Tom Booth Posted February 17, 2010 Posted February 17, 2010 No, I mean something quantitative. pV = nRT 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".
SH3RL0CK Posted February 17, 2010 Posted February 17, 2010 ...At any rate, I don't know as any useful information could be derived from pV = nRT. if you don't think this equation is useful then I don't think we can help you with your questions. Study up on this equation, Sayonora is showing you where you are wrong.
Mr Skeptic Posted February 18, 2010 Posted February 18, 2010 You should know how to convert units. A pressure times an area, gives you a force. A pressure times a change in volume gives you energy. PV=nRT gives you any of the variables in that equation, given the others. Add to that the specific heat, and you have all you need to do your calculations (well, if you know the math, that is).
Tom Booth Posted February 18, 2010 Posted February 18, 2010 ... Sayonora is showing you where you are wrong. Wrong about what ? Merged post follows: Consecutive posts merged...PV=nRT gives you any of the variables in that equation, given the others... 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.
Mr Skeptic Posted February 18, 2010 Posted February 18, 2010 You are making things far more complicated than necessary. Why do you insist on having a turbine efficiency lower than 100%, when your idea would still fail with a 100% efficient turbine, but the math would be easier? You can just extrapolate that the less efficient turbine will fail all the more. Why go through all the trouble? Just look at the work done on/by the gas, and the heat transfers. As for the variables, you should know all the others. For example you know the temperature, pressure, and volume of your input air. Whenever you change it, you keep track of how you change it. It's pretty simple.
Tom Booth Posted February 20, 2010 Posted February 20, 2010 ...For example you know the temperature, pressure, and volume of your input air. Whenever you change it, you keep track of how you change it. It's pretty simple. 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 ?
npts2020 Posted February 21, 2010 Posted February 21, 2010 TomBooth; Once you remove the candle from the Stirling compressor in your link, it will stop working. The same will happen in any compressor once the mechanism of whatever pressure/temperature differential you have is removed. You can have a gadzillion atmospheres of pressure differential to begin with but as soon as you take away whatever it is that you used to get that high differential to begin with, you begin losing that pressure and eventually the machine will stop.
Mr Skeptic Posted February 21, 2010 Posted February 21, 2010 (edited) 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, You can maintain generality by starting with a variable as your temperature differential. Eventually, you replace that with the temperature of your gas when it reaches that stage. If the temperature differential increases or stays the same, you win. 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 ? From what I understand, it will work, albeit inefficiently. Heating and cooling a solid is a good way to waste energy. However, feel free to assume that the solid takes no energy to heat/cool, ie is paper-thin, massless, frictionless, etc. Do not, however, assume that the gas is like this, since you can't "optimize" the gas in the same way. The amount of compression will depend on the cold and hot temperatures. The new pressure will, I think, be a increased by a percentage dependent on the the hot and low temps, from its original pressure which I guess is atmospheric. The volume will depend on the volume of your compressor. For example, we can look at two chambers of 10 liters each. You have a volume of 20 L at the cold temperature, Tc and at atmospheric pressure (1 atm). Now you increase the temp in chamber 1 to Th, while at maintaining the second at Tc (Or, leave the second chamber insulated, you will get less compression though). Use the ideal gas equation here; the increase in temp increases the pressure in chamber 1, forcing some warm gas into chamber 2. Now, the overall pressure will increase, and the amount of gas in each chamber will change (that's the n value of the ideal gas equation, stands for moles). So, first we can calculate the number of moles in each chamber, [math]n_0 = \frac{(1 atm)(10 L)}{RT_c}[/math]. By lowering the cold temp you get more gas to start off with. That will be in each chamber. Now, we increase the temp in chamber 1 while allowing gas to escape to chamber 2, so that the pressure between them is equal. The amount of gas is constant at [math]2n_0[/math], but it will no longer be evenly distributed. In chamber 1 and 2 you have, [math]\frac{P(10 L)}{n_1RT_h} = \frac{P(10 L)}{n_2RT_c}[/math], so that pressure cancels out. Now solve for [math]\mbox{compression}=\frac{n_2}{n_1} = \frac{T_c}{T_h}[/math]. So that is the amount of extra gas in the second chamber. If you have multiple chambers, you now close off chamber 1 and connect chamber 2 to chamber 3. After running this for some time, chamber 3 will be at the same pressure as chamber 2 at this stage. The same equations will apply, and your compression will be [math]\mbox{compression}_{total}= (\frac{T_c}{T_h})^{\mbox{number of chambers} - 1}[/math] (that's to the power of). Now, things will get a little complicated and you'll have to use calculus to calculate the heat flow because your variables are changing constantly. More heat flow will be needed in the latter chambers as there will be more gas in them, but the proportion should be the same. However I have to go now. I'll get this later. Edited February 21, 2010 by Cap'n Refsmmat fixed latex for you
Tom Booth Posted February 21, 2010 Posted February 21, 2010 TomBooth; Once you remove the candle from the Stirling compressor in your link, it will stop working... Naturally. But otherwise, given the heat input from the candle, you do believe this "compressor" would work. (?) Merged post follows: Consecutive posts mergedFrom what I understand, it will work, albeit inefficiently... 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
npts2020 Posted February 22, 2010 Posted February 22, 2010 Naturally. 1)But otherwise, given the heat input from the candle, you do believe this "compressor" would work. (?) Merged post follows: Consecutive posts merged 2)Now suppose we try this: Stirling Compressor II 1)Possibly, but to extract useful work will require far more input. 2)This will stop eventually from the ice melting for the same reason as removing the candle in the first one. I would have taken a very long time to come up with the equations Mr. Skeptic gave you but it would be well worth your while to at least understand how they are derived and what they are measuring. The only thing you have to be careful of is including all of the variables.
