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.

 

[hydtools]

Is the outlet of the small pipe opened or closed?

If the outlet is open, water is free to flow and the container will settle to the ground or the piston touches the inside of the container.

If the outlet is closed and the small pipe is full of water, the container will stay in place since no water displacement can occur.

Ted

 

[rb1957]

i'm havng a hard time understanding the question.

it sounds like you have a container of liquid sitting on top of a piston. if you raise the piston, won't the container of water rise with it ? what's restraining the container ??

if the upper surface of the container is attached to ground, then raising the piston will cause the liquid to displace. if the outlet is closed it'll cause the gas (trapped above the liquid) to compress.

You could consider work/energy ... work done by the piston = work done by the liquid and the gas in the container.

you could consider volumes ... volume due to the piston changing it's postion = volume change in the gas.

 

[EnquiringMind]

No no. I apologize. The small pipe is open at the top.

I should also note that the small pipe can extend upwards. I am not suggesting the water expel from the container... but would stay in the container... only moved to a greater elevation in the container.

 

[EnquiringMind]

rb1957. The Piston is attached to the ground. It is the container that is free to stay where it is, or fall downward onto the Piston (which would displace some of the water into the smaller vertical tube.

 

[MiketheEngineer]

Kind of depends on the weight of the container. If it is less than the weight of the height of the water column - it won't move. Otherwise it will descend and shoot water out of the top.

 

[EnquiringMind]

Here is a better illustration to clarify the example...

Under what circumstances will the container fall and displace the water into the upper chamber?

One person told me that, as long as the surface area surrounding the Piston is larger than the area of the Piston, the container would collapse.

 

[MintJulep]

One person is wrong.

 

[rb1957]

"the container that is free to stay where it is, or fall downward onto the Piston" ... this makes no sense to me. the container choose to fall.

is this like a balloon (flexible container) with maybe a straw sticking out of it sitting on top of a piston, perhaps balanced by the hand of a nearby experimenter. if th epoison is bigger than the container then i'd expect the container would spread out over the piston untill maybe the strain in the membrane restricts further spreading ... if the piston is bigger than this then the contianer should sit happily onto of the piston. if smaller, then the flexible container will spread over the edge and i'd expect it to collapse. sounds like a reasonable experiment to try ... somewhere you don't mind getting wet ...

 

[EnquiringMind]

The Piston cannot take more space than the bottom surface of the Container.

Even if the Piston WERE the bottom surface of the container, there would be an extremely small gap between the side of the Piston and the wall of the container... which is where the seal would occur. So technically, the Piston has to be smaller than the surface of the container.

In my example, I clearly made the smaller than the full surface area of the bottom of the container.


This is basically the difference between hammering a nail into someone's back, and securing a nail to the ground and asking someone to lay on it. My example is a bit more complex than that, but it is the same in flipping the problem around to the other side.


The confusion I have is that there are two different ways to look at the forces in play here.

First you have the water pressure at the bottom of the container, which is acting on the bottom of the container (which INCLUDES the surface of the Piston). That is what is happening inside due to the Hydrostatic Pressure, right?

But then if you step back and look at the whole picture, you see the entire weight of the water in the whole container, and the weight of the container itself.


What I am trying to find out, mainly, is what is the "breaking point" where the container will collapse around the Piston vs. where the container will stay as it is originally illustrated.

One person told me "if the area of the surface of the container is larger than the surface area of the piston, the container will collapse". I believe said that because there will be more total PSI pressure on a larger area of the container floor, than on the Piston surface.... therefore the container wall will want to go lower.

 

[EnquiringMind]

Should read...

"In my example, I clearly made the Piston area smaller than the full surface area of the bottom of the container."

 

[vpl]

Enquiringmind

Out of curiosity, do you have an engineering background?

Patricia Lougheed

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Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[rb1957]

when you put a balloon 1/2 full with water on a table, what happens ? the balloon spreads out untill there is equilibrium between the tension in the membrane and water pressure. yes?

if the balloon sides are very thin, it could break before it reaches this stable state.

now put the balloon on a smaller surface ... it spread over this surface and spill over the side. maybe this is "collapse". maybe there's another surface 1/16" below the original one, so the balloon will probably keep spreading.

 

[EnquiringMind]

I understand your balloon example, but this is a basically a Piston Pump with solid walls.

