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Correct.
Correct, and good on you to walk back on this:
Using a regulator in cold water.
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As said, these are perfect gas laws, but air, albeit being close to this ideal, is not entirely perfect.
So in a lamination valve the enthalpy stays ALMOST constant, resulting in temperature loss of just one-two Centigrade degrees (at moderate flow), while if the expansion was fully adiabatic-reversible (isoethropic) the temperature loss could be in the order of 100 centigrade degrees, or more.
So, a small temperature loss of 1-2 degrees is quite close to isotherm expansion, compared with the enormous temperature change which would happen in a truly iso-enthropic expansion.
Regarding the perfect gas law reported above, that describes EQUILIBRIUM conditions, it does not describe a fast adiabatic expansion, where the law is instead of the type p*V^n=constant.
The exponent n depends on the degree of reversibility (iso-enthropicity). For a biatomic gas, as air, a perfectly reversible (isoenthropic) adiabatic expansion will give you n=gamma=1.41. Instead, for a perfectly isoenthalpic expansion, n=1.000.
The real world is always in between. But at moderate gas flow, n is "almost" 1, something as 1.001, or perhaps 1.002, resulting in a minimal temperature loss.
Instead, when the gas flow is very fast (and a Scubapro MK25 can provide a VERY FAST gas flow, possible triple than a membrane reg) the exponent of the adiabatic expansion goes up, probably around 1.004, or perhaps even 1.005 (but still well below the maximum theoretical possible value of 1.41).
After a little math, one can rewrite the adiabatic equation in terms of p and T, instead of p and V, obtaining:
p*T^(n/(1-n)) = constant, which allows to compute the temperature after expansion, T2, knowing the temperature before expansion, T1:
T2 = T1 * (p1/p2)^((1-n)/n)
Assuming T=274 K, p1=200 bar, p2 = 8 bar, n=1.41 (isoenthropic adiabatic expansion), you get T2 = 107.5 K = -165.5 °C, that is way belong the freezing point!
Instead, with an almost-isoenthalpic expansion (which is what really occurs in your reg while breathing normally), the exponent is n=1.001, hence you get T2 = 273.1 K, which is still 0.1°C above the freezing point.
It is obvious that when you ask more air from the reg, you will get an expansion which is still almost isoenthalpic, but with an exponent which becomes, say 1.005. With such value you get T2=269.6 K = -3.4 °C, and you get freezing.
More info here:
Adiabatic process - Wikipedia
See the section titled "Adiabatic free expansion of a gas", where you can read:
"Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible."
So in a lamination valve the enthalpy stays ALMOST constant, resulting in temperature loss of just one-two Centigrade degrees (at moderate flow), while if the expansion was fully adiabatic-reversible (isoethropic) the temperature loss could be in the order of 100 centigrade degrees, or more.
So, a small temperature loss of 1-2 degrees is quite close to isotherm expansion, compared with the enormous temperature change which would happen in a truly iso-enthropic expansion.
Regarding the perfect gas law reported above, that describes EQUILIBRIUM conditions, it does not describe a fast adiabatic expansion, where the law is instead of the type p*V^n=constant.
The exponent n depends on the degree of reversibility (iso-enthropicity). For a biatomic gas, as air, a perfectly reversible (isoenthropic) adiabatic expansion will give you n=gamma=1.41. Instead, for a perfectly isoenthalpic expansion, n=1.000.
The real world is always in between. But at moderate gas flow, n is "almost" 1, something as 1.001, or perhaps 1.002, resulting in a minimal temperature loss.
Instead, when the gas flow is very fast (and a Scubapro MK25 can provide a VERY FAST gas flow, possible triple than a membrane reg) the exponent of the adiabatic expansion goes up, probably around 1.004, or perhaps even 1.005 (but still well below the maximum theoretical possible value of 1.41).
After a little math, one can rewrite the adiabatic equation in terms of p and T, instead of p and V, obtaining:
p*T^(n/(1-n)) = constant, which allows to compute the temperature after expansion, T2, knowing the temperature before expansion, T1:
T2 = T1 * (p1/p2)^((1-n)/n)
Assuming T=274 K, p1=200 bar, p2 = 8 bar, n=1.41 (isoenthropic adiabatic expansion), you get T2 = 107.5 K = -165.5 °C, that is way belong the freezing point!
