# Tom Booth

Senior Members

150

1. ## Heat engine experiments and 2nd law of thermodynamics.

In relocating the thread, an edit to the previous post was lost apparently. I had added: The 20% represents the potential drop in temperature for a "perfect engine" with no friction or other losses, in which case the working fluid heated to 212F can at the absolute maximum be brought down to the ambient temperature of 85F, not by rejecting heat to the sink, but by conversion of that heat to work. Below that point and heat would have to flow from the sink back into the engine. The remaining 80% represents the remaining heat, which if it could be utilized by the engine to produce work would bring the temperature of the working fluid down the rest of the way to absolute zero. My contention is, at or below the point of equalization, which in this case is 85F, no "heating the cold plate and the reservoir it's coupled to" is actually possible as this would require heat to flow across an equilibrium and simultaneously reduce the temperature of the engine to zero Kelvin, all in an instant. If that were to happen the engine would implode becoming a Bose-Einstein condensate or some such thing.
2. ## Stirling Turbine

That's fine. But can we please preserve a link to the continuation of this discussion, for anyone having followed it this far? https://www.scienceforums.net/topic/122721-heat-engine-experiments-and-2nd-law-of-thermodynamics/#comments
3. ## Heat engine experiments and 2nd law of thermodynamics.

That would be the conventional viewpoint. My hands on experience indicates otherwise. With the engine running on scalding hot water poured into a vacuum flask to prevent incidental heat loss, the cold plate remained cold. Then blocking the cold plate so heat could not escape to the reservoir, the plate, apparently became colder, though other possible explanations have been offered. As I said, I need to aquire some temperature probes to get an objective measure as I may simply be seeing what I want to see. If I fail to respond to any additional comments, I'm not ignoring them. This thread is moving too fast for me to keep up with and I need a break. I am very busy working on the house, pouring a new concrete foundation between posts. I will have to catch up here when that project is completed. I also have additional modifications to make and more engines to build and supplies to pick up, like the probes. Without proper "control" engines and such, and basic measuring equipment, discussing the "results" of experiments is inconclusive at best. I apologize for the interlude but I will be back. Many thanks to everyone for all the input, and feel free to carry on the conversation in my temporary absence. I'll catch up when I'm not so overloaded by other responsibilities.
4. ## Heat engine experiments and 2nd law of thermodynamics.

That is true, I apologize for any confusion, there were several different experiments, but a few posts back I suggested we focus on just my first experiment, where I ran the engine on hot water, then covered the sink or cold heat exchanger in an effort to block the "rejection" of heat to the sink. That post where I suggested focusing on that one experiment to begin with is linked above. My purpose is to report on the results of some experiments. I don't have questions about thermodynamics, though I welcome any suggestions regarding how the experiments might be improved. I have ideas about how Stirling heat engines work, from my personal experience in building them and doing independent research, which may be right or wrong, but which seems contrary to accepted theory. I predicted ten years ago that covering the cold plate of the engine would make the engine run better and improve efficiency which everyone I knew on the Stirling engine forum thought was preposterous. I wrote: "If more heat is extracted as work than what actually reaches the heat sink, then theoretically, insulating the cold end of the displacer chamber against the external ambient temperatures would improve engine efficiency." A few days ago I received my model Stirling engine in the mail and finally got around to running the actual experiment. The results were as predicted, which actually surprised me greatly. Generally, a theory that accurately predicts experimental results carries some weight in science, usually. But I'm not insisting my theory is correct. There may be other explanations, but my main intention is to simply put the results of the experiments out there and get some feedback.
5. ## Heat engine experiments and 2nd law of thermodynamics.

