sethoflagos

No it isn't. You can't play the victim card when you're the aggressor. I'm simply calling you out. No. You continually misdirect the discussion by ignoring central themes and cherry picking peripheral trivia to have a snipe at. You're bringing absolutely nothing to the table to support your views other than blind persistence. Call it belief, faith, or just plain trolling: the one thing it isn't is science. Your opinion carries less weight here than you imagine. I see no point in trying to share knowledge with someone who has no interest in it. What are you doing here, Tom?

A deliberate refusal to acknowledge the points of view of others when they challenge with your own 'faith'. Deliberate misdirection of the discussion and wrong. The vast majority of collisions are not head on. Kinetic energy gets shared out. There is a nett flow of heat to the cold sink. Conspiracy theory. Irrelevant historical detail. You really don't have the slightest interest in being enlightened do you? Why exactly are you here?

You could start by doing a literature search for current work referencing these citations: Milgram, J. H. 1965 Compliant water-wave absorbers. M.I.T. Department of Naval Architecture and Marine Engineering Report no. 65–13. Ursell, F., Dean, R. & Yu, Y. 1960 Forced small amplitude water waves; a comparison of theory and experiment. J. Fluid Mech. 7, 33–52.

Because it didn't "GO MISSING"! 500 kJ of heat @ 500 K has been transformed to 100 kJ of Shaftwork plus 400 kJ of heat @ 400 K It's still there but at a lower temperature. Add a second machine operating between 400 K and 300 K and that 400 kJ of heat @ 400 K can be further transformed into another 100 kJ of Shaftwork plus 300 kJ of heat @ 300 K On some space base for example with ready access to a cryogenic heat sink, you can potentially recover almost all of the original 500 kJ of thermal energy. Just recognise that extracting the kinetic energy from slow moving particles requires contact with particles that are almost stationary. I think one of the main issues is a very confusing terminology. In the above example we have one (ideal) machine with a 'thermal efficiency' of 0.20 coupled to another with a 'thermal efficiency' of 0.25. Normally when we combine various efficiencies in series we multiply them together and would expect an overall efficiency of 0.05. And yet here we get an overall 'thermal efficiency' of 0.40. It sends a very odd message. To a practical layman, a low efficiency figure implies some failings in the design that can be incrementally improved with a little attention to detail. This is an entirely understandable viewpoint to take. But the above examples are idealised, 'perfect' machines with no avoidable losses. By any reasonable definition they should be classed as 100% efficient. As they would be if instead of W/QH we accepted the Carnot limit for what it is and used Machine Efficiency = W/QH x TH/(TH - TC)

It is not supposed to be a practical machine. It represents a limiting case that all real world machines will fail to match in performance. Just as the speed of light is the limiting case for particle velocity. Your statement shows that you understand at least some of these words. And yet you persist in claiming that you can produce a machine that outperforms unity efficiency. Apparently you do not understand the meaning of the word 'limit'. Pity because you've reached one of mine.

Try not to get distracted by extraneous detail. Focus on the fundamentals such as this post.

A cold reservoir is as much a potential energy source as a hot one. It's the existence of a temperature differential that produces the opportunity to output shaft work. In some sense cold sinks are more useful in that they're associated with higher thermal efficiencies. Air cycles are not particularly part of my repertoire since we tend to look for high energy densities in our working fluids to keep the equipment compact. Phase change latent heats store far more energy per unit volume than just gas heat capacity as used in the typical air cycle HVAC systems used on aeroplanes. But aeroplanes have access to very low temperature ambients and ram air compression so horses for courses.

