Does Pascal's Law Apply to a Sealed Fluid System with Gravity?

[gloo]

If you have a structure like a large bike air pump (i.e cylinder, air hose at bottom, plunger etc) and you make the following adjustments:

1. insert the hose back at the top of the pumping cylinder (making it a cyclical system)
2. fill the cylinder with water so that there is no air in cylinder and hose...the whole system is filled with water
3. completely seal/enclose the unit so that it is enclosed fluid

Questions:

1. if the plunger is near the bottom of the cylinder, do you need a force large enough to move the equivalent mass of water in the cylinder above? I.e if the mass of water above the plunger in the cylinder is 10kg (it's a large bike pump) and the plunger is 0.5 kg, then the force needed to move the water even a little is over (10.5kg * 9.81ms) = 103N (ignore friction)??

Or is it the fact that there is a slight pressure differential coming from the full column of water in the tube that will have a piston effect? So what I am thinking is that even the slightest force that is just bigger than the force of the plunger (0.5kg*9.81) will result in the plunger moving up which results in the water at the top forced down the tube and back up below the plunger (large piston)

2. Would it be any different (answer) if the pump was not enclosed and large cylinder is open at the top (provided we did it slowly so as to not spill over top)?

 

[Simon Bridge]

Please see Museum of Unwrkable Devices - particularly the sections about buoyancy motors, to get a decent primer on the concepts involved.

1. you need to be able to supply energy to account for changes of energy in the system.
The pressure of the column of water does change with depth.

2. this would add air pressure to the top. If you were very slow, so the upper surface of water did not change, then it would make no difference.

If I understand your description - you have water above and below the plunger-head.
Pushing the head upwards will also draw water in behind it - the change in energy is determined by the change in the center of mass of the system (water+plunger) which you have to look at carefully since the plunger mechanism certainly displaces some water.

If the return tube is a smaller diameter to the plunger tube, there will probably be some back pressure as you need to force the water through a smaller tube.

Best practice is to use energy rather than force arguments when thinking about these sorts of things.

 

[gloo]

Wait, so does the plunger draw water in behind? or is it forced up because of the force i applied upward is pushing the water down the tube? Isn't it just the plunger moving up and forcing water in from below. Is this a pascal's law type application? The enclosed fluid receives the force of plunger being pulled up and transfers the pressure back up from below?

 

[Simon Bridge]

Both. Think of it as a chain that goes in a loop - when you push part of the chain, the whole thing moves. Does the chain move because you are pushing or because the link you push is drawing the chain up from behind?

 

[gloo]

Thanks for your help Simon :)

So, if we are talking about energy for the system, and ignoring the weight of the plunger and it's components, the plunger being right in the middle, it takes more work to go up because of gravity?

 

[mark1234]

My two cents:

If the top and the bottom of the cylinder are connected by a hose, the pressure in the water column is hydro static (in case of no movement . The pressure at the bottom and the top of the piston therefore balance each other out in a static situation (whatever the position, of height of the piston) and no net force is present on the plunger.

When you want to move the plunger water is forced though the system and experiences frictional losses, mostly in the assumed small hose connecting the top and the bottom. These frictional losses result in a pressure difference on the top and bottom of the plunger. The force required to move the plunger is this pressure difference x area of the plunger.

The force is not depending on the location of the plunger in the cylinder, the frictional losses to overcome will be the same whatever the location of the plunger.

 

[Simon Bridge]

Thanks for your help Simon :)

So, if we are talking about energy for the system, and ignoring the weight of the plunger and it's components, the plunger being right in the middle, it takes more work to go up because of gravity?

You cannot ignore the weight of the plunger and it's components because it displaces some water i.e. where the plunger is, there is no water... but let's pretend we have used a plunger etc which has the same density as the water but it's solid. That should suit the thought experiment you are trying for.

In that case it makes no difference which way you move the plunger: the same amount of water goes down the small pipe as gets lifted up by the plunger. Net change in gravitational PE is zero.

This is an idea that is explored in more detail in the link in post #2.