I've highlighted some statements here that are incorrect, or perhaps are using inapplicable terminology, that at any rate doesn't reflect the actual structure or operation of these engines. First of all, there is no diaphragm. I'm not sure what you assume to be, or are referring to as a "diaphragm". It is a piston engine. The large disk is not a diaphragm or piston but a "displacer". The displacer does not do any work "against atmospheric pressure", if by chance, by "diaphragm" you are actually referring to the displacer, which does not itself do any actual or appreciable work whatsoever. The engine does no work to expand and contract the volume of the working fluid. Heat itself, the heat applied to the engine effects expansion of the air trapped inside the engine. Heat expands the gas, or causes the gas to expand. The expansion of the gas drives out the piston which by means of a connecting rod, turns the crank. The "WORK", is performed by the expanding gas which is expanding due to the application of heat. "Working fluid", I believe, refers to the fact that the heated and expanding gas is what is "working" doing the work of driving the piston which in turn drives the entire machine along with any load. All of the work performed is performed by the expansion of the gas, or by atmospheric pressure after the gas spends it's heat energy and becomes cold. The displacer does virtually no work, rather it simply controls the intake of heat to the engine by periodically uncovering the hot plate heat exchanger. What moves the displacers position is again, the heat, expanding the gas and driving the piston which turns the crankshaft which controls the positioning of the displacer which brings the air or "working fluid" into periodic contact with the hot plate. There is no diaphragm. The Carnot heat engine efficiency formula does not take into account "work". Carnot had no concept of heat being converted into work. Nevertheless, the formula is still used and treated as accurate. (Just for the record, that does not reflect my own personal opinion). My personal opinion is that Carnot had no actual familiarity with how engines actually work. His Carnot engine, which is supposed to be "the most efficient heat engine possible", is an intellectual abstraction with no real world functionality, It is not "efficient". It's effective work capacity is zero. It requires outside assistance just to be moved from one reservoir to another and back, and outside assistance to apply weights and such, I have difficulty imagining how or why anyone ever took it seriously, never mind considering it "the most efficient engine possible". It's efficiency for converting heat into work is, taking a look at it myself, as an engine mechanic, exactly zero. It's completely non-functioning! The "science" of thermodynamics today, resembles a hodge podge of obsoleted nonsense. Hardly anyone alive today has any real practical knowledge of the actual inner workings and functionality of Stirling heat engines, and those who think they know are mostly wrong.
6. ## Heat engine experiments and 2nd law of thermodynamics.

The experiment in question is an engine running on boiling hot water, not ice, in an ambient environment at 85F. Let's not confuse things beyond what is necessary and treat one example at a time. Why? how can it? Heat flows from hot to cold. In a Stirling engine the working gas is sealed in the engine. Heat cannot just be exhausted as in an internal combustion engine. If the temperatures are equal there is no heat flow from hot to cold as there is no hot or cold. Heat engine efficiency is a mathematical calculation based on the formula Th - Tc / Th. at the temperatures in this experiment, that works out to 18.9% That maximum efficiency represents the point where heat is converted into work, bringing the temperature down to Tc. Tc is the temperature of the sink. It can't just be asserted that an 18.9% efficient engine is 100% efficient. heat flow cannot continue where the "heat rejection temperature" is equivalent to the sink. It is two different heat scales being conflated.
7. ## Heat engine experiments and 2nd law of thermodynamics.

Let's try this. In my experiment using boiling water (212F) at ambient temperature (85F), "maximum efficiency" can be calculated to be 18.9% if the engine "rejects heat" at a temperature of 85F. That is, it would have to extract enough work to bring the input temperature of 212F down to 85F output temperature. But, heat cannot be "rejected" at a temperature equivalent to the sink. Setting that fact aside for the moment we have the fact that my engine only needs to have a Carnot efficiency of 18.9% (not 100%) to no longer have a condition where heat is being "rejected". At 18.9% efficiency heat is no longer being rejected. Temperatures are equivalent, so no "flow" can result. An efficiency of a mere 19% would have heat flowing backward from the ambient sink back into the engine. Therefore, to say the engine must reject heat because no engine is 100% efficient is apples to oranges. It is not necessary to have 100% efficiency on an absolute scale, or bring the temperature down to absolute zero, rejection to the sink ceases at any efficiency >= 18.9%
8. ## Heat engine experiments and 2nd law of thermodynamics.

My point is that for an engine to be 100% Carnot efficiency it would have to cool the working fluid down to an impossible 0 K not Tc.
9. ## Heat engine experiments and 2nd law of thermodynamics.

