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Tom Booth

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  1. Hi, Earlier I posted this drawing of a simple Stirling-Type Air Pump: Theoretically, I figure that if this crude model can blow up a balloon then it should be possible for such a device to power a turbine. Anyway, I have finally gotten around to building a prototype of this component. I'll be testing this shortly, within the next few days. If possible I'll make a video demonstrating the success or failure as the case may be. Tom
  2. It might be said or observed that this system basically - very much resembles a simple "bootstrap" air-cycle cooling system where the energy produced by the expansion turbine helps reduce the load on the compressor which compresses the air to run the turbine. In such a "bootstrap" Air-Cycle system, however, the turbine can only REDUCE the load on the compressor by some relatively small percentage. In this case, though, the energy to run the "compressor" is derived from atmospheric or ambient heat. The "compressor" itself is a kind of hybrid heat-engine-heat-exchanger that requires no or very little assistance from the output from the turbo-generator. The "small percentage" that the bootstrap system normally would use to reduce the load on the compressor can instead be used for some other purpose remote from the system or as Tesla would say, this output from the expansion-turbine is "clear gain". The energy required to move a "displacer" in a conventional Stirling Engine is negligible. Of course, to get this thing started one would first need to "dig the hole" so to speak. That is, to get the cycle started would involve creating an initial heat-sink and heat source within the system. This might be accomplished by dropping some "dry-ice" down the exhaust shoot; applying some heat to the heating coils with a torch; adding an auxiliary compressor to get things started, or some combination of some such strategies. Once the cycle is started however, all that should be required to maintain it is to keep some form of external load on the turbo-generator to draw away the excess heat. This is where the energy is converted into a different form - from heat into electricity - so that the heat does not accumulate within the system - instead it goes out and re-emerges at some remote point far away from the system - such as, perhaps, the light and heat emitted from a light-bulb attached by some wires to the turbo-generator. This remote load is the means of converting the heat-energy into a different form so that - as in Tesla's illustration, the incoming "water" does not accumulate in the bottom of the underwater "tank" so that the tank remains empty and the "Sink" is maintained. Once set in operation, if the load on the turbine were removed then no conversion of heat-energy to a different form (electricity in this case) would take place and the "tank" would quickly fill up with "water". That is, the heat would accumulate in the system and the mechanism would loose its temperature differential which is its means of functioning. The "Stirling Engine" or "Heat-Engine" or "temperature differential engine" part of the system, without some electrical load, would "overheat".
  3. I'd like to address a few issues regarding this design: 1. Earlier someone expressed that a turbine does not or can not be used to remove heat in the manner described. I will take one reference regarding this, there are many others like it that I've read in studying Air-cycle refrigeration: --------------------------clip Refrigeration and Air-Conditioning Hundy, Trott and Welch Page 28: "Air cycle refrigeration works on a reverse Brayton or Joule cycle. Air is compressed and then heat removed; this air is then expanded to a lower temperature than before it was compressed. Heat can then be extracted to provide useful cooling, returning the air to its original state (see Figure 2.14). "Work is taken out of the air during the expansion by an expansion turbine, which removes energy as the blades are driven round by the expanding air. This work can be usefully employed to run other devices, such as generators or fans..." ------------------------------------ There is no doubt whatsoever that in expanding through the turbine the Air releases its internal kinetic/heat energy to drive the turbine and simultaneously the air becomes very cold as a result. It seems too good to be true but the reality is that not only does this provide very cold temperature air for cooling purposes but simultaneously produces usable energy, in fact, it MUST produce energy or rather convert the heat into usable energy in another form in order to produce the cooling effect. In other words it is a requirement in an air cycle cooling system that the turbine used for effecting such cooling be attached to some form of load - such as driving an electrical generator so that the energy is converted and so removed. The resulting cold is a result of the removal or conversion of the heat energy in the air driving the turbine. A second issue I've run across is the assertion that an Air-Cycle system is inefficient. The reason for such inefficiency however must be kept in mind. The air cycle system is not practical for most refrigeration purposes in that the temperatures produced are too extreme for most purposes. An Air cycle system produces cold as well as a great deal of heat which is usually wasted as there is no use for heat in most air-conditioning applications, however the following should be noted: From: Frozen Food Science and Technology By Judith A. Evans Pg 119: ---------------------------- "The established public opinion is that the air-cycle is inefficient as it has a low COP. It is fairly mentioned in almost every manual, that air-cycle possesses low thermodynamic efficiency when using it solely for cooling within a temperature range common for the commercial food refrigeration. Nevertheless, thermodynamic evidence shows (Kulakov et al., 1999) that air-cycle refrigeration may reach or exceed the performance of vapor-compression systems in two practically valuable cases: (i) at low refrigerating temperatures (approaching or falling within the cryogenic range), and (ii) when using the air-cycle for both cooling and heating, i.e. as a heat pump." --------------------------------------------- This "Stirling Turbine" utilizes both the extreme cold produced by the air-cycle system as well as the "waste" heat which are both utilized in driving the Stirling-Type "compressor". In this application, therefore, the Air-Cycle System should be about as efficient as any closed cycle heat pump using conventional refrigerants.
