volume , pressure , with constant temperature throughout. For simplicity, lets assume that is the ambient pressure. As has been pointed out, the pressure law yields , and in fact for any intermediate state we have . Now, suppose I start with a gas in a tube of cross section 1 at pressure and volume , so the tube has length . Lets put a plunger at one end. Initially, I put no force on the plunger (which has area 1), and the ambient pressure puts a force on the plunger of magnitude . Now I put an infinitesimal force on the plunger and begin to slowly compress the gas. At any intermediate state, with the plunger at position , so volume at , and pressure at , I maintain a force infinitesimally larger than what is required to maintain steady state. BUT, at such an intermediate state, ambient pressure always maintains a force of , so the force that I must exert is
Thus, the work that _I_ do in compressing the air (and this is the recoverable energy) is
Now, typically we are wondering what the (recoverable, ideal) stored energy is in a tank, i.e. we know , and , but not . Using (so, e.g. ), (changing sign to account for who's doing work on what...) we can write:
The first term agrees with what in on the web page, but the web page ignores the fact that the ambient pressure would do work on the gas that is not recoverable (except if the tank were taken to outer space...), i.e. it ignores the second term. Writing the result in this format:
makes at least on property clear: linear in (double the size of the tank, double the energy). If we want the formula in terms of gauge pressure, lets use for the ambient pressure, for the gauge pressure, and just for the volume of the tank. (By definition, ). Then the formula becomes:
For , this has a nice Taylor series expansion.
— Preceding unsigned comment added by DavidIMcIntosh ( talk • contribs) 02:28, 12 June 2014 (UTC)
Is CAES practical for home energy backups to keep a phone or computer running after a power outage? To keep freezers and refrigerators cold, I think bags of brine (for freezer) or some liquid with freezing temperature slightly above that of water (for refrigerator) is more practical.
This article should talk about the economics of CAES, e.g. how it compares to other energy storage technologies, and the levelised cost of energy when used with energy generators. — Preceding unsigned comment added by 101.191.212.184 ( talk) 07:03, 22 October 2016 (UTC) 74.111.173.193 ( talk) 19:16, 2 October 2016 (UTC)
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" For Adiabatic CAES systems, it is important for the compressor to be functional in high temperature/high pressure conditions. This challenge can be overcome by using an intercooling system along with the compressor." This is the definition of adiabatic "a process or condition in which heat does not enter or leave the system concerned." And this is the definition of intercooler "an apparatus for cooling gas between successive compressions, especially in a supercharged vehicle engine.". So where does our contributor think the heat from the intercooler goes? it goes to the outside environment. therefore it is no longer adiabatic. Greglocock ( talk) 06:45, 10 March 2019 (UTC)
The entry in the "History" section for the power plant in McIntosh, Alabama contains the words "Although the compression phase is approximately 82% efficient, the expansion phase requires the combustion of natural gas at one-third the rate of a gas turbine producing the same amount of electricity at 54% efficiency". I'm confused; is this some sort of hybrid system that combines a gas turbine with CAES? Or is the heat needed to provide energy to help the gas expansion? Whatever it might be, the article should make this clearer, and explain exactly how it works. — The Anome ( talk) 07:33, 15 May 2024 (UTC)
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The problem with your article on isothermal air compression is that it deals only in academics, not the day-to-day use of compressed air systems. Perhaps the greatest advantage of pneumatics is their safety: moving cutters can be safely cleared no other way. Material moved through pneumatic (pressure-vacuum) systems clear debris from shops so that fall accidents are far less likely. Air tools can be safely used in wet environments and explosive atmospheres where electric tools are out of the question. Air-powered vacuums can easily remove even the most dangerous radioactive materials and asbestos without the dangers of contamination seen with electrical power. Air agitated paint tanks give the finest finish ever seen. And the list goes on. The physics of compressed air use are by definition, appealing: I can think of no less efficient way to deliver power. But we use air tools constantly, particularly indexing lifts, jacks, nailers, scalers, and air-motor driven tools. Because rotary screw compressors are now on-par with diesel powered equipment in deep-shaft mines, using the exhaust for breathing air and to cool the hot environment is a real advantage. Why didn't anyone else think to mention that? —Preceding unsigned comment added by 209.244.31.59 ( talk) 21:05, 17 April 2008 (UTC)
The Physics article seems completely wrong to me. By allowing the system to compress isothermally surely much of the energy is being lost? I'll check my thermo books tonight Greglocock 04:25, 21 March 2007 (UTC)
What about air-depression (you might call it fission as compression´s fusion)?
