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Thread: Measuring water pressure in long flexible tube.

  1. #1 Measuring water pressure in long flexible tube. 
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    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 column, 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.

    I think that if we press on the bottom of the hose, the hydrostatic pressure on the bottom would prevent (up to a certain point) the volume of the system (hose) 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).

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

    Please tell me what you think
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  2. #2  
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    There is no overcoming of force. They add up. By squeezing the hose you are reducing the volume of a tightly closed hermetic container thus increasing the pressure.
    As such manometers at the bottom, top, middle or other positions would measure hydrostatic head plus the hose compression pressure.

    If the hose was not sealed at the top and had available extra height then the water column would increase in height. The manometers would then measure an increased hydrostatic head corresponding with the total new height of the column.

    A manometer at the bottom shows a higher presure than at the top because of the hydrostatic head.
    Just like a scuba diver diving deeper.
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  3. #3  
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    Quote Originally Posted by pikpobedy View Post
    There is no overcoming of force. They add up. By squeezing the hose you are reducing the volume of a tightly closed hermetic container thus increasing the pressure.
    As such manometers at the bottom, top, middle or other positions would measure hydrostatic head plus the hose compression pressure.

    If the hose was not sealed at the top and had available extra height then the water column would increase in height. The manometers would then measure an increased hydrostatic head corresponding with the total new height of the column.

    A manometer at the bottom shows a higher presure than at the top because of the hydrostatic head.
    Just like a scuba diver diving deeper.
    I disagree. It will be a lot more difficult to compress the tube at the bottom so there will be more force required closer to the bottom to get the same volume change in the tube.
    There will be no difference whether the manometer was at the bottom or top once it has been adjusted for the starting fluid pressure.
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    Quote Originally Posted by Robittybob1 View Post
    I disagree. It will be a lot more difficult to compress the tube at the bottom so there will be more force required closer to the bottom to get the same volume change in the tube.
    There will be no difference whether the manometer was at the bottom or top once it has been adjusted for the starting fluid pressure.
    I've been doing that stuff for a long time. Go try to discredit someone else.
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  5. #5  
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    Quote Originally Posted by pikpobedy View Post
    I've been doing that stuff for a long time. Go try to discredit someone else.
    What sort of stuff have you been doing?


    So you think I'm wrong? Well it is one or the other, makes for a great debate.
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  6. #6  
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    Hi, thank you for your answer.
    The answer the pikpobedy gave i also got on some other forum, the answer was like this:

    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.

    Actually i was testing it yesterday, but it showed that probably this is true. The error in measurements (probably because of not ideal conditions) were the same if i put sensor on the top or on the bottom.
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  7. #7  
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    Quote Originally Posted by brain2k View Post
    Hi, thank you for your answer.
    The answer the pikpobedy gave i also got on some other forum, the answer was like this:

    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.

    Actually i was testing it yesterday, but it showed that probably this is true. The error in measurements (probably because of not ideal conditions) were the same if i put sensor on the top or on the bottom.
    I still maintain that is wrong. If it is increase in pressure you need, as in the simple example of a water bottle, if it is compressed where the deflection is the easiest, is the place where you'll get the best change of pressure.
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    Quote Originally Posted by Robittybob1 View Post
    I still maintain that is wrong. If it is increase in pressure you need, as in the simple example of a water bottle, if it is compressed where the deflection is the easiest, is the place where you'll get the best change of pressure.

    I thought so myself till the last moment. But yesterday i did tests, several tests, with 5 meters long hose. And test showed i was wrong . I still don't get it, but it is how it is.
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  9. #9  
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    My first answer dealt with manometers and not squeeze point location.

    Squeezing the container at the top, bottom or middle. Replace squeezing by a device where it is easier to see and calculate its displacement. For example a piston or bellows device. A hydraulic piston for example.

    A flexible (ie elastic) hose will strectch / expand to counter the increased pressure thus storing energy in the hose material
    A steel pipe would al;so do the same but the young's modudulus is much stiffer as such expansion would be smaller at the same pressure change.
    With a gaseous fluid, it is so easily compressed that most of the stored energy will be in the gas and very little will be in the the tank walls.
    A liquid fluid is not easily compressed at low pressures. So most of the energy put into compression will be transmitted to the container walls.


    A manometer tap at the top, middle and bottom will give three obviously different readings in a liquid fluid tank. The denser the liquid the steeper the pressure gradient.
    With a gas at one atm the difference from top to bottom would be quite slight for ten meters.


    Now comes the part of your question. Would squeezing the hose at different hights cause different pressure increases.
    If you provoked and identical displacement at top, middle or bottom the manometer readings in the hose would all increase by the an equal amount.
    However the force to provoke the same displacement would be more at the bottom than the top.

