sethoflagos

Both low lying areas subject to offshore cold water currents/upwellings and predominantly dry trade winds? They are deserts for a reason. Could add the Atacama to this list. (Ekman transport can be an important mechanism in these cases) The lower few hundred feet of steady onshore winds are typically in approximate thermal equilibrium with the ocean, aren't they? So where is the energy to come from to evaporate the water? Even if the air is at lower relative humidity (such as a descending Hadley cell), evaporation is going to chill it further. This doesn't sound like a good recipe for creating a rising thermal. More a recipe for fog. If it worked, nature would already be doing it, I think. As it does here in southern Nigeria during the rainy season. But come november the rainbelt will have moved south to Angola, the Hadley cells will shift what little air movement there is to a north-easterly flow, and it will be both cool (for us) and bone dry. Trying to 'make' it rain here at the turn of the year would be an exercise in futility. Everything would be working against you.

When I was at school there were 30.48 centimetres to one foot, so maybe just multiply the result of your calculation by 30.48? Or is the input to your calculation not in millibars either?

Hydrogen-Oxygen combustion in a fuel cell with full heat recovery is currently capable of around 85% thermodynamic efficiency, so we should be looking at a nett electrical output of a smidgen over 200 kJ/mol. Conventional alkaline electrolysis has a thermodynamic efficiency of ~70%, but a lot of work is currently being done on Proton Exchange Membrane (PEM) technology and we should soon be able to achieve 85% on that too bringing the energy cost (electrical) down to about 280 kJ/mol.

+1 Just to complete the circle, in the combustion cycle only a theoretical maximum of 237 kJ/mol of the enthalpy change is thermodynamically available for conversion to eg electricity for electrolysis (this is the meaning of Gibbs free energy).

Other than the top 6 or 7 commonest crustal elements, it's more about the density of the rocks of which they are a minor/trace component. ie less than 1% by weight of the local rock formation and having negligible impact on the bulk density.

Interesting to note that the last 10 positions on that list include all the Group VIII precious metals at typically 1 ppb whereas Thorium and Uranium despite being considerably heavier nuclei have over 3 orders of magnitude greater crustal abundance. This is strong evidence that the dominant factor is not atomic weight but chemical affinity with the crust being heavily depleted in the siderophilic elements (those that have a chemical affinity for a metallic iron phase) relative to the lithophilic elements which are preferentially drawn to a silica rich phase. Almost as if the crust had been washed 99.9% clean of siderophiles by a descending wave of iron heading for the core.

I'm sure you're better acquainted with the current technical definition of 'standard enthalpy of formation' than I am, and all the caveats that go along with it. And yes, if all those caveats are observed then delta H in the forward direction equals minus delta H in the reverse. But don't those caveats require that conditions at start and finish are globally unchanged? In other words, the equality of enthalpies is dependent on there being no overall change in entropy? That was my understanding but then I'm Chem Eng, not a Chemist.

I take your point. But in the reverse reaction you are breaking bonds by electrolysis, so aren't you steering hydrogen to the cathode and oxygen to the anode thereby unmixing them? I think the piper will still need to be paid his due. Similar arguments can I think be made for all implicit irreversibilities in the process (in both directions). However, the constant making and breaking of bonds in an equilibrium mixture of products and reactants (or similar system) has no irreversibilities. So here, the OP's premise must be true. The energy released by the forward reaction is exactly balanced by that consumed in the reverse.

I think @studiot may have made a valid point. As an example consider that your initial mixing of 2 moles of hydrogen with one mole of oxygen increased the entropy of the mixture by about 1.89 R. In the absence of reaction no energy change occurred and hence there is none to recover in the reverse reaction. However, mixing is an irreversible process (in the thermodynamic sense) and unmixing the products of the reverse reaction back into their pure elemental states will require quite a bit of unbudgeted additional work input.

Perhaps @joigus put it more succintly: I draw your attention to the phrase 'We don't know and we don't care!' Only delta V is relevant in context.