79.210.183.160 ( talk) 13:19, 23 June 2009 (UTC)
66kw (~90hp) to move a average car is quite high. While 90hp may be required to accelerate quickly, modern vehicles in the US can maintain 60mph with approx 20hp or ~15kw. The quick math to prove this is simple: Assuming a standard highway fuel economy of 30mpg & an almost universally accepted bsfc (brake specific fuel consumption) of 0.53 lbs/(hr*hp) for modern gasoline engines, an engine traveling 60mph will consume approx 2 gal /hr of gasoline. Gas in the US has a specific gravity of 0.73-0.75 and thus weighs approx 6 lbs, thus the engine is consuming 12 lbs of gasoline per hour. Using a BSFC of 0.53, the a vehicle traveling 60mph @ 30mpg has an average power of 22hp or 16.9kw. If an air car compressor is a heat pump,could it supplement a homes central heating?-- 79.67.199.164 ( talk) 17:41, 12 July 2008 (UTC)
The calculations are helpful, but impossible to understand with undefined variables. What are P_A and P_B? The article needs to be self-contained or contain references where terms are defined. —Preceding unsigned comment added by 68.7.105.15 ( talk • contribs) 03:12, 27 December 2007 (UTC)
This article talks mostly about the feasibility of compressed air cars. I'd love to see some discussion and math about using compressed air as a large-scale means to store energy. For example, storing the energy created by solar power during the day for use at night. -- Gadlen ( talk) 23:23, 18 January 2008 (UTC)
Hey Greglock, this is happytrombonist16, the alter-ego of whatever ip address my computer was using when I posted the facts tag. I'm kinda new, so it's very possible that wasn't the right tag to use for that situation. However, it looks like you've fixed most of the things I had a problem with, so thanks! Happytrombonist16 ( talk) 05:32, 27 January 2008 (UTC)
At the moment air engine redirects to this page. There exists an article with the much better name Compressed air engine. Why don't we move the air engine stuff there, and leave the physics and storage side here? Greg Locock ( talk) 09:45, 7 February 2008 (UTC)
The original focus of this article was on the use of compressed air energy storage for utility level power generation or grid energy storage. In my opinion the current version of this article focuses too much on the use of compressed air to power vehicles and not on power generation. Details like power output and storage time for the Mclntosh and Huntofr hybrid plants are missing. The section on types of compressed storage is inadequate and depends too much on the "German AACAES project information" which understates the efficiency of the hybrid CAES plants. The hybrid plants use the stored cooled compressed air to directly feed air into a gas turbine. This replaces the compressor that is powered directly by a normal gas turbine. The efficiency of a hybrid plant should be calculated by comparing the energy required to compress the stored air versus the energy required to run the compressor inline with the expansion turbine. See http://www.eere.energy.gov/de/compressed_air.html for details. —Preceding unsigned comment added by 68.2.181.124 ( talk) 21:26, 27 March 2008 (UTC)
The equations given in two sections of this article do not match. In one the specific energy is given per mole, in the other per N-m3. Probably an "n" missing somewhere. This needs to be checked and corrected.-- Theosch ( talk) 15:27, 4 April 2008 (UTC)
I removed the "m³-N" units and values like "110 kJ per m³" (instead of 100), because it wasn't clarified why they were used at all, they were used inconsistently, and the differences aren't that significant. W=pV ln(p1/p2) doesn't depend on temperature, so the "110 at 24°C" value has no sense, unless we measure the gas' volume at 0°C and use it (decompress isothermally) at 24°C. Maybe this is required when talking about energy density (I don't see why the decompressed volume is used anyway), so please correct me if I'm wrong. Tokenzero ( talk) 12:27, 8 November 2008 (UTC)
