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Comment on Open thread by Rob Ellison

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‘PdV = -VdP/gamma’ governing equation?????

-P dV = m Cv dT
Using d(PV) = P dV + V dP
=> V dP – d(PV) = m Cv dT

But PV = nRT etc….

You are not really following this are you?

There is a pressure P at height z – constant at the height where PdV is evaluated. Quasi static equation.

The dynamic equation ironically based on hydrostatics.

dP – m ρ dz

Gives pressure P on integration.

P = m ρ z – as I said before.

The derivation doesn’t account for buoyancy – or you would need to include Eb in the energy conservation equation. It assumes a stable air mass at some height and pressure P. This is the P to use.


Comment on Open thread by its.not.co2@gmail.com

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Pierre is right. High pressure does not maintain high temperatures, Gravity forms a density and a temperature gradient. The pressure gradient results from these two gradients because pressure is proportional to the product of density and temperature. Pressure is not the cause – gravity is. If you made the effort to understand Kinetic Theory you would have far less trouble understanding these concepts. The Wiki article is a good starting point in this case – note the assumptions and the reference to gravity acting on molecules.

Comment on Open thread by Pierre-Normand

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Rob Ellison wrote: ” it comes from not seeing the jiggle jiggle.”

I think that’s your own blindness. Google: “Why does a gas cool when it expands” You will find very many layman explanation (some incorrect, this being the internet!) on physics fora and academic references. There are many variations of my truck/tennis-ball analogy that you so readily dismissed. Here is one from physicsforiums: “Imagine a gas inside a cylinder containing a freely movable piston. If the pressure of the gas inside the cylinder is greater then the pressure outside then the gas will expand and do work by pushing the piston out. In terms of collisions, the gas molecules colliding with the retreating piston will bounce off with reduced velocities. Whether you describe it in terms of work done or reducing velocities of the molecules the result is the same i.e. the gas cools. The cylinder and piston can be dispensed with and the expanding gas will still cool by pushing against the atmosphere.” (Some reformatting mine)

As a result, the work done corresponds to the total kinetic energy lost by the molecules (assuming an ideal gas for simplicity), and since the number of molecules in the cylinder remains the same, the average KE drops as well (and therefore, so does the temperature).

Comment on New presentations on sea ice by beththeserf

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Not joshua, surely? )

Comment on Open thread by Rob Ellison

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Pierre is right. High pressure does not maintain high temperatures, Gravity forms a density and a temperature gradient. The pressure gradient results from these two gradients because pressure is proportional to the product of density and temperature. Pressure is not the cause – gravity is. If you made the effort to understand Kinetic Theory you would have far less trouble understanding these concepts. The Wiki article is a good starting point in this case – note the assumptions and the reference to gravity acting on molecules.

P-N can’t possibly be right – it is just not a possibility. Pressure is the result of the weight at the atmosphere. Weight exists because of gravity. It is a bulk property and the kinetic theory of gases is not relevant to atmospheric pressures.

An air parcel under a weight is compressed. Clearly if you understood hydrostatics you would know this. With less weight it decompresses. This discussion is going even further backwards.

Comment on Open thread by Pierre-Normand

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“The derivation doesn’t account for buoyancy – or you would need to include Eb in the energy conservation equation.”

Not true. The reason the buoyant force is irrelevant is rather because the work if produces gets *entirely* converted to gravitational potential energy of the whole air parcel. But this energy isn’t part of the internal energy and hence has no incidence at all in the temperature or pressure variation within the parcel. Likewise, when you lift a solid mass (assumed incompressible, or in a vacuum, say) you also increase its gravitational potential energy, but this has no effect whatsoever on the temperature of the solid mass. This force just doesn’t contribute anything to the thermodynamically relevant dW (thanks to Pekka for this analogy).

Comment on Open thread by Rob Ellison

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Perhaps I should correct my hydrostatic equation before the quibbles start rolling in.

dP = =g ρ dz

Comment on Open thread by Pierre-Normand

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Rob Elisson wrote: “dV = (m Cv dT)/-P

You could think this through for a change.”

This still leaves dV indeterminate since you don’t know dT either. In order to constrain either dV or dT, you need at some point to relate those to dz. You will find that there is no getting around using the two equations that relate together dP, dV and rho*g. And when you set up those equations, it tells you that dP/dz != 0, i.o.w., pressure varies as the parcel of air is rising.


Comment on Open thread by Pierre-Normand

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Rob Ellison wrote: “dP = [-] g ρ dz” (typo corrected)

This means that dP/dz = – g*ρ.

What is dz, in the context of the lapse rate derivation, in your view? I am interpreting it as the vertical displacement of the adiabatically expanding parcel of air. Pressure therefore varies with a rate -g*ρ. What is your intepretation of dz? Isn’t the expanding parcel of air rising at all?

Comment on Open thread by Rob Ellison

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Not true. The reason the buoyant force is irrelevant is rather because the work if produces gets *entirely* converted to gravitational potential energy of the whole air parcel. But this energy isn’t part of the internal energy and hence has no incidence at all in the temperature or pressure variation within the parcel. Likewise, when you lift a solid mass (assumed incompressible, or in a vacuum, say) you also increase its gravitational potential energy, but this has no effect whatsoever on the temperature of the solid mass. This force just doesn’t contribute anything to the thermodynamically relevant dW (thanks to Pekka for this analogy).