Not really. The formula for heat engine efficiency is: Th - Tc / Th. On a scale that begins at absolute zero. So if a heat engine reduces the temperature of the heat from T-hot down to T-cold that is simply the percentage of cooling produced on the absolute temperature scale, or how far the engine has cooled the "working fluid" on the way down to absolute zero. It simply states that at best, the engine cannot reduce the temperature any lower than Tc. It derived from Carnot who conceived the "flow of caloric" as including all that caloric down to the absolute zero temperature. The "rejected" heat is the un-utilized percentage of caloric below Tc down to absolute zero, which Carnot conceived as also flowing through the engine. That is the actual original basis of this engine efficiency formula. That is, of course, my understanding after some rather extensive reading and research on the subject. So this "maximum efficiency" represents a potential for heat reduction or energy conversion which (theoretically), bottoms out at Tc.
10. ## Heat engine experiments and 2nd law of thermodynamics.

Recognizing this, I have put the word flow in air quotes. Thank you Bufofrog for that clarification. So can we all agree that "flow" is a figurative expression, that in this context refers to energy transfer?
11. ## Heat engine experiments and 2nd law of thermodynamics.

Let's look at this statement: "no conversion of heat to work will be 100% efficient." What does that actually mean in real terms? Can we agree that if the temperature of a heat engine is equal to the temperature of the surroundings, no heat will "flow" into the engine? Also, if the temperature of the heat engine is the same as the surrounding ambient temperature, then no heat will "flow" out of the engine?

13. ## Heat engine experiments and 2nd law of thermodynamics.

Of course. Possibly.
14. ## Heat engine experiments and 2nd law of thermodynamics.

I provided video ("worth a thousand words"). The engine is an unmodified (model JAJ 838 KIT) ordered here: https://www.stirlinghobbyshop.com/english/stirling-engine-solar-ltd-low-temperature-stirling-engine-ltd-stirling/ltd-stirling-engine-ja-828/#cc-m-product-9759217783 stock engine, except for replacing the steel bolts with nylon bolts to minimize parasitic heat loss. A comon practice with Stirling engine model builders. Hot water from the teapot about 212F presumably. It had just boiled The ambient room temperature was approximately 85F. Insulation consisted of Corning fiberglass, styrofoam packing the engine was shipped with, and on top, 1/4 inch foil face styrofoam "house wrap". Before placing the 1/4 inch foil faced styrofoam over the "sink" the RPM of the engine was a steady 162 RPM. Immediately after completely securing the top disk of styrofoam, the RPM climbed to 180 RPM and remained there for several hours. This is a very simple and straightforward experiment. I'm at a loss to imagine what more I could add that isn't already completely transparent. This is one typical course in thermodynamics: These engines are virtually hermetically sealed. The "working fluid" itself is entirely retained in the engine not thrown away or continually replaced in any way. Part Ii I cannot agree with this presentation because, if the engine were to convert enough heat into work to simply reduce the temperature 1 degree below ambient, heat would begin flowing backward. Heat always flows from hot to cold. In a room at 85 degrees F, that would mean the engine would only need to bring the temperature down from 212 F to 84 F by means of conversion to work, for the heat to begin reversing direction. (Not as presented, down to an impossible 0Kelvin or absolute zero where mater as we know it ceases to exist.)
15. ## Heat engine experiments and 2nd law of thermodynamics.