Reading between the lines as well as I'm able, I don't think they quite got this right. Right from day 1 of my chemical engineering course, we were bombarded with the mantra: Output = Input - Accumulation ... until it was ingrained into us as second nature. Until we didn't need to consciously think about it - we just sensed it automatically. The reason being that it was vital to understand that while an output can be maintained without input via a depletion of reserves (-ve accumulation) this cannot be maintained indefinitely. Sooner or later the reservoir will run dry. In a complex system, the balance between these three quantities can be impossible to judge without precise calculation, and it can be oh so easy to let a slightly imprecise mental image carry you off on a fool's errand. It seems to me that you could see that the system kept running through it's accumulation of a high thermal reservoir and thermal inertia, and didn't spot the depletion of resources that was obscured by the systems complexity. Not only is this quite understandable, but it is actually a breach of the 1st Law not the 2nd. The correct response would have been to yes, point out the 1st Law transgression, but then acknowledge the value of extending power output into nighttime or at least say that wasn't a necessary consideration. (It has serious global significance now). I don't know. Without full knowledge of all the details maybe there was a 2nd Law transgression, but by saying so it certainly seems to have made you see the 2nd Law as your enemy and in doing so they did you a great disservice. The 2nd Law is very much the engineer's friend. A proper understanding of it keeps you on the straight and narrow and saves wasting time on nonsense. Which is a good thing. Carnot was one of the good guys.

Maybe let's return to more familiar ground with Stirling type heat engines. Imagine a stack of six of them. The top absorbs 300 kW of heat from the atmosphere at 300 K (80.3 oF), the bottom has a cold side at absolute zero. In between, the cold sink of the upper unit is the hot source for the next one down (and vice versa). Let us say each is designed for a temperature differential of 50 K (90 oF) and performs at the full Carnot limit for their respective temperature band. Machine 1: Temp Range 300-250; Input 300 kW; Efficiency 1/6; Output 50 kW Machine 2: Temp Range 250-200; Input 250 kW; Efficiency 1/5; Output 50 kW Machine 3: Temp Range 200-150; Input 200 kW; Efficiency 1/4; Output 50 kW Machine 4: Temp Range 150-100; Input 150 kW; Efficiency 1/3; Output 50 kW Machine 5: Temp Range 100-50; Input 100 kW; Efficiency 1/2; Output 50 kW Machine 6: Temp Range 50-0; Input 50 kW; Efficiency 1/1; Output 50 kW Can you see the beautiful symmetry in this? 300 kW of heat goes in and becomes 300 kW of Work. If Machine 1 were some how able to extract 50.1 kW, the symmetry would be irredeemably broken. We could potentially get more energy out of the system than was there to begin with. You really are at odds with the most fundamental of our scientific understandings and the countless experimental datapoints that confirm those understandings. Edit: I've just seen your next post and sympathise. I can't comment on that particular application as I'm not familiar with it. But I've worked for that kind of employer (also in Iraq as it happens) and they can have a very nasty side to them.

It's more like a commercial freezer room than a domestic fridge, perhaps. They run typically from -15 oC (5 pF) down to maybe -35 oC (-31 pF). No. Heat is being exhausted into the (slightly cooler) atmosphere through the condenser. Yes. We are extracting QC from the (insulated!) freezer, adding the nett shaft work, and exhausting QH to atmosphere. Understood. My designs were more more scaled for larger industrial applications than perhaps you are familiar with. They all worked moreorless as intended. And the same principles hold at all scales. It's self-adjusting to a certain extent. In hot ambient conditions the cold side temperature will rise until a new equilibrium temperature profile is established. The 50K (90 oF) temperature differential is built into the equipment design, but the absolute temperatures will 'float' up and down as necessary on the external temperature seen by the condenser. In practice, we would design for the warmest anticipated ambient conditions, and provide additional controls (maybe an inverter variable speed drive on the compressor for example) to optimise year round performance. But this isn't the place to go into those details. Just read 'compressor' or 'expander' as appropriate. The exact type (turbine, positive displacement, screw etc.) is not relevant to this discussion.