  4. In viewing the first illustration (above) It might be noted that the so-called "compressor" consists of nothing more than a chamber with two check valves, one check valve for admitting ambient air and the other for "exhausting" the same air or rather compressing it into a narrow tube. The mechanism for "pumping" air through this chamber is almost identical to that of a hydraulic jack. All a hydraulic jack consists of, basically is a means of drawing in and pushing out a fluid - a plunger raised and lowered by the jack handle between two check valves so that the fluid is "pumped" from a reservoir into the hydraulic cylinder. The "plunger" in this case, however is not the "displacer" illustrated, but rather the air itself which in the process of expanding and contracting acts as a kind of pneumatic ram or piston to "pump" additional ambient air into the system. The thing that looks like a piston to pump the air is actually just a "displacer" to move the air from hot end to cold end of the chamber. The pumping action is accomplished by the resulting expansion and contraction (heating and cooling) of the air in the chamber not by the "displacer" which only serves to direct or "cycle" the air towards the hot or cold end of the chamber as is accomplished in a regular Stirling engine. The ambient air contains heat which, given a volume of compressed air can be extracted by more or less ordinary methods such as is commonly used in refrigeration, heat pumps, air conditioning etc. The compressed air being forced through a narrow coil or tube is made to give up its heat. The heat thus obtained being utilized to "pump" additional air. After the "pumping" has been accomplished the air is allowed to decompress through a turbine where it becomes very cold by two means simultaneously. First by ordinary expansion as in common refrigeration and secondly by being expanded THROUGH a turbine as it expands where it is made to give up additional "internal" heat/energy - the heat being converted into mechanical "work" -> the "work" used to produce electricity by the action of the turbo-generator. This is the point at which the heat is converted to a different form as described by Tesla. The heat - converted into electricity, leaves the system as a different form of energy so that the heat does not ever reach the "Sink" and so does not accumulate. The resulting very cold air is then used in the Stirling type "displacer chamber" for the contraction of air in the pneumatic compressor which contraction or temporary vacuum draws in additional heat laden air from the atmosphere and in the process the same cold air also scavenges out any excess heat that passes through the system.. I see a few possible problems with this very simple design such as there not being enough pressure generated by the pneumatic compressor to give much of a rise in temperature; The placement of the "intake" and "exhaust" ports or check valves may be critical so as to retain the most incoming heat for extraction and use, not enough heat converted into mechanical energy by the turbine so that there is not sufficient "cold" to scavenge out excess heat passing through the chamber, The diameter and length of the various heating and cooling tubes or coils would no doubt be critical just as in any other heat pump or refrigeration system, but these various incidental problems do not appear to me to be insurmountable. They are design problems rather than any actual flaw in the theory of operation. Having stumbled upon Tesla's description of his own theoretical "Ambient heat engine" has given me some renewed ambition to go ahead with building a small prototype engine. I recently paid a visit to the local scrap metal yard and have gathered what materials I felt would be needed; old refrigeration tubing, gas canisters and so forth. In reading what I could find regarding refrigeration, heat pumps etc. trying to get a clue as regards lengths of tubing needed and such, what I discovered is that even in the construction of a conventional refrigerator or heat pump the mathematics involved is just too complicated. Even a computer would only be able to arrive at an approximation. In practice it seem that most refrigeration systems, heat pumps ect. are actually built more by trial and error than by any kind of accurate mathematical calculations.
  5. I am currently working on building a model of the proposed "Stirling Turbine" discussed here. While doing some research I came across a description of the theoretical working principles of this engine written by Nikola Tesla and published in Century Illustrated Magazine, June 1900. I find it very interesting that Tesla believed this principle possible and for some time worked on some such engine himself though due to various circumstances - such as his workshop burning down - never completed it and/or moved on to what he considered projects more of a priority. Here are some extracts of the article: ---------------------------------------- ...HARNESSING OF THE SUN'S ENERGY. by Nikola Tesla ... The windmill, the solar engine, the engine driven by terrestrial heat, had their limitations in the amount of power obtainable. Some new way had to be discovered which would enable us to get more energy. There was enough heat-energy in the medium (The Air, Earths Atmosphere), but only a small part of it was available for the operation of an engine in the ways then known. Besides, the energy was obtainable only at a very slow rate. Clearly, then, the problem was to discover some new method which would make it possible both to utilize more of the heat-energy of the medium and also to draw it away from the same at a more rapid rate. I was vainly endeavoring to form an idea of how this might be accomplished, when I read some statements from Carnot and Lord Kelvin (then Sir William Thomson) which meant virtually that it is impossible for an inanimate mechanism or self-acting machine to cool a portion of the medium below the temperature of the surrounding, and operate by the heat abstracted. These statements interested me intensely. (...) DIAGRAM b. OBTAINING ENERGY FROM THE AMBIENT MEDIUM (...) Conceive, for the sake of illustration, [a cylindrical] enclosure T, as illustrated in diagram b, such that energy could not be transferred across it except through a channel or path O, and that, by some means or other, in this enclosure a medium were maintained which would have little energy, and that on the outer side of the same there would be the ordinary ambient medium with much energy. Under these assumptions the energy would flow through the path O, as indicated by the arrow, and might then be converted on its passage into some other form of energy. The question was, Could such a condition be attained? Could we produce artificially such a "sink" for the energy of the ambient medium to flow in? Suppose that an extremely low temperature could be maintained by some process in a given space; the surrounding medium would then be compelled to give off heat, which could be converted into mechanical or other form of energy, and utilized. By realizing such a plan, we should be enabled to get at any point of the globe a continuous supply of energy, day and night. More than this, reasoning in the abstract, it would seem possible to cause a quick circulation of the medium, and thus draw the energy at a very rapid rate. Here, then, was an idea which, if realizable, afforded a happy solution of the problem of getting energy from the medium. But was it realizable? I convinced myself that it was so in a number of ways, of which one is the following. As regards heat, we are at a high level, which may be represented by the surface of a mountain lake considerably above the sea, the level of which may mark the absolute zero of temperature existing in the interstellar space. (...) Heat, like water, can perform work in flowing down,(...) But can we produce cold in a given portion of the space and cause the heat to flow in continually? To create such a "sink," or "cold hole," as we might say, in the medium, would be equivalent to producing in the lake a space either empty or filled with something much lighter than water. This we could do by placing in the lake a tank, and pumping all the water out of the latter. We know, then, that the water, if allowed to flow back into the tank, would, theoretically, be able to perform exactly the same amount of work which was used in pumping it out, but not a bit more. Consequently nothing could be gained in this double operation of first raising the water