Instead, with an almost-isoenthalpic expansion (which is what really occurs in your reg while breathing normally), the exponent is n=1.001, hence you get T2 = 273.1 K, which is still 0.1°C above the freezing point.
It is obvious that when you ask more air from the reg, you will get an expansion which is still almost isoenthalpic, but with an exponent which becomes, say 1.005. With such value you get T2=269.6 K = -3.4 °C, and you get freezing.
More info here:
Adiabatic process - Wikipedia
See the section titled "Adiabatic free expansion of a gas", where you can read:
"Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible."
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Angelo, the expansion of breathing gas in a scuba reg is most certainly irreversible and thus not isentropic. And since the heat transport from the outside into the reg is seriously limited, we are much closer to an irreversible adiabatic process that to an isentropic or an isothermal process.
The gas expands irreversibly, and since heat transport into the system is seriously limited, the temperature drops noticeably (quite analogous to how our tanks become hot when we fill them quickly, just that we see the opposite phenomenon). Sometimes below the freezing point of water. That's why unsealed regs are prone to freezing, and that's why you can see pictures on the web of first stages covered in ice.
As an aside, I don't particularly appreciate that extremely specialized terms like "isentropic", "isenthalpic" or "adiabatic" are used in discussions on a board for non-specialists. It could easily be regarded as attempts of Argument from authority which don't belong in real, honest discussions.
OK, iso-ethalpic means that the enthalpy does not change (which is what happens to the gas expanding in the regulator, hence the temperature changes very modestly - if the gas was really perfect, the temperature would not change at all).
Iso-enthropic means that the enthropy does not change. Which happens only if the process is without heat exchange and reversible, which of course is not.
So we are saying almost the same thing: the process is adiabatic (which means no significant heat exchange), and strongly irreversible (hence almost isothermal), as the elastic energy is not converted to useful work.
If it had been reversible, the temperature had dropped by more than 100 °C, instead it drops just of a couple °C, which in proportion is almost nothing.
You see a lot of ice just because the process is happening very close to the freezing point, so a modest temperature loss (as said, by just 2-3 °C) is enough for going form slightly above to slightly below the freezing point.
But on an absolute scale, the process is "almost" isothermal, the loss of temperature caused by a true reversible adiabatic expansion would be much more relevant.
Compare to what happens, instead, when filling the cylinder. In the compressor, you are making a compression which is quite close to be reversible (usually it is almost 75% percent reversible). Hence you get an enormous temperature increment, to the point that you need to place the cylinder in water for avoiding overheating.
And this despite the process is much slower (hence not adiabatic, as there is time to exchange heat).
In practice, you have a relevant increase of temperature when filling the cylinder, and a minimal temperature decrease when you laminate the gas in the regulator, emptying the cylinder.
A strongly asymmetrical situation...
You cannot compare what happens when filling (an almost reversible process, slow, and not adiabatic) to what happens in a properly designed regulator (the process is fast, adiabatic, and completely irreversible, as 99% of the elastic energy is dissipated in friction losses caused by purpose by the geometry of the lamination valve).
Iso-enthropic means that the enthropy does not change. Which happens only if the process is without heat exchange and reversible, which of course is not.
So we are saying almost the same thing: the process is adiabatic (which means no significant heat exchange), and strongly irreversible (hence almost isothermal), as the elastic energy is not converted to useful work.
If it had been reversible, the temperature had dropped by more than 100 °C, instead it drops just of a couple °C, which in proportion is almost nothing.
You see a lot of ice just because the process is happening very close to the freezing point, so a modest temperature loss (as said, by just 2-3 °C) is enough for going form slightly above to slightly below the freezing point.
But on an absolute scale, the process is "almost" isothermal, the loss of temperature caused by a true reversible adiabatic expansion would be much more relevant.
Compare to what happens, instead, when filling the cylinder. In the compressor, you are making a compression which is quite close to be reversible (usually it is almost 75% percent reversible). Hence you get an enormous temperature increment, to the point that you need to place the cylinder in water for avoiding overheating.
