sethoflagos

If you work on the principle of not leaving your environment in worse condition than what you inherited, then it carries on indefinitely. Our generation has clearly failed massively in this regard despite those in power from the mid-20th century on knowing unambiguously the unsustainable nature of a hydrocarbon fuelled economy (not to mention nuclear proliferation). I wouldn't be the one to line up the energy sector major shareholders in front of a firing squad, but I might turn a blind eye ....

I think that 'Do as you would be Done by' is a pretty fair principle. If you would prefer not to be wilfully harmed by any deed of act or omission by others then it seems a contradiction not to reciprocate and it's not unreasonable to extend that duty of care to future generations. Without putting a moralistic value on it, one might accept the validity of one who was indifferent to the suffering of others more remote than his immediate neighbourhood so long as they made no complaint if say an electrician couldn't be bothered to make their domestic wiring safe and they got electrocuted. Swings and roundabouts. But in my experience, such people are usually the first to express their outrage at any perceived violation of their rights. I wouldn't be the one who threw stones at such self-centred wretches, but I might turn a blind eye to any who did.

No. I take care to line up my initial conditions in a consistent way so that I don't get caught in a circular sequence of robbing Peter to pay Paul (which your posts tend to be full of btw). Eventually the message sinks in that there are no free lunches and life's a bitch.

Option 1) Perform the simulation by long hand calculation - the way we used to do this stuff. Option 2) Depends on your precise contractual relationship with the software suppliers.

There is no heat flow into the engines because they at they same temperature as the hot reservoir. There is no heat flow into the engines because they at they same temperature as the hot reservoir. There is no heat flow into the engines because they at they same temperature as the hot reservoir. However, the air inside the engines has started to cool as they melt the ice sliver. The air contracts perhaps enough to pull the power piston down from TDC to BDC and move the displacer away from the hot side heat exchanger. Now at last heat can start to flow into the engine. We reheat the air back to the hot reservoir temperature and if no nett work is extracted and there air no friction losses we may have just enough energy to get the piston back to TDC. Then we can melt a bit more ice. You don't get access to refrigeration without inputting power.

Therefore: is answered by 'energy is energy'. And 'energy exchanges are energy exchanges'. No. Some of the water's potential energy is transformed into shaftwork just as some proportion of heat (thermal energy) can be converted into shaftwork. You ignored my questions relating to the limit (1 - ht/hm): The ratio is reject energy / input energy just as TC/TH is in the Carnot limit.

If you express the total potential energy of water in terms of its height above mean sea level, then the amount of it that you can extract from an overshot waterwheel is limited to (1 - ht/hm) where hm and ht are the heights above msl of the mill race and tail race respectively. Do you see anything familiar here? Can you imagine all that waste potential energy disappearing down the tail race? Why isn't waterwheel efficiency a part of this limit?

Exactly. The block can have any temperature distribution compatible with Fourier's Heat Equation. Regular opening and closing of the freezer door might produce a nice decaying harmonic oscillation with depth for example.

Off the shelf for sure. Then we have repeatability by other experimenters.

Any small known weight sufficient to turn the engine but not too fast. That should work. It must be an operable engine. We have no interest in performance data for ruined engines. Something like that. If you can perform a second experiment experiment with the displacer linkage disconnected (if this is possible without breaking the engine) then that would be useful too.

The only load on the machine is frictional losses, which should be very small and most of those are retained within the system. If waste heat is actually 4 times the work output (it will be a little more), simply ask yourself what 4*0 is equal to. It might be helpful to know what these no-load losses are: If you are able to secure a cotton bobbin squarely and centrally on the flywheel and a small weight to the free end of the cotton you can time how long it takes for the weight to drop through say 3 feet. Try and find a weight that's large enough to turn the machine, but small enough to take at least 10 seconds to cover the distance (for timing accuracy). Repeat until your results are consistent and you have at least 10 timings. Then post the timings here along with the distance of travel, size of weight used and the diameter of the bobbin would be interesting as well.