and then letting it fall down. This would mean that it is impossible to create such a sink in the medium. But let us reflect a moment. Heat, though following certain general laws of mechanics, like a fluid, is not such; it is energy which may be converted into other forms of energy as it passes from a high to a low level. To make our mechanical analogy complete and true, we must, therefore, assume that the water, in its passage into the tank, is converted into something else, which may be taken out of it without using any, or by using very little, power. For example, if heat be represented in this analogue by the water of the lake, the oxygen and hydrogen composing the water may illustrate other forms of energy into which the heat is transformed in passing from hot to cold. If the process of heat transformation were absolutely perfect, no heat at all would arrive at the low level, since all of it would be converted into other forms of energy. Corresponding to this ideal case, all the water flowing into the tank would be decomposed into oxygen and hydrogen before reaching the bottom, and the result would be that water would continually flow in, and yet the tank would remain entirely empty, the gases formed escaping. We would thus produce, by expending initially a certain amount of work to create a sink for the heat or, respectively, the water to flow in, a condition enabling us to get any amount of energy without further effort. This would be an ideal way of obtaining motive power. We do not know of any such absolutely perfect process of heat-conversion, and consequently some heat will generally reach the low level, which means to say, in our mechanical analogue, that some water will arrive at the bottom of the tank, and a gradual and slow filling of the latter will take place, necessitating continuous pumping out. But evidently there will be less to pump out than flows in, or, in other words, less energy will be needed to maintain the initial condition than is developed by the fall, and this is to say that some energy will be gained from the medium. What is not converted in flowing down can just be raised up with its own energy, and what is converted is clear gain. Thus the virtue of the principle I have discovered resides wholly in the conversion of the energy on the downward flow. FIRST EFFORTS TO PRODUCE THE SELF-ACTING ENGINE—(...)—WORK OF DEWAR AND LINDE—LIQUID AIR. Having recognized this truth, I began to devise means for carrying out my idea, and, after long thought, I finally conceived a combination of apparatus which should make possible the obtaining of power from the medium by a process of continuous cooling of atmospheric air. This apparatus, by continually transforming heat into mechanical work, tended to become colder and colder, and if it only were practicable to reach a very low temperature in this manner, then a sink for the heat could be produced, and energy could be derived from the medium. This seemed to be contrary to the statements of Carnot and Lord Kelvin before referred to, but I concluded from the theory of the process that such a result could be attained. This conclusion I reached, I think, in the latter part of 1883, when I was in Paris,(...) This work was continued until early in 1892, when I went to London, where I saw Professor Dewar's admirable experiments with liquefied gases. Others had liquefied gases before, and notably Ozlewski and Pictet had performed creditable early experiments in this line, but there was such a vigor about the work of Dewar that even the old appeared new. His experiments showed, though in a way different from that I had imagined, that it was possible to reach a very low temperature by transforming heat into mechanical work, and I returned, deeply impressed with what I had seen, and more than ever convinced that my plan was practicable. (...) Full Article can be found here: http://www.tfcbooks.com/tesla/1900-06-00.htm ---------------------------------- Tesla's idea of creating a "sink" or "Cold Hole" for heat energy to run towards affording a flow from which some energy could be extracted and that; by converting the heat flow into some other form of energy so that the "Hole" is never filled "By transforming heat into mechanical work" thus maintaining the "Cold Hole" or "Sink" in the process all describes in a nutshell what is accomplished by the "stirling Turbine". "The conversion of the energy on the downward flow" so described by Tesla is, I think, something accomplished on a rather routine basis today by means of an "expansion turbine" or "turbo-expander" used for liquifying gases, cryogenics and other cold temperature processes. As a compressed gas passes through the turbine the Heat-Energy within the gas is converted into mechanical energy - leaving the system. This is the point of conversion. Like Teslas decription of an underwater tank that is never filled because the water entering is converted to a gas which is allowed to escape. In a Turbo-expander the gas itself is left at an extremely low temperature. In the Stirling Turbine, (this cold gas the energy from which has been spent to produce electricity) is then utilized to "scavange" whatever heat "leaks" through or into the system without being converted. Again, here is a "working" illustration of the engine, though I've made a few design changes since this drawing was made it does illustrate the basic idea or principle that I'm working on: http://prc_projects.tripod.com/stirling_air_turbine.html I don't know as the configuration shown will work exactly as illustrated, but basically the flow of compressed air from the Stirling type "displacer chamber" will give off heat while compressed and become very cold when expanded through a turbine. The turbine generates more power than it takes to raise the displacer therefor producing an excess of energy while maintaining its own temperature differential. This is essentially. IMO, identical to how the "dippy bird" toy works in principle. Both are types of heat engine that produce enough excess power to operate their own heat exchanger so as to maintain a temperature difference. A slightly more elaborate design here: http://prc_projects.tripod.com/stirling_air_turbine_2.html And animated: http://prc_projects.tripod.com/stirling_air_turbine_anim.html BTW. I misspelled "Insulation" but I'm not going to go through the hundreds of frames of animation to fix it.
  6. OK, I get what you mean by ""cancel" now. This is rather obvious, I just wasn't familiar with the usage of the term. Thanks, I think I'm almost out of the woods.
  7. OK. I think I follow you. What I referred to as "absolute" above (second column) is Pascals. But as I understand it, if you use pascals you are automatically using a measure of absolute pressure, since pascal units start at zero pressure or a vacuum rather than at 1 atm or anything else. Similar to how absolute temperature measured in Kelvin... Or am I just lost in the woods ? Never-mind, I think I've got the picture. I was thinking in terms of the "atmosphere" scale starting at 1 atm. It just clicked. All the scales start at zero. so 0 atm is the same thing as 0 Pascal or zero anything else.
  8. OK, Thanks. I'm not entirely sure what you mean by "unit cancellation" yet, but maybe I'll figure it out when I get further along. At this point, no real work is being done. I'm just trying to calculate what maximum theoretical pressure could be built up in the chamber at any given temperature differential if the chamber only had an inlet valve but no outlet to the piping or the turbine. (i.e. starting at 1 atm, the volume of air in the chamber is cooled and contracts, more air is drawn in, the total air is heated in the now sealed up chamber - what would be the new pressure.) If the air were released, it would be back to 1 atmosphere at this point. I don't think a second cycle could result in any further pressure increase as there would now be more air in the chamber and until the air was released the pressure in the chamber would probably remain above 1 atmosphere even if cooled back down again (the greater inside pressure holding the inlet valve closed). Merged post follows: Consecutive posts mergedPS. I'm a little confused by your statement: I was going by the statement here: The modern form of the equation is: pV=nRT where p is the absolute pressure of the gas http://en.wikipedia.org/wiki/Ideal_gas_law I assumed that if p represents absolute pressure as stated there, would it not then be necessary to convert to those units before doing any calculations ?
