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Pumped storage of thermal energy

 How to build a model of it
  (+2, -9)(+2, -9)
(+2, -9) 		  [vote for,
against]

[Note to those who come along and vote against just because the description is not clear and everyone else voted against: you really ought to abstain unless you understand the debate. If you do understand, you should state your objection instead of just voting anonymously. So far I haven't conceded any serious flaws, although I have discovered some errors. I suspect a description of an existing power plant or chemical plant would also receive all negative votes...]

I'm working on a new design, but this is one which you can build a model of, and the model will work. It will be extremely cost effective when built on a large scale. It will refill the reservoir behind a hydroelectric dam from a reservoir below the dam, using only ambient heat, and increasing the amount of power available substantially. Using the excess electricity from many of these machines, hydrogen gas can be made to power cars. Carbon dioxide can be removed from the air, reversing global warming (if enough of these are built). There will be no need to use oil (which is good feedstock for making things) as fuel.

The machine consists of nothing but dams and reservoirs, tanks, pipes, valves, and some sensors and automatic control equipment. It uses water and air to store energy. The machine lowers raised water while compressing cold air. Then that air is heated as the day warms up. Then the machine uses the hot compressed air to raise water. We must show that the machine raises more water than it lowered.

It is well known that heating compressed air at constant pressure increases the volume. The machine will raise the same amount of air for a given volume of compressed air at a given pressure, regardless of the temperature of the air. So if you heat up your air, you can raise more water.

The real question is how much more water is raised than lowered? Is it just a few drops? That depends on how much you heat the air. If you double the absolute temperature, you double the volume. But if you raise the temperature from freezing to hot (0degC to 30degC = 273K to 303K), that is only about a 10% increase in the absolute temperature, so the volume will increase only by about 10%. So the machine will raise at most 10% more water than it lowered if the air goes from 273K to 303K. A real machine would be lowering the whole lake, and raising 10% extra which would be used to generate electricity. Instead of needing rain to refill the reservoir, the daily temperature variation will do it. Building a reservoir on the top of a mountain will work, because you don't need a river to fill it. You just need a reservoir at the bottom as well.

A model will be not so hard to build. Now I know why I was saving all of those soda bottles I have... You need two big plastic tubs of water, one at the top of an outdoor staircase (or better, a long hill) one at the bottom, hundreds (or at least 16) two or three liter soda bottles, a lot of cheap garden hose, and a way of making reliable valves cheap (you need one for every tank). You really need to be able to open and close a bunch of these at once. You need a T joint for every bottle, which screws into the top of the bottle (hopefully it is made from the original bottle cap) and which attaches to cut ends of garden hose.

The maximum pressure developed is equal to the "head" of water, i.e., 15 psi for every 30 feet of elevation. So if the equipment withstands ordinary water pressure, you are in excellent shape unless you are building a very big model.

You cut the bottoms off the soda bottles and effectively seal the top sections together so you get two threaded fittings at top and bottom of each tank.

The machine has two separate parts which can be demonstrated separately. The real machine uses one upper reservoir (open surface) and a large tank half full of water with a closed top that can hold pressure or a partial vacuum. But for one part of the demonstration, we don't need the large sealed tank. A set of connected tanks compresses air using a pair of upper reservoirs at different heights, or raises the level of the water in one upper reservoir while allowing compressed air to expand. For this, you connect one hose with a valve between the top T of every tank. Then you connect a hose with no valve between the bottom T of every even numbered tank (so the bottoms of all even numbered tanks are connected) and between every odd numbered tank as well. The last two tanks have only one bottom connection. The first and second tanks connect to the bottom of both upper reservoirs. That is, both reservoirs have T's on the bottom, and a hoses with valves connect one reservoir to both upper tanks, and the other reservoir to the other side of the T's of the upper tanks.

To compress air using this setup, first close all of the valves and let water flow from an upper reservoir down into both the even and odd tanks. The water will rise higher into the lowest tanks, compressing air there more. This sort of compression is called irreversible because a small change will not change it into expansion, and it isn't what the machine does after it is running. We are just setting it up initially.

Now that all of the tanks contain some air and some water, we want to move all of the water into say the even tanks by letting the compressed air in the odd tanks rise into the even tanks. So open valves between the lower odd tanks and the next higher even tank. Water will immediately flow upward from the bottom hoses into the odd tanks, and push most of the air into the even tanks. The exact geometry of the tanks is important here and will need some tweaking. Then close the valves. This is the normal working setup: half of the tanks are full of water, and the other half contain compressed air. The air in the highest tanks is compressed less than the air in the lower tanks. To compress the air more, make sure the higher reservoir is connected to the tanks with the air in them. Water will flow down from the highest reservoir into the tanks, compressing the air. Then, move the air downward by opening the valves leading to the water filled tanks. Since the reservoir which is slightly lower than the other is connected to these tanks, more water will flow into the upper tanks and the air will move into the lower tanks. It may be necessary to adjust the height of the reservoirs to achieve this.

Next, compress the air which has been forced downward. Close the valves and swap the reservoirs. More water will flow into the tanks with the air because a higher reservoir has been attached. When this stops, open the other set of valves and the air will flow down another step. Keep going, and the air will end up in the bottom tank compressed.

The other set of connected tanks pumps water from the lower reservoir to the upper reservoir, using low-pressure air to move the water upward a small distance each step, or using descending water to create low-pressure air. The setup is almost exactly the same, except it is turned over. The valves and hoses connecting each tank to the next neighbor goe on the bottom, while the jumper hoses connecting even numbered tanks and odd numbered tanks goes on the top. Water flows from one tank to the next, while the low pressure air connects either to the even or the odd tanks, while the others are open to atmospheric pressure. The valves from even to next higher odd tank are closed, and the extra air pressure is applied to the even tanks. The pressure pushes water from the even numbered tank down through the hose with the open valve and up into the next higher odd numbered tank. This happens to all pairs of tanks at once. Then all of the valves are closed, the valves which were closed before are opened, and the extra pressure is connected to the odd tanks. The water rises another step.

To get a small pressure difference from water flowing downward, just reverse the operation. Air will flow out of the hose connected to the lower tank of each pair.

—Archimerged, Apr 17 2006

Description on my blog http://archimerged....ent-temperature-ii/
Pumped storage of energy from ambient temperature II [Archimerged, Apr 17 2006]

Article on renewableenergy.wikia.com http://renewableene...anks_few_reservoirs
Pumped storage output multiple expansion tanks few reservoirs [Archimerged, Apr 17 2006]

a passive solar collecter for collecting heat energy http://www.thermomax.com/tech.htm
[jhomrighaus, Apr 18 2006]

Diagram http://renewableene...a/Pumpedstorage.png
Diagram of tanks, pipes, valves, and reservoirs [Archimerged, Apr 18 2006]

stirling info http://www.stirling...ope=public&faq_id=1
More important to this discussion than you think [jhomrighaus, Apr 18 2006]

Pumped Storage http://en.wikipedia...wiki/Pumped_storage
Wikipedia article on Pumped Storage hydro stations [Archimerged, Apr 19 2006]

Raccoon Mt pumped storage http://www.tva.gov/.../pdf/raccoonmtn.pdf
The water drops 990 feet. [Archimerged, Apr 19 2006]

Simplified System http://www.bimmerbo...lified%20system.JPG
Here is a system that does the same thing using one tank and a floating piston [jhomrighaus, Apr 19 2006]

Compressed gas electromagnetic potential energy http://en.wikipedia...i/Mechanical_energy
info on Potential energy [jhomrighaus, Apr 20 2006]

(?) Thermal cycles diagram http://www.geocitie...oy/cycle/cycle.html
Cases 1 to 5 [Ling, Apr 20 2006]

intermolecular potential Energy http://www.newton.d.../eng99/eng99082.htm
comments regarding Potential energy in compressed gasses by a PhD in the field. [jhomrighaus, Apr 20 2006]

A simple model and question http://www.bimmerbo...iginal/Diagram2.JPG
Where does the energy come from? [jhomrighaus, Apr 20 2006]

Hydrostatic Stirling cycle air compressor http://renewableene...ycle_air_compressor
Article where I am consolidating the ideas from this discussion. Includes discussion of potential energy of gasses. Feel free to participate, sorry for the google ads. I don't get anything from them. It makes compressed air instead of raising water but the principle is the same [Archimerged, Apr 21 2006, last modified Apr 24 2006]

Joule Thompson effect http://www.chem.ari.../jadjte/jadjte.html
Good discussion of Joule Thompson effect and potential energy or absence thereof in ideal gasses. "This is an important and useful result. It says that the internal energy of an ideal gas is not a function of T and V, but of T only." [Archimerged, Apr 21 2006]

renewableenergy.wikia http://renewableene...sure_compressed_air
My comments on high pressure compressed air. [Archimerged, Apr 21 2006]

(?) Diagram of pumping air cycle http://www.geocitie...umpair/pumpair.html
[Ling, Apr 21 2006]

(?) For Archimerged 23rd April http://www.geocitie...y/pumpair/pump.html
Note the easy way to compress. The water pump system, I think, is your idea. [Ling, Apr 23 2006]

Compression Power Estimate http://www.processa...otating/recip_s.htm
Calculator for estimating compressor power requirements [jhomrighaus, Apr 24 2006]

Carnot Cycle Efficiency http://hyperphysics.../thermo/carnot.html
A more efficient process that wont even come close to doing the job [jhomrighaus, Apr 24 2006]

[link]






       A picture is worth a thousand words.   

       Please illustrate, as I am sceptical and confused.
—Texticle, Apr 18 2006
   

       [Archimerged], is this not: Potential energy to potential energy, then back up to potential energy minus 10% (this 10% to be gotten from the Sun)?   

       Where is the energy output?
—methinksnot, Apr 18 2006
   

       No, it's not. It is gravitational potential energy to gas pressure, say pressure = P, volume = V, potential energy = PV + constant. Then absorb heat from the surroundings (which comes from sunlight absorbed by the atmosphere over many square miles). The result is about 10% more energy than was put in. That is, either the pressure is 110%P, or the volume has increased to 110%V, or in any case potential energy = 110%PV + constant. We convert the 110%PV back to gravitational energy, raising 110% of the water that was lowered. The machine will have some inefficiency, but because it has no solid moving parts, there is very little friction, very little leakage, and efficiency could easily be 99%, raising about 109% of the gravitational energy. But if the machine is built where there is a wider temperature range, we get more than 110%.   

       I'm working on illustrations.
—Archimerged, Apr 18 2006
   

       Can I take from your explanation that the machine will only be able to yield approximately 10% of the water's initial potential energy?
If so, this is not much even at large scales.		   

       The background concept of this idea, and that of your previous one, was harnessing energy in naturally-occurring phenomena. I think it would make sense to use something more reliable than temperature differentials. Generating energy out of tidal differences would be more reliable. Don't you agree?
—methinksnot, Apr 18 2006
   

       You need to look at the efficiency of the process in comparison to other systems for converting sunlight into power. if this is more efficient than solar panels then its worth the effort.
—jhomrighaus, Apr 18 2006
   

       More than comparing its efficiency with that of solar panels you need to compare whole-of-life cost/benefit ratio. It's not very environmentally friendly to produce and dispose of solar panels.
Furthermore, for this to be viable, it needs to favourably compare with all other energy harnessing sources. Believe me, you don't build a hydro plant if it's cheaper to use coal or nuclear at a specific location.
—methinksnot, Apr 18 2006
   

       I would disagree.   

       If the ultimate efficiency of this process is, just to pick a number, 5% and the ultimate efficiency of say solar panels is 50%(just picking numbers for sake of argument) then why should we continue to look and invest in something that has limited long term potential.   

       The idea is a bit hard to follow without illustration but it sounds interesting. What I do know is that you dont get something for nothing and this sounds like it could be a bit more energy intensive than you are thinking. This works great for those little hand bubbler toys but this is a whole different matter.
—jhomrighaus, Apr 18 2006
   

       It depends on your definition of ‘ultimate’. What I meant is that you need to compare the whole-of-life kW/hour cost.   

       I understand that [Archimerged]'s idea does not get something for nothing and I am not prepared to question its efficiency until I see some numbers, but at this stage I'd like to question its effectiveness and economical viability.
—methinksnot, Apr 18 2006
   

       Temperature differences are pretty damned reliable. They happen every day. Not every location is as good as another. This would be installed at existing hydro plants or pumped storage plants so you don't have to build the hydro part. Or it could be built on a tall mountain (so you get very reliable cold from the top, and gravity feed heat pipes are good at carrying heat upward) with hot temperatures at the base, giving a wider delta T so better than 110% of the total water lowered is raised back up. It might even be able to run continuously instead of one cycle per day. The total amount of water lowered each day is a constant determined by the size of the reservoir. The cost of the reservoir scales very nicely with size since the walls are 2D and the volume is 3D. So this sort of machine can create a new hydroelectric resource at a location where hydro power does not currently exist. You still have to pay for the hydro plant, unless you tap off the compressed air instead of using electricity. But with oil costs rising, nuclear fuel disposal still not done, and coal causing global warming, more hydro would be very nice.   

       The major question here is do I really get 99% efficiency in raising the water back up? Or are there gotchas such as losses due to tanks stretching and contracting, valves leaking, etc. etc. I'm hoping to hear from some experienced engineers on this.   