100kJ * ln(20MPa/100kPa)=230 kJ and not 530 kJ as written in the article. please check., Jan 25 2012 — Preceding
unsigned comment added by
Kommentier (
talk •
contribs)
14:16, 25 January 2012 (UTC)
No you have to use log e but I accidentally pushed the log base ten button instead of the log base e (the buttons are right next to each other on my calculator) !! That is the author of the calculation rightly wrote: "Nevertheless it is useful to describe the maximum energy storable using the isothermal case, which works out to about 100 kJ/m3 [ ln(PA/PB)]. Thus if 1.0 m3 of ambient air is very slowly compressed into a 5 L bottle at 20 MPa, the potential energy stored is 530 kJ." The formula is from above where you take the integral over 1/V dV, which is ln V, i.e. the log over e, so I think the formula is right. now the typical air pressure is about 100kPa so 20MPa/100kPa=200 and now finally my calculator gives ln 200= 5.3. — Preceding unsigned comment added by Kommentier ( talk • contribs) 10:24, 27 January 2012 (UTC)
At first sight this looks like an intelligent suggestion. But, I have never seen a reference to it, and certainly lake storage is rather odd. This is energetically equivalent to pumping water into the lake (to raise the water level) and then using it to generate power later on. But, it is more efficiewnt and cost effective to do this directly with water than with air. So, find some cites or I'll blow it away. Greg Locock ( talk) 01:00, 29 April 2008 (UTC)
I added citations, including one relating to a patent (expired or about to expire). The patent covers the same material: underwater storage bags and their advantages. As for pumping water into a lake, that's not the correct analogy for understanding the energies: highly pressurized air is useful stored energy, water sitting in a high dam is also useful, but water in a lake can't do anything. The criticism, however, may point to other problems: for example, at extreme depth, effects relating to hydrostatic equilibrium and pressure gradient force could create problematic heating and cooling and potentially catastrophic "hammer" effects (like "water hammer" in domestic plumbing). (I'll try to add some math on this - into this talk section - if I can remember the appropriate integral calculus equation to use or have the patience to rough it out with a sums table.)
Clearly, the deep water storage system has thermal efficiency disadvantages compared to adiabatic storage. (Though ... I suppose the air bags could be insulated with "wet suits" - good for perhaps 12 hours of storage.) Carnot efficiency can only be improved by complex staged pumping, possibly with large pressurized equalization chambers (or columns) containing the necessary heat exchangers. On the other hand, energy recovery is greatly simplified by the constant pressure of the source (the turbines can be optimized for a particular intake pressure and rate of flow). Also, the huge expense of mining is eliminated. Anthony717 ( talk) 04:58, 30 April 2008 (UTC)
As an outsider, I found these formula's to be surprising and completely non-intuitive to the point that it would be great if someone with a good handle on them could add some text on the significance to the apparent fact (based on the ln function in the equation) that as you increase the pressure, the energy storage becomes less and less dramatic.
Intuitively, if I have a handpump that pumps in 1 liter of low pressure gas per stroke into a 1 cubic meter tank it takes me 1000 strokes to double the pressure, and then 2000 strokes to double the pressure again (both times I pump slowly and all heat conducts away). Not only did the second doubling of pressure take more strokes with my handpump, each stroke was significantly harder that during the first doubling of pressure, and yet, the formula for energy content claims the second doubling of pressure only added as much energy as the first doubling.
Instead I would expect the second doubling to provide ~4X the amount of energy since it was 2X the number of strokes at an average of 2X the force.