It most definitely isn't. Far better to pick example thermodynamic processes whose paths are simple, well defined and relevant. Excellent point. But this result could have been obtained by considering the gradual inflation of a balloon for example. I pick this because in good teaching examples of PV work, it is usually arranged that the pressure exerted by the surroundings on the surface of the system is well defined and equal to the pressure exerted by the system on the surroundings. In the OP's example, this is not the case: P1, P2, and any possible thermodynamic path between them are complete and utter red herrings. I don't think I mentioned Z, did I? Bit strawmannish there, J 😉. Besides which, all the process simulators seem to be running Peng-Robinson these days. But that's for much later in the course, as you say. By 'out of equilibrium' I'm assuming you mean that the system has significant P, T gradients, ie parts of the system are not in thermodynamic equilibrium with other parts never mind the surroundings. Whether or not it's necessary to invoke N-S (thankfully not in most cases), a reasonable approach may be to shrink the system under consideration to a differential volume moving with the local mean fluid velocity. This unit cell can be assumed to be at some localised P, T equilibrium state with no mass interchange with the surrounding cells, and therefore the standard thermodynamic analytic techniques should be applicable to the interactions of the unit cell with its 'external environment' of neighbouring cells. And then the maths starts, using the equation of state to relate the pressure and density terms in the applicable flow and mass continuity equations. Pointless to go into the methodologies as these are highly case dependent. Sometimes you're lucky and end up with a simple ODE. Sometimes not.

Just to be clear, I have a different viewpoint on this. Modelling of expansion/compression processes assuming constant PV^k is just too good a predictor of real machinery performance to be ignored. Both P and V must be quite mappable through these highly dynamic changes of state irrespective of equilibrium considerations. It is unfortunate that thermodynamics and fluid flow are treated in most universities as separate topics since scenarios such as the OP are under the control of not just the thermodynamic equations of state, but also the appropriate form of Navier-Stokes equations. And the particular example of the OP is dominated by the most challenging form of the Navier-Stokes: that for flow of compressible fluids where both inertial and viscous terms are significant in all three spatial dimensions. Little wonder the OP is confused. And if he's doing one of the pure sciences, it's highly unlikely that he'll ever be given the analytic tools that might help him make sense of it. He would learn far more from looking at gas expansion through a porous plug. Same process of turning U to W but so little of that nasty kinetic stuff that it can be cheerfully ignored.

+1 Personally, I found it very difficult to develop an intuitive feel for PV diagrams. In power system design, an engineer is far more likely to use HS diagrams as they indicate changes in energy so much more clearly. So the path for the OP case would be a near vertical plunge south for the expansion phase followed by a steady drift north-east for the reheating phase. No need to worry about any mysterious goings on in the P,V,T planes.

Consider a tyre burst. Allowed near free expansion, the air in the tyre converts itself to a considerably colder, high velocity stream at a pressure suitable to interact with the external environment with balanced forces. It is this sense of internal self-readjustment that I intended. The reason neither P1 nor P2 appear explicitly in the final equation. This process, the conversion of internal energy to kinetic energy is a work term and isentropic to a first order engineering approximation. However, in the downstream end of the process, instead of driving a turbine or piston, the kinetic energy is simply allowed to return back to internal energy once more, a dissipative heating process that is most definitely irreversible. When the dust settles the leakage performs Pex*(V2-V1) Joules of work and absorbs Pex*(V2-V1) Joules of heat, restoring the original ambient temperature. (But not the original entropy). If I've read your posts correctly, then we have essentially the same picture of the context whereby the process is indeed irreversible. And we agree on the value of W. However, what is missing is any mention of Q, the dissipated heat that balances the 1st Law (Q=W) and creates the 2nd Law consequences of the tyre burst via dS=dQ/T. This seems to indicate a failure to appreciate that there are (at least!) two distinct processes going on here. And they really are as different as chalk and cheese. After many years of mentoring junior graduate engineers, the phrase 'irreversible work' speaks volumes.

Granted, but that leaves us having to guess the context in which the shaded area is indeed an irreversible quantity. Which requires an initial reasonably deep understanding of the nature of irreversibility. Which in turn makes it a very poor way to introduce the concept of irreversibilty to new students, doesn't it? And yet this is what we seem to present to them. Well this paragraph begins with a statement of fact that I've never observed. The concept of irreversible work is interesting - could this be a conflation of reversible work and irreversible heat transfer occurring more or less simultaneously? I think that by the end of the paragraph you've come to realise that it might be. Certainly, the OP has a nice simple equation to use to calculate a part of the work function. But why has he posted it here? And is it going to help him get to grips with the more interesting part of the integration above the shaded area? It seems that that isn't happening. Why? Like it or not, that is the nature of the field. But here I'm simply asking why the OP has labelled an area I would normally expect to be usable work as not so. And maybe elaborate both on the practical context of the example and on his understanding of what's happening in the unshaded area below the PV curve. I don't see fluid pressure acting on anything here but itself until it falls to P2. It's likely that my beef is more about the way the OP is being taught the subject.

