Compressed Air Cooling Joule Thomson Effect

It's the air, it doesn't have to go through a valve, it could be inside an expanding chamber

Thanks for that Passingtongreen, but my interest is in a breathing apparatus cylinder (fixed volume) at 300bar with a valve connected to a pipe at 10bar feeding a facemask at atmospheric pressure.

I need to assess the likely temperatures of the air in the cylinder and valve and the pipe and the facemask during the expansion process until the cylinder is empty (ie. 10bar).

Got anymore critical information to pass along, or do you expect the folks here at CR4 to keep throwing answers at you until we guess what the whole situation is?

At the slow rate for breathing, the change would be negligible...for air

Reply to Solareagle 4

You are probably right - but can you put a value to "negligible" please.

The JT coefficient produces a figure - circa 45°C - which seems a lot - but probably right in oerfect adiabatic conditions - but in my case although the instantaneous expansion might cool the air a lot - the surrounding metalwork heats the air - perhaps moreso at the point of cooling where the temperature is lowest .....and I wonder at what point it occurs.

I use an oxygen flask with a demand regulator when I am out, I have not detected any difference in temperature, and yes, I did try.

Visit a dive shop and take a look at some SCUBA tanks and plumbing. Then, buy one.

You seem to be re-inventing something that already works.

The temperature drop is not something to be dismissed offhand. Depending on specific conditions and equipment, the temperature drop can be significant.

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In a perfectly insulated ideal system, an immediate adiabatic expansion of an amount of dry 80 degree F air at 300 bar down to 10 bar would result in a temperature drop over 300 degrees Fahrenheit.

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It is easy to see why moisture in pressurized air can result in valves freezing.

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In most real systems, the pressure drop is not immediate, insulation is far from perfect, and the evolution is not really adiabatic, so you needn't worry too much about freezing your sinuses when you inhale.

If this is a experimental venture, I'd be much more concerned with the quality of the air (being free of oils and things like carbon monoxide) and safety of the regulators, i.e. working reliably and safely in all expected conditions... It isn't too hard to imagine a regulator (not designed specific to the purpose) potentially creating an overpressure condition that could result in an air embolism.

Reply to 6;

I have to query the 300F - 30F perhaps - but if 300F how does that arise?

You have made a direct hit with air quality. I have asked the question because international standards are being drawn up for air quality and the maximum permissible water content has to be set for the purpose of preventing internal freezing of the BA valves when used in low temperature. Hence the temperature drop is required.

Carbon monoxide is an issue, but that's another story.

300+F drop is a rough estimate of what will occur for an immediate drop in pressure with no heat transfer in or out of the air, based on the ideal gas law: PV=nRT

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In this case one unit of air at 80F and 300Bar undergoes adiabatic expansion to a little over 11.3 units at a little less than -256F and 10Bar

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For a quick check, we can verify that the Pressure x Volume / Absolute Temperature of the initial state is the same as in the hypothetical final state after adiabatic expansion.

(300Bar x 1 unit/540 Rankine) is not too far off (10Bar x 11.3 units/203 Rankine)

Reply to 10

Care is needed when juggling with PV = nRT. This is an excellent formula when dealing with before-and-after conditions. But more information and data is needed to deal with values during the transition between stages .

With P1, V1, T1, n1 and R known to start with, and only P2 known to finish with then V2, T2 are unknown except as a ratio. Enthalpy, adiabatic expansion and mass flow has to be taken into account to link P1.V1 to P2.V2 where the change in T creates the balance. Which for a perfect gas dT is zero.

Our atmosphere is not perfect and thus a temperature drop occurs. Similarly the processes are not adiabatic or constant enthalpy because there is heat transfer from outside the system to the cold air inside when there is a temperature difference. Thus any marked drop in temperature of the air inside the system will be limited to some extent to a figure less than predicted by JT .

It seems from other posts (thanks everyone) that the cooling is at the immediate outlet of the valve.

And Lyn - post 5 - dive shops have no idea how much the air cools - so why should I spend my money with people who can't answer my question - which incidentally is for their benefit.

OK.

