Is it possible to have water levels at different heights without a pump?

  • Thread starter Thread starter StefanBoon
  • Start date Start date
  • Tags Tags
    Levels Pump Water
AI Thread Summary
Air cannot remain trapped under a structure with different tube heights without a pump, as the pressures from the air and water must be balanced for stability. If the downward air pressure exceeds the upward water pressure, the water level will fall, and vice versa. The diameter of the tube does not influence pressure; only the height of the water column does. An air lock can occur in inverted U-shaped tubes, where trapped air supports the water columns, preventing a siphon effect. For stability, a free-moving piston is necessary to maintain a flat water surface, and narrow tubes are essential for effective operation.
StefanBoon
Messages
2
Reaction score
0
I wonder if air can remain trapped under a structure that has a deep and a shallow tube. I have already tested with a prototype whether increasing volume in the shallower tube works. Air bubbles came out on the short side, so this doesn't work. Is there any way to realize this concept.
1598878371913.png
1598877526956.png
 
Last edited:
Physics news on Phys.org
The quick answer to your question is no, but it is worth spending some time thinking about why.

Think about the forces acting on the water right at the surface: if the downwards pressure from the air above is not exactly balanced by the upwards pressure from the water immediately underneath, the level will either fall (air pressure is greater) or rise (water pressure is greater). What can you say about the air and water pressures above the two tubes?
 
  • Like
Likes StefanBoon
Nugatory said:
The quick answer to your question is no, but it is worth spending some time thinking about why.

Think about the forces acting on the water right at the surface: if the downwards pressure from the air above is not exactly balanced by the upwards pressure from the water immediately underneath, the level will either fall (air pressure is greater) or rise (water pressure is greater). What can you say about the air and water pressures above the two tubes?
Thank you, very much appreciated. What do you mean with the last sentence 'What can you say about the air and water pressures above the two tubes?'

Do you mean that it can work if the pressures of the air above is the same as de water at the surface of the short tube?
 
Last edited:
StefanBoon said:
I wonder if air can remain trapped under a structure that has a deep and a shallow tube.
First of all I have to point out a very basic fact that it is not the diameter of the pipe that affects pressure - just the height of the vertical column.

I'm not sure if the image is the exact layout where what you describe happens (or doesn't happen) It sounds as if you are describing what's known as an 'Air Lock'. In a situation where there is air in an inverted U, the 'bubble' of air can be longer than the differences in the two water levels. The section of trapped air floats and supports the water columns so you don't get a simple siphon effect. (Plumbers come across this problem all the time where there are long horizontal runs of pipe in low pressure systems.

Practical Note: your drawing shows a flat horizontal surface at the bottom of the upper section. This is not stable and one side or the other will fall after a very short time. You would need a free moving piston to keep that water surface flat and level. You need narrow tubes to make this work.
 
  • Like
Likes StefanBoon and Lnewqban
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top