  9. (continued from previous thread) I'm starting to make some progress with the mathematics. At least I finally think I figured out what "p" in pV=nRT means in terms I can understand. Please correct me if I'm wrong. p is absolute pressure. The scale of absolute pressure begins at a total vacuum. This is very different from what I had supposed earlier. Putting this in units that are more familiar I came up with this table: (figures are rounded out) 1 psi = 6895 p (absolute) = .06 atm = .0689 bar 5 psi = 34475 p = .34 atm = .34 bar 10 psi = 68950 p = .68 atm = .69 bar 14.7 psi = 101317 p = 1 atm = 1.01 bar 15 psi = 103425 p = 1.02 atm = 1.03 bar 20 psi = 137900 p = 1.36 atm = 1.38 bar 25 psi = 172375 p = 1.7 atm = 1.72 bar 30 psi = 206843 p = 2.04 atm = 2.07 bar 35 psi = 241325 p = 2.38 atm = 2.41 bar 40 psi = 275800 p = 2.72 atm = 2.76 bar 45 psi = 310275 p = 3.06 atm = 3.1 bar 50 psi = 344750 p= 3.4 atm = 3.45 bar etc. I'm most familiar with psi as that is the common measure for blowing up tires and what not, it is easily read on a gage - such as a common household compressor which I've used frequently etc. I put absolute pressure second (and bold-ed) as those are the numbers we need for our calculations. It is also nice to be able to relate all this to atmospheric pressure, which we all experience, so I've included that. Bar I just threw in for the hell of it. T - Absolute temperature I'm OK with (Kelvin). n - moles aren't a problem, that can be easily calculated. V - volume, I'm still not sure what measure is supposed to be used and R - the gas constant, I'm still fuzzy on. But I am making some progress I think. What I want to figure out to begin with is what happens if we start off with a temperature "x" (above ambient) for the hot end and "y" (below ambient) for the cold end.... (achieved by artificial means, such as applying heat and cold from a torch flame on one end and ice on the other) First we drive the air (at ambient/atmospheric temperature and pressure) in the chamber to the cold end where it contracts by whatever amount - drawing in more air, (which might also contract some once it enters the chamber). Then we drive that combined air to the hot end where it expands or tries to, increasing the pressure. What I want to figure out to begin with is if the exhaust port were blocked so that the air could not expand or escape the chamber what pressure would have developed by the above procedure ? I think this would represent about the maximum pressure that could be achieved at the given (x,y above) temperature differential. The air is not yet escaping and flowing through any tubes or turbines or anything so this should be relatively easy to figure.
  10. I'm quite sure it will, or would, take a great deal more work or energy to compress the air than what you would get out of the turbine. As I assume you know though, in any "heat engine" that energy comes from the heat applied to the engine, not from any external mechanical input. In my designs, I am, mostly for convenience and possible increased efficiency, borrowing a fraction of the electrical output from the turbine to move the displacer. probably 95% or more of the energy to compress the air is coming from the heat input from the condenser coils not the turbine. Moving the displacer without the temperature differential would produce no compression whatsoever. At any rate, I don't think any amount of debate will settle the issue. Hopefully in the near future I'll have the time and resources to just go ahead and build a small model engine and see what happens. I'll admit it is a pretty Iffy theory and the likelihood of it actually working seems next to nil but the engine is fairly simple, mostly just a canister some check valves and a lot of tubing. It shouldn't be too difficult to build.
  11. Well, that is where the turbine comes in, to extract additional heat (by making the gas do work) thus increasing the temperature differential. This seems to be a very difficult point to get across and I have some difficulty entirely understanding just how it works myself, but I offer the following references (there are many more but those online are mostly PDF downloads that require a paid subscription) These and many other references all read about the same as the above i.e. "turn as much heat as possible into work" - With the turbine, heat is being "turned into work". The greater the load on the turbine, the more heat is converted into work leaving the system at that point and leaving the gas, returned to the "compressor" that much colder, increasing the temperature differential. I'm skeptical myself, only because when I do work I get hotter. It goes contrary to human experience to say that when a gas does work it gets colder, but apparently that is the case, not just from the above references but many others including the trade literature from the companies that actually build and sell turbines for this very purpose (to cool gases to very low temperatures) as well as my own experience working with air tools. When you do work with an ordinary air tool, such as removing bolts with an air wrench, the air exiting the tools exhaust port gets freezing cold forming frost around the exhaust port. One guy who worked at the same shop I did years ago was careless and had his finger frozen as he was doing some heavy work with an air tool and left his finger by the exhaust port and went out on disability as a result. The air exiting the tool was so cold that his finger froze without him even noticing it.
  12. Granted. I just thought I needed to make a distinction between heat input and electrical or mechanical input. I certainly realize all that. All I wanted to point out is that the "compressor" (just this one part) can or could very well operate on a temperature differential alone, with no dependence upon any input whatsoever from a turbine. Provided with some source of heat (or cold, above or below ambient, or both) the "compressor" or pump can, in fact, stand alone without any turbine. What I am leading up to, (but lets not get ahead of ourselves) is that; generally speaking, a heat pump operates by means of a compressor. Generally speaking, compressors require electricity to run. However a heat pump using 1 kWh consumed power would provide the equivalent of 3.5 kWh of heat output. http://en.wikipedia.org/wiki/Coefficient_of_performance From what I've read (above and elsewhere). A heat pump consumes x joules or whatever of energy in the form of electricity to run its compressor but is able to output or move 3x joules. i.e. it is "300 to 350% efficient". http://en.wikipedia.org/wiki/Coefficient_of_performance It, of course, does not make or produce the heat/energy. That is already in the atmosphere which has been heated by the sun, it just moves or concentrates the heat that is available. Maybe I'm missing something, but if you had a "compressor" that could operate solely on a temperature differential and that compressor required the same amount of energy input as a conventional compressor running on electricity (x joules) then would you not have an energy surplus of 2x joules ? I'm not trying to say that this proves anything in regard to my contraption, but just on basic principles, if a heat pump operating on a compressor (which compressor runs on a temperature differential) can concentrate 3x or 3.5x units of heat energy and the compressor only requires x units of heat energy to run the heat pump... I'm sure you can see what I'm getting at. I don't think this would work in a closed system as you would have to expend energy to reheat your working fluid for each cycle. That is, if you separate out the heat from the air and then convert that heat into electricity you are left with cold air, which cold air, in a closed system, you would have to reheat, but in an open system where with each cycle you are drawing in fresh air that is already heated by the sun I'm thinking that you might just be able to get away with it. Merged post follows: Consecutive posts merged That would probably be fun I suppose. Not likely to happen at my age though I'm afraid. I don't really know enough about how their system is supposed to work to offer an opinion. They seem to have the same or a very similar idea i.e extracting energy from the air utilizing some sort of heat engine and turbine. I do see some potential advantages of using a closed system. you can avoid the problems with ice formation in the turbine due to humidity in the air. You can also use compressed gas which is generally more efficient in a heat engine than mere atmospheric density air. I would imagine also that using helium might have some hidden advantages I'm not aware of. I found recently that helium is commonly used as the working fluid in most cryocoolers. I would imagine there is some good reason for that. I've also thought that it might be possible to use a closed system with my own engine. If your heat source is the atmosphere then I'm not sure there is any difference between drawing in fresh warm air and using a passive heat exchanger (or solar panels). I appreciate your comment, though I'm not sure my idea has any more merit than theirs. I think if one is impossible than the other would be just as impossible. Still that damn "dippy bird" keeps bobbing. If my idea of an expansion turbine functioning LIKE a virtual heat sink has any merit then who knows ? Of course, I'm already well aware that no one believes that such is the case. But the way I look at it, and I may be wrong, but the "dippy bird" does not have any conventional "heat sink" either. The heat is absorbed by the evaporating water molecules during phase change. This is not IMO a "heat sink". It is not a cold reservoir that the heat is flowing to, rather on a molecular level the heat is being locked up somehow in the very structure of the molecules. But if in effect the heat is gone or made unavailable, one way or another then for this "dippy bird" it (the heat) is made unavailable and this functions as a heat sink for all practical purposes. In the turbo expander, the heat is changed in a different way, but it is still being made unavailable to the "system" and I don't see why this could not function as a "heat sink" in effect. P.S. I do suspect that although a turbo expander can generate extremely cold temperatures (For liquefaction of gases etc.), generally this is accomplished with very small "micro-turbines" (about the size of a silver dollar in diameter) which are operated within something like a thermos bottle that is itself within a "cold room". In other words, although the degree of cold produced sounds impressive, the actual quantity of cold produced is small, in a confined area and must be thermally insulated to the extreme. It is doubtful that this would be adequate to carry off the waste heat from a Stirling engine or "compressor" capable of any real power output.
