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A Hammer in Pipelines

  1. Jun 7, 2016 #1
    Assume a pipe which can be closed rapidly simultaneously from both sides and just one wave can propagates along it. (This model is completely subjective and can not be done actually. Because if both sides of a pipe close rapidly and simultaneously, then two waves propagate not one). The pipeline has friction throughout.

    1. Is any flow in this pipeline after closing both sides?
    2. Do Packing and Attenuation occur for this pipeline?
    3. Thank you; Mohsen
  2. jcsd
  3. Jun 9, 2016 #2
    What are your thoughts on this?
  4. Jun 10, 2016 #3
    Actually I think there is no any flow in the pipe (like what was before closing the valves), but compression of water (due to hammer phenomenon) creates flow in the closed system (which can make Packing and Attenuation). It means that friction can dampen the wave after a while.
    Am I right?
  5. Jun 10, 2016 #4
    Yes. Exactly.
  6. Jun 13, 2016 #5
    Let me use this opportunity to ask about 'Attenuation' phenomenon. Does Attenuation is called to the headloss of flow due to compression of fluid (regarding to my previous explanation)? I know what Packing is. But about Attenuation, am I right?
  7. Jun 13, 2016 #6
    It means that the magnitude of the pressure variations is decreasing with time as s result of viscous dissipation.
  8. Jun 13, 2016 #7
    Thank you very much.
    Attenuation is decreasing in head (pressure) which is due to viscosity OR Decreasing in head (pressure) which is due to density change (which results flow)? Which one?
    explain it more please.
  9. Jun 13, 2016 #8
    After the valves are closed, the original kinetic energy of the fluid will be converted into compressional and expansion energy of the fluid, and will also be partially present as kinetic energy. The compressional and expansion energy will exchange with the kinetic energy as time progresses, as the expansion and compression waves pass up and down the fluid in the pipe. However, if there is no viscous dissipation, then the total of all these energies will never change, and the exchange will continue forever. Viscous dissipation damps out all this mechanical energy and converts it into internal energy of the fluid. This will result in and increase in internal energy of the fluid and an accompanying slightly higher temperature of the fluid (if the system is adiabatic). And, in the end, the pressure will be uniform, the density will be uniform, and the velocity will be zero everywhere.
  10. Jun 13, 2016 #9
    Dear Chestermiller
    You are right, But I can not still comprehend why you focus on Viscosity not Density. Compressional and expansion, changes Density of the fluid. Why you focus on Viscosity? This is my question exactly. What is the effect of decreasing and increasing of pressure (Compressional and expansion) on viscosity and density separately?
    This phrase made me confused: 'if there is no viscous dissipation, then the total of all these energies will never change, and the exchange will continue forever'. Would you explain more about it? Do you mean pressure increasing makes higher viscosity and pressure decreasing makes lower viscosity? If so, what is the relation between this and attenuation?
    Thank you very much for your answer in advance.
  11. Jun 14, 2016 #10
    I'm afraid I haven't explained this very well, so I've gotten you terribly confused. So I'm going to try an entirely different approach. I'm going to introduce an analog system to our fluid in a tube that exhibits qualitatively all the key features of our fluid, including mass (inertia), compression/expansion, and viscous dissipation.

    Consider a horizontal tube, closed at both ends, with a series of n equally spaced frictionless pistons situated along the cylinder axis, each with mass m. Each piston is connected to its two immediately neighboring pistons by a massless spring of spring constant k. The two pistons at the very ends of the cylinder are connected to the closed ends of the cylinder by a spring of spring constant k. This is case A. The pistons of mass m are analogous to the distributed mass of the fluid. The springs are analogous to the distributed compression/expansion behavior of the fluid. Now, at time zero, each of the pistons is given an initial velocity of v0. This corresponds to the initial velocity of the fluid at the time that the valves are closed. The total kinetic energy of the pistons at this time is ##nm\frac{v^2}{2}##. As time progresses, the total elastic energy stored in the springs plus the sum of the potential energies of the n pistons will be equal to this same value, and it will never change. The way that the velocities of the pistons and the elastic energy of the springs is distributed at any time is immaterial. This corresponds to the case where the fluid viscosity is zero.

    Case B: Now consider the same model as above, but instead of just a spring being attached between each pair of pistons, there is also a damper. There are also dampers connected between the two pistons at the very ends and the closed ends of the cylinder. In this case, as time progresses, the viscous response of the dampers dissipates the sum of the kinetic energies of the pistons and the elastic energies stored in the springs. So, even though the total mechanical energy still starts out as ##nm\frac{v^2}{2}##, as time progresses the total mechanical energy decays to zero. At infinite time, the springs are at their unextended length, the velocities of all the pistons is zero, and the pistons are all equally spaced.

    During the deformations in these two models, there will be regions where some of the springs are compressed and some of the springs are extended. This corresponds respectively to increased density and decreased density. But, in the end, the density is again uniform.

    Are you able to understand these two cases and how they relate to your fluid problem?

  12. Jun 15, 2016 #11
  13. Jun 16, 2016 #12
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