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[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.
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 17 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 18 2006]
Raccoon Mt pumped storage
http://www.tva.gov/.../pdf/raccoonmtn.pdf
The water drops 990 feet. [Archimerged, Apr 18 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 19 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]
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A picture is worth a thousand words. |
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Please illustrate, as I am sceptical and confused. |
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[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)? |
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Where is the energy output? |
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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%. |
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I'm working on illustrations. |
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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. |
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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? |
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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. |
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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. |
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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. |
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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. |
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It depends on your definition of ‘ultimate’. What I meant is that you need to compare the whole-of-life kW/hour cost. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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) |
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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. |
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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. |
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//Temperature differences are pretty damned reliable. They happen every day. Not every location is as good as another.// [Archimerged]. |
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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. |
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[jhomrighaus]' numbers pretty much invalidate the whole concept from the economic point of view. |
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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. |
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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. |
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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. |
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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. |
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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) |
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Just to get the 1 cubic meter of water to its boiling point from would require more than 300000000 Joules of heat energy. |
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1 calorie is needed to raise temp of 1 gram of water 1 degree C. |
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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. |
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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. |
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Please someone do the numbers and check my calcs just in case I missed something. I still say its a giant Boondoggle. |
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[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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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? |
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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... |
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//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. |
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A diagram is now available. See the links section above. |
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[coprocephalous] There is nothing about boiling water in this idea. |
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[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. |
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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. |
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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? |
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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. |
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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. |
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//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) |
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It's late here [copro]. I'm having a whiskey and would not change it for a cuppa for all the tea in china. |
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I'm also very happy you did not 'introduce' the 'boiling bit'. If you know what I mean... |
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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 |
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edit:
7. How do you get the air back in the lower system of tanks? |
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the boiling water bit was simply a point to demonstrate that large amounts of energy are involved in making a heat engine run. |
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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. |
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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. |
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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. |
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[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. |
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[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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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%. |
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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. |
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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. |
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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%. |
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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. |
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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. |
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[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. |
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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. |
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[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. |
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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. |
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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. |
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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). |
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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) |
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All the extra tanks just confuse things and Im really not sure they contribute to the quality of the idea. |
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Something like this would be far more effective the greater the temperature swing. |
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[bigsleep] Thanks for the questions. They are _very_ helpful. |
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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. |
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See the pumped storage article on Wikipedia: http://en.wikipedia.org/wiki/
Pumped_storage |
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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. |
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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. |
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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. |
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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. |
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[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. |
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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. |
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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. |
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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. |
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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. |
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Perhaps this could be done over Lake Powell. It would be a very large pane of glass and a lot of plastic. |
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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) |
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Please help me understand why this wont work and all the tanks and lines are needed. |
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This is in essense a large sterling engine that uses air to heat and cool the cylinder as was stated in other Annos. |
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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) |
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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. |
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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. |
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Another reason for lots of tanks is every one of them has heat pipes running through it. Those heat pipes will be very busy. |
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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. |
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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. |
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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. |
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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. |
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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. |
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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) |
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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. |
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//Also, if the check valves are as shown in the diagram, it means water never flows into the lower reservoir except through the turbine.// |
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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. |
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//That means the tank above the piston has to be large enough to hold the entire contents of the reservoir.// |
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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. |
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//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.// |
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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. |
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//Your machine has the biggest piston every imagined (needed in order to make it reversible)// |
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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) |
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This is well established technology and the forces involved are actually very small when it comes to the piston itself. |
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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) |
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//See the Wikipedia article on Ericsson's ship which had huge pistons but failed after one day of operation.// |
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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. |
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//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.// |
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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. |
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[jhomrighaus] : Thanks for the continued discussion. I find it very helpful. |
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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. |
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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. |
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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. |
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[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. |
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/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. |
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/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). |
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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. |
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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) |
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How does your balance come out when you net out the water used to increase the presure of the air. |
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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. |
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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) |
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I will describe the energy and entropy flow: |
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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. |
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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...) |
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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. |
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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. |
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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. |
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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. |
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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. |
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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.) |
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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. |
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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. |
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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. |
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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. |
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[jhomrihaus], that was the point I was making when i said:
"2. The initial heat of compression will be lost." |
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This lost heat must be replaced, at least, in order to have a chance to use the excess energy to lift the water again. |
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