    Think of it uniquely like a tank without a top. All addition of liquid or squeezing of the tank will increase the column height.
    At the top adding a bit of liquid or squeezing the walls will be easy because there is little hydrostatic head there. Doing the same thing at the bottom is much more difficult because of the hydrostatic head is more.
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  10. #10  
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    Quote Originally Posted by Robittybob1 View Post
    What sort of stuff have you been doing?


    So you think I'm wrong? Well it is one or the other, makes for a great debate.
    Applied fluid dynamics, extractive metallurgy, process chemicals, engineering, quality management.
    Never said you were wrong. My first answer dealt with manometer locations not "squeeze point" location.

    Have you ever calculated the burst pressure of a pipe simple based upon the pipe's material science properties?
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  11. #11  
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    Quote Originally Posted by pikpobedy View Post
    Applied fluid dynamics, extractive metallurgy, process chemicals, engineering, quality management.
    Never said you were wrong. My first answer dealt with manometer locations not "squeeze point" location.

    Have you ever calculated the burst pressure of a pipe simple based upon the pipe's material science properties?
    In a practical way I do, we work with hydraulics and water pressures.

    You do venture into the squeeze point issue.
    There is no overcoming of force. They add up. By squeezing the hose you are reducing the volume of a tightly closed hermetic container thus increasing the pressure.
    OK it sounds near illogical. "There is no overcoming of force." well there could be if the opposing force is too great.
    "They add up." What adds up?
    "By squeezing the hose you are reducing the volume of a tightly closed hermetic container thus increasing the pressure." I agree with that but the point being made is it was a lot harder to squeeze the hose where the pressure is higher.
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  12. #12  
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    We may model this as a long elastic cylinder containing a massive fluid.
    At the top and bottom are equal diameter pistons.

    There is a sustaining force on each piston to keep it from moving.

    The force on the top piston is about zero.
    The force on the bottom piston is something like ahg*rho where rho is the fluid mass density.
    The exact equation is immaterial.

    Apparently a change in force on either piston will result in the same pressure change within the volume of fluid.

    Any confusion seems to result from considering absolute force rather than a change in force--or I have misanalyzed this
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  13. #13  
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    We may model this as a long elastic cylinder containing a massive fluid.
    At the top and bottom are equal diameter pistons.

    There is a sustaining force on each piston to keep it from moving.

    The force on the top piston is about zero.
    The force on the bottom piston is something like ahg*rho where rho is the fluid mass density.
    The exact equation is immaterial.

    Apparently a change in force on either piston will result in the same pressure change within the volume of fluid.

    Any confusion seems to result from considering absolute force rather than a change in force--or I have misanalyzed this
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  14. #14  
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    Quote Originally Posted by Useful Idiot View Post
    We may model this as a long elastic cylinder containing a massive fluid.
    At the top and bottom are equal diameter pistons.

    There is a sustaining force on each piston to keep it from moving.

    The force on the top piston is about zero.
    The force on the bottom piston is something like ahg*rho where rho is the fluid mass density.
    The exact equation is immaterial.

    Apparently a change in force on either piston will result in the same pressure change within the volume of fluid.

    Any confusion seems to result from considering absolute force rather than a change in force--or I have misanalyzed this
    What is applying the force to the bottom piston? There could be so much weight of liquid above it any additional force still will not overcome the downward force and hence there is no compression and hence no change in volume and pressure.
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  15. #15  
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    Quote Originally Posted by Robittybob1 View Post
    What is applying the force to the bottom piston? There could be so much weight of liquid above it any additional force still will not overcome the downward force and hence there is no compression and hence no change in volume and pressure.
    Before applying any change equivalent to squeezing the tubing the required force is pA, pressure times piston area. This is the initial static condition. Squeezing the tube is equivalent to adding additional force to the tube.

    The forces add; F=F1+F2. Pressures over the same area also add.
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  16. #16  
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    Quote Originally Posted by Useful Idiot View Post
    Before applying any change equivalent to squeezing the tubing the required force is pA, pressure times piston area. This is the initial static condition. Squeezing the tube is equivalent to adding additional force to the tube.

    The forces add; F=F1+F2. Pressures over the same area also add.
    I think you have simplified it too much. Tubing can be very hard to compress (variable compression forces are required just to distort the tubing) and this is before any addition pressure is added to the fluid. So I question the idea of just adding the forces together.

