Pressure Depth in a Cave

[H2Andy]

so in a normally occurring air pocket, if you surface into it, your gauge would read zero, right?

 

[lamont]

This is exactly the way I took to understanding the scenario also, and yes I agree these were the exact same two mistakes I made in this discussion. In my mind the air columns above A and E were as good as pressurized plugs that "closed" the system at A & E and contributed to introducing pressure into the system between points B and D.

the air columns are as good as pistons pressing down at A and E. there is nothing magical about the air.

you could also take an entirely sealed system and take a nipple and either compress or bleed off pressure until 1 ATA was reached at A and E and it would be an identical problem. if you then took that nipple and either pressurized or depressurized the system through that nipple you would see that the problem remained the same throughout only with an offset of however much pressure you were adding/subtracting.

the mistake you made was solely in applying pascal's law to fluids that were not at the same height in a gravitational field where it does not apply.

 

[lamont]

so a normally occurring air pocket, if you surface into it, your gauge would read zero, right?

well, if you've got an air pocket formed by scuba exhaust at depth then it would be under the same pressure as the water surface pressing up on it.

natural air pockets that have been formed by the level of water in the system dropping and pulling in air through cracks in the rock should be at zero.

 

[lamont]

Some of the confusion over Pascals law may be from reading it in this form:

"any change in pressure applied at any given point of a fluid is transmitted undiminished throughout the fluid."

The subtlety which is not easily apparent here is that for any small delta change in height of the fluid there will be a change in pressure applied to the liquid at that point equal to the weight of the water in that infinitesimal change of height.

So it is entirely correct that forces are transmitted undiminished throughout the fluid, but you must not neglect the gravitational force that acts at every point.

 

[H2Andy]

well, if you've got an air pocket formed by scuba exhaust at depth then it would be under the same pressure as the water surface pressing up on it.

natural air pockets that have been formed by the level of water in the system dropping and pulling in air through cracks in the rock should be at zero.

danke!

 

[radinator]

I've been enjoying this thread, but haven't noticed it until now so I haven't participated.

So, here is another fun one.

I have a 33' vertical pipe which is closed on the bottom and open at the top, nearly immersed in the ocean. The top sticks out just far enough so that the glassy surface ocean water doesn't spill in.

Inside, the pipe is filled with fresh water. Outside the pipe is the salty ocean water.

The pipe was closed at the bottom with a valve. I now open the valve. What happens?

 

[Blackwood]

I've been enjoying this thread, but haven't noticed it until now so I haven't participated.

So, here is another fun one.

I have a 33' vertical pipe which is closed on the bottom and open at the top, nearly immersed in the ocean. The top sticks out just far enough so that the glassy surface ocean water doesn't spill in.

Inside, the pipe is filled with fresh water. Outside the pipe is the salty ocean water.

The pipe was closed at the bottom with a valve. I now open the valve. What happens?

Ocean water will flow into the pipe.

 

[lamont]

Ocean water will flow into the pipe.

I was hoping that would take longer, I PM'd radinator since I didn't want to give it away:

answer...

33 fsw = 34 fsw. you would need a 34 foot long pipe (1 foot above the ocean) to get hydrostatic equilibrium at the bottom of the pipe. since you don't have that sea water will flow in. since water is incompressible that will push freshwater out the top. in the ideal case all the freshwater gets pushed out (EDIT: or saltwater flows into the tube until the salinity of the tube equals the salinity of the ocean...).

 

[radinator]

I knew it wouldn't take long. I just walk away from the keyboard for a few minutes...

An interesting variant of this is if you could theoretically put a salt-removing filter at the bottom of the pipe, you'd have a perpetual fountain. I've heard this proposed (but with a thousand foot long pipe to overcome the resistance of the filter) as a fresh water supply. Of course, someone will have to change the filter frequently...

I can't declare a winner, since both Lamont's PM and Blackwood's post were sent at 6:56 pm (PST). So it's a tie.

 

[Blackwood]

I was hoping that would take longer, I PM'd radinator since I didn't want to give it away:

Whoops. Sorry. I guess I'm a read-and-reply kinda guy.

If you want one that will take longer, I can probably oblige.

33 fsw = 34 fsw. you would need a 34 foot long pipe (1 foot above the ocean) to get hydrostatic equilibrium at the bottom of the pipe. since you don't have that sea water will flow in. since water is incompressible that will push freshwater out the top. in the ideal case all the freshwater gets pushed out (EDIT: or saltwater flows into the tube until the salinity of the tube equals the salinity of the ocean...).

Also of note: The water in the pipe will move (much) faster than the water in the ocean, further lowering the pressure in the tube (and increasing the pressure differential).