Using a regulator in cold water.

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Just two points, then I give up.
1) In refrigerators, heat pumps, etc. the fluid passing through the lamination valve is not a gas, it is a liquid which changes phase, becoming steam. The temperature drop is not given by the gas expansion, but is given by the change of phase, which absorbs a quite relevant amount of energy, called the "latent heat of vaporization". Had it been a gas, with no phase change, the temperature drop had been minimal (as it happens with air, which is "almost" a perfect gas). And if the gas was truly perfect, the temperature drop had been zero.
2) When filling the cylinder the compression is almost reversible, hence an enormous quantity of heat is generated. Approximately 100 times greater than the quantity of energy causing cooling during free expansion in the regulator. Of course a significant part of it is removed already inside the compressor, or in the finned pipes which connect one stage of compression to the next. In fact, this is not an adiabatic process, and the more heat you manage to remove during the compression, the less power is required by the compressor engine.
So the compression is an entirely different process than the expansion: during compression a lot of work is applied by the engine to the air being compressed, and also a lot of heat exchange for avoiding the temperature increment (which, without cooling, would exceed 100 °C).
Instead expansion happens with no work being released, and with minimal heat exchange. If the gas was truly perfect and the output kinetic energy is negligible, the expansion will be perfectly isotherm. As the gas is not exactly perfect, and there is small kinetic energy in the output flow, there is a small temperature loss.

At this point I give up with explanations, because the point for me remains the same: avoid flowing a lot of air when close to the freezing point, as this would increase the amount of kinetic energy subtracted to the gas, causing a further, significant cooling effect. The most of the cooling effect is due to the air speed, so minimizing the speed minimizes the cooling effect. This, in the end, is all what we need to know...
 
the point for me remains the same: avoid flowing a lot of air when close to the freezing point
On this, we agree completely.

The most of the cooling effect is due to the air speed
Joule-Thompson has been mentioned upthread. I'd really like to see calculations and numbers supporting this claim (that kinetic energy is responsible for the majority of the cooling, and, as a corollary, that the Joule-Thompson effect is negligible).

Just a rough estimate:
The J-T coefficient for N2 at 25 degrees C and roughly 200 bar is 0.76980 K/MPa (Dortmund Data Bank).
The J-T coefficient for N2 at 25 degrees C and roughly 15 bar (close to IP) is 1.29287 K/MPa (Dortmund Data Bank).
The J-T coefficient for N2 at 0 degrees C and roughly 15 bar is 1.08562 K/MPa (Dortmund Data Bank).

J-T: μ = (T1 - T2) / (P1 - P2)
T2 = T1 - μ(P1 - P2)
T2 = 25 - 0.76980(20-1.5) = 10.7.

So if we have no heat transport into the 1st stage, we see roughly a 14 degree C temperature decrease solely from the J-T effect. For me, that isn't negligible.

Disclaimer: this calculation is a very rough estimate, assuming that air = pure N2, and not taking into account the change in J-T coefficient with pressure and temperature. For an accurate estimate, I'd have to check my thermo diagrams for air, which I don't have with me right now. But it ought to give a ballpark number. Which probably is on the conservative side, as the J-T coefficient for 15-ish bar is higher than that for 200-ish bar. If I just average the J-T coefficients for 25C/200 bar and 0C/15 bar, I get a temperature drop of some 17 degrees C. Still in the ballpark and, to me, far from negligible.
 
https://www.shearwater.com/products/swift/

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