Very funny. The flows in rising air masses are all turbulent down to micro scales. The potential energy of the buoyant parcel is converted to kinetic energy at the molecular level and bulk level. The potential energy of compression is converted to kinetic energy as molecules are thrown outward in decompression.

The air rises to a stable point – where it will stay in a local thermodynamic equilibrium. Unless it is warmed or cooled it is essentially weightless – if it cools or warms it develops a buoyancy potential. Think of air as a fluid.

Comment on Open thread by Pierre-Normand

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“The potential energy of the buoyant parcel is converted to kinetic energy at the molecular level and bulk level.”

If you really believe this, then *why* are you setting this energy to zero (as you should!) in the lapse rate derivation?

Comment on Open thread by Rob Ellison

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We assume for the sake of the calculation many things that are not true – this is why the environmental lapse rate is so different. The dry adiabatic lapse rate – as I said before – is useful only for providing an indication of atmospheric stability.

You confuse gross approximations with reality.

Comment on Open thread by Pierre-Normand

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Of course, true to form (thanks Jim D!), you are mocking and rejecting the exact same claims that you previously were acknowledging, saying to have made yourself. Here is what I had written:

(Pierre-Normand): “Sure, but the energy source for performing this work — the work of the buoyant force — entirely is *external* to the parcel of air that is being raised some distance dz and hence has *no* effect at all on the kinetic energy of the constituent molecules of the raised air parcel. (Just write down a simple force diagram). The decrease in kinetic energy of the constituent molecules only is due to the adiabatic expansion and the internal energy expended as a consequence of this work W= PdV that the parcel exerts on its surrounding while expanding. [snip Carnot cycle second stage analogy]”

And here had been your reply:

(Rob Ellison): “That’s what I have said a number of times now quite explicitly. Are you so cretinous you reply to what you think my comments are rather than actually reading them?

As for the rest I see much arm waving about things that have been discussed endlessly – but nothing more than the terms of the dry adiabatic lapse rate I have given several times as well. Do you take pleasure in repeating the obvious?”

Comment on Open thread by Pierre-Normand

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Rob Ellison: “We assume for the sake of the calculation many things that are not true – this is why the environmental lapse rate is so different. The dry adiabatic lapse rate – as I said before – is useful only for providing an indication of atmospheric stability.”

But we are not discussing the environmental lapse rate. The ELR mainly departs from the dry adiabatic lapse rate because of multiple local factors that we haven’t discussed at all: condensation, overturning that isn’t fast enough, IR absorption, limitations of the adiabatic assumption, etc. We have been discussing the *dry adiabatic lapse rate* and disagreeing about *it*, not about its conditions of perfect realization in the real atmosphere.

Comment on Open thread by Pierre-Normand

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“You confuse gross approximations with reality.”

Again, we can certainly agree that it merely is an approximation (pretty good actually, far from the dew point, at daytime, under calm weather conditions) while strongly disagreeing, as we are, about the nature of the process that underlies the idealized physical process itself and the interpretation of its mathematical derivation. If we don’t agree about *that*, then it isn’t productive to discuss small higher-order corrections (with e.g. non-ideal gases) or deviations in special circumstances.


Comment on Open thread by its.not.co2@gmail.com

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What you guys don’t realise is that the air molecules move between collisions at a speed of about 500 metres per second. That translational kinetic energy totally dominates all your calculations pertaining to the bulk kinetic energy which is orders of magnitude less, even in a hurricane.

Source: http://www.insula.com.au/physics/1221/L8.html

Comment on Words of wisdom from Charles Lyell by rmdobservations

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I always enjoy reading texts from almost 200 years ago. It’s a nice example of how difficult it is to see the “big picture” when one is sitting inside the frame.

I did not get through all the responses. I did experience a chuckle from the “vacuum” thread at the top. I appreciate the book references and will look up the Doug Macdougal book. I went to University in the 80s and had little money to spend on the few books that were written. I had one great course by an Australian sea-level specialist Rhodes Fairbridge. He asked the students to compare what geologists are saying about climate compared to what atmospheric scientists were. I think that course shaped my thinking.

I lean towards the pragmatic side of climate science. Keep measuring, keep improving the models and by all means keep monitoring. And maintain funding for these activities.
Rose

Comment on Open thread by its.not.co2@gmail.com

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Air molecules move between collisions at a speed of about 500 metres per second. This is orders of magnitude faster than any bulk air movement.

That should give you an idea of the rate at which gravitational potential energy changes for any particular molecule, because Newton’s laws can be applied for all practical purposes.

At thermodynamic equilibrium the mean sum ….
(KE + Gravitational PE) = constant at all altitudes.

When a molecule with mean KE for its initial altitude rises (passing on average about 30 molecules in its mean free path) its kinetic energy is partly converted to extra gravitational potential energy and so, when it next collides its new KE is the same as the mean KE at that level, and so it has no warming or cooling effect on the molecule it strikes. Hence there is a temperature gradient, because the mean KE per molecule is proportional to the absolute (K) temperature.

Comment on New presentations on sea ice by rmdobservations

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Yes. See <a href="http://www.arctic.noaa.gov/reportcard/black_carbon.html" rel="nofollow">arctic.noaa.gov</a>. Rose

Comment on Open thread by its.not.co2@gmail.com

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No rising gas, no expansion or contraction, no surface, no direct solar radiation and no internal heat generation or long-term cooling is needed for there to be a temperature gradient which is always found in every planetary troposphere.

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