  13. I have a lot to learn about a lot of things. I was there referring specifically to the "compressor" (or pump) being able to operate on a temperature differential, which if I read them right, both Mr. Skeptic (directly: i.e.: "From what I understand, it will work, albeit inefficiently..." and npts2020 indirectly: i.e.: "Once you remove the candle from the Stirling compressor in your link, it will stop working..." implying that it will work provided some heat input is maintained) Seem to believe that this one aspect of my idea should work. Mr Skeptic even provided the mathematics. This however has nothing to do with Kender or their engine design as from the information on their website, their intention is to use a conventional compressor driven by electricity, not a temperature differential. I've stated before that it might be a good idea to start a new thread to avoid this sort of confusion. Again, when you say "this (will not work)" or "...it (will work) you are lumping two entirely different things together. I am not saying that just because a "compressor" might run on a temperature differential that this proves that the rest of my idea will work and it certainly has nothing to do with the Kender engine as they have proposed no such thing.
  14. By that I assume you mean: increasing the temperature differential not "input" in regard to the hand operation. By using "dry ice" ( -109.3 F ) for example - or by heating the opposite end simultaneously. Using a stronger guy to work the displacer up and down would have no effect. Regardless of the temperature differential the pressure on the displacer will be equal on all sides so that the actual mechanical input remains the same. Of course. As far as the actual math is concerned, I don't think I have a problem. Looks like just simple multiplication and devision to work out the equations. What I'm having some trouble with is the symbols. In particular, what is "R" and what would be its value in this context ? I'm assuming this is in reference to the "ideal gas constant": http://en.wikipedia.org/wiki/Ideal_gas_constant Unfortunately, reading that Wiki article has not enlightened me in regard to what value to use. Here it states: R is the gas constant (which is 8.314472 J·K−1·mol−1 in SI units) http://en.wikipedia.org/wiki/Ideal_gas_law Do we just leave R alone ? The deeper I dig trying to find out more about it the more complicated it becomes. Merged post follows: Consecutive posts merged Here, I'm not sure what you are referring to as "a solid". Are you referring to where the candle is heating the bottom of the chamber (i.e the chamber itself is the "solid") ? I'm also having a little difficulty understanding what "chambers" you are referring to, and if you mean this literally or hypothetically i.e. (as-ifthere were more than one chamber) Here again, in this simple "compressor" there is only one camber. It might be assumed that we would want to add a tank of some sort to hold the air being "compressed" or pumped, but that is not necessarily the case. Here you have really lost me. In my drawing: http://prc_projects.tripod.com/Stirling_Compressor.html there is only one chamber. I'm not sure if there is a misunderstanding here or not. Maybe if I make this little modification to the drawing it will be more clear: http://prc_projects.tripod.com/compressor_balloon.html Between the intake check valve and the exhaust check valve there is but one chamber. Surrounding the chamber you would have atmospheric temperature and pressure (inside the chamber as well, until we move the displacer). The displacer does not divide two chambers, it moves freely up and down within one chamber not contacting the walls of the chamber at all, the air just moves around it. When the displacer is pulled up, the air moves to the bottom. As the air moves down around the displacer it creates some turbulence causing the air to come in contact with the hot bottom of the chamber as it swirls around, heats up and expands. As the air expands, the increase in pressure forces some of the air out through the exhaust check valve into the balloon. When the displacer is pushed down, the air is forced up where it contacts the relatively cold (ambient) top and sides of the container. (Contact with the hot bottom is now blocked by the displacer). As a result the air cools back to near ambient temperature and contracts - drawing more air into the chamber through the intake check valve to replace the air that went into the balloon. The question is; if we continue moving the displacer up and down as described, will this "compressor" or pump blow up the balloon ? I don't really think we need any complicated mathematics to figure this out. Merged post follows: Consecutive posts mergedI thought that I might also mention that it is possible to construct a Stirling Engine that does not have any external mechanical input whatsoever. This seems to be a point of contention in regard to the turbine having sufficient power output to move the displacer. I've been trying to get across that in any Stirling Engine the displacer requires very little power input. It is very light and easy to move. However, what I didn't mention is that it is also possible to design a Stirling Engine where the displacer actually has no mechanical power input at all. Rather it functions on the basis of the displacer chambers own pressure changes. In other words, the displacer really does not require ANY power input. It can operate on heat alone. Here is an example: This type of "Free Piston" Stirling Engine is commonly used such as on these 5 kW solar collectors: http://www.stirlingengines.org.uk/sun/sola4.html I played around with this idea also (i.e. using a "Free Displacer" with a turbine instead of a piston) but I thought that for experimental purposes I would like to have more control over the displacer movement as I considered that with the turbine arrangement, it might be possible to get more power output by increasing the "dwell" (the amount of time the displacer spends at the ends of the chamber) which theoretically might increase heat exchange and result in higher pressure, more than compensating for little bit of power that would be used to gain more control over the displacer movement. I think this is an important point to get across. The displacer does not require any mechanical power input to operate and even where some mechanical power input is used it is negligible.