@Tom Booth The working fluid, propane in this case, flows around a circuit ABCD It is arbitrary which point we start at, but let's say it's the inlet to the compressor. At this point A I picked a pressure of 2 bar and an enthalpy of 550 kJ/kg. This corresponds to a temperature of -22.7 oC, entropy of 2.427 kJ/kgK and quality q = 1 (mass fraction in vapour form) Path AB is isentropic compression to 10 bar. The entropy at point B is 2.427 kJ/kgK when the enthalpy is 626 kJ/kg, temperature 38.5 oC and q = 1. The work input required for this stage is WC = 550 - 626 = -76 kJ/kg Point C is after heat exchange to atmosphere, still at 10 bar, temperature now 26.9 C, enthalpy 275 kJ/kg, entropy 1.257 kJ/kgK and q = 0.014 (nearly all liquid). Path BC implies an ambient temperature of a little less than 26.9 oC which is the fixed point for the system. It will ride up and down on this according to climatic conditions. Under the stated conditions we have an exhaust to the hot sink QH = 275 - 626 = -351 kJ/kg. Path CD is isentropic expansion down to 2 bar, so entropy remains 1.257 kJ/kgK, temperature -25.5 oC, enthalpy 260 kJ/kg and q = 0.298. This stage generates a work output WE = 275 - 260 = 15 kJ/kg. Path DA is a constant pressure (nearly isothermal) cooling of a refrigerated space. The temperature at A constrains this space to be a little above -22.7 oC. QC = 550 - 260 = 290 kJ/kg. You seem to have difficulty with my use of the word 'cycling'. It refers in this case to the changing state of a 1 kg mass of propane cycling around and around the system. In a real practical system, we would probably lose the turbo expander in path CD (too complicated and problematic for such a small efficiency saving) and replace with a simple adiabatic flash nozzle.

There's an interesting interactive site at https://www.flycarpet.net/en/phonline where you can study refrigeration cycles for various refrigerants. I had a little play this morning with a propane (R290) based cycle between 300K and 250K, and pressures 2 bar and 10 bar. Assuming isentropic compression/expansion I got QH = 351 J for a work input of 61 J (scaled to 1 kg of propane traversing the full circuit). ie a thermodynamic advantage of very nearly 6:1. To put this in context with the OP, we can see that the Carnot limit for a heat pump operating between these temperatures is 0.1667 or 1:6 exactly. This is no coincidence. If we operate an ideal Carnot cycle between hot and cold sinks generated by the refrigerant cycle, from QH = 351 J we can potentially generate a work output of 351/6 =58.5 J. Not quite enough to run our refrigerator. So no free power. If the Carnot limit was only a little bit of an underestimate, we'd have over unity-machines and all that nonsense. Have a play. See if anyone can get a thermodynamic advantage above 6:1 (and break the universe as we know it)

I remember the Brownian ratchet being comprehensively debunked by Magnasco 30 years ago, which was 30 years after Feynman's original debunking. Is someone trying to resurrect the long dead?

'Heat utilisation' is a bit woolly and ambiguous, but arguably okay. What do you mean by 'above the ambient baseline'? You appear to be confusing 'heat' and 'temperature'. They can certainly be related but they are not the same thing - their units are never interchangeable. The amount of heat absorbed from the hot sink in the ideal Carnot cycle Is represented by QH = TH(SB - SA) and at the cold sink QC = TC(SA - SB) - These arise from simple application of the 2nd Law for reversible heat exchange. Neither of these terms reference any 'ambient' condition. From W = QH - QC we then get Carnot cycle efficiency = W/QH = (QH - QC)/QH = 1 - QC/QH = 1 - TC/TH There is a huge body of work and experimental data that is in agreement with the absolute nature of this limit. Naturally. It's one thing to say something. It's another to understand what you are saying.

You refuted something by this means, but it wasn't Carnot. Well, I tried.

I listed them for you in this post. Read it carefully from the beginning.

So you are just going to ignore the true context intended: Which would be a pity, because all your misunderstandings appear to stem from this single root cause.

That's not what I said at all. I said that you persistently confuse actual machine efficiency with the Carnot limit. However, hot surfaces transmit more momentum than cold surfaces. So although it is entirely possible for a cold body to transmit a quantum of heat to a hot body, it is overwhelmed by the momentum flow in the opposite direction. The nett direction of heat flow is determined by force of numbers.