Mr Skeptic Posted February 22, 2010 Posted February 22, 2010 Right, and the equations I just did don't by themselves tell you anything about the efficiency. For that, you'd need the heat flow in addition, which if done as described would be calculus based. If you had trouble with the simple equations, I guess you wouldn't have a clue about calculus. If you want simpler math, you can modify your compressor to work on closed chambers, which are then opened and quickly closed with no heat transfer occurring in between. This will maintain the variables constant and so have easier math, but will be more inefficient.
Tom Booth Posted February 23, 2010 Posted February 23, 2010 1)Possibly, but to extract useful work will require far more input. 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. 2)This will stop eventually from the ice melting for the same reason as removing the candle in the first one. Of course. I would have taken a very long time to come up with the equations Mr. Skeptic gave you but it would be well worth your while to at least understand how they are derived and what they are measuring. The only thing you have to be careful of is including all of the variables. 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...From what I understand, it will work, albeit inefficiently. Heating and cooling a solid is a good way to waste energy. However, feel free to assume that the solid takes no energy to heat/cool, ie is paper-thin, massless, frictionless, etc. Do not, however, assume that the gas is like this, since you can't "optimize" the gas in the same way. 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") ? ...For example, we can look at two chambers of 10 liters each. You have a volume of 20 L at the cold temperature, Tc and at atmospheric pressure (1 atm). Now you increase the temp in chamber 1 to Th, while at maintaining the second at Tc (Or, leave the second chamber insulated, you will get less compression though). Use the ideal gas equation here; the increase in temp increases the pressure in chamber 1, forcing some warm gas into chamber 2. Now, the overall pressure will increase, and the amount of gas in each chamber will change (that's the n value of the ideal gas equation, stands for moles). 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) Now, we increase the temp in chamber 1 while allowing gas to escape to chamber 2, so that the pressure between them is equal. 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. The amount of gas is constant at [math]2n_0[/math], but it will no longer be evenly distributed. In chamber 1 and 2 you have, [math]\frac{P(10 L)}{n_1RT_h} = \frac{P(10 L)}{n_2RT_c}[/math], so that pressure cancels out. Now solve for [math]\mbox{compression}=\frac{n_2}{n_1} = \frac{T_c}{T_h}[/math]. So that is the amount of extra gas in the second chamber. If you have multiple chambers, you now close off chamber 1 and connect chamber 2 to chamber 3. 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.
SH3RL0CK Posted February 23, 2010 Posted February 23, 2010 Naturally. 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. If you are serious about this idea, it behooves you to study and learn this science at least as well as Mr. Skeptic. I would suggest you enroll in an engineering college somewhere. That you lack an understanding of PV=nRT indicates to me that you have a lot to learn about this field. 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. Other than from yourself, who exactly is confident this idea will work? Everyone except yourself (and the fine people at Kender, apparently) on this thread are quite confident this will not work. The people thinking this will not work have backed up their beliefs by mathematical equations. The people thinking it will work have a more nebulous arguement that I, for one, am not buying. The math does not lie.
npts2020 Posted February 23, 2010 Posted February 23, 2010 TomBooth; Increasing temperature/pressure differential is input. Whether it is more or less is not what is relevant. What is relevant is that you account for that input in your equations. This is what will cause you to get less energy out than you put in, like any other engine. I did not watch the video for lack of patience (vids generally, not just that one) but the last reference is using solar energy as the input I have just described, look at the picture and description. It is much different from what we have been discussing.
Tom Booth Posted February 23, 2010 Posted February 23, 2010 ...That you lack an understanding of PV=nRT indicates to me that you have a lot to learn about this field. I have a lot to learn about a lot of things. Other than from yourself, who exactly is confident this idea will work? Everyone except yourself (and the fine people at Kender, apparently) on this thread are quite confident this will not work. 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. The people thinking this will not work have backed up their beliefs by mathematical equations. The people thinking it will work have a more nebulous arguement that I, for one, am not buying. The math does not lie. 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.