  15. Naturally. But otherwise, given the heat input from the candle, you do believe this "compressor" would work. (?) Merged post follows: Consecutive posts merged I certainly envy your command of numbers and formulae Mr Skeptic. I must confess that much of what you posted is a bit (or more than a bit) over my head. Nevertheless, it seems there is some confidence all around that this heat compressor, or compressor operating primarily on a temperature differential will work, to one degree or another. Now suppose we try this: Stirling Compressor II
  16. Beyond drawing in air at atmospheric temperature and pressure, I don't really see how I can know much more than that without building the thing. Let's just say we set up an artificial temperature differential to start with, Will the "compressor", given that theoretical temperature differential work at all ? Will it be able to compress the air to 2 or 3 atmospheres 5, 10 or what ? Then, ow hot will the air get traveling through the tubes after being compressed ?That has to do with things like the boundary layer effect and velocity. Already we need different equations. Suppose we start with this "compressor" by itself: Stirling Compressor Can we make any meaningful calculations without knowing if it will do anything at all ? Lets just leave the coils and turbine out of it for the moment. Would such a compressor, operating on a given temperature differential be able to compress any air at all ? I mean, using a different example: You can say a2 + b2 = c2 "proves" that my ladder won't reach the top of the house. I can say a2 + b2 = c2 proves that my ladder WILL reach the top of the house. Both statements are meaningless without some actual measurements. How long is the ladder and how high is the house ?
  17. Wrong about what ? Merged post follows: Consecutive posts merged Well, I think that's the problem. You say "given the others..." (or one of the problems). Maybe someone can tell me if h1 and h2 refer to entropy, enthalpy, or heat - or what. I'm finding similar equations in an old gas turbine book that says h is entropy and elsewhere h is heat ? How about I just build the thing and stick a pressure gage on it ? You said you guys could "help" with the math. Saying glibly "pv = nRT" as if that is supposed to mean something isn't much help.
  18. I've been working on trying to figure out some of what appear to be the applicable equations. The one that seems most applicable to the turbine end of this thing (for an open system) is: W = m(h1 - h2) = mcp(T1 - T2) Power Output = mass(initial enthalpy[not entropy] - final enthalpy)... enthalpy... Watching the You Tube video about Air Cycle Refrigeration systems... the Indian guy giving the lecture was a bit difficult to understand. I believe the "h1" and "h2" refer to enthalpy not entropy as I had said earlier. At any rate, I don't know as any useful information could be derived from pV = nRT. I don't see how static variables would apply in connection with a dynamic system. Except for the compressor, this seems to most resemble a reverse Bryton Cycle, with some elements of the regenerative type air cycle. This video covers the math for those: I'm still grappling with the terminology though. All in all this thing is about 90% Air Cycle Refrigeration System. About the only thing different is the Stirling Type "compressor".
  19. I'm not saying that the turbine extracts more energy than the gas does work. I'm saying basically that the heat/energy added to the system as a whole, by in-drawing fresh warm (ambient temp) air should be roughly equivalent to the work/energy (electrical) output of the turbo-generator. You have taken the heat out of the air by making it do work as it expands through the turbine. The resulting cold air is then used as a heat sink (or heat scavenger) for the heat/engine/compressor. The compressor draws in warm air (energy added to the system) and compresses the air. as energy is being added to the air it gets hotter. The heat is driven off in a heat exchanger (at the top of the heat/engine/compressor). The air then decompresses through the turbine extracting more heat/energy (energy leaving the system as electricity). The air leaving the turbine is now very cold and is passed through another heat exchanger (where the cold air scavenges heat from the heat/engine/compressor on its way out) . The heat/engine/compressor uses the temperature differential between the hot and cold heat exchangers to draw in and compress more air. If there is some failure point in all this, I would suspect it would be at the compressor, or between the compressor and the turbine, as this is where you are trying to force compressed air into hot coils via the heat derived from the same hot coils and I can imagine that there would be some build up of back pressure there which might result in the compressor valves failing to open or the coils wouldn't generate enough heat to power the compressor or both. However, if the pressure is relieved through the turbine, then the valve would be able to open again, the air would move through the coils generating heat... I'm imagining that the displacer action could be controlled by a pressure switch. In any event, the "compressor" would have to have the potential to pump more air than the turbine needed. This seems a bit more questionable than whatever the alleged issue with the turbine is supposed to be, which I just don't seem to get. Given a supply of compressed air, a turbine will output work and cold air.
  20. This would be true of the turbine as well. If you take away the load on the turbine then the turbine will just freewheel and the gas will pass through without doing any appreciable work and there would not be much cooling effect. I'm going to assume you meant: efficiency <100% My supposition is that the evaporative cooling on the dippy birds head results in a minuscule temperature differential. The bird, functioning as a heat engine can't extract much energy from such a small temperature differential. If you increase the temperature differential the bird will move faster. Replace the water with alcohol for instance. The more rapid evaporation of the alcohol will result in a greater temperature differential and therefore the heat engine can extract more energy. An air cycle cooling system, or similar system using a turbine for refrigeration, creates the most extreme temperature differentials of any heat exchange system. Not the few degrees difference resulting from the evaporation of a few drops of water but potentially hundreds of degrees difference. Wider temperature differential, greater potential for extracting energy with a heat engine. I mean, this thing probably won't work. Granted. But it won't be because turbines or turbo-expanders or expansion turbines don't work. That aspect of it is proven IN USE technology all over the world. Merged post follows: Consecutive posts merged The turbine itself doesn't draw much energy... a bit of friction on the shaft bearings. The real energy is being drawn by the ELECTRICAL LOAD on the generator coupled to the turbine. The energy to turn the turbine and attached generator against the electromagnetic force created by the load on the generator is being taken from the air or gas driving the turbine. As a result, the gas looses internal energy (i.e. the temperature of the gas drops in proportion to the work performed) "for an ideal gas Joule showed that internal energy is a function of temperature only" http://www.ent.ohiou.edu/~thermo/property_tables/gas/specific_heat/Cp_Cv.html Google results for: "the internal energy of an ideal gas is a function of temperature only" There is nothing "magical" about the turbine. That a gas drops in temperature as a result of doing work, (in a turbine or otherwise), is basic thermodynamics.
  21. Just the turbine ? You mean something like this: Diagram You have a mass of compressed air, leaving the nozzle at high velocity driving the turbine - regardless of the air temperature or any temperature changes. The turbine is driven by a pressure differential not a temperature differential. A temperature differential results from the air doing work against the load on the turbine. The temperature differential does not power the turbine. The pressure differential powers the turbine. I don't really understand the problem, or why the focus on the turbine. The turbine is the last thing I thought there would be any debate about. These things are IN USE and have been for many years by industries all over the world for cooling and refrigeration. In transferring energy to the turbine, the gas looses energy and gets cold. Simple. Heat is being "consumed" by the water. The water needs the heat to liberate itself (evaporation/phase change). The evaporating water absorbs energy resulting in a drop in temperature (heat sink). In the turbine, energy is ultimately being "consumed" or absorbed by the load on the turbo-generator - resulting in a drop in temperature (heat sink). I don't see anything too vastly mysterious or "magical" about this.
