Measuring water pressure in long flexible tube.

[brain2k]

Hello dear sirs/madams,

I would ask you to answer the following question.

I have a flexible hose with a diameter of 5 cm, a length of 10 meters, it is in a vertical position and is filled to the top, with water.

The hose is sealed on the top and the bottom.

At the top of the hose, there is a pressure sensor.

Now the question :

Does, the pressure sensor, always shows the same value, if i press on the hose with a finger with a force of 10 Newtons (force is not important really):

1. ) on the bottom of a hose, for example 10 cm from the bottom
2. ) in the middle of the hose, for example at a height of 5 m
3. ) at the top of the hose, for example 10 cm from the top

So would the pressure sensors show the same value in all three experiments?

Personally, i am thinking like this: because in 10 m water columnt, the hydrostatic pressure is 1 bar. Our finger has first to overcome this force, then the pressure in the liquid began to increase. Am i thinking right

So I think, it can even happen that, this sensor will not detect any pressure change, if we press on the bottom of a hose.

And a sub-question:
What if pressure sensor is mounted on the bottom, does this change anything?

Please tell me what you think

 

[euquila]

It is a closed system and an external increase in pressure or force applied will correspond to an equivalent increase in pressure inside (indicated by your sensor).

Assuming the hose is RIGIDLY supported, the direction of the force makes no difference. (Note the word that I capitalized, I think you need to think about what this word means and how it might / might not apply to various physics problems.)

 

[brain2k]

Thank you for your answer euquila :)
So hydrostatic pressure (or height of the hose) has really no inflluence in this case?

best regards.

 

[euquila]

Well, I thought we were talking only about changes in pressure. Of course, if you move the sensor down the hose, the pressure reading will increase because of the weight of the column of water above the sensor (due to gravity). In the absence of any gravitational field, then the position of the sensor along the hose would no longer matter. You can think of the gravity field as creating an asymmetry in the system where the pressure increases in the direction of increasing strength of gravity field.

However, if the position of the sensor is fixed and the boundaries do not exchange heat, the pressure difference in the closed system is due entirely in response to the applied external force, which in turn decreases the volume of the system.

Incidentally, if the boundary / hose walls were rigid themselves, then the applied force would NOT increase the internal pressure because the walls would not deflect.

 

[brain2k]

Yes we were talking only about changes in pressure.
Thank you for your answers.

 

[euquila]

My pleasure!

 

[brain2k]

Euquila i would like to ask one more thing. You said:

However, if the position of the sensor is fixed and the boundaries do not exchange heat, the pressure difference in the closed system is due entirely in response to the applied external force, which in turn decreases the volume of the system.

But if there is big water column (lets take 100 instead of 10 meters ) than the hydrostatic pressure would prevent (to a certain point) the volume of the system to decrease (wall of the hose to deform), and pressure sensor would really show less than if we would apply external pressure on the top of the hose (because there is no hydrostatic prerssure, and wall of the system would easely bent in - deform)

I have to test it, because i am doubting Thomas

 

[euquila]

Yes, it is true that deflection of the walls in response to an applied external force at the bottom would be less than the same force applied at the top because, as you pointed out, there is a much larger counteracting force at the base. However, the CHANGE in pressure inside the system would still be the same. This is because the density of the fluid at the bottom is also proportionally larger. So a small deflection + large density or large deflection + low density actually amounts to the same change in pressure. It is analogous to how the momentum of a bullet fired from a gun can be close to that of moving truck.

 

[brain2k]

Ok this explains it, i am really thankfull to you!

 

[euquila]

Thank you for the good questions.