At this point, the focus of the discussion is, though the caloric theory is recognized as obsolete, there is a persistent remnant stated in a somewhat vague way in courses on thermodynamics that some heat must always be rejected to the sink in a heat engine. My insulation is not perfect, so indeed, it is within the realm of possibility that "some" heat is still getting through, but if the functionality of a heat engine depends in any way on such "flowing through" of heat, then insulating the sink is roughly equivalent to stuffing a potato in the tailpipe of a car. Or shutting off the outlet in the water turbine of Niagra falls. The. Expected result SHOULD BE, if flow through of heat is a functional necessity of the engine; a slowing down or stopping of the engine should be observed. We are not talking about MY engine at this point, the focus is just a stock model "off the shelf" Stirling engine readily available from several internet markets. We are talking about expected results of an experiment. How can it be explained that blocking the path of heat flow to the sink to any degree results in higher RPM and a longer run on a finite heat source? In this case a vacuum flask containing hot water. My story is just some background. The topic is the experiment. The results suggest there may be something wrong with the general "flow through" theory of heat engine operation. If not, why not? It seems a peculiar result, does it not? My supposition is that since a Stirling engine's efficiency increases with increased temperature differential, blocking the so-called "sink" in some way increases the temperature difference. My tentative explanation is that contrary to popular opinion, a Stirling heat engine does not actually always reject heat to the sink, but rather, under some circumstances; infiltration of heat into the engine (in reverse) through the sink reduces the temperature difference that the engine itself is maintaining by converting incoming heat from the source into "work", including momentum and then using that momentum for cooling by expansion, which brings the cold side temperature down below ambient. So an engine with an insulated cold side runs better by blocking reverse heat flow. For now the focus is on this experiment: Heat is delivered into the engine through the bottom. I then insulated the sides and replaced the bolts in an effort to minimize loses, so the heat can only escape the Dewar by passing into the engine and up and presumably out through the top, That way most of the heat must pass into or through the "working fluid". Now insulating the top, should block the flow of heat, causing a slow down, logically, shouldn't it? But it doesn't. Instead, my friend and I using a stop watch and counting the revolutions, fount that the engine ran at an RPM of +18 with the top insulated. The RPM increased by 18 revolutions per minute.
16. ## Heat engine experiments and 2nd law of thermodynamics.

To offer some context: growing up in the country there was always machinery breaking down which was very expensive to get repaired all the time, so in 10th grade (high school) my dad suggested I enroll in a vocational course for engine repair. There was some basic theory, but mostly we just tore down and rebuilt engines all day. After graduation I opened a repair shop out of my parents garage, then spent most of my life working in engine repair shops Fast forward to about 20 years ago, I was asked to help with designing a solar Stirling engine for a semi retired government contractor. I worked hard on this because it seemed like a big opportunity at the time, and I came up with what I thought was a great design that exceeded specifications. He ran it past some associates in, I think, the Dept. of Energy and their verdict was that it was impossible because it violated the second law of thermodynamics. So, instead of a plane ticket to California and a lucrative job, I was left with a rejected plan for a Stirling type heat engine. So, not knowing the reason why my design was supposed to be "impossible" and in violation of some law of physics I never heard of I casually began doing research on the subject in my spare time as a kind of hobby. Carnot's theorem was, as far as I know, the earliest formulation of the second law: https://en.m.wikipedia.org/wiki/Carnot's_theorem_(thermodynamics) Derived from the Caloric Theory of heat: https://en.m.wikipedia.org/wiki/Caloric_theory After that we have Clausius, Lord Kelvin (William Thompson) etc. This Second law (specifically as it is supposed to apply to heat engines) has gone through so many iterations I am not really able to pin down exactly what it is supposed to be saying. But, the long and short of it is that my engine design was a fusion of a Stirling engine and an air cycle heat pump. It did not use solar directly, but rather drew in atmospheric air to run through a compression/expansion refrigeration cycle which would produce the temperature difference to power the Stirling engine. As a combined machine, it had few moving parts and several shared functions. The thing that seems to have gotten under someone's craw was that it ran on "one reservoir", or a single heat source. Solar heated warm air, directly from the atmosphere. Such a thing has been declared "impossible" at various times in various ways, all such statements having been wrapped up together as valid statements of the second law. Specifically, this second law, in its early formulations seems very slippery, and I don't have a clear picture as to, does it apply to this "open system" or not. Now that I have some finances available, I've just gone and purchased six model engines to run some common sense tests and experiments to see if, in principle, my engine design is impossible or not. The first order of business, I thought, should be to settle this Caloric theory that a heat engine works by heat flowing through it from a hot reservoir THROUGH to a cold reservoir. Blocking the path or outlet from the engine is, according to Carnot's theory equivalent to stopping up the outlet of a fluid turbine. Without an outlet for the heat, according to Carnot, the engine should quit. In my experiment however, insulating the path or outlet of the model Stirling engine had the opposite result. The engine, rather than overheating and stopping or something, ran measurably faster and longer.

