Column Pressure and conservation of energy

[genergy]

Mwater g(H+h)

If you have a column of water 100 meters tall (10 atm) and you insert an object in at the bottom of the column you have to use enough energy to displace the volume of the object.
The pressure times the volume is your energy requirement: correct?

But suppose you were able to put a solid barrier between the top of the water at a height of only 10 meters (1 atm). None of the pressure of the upper 9 atm is allowed to transfer into the bottom 1 atm environment.
Would the energy requirement to insert the object be the volume times 10 atm or 1 atm?

Are you violating the Law of Conservation of Energy if you say 1 atm?
Why or why not?

 

[Simon Bridge]

There have been perpetual motion machines based on the idea you just showed.
The idea is, usually, that you close off the column with a valve, insert a buoyant object at the bottom, open the valve, the object shoots to the top with such force it pops out of the surface and falls back to the ground... where it uses it's momentum from falling to reinsert into the bottom.

Lynchpin: When you insert the object you also have to displace the water you are pushing it into.

When you insert into the unbroken column you have to get the displaced volume of water to the top. When you put the barrier in, you only have to get the displaced water to the top of the lower part ... as described, and if your seals were perfect, then you would not be able to insert it ... otherwise the displaced water will have to exit around the seals.

It's more fun if you are using a heavy, but compressible, gas.

To answer your question: you have not described anything that violate conservation of momentum - but it cannot be turned into a loop.

http://www.lhup.edu/~dsimanek/museum/unwork.htm
... scroll down to "Buoyancy motor #4".
The page on buoyancy misconceptions is also useful.

 

[genergy]

Would you be violating conservation of momentum if you displaced the water into an empty tank? Obviously, at some point the tank will fill up!
But, until the tank fills would there be any problem with conservation of momentum?

 

[Simon Bridge]

Nope. No problem. Having to do this is what makes the ppm's fail.
What is this in aid of?

 

[PJC01]

I would like to clarify a few points regarding this scenario. First, the equation being used here is based on the hydrostatic pressure equation, which states that pressure (P) is equal to the density of the fluid (ρ) multiplied by the acceleration due to gravity (g) multiplied by the height of the column (H). Therefore, the equation should be written as P = ρgh, where h is the height of the column in meters.

Now, to address the question at hand, the energy requirement to insert an object into the column of water would indeed be the volume of the object multiplied by the pressure at the bottom of the column. This is because the pressure at the bottom of the column is the total pressure exerted by the entire column of water above it, regardless of whether or not there is a solid barrier at the top limiting the transfer of pressure.

If a solid barrier is placed at the top of the column, the pressure at the bottom would still be 10 atm. However, the pressure at the top of the barrier would be 9 atm, and the pressure at the bottom of the barrier would be 1 atm. In this case, the energy requirement to insert the object would be the volume of the object multiplied by the pressure at the bottom of the barrier, which is 1 atm.

No, this does not violate the Law of Conservation of Energy. The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this scenario, the energy is being transferred from the water to the object being inserted. The barrier simply limits the transfer of pressure from the top of the column to the bottom, but the total energy required to insert the object remains the same.