  22. Well, I certainly don't claim to be any turbine expert. I'm talking specifically about a Tesla Turbine, which, as far as what I've read about it and seen from You Tube videos and such, it has closely spaced perfectly smooth parallel disks with no tapering one way or the other. This is also a different application. Most turbines are not insulated so as to prevent the gas entering in from picking up heat to do the expanding. Here the idea is to keep the turbine cold so that the gas would have to draw on its own internal energy to expand. and yes, I did say expand, but I'm thinking that this expansion might be limited and before the gas finished spiraling to the center of the turbine (as it does in a Tesla turbine - delivering power all the way, as it spirals inward, loosing energy), I would imagine that this expansion would be reversed. In liquefaction processes using the same method (pre-cooling the gas and insulating the turbine) the gas liquefies, condensing on the inside of the turbine housing. Prior to condensing into a liquid, it seems reasonable to assume that the gas contracts before it condenses - before leaving the turbine. Anyway, that idea was just a thought on a possible way to tweak the power output of a working engine. We don't have a working engine as yet so the point is mote either way. Well, technically, in this engine, (IMHO that is), the heat SINK is not the atmosphere, it is the kinetic/heat/energy being extracted by the turbine being fed to some remote electrical load. In a kind of convoluted way, the electrical load on the turbine is the heat sink. At least as far as I can figure, in an expansion turbine of this sort, that is where the heat is "disappearing" to. That is how a cryo-cooler using an expansion turbine works, from everything I've read on the subject, the heat is transferred to the load on the turbine. That load can be just a water cooled brake or an electrical generator. Eventually the heat would get back to the atmosphere one way or another I suppose - through heat from incandescent lamps or heating elements or whatever - but maybe these will be far enough away so as not to cause any major concern - I don't know. How else do you explain the EXTRA cooling effect of an expansion turbine, above what would be expected from expansion alone if the HEAT in one form or another is not being transferred to the break on the turbine ? Where is it going then ? This is what the literature on these things says quite plainly - i.e. the heat being extracted by the turbine "re-emerges" at the braking system and is carried of by the water used to cool the braking system on the expansion turbine. Sounds kind of like one of those "spooky action at a distance" things in quantum physics that Einstein couldn't get used to - but as far as I can see, that's the way these things (expansion turbines or turbo-expanders) work. That looks like a "heat sink" to me, for all practical purposes. But we've already been through all that. Merged post follows: Consecutive posts merged I think that for the most part, I understand most of what is being said. I'm just thinking that most of what is being said doesn't seem to apply to this engine - other than it can't work because it would be "perpetual motion" or that it would violate "the second law of thermodynamics". Those reasons in particular look to me like beliefs or prejudices rather than specific scientific explanations as to why some aspect or detail of this engine just won't work. clip----------------- Even though they fully respect the laws of thermodynamics, there are a few conceptual or real devices that appear to be in "perpetual motion." Closer analysis reveals that they actually "consume" some sort of natural resource or latent energy, such as the phase changes of water or other fluids or small natural temperature gradients. In general, extracting large amounts of work using these devices is difficult to impossible. wiki - Perpetual motion The drinking bird is given as one example of this apparent "perpetual motion". As far as I can figure, this engine takes advantage of the same principles as this toy bird. i.e. converting a temperature differential into a pressure differential so as to extract work from the pressure differential. clip----------- "The drinking bird is a heat engine that exploits a temperature differential to convert heat energy to a pressure differential within the device, and perform mechanical work. Like all heat engines, the drinking bird works through a thermodynamic cycle" wiki Drinking bird I'm not saying that this bird proves that this engine will work. but I think it does prove that it is not impossible for something like this to work. About the only difference in principle is that the bird uses a very inefficient cooling system to establish the initial temperature differential. Instead of evaporative cooling, this engine uses a turbo-expander. Do you suppose the engine would work if it used evaporative cooling instead ? IMO the "latent energy" being "consumed" (as described above) is the extra energy being extracted by the turbo-expander. In a sense it (this extra heat/energy) is being "consumed" as it leaves the system to power some remote load.
  23. I don't have you guys stumped do I ? I had a few ideas about possible ways to increase the efficiency of the turbine: First of all, thinking some more about this entropy thing and the fact that the application for this turbine is power production rather than liquefaction or JUST cooling alone, It crossed my mind that it didn't make sense to pre-cool the gas in the line as this would rob the gas of some of its energy before reaching the turbine - wouldn't it be better and make more sense to pre-cool just the turbine housing ? The gas would still have to draw on its own internal energy to do the expanding - more so, I should think, but would have more internal energy to work with and therefore could deliver additional power to the turbine rather than having that power extracted before reaching the turbine. So I came up with this idea for using the turbines cold exhaust gases to pre-cool the turbine housing itself instead of pre-cooling the gas before it enters the turbine: Turbine Details On the same page (scroll down) I've included another idea for increasing the efficiency of the Tesla Turbine (theoretically). I was thinking that in the Tesla Turbine, perfectly smooth and uniform flat plates are used. Normally this would probably be OK as the gas would be picking up heat from the turbine as it spirals towards the center. But in this application, the gas would be loosing internal energy and cooling and presumably contracting as it spirals towards the center of the turbine, which I think would tend to create some degree of turbulence or drag on the turbine disks as in effect, the contracting gas would have too much room between the disks and so would loose cohesion with the sides of the disks - slow down and create turbulence - reducing the power output of the turbine. So my idea was to TAPER the disks. In this way, the passage the gas is using as it spirals inward would become narrower and narrower as the gas simultaneously cools and contracts. Theoretically, the narrowing of the passage would result in the gas maintaining speed as it spirals in and cohesion with the turbine disks. wadayathink? By the way, I'm starting to get somewhere with those equations. Mostly I'm just having some trouble with the terminology. Like "the gas constant". What is the "absolute pressure of the gas". How is this different from regular pressure ? The equation for an isentropic process in an open system (the turbine): W = m(h1 - h2) = mcp(T1 - T2) W = Power Output; T = Temperatures; m = mass; mcp = specific heat at constant pressure (?) h1 and h2 represent entropies. How do you determine the before and after entropy ? The idea of entropy seems rather abstract to me at this point and I'm wondering how this can be reduced to numerical values. Or is that what we are supposed to be solving for ? Anyway, without some actual values to plug in to the equations, it would all be guesswork anyway. What I need to know is how much actual pressure can the "compressor" really deliver - and since this is only a theoretical compressor that hasn't been built yet, who knows ? It might not work. It seems that the only way to find out would be to build one. If it does work, then the actual pressure it can develop can be measured and then that can be plugged into these various equations and formulas to arrive at some guestimate about potential power output (or lack thereof). But if it is 60 F out and the "compressor" can maintain about 30 PSI (absolute i.e. 2 atm or about 15.5 PSI above 1 atmosphere) then from what I've figured so far, the "heating coils" might be able to develop a temperature of around 550 F. If 60 PSI 1,500 F, if 100 PSI 3,000 F I think that just refers to the temperature reached from pressure alone. I'm not sure if this applies to passage through the coils. It also doesn't account for any regenerative effect. Worst case, the lowest pressure that could turn a Tesla turbine is about 3 PSI (above 1 atmosphere = about 17.5 PSI absolute (?)) which might generate 140 F (?) Given a "heat pump" averaging 300% efficiency and a "Stirling Engine" averaging 20% efficiency, combining the two in one engine should produce a 60% overall efficiency (20% x 3 ?).