18. ## Help needed for pv=nrt calculation

On second thought, a 9 degree change on the C scale is more like 15 on the F. That seems a bit more significant somehow. Especially when taking into account the humidity factor along with the possible cumulative effect of repeated iterations in a running engine addressed above, it seems to me that the cooling effect could be quite pronounced. This video helps to explain the process of "Throttling" and why it results in cooling. Essentially kinetic energy is reduced and potential energy is increased as the air molecules are drawn further and further apart.
19. ## Help needed for pv=nrt calculation

If this is the case, such a slight drop in temperature seems rather insignificant. There may be another factor involved though: Humidity. I came across this discussion while looking up random references to Joule-Thompson Effect. http://www.clubfrontier.org/forums/f11/throttle-body-icing-74415/ It seems that under certain conditions automotive carburetors can ice up at the throttle due to Joule-Thompson Effect resulting in a run away engine - even under the hood where temperatures are rather hot. The result being that some manufacturers have installed heaters to circumvent the problem. One commentator on that thread states: "Carb icing isn't really a hot or cold weather issue, but rather a humidity/dew point issue. so keep in mind that as air flows thru the venturi part of the throttle body, there is a drastic localized drop in temperature that goes along with the drop in pressure." The line along which I'm thinking at this point is that although the temperature drop might be initially rather slight. Just a few degrees, the effect, as the engine continues to operate drawing more and more air through the valve, the effect might be cumulative. That is, a few degrees of localized cooling with the first stroke of the piston, a few more degrees with the second stroke, a few more with the third etc. until the cooling effect becomes significant. In other words, not just the air gets cold but also the valve itself along with all the surrounding metal in the vicinity of the valve, the valve housing, cylinder, top of the piston etc. which might get progressively colder the longer the engine runs. Correspondingly the air entering through the valve would tend to have more and more heat drawn out of it with each cycle. If the temperature drop were only .5 (1/2) degree per cycle, what might it be after 1 minute of operation in an engine running at say 500 RPM ? This is, of course, mere speculation, but if a carburetor in a car engine, under the hood, in a relatively hot environment with what could conceivably only be a very slight pressure drop at the throttle could freeze up the entire throttle body I would assume this would have to be due to a cumulative effect in a running engine. The temperature drop with each cycle might be nearly imperceptible but the cumulative effect over time could be significant. So, was I hoping for more of a change? I suppose. I'm mostly just trying to find some logical explanation for how this so-called "compressor" could operate as described in the patent. I would tend to simply dismiss the whole thing as manifestly impossible, if not for the fact that in an interview with the inventors son, he stated that his father did indeed have a working prototype as well as a small working model which he saw operating when he accompanied his father to the patent office. This Wiki reference was also cited in that thread: http://en.wikipedia.org/wiki/Carburetor_icing What I'm getting at is that in most scientific discussions about Joule-Thompson, such as you have cited above from "Chemical Thermodynamics" the discussion is limited to a strictly one time event. That is, air being passed through some throttling device from one chamber to another where there is no possibility for any cumulative effect as might be seen in a running machine where the air flows through the valve continuously or repeatedly in quick succession.
20. ## Help needed for pv=nrt calculation

This would be true. However this so-called "compressor", according to the patent, does not use high pressure compression. Supposedly the air is still at atmospheric pressure AFTER "compression". In other words, it seems, according to the patent that the compressor does not really compressing the air much at all. It mentions that the air, after being delivered to the tank is below freezing. Electric heaters are used to heat the air from the tank for use to run an air motor. The implication, though this is not explicitly stated in the patent, is that the air is being reduced in volume by cooling rather than compression. Water cooling could conceivably bring the temperature down to somewhere above freezing I would suspect, but some additional cooling would have to come from somewhere to get it down below freezing. This "throttling valve" is all I've been able to find that looks like it might have some potential to do that. At any rate, the patent does explicitly state that the pressure does not go above about 15 psi or atmospheric pressure.
21. ## Help needed for pv=nrt calculation