    Like if the tube is made from iron you might even be able to run it over with a 10 ton truck and no addition pressure is experienced in the liquid.
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  17. #17  
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    Forces add.
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  18. #18  
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    Quote Originally Posted by pikpobedy View Post
    Of course they do. You know that, I know that. BTW you are correct all over this thread.
    Well you show me how the the weight of the 10 ton truck is added to the pressure inside the tube?
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  19. #19  
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    Quote Originally Posted by Robittybob1 View Post
    Well you show me how the the weight of the 10 ton truck is added to the pressure inside the tube?
    You can figure that out.
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  20. #20  
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    Quote Originally Posted by pikpobedy View Post
    You can figure that out.
    I wanted you to help me understand the problem.
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  21. #21  
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    I've seen your posts and you seem to have side to you that is not inquisitive but is rather trying to amuse yourself at the expense of others which makes you a complicated member. The question you should ask yourself is why would a mature person answer if they did not know the answer. We are in a physics forum here, not in some kitchen or bar.

    Assume a certain sized wheel and certain sized contact area. The force on the contact area becomes a pressure on that contact area. That pressure will deform the hose until equilibrium is reached. If equilibrium is not reached the hose has flattened and the volume of the hose has been reduced to zero for the width of the wheel (tire foot print).
    A ten ton truck with six wheels, the force is 1.67 tons per wheel. Assume tires with high pressures and thick reinforced rubber that don't deform when going over a garden hose with only a 10 meter head pressure. If the hose is ten meters long with a ten meter head and you flatten it over say one meter you have removed 10% of the volume.
    pv initial = pv final
    10 * v initial = p final * vfinal
    p final = 10 * v initial/v final
    p final = 10 * v initial / (0.9 v initial)
    p final = 11. 1 meters of head.

    Now how do I know a garden hose with a 10 meter hose will be flattened by a ten ton truck.
    I can calculate that.
    I will assume that six tires are the aforemention 1 meter width. The total pressure on the hose will according to the affected surface area which is a function of the length and width of the hose affected. The length is the aforementionned 1 meter and the width can be maximum half the circumference of the hose.
    A very light vehicle will not crush the hose but will compress it until equilibrium is reached. This can be calculated.


    A cement mixer will flatten a fire hose connected to a hydrant. A car will not. Two different conditions in two different regimes. In one case the pressure in the hose increases because it reverts to static hydrant pressure and in the other it increases to equibrium of reduced flow rate and pressure drop caused by the restrictions in the flow.

    Similarly in a fixed size hose with a fixed volume of water. Any squeezing of the hose will create two diffrent conditions. Completely crushing the hose will stretch the hose to a maximum possible value. Anything that does not crush the hose will create a pressure inside in equilibrium with the weight of the copressive forece transmitted via pressure in on the affected surface areas.
    Forces add up. They have to be added correctly. Thus the time spent teaching about pressure, moments, torques, levers, pulleys, gears, beams, columns, rigid members, hydraulics, fluid dynamics etc.
    As for a resilient hose or a steel pipe. I have already stated that the difference is young's modulus.
    Last edited by pikpobedy; 04-13-2014 at 08:46 PM.
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  22. #22  
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    Quote Originally Posted by pikpobedy View Post
    As for a resilient hose or a steel pipe. I have already stated that the difference is young's modulus.
    Well I'm sorry but I missed any post about Young's Modulus.

    I was originally thinking of alkathene piping which is sort of flexible in that it can be bent, but it is not possible to compress it with your hand, so no matter how much hand pressure you applied to the outside you would not affect the pressure on the inside.
    No doubt it is dependent on the definition of "flexible", and really hoses like "fire hydrant hoses" etc are flexible to compression but resist expansion. There would be other flexible hoses that might expand at another place as soon as pressure is applied.
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  23. #23  
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    Always remember that for every action there is an equal opposite reaction. When you hand squeeze a hose, tube, pipe whatever the material will push back with an equal and opposite force. It gets this force from the distortion of its shape and material.
    If I hang a weight from a bungie cord, ten parallell bungie cords, a chain, a hemp rope, a steel cable or a nylon cable all those will stretch elastically according to hooke's law or (and in a profound way to young's modulus) and return to its original length if the weight is within the elastic limit. So elevator cables stretch. This is tensile stress versus strain behavior. The same goes for compression, torsion and shear.

    So whether you hand squeeze a delicate hose, a stiff hose or a steel pipe the distortion caused is a question of magnitude and not a yes or no thing. Realistically steel pipes are designed not to be affected by the human hand so all squeezing would be almost imperceptible to the water because (almost) all the force reaction would be from the pipe with (almost) none required from the water pressure.
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  24. #24  
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    A bit related is the Cartesian diver where we see that squeezing force on the comtainer walls being transmitted to the contained water and via the water to the diver (and to all surfcaes inside).

    Cartesian diver - Wikipedia, the free encyclopedia
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  25. #25  
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    Quote Originally Posted by pikpobedy View Post
    A bit related is the Cartesian diver where we see that squeezing force on the comtainer walls being transmitted to the contained water and via the water to the diver (and to all surfcaes inside).

    Cartesian diver - Wikipedia, the free encyclopedia
    We used to make them as kids - years ago!
    Thanks
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