       Regarding the cost of the heat engine, including tanks and pipes and valves: the amount of material scales as the surface area while the energy handled scales as the volume. The water raising and lowering portion runs at low pressure so can be built of cement and does not need heat pipes. The air compression part does not need to operate at extreme pressures since we are not aiming to make high pressure compressed air. We balance the cost of storing a large volume at lower pressure against the cost of storing a smaller volume but needing more expensive tanks. Ideally, the compressed air would be stored in an existing mine. For example the Homestake mine in ND might be a good location for this, since a lot of heat is available inside the mine and cold is available on a nearby mountain.
—Archimerged, Apr 18 2006
   

       You would need to increase the temperature of a fixed volume of air about 600 degrees K to gain 15psi increase in presure with a 100% increase in volume. For 1 cubic meter of air you would be able to displace 1 cubic meter of water. So if you had a box with a Sliding air tight divider at bottom, 1 cubic meter of air at bottom and 1 cubic meter of water at top open to air on top and the air inside at 15psi water level in box would rise 1 meter when the air is heated 600 degrees Kelvin. This would require about 8000 joules of heat energy. With perfect heat transfer of Solar Radiation you would need 8 to 10 sec to achieve this amount of energy input.   

       So maximum potential(in lossless system) is about 8000 joules per second of energy per square meter of collector or about 8 watts per meter squared. Assuming you could recover the energy with a lossless turbine(in reality its about 80% for steam turbines) then you would need about 10 m2 of collectors to power a light bulb.   

       A much simpler aproach would be to simply connect a circulator to a presurized tank and heat the water(kind of like a presure cooker)   

       Bottom line This is nothing but an overly complex passive solar collector. I say Bone as there are better more direct ways to utilize the energy see link for an example.
—jhomrighaus, Apr 18 2006
   

       Just a note the maximum difference in temperature in any given day on earth probably doesnt change by more than say 30 degrees Centigrade. as I stated above you would require more than 600 degrees to gain even a modest increase in the height of the water.
—jhomrighaus, Apr 18 2006
   

       //Temperature differences are pretty damned reliable. They happen every day. Not every location is as good as another.// [Archimerged].   

       Not so. What you need for your model to work is extreme temperature differences that are not only reliable but also above 0 deg C and with specific topographical characteristics (not to mention an abundance of water). This does not happen often.   

       [jhomrighaus]' numbers pretty much invalidate the whole concept from the economic point of view.
—methinksnot, Apr 18 2006
   

       These numbers are just wrong. If 600K were required to operate a heat engine, there would be no industrial revolution and James Watt would have failed to pump any water out of mines.   

       Revised: you said "600K increase" not 600K temperature. That means 300K cold reservoir, 900K hot reservoir, which is 627C, about 1161F. I believe it is possible to operate a steam engine at lower temperatures than that.
—Archimerged, Apr 18 2006
   

       600k is required to displace 1 cubic meter of water using 1 cubic meter of air starting at 15psi. This implies the only force in use being the expansion of the contained and heated air.   

       Watts engine used steam which is a whole other ball of wax as you are dealing with a phase change with an expansion ratio of 1600 to one. This requires a tremendous amount of energy as shown below.   

       Stiring engines on the other hand do use only heated air to operate and those units that are powerful enough to do any real work utilize a temperature diferental of hundreds of degrees(i think I saw 500 degrees plus on one design)   

       Just to get the 1 cubic meter of water to its boiling point from would require more than 300000000 Joules of heat energy.   

       1 calorie is needed to raise temp of 1 gram of water 1 degree C.   

       There are 100x100x100 cm in 1 cubic meter or 1,000,000 cubic centimeters. That means 1,000,000 calories are required to do the job. there are 4.19 Joules per calorie so that woud bring us to 4,190,000 Joules. Next we need to raise the temperature by about 80 K (20C to 100C) 80x4190000 is 335200000 joules or 335300 watts. That would be 42 1meter perfect energy collectors just to boil the water.   

       All the needed formulas are available on line to calculate this for yourself. Im sure I could have made a mistake in all this calculating but I think if you look at the temperatures that occur in any old engine you will find it is not all that far fetched.   

       Please someone do the numbers and check my calcs just in case I missed something. I still say its a giant Boondoggle.
—jhomrighaus, Apr 18 2006
   

       [jhomrighaus] As I said on another page, I do know a little about thermodynamics. I don't doubt any of your calculations. They just are beside the point.   

       I am trying to reproduce your calculations. 15 psi is close to atmospheric pressure. You describe a piston ("sliding airtight partition") with 1m^3 water on top and 1m^3 air (pressure unspecified) inside the cylinder. Assuming this means a column of water 1 m high, then the pressure of the air is 1.422 psi above ambient pressure. You want to heat the air to make the piston rise 1 m, so the volume of air doubles from 1m^3 to 2m^3. The pressure must remain at 1.422 psi above ambient, or the piston will move, so to double the volume, you have to double the temperature. Ah, I see where 600K comes from. You are assuming we start at 300K (near room temperature) and double the temperature so that the piston rises exactly 1 m. There is simply no requirement that the piston should rise 1 m, or that the volume should double, or anything like that.   

       And if we started at 280K, then we would need to increase the temperature to 560K to double the volume. The change in temperature is 300K or 280K, not 600K.   

       The heat engine I describe follows a Stirling cycle. It is not a Stirling engine, which is different from using a cycle with isothermal compression and expansion and isochoric heating and cooling. Except for losses due to leakage around valves, expansion and contraction of solid things that shouldn't expand or contract much, etc., the machine operates with near perfect efficiency. It doesn't have any solid moving parts except for the valves.   

       The requirement is that the machine starts with brine in an upper tank and a lower tank. When the temperature is 273K the water is allowed to descend to the lower tank by opening and closing valves in the machine in the proper sequence, always waiting for flow to stop before continuing to the next valve setting. Then the temperature rises to 303K. Valves in the machine are opened and closed in a different order, and water rises from the lower tank to the upper tank. However, when the same number of steps have been completed, and the same volume of water has been raised, there is still excess air pressure in a tank. If we continue operating the valves until nothing more happens, there will be about 10% more water in the upper tank. than was there when we started. At a real pumped storage facility, this would be converted to electricity via the existing water turbine.   

       I am working on diagrams which will show the order the valves are opened in and how the air and water move through the machine on the way between the water reservoirs.
—Archimerged, Apr 18 2006
   

       Archimerged, I am looking forwards to your diagrams, but however your system works, I can see perhaps one problem: How do you seperate the air and water so that the air can heat up through the course of one day, and yet not lose the heat to the water?
—Ling, Apr 18 2006
   

       Yes, the fact that a large body of water does not heat up as rapidly as the air is a problem. However, the machine has completely separate compartments devoted to pumping the cold water vs. letting the hot air expand. The water which serves as liquid piston for the compressed air is a much smaller volume and will get heated up long before much of the lower reservoir is pumped back up. The force generated by the expanding high-pressure air is transmitted via columns of water (which don't flow much at all) to low pressure air at the top of the columns which then acts on the cold water. The air in contact with the cold water will contract initially, but more air gets added and it doesn't circulate but just moves back and forth. Thanks for the interest. Back to the diagrams...
—Archimerged, Apr 18 2006
   

       //Next we need to raise the temperature by about 80 K (20C to 100C) 80x4190000 is 335200000 joules or 335300 watts. That would be 42 1meter perfect energy collectors just to boil the water// Actually, rather more - you forget the latent heat of boiling, which is about 2260000 j/kg.
—coprocephalous, Apr 18 2006
   

       A diagram is now available. See the links section above.   

       [coprocephalous] There is nothing about boiling water in this idea.   

       [jhomrighaus] The link about the passive solar collector has nothing to do with this idea. This machine does not operate on direct solar energy. It extracts heat from the ambient air, regardless of how the heat got there.   

       A full sized machine has finned "heat sinks" above and below the expansion/compression tanks which collect or reject heat from the air, and gravity feed heat pipes extending inside the tanks which carry heat upward and "cold" downward due to evaporation and condensation of a refrigerant inside the sealed copper tubes.   

       If you want to raise valid questions about this design, you should calculate what mechanical efficiency I need when using a 273K to 303K temperature difference to ensure that more water is pumped upward than downward. You seem to think that even perfect efficiency would not permit a machine to pump water back up. So tell me, what efficiency will result in the machine pumping exactly the same amount up as went down? What efficiency would result in it pumping up 90% of what went down?
—Archimerged, Apr 18 2006
   

       I think the boiling water argument is obfuscating the actual point. As far as I can tell from [Archimerged]'s description, this energy harvesting machine does not rely on collecting that much energy that water boils. Rather, subtle-yet-regularly-occurring temperature changes are cleverly used to attempt to pump some of the water uphill.   

       In my opinion, the discussion that needs to be had is whether the cost per unit of energy harvested by this machine compares favourably to that of a hydro plant using the same vertical drop and water flows. I would be prepared to disregard the fact that finding a suitable location for this is, in my opinion, pretty much impossible. I remain unconvinced of its practicality but less so of its plausibility.
—methinksnot, Apr 18 2006
   

       //energy harvesting machine does not rely on collecting that much energy that water boils// If you can't make a decent cup of tea with it, then what incentive is there to build it? (BTW, I didn't introduce the boiling bit)
—coprocephalous, Apr 18 2006
   

       It's late here [copro]. I'm having a whiskey and would not change it for a cuppa for all the tea in china.   

       I'm also very happy you did not 'introduce' the 'boiling bit'. If you know what I mean...
—methinksnot, Apr 18 2006
   

       Well, I printed it all out, and spent 1/2 hr trying to follow what you meant. Some comments:
1. The volume of pipes compared to the volume of tanks should be small, otherwise the compression will take place in the pipes...
2. The initial heat of compression will be lost.
3. The final pressure, when you say "the air will end up in the final tank compressed", will not be any more than the upper water reservoir head. However, the volume of air will be higher, and mostly in one place.
4. To get one tank 'full' of water, and the next 'full' of air, the initial compression should be arranged so that the volume of air in the tank pair changes to exactly half. But this cannot happen in every tank pair, since the pressure in each pair will be different. But I think the pumping will happen anyway.
5. I can follow your explanation of the pumping system, but you neglect to mention where the heat input and output is. Perhaps the bias tank is heated? Where does the heat flow out?
6. Interesting idea, none-the-less		   

       edit: 7. How do you get the air back in the lower system of tanks?
—Ling, Apr 18 2006
   

       the boiling water bit was simply a point to demonstrate that large amounts of energy are involved in making a heat engine run.   

       In order to make a hot air piston displace a cubic meter of water you need to rise that meter of water (ie do work with the piston) this would require you to do more than just double the volume you need to increase the pressure as well.   

       After reviewig everything I have to say that I may have an error in the 600 degree diference but Its not as far off as you might think. As to this Machine in order to generate anything but token you would need a VAST quantity of air to act on a VERY SMALL quantity of water for such a small increase in temperature. If I had to guess you would lose more power just actuating all of the valves and the computer to run it all than you would ever get from its use.   

       You seem to forget that there is more to do than just raise and lower the water. You want the water to do work on the way down. This means you need to increase the potential energy of the water which means you need to add energy to the system to do this work.
—jhomrighaus, Apr 18 2006
   

       [Ling] Thanks very much for taking time to work on this. I'll address your remarks in an edit to this annotation. Some may be addressed already.   

       [jhomrighaus] Thanks for that solar collector link. I thought it was one I had seen before, but it is new to me, and seems good quality. (I haven't looked at in detail but plan to). And thanks for continuing to discuss the idea. It helps me to improve my explanation and check for errors.   

       The stirling info link is of course interesting. "A good general guideline is that if the hot side of the engine is not at least 500 deg. F. (260 deg. C) the engine will be too bulky for the amount of power it puts out." That is true for an engine which is intended (like almost everything called an engine) to produce motion and operate machinery.   

       The machine I describe is intended to collect energy, not to produce work. It is naturally very large and occupies otherwise useless space like a mountainside. The amount of energy it handles scales as the volume of the machine but the materials required to build it scales as the surface area.   

       In the text posted in this idea, I didn't describe the heat collection and rejection apparatus (which I expect is needed only on the hydrostatic expansion / compression tanks, not on the pumping tanks or the bias tank). It includes gravity-feed heat pipes between the interior of the tanks and the heat sink (above the tank) and heat source (below the tank), and provision for air flow: the heat sink heats air and a chimney carries the heat away because hot air is less dense than cold air. And the heat source cools air, which then falls into a downward chimney and carries the cooled air away.   

       BTW do you know the T-S diagram for the Stirling cycle? (Recalling Stirling engines don't quite use that cycle). I'm looking at the Carnot_heat_engine article on wikipedia, where the T-S diagram is a square, but a Stirling cycle exchanges heat in all phases so its T-S diagram has no vertical lines. I haven't needed that diagram because I don't care much about how much heat goes into and comes out of the machine, since both are supplied by the surroundings. The amount of heat which flows is determined by the heat capacity of the working gas (wet air) and the number of moles stored, and by the requirement that the compression and expansion be isothermal. The number of moles of working gas has to be adjusted in order to be able to pump the desired amount of water.   

       So one good question to ask is how long will it take for the required amount of heat to flow across the available temperature difference. It is true that heat flows more slowly the smaller the difference. I admit I haven't calculated this yet and it remains a potential gotcha. You haven't calculated it either.   

       Another good question is how much effect the water vapor has. The amount will change when the temperature changes. I haven't worked that out for this machine but I don't expect it to be a big problem. It is another potential gotcha.   

       I'm calculating the answer for required energy conversion efficiency so that this machine can pump more water up than it lets down. I want to use a single fraction, x, for the conversion efficiency from gravitational potential energy to gas pressure potential energy. This is easy enough to explain (but it took me a while to figure out how). Suppose the ideal cycle puts out K times the work it takes in. The inefficient machine converts gravitational potential W to work x*W. Then the ideal cycle operates using that work plus whatever heat it absorbs and rejects, and puts out work K*x*W. Finally, the inefficient machine converts the work back to gravitational potential energy K*x^2*W. So, an ideal cycle which multiplies the input work by K will yield more gravitational potential than it takes in only when K*x^2 > 1. So if K is 1.1, then x must be > sqrt(1/1.1) = 95.4%.   