Are you suggesting then that the bottom intake valve is stronger to compensate for the poor seal around the connecting rod, and that the top has no connecting rod so can make use of a weaker spring. If so, I would say that makes no sense. A stronger spring on the bottom chambers intake would only create more vacuum aggravating the problem of air leakage through the connecting rod packing. I've work as an engine mechanic most of my life. In my experience, check valves using a little ball have very weak springs that can be easily compressed between the fingers. The valve at the top looks exactly like a typical automotive type or internal combustion type valve. Typically these valve springs are so strong a special valve spring compressor tool is required and they can be rather dangerous if they slip out of the valve spring compressor. One could put a hole in the ceiling or put out an eye. If you are saying that a stronger spring is needed at top because this is the main compressor and a better seal is needed to prevent leakage back through the valve. That would make more sense. What I'm generally looking at is references such as: "The throttling of the gas at this stage produces a forced expansion of the gases and condensable vapors, and simultaneously lowers the temperature in proportion thereof. The fall in temperature and pressure causes the entrained vapors to condense to liquid, and collect this condensable matter, the gases are caused to discharge into the separator b through nozzle c." From "Industrial fuels" by Joseph Stephenson pg 33. Or this extract from Wiki: " Helium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one atmosphere are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up when expanded at constant enthalpy at typical room temperatures. On the other hand nitrogen and oxygen, the two most abundant gases in air, have inversion temperatures of 621 K (348 °C) and 764 K (491 °C) respectively: these gases can be cooled from room temperature by the Joule–Thomson effect.[1] For an ideal gas, is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy." In other words, for an "ideal gas" which is what pv=nrt applies to, there would be no change in temperature from such "throttling" or forced expansion through a valve. Some gases, like helium would actually heat up but regular air being mostly nitrogen and oxygen would be cooled down. I was hoping there might be some way to get an idea how much cooling would be possible, or how much cooling could reasonably be expected, but it seems pv=nrt isn't going to be of any help as air is not an ideal gas. Perhaps there is some other formula or equation that would be applicable from some HVAC manual but I'm guessing that any equation relating strictly to an "Ideal Gas" is going to show no change in temperature.
22. ## Help needed for pv=nrt calculation

"Nobody said it was a two stage pump." OK, my mistake. I thought that was what you were implying. Because the valve on the lower left of the bottom part of the cylinder is also a (fresh air) inlet valve. "It may look that way to you, but it doesn't to me." No problem. you may very well be right. I appreciate your input. Thanks! OK, thank you all very much. I'll accept your judgements. What I don't get though is then why this elaborate description of adiabatic cooling on expansion from this website: http://hwstock.org/adiabat.htm And I think I've read other references similar to this over the years. Could it be because Air is not an "Ideal Gas" ?
23. ## Help needed for pv=nrt calculation

Really ? I was actually getting that result at one point using an online calculator, but I thought, that couldn't possibly be right. Increasing pressure raises temperature right ? So doing the opposite should result in cooling. Then I was reading how if you have a balloon with air in it and you carry it up a mountain, the balloon will expand and the air inside will cool. Bring it down and the opposite happens. Somehow this just doesn't seem like it could be right. Like a violation of the first law of thermodynamics not right. If I expand a gas to twice the original volume and its stays the same temperature, then what happens when I compress it back down again ? It would get hot right. If I just kept expanding and compressing the same air I could make it as hot as I wanted. BTW, I'm not doubling the volume by adding more air. Just to be absolutely clear on this and make sure we are both thinking in the same terms. What I'm talking about is this. Words can be tricky sometimes. Like by "pressure" I could mean inside pressure or outside. "Volume" could mean amount or size etc. so just to be clear. these are the conditions: The cylinder contains a piston with an air tight seal. No air can get in or out. There is a fixed amount of air in the cylinder. Press down on the handle to raise the piston. If the starting conditions are known, is it impossible to determine the result ? The gas inside the cylinder is 1 atmosphere, (about 14.5 Psig) and 60 F. call it one pint. That is before pushing down on the handle to expand the air to twice the volume. There is no air being added. Just mechanically expanded to increase the volume. Does it really stay the same temperature ? Just double checking. Well, none has been compressed yet in the other half of the cylinder either. There, just check valves are used. The compressor is double acting NOT two stage. That is it compresses air on both the up and down stroke. but always new air. Anyway, this one valve is accessible, The check valves are not. So, it could be made adjustable, or could be replaced. Perhaps the inventor built it that way so he could experiment with different spring strengths or make adjustments while the engine was running.
24. ## Help needed for pv=nrt calculation

