Pressure Depth in a Cave

[Melicertes]

It doesn't really matter where the system is open, either through cracks in the rock or through the mouth of the cave at point B. As long as it's all one body of water the pressure at C only depends on its depth below the surface at A.

Well said.

 

[Drewski]

Respectfully, I'd be careful about calling a cave system "closed" and the water “stagnant.”

Remember, some Florida caves put out MILLIONS of gallons per hour and if you are swimming through an entrance with an OUTFLOW, there is a "residual" water pressure to be found in the cave, otherwise there would be no outflow.

Think about this. A water main is pressurized to about 80 PSI. A large valve at the end of the pipe is opened and water starts flowing out. At the supply end the water pressure is still 80 PSI. It drops rapidly as it approaches the valve and might drop by more than 60% in the stream because the container walls have been lost.

I'd argue that there IS a "residual" pressure in caves caused by outflow but that it is NOT a function of the "position" of the water column, the diver or anything else "invisible." I'd be REALLY surprised if it was more than 2 PSI, just based on the outflow I've experienced at cave openings. If there are engineers on this board who remember fluid dynamics homework questions from college, this outflow pressure question would be an interesting exercise.

Nevertheless, the "gauge" pressure divers measure in caves ACCOUNTS for this "residual" outflow pressure - if it really exists - because depth gauges measure ABSOLUTE pressure surrounding the wearer. So, in short, don't worry about it...

 

[ClayJar]

Respectfully, I'd be careful about calling a cave system "closed" and the water “stagnant.”

Only in the context of the diagram on post 1.

If there are engineers on this board who remember fluid dynamics homework questions from college, this outflow pressure question would be an interesting exercise.

I'll have to find my Perry's, but if I can get a little time, I'll see what I can do.

Of course, there is that thing about what millions of gallons means. One million gallons in a day is less than 1.55 cubic feet per second, after all. That said, I've been to Ginnie's springs, plus Cypress and Morrison (and Vortex, of course).

 

[Drewski]

That said, I've been to Ginnie's springs, plus Cypress and Morrison (and Vortex, of course).

Yep, that "blast" coming out of the ear on an average day is good flow. Imagine if someone put a cork in it suddenly...

 

[boat sju]

Drewski,

If Ginnie Springs is at the upstream side of our cave system and the ocean is the downstream outlet, and we're diving in an underground stream that connects them, then yes the pressure read by your gauge will be greater than your depth below sea level at the outlet of the stream where you entered the cave. Draw a line between the free water surface at Ginnie Springs and sea level above the entrance to the cave. Your depth will be measured down from that line as you move about in the underground stream/cave system. That make any sense? If not, that's why I'm an Engineer and not a teacher.

 

[lamont]

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?

Doesn't matter if you seal it or stick pistons on it. If those pistons are pressing down at A and E with equal force then the pressure at B and D will be 3 ATA greater due to the addition of the force of the pistons and the weight of the water. The pressure at C will be 1 ATA greater than A and E.

If you check the definition of Pascal's Law on wikipedia (or grab a physics textbook of your choice), http://en.wikipedia.org/wiki/Pascal's_law

In the physical sciences, Pascal's law or Pascal's principle states that the fluid pressure at all points in a connected body of an incompressible fluid at rest, which are at the same absolute height, is the same, even if additional pressure is applied on the fluid at some place.

The bold text is the bit that you're missing and is where you're going wrong.

 

[lamont]

But all this raises a question that, frankly, I can not answer intuitively. Let's say we have two vertically identical, very large aquaria tanks that are open on the top and are 33 ft deep. These tanks are connected together by two "true siphons," one at just below the surface of the water and at the bottom of the tanks. Each of these siphons go up above the surface of the aquaria by 33 feet. What is the pressure at the top of the siphons? Why? My first guess would be that the shallow origin siphon is approx 1 ATA all the way through (ignoring the fine point of siphon's diameter) and the deep one would be approx 2 ATA all the way through (with the same stipulation) but my knowledge of fluid dynamics is inadequate to either be confident of that answer or to explain it. Can anyone help?

Nope, top of the siphons are 1 ATA lower pressure than the bottom of the siphons.

Like someone else said, the top siphon would try to pull a vacuum, and the water would boil and you would get water vapor at the top of the siphon -- the pressure of the water vapor would equal the vapor pressure of the water at its temperature (and salinity would affect that, etc) -- the vapor pressure of water at 100 deg C is 1 ATA which is why water boils at that temperature.

When you go up a water column (in a gravitational field, space shuttle doesn't count) then you must have a reduction in pressure.

 

[H2Andy]

When you go up a water column (in a gravitational field, space shuttle doesn't count) then you must have a reduction in pressure.

could you ever have a "dry" portion of a cave where your gauge would read 33 feet depth? for example, an air pocket?

(or actually, any depth below 0)

 

[lamont]

I'm assuming, because of "b" above, that if I had "sealed" the water at D, C would still be 2 ATA -- am I getting closer to understanding?

Yes, sealing the system does nothing to the problem. You do have to specify, then, what a pressure gauge placed at A would read to figure out the problem, but if it reads 1 ATA then the answer is the same.

 

[lamont]

could you ever have a "dry" portion of a cave where your gauge would read 33 feet depth? for example, an air pocket?

(or actually, any depth below 0)

You'd have to have an air-tight bell and the air would be pressurized to 2 ATA. This is the situation in GUEs habitats and troughs where they do O2 at 30 feet to deco.

I think you might have a 'mechanical' issue since the depth gauges read water pressure and are water activated, but if you just took an air pressure gauge in there it would read 2 ATA, which is 33 fsw.