Less psi in chilled tanks = less air?

[willydiver]

OK, let's throw this one out just for kicks and the sake of interesting conversation.

Imagine your tank/compressed air are at 50F due to the temperature of the water your are diving in. So, you are breathing from a "cooled" tank. Assume the breath of air delivered from your regulator to your lungs is warmed during the path from your mouth to your lungs, to around 98.6F.

Does this mean that the air has expanded by the time it reaches your lungs, thus a slight increase in pressure (inside the lungs) is realized? Will this additional expansion of air result in the taking of smaller breaths from the regulator? Won't this counteract some of the "lost pressure" syndrome encountered with a cooled tank? After all, the important thing with breathing at depth is the pressure inside your lungs to counteract the surrounding water pressure.

Think about this, too. Cooled air (outside of a fixed volume) is more dense than warm air. If you are breathing cooled air, aren't you taking in a higher concentration of oxygen molecules per breath? Won't this translate into your body requiring slower breathing (less air demand) for a given activity. Will this also counteract some of the "cooled tank" syndrome?

I turned off my common sense switch while blurting this out, so don't blast me too bad if it doesn't make sense at all. In reality, I think the changes in volume/density are so small as to not matter.

WD

 

[Charlie99]

Think about this, too. Cooled air (outside of a fixed volume) is more dense than warm air. If you are breathing cooled air, aren't you taking in a higher concentration of oxygen molecules per breath? Won't this translate into your body requiring slower breathing (less air demand) for a given activity. Will this also counteract some of the "cooled tank" syndrome?

Removal of CO2 is what determines how much air you need to breathe underwater. You have more than enough oxygen at depth. This is also why using nitrox doesn't materially affect SAC.

To the first approximation, you need constant volume of breathing to get rid of CO2 ---- that's why you breath about 4 times as much mass of air at 100' as at the surface.

To put the discussion of tank air temp into perspective, consider that cooling a tank from 104F/40C in the hot sun to 68F/20C in the water reduces the pressure by about 20K/300K or 7%. This does cause a noticeable reduction in tank pressure right at the beginning of a dive.

 

[WJL]

....
Think about this, too. Cooled air (outside of a fixed volume) is more dense than warm air. If you are breathing cooled air, aren't you taking in a higher concentration of oxygen molecules per breath? Won't this translate into your body requiring slower breathing (less air demand) for a given activity. Will this also counteract some of the "cooled tank" syndrome?
.....

Remember that you are only using a small percentage of the oxygen in each breath you take in. Increasing the oxygen concentration slightly won't affect the rate at which your body extracts oxygen out of the air you're breathing.

That's why rebreathers are so much more efficient - they recapture the unused oxygen that otherwise is "wasted" when using open-circuit scuba.

 

[willydiver]

That's right... The build up of carbonic acid in the blood/lungs is what gives you the urge to breathe. Thanks for the physio, refresher.

Great point, also, on the small use of the available O2 in the gas you are breathing. I like the rebreather analogy.

This site is great for information.

WD

 

[WarmWaterDiver]

Also, the pressure inside your lungs while inhaling cannot be greater than the pressure at the second stage regulator mouthpiece. Flow occurs in direction of higher pressure to lower pressure. So you actually have a bit of additional 'expansion' during inhale.

 

[-hh]

OK, let's throw this one out just for kicks and the sake of interesting conversation.

Imagine your tank/compressed air are at 50F due to the temperature of the water your are diving in. So, you are breathing from a "cooled" tank. Assume the breath of air delivered from your regulator to your lungs is warmed during the path from your mouth to your lungs, to around 98.6F.

Does this mean that the air has expanded by the time it reaches your lungs, thus a slight increase in pressure (inside the lungs) is realized? Will this additional expansion of air result in the taking of smaller breaths from the regulator? Won't this counteract some of the "lost pressure" syndrome encountered with a cooled tank?

Great question.

I think that the _general_ answer is the air is effectively "preheated" by your body before it gets to your lungs. As such, it has already expanded as far as your body and diaphram are concerned for purposes of what its volume is.

However, I have noticed that my SAC is higher on coldwater dives than on warmwater ones, which could infer that the above is wrong. But I don't attribute this higher SAC due to "denser air", but instead due to an increased workload to stay warm, which has increased metabolic demand.

One thing that we need to remember through all of this is that the body is pretty inefficient - - this is why mouth-to-mouth receccitation works to rescue people - - so a lot of these "5 percent" issues are lost in the noise.


-hh

 

[-=>Larry<=-]

Imagine your tank/compressed air are at 50F due to the temperature of the water your are diving in. So, you are breathing from a "cooled" tank. Assume the breath of air delivered from your regulator to your lungs is warmed during the path from your mouth to your lungs, to around 98.6F.

Interesting to consider - aside from the above comments - the expansion ratio is only 50 degrees Rankine / 475Rankine - or about 10% - - Similar to the amount the pressure drops in cooling tanks.
If you consider the metabolic work that goes into warming the air - I'm guessing you could barely see the difference.

 

[GDI]

Without reading all the threads

P1V1/T1 = P2V2/T2

Ok the physical dimensions of the cylinder will not change. So let us introduce Charles Law into the Bolye's Law equation. aka the Ideal Gas Law. In 1610 Robert Boyle reported that if a given quantity of gas is held at a constant temperature and the volume varied over a wide range, the absolute pressure also varies in such a way that they remain constant. in studying the pressure -volume relationship Boyle noted that as the value of the constant was increased so was the pressure. Joseoh Gay-Lussac published observations in the late 19th century stating that if a given mass of gas was maintained at a constant pressure, the volume would vary directly with the temperature. In 1787 Jacgues Charles had shown this relationship to be:

Volume/Temperature= Constant

Combining this with Boyles Law we have:

(Pressure x Volume)= (constant x temperature)

 

[AggieDad]

[

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.

This statement made me laugh but not because it isn't correct...it is, I just had this picture come into my mind of a kindergarten classroom with the teacher being the pressure gauge and the kiddos all bouncing around and her pressure rising. When the kids "cool off" there are still the same number of kids in the same amount of classroom but the teacher's pressure goes down.

Thanks, algebra and any math over 2 + 2 makes me feel narced but now even I understand this concept.

 

[peterf]

...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[/QUOTE]

True to an extent but you also need to remember that the pressure (P) in the equation must be Absolute NOT guage pressure. Therefore to calculate exactly you would need to add one atm to the starting guage pressure and then deduct one atm from your final answer.