Okay, let's try and work our way through this step by step. No it wouldn't. 100% of the enthalpy of the hot working fluid is thermodynamically available for conversion to work, but all real machines have their inefficiencies. So in practice, you may only recover, say, 80% of this as nett work output. We would call this figure the isentropic efficiency of the machine (as opposed to the thermal efficiency). The remaining 20% of the energy input would be discharged to the cold sink with the working fluid at a significantly higher temperature than the cold sink. I've underlined that last part because it is crucially important. It's where the excess entropy is being generated. This is were we need to be extremely careful about which efficiency were are talking about - the theoretical Carnot limit or the real world isentropic efficiency. In the theoretical world there could be a near zero heat flow into a cold sink at absolute zero. In the real world, there would be a significant heat flow. .Only for a machine with an isentropic efficiency of 100% which is not a practical proposition. For a real world machine impeding the heat exchange is equivalent to heating up the cold sink. Less of the input energy is now available for conversion to work. The two scenarios are not equivalent. In fact they are polar opposites. Absolutely not. They are diametrically opposed. From the extreme difference in temperatures on the cold side of the machine. Correct. The work produced is limited by the isentropic efficiency. Reread my first point in this post. You're conflating the Carnot limit with actual machine isentropic efficiency. They are entirely different concepts. Confusing the two leads to absurd conclusions especially at absolute zero. Again focus on the phrase 'maximum efficiency'. Can we agree that this is different from 'actual efficiency'? Here you are allocating an isentropic efficiency of 100% to the machine. I'm finding it difficult to picture 4% of the output of a typical hair dryer as a 'massive heat transfer' and indeed how one would see it. I see no science here. You've gone a long way down a rabbit hole and need to find your way back. It's taken me quite a while to wade through all the steps in your thinking so I'd be grateful if you spent a similar amount of effort in trying to understand what I've presented here. Obviously, I'm only too happy to assist with further clarification.

Refresh your memory of what a heat engine is at https://en.wikipedia.org/wiki/Heat_engine Other than demonstrate that industrial machines are tested by professionals up to their thermodynamic limits on a daily basis contrary to your claims.

And yet your engine still runs. Therefore there is still a significant thermal gradient across the device. Because they have significant losses over an ideal isentropic process. In addition to the usual friction losses, they feature approximately isothermal expansion and compression stages that are a lot less efficient than the corresponding approximately adiabatic stages employed in turbines for example; mixing of warm and cold working fluid occurs around and through the displacer piston; and the kinetic energy of your aforementioned convection currents has to come from somewhere.

You are not reducing the cold side overall heat transfer coefficient as much as you think you are. The machine as supplied is designed to run despite the cold side heat rejection passing through two 'insulators': acrylic and air. Adding a layer of aerogel only retards convection. The machine clearly is able to run on 'conduction only' mode. So aerogel is just a different type of air as far as the machine is concerned. Designs of this type do not approach the Carnot limit in any way shape or form so trying to draw any conclusions about the validity of Carnot efficiency is kind of crass.

What design of Stirling engine is it? Alpha? Beta? Gamma? Something else? Manufacturer/Model number? Is there a schematic diagram available so we can see the piston arrangement, heat exchangers locations, regerator (if any) just to give us some idea of what you're asking us to comment on. What's the process gas? Hydrogen? Helium? Air?

I thought that you might care to review your claim that information entropy was not subect to 2nd Law constraints after browsing through this paper: https://www.physik.uni-kl.de/eggert/papers/raoul.pdf I've attached a copy for your convenience. Other relevant references are: https://en.wikipedia.org/wiki/Maxwell's_demon https://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_information_theory I think some of the confusion lies in a tendency to think of information theory in purely abstract, mathematical terms when its application is very much a real world phenomenon. Shannon definitely framed it in terms of a physical link between sender and receiver. In fact there seems to be a growing view that classical Clausius entropy and von Neumann entropy are simply special cases of the more general Shannon entropy. And the 2nd Law rules them all. raoul.pdf Markus, Can you briefly explain why we shouldn't expect to find a quark-gluon plasma at the heart of a black hole. There should be no problem in storing a huge amount of entropy in a small core of that if the uncertainty principle and extreme temperature is sufficient to resist collapse.

I'd say not, but it does rather depend on what you mean by 'state'. A simple example maybe?