It does depend I think on how 'closed' the test tube is. A reaction involves for instance some change of chemical bonding energy resulting in creation or absorption of heat. If there is no heat loss or gain through the test tube walls, there is no mass change as the total energy inside the test tube is unchanged.

A good and important point @Ghideon An ice block fresh out of the freezer should be oto -18 oC (0 oF) throughout. Even it has been left a while, the surface may well be at freezing point, but the bulk of the inside may be considerably colder. When it is placed under the cold plate of a Stirling engine, it is effectively insulated from ambient air while it continues to lose heat to the inside of the block. It is entirely consistent for the engine to appear to run well, rejecting a certain amount of heat to the ice interface, while that interface freezes and cools further due to a greater heat loss to the core. The cold reservoir is in this case the core of the ice block and the distribution of its temperature is unknown.

Me neither. Too closely associated in my mind with the Aztec two-step to give the right picture. But we should have a word for it really. In my own internal dialogues I call them 'condensations', but this is the first time I've communicated that to anyone. One person's mental picture doesn't necessarily work for others. This ifdea covers so many different processes and interactions that it's very difficuly to give a general answer that covers everything without quoting most of a thermodynamics textbook. So simple examples are probably the nest way to go. If we take too simplistic a view of entropy, many of the things we see happening around us don't seem to make sense. Take for example a glass jar of muddy water. The silt particles are all over the place so the entropy is relatively high. If we leave it to stand for a few hours most of the particles will settle to the bottom of the jar leaving the water much clearer. This is a lower entropy state and appears at first glance to contradict the 2nd Law of thermodynamics. We've produced 'order' out of 'chaos'. It only makes sense when you realise that as the silt particles settled, they created a bit of heat due to friction. That heat was lost to the surroundings, which raised the entropy of the surroundings by at least as much entropy as was lost from the contents of the jar. So the world starts making sense again. Understanding such simple pictures help you to understand how much more complex processes like photosysnthesis can build trees for example out of a few simple molecules and sunlight.

It's one of Brian Greene's catchphrases. It seems like a catch-all for all processes similar to the crystallisation of a low entropy solid phase out of a higher energy solution. It crops up about a minute or so into this.

You can 'age' exposed concrete structures in the garden (planting troughs for example) by painting the surface with yoghurt. It rapidly incubates a collection of mosses and lichens to help it blend in. A little tip I got from my mother many years ago.

There are, but they are usually behind paywalls. I'd be wary of using viscosity as your parameter. A particularly important viscosity behaviour is that of a binary mixture of water and ethanol. It is highly irregular in certain proportions and certain conditions due to specific interactions between the components. You can get the experimental data easily enough, but then the question is whether Ansys contains an appropriate mixing model for that particular system. Empirical data always trumps predictive modelling. It comes down to what you're exactly trying to achieve. Could you pick entropy instead? It follows a much more predictable path that's quite simple to calculate. Look for a wikipedia page on Entropy of Mixing.

I could do with one of those. We've not had mains supply for 48 hours.