SH3RL0CK Posted February 23, 2010 Posted February 23, 2010 TomBooth, Thanks for clarifying the consensus on workable and non-workable designs. I should add the design of heat pumps and the like are quite complex. More complex than can be adequately addressed here. Again, if you are serious about this, enrollment in an engineering college would be advisable. I think perhaps I did get your idea confused with the Kender engine...your plan certainly has more merit than their design.
Tom Booth Posted February 23, 2010 Posted February 23, 2010 (edited) TomBooth; Increasing temperature/pressure differential is input. Granted. I just thought I needed to make a distinction between heat input and electrical or mechanical input. Whether it is more or less is not what is relevant. What is relevant is that you account for that input in your equations. This is what will cause you to get less energy out than you put in, like any other engine. I did not watch the video for lack of patience (vids generally, not just that one) but the last reference is using solar energy as the input I have just described, look at the picture and description. It is much different from what we have been discussing. 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 mergedTomBooth, Thanks for clarifying the consensus on workable and non-workable designs. I should add the design of heat pumps and the like are quite complex. More complex than can be adequately addressed here. Again, if you are serious about this, enrollment in an engineering college would be advisable. That would probably be fun I suppose. Not likely to happen at my age though I'm afraid. I think perhaps I did get your idea confused with the Kender engine...your plan certainly has more merit than their design. 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. Edited February 24, 2010 by Tom Booth Consecutive posts merged.
npts2020 Posted February 24, 2010 Posted February 24, 2010 TomBooth; It is not really correct to equate Coefficient of Performance with Efficiency. However, I see what you are getting at about a heat pump or drinking birdy (which still requires a heat sink and source to work). The problem is that they both operate at such a narrow range of relatively low temperatures that extracting work in this manner is always a pretty inefficient enterprise. I say build a model. If a small scale machine works, it can always be scaled up. Even a slightly more efficient motor would be a worthwhile thing to have invented but I am very skeptical about being able to do it using the method described.
Tom Booth Posted February 24, 2010 Posted February 24, 2010 TomBooth; It is not really correct to equate Coefficient of Performance with Efficiency. However, I see what you are getting at about a heat pump or drinking birdy (which still requires a heat sink and source to work). The problem is that they both operate at such a narrow range of relatively low temperatures that extracting work in this manner is always a pretty inefficient enterprise. 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) "Any gas, when compressed, rises in temperature. Conversely, if it is made to do work while expanding, the temperature will drop." Reference - Refrigeration and air-conditioning By A.R. Trott, T Welch - pg 26 "The process" (of gas liquefaction) "has been considerably cheapened by the work of Kapitza. In the ordinary process air is highly compressed and cools itself by suddenly expanding. Kapitza saw that it would lose much more heat if it was made to do work while expanding, and designed a turbine to operate at liquid air temperatures to turn as much heat as possible into work. Today the Soviet Union leads the world both in the theory and practice of gas liquefaction." SCIENCE ADVANCES by J. B. S. Haldane pg 209 The story then is one of triumph after triumph. Oxygen was liquefied by Olszewski in 1883, and two years later nitrogen and carbon monoxide. The list of lowest temperatures as they were reached, reads like the records of athletic achievements; in both the next step is always the harder to achieve than the one before. A new principle was applied to produce the cooling. If a gas under pressure, below the inversion temperature, is allowed to expand freely through a fine nozzle or if it is allowed to do mechanical work while expanding, it will cool, and may reach a temperature at which it liquefies. Thus we get two methods for liquefying the gases. One employs the Joule-Thomson effect, which allows the gas to expand suddenly and so produce the necessary cooling by the performance of internal work; the linde and Hampson processes depend upon this principle. The other processes employ the reversible expansion of the gas with the performance of external work; the Claude and Heylandt processes are of this nature." Production and Measurement of Low Temperatures By K.R. Dixit No. 12 June 1938 "Another way to cool a gas is to have it do work adiabatically" (without heat transfer) "against a piston in an engine, and this has no temperature boundary like the Joule-Kelvin effect. Gas liquefaction using such engines is now common not only for helium, but also for air. These liquefaction plants are small and easily handled, so that every lab can have its own source of liquid air" Helium - http://mysite.du.edu/~jcalvert/phys/helium.htm 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 say build a model. If a small scale machine works, it can always be scaled up. Even a slightly more efficient motor would be a worthwhile thing to have invented but I am very skeptical about being able to do it using the method described. 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.
npts2020 Posted February 26, 2010 Posted February 26, 2010 TomBooth; A turbine puts out a constant amount of work for a given gradient across it, that's why they usually have some form of throttle. The killer in the design, however, is the compression (air?) to run a turbine. In every design you have shown, there is still an external input to compress the gas going across the turbine blades. The laws of thermodynamics are pretty persistent about not letting you get more work out of that compressed gas than it took to compress it in the first place.
