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Circulating a fluid at a very high speed in a spiral pipe

  1. Mar 28, 2013 #1
    Through a spiral pipe is circulating a fluid with high speed, this will lead to many effects and forces like inertia and gyroscopic effect.


    I do not know how to approach the problem in order to find the distribution of forces.


    In this process of circulating the fluid in spiral pipe will appear lateral forces, or spiral pipe will remain inert because forces cancel each other at all levels?
  2. jcsd
  3. Mar 28, 2013 #2


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    Interesting question.

    Just looking at the inlet and outlet pipes shows you have mass that enters the system with one velocity and comes out with another velocity (the speed component is the same but the direction component changes).
  4. Mar 28, 2013 #3


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    I think the force will be just down, like shown in the picture, due to momentum change. It is the same force as with a simple U-piece. To find the distribution go to the rotating rest frame of the water in the turn and consider the pressure gradient caused by the centrifugal force.

    For the gyroscopic effect the number of windings does matter, because you have more mass and thus more angular momentum.
  5. Mar 28, 2013 #4
    Thank you very much.
  6. Mar 28, 2013 #5
    Google Fluid Flow in a Curved Pipe. Such a flow is known to exhibit a secondary flow at each cross section.
  7. Jul 26, 2013 #6

    I did a practical experiment to check the distribution of forces and I noticed the appearance of a predominant force which is the result of trajectory change of the fluid.


  8. Jul 27, 2013 #7
    Any feedback is appreciated, thank you.
  9. Aug 1, 2013 #8
    It is assumed that a mass which travels in a closed loop will not generate a linear motion and all forces will cancel each other. Such a system would necessarily violate the law of conservation of momentum.

    Imagine that the spiral pipe is a closed loop system, do you think that building a thruster of this kind it is impossible? I think I can demonstrate the opposite. I wait for your opinions.

    Thank you in advance.
  10. Aug 2, 2013 #9


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    If the fluid goes out at the same velocity as it comes in, there should be no net force on the pipe. But this is not the case in the video:


    The fluid comes back where it came from, so it changes momentum. There is a reaction force on the float, which interacts with the elastic pipes, causing the swinging. Even if you make the net force zero, you still eventually have a net torque which will cause some movement. There is also possibly a "kick" when you switch the pump on, and the higher pressure straightens the pipes.
    Last edited: Aug 2, 2013
  11. Aug 2, 2013 #10
    What if the fluid is accelerating?
  12. Aug 3, 2013 #11
    The coil causes a pressure drop so the inlet pressure will be higher than the outlet pressure. As pressure over an area relates to force there will be an imbalance. Also the friction will allow heat to dissipate along the coil and so an energy transfer happens there. The differential pressure within the pipes will be difficult to quantify as they will want to straighten under pressure, this force may corrupt your test results.

    Also the coil you have drawn has a distinct seperation from inlet to outlet but your test apperatus has the pipes close together, you may like to try the test with the pipes as drawn. Acceleration and de-acceleration will play a large part in your test as you start and stop the pump.

    It is a rather elagant test but pipe straightening, P*A forces, heat dissipation and pressure loss make it tricky to deduce meaningful results. Not to mention the pipe shortening due to pressure increase, twisting of the coil as a result and gyroscopic effects causing motion.
  13. Aug 4, 2013 #12
    Action and reaction

    From a simplified perspective I will try to present my point of view on the phenomenon.


    Reaction is represented by the force exerted by the pipe walls on the forward path of the fluid, and from there result a change of trajectory equivalent with the distance "d" between the point of entry of the fluid "1" and the output point of the fluid "2". From this change of trajectory of the fluid results a linear force which occurs simultaneously and opposite in direction, according to Newton's third law of motion.
  14. Aug 4, 2013 #13


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    No. Translating the trajectory, while preserving the velocity vector, doesn't change the momentum of the fluid, so there is no linear force.
  15. Aug 5, 2013 #14
    So you are saying there is no force involved here? It seems moving a mass from plane to another needs a force otherwise this this neat little tube could represent, say, a tube train tunnel where we get free transport. Problem of mass transport solved!