By "doubling the volume" I'm just adapting an arbitrary scenario. The cylinder holds X volume of air. It admits only 1/2 that potential due to the spring being "medium" strength. Yes that depends on the strength of the spring for sure. However, this is purely hypothetical so I could make the spring tension adjustable and turn the tension wing nut or whatever until that's the way it is. I could increase the tension and then say tripling the volume and work it out from there. In that case the even stronger or tighter valve would only allow 1/3 X into the cylinder. I'm treating this AS - IF the amount of air entering the chamber were controllable and the action of the valves opening and closing were controllable. In that case, you open the valve fully. Draw the piston down 1/2 way then close the valve completely and draw the piston down the rest of the way. Now you have expanded the air to twice the volume. I think it can be assumed that under such hypothetical controlled circumstances, doubling the volume of a gas would halve the pressure. By looking at it. (in the patent drawings above) By comparison, the spring on the inlet valve (#44) is at least 4 times bigger than the springs on the other valves (#43). Bigger does not necessarily mean stronger but I think it's a fairly safe assumption. This is a double acting compressor. The other three valves are all regular "check valves". Check valves are generally designed to offer as little resistance as possible so have very weak springs, just enough to hold the valve closed. In a normal compressor, there would be no need for anything else. The one valve on top however is obviously different. Much bigger with a much larger spring. I think it can be assumed that this one valve serves some special purpose. Of course I can't take it out of the patent drawing and test it to see if it is stronger than the other springs but it sure looks stronger to me judging by what I can see in the drawings. Don't you agree ?
25. ## Help needed for pv=nrt calculation

Well, using various online sources / calculators / charts etc. I've come to the conclusion that for Air, there would be a 10 degree F drop in temperature (approximately) for every loss of 1 psi due to expansion (or pressure reduction). Starting at 14.5 psi at sea level (14.6488 - 14.6950 actually, according to different sources). Would not expanding that to twice the volume (whatever the original volume) result in also cutting the pressure in half ? i.e. to about 7 psi ? So expanding air to twice its volume would result in a drop in temperature of about 7 X 10 or 70 degrees F. So starting at 60 degrees Fahrenheit, in the above example, (60 F 1 atm expanded 2X) the resulting temperature would be about -10 Fahrenheit after expansion to double the volume. I'm not arriving at any of this from pv=nrt directly, though some of the sources used it. Rather this is based on pressure changes due to elevation in the atmosphere. As you climb a mountain the pressure drops. About .5 psi per 1000 ft of elevation. (or 1psi per 2000 ft) and according to the sources, a corresponding temperature drop of about 5 F per 1000 ft. Primary source: http://hwstock.org/adiabat.htm also: http://www.unitarium.com/pressure The thing at this point I'm not really sure about is, am I correct in assuming that doubling the volume of a fixed amount of air would halve the pressure ? Seems to follow from basic logic but... Is it necessary to use "absolute pressure" rather than psi ? Is there a difference ? I'm a little fuzzy there. I'm also assuming that pressure and temperature drop due to elevation would correspond to temperature and pressure drop due to mechanical expansion, which may be wrong but I have little else to go by at this point. Anyway, if anyone can come up with a refutation, correction or improvement in regard to these basic conclusions of rough estimates I would certainly appreciate any additional help or insight. Thanks!
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