If the ln function is correct, I could also do some surprising things like take a 1 cubic meter tank that is at 3 atmospheres connected via a closed valve to a tank that at one atmosphere and calculate the energy. Then I could open the valve and hear a hiss and even drive a tiny turbine and power a flashlight while the tanks adjusted to be 2 tanks of 2 atmospheres each. Then if I recalculate the stored energy I would find the energy stored went up, not down because ln(3) < 2*ln(2). Are the formula's correct or am I wildly off the rails. If the formulas are correct, then some text about the implications of this would be a great addition.
Johnkoger (
talk)
04:08, 1 August 2009 (UTC)
While tidying the "Types" section I removed the following unreferenced text: "see CHARLES HODGES AND THE PORTER COMPOUND AIR LOCOMOTIVES". If anyone knows what this relates to and can add a ref please do so. Cheers -- Timberframe ( talk) 10:52, 21 April 2009 (UTC)
A similar proposal to the caisson method is being funded by EON - [1] The reference quotes the energy content at a depth of about 1,970 feet,as some 6,945 MWh of energy for every cubic meter, however this is clearly wrong, since 1 cubic meter of air at that depth (600 meters) would contain 600m*gravity*1000kg/m^3 = 600*9.8*1000/(3600*1000000) = 0.0016 MWh
bear in mind, it is not the energy content of the air you are interested in. It is the energy contect of the 1 m3 of water than you have raised by 1 meter - this contains about 10 times the energy of the compressed air per se.
Are you suggesting that in an adiabatic process there would be any difference in the energy stored, whichever way you calculate it? Greglocock ( talk) 02:40, 2 May 2009 (UTC)
References
Hi, I don't understand the formulas. The article says Energy W is a volume integral from V_A to V_B. But the volume of the vessel is constant all the time, may V means the amount of air at atmospheric pressure? 91.8.113.69 ( talk) 21:56, 28 November 2010 (UTC)
Compressed Air Energy Storage is a very specific term, applying to a particular approach to large scale energy storage within the utility industry. It's relatively simple. But the article as written dives right into all the talk about adiabatic, isothermal, etc. without first explaining the basis of how CAES projects (and there are currently two specific CAES projects operating) work. These more obscure engineering references might be discussed much further down, in the context of some of the R&D going on for CAES. — Preceding unsigned comment added by Onefinity ( talk • contribs) 08:13, 5 January 2011 (UTC)
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I am looking to different energy storage process, and I think a french version of you article will be welcome, how can I proceed ahead ? — Preceding unsigned comment added by Mbariou ( talk • contribs) 10:33, 7 November 2011 (UTC)
I added some material to the history section with cites in response to the expand section tag placed there. After making the additions, I removed the tag. If anyone feels further expansion is needed, feel free to restore it. There certainly have been many more announced projects and these were not mentioned. J JMesserly ( talk) 17:45, 14 April 2012 (UTC)
Note on a retracted bit of info regarding salt dome capacities. In the Times article I cited, Wald wrote that the salt cavern could handle 5 times the 1100 psi pressure that McIntosh now uses. Others may see this and want to mention it as I did. I wound up retracting it because of an authoritative report in Gas Turbine World, Sept-Oct 2009, editor de Biasing states that the max allowable pressure is calculated at .9 of depth. So the ceiling of the McIntosh cavern at 1500 foot depth allows a max of only 1350psi. a reprint may be found at www.espcinc.com/library/GTW%20 ACAES%20Article.pdf. I suspect Wald is in error, but I don't know. Since it is contested, I pulled it. - J JMesserly ( talk) 10:30, 17 April 2012 (UTC)
I wonder why there is no mention of a trombe as a means of storing compressed air. — Preceding unsigned comment added by 83.240.200.90 ( talk) 18:15, 3 December 2012 (UTC)
Please see http://www.lightsail.com/tech.html and their CEO's Google Solve for X talk from a few days ago for the details, but I'm not sure which category (e.g. hybrid etc.) this fits into. so could someone please add it to the article? Apparently using water mist as a heat exchanger is a completely new idea bringing efficiency from 30% to 70%, and LightSail has patents on it. I'd feel more comfortable if an expert took a look at it before it was added. Neo Poz ( talk) 23:15, 12 February 2013 (UTC)
Just came upon that article, interesting. I'm just very surprised to hear about Dresden, which is my hometown - I never heard of a pneumatic storage system installed there anywhere and I'm very tech interested. Can't find anything with Google as well. So is it possible to have a source for that??