    For sure there is a force.
  16. Aug 6, 2013 #15


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    Moving a mass does not require a net force. Accelerating a mass requires a force. The mass in this case (the fluid) comes into the system and leaves the system with the same momentum, so no force is required.
  17. Aug 7, 2013 #16
    OK so lets rotate the coil by 90 degree in such a way that the exit is above the entry. Now the fluid leaving is higher and so has potential energy increase, that surely requires a force.
  18. Aug 7, 2013 #17


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    We are discussing the net force on the pipe by the fluid here. This is only non zero when there a net momentum change of the fluid in the pipe. The force that moves the fluid upwards is not coming from the pipe, but from the pump.
  19. Aug 10, 2013 #18
    Linear force from inertia

    Let's suppose we have a closed loop, as shown in the drawings. From the earlier discussion I concluded that to produce a linear force, acceleration is needed F=m[itex]\ast[/itex]a, but to produce a linear force in one direction, irrespective of the direction of circulation of the fluid in the system, another arrangement is necessary.


    From my point of view, irrespective of the direction of fluid circulation, linear force will be generated in one direction and there is no other force to cancel it.
  20. Aug 10, 2013 #19


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    No matter how many funny shapes you invent: If the fluid makes one cycle, and arrives at the start point with the same momentum vector, then there is no net momentum transfer to the pipe.
  21. Aug 11, 2013 #20

    Now it's open:

  22. Aug 11, 2013 #21


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    Last edited by a moderator: May 6, 2017
  23. Aug 13, 2013 #22
    Funny shape and momentum

    Now I have to explain why I made this funny shape. For example, if we assume we have a pipe semicircle shaped and assume that at both ends we have a piston that works in tandem with the other one, one pushes and the other pulls the fluid inside the pipe, as represented in the picture:


    From here we see that all the forces are manifested in the same plane, interacting and canceling each other. Further I sought for a solution to break the forces in two perpendicular planes and the best solution found for this is a spiral pipe bent at an angle of 90 degrees:


    As I stated in previous posts, I looked for a solution to get a directional force only in one direction, irrespective of the direction of fluid circulation:


    If the fluid has variable acceleration and variable speed, there is net momentum transfer to the pipe?
  24. Aug 14, 2013 #23


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    If the flow rates at inflow and outflow change in the same way, so they are equal at any time point, then no.

    If you get into compressibility and pressure wave propagation, they you can have short term instantaneous net forces on the pipe, but they will average to zero, if you accelerate/decelerate the flow cyclically.
  25. Aug 15, 2013 #24
    Impulse, momentum, change in momentum

    When a fluid speeds up or slows down, inertial forces come into play. Such forces may be produced by either a change in the magnitude or the direction of the velocity since either change in this vector quantity produces acceleration.

    The change in momentum of a mass is equal to the impulse given to it. (Newton's 2nd law of motion)

    Impulse = Force x time
    Momentum = mass x velocity
    Change in momentum = Δmv

    Newton's second law: Δmv = Ft
    Δmv/t = F
    Since Δv/t = acceleration "a", we get form of the law F=ma

    The force is a vector quantity which must be in the direction of Δv. Every force has an equal and opposite reaction so there must be a force on the bend equal and opposite to the force on the fluid.
  26. Aug 21, 2013 #25
    Inertial propulsion system

    I will explain the whole system from my point of view, please correct me if I'm wrong.
    It is widely accepted that an propulsion system which do not expel reaction mass, but it is not reactionless may be impossible due to some factors. So, if a mass makes one cycle on a trajectory and arrives at the start point, then there the sum of all forces generated will be zero. I agree with this, because all attempts from field of "inertial propulsion" have one thing in common: cyclical motion of the mass and trajectory, takes place only in one plane.

    But what happens if in an inertial system the cycle of the mass it is splitted in two or more planes? A mass makes one cycle on a trajectory which is splitted in two or more planes and arrives at the start point. Generated forces will be different from zero on each plane and if planes are perpendicular, generated forces will no longer cancel each other. Solution found to split the forces in two perpendicular planes is a spiral pipe bent at an angle of 90 degrees.

    Now I return to my closed loop pipe and I imagine that is containing inside a fluid that can be controlled by electromagnetic fields. It is known that electromagnetic fields can carry energy and momentum. If there will be momentum exchange between electromagnetic fields and fluid, the system is not closed.


    From my point of view in this particular case, this system can become an inertial thruter which do not expel reaction mass, it is not reactionless and I want to know if there are factors that may make it impossible. I know scientists are not in the habit of trying things they know won't work, but if someone will test the system, it is very possible to have a surprise.
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