And this despite the process is much slower (hence not adiabatic, as there is time to exchange heat).
In practice, you have a relevant increase of temperature when filling the cylinder, and a minimal temperature decrease when you laminate the gas in the regulator, emptying the cylinder.
A strongly asymmetrical situation...
You cannot compare what happens when filling (an almost reversible process, slow, and not adiabatic) to what happens in a properly designed regulator (the process is fast, adiabatic, and completely irreversible, as 99% of the elastic energy is dissipated in friction losses caused by purpose by the geometry of the lamination valve).
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It's "enthalpic".
It's "entropy".
I've had my 2nd stage freeflow in room temp when I've bled down my tank via the reg set. That's a more than 20 degree C temperature drop.
Well, tell that to anyone who has had a serious freeflow or has bled down their tank by opening the valve. The process is most definitely not isothermal, which is obvious to anyone with a modicum of practical experience.
Opening the valve, without the reg doing his job of laminating the flow, of course results in a big temperature drop, as a signicant part of the original potential energy is converted to kinetic energy, which carries that energy away.
If the flow is correctly laminated, that energy converts back to heat almost entirely, which keeps the tempearature of air almost unchanged.
Depending on the design of the first stage, this massive energy conversion from potential to kinetic can also occur during free flow (induced, for example, by pressing the purge valve).
Some regs, such as SP piston ones, are designed for allowing a massive flow, hence the speed is high, and this subtracts energy, hence the gas' temperature drops.
If instead the first stage does not allow for such a large flow, most of potential energy is dissipated, keeping the air more warm. Of course there is always some air speed, hence some kinetic energy is lost.
But what must be understood is the cause-effect relationship. It is the fast air flow which subtracts energy and makes the air to cool significantly. Then some ice can lock the reg and maintains the free flow.
But all this does not start if one only ask for a small quantity of air, at a very slow rate.
This is the reason for which it is recommended to not press the purge button, and to use the BCD control very gently in very cold water.
If the flow is correctly laminated, that energy converts back to heat almost entirely, which keeps the tempearature of air almost unchanged.
Depending on the design of the first stage, this massive energy conversion from potential to kinetic can also occur during free flow (induced, for example, by pressing the purge valve).
Some regs, such as SP piston ones, are designed for allowing a massive flow, hence the speed is high, and this subtracts energy, hence the gas' temperature drops.
If instead the first stage does not allow for such a large flow, most of potential energy is dissipated, keeping the air more warm. Of course there is always some air speed, hence some kinetic energy is lost.
But what must be understood is the cause-effect relationship. It is the fast air flow which subtracts energy and makes the air to cool significantly. Then some ice can lock the reg and maintains the free flow.
But all this does not start if one only ask for a small quantity of air, at a very slow rate.
This is the reason for which it is recommended to not press the purge button, and to use the BCD control very gently in very cold water.
It has been a while since I created that reference, but I believe I based it on this portion from the Dive Lab document I linked to earlier: "When ice forms over the sensing ports of a piston first stage or over the diaphragm on a diaphragm first stage, the first stage will lose pressure sensing and the ability to regulate the outlet pressure to the second stage. In some cases this can cause a massive free flow of the second stage due to the first stage over driving. In other cases it can result in a lower intermediate pressure resulting in inadequate 2nd stage supply pressure."
https://divelab.com/download/procedures-and-guidelines/ScubaRegulatorFreezing-ChillingFacts-4-9-14.pdf
The author does not mention or differentiate between downstream and/or upstream valves, and appears to be saying either scenario is possible with the same regulator or type of regulator. None of this is as simple as we wish it to be.
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Yes, and it also explains why refrigerators, heat pumps and air conditioners work. Because if the expansion of the medium in those devices had been '"almost" isothermal', they wouldn't have worked.
And while I'm at it:
This is patently wrong. Because compressors are equipped with cooling between stages to avoid excessive heating of the gas. Yes, the tank gets warm, but far from to the degree you're claiming.
Which everyone who has filled a tank themself can confirm.
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