Erratum For 'if V2 = Vex' read 'If P2 = Pex'

What makes this an irreversible process? As @joigus states, (if V2 = Vex) then P2*(V2-V1) can represent the work done on the environment (hence the minus sign) in making space for the expansion from V1 to V2. However, this is not in principle an irreversible process. The PV path you indicate represents at least in part, a conversion of internal energy to bulk kinetic energy, possibly involving a frictionless piston. What happens when the system reaches the pressure of the external environment? Does it just stop in its tracks? Or does its own momentum carry it into the vacuum zone enabling the environment to reverse the piston and return in full the energy that it borrowed? If you want to establish irreversibility, you must demonstrate a global entropy increase involving the transfer of heat across a non-trivial temperature difference. Nothing else is relevant. Certainly as regards teaching the subject. I do remember sitting through this lecture nearly 50 years ago and being confused by its weird assumptions. The damage done was fortunately reversible. Also bear in mind that if you want to understand where the energy in this system is going, you need to integrate the VdP side of the graph in addition to the PdV side.

A non sequitur fit for yesterday's Westminster PMQs!

Perhaps you could thicken your reagents by adding - 15% by weight plain tissue paper and grinding it into a thick slurry. This should adhere pretty well to your surface and keep the surface evenly wetted via capilliary action. When you're done, a quick rinse under the tap will remove it immediately. There shouldn't be any adverse reactions since the cellulose fibres will almost certainly have seen both high alkalinities and aggressive bleaching during manufacture.

The Baalbek quarry is close by and a little higher than the site where the trilithons were placed. No need to lift. Other than people power, placement would appear to need little more than a well-levelled road, rollers, rope and capstans, all of which the Romans could manage quite routinely, yes? Not saying that it was an easy job, they seem to have given up after installing three (up to 800 tonnes each), leaving at least three more unfinished back in the quarry. However, with a simple solution staring us in the face, why go looking for a more far-fetched explanation?

The OP has clearly gone for an extreme scenario to gain attention. He could just as easily have loaded his megalith onto an adequately large barge and pulled it along a conventional canal or pre-existing waterway. This technology was not beyond our forebears and the underlying physics is moreorless identical. It's just a little less clickbaity.

To be fair, I never actually mentioned Archimedes. Whether his principle 'applies' or not depends on how you visualise it, and it is clear from the posts here that there is quite a variety in these visualisations. Rather, I simplified by picking a convenient horizontal reference plane (the base of the monolith is an obvious candidate) and checked for isostatic equilibrium. ie that any given area (A) on that plane supported a vertical column of mass m such that m/A was constant. (and where mg/A equals gauge pressure) Hence a small mass m1 acting on a small area A1 will be in isostatic equilibrium with (ie 'balances' or 'floats') a large mass m2 acting on an area A2 iff m1/A1 = m2/A2 This approach also immediately identifies @swansont's vertical wooden pole as a significant mass acting on a small area. Hence it requires a much larger liquid depth to establish isostatic equilibrium than a horizontal pole would. Beats worrying about individual buoyancy forces anyway.

I think I'm going to treat this as 'information only'. m1/a1 = m2/a2 is the relation to focus on. I refer and defer to @J.C.MacSwell

No. Hydraulic pressure is determined by the depth of the liquid column, irrespective of its cross-section. So a 2" depth of mercury will float a 12" depth of rock (using your figures). Even if the mercury connects to the atmosphere via a pinhole. However, the rise of the rock vs the fall of the mercury level is in proportion to their respective cross-sectional areas. This ratio is also called the 'mechanical advantage' which can permit say 1 kg of mercury to lift say 1 tonne of rock.

Static pressure is all you need. A 1 metre thick slab of rock exerts of the order 3,000 kgf per m^2 on the surface supporting it. Mercury could provide that supporti at a depth of 3,000/13,600 = 0.221 m If the mercury surface level is >0.221 m above the bottom of the channel, the rock floats. Increase the mercury level, the rock slab rises accordingly. The pump wasn't introduced in response to the OP, it was introduced in response to Hence my suggestion that if a rock is being raised, an external source of energy input is required. The OP seems to imply it may be via a bloke with a bucket (which I tend to view as a low technology pump anyway).