       Let me consider this in more detail, since usually ideal cycles are discussed in terms of the heat in and out rather than the work in and out. The ideal Stirling cycle starts with its working gas at temperature H and volume V. It cools the gas to temperature C volume V putting out the necessary heat to do this. Next it takes energy W in as work, compressing the gas to volume Y while still at temperature C, and putting heat out at temperature C. Then it takes some heat in at temperature H (warming the working gas and keeping it warm) and puts some work out by letting the gas expand to volume V and temperature H, the original state.   

       How much work does it put out? Can we figure this without knowing the amounts of heat? Using an ideal gas for the working gas, we know that the work is for volume change from V to Y is integral Pdv, with P = nRT/v. This works out to nRT integral dv/v = nRT*log(Y/V). So the work W done compressing gas from V to Y at temperature C is W = C*nR*log(Y/V). The work extracted when the gas expands at temperature H is H*nR*log(Y/V) = W*H/C. The work out is the work in times H/C. Thus, K = H/C. The ideal Stirling cycle machine does multiply work in by a constant. That's why a Stirling engine can't start up by itself.   

       But you are right, I did not calculate how much heat is involved. That's because I just want to know whether or not the machine operates at a profit or at a loss. For H/C = 1.1, the efficiency has to be over 96%. Below that, and the machine must operate at a loss, and you can't "make it up on volume" as the retail sales joke goes. It seems quite plausible to me that the liquid piston no-moving-parts (except for valves) mechanism I describe could operate at efficiency over 96%.   

       Next I go over the cycle in terms of the new diagram of pipes, valves, tanks, and reservoirs. The machine converts gravitational potential of water in the pairs of pumping tanks to low-pressure compressed air in the bias tank. That pushes water downward through the hydrostatic manifold and up into half of the hydrostatic tanks (the others are full of water), with a net increase in the air pressure in the hydrostatic tanks relative to atmospheric pressure. Each hydrostatic tank is at a different pressure, but the same bias pressure raises the water level in each tank a little. That increases the pressure in the tank because the volume decreases. It also increases the temperature a little, but a heat pipe carries the extra heat away. Next, the valve to the next lower reservoir is opened. Water flows out of the bias tank down the manifold and into all of the hydrostatic tanks, displacing the air into the next lower hydrostatic tank. This flow stops when pressures balance. Meanwhile, air flows into the bias tank from the large number of pumping tanks, and water flows downward in each pair of pumping tanks. Water from the lower hydrostatic tank of each pair flows into the other manifold and up into the upper reservoir. This completes a half-step, and all valves are closed. Water from the upper reservoir is let into the bias tank to replace the water which flowed into the hydrostatic tanks to displace the air downward into the next lower hydrostatic tank. Also, water is let into the uppermost pumping tank to replace the water which flowed downward into the second pumping tank. This flow has to be metered to prevent flooding of the pumping tanks -- it does not stop automatically. The refill valves are closed, and the other set of pairing valves is opened to create the opposite pairing (odd-even instead of even-odd) of hydrostatic and pumping tanks, and the manifolds are connected in the opposite sense, so that the odd tanks are connected to whatever the even tanks were, and vice-versa. More air flows out of the lower tanks of each pair of pumping tanks, and into the bias tank. I believe this is where we came in.   

       The above describes a step of one quarter (isothermal compression) of the Stirling thermodynamic cycle executed by this machine once each day. This happens in the early morning during a period of low temperature. The second quarter (constant volume heating) happens all morning and half of the afternoon while the temperature increases. The air in the lowest hydrostatic tank increases in pressure and the water level decreases, but the volume changes very little because the tank is tall and thin (not shown). The third quarter (isothermal expansion) of the Stirling cycle occurs when the temperature is at a maximum, in the late afternoon. The process is almost identical to that described above, proceeding in reverse. The main difference is the water level in the bias tank. It is above the level in the upper reservoir instead of below, and the bias pressure is slightly below atmospheric. This causes water to be "sucked up" into the lowest pumping tank. Actually, atmospheric pressure pushes down on the surface of the lower reservoir and pushes the water up into the lowest tank where the pressure above the water level is a little below atmospheric. Similarly, atmospheric pressure above the water in each lower tank of a pair of pumping tanks pushes water upward into the upper tank of the pair where the air above the water is at the bias pressure. Some air flows into the bias tank.
—Archimerged, Apr 18 2006
   

       [Ling] Here are my replies: 1. The volume of pipes compared to the volume of tanks should be small, otherwise the compression will take place in the pipes... -- Yes. 2. The initial heat of compression will be lost. -- Yes. It goes up the heat pipe into the heat sink and up the chimney. 3. The final pressure, when you say "the air will end up in the final tank compressed", will not be any more than the upper water reservoir head. -- Yes. However, the volume of air will be higher, and mostly in one place. -- I think you mean the _pressure_ of air will be higher, the volume will be lower. This machine uses a constant amount of working gas (except for the water vapor which comes and goes with the temperature). 4. To get one tank 'full' of water, and the next 'full' of air, the initial compression should be arranged so that the volume of air in the tank pair changes to exactly half. But this cannot happen in every tank pair, since the pressure in each pair will be different. But I think the pumping will happen anyway. -- Remember that the manifolds have no valves. The water flows in and out of the tanks to make the hydrostatic pressure match the gas pressure. When the valve is opened from the tank with gas to the lower tank with only water, the hydrostatic pressure in the lower tank is less because it is connected to the bias tank, and the water level in the bias tank is below the level in the upper reservoir by more than height distance between tanks. So water has to flow into the upper tank. Air will have to flow out the top. No water can flow out the top until all the air is gone. But once the air gets into the next tank, any water following will flow around the air and out the bottom if necessary. Hmm, there might be an inefficiency here, a way for water to flow without compressing gas. Thanks for the question. 5. I can follow your explanation of the pumping system, but you neglect to mention where the heat input and output is. Perhaps the bias tank is heated? Where does the heat flow out? -- See the previous annotation. 6. Interesting idea, none-the-less -- Thanks. 7. How do you get the air back in the lower system of tanks? -- If you mean the tanks on the bottom of the drawing, that is the hydrostatic tanks, the air in there never leaves. If somehow the air is lost, you open some extra valves and put it back in, but that doesn't happen normally. If you mean the bottom pumping tank, the bias tank pressure is lower than atmospheric and water from the lower reservoir is pushed down through a pipe and up into the lowest pumping tank.
—Archimerged, Apr 18 2006
   

       Correction: I may have described the level in the bias tank incorrectly. And there is a missing valve on the diagram: you have to be able to open the tank top and bottom to atmosphere and upper reservoir to restore the level. When those two valves are closed, the pressure in the bias tank can't be read by comparing the level to the reservoir level, since the two are not connected by a U tube.   

       [bigsleep] Air does just fine. Other things on the earth absorb sunlight and then transfer the energy to the air. And there are quite a few cubic miles of air available near the machine. Remember we have all morning to warm it up and when heat gets transferred to the machine, the air cools and falls down the lower chimney causing a draft which brings more warm air. The volume of the machine itself has nothing to do with the heat capacity of atmospheric pressure air. I believe the substance you refer to is copper, but since the heat isn't stored there very long you don't need so much. The heat is stored in a large amount of highly compressed air during the morning, and then used to raise water. This device is not interested in the pressure fluctuations of weather, only the temperature.
—Archimerged, Apr 18 2006
   

       Why are all the individual reserviours needed. Wouldnt one large reservoir serve the same purpose. Seperate the water from air with a diaphragm and compress with falling water.   

       Be aware that the maximum pressure attainable will be that of the head pressure of the highest point of the water column that feeds the machine. Another point to note is that the gas law is linear at about 33% increase in pressure for any given starting pressure with a rise of 100 degrees. This means that the higher the pressure the greater the useful change in pressure. so charging the system to the maximum allowable will improve the effectiveness of your machine.   

       In the grand scheme of things this plan will work but it will provide limited effectiveness. The only gain will come from the pressure above and beyond the hydrostatic head pressure of the system. so for a system with a 33 foot head you would be starting with 15psi Raise the temperature by 100 degrees and you will increase the presure to 20 psi. this will push water up until the presure returns to 15 psi then you need to start over. Im not sure but this fact alone may limit the effectiveness of the machine to the maximum temperature differential gas expansion irregardless of the height of the head of the machine.(ie a 100 foot machine will move same amount of water as a 1000 foot machine given the same temperature differential).   

       I would think that based on this info that you could expect from 5% to 10% increase in Pressure for the 30 degree temperature difference you are siting. Heat transfer will definatly be an issue as Air is generally a pretty decent insulator(poor heat transfer)   

       All the extra tanks just confuse things and Im really not sure they contribute to the quality of the idea.   

       Something like this would be far more effective the greater the temperature swing.
—jhomrighaus, Apr 19 2006
   

       [bigsleep] Thanks for the questions. They are _very_ helpful.   

       I need to correct the statement that the machine stored the uncompressed air. It has to bring air in for compression from the surroundings just as it brings water in from the reservoirs. Otherwise yes the machine would be very very big. I don't believe that changes anything, because it's not hard to get air to flow into a tank where descending water is creating a vacuum.   

       See the pumped storage article on Wikipedia: http://en.wikipedia.org/wiki/ Pumped_storage   

       I am right now trying to find out how many joules are stored in a pumped storage reservoir. But 1 GW plant running for 24 hours produces 86,400 GJ. These reservoirs must contain something like that amount of gravitational potential energy.   

       Thanks very much for asking this question. I have a better understanding of what it means to say the pressure times volume product is energy. A given number of moles of gas at a given temperature always contains a certain amount of kinetic energy, regardless of the pressure. This seemed a big mystery for a while just now. I thought I was storing energy _in_ the compressed gas. It turns out I was wrong. Actually, temperature is a measure of how much compressed air is _actually in_ the gas. When attached to a temperature reservoir, compressed gas can do more work than the energy it actually contains.   

       To collect energy from ambient heat, I use gravitational potential energy to create pressurized gas at temperature C. In fact, the energy isn't staying in that gas. It is going away into the surroundings in the form of the heat I had to remove in order to keep the gas at temperature C, but I get it back when I let the gas expand slowly, since the energy will come back into the gas from the surroundings. After storing the energy in the surroundings, I warm the gas to temperature H (picking up part of the energy gain), and finally I let the gas expand slowly while more energy returns from the warmer surroundings than dissipated earlier when the gas was being compressed. As the gas expands, it does work on water in the pumping tanks, which ends up in the upper reservoir.   

       This has some implications regarding the amount of heat which has to be dissipated as the working gas is being pressurized. In fact, _all_ of the gravitational potential energy extracted from the falling water gets converted to heat and dissipated. So this machine is going to be putting out a substantial amount of heat in the early morning, and gathering a substantial amount of heat in the afternoon. I have to think about this a while but I want to put the comment up now.
—Archimerged, Apr 19 2006
   

       [jhomrighaus] The extra tanks really do serve a purpose. I'm not exactly sure why you say they are unnecessary. The machine cannot be nearly reversible without extra tanks, and irreversibility causes additional heat transfer at least. Some of the pumped storage facilities have a head of 1000 feet, see the link to Raccoon Mt.   

       Ok, you propose putting all the air in one big tank and letting water flow into the tank. There is no need to separate the water, just use a U tube to feed it into the bottom. The problem with this is it is not reversible. Maybe that's why you don't see how I can pump water upward as well as I do. The irreversible compression can't be reversed: it won't pump the water back up. The reversible compression which requires lots of different tanks can pump the water back up. Think about it some more. Thanks for the comments. And yes there is a problem with heat transfer, but it might be solvable. I haven't started working on it yet.
—Archimerged, Apr 19 2006
   

       Consider Wind Cave. The wind moving into and out of the cave is generated essentially by the forces you describe here: temperature (and barometric pressure) changes in the outside atmosphere. The wind is produced because the inside atmosphere of the cave wants to equilibrate: essentially what you propose, except the inside atmosphere is in these reservoirs.   

       This energy could be captured with a windmill at the mouth of the cave, or between your reservoirs and the world at large. It would be simpler. The appeal of your setup is that it could be vast - maybe really vast, and the energy capture would be more effiecient.   

       The trick is how to make it vast but use, as far as possible, naturally occuring structures. Once could place a glass tank over a lake with a rigid clear top and opaque flexible bottom. On heating up the bottom would push down into the lake, and could be used to force water up a pipe. On cooling at night the reverse would happen. You would need strong girders to reinforce this thing.   

       Perhaps this could be done over Lake Powell. It would be a very large pane of glass and a lot of plastic.
—bungston, Apr 19 2006
   

       See the attached link for simplified system. The inital pressure charge can be obtained by pumping in Air(or whatever medium is found to be most effective) until the piston is at the proper location(low in stroke when cold, high in stroke when hot) From there Ambient air is moved through the Air to Air(or other medium) heat exchange apparatus(Tubes or Fins etc) If Sufficiently large I would think that a natural draft would do the job. Also this could be combined with a Hydrothermal Loop to heat the chamber and a draft air flow to cool it. This would allow for larger temperature differentials and more effecient operation in different seasonal conditions. Also creating an aresol mist at the entry to the cooling section would allow for additional cooling through evaporation with no mechanical cost(pressurized water from upper reservoiur)   

       Please help me understand why this wont work and all the tanks and lines are needed.   

       This is in essense a large sterling engine that uses air to heat and cool the cylinder as was stated in other Annos.   

       For 30 degree difference I would guess that you could move about 10% of the total air volume of the air chamber(1million gallon capacity air chamber would move 100000 gallons of water)   

       Eager to hear comments
—jhomrighaus, Apr 19 2006
   

       My immediate answer is that the air pressure in the tank never gets very low. It certainly never gets down to atmospheric pressure. My machine runs isothermal expansion and compression between atmospheric pressure and high pressure. It does not need to confine the atmospheric pressure air, but captures it anew each morning. This is possible because of the many different tanks. I hever have much low pressure air in confinement at any one time. While it is being compressed, space is being made for it in the lower tanks. So the total volume of my many tanks is a _lot_ less than the total volume of a tank that can hold all of the air at low pressure.   

