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sethoflagos

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Everything posted by sethoflagos

  1. You're joking, surely! The nearly-fully conical brass instruments: flugle horn, euphonium, and tuba are hardly rare, and they present a by far greater challenge to in-tune harmonics compared to those instruments with a conventional Webster horn profile. Hence the prevalence of more-than-three valves on these instruments with additional compensating loops, especially on the tubas. Really beautiful sounds in the hands of professional performers, but in the hands of amateurs? Sheesh! Worse that trying to keep in tune with herd of great Highland bagpipes! A friend of mine, Rowuk on Trumpetboards.com is also a professional zink player when he's not tootling his baroque trumpets. I think he'd have a word or two to say about your assertion that 'the playing technique is lost'. Sat through a concert by a zink trio myself in York minster not so many years ago. They seemed to know which end to blow down.
  2. I've provided an analysis of your heat pump system and have thus answered your question. However, I need you to explain to me why this provides a solution to the OP. You could perhaps help by providing me with a simple expression for T2 in my worked example. After all, you do understand your pet a lot better than I do (plus you have the textbook it came from).
  3. The congratulations really aren't necessary, sincere or otherwise. I provided you with a standard 1st year BSc thermodynamic calculation and in return you provide some ontological classification of my methodology? What's the point? This is a non sequitur. How many versions of the truth can there be? Neither of you need any help from me in that. IMHO you both ensnare yourselves in petty formalistic rituals that simply obscure. You're no nearer to answering the OP paradox now than you were when you first posted at 9.51 pm on Friday. If you don't know then just say so. Frankly, with so many of you stumbling over the 1st Law constraints, there seems little point in discussing the 2nd.
  4. The gas has increased in entropy by W(T2-T1/2)/T1T2 (= W/T1 - W/2T2) Reservoir 1 has decreased in entropy by W/T1 Reservoir 2 has increased in entropy by W/2T2 W/T1 - W/2T2 - W/T1 + W/2T2 = 0 Hence no nett change in entropy. Your examples don't come into it. Your assertion (paraphrasing) "it is reversible therefore it is isentropic" is a clearly flawed assumption. My version of the 2nd Law is that there are no versions. On more than one occasion I've answered your question quite fully with the statement that "The 2nd Law is the 2nd Law". If you're unhappy with that answer then I'm afraid that's your problem. If there is an alternative version to consider, then either it yields precisely the same results (and is therefore identical in all but name and is therefore superfluous), or it is wrong. It's not rude to evade semantic entrapment. Simply prudent.
  5. Both in the text and on the associated sketch you refer to Stage 2-3 as 'adiabatic compression'. Adiabatic processes may be isentropic but not necessarily so. Because that's how you defined those two isothermal stages. So you do acknowledge that the entropy changes are precisely as I defined them in my last post. You had not mentioned them previously. This appears to be in direct conflict with your earlier statement: ... which I think you need to gracefully withdraw. No shame. Just own the error. Not in dispute.
  6. Some unnecessary complications in here aren't there? The 2nd Law is the 2nd Law. Your process extracts a certain amount of work from a pressurised gas in a W=Q process so W = Q1 and dS1 = W/T1 ( - W/T1 to reservoir 1) Let us say half this shaftwork is used to isentropically compress the gas in stage 2 where dS = 0 by definition The remaining W/2 is spent in a further W=Q process at T2 so Q2 = W/2 and dS2 = - W/2T2 ( +W/2T2 to reservoir 2) So what's happened to the gas overall? We've heated it increasing it's internal energy by W/2 and increasing its entropy by W(T2-T1/2)/T1T2 So okay, you've moved W/2 worth of heat from T1 to T2 but where is the nett change in entropy? There isn't one. 2nd Law is good. dS = dQrev/T by definition. Your process may be ideally reversible, but it absolutely is not isentropic other than the stage you call 'adiabatic compression'. If you also called this stage 'isentropic compression' it may help alert you to the fact that isothermal compression processes are very far from isentropic. So if you've drawn inferences from this line of thinking that you believe will help me with my box problem (I no longer have one), I fear that you have managed to confuse yourself. In passing, I would strongly recommend sketching out thermodynamic processes on HS diagrams rather than PV. On an HS diagram ideal isothermal processes are horizontal lines, ideal isentropic processes are vertical. With practice, you can see at a glance whether there's a 2nd Law infringement. Invariably, such infringements arise from human error.
