Less psi in chilled tanks = less air?

[MyDiveLog]

Yes, the PSI drops when a tank is cooled. No, the amount doesn't change (it can't because the tank is sealed shut); it must gets more dense.

Think of it like this: You have a bunch of atoms of gas bouncing around inside the tank. When the tank is heated, they get more energy and bounce harder (space farther apart), which makes them push against the inside of the tank harder ... pressure rises. The opposite happens when the tank is cooled; they lose energy, slow down, get closer together, and press less hard against the inside of the tank, so pressure drops.

When you refer to a tank's "Rated Capacity," you actually use a shorthand phrase. You say it's an "AL 80," meaning it contains 80 cubic feet of air when the pressure is 3300 PSI inside it ... implicit in that, but unstated, is an assumed temperature (basically room temp). Why do you care about this? If the shop fills the tank very fast the tank will heat up (e.g. gives you a "Hot Fill"). The pressure gauge will say 3300 PSI but, when the tank cools down, the pressure will drop; you didn't really get 80 cubic feet of air. Go from room temp into the cold CA water, and it will drop even further.

How much does Pressure drop with temperature? As a quick rule of thumb figure about 5 PSI for every degree F....if the tank feels warm, but not burning, immediately after filling that's probably about 20 degrees above ambient, or 100 PSI.

If you are more comfortable with algebra, all the gas laws (Charles, Boyles, etc.) are variations of the ideal gas law (PV = nRT). "n" and "R" are constants that cancel out when comparing two different states, so it becomes (P1 x V1) / T1 = (P2 x V2)/T2. T isn't in Farenheight or Celcius, but Rankine...which is F + 460.

Simple enough?

 

[lamont]

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.

The warmer the gas the higher the kinetic energy of the individual gas molecules (actually temperature is the direct result of higher kinetic energy). The pressure is due to the collisions on the walls of the container. To make a higher pressure you can either increase the number of gas molecules colliding with the walls (increase the mass of the gas in the container) or else increase the energy and momentum of the individual particles (increase the temperature of the gas).

What you're saying is true for a constant volume and pressure of gas that a higher temperature results in fewer gas molecules and more space between gas molecules (i.e. this is true in the case of air that you breathe into your lungs). Cold, however doesn't necessarily imply denser. The colder scuba tank is still just as dense, but with the mass of gas constant and the volume constant what varies is the pressure.

What really matters in the case of breathing off a scuba tank is the temperature of the gas which you are inhaling. If you're inhaling colder gas then the constant volume of your lungs (and at a particular constant pressure based on a constant depth) will result in more air mass in your lungs, and you should go through air faster. I have *no* idea how the temperature of the tank relates to the temperature of the air in your lungs. It probably has some kind of effect, but i bet its much less than the temperature drop caused by the adiabatic expansion of the air dropping from tank pressure to the IP and then to ambient pressure.

 

[-=>Larry<=-]

We've got the perfect gas law working well [although in Imperial units and Rankine...Wow!] - and the Temp. vs pressure relationships

but there's still this

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.

I didn't want to let Desa off.... Think about this - Closed tank [same number of gas molecules] and relatively constant volume. - This means that he space between gas molecules will also be relatively constant. What happens is that with lower temperature - their velocity will be lower - and therefore the pressue.

Now - don't go opening the tank and/or breathing off of it - Then suddenly we get into ambient pressure conditions and adiabatic cooling and....Oh my!

 

[SDAnderson]

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

Thanks!

Angie

I'll bet you're glad you asked...

 

[Angie S]

Thanks for everyone's replies. I still need to noodle over this some more, but you haven't completely lost me in your explanations. There's a brain-tease aspect to it that's kinda fun to think about.

Angie

 

[lamont]

really its pretty simple. you have a constant mass of gas in your tank no matter what the temperature/psi is.

the important thing is the mass/time going through your lungs. if you're breathing steady that is going to be a constant volume of gas per time. so the important variables are ambient pressure (how deep you are) and temperature of the gas in your lungs.

n = PV / RT

Pressure = ambient pressure at depth
Volume = (constant) volume of lungs
n = moles of gas (what we're looking for)
R = (constant) gas constant
T = temperature of air in lungs

 

[willydiver]

Assuming the volume of the tank stays constant (although it does expand/contract a microscopic amount with pressure changes), there is actually no more space between air molecules in a tank of 160F air than one of 80F air. The volume (space inside the tank that the air can occupy) remains the same but the velocity, that the individual air molecules travel at, does increase.

Hot air molecules zig-zag around faster than cold air molecules. As you add energy (heat) to the air molecules', their velocity increase is measured indirectly as a pressure increase. Think of the molecules as saying "Hey man, it's getting hot in here!" So they bounce off the tank walls faster and harder due to their increased velocity. In a fixed volume, the air's density cannot change unless more air (air molecules) are added.

When you get a "hot" air fill, the hot captured air molecules are agitated and doing their thing moving and bouncing around very quickly. As they cool and "calm down" their reduced energy state translates to a lower pressure.