The problem with the balloon example is that the pliability of the balloon walls may be able to prevent any water from being displaced up into the straw. It is not similar enough to my liking.

.... however, you do seem to be agreeing with that "someone" who told me the Piston surface area (relative to the area of the entire container floor) plays a big factor in if the container would fall around the Piston.

 

[Yagonyonok]

What exactly are you trying to make (tidal/wave pump)? Or are you looking for a perpetual pump that uses water pressure to pump the water, or just trying to win a debate with "one person?"

Assuming you have a rigid container sliding over a fixed piston then you will be able to create head pressure based on the buoyancy of the container. The effects of water pressure on the surface of the container alone will not create a pressure differential that would allow you to pump water.

If you can provide more info into what exactly you are trying to determine then somebody here may be able to help. However, if this is for a real world problem, you should get an engineer to work on it. Nobody here is going to design it for free for you, and without an engineering background it will be difficult for you to come up with a workable design.


Use of product voids warranty.

 

[Compositepro]

The question and its answer are fairly simple, but there are undefined factors. Neglect seal friction and container mass as this is a theoretical question.

A1= piston area
A2= container area
A3= pipe area

H1= height of container
H2= height of water level in pipe (the large reservoir at the top of the pipe adds a complication which I don't think was intended. It will behave the same as a uniform diameter container with no pipe).

I'm not going to work out any equations as there are too many undefined factors.

The key factors are:
1. The weight of water outside the piston diameter ((A2-A1)H1) will push the container down and displace water up the pipe.

2. The head of water in the pipe above the container will push the container up by acting on an area A1-A3.

 

[rb1957]

is it a rigid container ? if it is, what could be the question ??

 

[prex]

Easily solved in terms of energy.
A[sub]1[/sub]=piston area
A[sub]3[/sub]=upper container area
s=container displacement
h=level change in upper container
H=initial distance of level in upper container to piston top
[ρ]=liquid density
M[sub]c[/sub]=container mass
Now:
A[sub]1[/sub]s=A[sub]3[/sub]h
Change in potential energy of liquid
[Δ]U[sub]l[/sub]=[ρ]A[sub]1[/sub]s(h/2+H-s/2)
Change in potential energy of container
[Δ]U[sub]c[/sub]=M[sub]c[/sub]s
By equating the changes in potential energy one can easily calculate s.
Limit cases:
-if H=M[sub]c[/sub]/[ρ]A[sub]1[/sub] the container won't move, if H is larger the container will go upwards
-if A[sub]1[/sub]>>A[sub]3[/sub] like in the first sketch, container movements will be small (the container behaves rigidly when you try to displace it)

prex

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[Denial]

Even easier solved by simple geometry.

Let A be the area of the piston head.
Let V be the total volume of water.
Let w be the unit weight of water.
Let H be the vertical distance from the piston head to the free water surface.
Let W be the weight of the (empty) container.

Consider the movement of water when the container moves down a small amount d.

This can be viewed as the entire "shape" of water (whose volume is V) moving downwards by a distance d, on top of which is "superimposed" the upwards movement of a volume Ad by a distance H.

Problem solved. The net effect of these two movement components is an increase of
(HA-V)dw
in the potential energy of the water. The potential energy of the container has decreased by
Wd

The container will be in equilibrium if these two quantities are equal. (I think this is a stable equilibrium, but do not have time to verify my feeling.)

It will tend to move up if
H > (Vw+W)/(Aw)

This is NOT the same result derived above by Prex. I think Prex has omitted the mass of the overall body of water.

 

[prex]

Correct, Denial, I forgot the water mass. However your solution isn't complete, as it doesn't allow to calculate the displacement starting from an arbitrary initial state.
Let me try again.
When the container goes down by s, the net effect on the water mass is that the volume A[sub]1[/sub]s goes up by H-s/2+A[sub]1[/sub]s/2A[sub]3[/sub]=H+(A[sub]1[/sub]/A[sub]3[/sub]-1)s/2 , and the rest of the water goes down with the container.
So the energy balance gives, after some rearrangement (V= total volume of water):
M[sub]c[/sub]+[ρ](V-A[sub]1[/sub]H)=[ρ]A[sub]1[/sub](A[sub]1[/sub]/A[sub]3[/sub]+1)s/2
At equilibrium H is as per Denial's solution, and again, s will tend to be smaller when A[sub]1[/sub]>>A[sub]3[/sub].

prex

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: Online engineering calculations

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