  7. My understanding is that the difficulties of applying the microcanonical ensemble to the real world are pretty well documented. Following this thread has highlighted to me some of the more blatant pitfalls that people can fall into. I didn't name and shame the Youtube channels concerned in the OP since actually, some of them are pretty well-informed - their message doesn't crumble under research, but in this instance ... I'm 61. I'm not blind to my own limitations. Those I forget, my better half reminds me of quickly enough.
  8. It's late and a combination of your obscurantism and swansont's bloody-minded negativity has exhausted my patience (which to be frank, I've never had in great excess). It must be obvious to you by now that I've been clear in my own mind since way back on page 1 of this thread where the Youtube presentations break down. The key lies in conservation of angular momentum which is a topic I usually shy away from. Turns out, it can be quite useful on occasion. Do we have anything more to discuss? If your earlier posting was a joke, then I'm sorry I didn't get it and took its meaning at face value.
  9. I'm sorry that the standard engineering techniques for analysing thermodynamic systems cause you such consternation. Better engineers than me have approved my approach and given me repeat contracts to lead the detailing out their power design projects since I turned thirty, and that was a very long time ago. Your pride blinds you. I'm out of here. Just stick a heat engine between T2 and T1 heat reservoirs and you've got free energy. Well done! You've solved the worlds energy problems! Have a ball!
  10. Do you think this is the first time I've ever seen a sketch for a heat pump that claims to break the 2nd Law? Really? I've just worked through the expression for the shaft work of your blasted adiabatic compression stage wondering why the hell am I spending my time on this hair-brained stuff when I'm not being paid for it, and you have the gall to say I'm messing around wasting everyone's time. HOW DARE YOU! Let's be clear on this - you are the one who has come up with an invention that any patent office would chuck in the waste-paper basket unread! I had about 12 hours work left to properly research and analyse your scheme so that I could help you through your misunderstanding. I've been helping my junior engineers overcome such hurdles for many years. But now shall I bother? What the hell do you think. Grow up!
  11. I've not changed the conditions one iota. Even ideal systems have to observe basic symmetries. Better. Now sum all those microstates where the container remains unchanged from its original state, and we have a workable thermodynamic ensemble. The mean free path in air is what? 100 nm? There may be a lag between arrival and collision, but it's trivial in a macroscopic system. We don't need a 1-1 match up. The entire reaction vector to the momentum of a single particle can be split anyway you like between the remaining N-1 particles in the system. In a sense it is. So long as motion of one particle is exactly matched by some ensemble in contrary motion to fill the void, then sanity is preserved and you have the restoring force you've been asking for. Not when you've transferred nett momentum into your box. But now you have a non-canonical microstate. Do you really think you can both have your cake and eat it? They all require work in a sense. However small fluctuations are reversed almost immediately by the inertial reaction mechanisms discussed above, and are inherently reversible in nature. Larger fluctuations imply a long term resistance to those reaction forces that increase in direct proportion to the scale of the disturbance and I see no internal mechanism that could account for such a resistance.
  12. Looks to me like you've just heated up some high pressure gas from T1 to T2 at roughly constant pressure. Close the cycle, put some preliminary numbers to it and we can discuss.
  13. However all the particles are in one side of the box. So lets drop in a partition isolating a system of N particles on one side from an absolute vacuum on the other. Please don't reply 'there is no partition'. Yes it is. Ball and box is a two-body system with regular interchange of (at the very least) momentum between them. If you think that this is awkward, just wait until we start considering the regular interchange of torque. Every action has an equal and opposite reaction. Every particle arrival or departure within a space brings with it changes in mass, linear momentum, orbital angular momentum, axial spin etc and each must be balanced precisely by an equal and opposite external action. Sit it stationary at the CoM and forget about it. It isn't a problem until something bumps into it. I'll ask the mice. The box is a mathematical artefact introduced solely to define the system geometrical constraints. A condition of it being considered non-interacting is that we must be able to delete it from the snapshot and see no nett change to the system. Therefore the only states it is admissible to consider are those where every single particle linear momentum vector is exactly balanced by an equal and opposite vector. Just as the box could transfer momentum to and from the system, so can the octants transfer momentum between each other. While it must be possible for one octant to momentarily grab an extra particle or two (I'll agree with you that far), there are one or two particles on the imminent verge of leaving it probability one ensuring that the system never departs by a measurable amount from equilibrium. Serious departures demand serious external work.
  14. Every fibre of my being feels this principle to be true. Though the name is unfamiliar to me. Thank you. I'll try to remember the spelling! Fire away, I'm all ears.