I remind members of the opening few paragraphs of the OP Exactly how much of this are we supposed to accept without question? Given the extraordinary nature of the claims in this OP, I think it quite right and proper that we take a very close look at exactly what it is the OP is trying to do. I note the OP's reluctance for us to do so, and that in itself tells a story. So what exactly should we expect to happen when we fully decouple a Stirling engine from its cold sink. Despite the OP's earnest protestations, I'm going to start with the idealised model because that's how it's done. And let's put some numbers in: Stage 1: Isothermal expansion @ TH The power available from isothermal expansion of an ideal gas from compressed volume VC to expanded volume VE can be expressed as WE = nRTHln(VE/VC) where n & R have their normal IGE meanings. To keep matters simple we can assign it the value of 1 kJ over a certain number of cycles. The source of this energy comes entirely from heat input from the hot source hence QE = WE = 1 kJ Stage 2: Isochoric cooling Adiabatic null process by design. See Stage 4. Stage 3: Adiabatic compression from VE to VC Here we must introduce the ratio of specific heats k (1.40 for air), and can derive: WC = knRTH/(k-1).((VE/VC)^(k-1) -1) >= knRTHln(VE/VC) >= kWE >= 1.4 kJ The approximation tends to equality as VE/VC tends to unity. Since I've introduced inequalities, I'll use absolute values for Q & W to avoid confusion. Stage 4: Isochoric cooling from TA to TH The balance of energy, heat of compression, is returned non-reversibly to the heat sink at TH which is now functioning as a cold sink. QC = WC >= 1.4 kJ Non-ideal behaviour OP is of course correct in asserting that idealised processes can rarely if ever be fully realised in practice. However, that does not make them intractable. Stage 1 may be allowed to have a reasonable adiabatic element by allocating a k value of say 1.04 rather than the implied default of k=1 for ideal isothermal behaviour. This will have the effect of allowing some expansion cooling below TH and slightly increase power output on the expansion stroke. Similarly reducing the k value for Stage 3 from 1.4 to say 1.36 will introduce some isothermal behaviour, reducing TA and slightly reducing the power absorbed by the compression stroke. Which leads us to consider whether the machine is truly decoupled from its cold sink, since if it isn't this will reintroduce Stage 2 cooling and drastically reduce the adiabatic nature of Stage 3. We will return to this. Summary 1) While the Stage 3 compression phase is more adiabatic in nature than the Stage 1 expansion phase, the machine would require a significant work input to continue running for more than a couple of cycles. To this extent the OP prelinary assumption is true. 2) In response to the semi-rhetorical question posed by the OP It is clear that Remains by far the most credible explanation. The OP neglected to respond to this specific comment. What was the goal of the experiment? We need look no further than here Throughout this thread, the OP has treated the following statements as logical consequences of each other: A 100% efficient heat engine transmits no heat to its 0K cold reservoir. A heat engine connected to a 0K cold reservoir is 100% efficient. A heat engine transmitting no heat to a cold reservoir is 100% efficient. By disproving any one of these propositions he imagines he disproves them all.

Many thanks for that Joss. I've been a subscriber to that channel for a couple of years and it rarely disappoints.

An 'avatar' of the local Cernovii (possibly 'horned ones') iron age tribe as suggested seems quite credible. Can't find any info on its physical size, but similar stones are often boundary markers carrying the implicit message 'Our land. Trespassers may be poked with a pointy thing and ritually garrotted.'

Then I suggest you don't waste your valuable time in composing one of your usual 500+ word rambles. I summarised your proposed cycle in about 50 words. Your alternative proposed wording should be similarly concise. Don't link to external sources. Just plain simple wording for all to see.

Fish are pretty isothermal. As is a bucket of iced water. Why are you running scared of the isothermal assumption, Tom? If anything it works to your favour by maximising the efficiency of your machine. I'm perfectly happy with 'nearly isothermal', or 'not even nearly isothermal' or even 'adiabatic' if you want the lowest possible efficiency. Your choice. But I do insist that you make a choice. It's spelt 'isochoric' and means 'conducted at a constant volume'. I'll only accept 'very nearly isochronic' since it's pretty easy to ensure machinery isn't too elastic. But again, the theoretical ideal works to your advantage. Your choice. But I do insist that you make a choice. 'Isochronic' is something entirely different.

Just to clarify, the intent is to make the following changes to the Stirling Cycle. 1. The isothermal expansion stage at TH stays as is outputting power WE. 2. Isochoric cooling stage becomes a null event since it is adiabatic. QC = 0. 3. The isothermal compression at TC stage becomes an adiabatic compression from TH to TA absorbing power Wc. 4. The isochoric heating stage from TC to TH becomes an isochoric cooling from TA to TH. Are we all in agreement? I do know that you've read this, so I'll take your failure to respond as concurrence.