Rheinisch-Westfälisches Elektrizitätswerk AG is developing 2 new systems; see ALACAES and ADELE. Ref 2: http://content.lib.utah.edu/utils/getfile/collection/etd2/id/86/filename/1430.pdf Mention in article KVDP ( talk) 17:36, 24 July 2013 (UTC)
Maybe its just late at night and I'm being stupid, but isn't the formula for the stored energy ignoring the work done by the ambient pressure, which should not be included in the calculation of the "stored" energy, as it is not recoverable?
Let's look at the situation of moving from state A: volume , pressure , to state B: volume , pressure , with constant temperature throughout. For simplicity, lets assume that is the ambient pressure. As has been pointed out, the pressure law yields , and in fact for any intermediate state we have . Now, suppose I start with a gas in a tube of cross section 1 at pressure and volume , so the tube has length . Lets put a plunger at one end. Initially, I put no force on the plunger (which has area 1), and the ambient pressure puts a force on the plunger of magnitude . Now I put an infinitesimal force on the plunger and begin to slowly compress the gas. At any intermediate state, with the plunger at position , so volume at , and pressure at , I maintain a force infinitesimally larger than what is required to maintain steady state. BUT, at such an intermediate state, ambient pressure always maintains a force of , so the force that I must exert is
Thus, the work that _I_ do in compressing the air (and this is the recoverable energy) is
Now, typically we are wondering what the (recoverable, ideal) stored energy is in a tank, i.e. we know , and , but not . Using (so, e.g. ), (changing sign to account for who's doing work on what...) we can write:
The first term agrees with what in on the web page, but the web page ignores the fact that the ambient pressure would do work on the gas that is not recoverable (except if the tank were taken to outer space...), i.e. it ignores the second term. Writing the result in this format:
makes at least on property clear: linear in (double the size of the tank, double the energy). If we want the formula in terms of gauge pressure, lets use for the ambient pressure, for the gauge pressure, and just for the volume of the tank. (By definition, ). Then the formula becomes:
For , this has a nice Taylor series expansion.
— Preceding unsigned comment added by DavidIMcIntosh ( talk • contribs) 02:28, 12 June 2014 (UTC)
Is CAES practical for home energy backups to keep a phone or computer running after a power outage? To keep freezers and refrigerators cold, I think bags of brine (for freezer) or some liquid with freezing temperature slightly above that of water (for refrigerator) is more practical.
This article should talk about the economics of CAES, e.g. how it compares to other energy storage technologies, and the levelised cost of energy when used with energy generators. — Preceding unsigned comment added by 101.191.212.184 ( talk) 07:03, 22 October 2016 (UTC) 74.111.173.193 ( talk) 19:16, 2 October 2016 (UTC)
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" For Adiabatic CAES systems, it is important for the compressor to be functional in high temperature/high pressure conditions. This challenge can be overcome by using an intercooling system along with the compressor." This is the definition of adiabatic "a process or condition in which heat does not enter or leave the system concerned." And this is the definition of intercooler "an apparatus for cooling gas between successive compressions, especially in a supercharged vehicle engine.". So where does our contributor think the heat from the intercooler goes? it goes to the outside environment. therefore it is no longer adiabatic. Greglocock ( talk) 06:45, 10 March 2019 (UTC)
The entry in the "History" section for the power plant in McIntosh, Alabama contains the words "Although the compression phase is approximately 82% efficient, the expansion phase requires the combustion of natural gas at one-third the rate of a gas turbine producing the same amount of electricity at 54% efficiency". I'm confused; is this some sort of hybrid system that combines a gas turbine with CAES? Or is the heat needed to provide energy to help the gas expansion? Whatever it might be, the article should make this clearer, and explain exactly how it works. — The Anome ( talk) 07:33, 15 May 2024 (UTC)