  15. The 2nd Law is the 2nd Law. Let us assume dS/dt >= 0 for an isolated system. You're in the same trap as swansont. At each bounce, there is an interchange of momentum with the box. The box is actively participating in the thermodynamic process, so you now have to consider it's inertia, thermal capacity, temperature and entropy. Your system is an almost perfect vacuum. You want to discuss vacuums? They're pretty well defined, I think. It's rapidly becoming irrelevant in your idealised case. I'm more interested at this point in what the temperature of your box is, since this is now setting the conditions necessary for thermal equilibrium. Is it? Never heard of it. Your box is an active part of the thermodynamic system, which is why I ask what its temperature is, because that will define the equilibrium state. I suspect with only one single particle the equilibrium temperature is to all intents and purposes absolute zero, so your particle will shed its last little shred of momentum to the box and become stationary wrt the system CoM. Equilibrium has been reached. Real world processes tend to follow non-analytic PVT paths, which is why we characterise them by a sequence of analytic ideal steps, often an infinite number. There may well be no physical piston or heat exchanger involved but is entirely appropriate to fabricate a few for calculation purposes. See https://en.wikipedia.org/wiki/Thermodynamic_process_path#:~:text=A thermodynamic process path is,-entropy (T-s) diagrams. "A thermodynamic process path is the path or series of states through which a system passes from an initial equilibrium state to a final equilibrium state and can be viewed graphically on a pressure-volume (P-V), pressure-temperature (P-T), and temperature-entropy (T-s) diagrams. There are an infinite number of possible paths from an initial point to an end point in a process. In many cases the path matters, however, changes in the thermodynamic properties depend only on the initial and final states and not upon the path." At last! Conservation of linear momentum. It's not permissible in this simple NVE case to conveniently transfer some to the box because that is the equivalent of introducing external Q and W terms. Precisely! A low one. You objected to my earlier suggestion of 42, but it wasn't entirely flippant. Enough to constitute a system that isn't too swamped by quantum effects. Having said that, if you take the st. dev. of momentum from the Maxwell-Boltzmann distribution and plug it into the Heisenberg Inequality, some very interesting expressions arise that look very 2nd Law-ish to me. I suspect a very deep link in there somewhere, which actually strongly enforces my trust in the 2nd Law.
  16. No, the CoM of the ball (wrt the system CoM) is not fixed. It was you who introduced the single ball case Needlessly offensive phrasing. My 'agenda' is simply to respond to points you raise in support of arbitrary random changes of state. There's a principle we use in Chem Eng that the transition between any two thermodynamic states is path independent i.e. when a body of gas transforms from one PVT state to a new PVT state, it is perfectly admissible to break it down into a sequence of smaller idealised steps, knowing that the overall changes remain fixed for purposes of calculating the enthalpy changes for instance. Hence, in the case in question, I analyse the change as an ideal isentropic compression (external work performed with no change in entropy) followed by a possible reversible cooling stage (shedding heat and entropy into the environment). The Youtube presenters haven't specified the final temperature of the system, but they have stated that the entropy has fallen, therefore the cooling stage is a valid means of estimating the minimum transfer of entropy. My 40 years work experience screams out 'THE ENTROPY OF THE ENVIRONMENT HAS RISEN BY AT LEAST AN EQUAL AMOUNT'. Must have. Otherwise all my design calculations over the last 40 years have been wrong and numerous production facilities around the world are fundamentally unsafe. A little over-dramatic maybe, but my hackles have been raised.
  17. I do not believe it is possible for all particles to be on one side of the box in the absence of shaftwork or equivalent energy input. We're not able to compute system evolution collision by collision, but what we do know for sure is that the outcomes of each collision are very far from random - the entire discipline of physics is based on the symmetries of conserved quantities through such events. That deterministic principle does not suddenly vanish simply because we have insufficient computing power to handle the long term dynamics of macroscopic systems: those single event symmetries remain intact, instant by instant, eon by eon. In my experience, truly random behaviour is rarely observed and then only in a very few, very special scenarios. The rest is merely 'complicated'.