       I'm working on calculating how much working gas is required for a large plant, in order to know how much high pressure air storage I need. I now know that the heat to be disposed of up the chimney in the morning is equal to all of the gravitational potential energy of the water being lowered.   

       Another reason for lots of tanks is every one of them has heat pipes running through it. Those heat pipes will be very busy.   

       Recall that the energy handled by the machine is the area enclosed in the cycle on the P-V chart. By restricting the cycle to high pressures, you are eliminating most of that area.   

       Another problem with your machine is it has a solid moving part, the piston, and I really don't see how you are going to make it 95% efficient.   

       Also, if the check valves are as shown in the diagram, it means water never flows into the lower reservoir except through the turbine. That means the tank above the piston has to be large enough to hold the entire contents of the reservoir. My machine is intended to pass nearly the whole upper reservoir down to the bottom in the morning and pump it back up at night. Because of all of the tanks, I need only a small volume in the pumping tanks, not the whole reservoir. Your machine has the biggest piston every imagined (needed in order to make it reversible). See the Wikipedia article on Ericsson's ship which had huge pistons but failed after one day of operation. Because the piston rises and falls the entire distance, the last bit of water lowered is not lowered all the way down, but only a small amount. The simplified system is not at all equivalent.
—Archimerged, Apr 19 2006
   

       The machine has one moving part. Yours has more than 30 valves alone. I donot understand the benefit of pulling in new air every morning. The higher the pressure of the air at the starting point the greater the displacement that can be achieved.   

       The machine I show I would think would be extremely efficient. It would require literally no input of energy beyond the initial compression of the chamber. beyond that it is a completely passive process. the air heats up and expands forcing water upwards. As it cools the piston retracts and more water is pulled into the piston and the cycle repeats.   

       In many things the simplest approach proves to be the most effective(look at Jet engine technology for example, the first type was a tube with a fuel injector, then came compressors and turbines and afterburners now we find that the most efficient and powerful jet technology is a scram jet which is a tube with a fuel nozzle)   

       I am still totaly unclear as to the advantage of the tanks and pipes. My machine very clearly and simply lifts water from a low point to a high point using only ambient thermal input. Isnt that what your machine does? Why would I build your machine when mine does the same thing? Whats the benefit? You are still at the end of the day limited to the head pressure of the reservoir and the expansion of a fixed volume of gas. If the volume of the air reservoir of your system equals the volume in mine and the thermal heat transfer efficiency is the same then bothe systems will do exactly the same amount of work. The Physics are pretty clear on this point. You cant get something for nothing.
—jhomrighaus, Apr 20 2006
   

       //Also, if the check valves are as shown in the diagram, it means water never flows into the lower reservoir except through the turbine.//   

       It is very clear that your understanding of this sort of system is very limited. This is a pump. The water in check valve pulls in water from the lower water body then the water is pumped up through the outlet check valve. The pressure in the air chamber is equivilent to the hydrostatic head pressure of the upper reservoir any expansion that occors in the air chamber will result in water moving up into the upper reservoir. Of course water only flow through the turbine, isnt that the whole point of the system is to generate power from the water moving down through the turbine. This system allows you to move water upwards just like a pumped storage system except that you get to lift the water for nothing.   

         

       //That means the tank above the piston has to be large enough to hold the entire contents of the reservoir.//   

       This makes no sense. The piston stroke only needs to be large enough to move a volume of water equivilent to the maximum expansion of the air in the chamber.   

       //My machine is intended to pass nearly the whole upper reservoir down to the bottom in the morning and pump it back up at night. Because of all of the tanks, I need only a small volume in the pumping tanks, not the whole reservoir.//   

       So your machine just moves water around in a circle. Isnt that kind of a waste. Also why is that different than what im doing except that I dont start with a new batch of air every day.   

       //Your machine has the biggest piston every imagined (needed in order to make it reversible)//   

       Your idea of what constitutes a piston is obviously very narrow. Floating covers are used extensively in the petroleum and waste water industries the world over. This is nothing more than a seperation of the water and the air. It could as easily be a diaphram(like a big ballon)   

       This is well established technology and the forces involved are actually very small when it comes to the piston itself.   

       As to that the piston is needed to make the flow reversible this is also incorrect. The water outlet and inlet could be located at the bottom of the air chamber and no piston would be needed at all to make the machine work, however air would tend to dissolve into and out of the water and so regular adjustments would need to be made to the air volume to maintain long term proper operation(a layer of something like a silcone oil or even a plastic membrane would also work to eliminate the effect of diffusion)   

       //See the Wikipedia article on Ericsson's ship which had huge pistons but failed after one day of operation.//   

       Again this does not apply in any way to the machine in question. Massive pistons in an engine or powerplant experience massive force loadings do to the rapid direction changes encountered in any high speed reciprocating assembly which this has nothing in common with.   

       //Because the piston rises and falls the entire distance, the last bit of water lowered is not lowered all the way down, but only a small amount. The simplified system is not at all equivalent.//   

       I dont even know what you are trying to say here. The systems are equivilent and perform the same functin using the exact same principle of operation. It only matters how much water is displaced by the expansion of air in the chamber. Thats it nothing else really matters.
—jhomrighaus, Apr 20 2006
   

       [jhomrighaus] : Thanks for the continued discussion. I find it very helpful.   

       Your machine operates on a completely different principle than mine. You never use the gravitational potential energy of the upper reservoir to compress air. You compress it initially (with an electric air compressor, for example) and then just let it expand and contract due to the temperature. So you can pump the volume of water equal to the volume change of the high pressure gas, and no more.   

       My machine pumps vastly more water than the volume of the high pressure gas. In fact, it pumps the volume which that gas expands to when at atmospheric pressure.   

       A full sized machine would wrap the hydrostatic column (a column of air at max pressure goes back up to the top of another column of water so the hydrostatic pressure at the bottom of the second column is twice that at the bottom of the first). For instance, if the upper reservoir is 1000 feet above the lower, then a full sized machine with 300 atm high pressure tanks would wrap the column 10 times, in order to get a hydrostatic head of 10,000 feet on the high pressure gas. Now air at 300 atm expands 300 times to reach atmospheric pressure. So I can pump 300 times the volume of the high pressure reservoir.   

       [jhomrighaus]: Reply to the later comment. I didn't understand your pumping system because mine is completely different. It lets water downward in the morning and pumps it upward in the afternoon. Your system with check valves is not capable of letting water down from the upper reservoir. You only let it down from the lower reservoir into the pressure chamber.   

       /The piston stroke only needs to be large enough to move a volume of water equivilent to the maximum expansion of the air in the chamber./ Right. And that is all it can pump. But it is required to be able to pump the entire reservoir. So you need a chamber 10 times the size of the reservoir.   

       /So your machine just moves water around in a circle/ Well, it pumps 10% more up than it lets down. That 10% comes down via the turbine. But remember the volume of my high pressure tank is only 1/300 of the volume of the reservoir, while yours is 10 times the volume. You raise 10% of the volume of the tank and let the turbine bring it down, while I pump 30 times the volume of the tank as extra water for the turbine, in addition to 270 times the volume as the amount which came down in the morning (since the max pressure is 300 atm, I can't let it get to 300 atm while cold).   

       And note that the volume of the many individual hydrostatic pressure tanks which are at all intermediate pressures from 1 atm to 300 atm (at least 300 of those tanks at 1 atm intervals, e.g.) -- each of those tanks is very much smaller than the bank of high pressure tanks at the bottom. On the diagram I showed the bottom tank as the same size, but it is much bigger.
—Archimerged, Apr 20 2006
   

       Im not sure I follow but I think you have a fundemental flaw somewhere in there. You can never lift more water than the equivilent amount of energy as was injected in to begin with. You are using energy from the potential of the water and heat to move water up. You may be able to move more water but when you do the math for water in and water you still have to equal out in the end. You move all the water you dropped plus about 10% more. I move about the same amount as your 10% more. unless Im really missing something you are violating the consevation of energy law(energy in = energy out)   

       How does your balance come out when you net out the water used to increase the presure of the air.   

       How does you system work when utilizing one or two tanks(The principle of operation should work on a small scale same as a large scale, with few tanks or many tanks.   

       As to using Hydrostatic Head Presure to charge system you could absolutely use the hydrostatic presure to do the job, its just easier and faster to do it one time(or use the turbine generated electrical power)
—jhomrighaus, Apr 20 2006
   

       I will describe the energy and entropy flow:   

       Gravitational potential energy in water in the upper tank of each pair of tanks of the pumping chain does work on air in the pumping manifold, pressurizing it to slightly above atmospheric pressure. There are hundreds of these pumping tanks, leading from the upper reservoir down to the lower reservoir. All of them are pushing on the same air to increase its pressure a little. This bias tank air does work on the water in the column which leads down to one tank of each pair of hydrostatic tanks. The water in the columns does work on air in the hydrostatic tanks, compressing it.   

       Note this well: the energy which came from the gravitational potential of the water falling from one pumping tank to its next lower neighbor is now ALL dissipated as heat. Interestingly, there was no entropy added to the working gas as it was compressed (no heat flow) but as it gets cooled back down to 275K, the amount of entropy removed is w/275K, where w is the work done on the air. I never quite understood this until yesterday (and an idiot is someone who doesn't know today what you learned yesterday...)   

       So, after all of the water in the upper reservoir is lowered, all of the gravitational potential energy has gone up the chimney above the heat sink.   

       And w/275K of entropy has gone up that chimney as well. The disorder that was in the 1 atm air we just collected and compressed has been returned to the atmosphere and is not in the small compressed air tanks.   

       Now we wait until the atmosphere warms up to 305K. The high pressure air also warms up, but it turns out that that is a very minor part of the process. High pressure air is very good at extracting heat from the surroundings as it expands. When high pressure air does work on water in the tanks, it cools. The more it cools, the faster heat from the copper heat pipes will warm the air, and the faster propane inside will condense on the copper to warm the copper back up, and then the propane flows back down to the big copper "heat sink" with lots of fins (actually heat source) and with 305K air flowing by. (Actually, say 310K air). The air gets cooled by the 305K heat source, and falls into the lower chimney creating a draft which brings more 310K air in. So we are extracting ALL of the energy we need to raise 110% of the water that was lowered in the morning from the atmosphere.   

       Well, PV and nRT are both energy. The PV product of the gas increased when its temperature increased, but essentially none of that energy was removed during the time the water was raised. When we release the gas back to the atmosphere, it is still at 305K, so the PV product is the same as it was when the pressure was 300 times more and the volume was 300 times less.   

       Compressed air doesn't contain a lot of energy. It is just good at extracting energy from the environment because when it expands, it cools, and the more it cools, the faster energy flows into the air. And I just learned this yesterday. By replying to questions on this forum.   

       Note that these hydrostatic tanks are all at different levels. For 1000 feet drop and 10 feet of water head between each pair, that would be 100 pairs, 200 tanks. (And I need 10 1000 foot chains of these, so there are 2000 of these hydrostatic tanks). Each has a different pressure, from 5 feet of water to 10,000 feet of water in steps of 5 feet. (But during any 1/2 step, half of the tanks are full of water.)   

       And one other thing. You can burn biomass underneath the big copper "heat source" with heat pipes leading up into all of the hydrostatic tanks, and get somewhat hotter temperatures. If you get only 330K, it makes a big improvement. In any case, it makes it much more likely that you get 305K after all of the temperature drops across the interfaces.
—Archimerged, Apr 20 2006
   

       By wasting away what heat is generated you are bleeding off all the energy you have gained. This will greatly decrese the efficiency of the system.   

       Also you are assuming that each step in the system will impart a sequential whole amount to the total presure. You are thinking that the water will continue to do additional wrk each time it falls downward. however at each step of the way the head pressure is limited to the water column in the tank plus the atmispheric presure within the tank. The moment you open the tank pair to the next pair you will equalize the presure over the whole set. Then you vent the other half of the system as the water contiues down the cascade.   

       Your comments on compressed air not holding a lot of energy are very wrong. The potential energy stored in a compressed gas is enourmous for the exact reason that it expands so much when it returns to atmospheric pressure. Your system will not return to its full volume while doing work on the water.(in the scenario of a gas bottle expolding all energy is expended in one shot so big expansion 0 latent presure. When lifting water up you will have a latent pressure equal to the hydrostatic head pressure. of the water column being lifted otherwise water would flow downward to equalize4 the pressure. The only energy you get for lifting comes from the expansion of the enclosed air. The energy gained by dropping the water is netted back out to lifting an exactly equivilent amount of water back to the top(assuming a lossless system) You cannot get around this, conservation of energy laws prohibit this from happening.
—jhomrighaus, Apr 20 2006
   

       [jhomrihaus], that was the point I was making when i said:
"2. The initial heat of compression will be lost."		   

       This lost heat must be replaced, at least, in order to have a chance to use the excess energy to lift the water again.   

       Perhaps I'm too subtle.   

       The drawing that you sketched would be better if the bottom tank was 'upside down' so that the water is resting in the bottom with no piston. I think you may have already mentioned that somewhere up there ^.
—Ling, Apr 20 2006
   

       [jhomrighaus] I do hope you will take the time to think before responding. I do know a little about thermodynamics. I may not have much practical experinece in engineering, but I understand the principles. I am very aware of the laws of thermodynamics, including the conservation of energy, and that irreversible processes increase entropy. By gathering heat energy from the atmosphere during the expansion of the compressed gas, I am not violating conservation of energy.   

