Less psi in chilled tanks = less air?

[Angie S]

I'm hoping someone here can help me get my brain around something -- in thinking about the effects of temperature on a gas. I understand why, if you take a cylinder of air into cold water, it will lower the psi in the tank. What I'm not sure about is: Does that mean that you now have less air to breathe? That would seem to be the case, but it's not like you have lost any air from the tank, it's just at a lower pressure... I'm trying to make some sense of this.

I'm no physicist, so if anyone can explain it to me in simple terms, I would so appreciate it.

Thanks!

Angie

 

[utnapistim]

Hi Angie,

if I remember correctly from highschool physics, Boyle's law states that for a given mass of gas the temperature, pressure and volume are in a direct relationship. This means that if you change either of them, the other two are bound to change.

In this case, the volume is forced to be constant (the tank holds the volume constant), and when you change the temperature, the pressure drops.

It's the same law that assures you that your lungs might explode in uncontroled ascents: in that case the air in your lungs is at the same temperature, but as the pressure drops when you ascend, the air expands; being the same amount of air).

just hope I didn't confuse you completely

 

[Desa]

I might be off track here, but I think it has to do with: the warmer the gas, the more space between molocules. The colder and therefore the more denser the gas, the less space between molocules. If you are breathing a given quanity of air, if it is more dense, you are infact taking in more molocules there for there will be less gas available for breathing.

If I am way off track, someone will be sure to let you know.

 

[SDAnderson]

Short answer, yes. Though you haven't altered the number of molecules of gas in the tank, by reducing the temperature you have increased the number of molecules that will occupy a given space. Thus, the reduction in pressure corresponds to a reduction in the number of breaths you will be able to take.

I'm hoping someone here can help me get my brain around something -- in thinking about the effects of temperature on a gas. I understand why, if you take a cylinder of air into cold water, it will lower the psi in the tank. What I'm not sure about is: Does that mean that you now have less air to breathe? That would seem to be the case, but it's not like you have lost any air from the tank, it's just at a lower pressure... I'm trying to make some sense of this.

I'm no physicist, so if anyone can explain it to me in simple terms, I would so appreciate it.

Thanks!

Angie

 

[TwoBitTxn]

You can have an 80 cuft steel cylinder that operates at 2250psi and an Aluminum 80 that operates at 3000 psi. You still have 80 cuft. The pressure is different. That's all.

Desa said it right. As the temp of a gas increases so does the pressure. The volume stays the same (unless there is a leak). As you put gas under pressure it heats up.

This may be what is confusing you...

As you fill a cylinder, if you fill it too quickly it gets hot. The shut off is pressure not volume dependent. You can quickly fill an 80cuft cylinder to 3000psi and only get 75cuft because of the temperature/pressure relationship. As the cylinder cools the pressure drops. Not the volume. Or, you can slowly fill a cylinder, moderate the temperature increase and accomplish a more complete fill.

Make sense?

TwoBit

 

[simbrooks]

When we do cavern/cave dives we give our tanks a little cooling off time in the water (watch a 100psi drop or so), we do this to work out turn pressures and in a round about way the amount of gas we have (in case of dissimilar tanks ie 80 vs 130).

Every tank is rated at a certain volume and pressure at a certain temp (although i cant recall what that temp is - 75/80F?). So from these base points say 77.4cuft @ 3000psi (for an AL80) you can work out your volume in the tank from the SPG reading (ie SPG/3000*77.4=actual vol). As reefraff stated you are in fact (in theory) losing gas, but IMO its not really there anyway as it was only registered as a pressure increase from the filling and once it gets back to the global environmental temp (the hot tank will try to equilibrate to that temp) that gives you a pretty close to accurate idea of how much gas you have.

Ex. If you fill to 3200, you might lose 2-300psi off a fairly common fairly fast fill, on a slow fill maybe only 100psi. The number of molecules in there is the same, but due to the heat they had - giving them energy to move around - making the higher pressure, you would have a "misreading" of volume of volume if you didnt correct for temperature (see below) - ie. dont plan your dive bottom time assuming 3200 (82.6cuft on an AL80) when in fact its closer to 3000 (77.4 cuft) once it has cooled down to the air/water temp.

There might be a small correction due to the standard temp associated with the rated pressure and volume, but the tanks way exceed that when they are filled, and come close to the rated temp when they are in water (at least around here). When you think about these temps (and pressure and volume), remember that they are absolute, that is Rankine (~492R=32F (i think) when using Farenheit - tedious imperial measurements) or Kelvin (273K=0C, when using Celcius/centrigrade), so the tank being 80F degrees hotter (160F) during filling is not twice the pressure of one at equalibrium with the environment 80F, but more like 14-15% higher pressure - which might be just over 400psi on a 3000psi rated tank. I hope that makes sense to you. The ideal gas law is simplified to P1V1/T1=P2V2/T2 or PV=nrT where n & r are constants for all intents and purposes. Hope that helps, although i have probably just muddied the waters

 

[Desa]

Thats a very impressive explanation. Heck, it even makes sense. But,,,,I think Rankin is 460 plus the temp which would make 32F =492R. I could be wrong. Infact if I am remembering correctly, Rankin is actually some rediculous number like 459.something?

I really did like the "real world" analogy. IE putting the tank in the water for a while to acclumate it. While that would work in the caves and springs, it isn't quite as effective up here where the water temp can very as much as 45 (F) from surface to depth.

Good job Simon, Oh, and congrads on mod

 

[simbrooks]

But,,,,I think Rankin is 460 plus the temp which would make 32F =492R. I could be wrong. Infact if I am remembering correctly, Rankin is actually some rediculous number like 459.something?

I really did like the "real world" analogy. IE putting the tank in the water for a while to acclumate it. While that would work in the caves and springs, it isn't quite as effective up here where the water temp can very as much as 45 (F) from surface to depth.

Thanks for the correction, have just edited my post for clarity - sorry not used to dealing in imperial units, the Brits dont use them that much anymore, i went through school on metric, only when i got here did they insist

I checked out the value on the convert program (use it all the time for engineering stuff) and it said 492R=32F or 460R=0F, so i concur on that one. Acclimation of the tank to various temps might be about 7-8% (rough guess) for 45F water in terms of a psi difference between environment and rated temp. So taking your 80F take out at 3000psi would cool to 2800psi in 45F water, but still 80cuft in theory if it were 80F again.

 

[texdiveguy]

Can you guys repeat all the above,,,,I missed something....

 

[-=>Larry<=-]

-- in thinking about the effects of temperature on a gas. ...

I'm no physicist, so if anyone can explain it to me in simple terms, I would so appreciate it.

Angie

There's a very nice illustration you can do - where you can relate pressure directly to the speed of all the little gas molocules bouncing off the sides of the tank - - - When you cool the gas - It slows it down - so the pressure you measure is less. You still have the same number of molocules - - They just press less.

If you prefer the perfect gas laws and algebra:

pV=nkT -> keep V,n,k as constants - - so then you have p=Constant*T

and yes - temperature must be measured in Kelvins or Rankine [note spelling] - to make the math work out.