Pressure Depth in a Cave

[Charlie99]

You run into the same problems that NASA does with astronauts going EVA. You'd want to prebreathe a bunch of pure O2 to wash the N2 out of your system first.

Exactly. Or if you don't prebreathe O2, you'll bubble enough to get bent after 10 or 20 minutes at the greatly reduced pressure.

You can get some pretty counter-intuitive stuff going in cases like this. Another really strange thing my father was involved in many years ago was a large storage tank that needed to be repaired. It was a typicay industrial tank for liquids a few 10's of feet in diameter, maybe 15 feet tall, and a top that was close to flat. The operators had managed to suck in the top of the tank by running a pump until the tank was completely emptied. Tanks like that are pretty strong in one direction, but are easily destroyed by just pulling a small vacuum.

The on-site engineer was going to pop the dished in tank top back out by using compressed air to pressurize the tank. That of course, is going to put a lot of energy into the tank, which can possibly explosively fail and kill people. My father's solution was to simply fill the tank with water up to the dished into tank top, then hook about 15' of garden hose to a fitting on the tank top. Then he took another garden hose and just starting dribbling water into the open hose end held a few feet above the tank top. Just a few feet of water pressure, when multiplied by the area of the tank top is a tremendous force. The tank top just slowly popped, crackled, and came back up to the correct shape.

 

[captain]

nope... gas has mass and weight as well -- otherwise we wouldn't have 1 ATA at the surface caused by the column of air over the divesite. the density of gas makes the weight of the column of gas negligible though for all practical purposes with containers of reasonable size (although the density of gas in a scuba tank should mean that the pressure differential between the top of the tank and the bottom would be much greater than an empty scuba tank -- 10 lbs over the cross sectional area of the inside of the tank in square inches would give the psi differential for an HP130).

Agree, but as you stated the weight of the gas is negligible. Spitting hairs so to speak.

 

[Blackwood]

My argument is that in a lab environment, and yes clearly non-practical and totally theoretical, if we were to construct a pipe system in the shape of a giant W where the outer points of the W is at 1ATA (A&E) and the bottom of the legs where the W touches the ground is at 4ATA(B&D,) the inner peak in between the legs that point upwards (C) will then be at 4ATA also because the pipes are providing an impermeable, in my words (perhaps inaccurately) "sealed", environment. This will be the case because Pascal's laws state that in a sealed container an increase in pressure introduced into a liquid will result in that pressure force to be equally exerted everywhere throughout the whole "sealed" system at right angles at every point where it encounters a barrier - in this case the sides of the pipes. The atmospheric pressure of 1ATA then at points A&E will act as plugs that could just as well have been watertight weighted plugs that exert 1ATA pressure on the water at A&E (hence my thoughts about a "sealed" system - because the water cannot go anywhere outside the pipe system.

The W isn't a sealed container. Neither is the cave. It's open at A&E.

 

[Charlie99]

My argument is that in a lab environment, and yes clearly non-practical and totally theoretical, if we were to construct a pipe system in the shape of a giant W where the outer points of the W is at 1ATA (A&E) and the bottom of the legs where the W touches the ground is at 4ATA(B&D,) the inner peak in between the legs that points upwards (C) will then be at 4ATA also because the pipes are providing an impermeable, and in my words albeit being a somewhat inaccurate choice of wording, "sealed" environment. This will be the case because Pascal's laws state that in a sealed container an increase in pressure introduced into a liquid will result in that pressure force to be equally exerted everywhere throughout the whole "sealed" system at right angles at every point where it encounters a barrier - in this case the sides of the pipes. The atmospheric pressure of 1ATA then at points A&E will act as plugs that could just as well have been watertight weighted plugs that exert 1ATA pressure on the water at A&E (hence my thoughts about a "sealed" system - because the water cannot go anywhere outside the pipe system.

Think through what you just said. Why are points B an D special? If point C is going to be at 4ata because B and D are, why can't you just say that everyplace in the pipe is going to be at 1ata because A and E are.

You are misapplying Pascal's law. even as you stated above it refers to the INCREASE in pressure. If you were to increase the pressure at A and E from 1ata to 10ata, that INCREASE would should up equally everywhere. The pressure differences between the various points (for example 2ata difference between B/D and C) will stay the same.

 

[H2Andy]

Think through what you just said. Why are points B an D special. If point C is going to be at 4ata because B and D are, why can't you just say that everyplace in the pipe is going to be at 1ata because A and E are.

they are ALL going to be at water pressure atmospheres plus one air pressure atmosphere

take your depth, divide by 33, add 1, you got it

 

[Melicertes]

The W isn't a sealed container. Neither is the cave. It's open at A&E.

Okay so plug A&E with two piston heads that fit perfectly into the diameter of the openings at A&E and weight them to exert 1ATA pressure on the liquid. You've just capped the water in the same way the air columns above A and E did and which exerted the same pressure of 1ATA on the water - incidentally this 1ATA is accounted for in the 4ATA pressure value at B&D because the water weight alone would only exert 3ATA. How is this not a sealed system?

 

[reefnet]

In the case of a closed cave system where C is not exposed to the atmosphere my answer is bubbled in at 4ATA.

This thread is absolutely shocking. Anyone who is convinced that the answer is 4 ATA should be flogged with his C-card.

The original poster's question is really just a convoluted version of the Hydrostatic Paradox.

You mustn't let looks deceive you...the pressure at point "C" has NOTHING to do with the solid rock above. It is defined solely by the height of the free water surface, and is therefore 2 ATA. You could replace the rock with marshmallows or pixie dust and the answer would be the same.

If you are still not convinced, consider the point 33ft down from the cave system entrance. The pressure there is certainly 2 ATA. If the pressure at C were anything other than 2 ATA there would be a pressure differential and the cave system would become an underwater river. For the system to be in equilibrium, point C MUST be at 2 ATA.

Class dismissed.

 

[Melicertes]

Man I love a good clean argument. Keeps one on one's toes

 

[Melicertes]

Think through what you just said. Why are points B an D special? If point C is going to be at 4ata because B and D are, why can't you just say that everyplace in the pipe is going to be at 1ata because A and E are.

Because A&E do not have the weight of the 99ft of water bearing down on it the way B&D does - that kinda makes them very different in terms of pressure, and I'll call them special in your words

You are misapplying Pascal's law. even as you stated above it refers to the INCREASE in pressure. If you were to increase the pressure at A and E from 1ata to 10ata, that INCREASE would should up equally everywhere. The pressure differences between the various points (for example 2ata difference between B/D and C) will stay the same.

I'd call adding the weight of a water column of 99' (A&B and E&D) plus 1ATA atmospheric pressure a pretty nice increase in pressure into the system between points B and D.

PS: I'm not trying to stir a fight here guys. I think we are drifting off the topic slightly (or not so slightly) because I am arguing a perfect lab environment model as opposed to a real life cave system model and there are differences between the two which allows me to see a possible 2ATA answer to the real cave model and I maintain a 4ATA answer to the lab model. I'd love to continue the discussion, just know I'm not trying to trawl here

 

[reefnet]

How is this not a sealed system?

It is a system in hydrostatic equilibrium. Period.

I'm not even sure what you mean by "sealed system", nor how that has any bearing on this problem. Every version of your analysis so far is flawed.