       /The only energy you get for lifting comes from the expansion of the enclosed air./   

       No, as the enclosed air expands, it does work on the water it is pushing downward. Molecules of the air collide with the water, and the average kinetic energy of the air decreases. The PV product decreases because nRT decreases because T decreases. But my enclosed air is not doing adiabatic expansion. It is doing isothermal expansion. This happens because there are heat pipes inside the high pressure tank leading downward to a heat source. Propane inside the heat pipe condenses on the cool surface inside the tank (cooled because expansion of the air reduced its temperature because the air delivered energy to the water it pushed down). That liquid propane flows downward to the heat source, where it absorbs energy and evaporates. When it collides with the cool wall again, it tends to stick (more molecules stick than come off) and it delivers its energy to the copper wall, which delivers the energy to the expanding air inside the tank which keeps that air at constant temperature.   

       Essentially all of the energy used in raising the water comes from the 210K atmospheric pressure air which is cooled a few K before it falls down the lower chimney and is replaced by more 210K air from the cubic miles of nearby atmosphere. (And it might need to be 220K or whatever to get the heat to flow fast enough, I haven't calculated this to find out how much energy I can run through a machine of a given size). The only time the gas itself could contribute energy would be if its temperature decreased. After all, the temperature is a direct measure of the energy content of the air.   

       /Your comments on compressed air not holding a lot of energy are very wrong/   

       Have you done the calculations? The energy present in compressed air is equal to its pressure volume product. So long as its temperature is constant, this energy remains the same. The energy in compressed air is kinetic, not potential. It comes purely from the motion of the molecules. The temperature measures 1/2 the average kinetic energy of one degree of freedom for each molecule of the gas. When you cool the gas, that removes energy from all of the degrees of freedom of the molecules. The temperature is directly proportional to this: kT is 1/2 the kinetic energy. Boltzmann's constant convert kelvins to joules.   

       When compressed air expands, it cools. If it can't get any heat from the surroundings, its PV product decreases. The energy an exploding cylinder delivers comes from the PV product. You get a lot more work out of some gas if you keep it at a constant temperature as it expands than if you let it cool. This is clearly shown on the PV diagram for the process. The total work is the area under the PV curve. If you add heat to keep the temperature constant, the PV curve is a hyperbola. If you don't, the curve is an adiabat, which has a much steeper downward slope, and hence the area under it is smaller.   

       When a Stirling engine compresses air at constant temperature, it keeps removing energy as heat and disipating the heat. If it didn't, it would have to use more work to compress the air. And when the engine lets the air expand, it keeps adding energy as heat. Otherwise, the temperature of the gas would decrease and the pressure would decrease faster than the volume increases. PV = nRT.   

       [Ling] The lost heat is replaced. This is the whole idea behind the Stirling cycle. If we didn't remove the heat during compression, a lot more work would be required to compress the gas and it would get very hot. The Stirling cycle relies on being able to heat up the compressed air after compressing it, not before. The net work produced by the engine (the amount produced during the power stroke minus the amount removed from the flywheel during the compression stroke) is equal to the area inside the closed curve on the P-V chart. The compression process is the hyperbola on the bottom of the curve. If you replace this hyperbola with an adiabat, the area is inside the curve greatly reduced.
—Archimerged, Apr 20 2006
   

       Your manipulation of the gas laws while certianly impressive do not make the system work better. You are inputing energy which is what drives the expansion of the gas. whether putting a little energy into a lot of gas or a lot of energy into a little gas you still are inputing enrgy. Your compression using the hydrostatic pressure is a fixed quantity. you cannot get more out of it than that.   

       The amount of energy you can absorb from the air or from your alternate heat source is the only variable in the equation. Extract more energy move more water. You can do this slowly over time or all at once but for the same energy in you will get same work out wheter is your machine, my machine, or a big raincloud in the sky. If all are lossless systems then all will move same amount of water.   

       As to Potential and Kinetic Energy water in a high reservoir has high potential energy and low kinetic energy. Compresses gas is acting on the cylinder with kinetic energy but once gas is cool it does not continue to cool or heat up it reaches equalibrium. There is now potential energy stored within the compressed volume of gas.   

       You absolutely can get more energy out of an expanding gas if you keep it at the same temperature because you are imputing energy throughout the expansion which, surprise surprise, means the gas does more work. This is not some special new thing it is the potential energy of the compressed gas plus the heat added to maintain the temperature of the expanding gas.   

       See link for info on potential energy of compressed gases.
—jhomrighaus, Apr 20 2006
   

       [ling]   

       The arrangement you propose is marginally more efficient in that it does not need to over come the Inertia of the "piston" or its friction. Other than that(which is very small in overall system) they would work exactly the same way with exactly the same results. The big advantage of either arrangement is that you do not need to deal with heat from compression of the air after the first charge of the system. From there it simply exctracts heat from the ambient air(or allternative heat source) and returns it to ambient air. Both of which can be highly optimized for maximum gain in temperature.   

       [Arch]   

       A stirling engine uses a fixed volume of air in its operation the arrangement of the pistons helps to optimize the heat transfer process but it is still the same fundemental process no matter how you manipulate it.
—jhomrighaus, Apr 20 2006
   

       [jhomrighaus] Thank you for pointing out an error in Wikipedia. I have corrected it. The explanation there is now applied to compression of a solid, not of a gas. When you compress a solid, you do store energy as potential energy. Not so for a gas. A new explanation for compression of a gas is given, including an explanation that the kinetic energy of a gas molecule does get stored as potential energy during the short time it is stopped during a head-on collision with the walls of the container.   

       //The amount of energy you can absorb from the air or from your alternate heat source is the only variable in the equation.// Yes. I can absorb more heat because I allow the gas to expand more. The more the gas expands, the more heat I can absorb, because whenever the gas expands, its temperature drops, and the rate of heat flow goes as the temperature difference. I allow the gas to expand 300 fold. You allow it to expand 1.1 fold.   

       A Stirling engine is different from a Stirling cycle. Anyway, my machine does always use about the same amount of gas, it just saves the expense of building a huge container by releasing it into the atmosphere.   

       The reason a Stirling _engine_ uses a constant amount of gas is that it needs to be sealed up carefully and it operates at many cycles per second. My machine operates one cycle per day. (Or it could do more than that if operated with a heat source such as burning biomass).
—Archimerged, Apr 20 2006
   

       Note: I started this before the previous one, but am working out exactly how the pressures work out in the pumping tanks vs. the expansion and compression tanks. I was using an approximation but am now figuring it out exactly. Getting closer to having it right. It still isn't exactly right but I will put further corrections below.   

       [jhomrighaus]/Also you are assuming that each step in the system will impart a sequential whole amount to the total presure. You are thinking that the water will continue to do additional wrk each time it falls downward/   

       You are absolutely correct. I am assuming that, because that's how the system works. It does not work like yours. It is different.   

       However, in writing this I have encountered an error, which is corrected below but not elsewhere. During compression, the pumping tanks have to be spaced with a larger head between each pair than the head between the hydrostatic tanks. During expansion, the opposite is the case. So I expect the best solution to this problem is to have separate sets of hydrostatic tanks: 1250 pairs of compression tanks 1 foot tall with 4 feet spacing from bottom to bottom of each tank, 5 feet from bottom of the lower tank to top of the upper, and 834 pairs of expansion tanks with 6 feet between each tank.   

       There is another reason for using separate sets of tanks: the compression tanks need connection by heat pipe to a heat sink above the tanks, while the expansion tanks need to be connected to a heat source below the tanks. Finally, given separate sets of tanks, compression and expansion could go on simultaneously if both heat source and heat sink are available simutaneously, as when it is cold out but heat is applied to the heat source by combustion of biomass.   

       (And by the way, pairs of both kinds of tanks go odd even for the first half step, and then even odd for the second.)   

       Also, the diagram is a little misleading, in that the pumping tanks are the wrong shape. They are wide and short, and do not overlap vertically. As shown, they are never filled higher than the bottom of the previous tank.   

       For 5 feet between pumping tanks and 4 feet head required when flow stops, the tank never fills higher than 1 foot from the bottom. Then when the upper tank is just empty, the lower tank surface is 4 feet below the upper tank surface. When pumping, to get all of the water out of the bottom tank and into the top tank (neglecting the piping which I guess must be well under 1% of the volume because we need 99% efficiency), the bias tank head will need to be 6 feet so that the surface of the water in the upper tank is 1 foot above the bottom, and the surface in the bottom tank is just below the bottom. If the bias tank pressure is a little too high, it won't do much work because it will just push the water lower into the pipe which is a small volume so not much mass is raised.   

       Let me explain how each step does impart an amount to the total pressure. Note that the air pressure in all of the even numbered pumping tanks is equal, and also in all of the odd numbered tanks. The spacing of the pumping tanks is 5 feet, and the water level in a tank is at 1 foot when full. The hydrostatic head between a pumping tank and its neighbor is 6 feet when pumping water up, and 4 feet when compressing air. There is no water anywhere in the pumping system under hydrostatic pressure exceeding 6 feet. (The water under high hydrostatic pressure is only in the compression and expansion systems, and that water never moves more than a few feet and always goes back and forth). Water flows downward from the upper tank into the lower tank, temporarily raising the pressure in the bias tank. This pressure is applied to the _top_ of the hydrostatic columns, which are sealed and not at atmospheric pressure. They are at the bias tank pressure which has just been increased from atmospheric by water in 200 tanks all falling 5 feet.   

       Just before the valves are opened, the bias tank is at atmospheric pressure (its level was just restored to the upper lake level while disconnected from the hydrostatic columns, and atmospheric pressure air was allowed into the space above the water). The upper tank of each pair is full of water up to a foot below the bottom of the previous tank and connected to the top of the bias tank, and the lower tank is full of air and connected to atmospheric pressure. All valves between pumping tanks are closed. Then, all 100 (or 99 if the top and bottom tanks are unpaired) valves between the upper and lower tanks are opened, and water starts to flow down out the bottom of the upper tanks, through the U, and up into the lower tanks. Atmospheric pressure air flows into the upper tanks to replace the water. Air from the lower tanks flows into the bias tank. The flow stops with the pressure about 5 feet. Then the valve connecting the bias tank water with the hydrostatic manifold is opened, and water starts to flow again. It stops when the pressure is 4 feet. The bias tank and manifold volume and the hydrostatic expansion/compression tank system is designed so that flow will stop when the surface of the water in the lower tank is 4 feet below the surface of the water in the pipe leading down from the upper tank. The pairs of tanks are all configured as U tube manometers, and the difference in water surface heights gives the difference in pressure between the bias tank and atmospheric. The height differences will all be the same, since the tanks are all connected either to the bias tank or open to the atmosphere. (The tanks shown in the diagram are too tall and would never be very full, but would work just the same).   

       The pressure in the compression hydrostatic tanks at the bottom of the column of water attached to the bias tank all increase by the difference between the bias tank and atmospheric. This is designed to be a tiny amount more than the spacing between the compression tanks. There is flow going on, and the bias tank pressure will start at 5 feet but decrease to 4 feet before the flow stops. When the flow stops, the valves connecting the biased tanks with the unbiased tanks are opened. The unbiased tanks contain water only. Water will flow up into the biased tanks pushing air up into the pipe through the valve and down into the unbiased tank until the water level in the unbiased tank is exactly 4 feet below the level in the biased tank. The amount of work involved in this step is minimal. No water flows up or down the columns. The water flow is simply between the two tanks, and gravity supplies the energy to move the water.   

       Any deviation from 4 feet (i.e., the spacing of the compression tanks) represents inefficiency that must not exceed about 2%. If it is not possible to make the bias tank pressure end very close to 4 feet, the machine will operate at a loss.   

       There are 100 pairs of pumping tanks running up 1000 feet to the upper reservoir. There are 1250 pairs of hydrostatic compression tanks at 4 foot heads, and 888 pairs of expansion tanks at 6 foot heads, in 10 chains running up 1000 feet, with an air column connecting the bottom of one chain with the top of the next. The pressure in this air column varies by at most 5 feet of water from its nominal value, which would be 1000 feet in the first column, 2000 feet in the second, etc.   

       So, there are 1250 pairs of compression tanks. These tanks have nominal pressures of 4 feet, 8 feet, 12 feet, all the way to 10,000 feet (300 atm, 4500 psi). The pressures will increase when the water falls and the pressure in the bias tank increases to about 4 feet above atmospheric. So the 4 foot tank will have a pressure of 8 feet, because there is 4 feet of water and 4 feet of pressure in the bias tank. The 8 foot tank is full of water. The 12 foot tank will have pressure of 16 feet, 12 of water and 4 of bias. The 16 foot tank is full of water. Etc. etc. up to the 9994 foot tank which will have 9994 feet of water and 4 feet of bias. After the flow stops and these levels are reached, valves are closed and the next half-step begins. The manifolds are swapped so that the hydrostatic column with atomospheric pressure above it is connected to the   

       The valves to the next lower tanks are all opened (i.e. 1250 or 1249 valves).
—Archimerged, Apr 20 2006
   

       Thermal cycles link.   

       Archimerged, you are trying for case 4? But if you are not careful with the design, you might get case 3, or worse, case 5.   

       jhomrighaus, about the piston idea: yes, you are right, and I would like to add that a piston needs to be relatively leak-proof. If the water is at the bottom, then there is no fear, unless someone switches off Earth's gravity.
—Ling, Apr 20 2006
   

       (I'm still editing the previous annotation which is not correct yet.)   

       [Ling], thanks for the diagram. My design is case 2. (BTW the curves should be hyperbolas, they represent PV = constant). When drawn to scale they look quite close together. The question is how slow it has to run. Most of the heat flow occurs along the hyperbolas. The amount of heat flow along the constant volume heating and cooling is minimal. This is because I compress to such extreme pressure of 10,000 ft water, 300 atm, 4500 psi. Because the large number of separate tanks each with its own collection of heat pipes and heat sink, the heat flow can be very high. It has to be. But remember that the budget for a gigawatt plant at $1000 per kw is one billion dollars. You can buy a lot of pipe for that.
—Archimerged, Apr 20 2006
   

       //BTW the curves should be hyperbolas, they represent PV = constant// Yeah, I can't seem to draw so well with a mouse...but in my defence (UK), it is only a sketch. BTW: In some cases, T is varying as well.   