SO, A "cooled off" tank will not last as long on a dive (as was the general concensus, anyway...). Lower pressure for a given volume translates to less breaths. It's been a few years since Thermodynamics class.

WD

 

[-hh]

really its pretty simple. you have a constant mass of gas in your tank no matter what the temperature/psi is.

Well, you have a constant mass until you use some

n = PV / RT

Pressure = ambient pressure at depth
Volume = (constant) volume of lungs
n = moles of gas (what we're looking for)
R = (constant) gas constant
T = temperature of air in lungs

Absolutely correct, although I personally prefer the same equasion in a different form to assist in understanding it, namely:

nR = PV/T

the "nR" functionally represents the total amount of substance you have.

and the PV/T half represents what that mass "looks" like.

One reason I prefer this form is because the "R" gas constant isn't exactly a constant: it also includes unit conversions and other messy stuff and changes as you move between Metric and different flavors of Imperial Units. As such, this form makes it easier for me to eliminate it..all it takes is the assumption that you're dealing with an unchanging, fixed mass, so that you can think of this equasion in this form:

nR = constant = PV/T

Now, as the tank is taken to different temperatures and the like, you apply it as:

nR = constant = PV/T = P(1)V(1)/T(1) = P(2)V(2)/T(2) = etc

...where the (1), (2) stuff is used to represent stuff like "before" and "after" changes in conditions.

For example, how much would we expect the pressure in a 3,000psi tank @ 120F (a hot car trunk, or a hot tank fill) to change when we then do a dive in 60F water?

This means:
Before:
P(1) = 3000psi
V(1) = whatever the tank's interior volume is
T(1) = 120F

After:
P(2) = X
V(2) = ditto...and no real change from V(1)
T(2) = 60F


First, since the tank's volume's not going to change, we can drop it from the equasion, so it becomes P(1)/T(1) = constant = P(2)/T(2).

If we solve from here, we're going to say:
(3000psi)/(120F) = (X)/(60F)
(3000psi) * (60F)/(120F) = X
(3000psi) * 1/2 = X
X = 1500psi

...but this is wrong (and a classic trick question on Exams)

Our error was in not realizing that the Fahrenheit and Celcius scales allow negative temperatures, and they would result in a negative pressure (or negative volume), which is impossible. This is why the Absolute temperature scales (Kelvin & Rankine) exist, and they must be used in the Ideal Gas Law formulas.

R = F+459.67, so trying again:

P(1) = 3000psi
T(1) = 120F ---> 579.67 R (let's call it 580R)

P(2) = X
T(2) = 60F ---> 520R


(3000psi)/(580R) = (X)/(520R)
(3000psi) * (520R)/(580F) = X
(3000psi) * (0.897) = X
X = 2691 psi

...which is just over a 300psi drop (10% of tank) in indicated pressure from a "big" 60F temperature drop (because its actually only a 10% change on an Absolute temperature scale...makes sense, eh?).


-hh

 

[WarmWaterDiver]

I prefer

PV = ZnRT

Where Z = compressibility factor, function is compensation for actual behavior vs. ideal gas behavior

For truly ideal gas behavior, Z = 1

The closer Z is to equaling 1, the more closely the gas behavior approaches ideal gas behavior.

And, for the original poster, don't feel intimidated by this. Itried giving an example months ago of two different tanks, all equal in size and internal pressure, with the only difference being temperature, so the tank with the lower temperature has more gas inside (there's that mass thing again in a closed container). I got replies from folks including moderators that somehow lower temperature makes for instant more gas in the same tank as a closed system - 'poof!' and this was even in replies from a board moderator! There are disconnects in even the concept of 'two tanks' not meaning 'a tank, and another tank' - some interpret 'two tanks' as 'one tank, and the same one tank'. If folks aren't even on the same basis for math, more detailed discussions like this thread are pointless. My hats' off to you!

 

[WJL]

...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....

Angie,

As previous posters have pointed out, if you have a closed cylinder of gas, heating it up or cooling it down makes the pressure go up and down, but doesn't change the number of gas molecules inside the tank. Cooling a hot tank decreases the pressure, but the amount of air in the tank stays the same.

So, cooling the tank does not mean you have less air in the tank, but you asked a slightly different question - you asked if cooling the tank means you have "less air to breathe." Since we are talking about this in the context of breathing from a pressurized tank through a scuba regulator, the answer to that question is, you will have a tiny bit less air to breathe from a cooler tank, I'm guessing on the order of a few breaths.

This is due to the way a scuba regulator works. The first stage steps down the tank pressure to an imtermediate pressure and delivers that to your second stage. When the pressure in the tank gets below the intermediate pressure for the regulator, the regulator stops working. So all things being equal, if you are getting down to the last bit of gas in your tank, having the tank hotter will increase the pressure slightly and keep it working longer than with a cooler tank. I haven't done any calculations to measure this effect, but it would be tiny at best.

This is not going to have any practical effect on your diving.