It seemed a good idea at the time.

'...Care is needed when juggling with PV = nRT...'

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Because I was only trying to give you an idea of the change in temperature possible due solely to reduction in pressure, exclusive of other influence, and because this is not exactly an adiabatic process, a rough estimation seemed warranted.

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While I did not take time to detail each step, I also didn't just stab at a guess. I used a ratio of specific heats of 1.4 for air to estimate the work done, which allowed V2 and T2 to be determined independently.

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If you take the time to do a more accurate calculation, my rough estimate won't be far off. As this was intended only to illustrate the magnitude of changes possible and because it isn't exactly what is happening, delving into the minutia seems misguided.

Hello Truth is not a compromise - can I call you 'Tinac' - it saves time typing!

My comment on PV=nRT was largely for general consumption by other followers of this thread.

Yes you are right, we don't want to get bogged down by trivia. As an engineering tool I have made assumptions about the conditions, as you seem to have done, where we apply general principles to predict an outcome. Which ever way you look at it, there appears to be a temperature drop when the pressure drops.

In my case the exact temperature drop is important because this has a direct bearing on the Dewpoint (or icepoint as is likely). Using PV=nRT by itself only tells us V/T is a constant - without telling us what V or T is. Alternatively using PVⁿ=k, a constant, might help, if we knew V.

It is necessary to know the temperature drop, where the nearest I can get is to use the JT cooling coefficient (circa 45C) where we only need to know the initial temperature and pressure drop.

The object of my exercise is to establish what the cooling is likely to be, and where the actual cooling is greatest, to assess the liklihood of ice particles forming in the air, the possible size and distribution, and the possibility of these particles internally blocking the demand valve to stop the air flow, or maybe blocking the mechanism to stay open thus to cause overpressure in the mask, where either situations could be fatal.

Specifying very low dewpoints would appear to solve the problem, but not based on the maximum possible temperature drop, because the would be impracticable.

Dear Noddy Timbs, Please research a device known as the "HILSCH TUBE" It is an interesting device that uses only compressed air to cool and heat.

Dragon

Whilst adiabatic phenomena help us to understand the relationship between pressure, volume and temperature it is important to note that this is not entirely an adiabatic system.

When the bottle is venting it is open to the atmosphere. The mass of the volume of stored air is therefore reducing.

The valve body is the only place in this system where there is an opportunity for anything adiabatic to occur. Adiabatic expansion occurs at the valve orifice where the air is escaping. The air cools down at that location. The heat in the escaping air is converted to kinetic energy at that location.

'....Adiabatic expansion occurs at the valve orifice where the air is escaping. The air cools down at that location.....'

...Yep, 'cooling' occurs there for the majority of the air (except the small amount leftover in the tank). As such, freezing could be a concern in that area.

Try this website. They use the principle in their products.

http://www.vortec.com/

It is on the cylinder; in 1994 I automated a piece of equipment that included a hybrid cylinder, actuated by compressed air, but displaced oil to achieve smooth motion of the mechanism. In order to settle the gides, I wrote a routine to retract and extend the slide at a rate of about 4 seconds per cycle, left it running overnight, and when I came to work in the morning I was surprised that the clear chamber of the cylinder was cold as a beer; hint: the exhaust was quicker than the charging.

Regards

ACCORDING YOUR QUESTION

-EXIST TWO POSSIBILITY

a) EXPANSION ISOENTHALPY

b) ISOENTROPY

CASE a) WHEN YOU OPEN VALVE, THE GAS WILL BE EXPAND AND DE TEMPERATURE FELL DOWN FEW GRADES, IT DEPEND OF HOW HIGH ARE THE PRESSURE

IN YOUR QUESTION DE GAS INSIDE AND OUT SIDE, FELL DAWN

CASE b) IN THE PROCESS PLANT USED EXPANSION MACHINE O TURBINE MACHINE, BECAUSE DE TEMPERATURE FELL DAWN ABOUT 40 / 70 C CENTIGRADE

FOR MORE INFORMATION SEE DE DIAGRAM T-S FOR EACH GAS

HAPPY NEW YEAR