Tom Booth Posted February 26, 2010 Posted February 26, 2010 TomBooth; A turbine puts out a constant amount of work for a given gradient across it, that's why they usually have some form of throttle. The killer in the design, however, is the compression (air?) to run a turbine. In every design you have shown, there is still an external input to compress the gas going across the turbine blades. The laws of thermodynamics are pretty persistent about not letting you get more work out of that compressed gas than it took to compress it in the first place. 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.
lazygamer Posted March 24, 2010 Posted March 24, 2010 what all of you seem to have forgotten is that zero Kelvin is the actual bottom temperature, everything above that is kinetic energy regardless of if its below or above the freezing point of water. the point about the kender-engine isn't the pressure, but the pressure/temperature relation, ie, you put 20psi gas at 73 degrees kelvin or so, lets it heat up to over three times the temperature, that is 273 kelvin or whatever temperature there is in the air (plus what the sunshine provides, though it heats up the atmosphere 24/7 so its ultimately solar nevertheless), and the pressure goes up in relation to the temperature to 200psi or something like that. then you run it through a turbine and a sort of a pressure-decrease valve and the temperature goes down to 1 or 5 or 20 psi or whatever and an equally low temperature, then it gets pumped into the radiator again at 20psi at 73 kelvin and heats up to the temperature outside and the pressure goes up. all you get is a little cool air (or cool to us, since there's absolutely no difference in the type of heat be that at 73 kelvin or 373 kelvin (373K=boiling point of water)). btw, heat only travels to molecules where there are less heat, a cool wind doesn't cool you down, you warm it up. (the kinetic energy in your molecules hit the molecules in the air that have less kinetic energy and transfers some of its kinetic energy to the molecules in the air) so no matter how you look at this, its far more effecient than any heat-engine that uses temperatures above the surrounding temperature to function (water needs to be aproximately 100 degrees above 273 kelvin to boil, so loads of the heat is going to warm up the air around the system, but if you use a medium that is liquid or gasious 100 degrees below the temperature around it, you can create loads of energy because everything around the system heats the system up). I'd think this forum would know these things...
Sayonara Posted March 24, 2010 Posted March 24, 2010 I'd think this forum would know these things... Do you have any idea how much work is required, in energy terms, to cool a fluid to 73 Kelvin? You are talking about -200°C. When you expend energy to cool down the fluid you have to take that as being a cost of the system you are using the fluid in. In other words, while you are correct that temperature differentials can drive turbines, in this case it would be additional energy supplied from an external source in order to force the system to work. That somewhat defeats the claims made by Kender.
lazygamer Posted March 24, 2010 Posted March 24, 2010 liquid nitrogen costs the same as milk these days, and once you put nitrogen or helium or whatever in it at 73 kelvin you don't loose the energy, you top it up maybe once or twice a year with more gas, but once you put it in at 20 psi and 73 kelvin and it heats up, the pressure goes up to 200psi, which doesn't escape (once you have put the cool gas in it you can allways take out the same gas at the same temperature if you want to). you don't need an external source, you only need liquid nitrogen, and if you take 39kw and multiply it by 8760 (number of hours in a year) you produce so much energy you could produce thousands of litres of liquid nitrogen. (and it isn't particulary energy-consuming to produce nitrogen, its 80% nitrogen in the air so you only use air, run it through a pressure-decrease valve and use the cooler air to cool down the hose before the pressure-decrease valve etc. dont forget that the air we breathe is at 14 psi, so no problems making cool air, its just that such a system is somewhat too big to fit in a refridgerator) btw, also take the 39x8760 answer and multiply it by the electricity-prices around where you live, then take out the cost for a litre or ten of milk (nitrogen), not exactly a huge expense. Most of the expense is probably going to replacing the valves and turbine etc, its not a 25 year-do-nothing like PV, but PV doesn't produce energy 24/7 in any weather. (btw, technically you dont need energy to cool something down, you just need to let the molecules you are trying to cool to give their energy away to other molecules)
npts2020 Posted March 24, 2010 Posted March 24, 2010 lazygamer; Have you ever tried to cool something down to the temperatures required to have liquid nitrogen? It requires a *lot* of energy, way more than an air conditioner uses and if you understood how an air conditioner worked, you would see why it takes so much energy to cool something. I have never seen a process that will cool anything to the temperatures you are talking about that did not require major expenditures of energy, despite the "simple" process of only having to lose internal molecular energy. The system you are describing will require more energy to operate than you will ever get out of it.
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