Trying To Figure Out A Hydrostatic Pressue Problem

[EnquiringMind]

I have attached a picture of a problem I am trying to understand. Basically I have a Piston attached to the ground, and a container of water "free standing" above most of the Piston.... meaning, the container of water could collapse onto the Piston if the Physics say it will (and thus displace water up into the small pipe).

I'm trying to figure out what values I need to know to figure out if the container of water will collapse onto the Piston, or if the container of water will just stay where it is.


If I'm trying to push a Piston upwards into a container of water, I understand that I need to overcome the PSI of the water. However, in this problem there is not only PSI but the weight of the container itself.... and I'm not sure what to do.

Any advice would be appreciated.

 

[zekeman]

Look at the container which I will assume weightless, for the moment. I also consider the case where the overfill in the upper reservoir is negligible, or non existent; the case where water is in the upper vessel can be easily handled, but, later.

The annulus of water + containment are always in equilibrium since no matter, the configuration the differential pressure top and bottom x the annulus area is the weight of the water in the annulus.

So we are left with a piston the same diameter of the remaining "containment" whose outside diameter is equal to the diameter of the piston. Now, the vertical forces on the "containment" are always net upward and will continue until all of the water is inside the lower vessel. If there is excess water beyond this, shown in the neck section then ,it will lift above the piston.

My conclusion is, there is no possibility of collapse, but only upward motion of the container, so long as the neck section still holds water; otherwise, static equilibrium will be achieved
at the moment there is no water left in the neck.

 

[Compositepro]

Zekeman, I don't understand your analysis. The weight of water in your "annulus", is a force that pushes the container down. Read my first post. The container will drop if the "pipe" diameter is not smaller than the piston diameter by an adequate amount.

 

[zekeman]

I read your post. You forgot the pressure from the top of the container pushing down on the annulus of water
The forces on the annulus of water are
1)downward pressure from the top of the container*A, A= annulus area
2) upward pressure from the bottom of the container*A
3) weight of the water h*rho*A, --h=height

but h*rho=pottom-ptop
in any configuration
which proves that that mass of water is always in equilibrium.

 

[Compositepro]

What do you mean that the mass of water is always at equilibrium?

I agree that it is always the same.

You seem to be stating that pressure at the top of the annulus plus the pressure due to the mass of water in the annulus is the pressure at the bottom of the annulus. This is true. But then you say its all in equilibrium. This is not true. The mass of water in the annulus results in a downward force. The only thing that can resist this is water pressure acting on the top of the container over the piston area (A1-A3 in my earlier post).

You say: "You forgot the pressure from the top of the container pushing down on the annulus of water".

Pressure created by the column of water in the pipe has no net effect in the annulus area. It pushes up and down by the same amount.

 

[hydtools]

In my view, the water in the annulus just goes along for the ride. If the water pressure at the bottom of the column x the piston area is a force that is less than the weight of the container plus that of the annulus water, the container will fall and water will be forced up the column/pipe.

Ted

 

[zekeman]

My bad,
Looked at this again and am now in agreement with the force of the annulus water pushes down on the container,and while if there is water in the pipe, then there is an upward force on the container due to the height of water in the pipe. Depending on the initial height of water in the pipe and the areas projected by the piston-pipe area, the container will move up or down.
For the case where there is no level in the pipe at first, then the container will move downward as Compositro says, raising the water level in the pipe and increasing the upward force on the container, and eventually, if there is sufficient pipe height, will reach equilibrium at which time

annulus water weight= rho *h*(piston area-pipe area)

h= height in pipe
rho water density