Quote:
Originally Posted by Chris
Yes, I saw that the other day. I'm still trying to decide if the article is demonstrating journalistic ignorance on a massive scale or whether I'm missing something fundamental, because it seems to me that if you super-cool air down to a liquid you're removing the energy from it, rather than storing any energy in it. What you're actually doing, I believe, is creating a source of potential energy, because of the useful work that can be done when the liquid air is allowed to recirculate with other components at 'room temperature'. The actual energy in that system comes from the environment, rather than the liquid air, but it is the liquid air that allows the energy to be extracted.
Any scientists about to enlighten us further?
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Well, as I do this stuff in my line of work I suppose I can contribute something here. I use calculators occasionally for calculating carot and open rankine efficiencies so I can understand that the engineers where saying, though I really don't think the journalist who wrote the SKY article has any idea what they are talking about.
http://news.sky.com/story/991949/liq...-of-the-future
The SKY news article really made me cringe, as it is not the fuel of the future at all, clearly this is just an energy storage method. A temporary one at that. The comments to the article were even worse, and when peeps started saying you can "store liquid air" at room temperature, I had to comment in the comments section. No such thing as liquefied air at room temperature, it must be maintained below critical temperature at cryogenic temperatures.
Anyway, energy is indeed being stored by the liquefied air. A great deal of energy in fact. The reason it's easier to store energy as liquid air is rather complicated and drawn out. It could be instead stored as heat absorbed within some other medium (say a material with a high heat capacity) and used to expand air to drive a piston, but there are limitations to that as I'll try and illustrate.
Thermodynamically speaking, work is required to remove the heat from the air to cool in the first place (like compressing a spring) and then the liquefied air stored in a thermally insulated vessel (such as a nitrogen dewar).
Although the temperature of the cooled air had of course cooled, the air temperature of the hot side of the carnot cycle had in fact increased. The cooled air possesses decreased entropy (the hot side of the refrigeration system has increased entropy) and of course by the second law of thermodynaics, entropy of a system will always try to increase when it can. Which it does when the liquefied air is heated.
This work (required from the refrigeration pump/stirling cryopump to cool the air) is easily calculated using long established engineering principles. However, when the liquid air is later heated (say by passing through a small heat exchanger), the thermodynamic energy that was used to create it in the first place (assuming 100% efficiency), is then released in the form of pressure through a process called isobaric expansion. Its just a release of stored potential energy.
Bear in mind of course, that the refrigeration system used to produce the liquid air in the first place, will not have been 100% efficient. There will be a small margin lost due to friction. So the energy recoverable from the liquefied air will not be quite as high as the energy consumed to produce it in the first place. Also, the pneumatic motor used to extract the energy, won't be 100% efficient either. Since the liquefied air system is proposed to store off peak energy (cheaper per kWh), its only viable so much as a profit is made (the balance between normal grid supply and off-peak liquid air storage being determined by the thermodynamic loss margin, mentioned above). Too high a loss means a reduced amount of air can be produced and still remain a profitable/viable option. Of course that's just common sense really.
As for the amount of energy that can be recovered from liquefied air, it really depends on how it is achieved.
In all likelihood it will be combination of isobaric and isothermal processes.
So to give an approximate indication, for calculation purposes its easier to simplify the process by assuming 1 litre liquid, undergoing isobaric expansion (constant pressure) to its critical point (max temp at which liquid an exist irrespective of pressure), and then confined and heated to its final temperature (298K, or 25 deg C) at constant volume, and then allowed to isothermally decompress whilst driving a piston.
Summing up the work done via isobaric decompression and isothermal decompression is a simple approach to estimating the amount of work that can be extracted.
Air is 78% nitrogen, so to simplify things lets assume the liquid air is liquid nitrogen (otherwise calculations will be duplicated for oxygen too, the extrapolate between to give the actual energy calculation for air, ignoring the 1% noble gases of course).
Density of liquid nitrogen is 807 kg/m3. Which is 807g per L.
All gas molecules at the same temperature occupy approximately the same volume due to kinetic motion. One mole (6.02 x 10^23 molecules) equals about 22.4 litres at STP (standard temperature and pressure, which is 0 deg C / 273.15 Kelvin and 1 atmosphere pressure).
Nitrogen is diatomic (N2) so one mole is 28 grams. So there are 28.8 (807/28) moles per litre of liquid nitrogen. Which at STP will occupy a volume of 645.6 litres (22.4 x 28.8).
The mechanical work done by expanding 1 litre of N2 at its boiling point to a final volume of 645.6 litres at STP is given by:
For isobaric process, expanding from 1L liquid nitrogen, to its critical temperature and pressure. This is 126.19 deg K, and 3.3978 x 10^6 Pa.
To calculate the volume at the critical point, use ideal gas law.
PV=nRT. Rearranging, V=(nRT)/P.
n=28.8 mol
R=8.3145 J/mol/K
T=126.19
Solving for V, volume is 0.008893 or 8.893 litres (volume at critical point).
Heat to 298K, then assume isothermal expansion to drive piston to do mechanical work.
isothermal expansion work is given by: w=nRT ln(final voume/initial volume)
Final volume is 28.8 mol N2 at 298 K, use ideal gas law. Gives 0.7044 m3.
w= 28.8 x 8.3145 x 298 x ln(final volume/initial volume).
w = 311985 J or 311.985 kJ per litre of liquid N2 or 386.6 kJ/Kg
Of course it won't be possible to extract all work as usable energy so the actual value will be somewhat less than this.
Here's a calculator that can be used to double check the sums.
http://hyperphysics.phy-astr.gsu.edu...rmo/isoth.html
So for 1 litre, *most* of 311 kJ should be extractable as usable work.
One horsepower is 746 Joules/sec. At 100% efficiency, 311 kJ will produce one horsepower for 418 seconds or about 7 minutes or 100 horsepower for 4.18 seconds.
The actual figure will depend on the pneumatic motor, which would be tested to yield a figure. Assume the efficiency is a reasonable 85%.
Whereas, if the energy were *instead* being stored in the form of heat rather than liquefied air, due to the low thermodynamic efficiency of expansion, in order to match 85% efficiency, thermodynamic efficiency is given by:
http://en.wikipedia.org/wiki/Thermodynamic_cycle
Ambient temperature will be assumed to be 25 deg C (298 deg K).
efficiency, n, given by: n = 1 - (Temp L / Temp H)
So to achieve 85% efficiency:
0.85 = 1 - (298 / hot temperature)
Hot temperature = 1986.67 deg K or 1713.52 deg C
Any temperature below 1713 deg C cannot be used to yield a thermodynamic efficiency of 85%, it will be lesser. So in practise, you would need a hot sink that is thousands of degrees and cannot be cooled below 1713 deg C without compromising efficiency.
Therefore, you can see its far easier to store liquid air, than to thermally insulate a heat source to thousands of degrees. Even then, it would only be with the limitation of only allowing the heat source too cool to 1713 deg C before sending it back to the furnace for reheating. Not to mention that materials cannot withstand these temp ranges and the heat could never be contained/insulated using current technology. Using this cryogenic process not only makes the technology simpler, but thermodynamically its more efficient because of the extremely high gas expansion ratio of expanding liquid nitrogen or air.
Anyway, I didn't intend for the wall of text but that is the simplest way I could think of to give a run down of the theory and calculation(s).