  24. I imagine what you say here is true in a closed system. But this is vented to the atmosphere. How could it possibly build up much pressure ? This is like saying that if I hold a blow torch to my cars tail pipe, the heat will create back pressure inside the tailpipe and the car will stall. Probably this would also be some problem with a reaction type turbine, as a reaction turbine is in the flow right inside the pipe or duct-work and relies on the pressure head in the line directly. I'm using an impulse turbine not a reaction turbine. The power for an impulse turbine comes from the high velocity jet of air coming out of the nozzle. It doesn't care about some pressure fluctuations in the exhaust. Thanks, I have no expectations. I more or less stumbled upon this particular arrangement by accident. I was not out to create any "perpetual motion machine".
  25. No, not yet. I spent most of yesterday on that animation. (And most of today working on making it a little more realistic. In particular the pressure gauge. There should be a lot more pressure drop during the intake phase as air would still be decompressing through the turbine at that point, I also got the turbine turning, but there are a few more doodads I need to work on, like the temperature changes in the regenerator -- that mesh in the center of the displacer) I've referred to this as an "Air Cycle" and I guess it is, but I think it has more in common with gas liquefaction methods. The Air Cycle used on aircraft generally uses ram in flight. High velocity, large duct-work, low pressure differential. Liquefaction is basically the same principle but higher pressure, smaller pipes, lower velocities. I figure there is a relation between velocity and pressure, similar to amps x volts = watts. Pressure x velocity = ? Yeah, more than likely the "jump start" would have to continue indefinitely - LOL... Originally I was just working on trying to design a more efficient engine so as to be able to reduce the size of the parabolic dish that would be needed for a Stirling solar generator. Something more suited for a residential application. I was told that the engines used on those big Stirling solar farms in the desert require a refrigeration system to keep the engine cool and throw off the excess heat. Sounded very inefficient. I thought if I could find a way to improve the efficiency of the Stirling Engine then the size of the dish required could be reduced. Well,... It seemed I reduced it alright. Reduced it right out of existence. I spent a lot of time yesterday also boning up on Thermodynamics. There were some points I found interesting. Like, the way these Laws a written. i.e. "in an isolated system...", "in a closed system..." Well, this is an open system. So are these laws that apply to an isolated system actually applicable to this engine ? How is the entropy of the atmosphere effected when this thing starts drawing in warm air and ejecting cold air, or when it starts simultaneously powering various remote loads ? You know I'm looking for a loophole. I'm also wondering about how "change of state" works into these laws. Does some of the air change state in an air-cycle system ? Maybe condensing into a kind of vapor cloud temporarily. I hate to get into this on a "classical science" forum, but aren't these laws also based on the assumption that heat IS NOT a particle ? Yet isn't infrared light supposed to be the same thing as heat, and isn't a "photon" a particle ? I'm referring to what I read here: clip---------- "Perpetual motion of the second kind "Before 1850, heat was regarded as an indestructible particle of matter. This was called the “material hypothesis”, as based principally on the purported views of Sir Isaac Newton. It was on these views, partially, that in 1824 Sadi Carnot formulated the initial version of the second law. It soon was realized, however, that if the heat particle was conserved, and as such not changed in the cycle of an engine, that it would be possible to send the heat particle cyclically through the working fluid of the engine and use it to push the piston and then return the particle, unchanged, to its original state. In this manner perpetual motion could be created and used as an unlimited energy source." ------------- http://en.wikipedia.org/wiki/Second_law_of_thermodynamics Isn't this recycling of heat what is behind the idea of the Stirling Engine "regenerator". It is also (apparently) an element in this engine in another way. i.e. the heat used to expand the gas in the displacer chamber one time is then reused as the same gas just heated and expanded is compressed to provide heat to compress the next batch of air, which next batch of air is then compressed to extract the same? heat again in order to compress more air etc. etc. etc. My thought was that the "kinetic" energy or movement of the air, being forced through the narrow tube was being converted into heat due to a kind of friction - giving off infrared radiation that gets transmitted to the incoming air. Particle or radiation, It seems it would amount to the same thing. The "heat" whatever it is is in effect being transferred and reused over and over again. Merged post follows: Consecutive posts merged Probably true, especially if for some reason the so-called "compressor" doesn't compress, then there will be NO energy. But if there is ANY, the displacer is very easy to move. I'm not sure what "vacuum" you are referring to. The turbine, theoretically, will run on compressed air and/or expanding air not a vacuum. I guess, relatively speaking, one atmosphere is a partial vacuum as compared with air in a compressor tank, but somehow I don't see this happening. If I take a compressor hose from an ordinary compressor tank and direct the nozzle through a heated tube, the heat on or in the tube is not going to cause the air to stop coming out of the nozzle - no matter how hot the tube gets. Well, the only real innovation here, as far as I can see, is using a temperature differential to operate a "compressor" with very little energy input, other than heat. If such a compressor works, I should think it would have numerous applications besides acting as the compressor for an air-cycle heat exchanger or refrigerator. An air compressor that runs, instead of on 20 Amps of household current - a flashlight battery and a heat source... It think that in itself would be something. But if that DOES work. Then the compressor output should be comparable to a regular compressor running on 110v or any other energy source. How could something with that kind of power output NOT be able to power something that could otherwise run on a flashlight battery, even if that other "something" is itself ? I don't know. But I'm not yet convinced that this couldn't work.
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