  18. The overall CoM is essentially fixed by your constraint of infinite box mass. The total linear momentum of the system with respect to the total CoM is equal to the linear momentum of the ball wrt the total CoM. What happens at collision with the box wall is observer dependent. An observer travelling with the ball will see the ball suddenly accelerate from rest and deduce that work has been done on the ball providing it with a kinetic energy it did not previously have. An observer stationary wrt the system CoM may (per your assumption) see a specular collision where the kinetic energy of the ball is preserved. A third observer may well see the ball suddenly come to a standstill. All intermediate variations are possible. Single particle thermodynamics really has no meaning due to this observer bias. It isn't until we get into multiple particle systems where particle-particle collisions are possible, that we can shake off this observer bias and establish the concept of thermodynamic equilibrium - one requirement of which will be to set constraints on the motion of the centre of mass of the particles in the system (excluding the box). Maybe I should give you a heads up of where we are heading with this. I'm using a combination of the 1st Law and conservation of linear momentum to establish the necessity for a stationary gas CoM as a priliminary condition of thermodynamic equibrium in, for now, a defined NVE state. The next stage will be to recruit the conservation of angular momentum to establish that there can be no nett flow towards or away from the CoM at equilibrium ie that in the absence of input of external work or torque, the gas remains evenly distributed throughout the available volume. After that, well I trust that any residual ideas of macroscopic systems hopping willy-nilly between unrelated, random microstates will have dissipated. I thank you all in advance for forcing me to think through all this stuff carefully stage by stage. It's over 40 years since I sat through my last thermo lecture!
  19. I specifically stated the centre of mass of the gas because this is precisely the quantity that you are trying to move around relative to the CoM of the box to establish your case.
  20. If the centre of mass of the particles remains stationary at the centre of the box, then it's possible that no nett work has been performed on the gas. Now explain to me how that centre of mass can move towards the sides of the box to any reasonable degree without the box performing work on the gas.
  21. Another interesting rabbit hole. By coincidence, I asked a question the other day about the strength of coupling between a CMB photon emitter and its TV aerial absorber. Mordred informed me that contrary to what I'd inferred from what I'd read of GR, they weren't actually touching since the photon wasn't a valid frame of reference. Obviously, I've no grounds whatsoever for disputing this, and too many Minecraft projects planned to try and get to grips with it. I just accept that for now and the foreseeable future, such fields are beyond my understanding. Not my system. You define your system as you wish. If you can. It's not easy, and perhaps will give some idea of why the microcanonical ensemble has given statistical mechanics some headaches in the past. The canonical and grand canonical ensembles find easier application in the real world and have no aberrant conflicts with classical thermodynamics. There is now an exchange of work between particles and box. It has become your very own 'undisclosed piston'. Do you now see and understand the issue?
  22. An interesting rabbit hole! Firstly, consider what is happening to your system centre of mass, and account for its apparent irregular motion. I look forward to your well considered response (which I'll not preempt here)
  23. My point. We're agreed. Agreed. The V/2 state perforce requires some combination of W and Q which must be reflected in a balancing change in U. 2nd Law implies a minimum increase in U (and hence T) under simple compression. They said it. Obviously. Absolute zero microstates are allowed into their ensemble by their reasoning. How do you propose to explain all the particles appearing in one half of the box? My inference is that there must be something equivalent to an undeclared piston compressing the system which, as you say, violates the conditions of the problem. That in itself is a clarification: the hypothesis is probably BS.
  24. Your point that classical thermodynamics largely time independent, I accepted without seeing the need for further comment. Which was the other point? I'm 100% with Mordred on this issue. Would you be happy to simplify the thread and leave it at that? Then we are in agreement. Obviously. Hope all is clear
  25. You do like to nit-pick! Depends a little on context. Formally, in my day job, it usually infers that the system is in a state of minimum Gibbs Free Energy - e.g there are no bulk convective processes going on within it. For a constant V, T system (I rarely encounter these) it would be a state of minimum Helmholtz Free Energy. In the OP scenario, the presenters start with an equilibrium V, T condition and claim that it can evolve spontaneously to occupy only V/2. This requires a bulk convective flow (eg a piston compressing it) and represents a fundamental change of state. The presenters concentrate on the position distribution of their system and fail to mention any impact on the momentum distribution. Do we infer the temperature has remained constant (breaking the 1st Law and the 2nd)? Has it increased as it would if it had been compressed by a piston (2nd Law preserved but not the 1st)? Or indeed has it decreased. We are left to guess. The only clue we have is that the presenters claim to have 'proven' evolution to a low entropy condition. If we believe them, this eliminates the higher temperature case from consideration. What we are left with is a proposed sudden and significant random change of state breaking both 1st and 2nd Laws. It's perhaps a personal flaw, but I've a habit of ridiculing such proposals by highlighting an extreme case that becomes allowable if their assumptions are correct. Such as a spontaneous jump to absolute zero. Of course the presenters do not state this inference explicitly as it would make them appear very foolish. But I'm quite happy to point out a logical extension of their false reasoning.
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