       I suggested case 4, since your temperature variations are small (you mentioned 30 degrees), and your pressure swings are high (you mentioned 300 bar).
—Ling, Apr 20 2006
   

       Right, case 4. The pressure variation due to isochoric heating is 10% while the variation due to volume change is 30000%. In principle the temperature fluctuation can be held under a K, maybe requiring some automatic controls attached to the heat pipes to control heat flux. The delta T across the heat sink to the gas would have to be several K (even tens of K) to get decent heat flow, but it could be held pretty constant if the flow of refrigerant in the heat pipes is controlled.
—Archimerged, Apr 20 2006
   

       Quick note but when you open all your tanks to the bias tank they will not all put there little bit of pressure into the system. The System will equalize out over the entire system so that air will flow from the higher pressure tanks back into lower pressure tanks and the entire system will net out at a pressure at a lot less than the maximum possible pressure. Essentially the water will find its level in the system and that will be somewhere slightly above half the total height of the system.   

       This is very clear from your manometer example that though there will be a lot of moving around of water I do not believe it will do what you are thinking. one tank at 5 psi connected with another at 100 psi do not create a pressure of 105 PSI in a third tank. They equal something like 48psi overall.   

       See link for additional discussions of potential energy in compressed gasses by a PhD in the field. Whatever its mode of operation(kinetic, or chemical etc.) Potential energy is energy that is stored at a higher level than it would be when at rest. A compressed gas clearly rempresents stored energy that has the potential to do work(we use compressed gas to power machines in the lab)
—jhomrighaus, Apr 20 2006
   

       [Vernon] meets [pashute].
—RayfordSteele, Apr 20 2006
   

       [jhomrighaus]: Thanks again for continuing to read this stuff. I haven't communicated it well enough yet but we are getting there.   

       Regarding the link: yes, a non-ideal gas expanding into a vacuum gets a temperature change up or down due to release of energy from or storage of energy into intermolecular forces. An ideal gas expanding into a vacuum does not change in temperature. I have been talking about ideal gasses, and all expansion is against a force. And yes, if it turns out that effects of the working gas being non-ideal add up to more than 3% or 4% inefficiency, this machine is dead in the water. Or at least the design has to be corrected to account for those effects, or the delta T has to be increased.   

       //one tank at 5 psi connected with another at 100 psi do not create a pressure of 105 PSI in a third tank.//   

       Yes of course, if the tanks are connected by a gas-filled tube. That is an irreversible change. In the case I am talking about, I have one tank at 5 psi and one at 0 psi at the top of a column of water 100 psi (around 200 feet) high, with a U tube into a tank at 100 psi relative to atmospheric. The tank at 0 psi has negligible volume compared to the bias tank at 5 psi. A relatively tiny amount of air flows into the tank and the pressure ends up around 4.9 psi. Also some more water flows downward in the pumping tanks to restore the pressure to 4.99 psi. Now the hydrostatic tank is 104.99 psi relative to atmospheric.   

       A good question to ask about this design is "what is the effect of changes in barometric pressure?"   

       //Essentially the water will find its level in the system// True. But each pair of pumping tanks is a separate system of water. Note that every other valve is closed on the bottom manifold. So I have 100 separate water levels, each 10 feet apart from the others. They are connected only by air-filled tubes.   

       There are two other water levels in the system (not counting the folded hydrostatic columns used to get more than 1000 feet of heat). Those are connected to the even hydrostatic tanks and to the odd hydrostatic tanks. The top of those columns are alternately connected to atmospheric pressure or to the bias tank.   

       [RayfordSteele]: No doubt those two are famous debaters on this site. I like to argue, and it helps me keep focused on figuring out how and whether this system really works, and on finding errors in the design, and sources of inefficiency.
—Archimerged, Apr 20 2006
   

       /Essentially the water will find its level in the system// True. But each pair of pumping tanks is a separate system of water. Note that every other valve is closed on the bottom manifold. So I have 100 separate water levels, each 10 feet apart from the others. They are connected only by air-filled tubes.   

       If all the tanks are connected by a manifold then the air in the lowest tank(highest Pressure) and the air in the highest tank(lowest pressure) have a connection to one another. So assuming an even distribution of tanks the pressure accross the entire manifold will equalize even if the water doesnt move at all. If the water side is connected then water will flow to the lower part of the system and air to the top but again pressure will equalize accross the entire system and you will lose your pressure gradients. at that point the water will simply move downward. with no net change in pressure through the system.
—jhomrighaus, Apr 20 2006
   

       How are you stepping down from one tank to the next. if a lower tank is higher presure you can never get something at lower pressure to go into it.   

       Also when you are dropping the water in the many tanks to build pressure all the tanks will be at the same pressure if there is any connection between them.   

       Another way to look at it is if you have a 2 tank system(one high one low) and you let the water do its thing and increae the pressure you will end up with a certian system pressure at that time. If you lower the entire system 10 ft and repeat the experiment you will get the same result.
—jhomrighaus, Apr 20 2006
   

       // After storing the energy in the surroundings, I warm the gas to temperature H (picking up part of the energy gain), and finally I let the gas expand slowly while more energy returns from the warmer surroundings than dissipated earlier when the gas was being compressed. As the gas expands, it does work on water in the pumping tanks, which ends up in the upper reservoir.//   

       This doesnt make sense. if you are holding the system isothermal and this is an ideal gas and there a no thermal tranfer penalties then the gas would reabsorb exactly the same amount of energy that is released while being compressed. This being the case the ONLY way you can achieve additional work is through an increase in temperature within the volume of gas.   

       So in essesnce you have gained "mechanical Potential Energy"(we will use this term to quantify the ability of the compressed gas to do work)equal to the amount of energy imparted by the hydrostatic head presure of the system(height of water column)less the amount of heat released up the chimney(irregardless of the mode of heaat transport tubes or air or propane) When you switch and start to pump upwards you start with the potential that was stored in the compressed gas. As the gas expands you recover exactly the same amount of energy as was released to net out at the end.   

       I think the major flaw here is that this scenario only works when the gas expands while doing no work.(ie a 300atm air tank is connected to a water column 1000 feet high nothing would happen, if the same 300atm tank was connected to a 500 foot column then water would be pumped upwards at a very high rate until the hydrostatic pressure of the water column matches the lower pressure but higher volume of air in the tank.   

       //I allow the gas to expand 300 fold. You allow it to expand 1.1 fold.//   

       This is correct in one way but it doesnt acount for the fact that you compressed your gas to 1/300ths of its original volume. My gas is already compressed and is not compressed any further. If we both put an equivilent amount of energy into our systems they would both expand in volume proportionaly to thier volume.(about 10% in our model) So for your system and my system with exactly the same static air volume(yours at atmospheric mine at 300atm) and an input of 30K my air would expand at Static Volume x 1.1   

       yours would expand at   

       Static Volume/300 x 1.1   

       Now your machine replaces the energy lost in compression so for my system the energy is   

       30K   

       for your system it is   

       Replaced heat + 30K   

       which returns both systems to exactly the same volume.   

       They both did the same amount of work on the water that was lifted(equal to the .1) but you moved a lot more energy to achieve the same goal, in a lossless system this would not matter. the systems would be effectivly the same for performance.   

       This is kind of like setting one of those old fashioned snap mouse traps next to a 1000 pound rube goldberg type machine. They both do the same thing but one is much more complicated than the other.
—jhomrighaus, Apr 20 2006
   

       [jhomrighaus]//This doesnt make sense. if you are holding the system isothermal and this is an ideal gas and there a no thermal tranfer penalties then the gas would reabsorb exactly the same amount of energy that is released while being compressed.//   

       I am not holding the system isothermal for the entire cycle. You need to reexamine your argument after removing that assumption. The net work done (work done raising water minus work done earlier by the lowered water) is equal to the area between the two hyperbolas PV = nR(275K) and PV = nR(305K). That area increases with increasing change in presure. The compressed gas absorbs more energy from the 305K surroundings than it gave up to the 275K surroundings. This is just how the Stirling _cycle_ (not engine) works. The Stirling engine generally works over a relatively narrow pressure range, and the presence of a "regenerator" is very important because the gas is being heated to Thot and cooled back to Tcold many times per second. My machine does this once per day, and the energy invested there is minor because delta T is a lot less than delta P.   

       It is extremely important that the machine run between a wide pressure difference. Ok, let's evaluate the integral and find the area. Integral from 1 atm to 300 atm of V(P)dP = nRT Integral 1 to 300 dP/P = nRT ln(300/1). The area within the cycle is (305K-275K) n R ln(300) = 30K n R ln(300). Looking at values, ln(300) = 5.7, ln(150) = 5.0 -- of course, the log curve starts ascending rapidly because it always has slope 1/x at (x,ln(x)) -- so maybe the investment in the last half of the pressure is ill advised. After all, the last half of the pressure only cuts the cost of the storage tanks in half while doubling the cost of the compression apparatus. If we already need the larger compression apparatus because of heat flow reasons, we may as well operate it at higher pressure and save on the tanks. Or if we decide to market compressed air, we might want to be able to produce it at 300 atm. But if we don't need added heat flow and aren't in the air business, we should pay more for tanks and less for compression. The design point depends on the relative costs of this equipment, and I don't know that yet.   

       Anyway, the bottom line is the work output by this machine is (Thot - Tcold) nR ln(Pmax/Pmin). More gas means more work. Higher temperature means more work. Higher pressure means less money spent on the storage tank but more money spent on the compression apparatus. Pmin has to be one atmosphere in order to avoid buying a huge storage tank for gas at low pressure other than one atmosphere.
—Archimerged, Apr 20 2006
   

       See Link for question and simplified diagram.   

       Where does energy come from to move rest of water used for compression.
—jhomrighaus, Apr 20 2006
   

       [jhomrighaus]: A very good question. If you understand the answer, you will see how the machine works. As arranged, there is too much hydrostatic pressure for the gas to expand. But if the arrangement is changed in a carefully designed system, then when a valve is opened, some work will be done and some water will flow until the pressure balances again, at a slightly lower pressure _and temperature_. That is the crucial point.   

       When a gas expands against an exactly balancing force, i.e., very slightly less hydrostatic pressure than gas pressure, and the pipes are carefully arranged so as the gas expands, the hydrostatic pressure increases, _not_ decreases (since otherwise the expansion would run away and cause a water fountain but lots of water would be left behind in the tank), then the temperature goes down. The gas did work and the energy for the work came from the temperature decrease. Agreed?   

       Now, when the temperature of a gas is lower than the surroundings, energy will flow out of the surroundings and into the gas. Still with me?   

       Assuming you agree, then you see where the energy to raise all of that water is going to come from. It comes from the surroundings. It is necessary to repeatedly decrease the hydrostatic pressure applied to the gas to match the gas pressure, such that as the gas expands, the hydrostatic pressure will increase so the flow will stop by itself. This can be done without expending significant energy, if the piping is arranged properly, by opening a valve and waiting for the flow to stop. The temperature goes down and heat flows in and a little more water flows but it stops pretty soon. Then a few other valves are adjusted, the crucial valve is opened, and the same thing happens again in a slightly different place with slightly lower pressure.   

       All of those "extra" tanks and valves are needed in order to arrange for the energy from the surroundings to flow into the gas and raise the water. This process is shown on the PV diagram (P is vertical axis, V is horizontal axis) as the two long hyperbolic curves which plot PV=nR(275K) and PV=nR(305K). The process of lowering all of the water is represented by the lower hyperbola, running from the lower right hand corner point P = 1 atm, V = nR(275K) / 1 atm to P = 300 atm, V = nR(305K) / 300 atm, in the upper left but below the next point. The process which raises the little bit of water shown in your diagram is represented on the PV diagram as the (very short) vertical line running up from the point with P = 300 atm, V = nR(275K) / 300 atm, to the highest point in the upper left corner with P = 300 atm, V = nR(305K) / 300 atm. The process of raising the remainder of the water and the extra water is represented by the upper hyperbola running from upper left to lower right (but above the lowest point). The process of cooling the gas is represented by the short vertical line at the right side running down to the lower right corner, in which P and T decrease and V is constant. The net work done by a complete cycle is the area inside the closed curve formed by the three curves mentioned, and the fourth which runs from P = 1 atm, V = nR(305K) / 1 atm to P = 1 atm, V = nR(275K). The area between the X axis and the lower hyperbola is the work extracted from the water and then redone to the water to pump up the same amount as was dropped. The area between the two hyperbolas is the work done to raise the extra water. n is the number of moles of air in the tank, and R is the gas constant. This area is equal to nR(30K) ln(300 atm/1 atm).
—Archimerged, Apr 20 2006
   

       every time you extract a small amount of the gas to lift a small amount of water you reduce the overall volume of gas available to do work on the next volume of water, so for the first few cycles the system will probably raise the water but then the pressure in the reservoir will fall and the volume of water that can be moved by each succesive cycle will decrease till eventually around the 110% of water dropped having been lifted you will run out of steam and the process will stop.   

       Each time the system move up it is losing some of its pressure. Half to the new tank pair and half to the old tank pair through many cycles you will eventually run out of air of sufficient pressure to cycle the system.
—jhomrighaus, Apr 21 2006
   

       In the end sitting in a big tank(ala my design) or broken up into little tiny bits you are ultimately limited by the amount of air compressed as to the total work you can do. My machine does less work per cycle but has a large amount of stored mechanical energy so it net energy never falls below a certian point. Your system starts at 0 cylces up to the same maximum potential then cycles back to 0 again then starts over. Your machine does amount of work X my machine does amount of work X - stored potential energy. At the end of the day Archmachine Max energy =jhom machine max energy .   

       You are limited by the hydrostatic head presure and the volume of gas under compression.   

       If I dont miss my guess your system would only compress a small volume of gas which means you are limited to about 10% of the total gas volume of excess water moved. If the compressed volume of gas for both machines was equivilent then both would move the same amount of water up gradient, all at once or a little bit at a time. You can not get any more expansion than I can which means you cannot do any more net work than I can.   

       Have you sat down and calculated the pressures and volume for each step in your cycle? i think you will find that in the end they net out to the same amount of water moved in the end.
—jhomrighaus, Apr 21 2006
   

       [jhomrighaus], First, note that your large tank system does not work. You would have to divide it up into many small tanks each separated by check valves. The pressure head leading up the top reservoir is so high that the pressure inside the tank will never be able to open the check valve. The additional temperature will increase the pressure, but that is not enough to overcome the additional hydrostatic head of the upper reservoir vs. the lower reservoir.   

       //every time you extract a small amount of the gas to lift a small amount of water you reduce the overall volume of gas available to do work on the next volume of water// That is not how the system works. The hydrostatic tanks stay at constant pressure. The amount of air in the tanks is reduced, but the pressure is determined by the fixed height of the water above the tank. The work done on the next volume of water comes from the surroundings, after the gas cools and heat flows in.   

       [[Edited to apologize for the following, which I'm not going to delete because it has been here for some time. There is a problem with hydrostatic pressures in the single tank design, however. Anyway, I do plan to build a model one of these days, and experiment is the proof of the pudding.]]   

       I doubt that you are going to believe this until you see one working. Then you might give up the preconceptions which keep you from figuring it out. I know you have spent a lot of time trying to figure it out, and I am sorry that you are not successful. Your machine doesn't work at all. It is not equivalent to mine.
—Archimerged, Apr 21 2006
   

       Votes to date: 0 for, 7 against.   

       As you were.
—Texticle, Apr 21 2006
   

       //I believe it is possible to operate a steam engine at lower temperatures than that.//   

       It is because you are dealing with a phase change of water to steam. The amount of energy that is input however is much higher.
—jhomrighaus, Apr 21 2006
   

       <aside> jhomrighaus, about your "intermolecular potential Energy" link:
This link mentions that in an expanding gas the cooling effect is a consequence of the molecules of gas attracting each other, and reducing the kinetic energy. I hope I summarised it correctly. I have difficulty with accepting that gas molecules "clump together" when the gas expands.
My conceptual model of the world thinks about it this way: as the gas expands, the average distance between the molecules *increases*, and the average kinetic energy in the same volume decreases. In my conceptual model, that's the same as reducing temperature.
I may be completely wrong, but I am always wary when someone uses "PhD" as a symbol for universal expertise, when perhaps the holder has studied a finite list of subjects.
—Ling, Apr 21 2006
   

       you may be right my machine may not work the way I thought. Build your model and show us that yours will lift more water than it uses.
—jhomrighaus, Apr 21 2006
   

       [ling] the point was more that upon compression the Kinetic energy level of the gas remains the same however the "potential" energy of the gas has been increased. This is the definition of potential energy, whether chemical, gravitational or pressure they are all forms of Potential energy.   

       I agree that I found the explination a little odd but I suppose it is probably about as accurate as any other description just a different point of view.
—jhomrighaus, Apr 21 2006
   

       //upon compression the Kinetic energy level of the gas remains the same //   

       In my mental model of the world: Yes, the total Kinetic energy remains the same(*), if heat is not lost. The Kinetic energy per unit volume increases. That is a higher temperature.
   

       (*) I have also imagined that the very act of pressing gas molecules together is equivalent to accelerating them i.e the Total Kinetic energy will increase. But this effect might be small, since I can accelerate air molecules quite easily with,say, a piece of paper.   

       Anyway, whatever, carry on...
—Ling, Apr 21 2006
   

       [Texticle], I fear this reflects a little on the quality of the analysis on halfbakery. This idea differs little from the one which got only positive votes. But I didn't post it here to get votes and I really don't care (much) about the votes. I posted it to get discussion, to light a fire under myself to get the work done and the problems figured out. I suspect some people voted against the presentation, or against the long and sometimes loud debate, or just because I put a note asking people to abstain if they don't understand it. After the defects are fixed, I'm going to post a new separate idea and see what happens next time. (Rather than deleting lots of stuff on this page. Besides, I want a slightly different premise so the new idea will be different as well as corrected).   

       [Ling], Re: Joule Thompson effect. The experimental evidence is that a real gas expanding into a vacuum can increase or decrease in temperature. An ideal gas expanding into a vacuum has zero temperature change. No work was done, there was no opposing force. In the real gas, there are forces between the molecules beside the hard sphere model of the ideal gas. Those forces can be attractive or repulsive.   

       Regarding conceptual model of temperature: There is no doubt that temperature of a gas is a measure of the average kinetic energy of a degree of freedom of motion of the molecules of a gas. The differences in heat capacity between the noble gasses and the diatomic gasses show this clearly. As a rule of thumb, kT is energy. k is joules/kelvin, T is kelvins. In fact, kT is 1/2 the average potential energy of a degree of freedom of "the average" molecule of a gas.   

       A piston in a closed cylinder feels a lot like a spring and so it seems reasonable to call the energy "potential energy stored in the gas." But the careful experiments done long ago show that most of the energy just isn't in the gas, and the energy that is "in" the gas is kinetic. When you get to really sensitive experiments that can tell the difference between an ideal gas and a real gas, there are potential energy effects reflecting energy storage in the position of objects: work, force x distance was expended pushing the objects together and they repel each other. Or there are attractive effects (electrostatic perhaps) between objects and work was expended pulling them apart.   

       See the link for some discussion on this. Feel free to edit the page or leave comments on the talk page. If you don't create an account your IP will be displayed.   

       [jhomrighaus] I am sorry I lost patience earlier. The diagram I posted is not very good and needs revision. The descriptions are scattered and need to be consolidated and revised.   

       I'm doing that on the "wikia" link. (btw anyone can edit there. It has google ads, which are annoying, but I don't have to pay for hosting. I was looking around for how halfbakery is supported and it seems to have no visible means of support... But you can't post TeX formulas here and you can on wikia.com). Then when I have a clear set of diagrams and a clear explanation, I will start a new idea here. Meanwhile, maybe you are still willing to read explanations? And I see a comment crossing during editing. Yes, I will be building a model.   

       //It is because you are dealing with a phase change of water to steam. The amount of energy that is input however is much higher.//   

       Let's just leave steam out of this... However I think you did acknowledge that the toy Stirling engines work off very little delta T.   

       //If all the tanks are connected by a manifold then the air in the lowest tank(highest Pressure) and the air in the highest tank(lowest pressure) have a connection to one another. So assuming an even distribution of tanks the pressure accross the entire manifold will equalize even if the water doesnt move at all.//   

       Every other hydrostatic tank is connected to one manifold connected as a U tube to the bottom of the tank. The even numbered tanks to one manifold and the odd numbered tanks to the other manifold. I've got to get a better diagram...   

       The pumping tanks are connected upside down: the manifolds are on the top and the valved connections between neighboring tanks on the bottom.   

       // If the water side is connected then water will flow to the lower part of the system and air to the top but again pressure will equalize accross the entire system and you will lose your pressure gradients. at that point the water will simply move downward. with no net change in pressure through the system.//   

       I suspect you missed the fact that there are two manifolds connecting every other tank? And the hydrostatic manifolds come in from the bottom while the pumping manifolds come in from the top? I really need a series of diagrams showing the water levels as the machine works. Guess I better get busy on that...   

       //This is the definition of potential energy, whether chemical, gravitational or pressure they are all forms of Potential energy.// There are different levels of abstraction when discussing energy storage. If you look at a sealed piston as a black box, yes it stores potential energy just like a spring. Especially if it is thermally isolated from the world.   

       But a piston which is attached to an infinite temperature reservoir with infinitely good thermal contact moves the energy into the temperature reservoir. The ideal gas inside the piston stays at a constant level of internal energy defined by nRT. When you look at an ideal gas, with billiard balls bouncing around having elastic collisions with container walls and each other, the only potential energy involved is seen during collisions when the balls are accelerating each other. (And if the balls are heavy, there would be gravitational potential energy keeping them in the bottom of the container, sort of like a column of gas miles high).
—Archimerged, Apr 21 2006
   

       [Ling]// //////upon compression the Kinetic energy level of the gas remains the same ////// In my mental model of the world: Yes, the total Kinetic energy remains the same(*), if heat is not lost. The Kinetic energy per unit volume increases. That is a higher temperature.//   

       You must state whether the compression is adiabatic or isothermal.   

       When heat is _not_ lost the total kinetic energy definitely changes. How could it not? Work has been done on the gas molecules, speeding them up. They were exerting force against a piston (so the kinetic energy of that individual molecule was temporarily stored as potential in an elastic interaction with the piston) while the piston was moving, and the piston accelerated the molecule while it was "stopped" with respect to the piston. Then the elastic collision finished, and the molecule was moving faster.   

       So upon adiabatic compression of an ideal gas, the total kinetic energy increases. Upon isothermal compression of an ideal gas, the total kinetic energy (= PV = nRT) remains the same.   

       Do you guys agree with this, or what? It seems like you do not. By the way, this is not a matter of opinion, at least in the rest frame of the gas and its container.
—Archimerged, Apr 21 2006
   

       //When heat is _not_ lost the total kinetic energy definitely changes. How could it not? Work has been done on the gas molecules, speeding them up. They were exerting force against a piston (so the kinetic energy of that individual molecule was temporarily stored as potential in an elastic interaction with the piston) while the piston was moving, and the piston accelerated the molecule while it was "stopped" with respect to the piston. Then the elastic collision finished, and the molecule was moving faster.//   

       Not that I want to argue, but I want to understand:   

       Then, if I moved the piston faster from A to B, the Kinetic energy stored in the gas would be more, since the molecules would bounce off the piston faster?   

       Or looking another way, if I accelerated the same air in an open environment, with the same piston, would I do the same work?   

       <thinks>There must be some force between molecules, and when they are packed closer together, they try to push apart: or is it that they bounce of each other more often, and hit the container walls more often, therefore giving more pressure (i.e. not only related to the speed or Kinetic energy of the molecule).   

       Sorry to divert your discussion. Back on track: Could you think about drawing the sequence of operation of one pair of tanks of each duty (water pump, air pump), in several steps, over several sketches? The sequence should show the 'as built', first charge, pumping steps, and reset cycle for the next pumping steps.
Then we could see the relationship between heights of adjacent tanks, the pressures, levels etc.
Then we could put several tank pairs together to get the bigger picture.		   

       I thought I understood the pumping, but now I'm not so sure. By the way, I haven't voted, but that's not important.
—Ling, Apr 21 2006
   

       [Ling]://Then, if I moved the piston faster from A to B, the Kinetic energy stored in the gas would be more, since the molecules would bounce off the piston faster?//   

       You would have to apply more force to make the piston move faster, so you would be doing more work, so naturally the energy added is more. The microscopic model is consistent with the accounting for work.   

       //Or looking another way, if I accelerated the same air in an open environment, with the same piston, would I do the same work?// In an open environment where the air can go around the piston, it will. The opposing force will be less and you will do less work. But yes, waving a paddle does warm the air slightly.   

       Good idea about the diagrams with just two tanks, and with water levels. I'm working on them...
—Archimerged, Apr 21 2006
   

       I suspect it might work, but it's very complicated so it's hard to be sure. Your other idea was easier to understand, maybe that's why it got more votes. Also the title is an interesting concept, (buy compressed air low, sell high).   

       About kinetic energy vs potential energy: calling what's stored in compressed air potential energy is inaccurate on a microscopic level, and can be confusing if heat transfer needs to be taken into account.   

       I think Gibb's free energy might be the correct term, but I haven't seen it used for compressed air, only in chemistry. It's not an additional form of energy but a way of saying how much is available to convert to work.   

       [Ling] //(*) I have also imagined that the very act of pressing gas molecules together is equivalent to accelerating them i.e the Total Kinetic energy will increase. But this effect might be small, since I can accelerate air molecules quite easily with,say, a piece of paper.   

       On a microscopic level, that's how I imagine compression heating working. Moving paper through air is more complicated since air currents get set up.   

       [archimerged] //So upon adiabatic compression of an ideal gas, the total kinetic energy increases.   

       Thermal motion is sometimes ignored for kinetic energy, and I wonder if that's what's causing disagreements, but otherwise, sure.   

       Here's how I'd describe the advantage of your system over various others proposed. It's doing "buy air low, sell high". It sells all the compressed air at the high point each day, unlike one of the alternatives which leaves the air partly compressed. I don't completely understand it so I'm not sure about the rest, but being optimistic: the volume compressed air is roughly the size of the compressed air tank, not smaller by the amount of compression, so the amount of potential energy from the water in exchange is also larger, and the water traded each day doesn't all fit in the machine at once. The large volume of water is less of a storage problem than having a pressure vessel that would fit both the air and the water.   

       Also, I saw someone refer to cycles. The device has cycles within cycles: each day it pumps air in and out. There's the daily cycle, and the cycle within the compression part that which pumps successive volumes of air in and/or raises pressure step by step of internal air.
—caspian, Apr 21 2006
   

       [ling] //Then, if I moved the piston faster from A to B, the Kinetic energy stored in the gas would be more, since the molecules would bounce off the piston faster?   

       If you moved the piston for the same length of time, yes. If you moved it for a shorter time so as to go the same distance, no, assuming the idealised formulas work. Fewer air molecules will be bouncing off the piston in the shorter time, and that balances out the increased amount of acceleration given to each one.
—caspian, Apr 21 2006
   

       Link for air pumping cycle. How does the air get back in?   

       Caspian, I find it difficult to believe that the velocity of the piston is increasing the Kinetic energy by increasing the speed of the molecules. I'm just getting confused, now.   

       Pressure is due to what? Number of bounces in a certain time? Temperature is due to what? Speed of bounces?   

       Anyway, I will rest easy knowing that whatever happens, my car engine will still start up and get me home tonight.   

       Regardless of what I may think.
—Ling, Apr 21 2006
   

       A whole lot of approximation follows:   

       [Ling] Pressure: number of bounces times momentum of each bounce (per area, per time). Momentum is molecule weight times speed (roughly), and momentum change is twice that.   

       Number of bounces (per area per time) is roughly number of molecules per volume, times speed.   

       So pressure is roughly speed squared, times molecule mass, times molecules per volume, or temperature times density.   

       Temperature: Energy of each degree of freedom. Roughly, mass (of a molecule) times square of speed (of a molecule), or square of momentum divided by mass.   

       All approximate and maybe a bit simplified too.
—caspian, Apr 21 2006
   

       ...and hence why PV/T is a constant. I like simple, very much.
—Ling, Apr 21 2006
   

       [Ling] I've started a design with some changes, which is why I haven't made diagrams.   

       The new design has faster moving water, moving as a simple harmonic oscillator in the pumping tanks and the compression tanks. To pump water, we excite the oscillator in the expansion tanks by letting air expand by removing some water from the top of the tall water column. This raises water, passes the equilibrium point, and then falls back down, re-compressing the air. We couple this oscillator to another in the pumping tanks. When the energy moves from the one to the other, the air has ended at the proper equilibrium pressure, different in each of the expansion tanks, and we have water flowing back and forth in the pumping U tubes at all levels of the way up to the reservoir. We simply close valves and stop the flow with the water at maximum height and move it to the next U tube. Then repeat the process to raise some more water. No need to worry so much about exactly matching forces and making things move slowly. I'm kind of excited about that design, so I haven't made diagrams...   

       But to answer your question from the diagram, in fig 5, there should be an open valve to atmospheric pressure on the right side of the uppermost tank. Pressures should be such that air will flow into that tank.   

       Wait, it looks like you are showing the deeper tanks with lower pressures. The head is the vertical distance from the water surface inside the tank to the surface of water in the open reservoir connected by a U tube.
—Archimerged, Apr 23 2006
   

       //But to answer your question from the diagram, in fig 5, there should be an open valve to atmospheric pressure on the right side of the uppermost tank. Pressures should be such that air will flow into that tank.// Impossible, if the bottom of the tank is always connected to one reservoir, or another. Air cannot flow in.   

       However, look at my latest link. How about that?
—Ling, Apr 23 2006
   

       [Ling] The bias tank water level starts matching upper reservoir, and then more pressure is added above the water. If the top tank is higher than the upper reservoir, and bottom is connected to manifold attached to upper reservoir, then opening the right hand valve to atm will let air in and water out and down into manifold and from there to reservoir. Closing valve to atm while connecting the manifold attached to bottom to bias tank will compress the air so captured. This will be the biggest flow of water out of bias tank into compression tanks b/c the others are at higher pressure.   

       Re pumping diagram: pump some water up to highest odd numbered tank, leave it there, start again and pump up to second highest odd numbered tanks, etc until all odd numbered tanks are full. Then you can pump odd to even all at once. Note also that this diagram also applies to lowering water to generate low pressure air to feed bias tank.   

       Compression diagram: it wastes a lot of energy draining water farther than necessary. The water involved in compression (or expansion) only moves back and forth a little between top reservoir and bias tank. Note that the layout for compression is also used for expansion with very little change.
—Archimerged, Apr 23 2006
   

       but does it make tea?
—xenzag, Apr 23 2006
   

       Grossly over-complicated machine for pumping water uphill,   

       jhomrighaus and author disagree on weather or not system will work as intended(raise more water than it uses), Ling asked some questions and made some diagrams, some other people added quips, general consesus says its a dud.   

       jhomrighaus working on response, everyone else attempting to understand what on earth this idea is about.   

       hows that?
—jhomrighaus, Apr 24 2006
   

       Summary: jhomrighaus still seems to believe that the Stirling cycle does not work. Author believes that it does. The stimulating discussion above has increased the author's understanding of high-efficiency heat engines.   

       (Author has not read jh's last link on compression power estimate). But it is just a calculator for mechanical compressors using engines and pistons. They are much less efficient than this machine. And much smaller. And they require fuel or electricity.   

       See Hydrostatic Stirling cycle air compressor link above (now properly titled, described as "where I am consolidating ideas from this discussion"). There I (no one else has edited the article but anyone can) discuss a machine which doesn't need a huge water reservoir but actually the mechanism can be used to make compressed air or to raise water, depending on if you want to install it at an existing pumped storage plant, or if you want to make compressed air. It will do either.
—Archimerged, Apr 24 2006
   

       Link can be used to esitmate amount of heat generated on compression of a gas. If the case of 1000 cubic meters of gas the temperture change is from 200K to 990K or 790 degrees Kelvin change in Temperature.
—jhomrighaus, Apr 24 2006
   

       That is why there are heat pipes inside the compression tanks. This machine carries out isothermal compression. The compression must go no faster than the heat can be carried away. (One reason why the compression is done in small steps, in a different tank each time). That rate depends on the amount of propane in the heat pipes and the temperature difference between the heat sink and the gas. I never said the machine would be cheap to build. But once built, it gathers energy from small delta T and stores it temporarily in a large water reservoir for use in a hydro plant, or if you prefer, produces lots of high-pressure compressed air.
—Archimerged, Apr 24 2006
   

       The Stirling Cycle works just fine. Your use of it however does not.   

       The efficiency of the stirling cycle(which is cited here) is less than that for the Carnot Cycle the most efficient cycle possible. See link for Efficiency Calculator. Based on that calculator the maximum efficiecy of the system is 11% based on a 30 degrees F difference(the most reasonable maximum temperature differential one can expect)   

       Your machine just wont do what you think.
—jhomrighaus, Apr 24 2006
   

       Since the operating cost of the machine does not include any cost for heat going in or out, the thermal efficiency does not matter. It is important to do as well as possible, i.e. around 10%. And the conversion of work to gravitational potential must be very efficient, more than 96%. Where do I fail to do that?   

       Carnot's theorem: "All reversible engines operating between the same heat reservoirs are equally efficient." (Wikipedia article on "Carnot heat engine", efficiency is work done divided by heat removed from the hot reservoir, and remember both machines are reversible). If the two cycles are not equally efficient, run the more efficient one (more heat from hot reservoir to do the same work) as an engine to power the less efficient reversible cycle in reverse (more heat from its "hot" reservoir which is the cold sink, same work) as a heat pump moving heat from the cold sink to the hot source. Then the heat pump dumps more energy into the hot source than the engine removes from it. You have perpetual motion with no input of energy because you can run another machine off the excess heat in the hot reservoir. So the Carnot cycle (using adiabatic steps in place of isochoric steps) has the same efficiency as the Stirling cycle. A Stirling engine is a completely different beast. (Edited to correct the proof. Carnot was a clever man...)   

       [jhomrighaus]: Wait a minute. Are you saying you admit that you think I need a cycle with efficiency greater than 11%?
—Archimerged, Apr 24 2006
   

       I understood Carots theorum to be that the maximum achievable efficiency of a heat engine operating between a hot and cold reservoir could not exceed that of the ideal cycle(the Carnot Cycle)   

       Put another way. The highest possible efficiency of any process when acting between two reservoirs is the Carnot Cycle(which cannot exist in the real world) So the Maximum theoretical efficiency of any other process cannot be any higher(cannot have a Stirling engine with a higher efficiency rating than that of the Carnot)   

       The best you can expect is less than Carnot Cycle would allow which is 11% for a 30 Kelvin difference(few places on earth experience this amount of temperature flucuation) I think this drops below 8% for a 30 Farenheit difference.   

       The entire concept here relies on very small steps to achieve the goal. For every step thats added the inefficiencies add up. If you lose 1% to friction and heat loss at every step of the process(a VERY LOW ESTIMATE) you only need 10 steps before you have begun to run into a negative energy budget. Just looking at the valves alone the energy required to overcome the torque of the valve is like 20 joules or so. There are a LOT of valves in your design and 3 or 4 at a time need to be accuated at each step with some steps requireing hundreds of valves be accuated.   

       This is the difference between theory and engineering. Theoretically the bigger the engine the more power it can make, but in reality the exotic materials and technologies required to make it happen are non existant. They are only fantasy. For this to work you would need valves which require 0 energy to accuate, materials with 100% heat transfer efficiency, A location with a greater than 30 degree temperature change every day, a control system that consumes 0 energy, Seals and Materials that allow 0 leakage, and Materials that have zero Elasticity or deflection under pressure and gravitational loading. You have not in any way addressed gas dissolution at the Water/Air interface which will play havoc on the Air system and water volumes(compressing air and water in same system leads to air dissolving into water and then boiling out when pressure is reduced) This is a dream fantasy that will not do what you think it will.
—jhomrighaus, Apr 25 2006
   

       [jhomrighaus], Now you are being more reasonable. The actual problems with the design are in friction and leakage, not in "where does the energy come from." I agree they are difficult problems, but I disagree that they aren't worth persuing. The energy required to open a valve cannot scale with the volume of the valve, only with the surface area. So bigger valves will be more efficient. No?   

       By the way, just out of curiosity, do you believe that CO2 causes global warming and will if not fixed cause major damage to low-lying areas?
—Archimerged, Apr 25 2006
   

       Bigger valves are heavier and have more inertia to overcome. The surface area of the vavle is one factor. Pressure exerted on the valve is another. The higher the pressure the valve must contain the stronger it must be(weight) the tighter its seal must be(friction) and the harder it will be to turn.   

       I will wait to see your numerical proof of concept(amount of energy exhausted in compression + volume of Water + energy to be reabsorbed from the environment. Where is the threshold for volume of water that can be lifted and the distance it can be lifted given the limitations on the system(temperature differential) Exactly how many tanks will be required for any given height, How long does each step take(remember that you only achieve your extremes of Temperature at 2 points during any given day)   

       Incidentally the Fix for my machine is quite simple. A big Rubber band(sping, gas strut whatever) that is calibrated to match the cold temperature PSI to retract the piston to its starting position after each stroke. Some energy will be lost in loading the band but the efficiency of that process would have a minimal effect on the overall operation of the system.
—jhomrighaus, Apr 25 2006
   

       But none of the valves have high pressure across them. The pressure difference is never more than about 5 ft of water.   

       If the 30K doesn't do the job, there are additional sources of heat, such as burning biomass. So don't harp on the 30K or the time of day. That's not the point. Also, this machine could be placed to use waste heat from existing coal and nuclear plants. If compressed air is harvested instead of electricity you don't need such a big upper water reservoir.   

       //amount of energy exhausted in compression + volume of Water + energy to be reabsorbed from the environment//   

       Work done to gas during compression is n R Tlow ln(Pmax/Pmin). This comes from the gravitational potential and is rejected as heat to the cold sink. Compressing the gas at constant temperature does not change its energy content.   

       Work done by gas during expansion is n R Thi ln(Pmax/Pmin). This energy is absorbed from the hot source and goes into raising water.   

       Volume of water: if built at existing pumped storage, there is no problem with water volume. Otherwise, you need simultaneous access to hot and cold and run compression and expansion at the same time.   

       Amount of gas: for 100KW net work, 1 megawatt heat flow, is 162 mol/sec, about 3.9 m^3/sec ambient air, 13 l/sec at 300 atm. I never said this was a small machine. The biggest air tank is 4 m^3, and they get smaller at each step down to 13 l at the second last one. The last tank is large, preferably constituting a high pressure pipeline. About 1.3 l/sec of 300 atm air is produced for output if no water is used to run a turbine.
—Archimerged, Apr 25 2006
   

       Well Now that is a whole different ball of wax when you add in alternate heat sources. The question then is not weather it will do what you say(which WAS raise water using nothing but falling water and ambient temperature changes) But rather if this process is more efficient than say hooking up a big pump or some other mode of lifting water. Even a big counter balanced water wheel would do the job very quickly and efficiently and be far simpler.
—jhomrighaus, Apr 25 2006
   

       But where did you get the energy to run the pump? My energy came exclusively from existing ambient heat or from heat which is otherwise useless. The point is if I burn something, I would also run a turbine off it before using the waste heat for this. I only need to get maybe another 30K, not 300K. And the numbers I quoted do not require burning biomass.   

       Ok, here is a question: do you agree that _absolutely any_ compressed air operated water pump would raise more water with the same number of moles of compressed air if the air were 30K hotter? Or not?   

       revised: ok, I see part of the problem. Some air powered water pumps might not keep the air at the higher temperature, but would allow it to cool while pumping. In that case, they won't raise very much more water, only a tiny amount. So I have to revise the question: will any compressed air operated water pump which keeps the air at constant temperature while pumping pump more water when the temperature it is operating at is higher?   

       A big water wheel has bearings and friction and other losses. And my machine is not complicated. It is just fixed pipes and tanks which never move, and valves which don't have high pressure across them, and a computer and actuators and sensors. The only things which move are the air and water and the valves. The energy consumption of the actuators and the computer will not change much if you make the tanks and pipes bigger.   

       The important thing the machine does is produce energy without burning fossil fuel or using nuclear fuel. Whether it raises water or compresses air is completely immaterial. That depends on what the market is. Raising water means already having the reservoirs. So at those sites you do that. Elsewhere you compress air.   

       You can burn biomass, use solar heat, or use big heat differences at nearby locations like mountains with cold top and hot base.   

       But remember the numbers I quoted are for 30K. The important question is can I get the mechanical efficiency. Using water as a piston is a big help there. Very little friction. And the valves don't have a big pressure across them so they don't need a lot of energy to operate. The control system uses the same energy whether it is a 1kw plant or a 100 MW plant.
—Archimerged, Apr 25 2006
   

       OOH this is a lively discussion. I am voting for you because I think this kind of energy development should be encouraged.
—Cesiii, Jun 26 2006
   
      
[annotate]
   


 

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