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Consider a long pipe filled with some newtonian liquid, with piston at both the ends. You push the piston at end A, and observe it at the other end B. How much time does it take for piston at B to move ? Or what is the speed at which this transfer takes place ?

If we consider a perfectly rigid pipe, and a perfectly incompressible liquid (without any dissolved air in it), what speeds can it reach ? Will it break the sound barrier ?

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  • $\begingroup$ The speed of sound is relative to each material. If you mean would an ideal incompressible fluid have a higher speed of sound than that of air, then the answer is yes. $\endgroup$ – honeste_vivere Oct 25 '14 at 15:45
  • $\begingroup$ All the answers already told you that the speed of sound is the rate at which the disturbance passes. So if you take an incompressible fluid (which means infinite sound speed) it would be instantaneous, but the fact the sound speed is finite is exactly why you can't assume that. However: fluids can move faster than the speed of sound if you have shockwaves (for instance, if you push the button faster than the speed of sound in the liquid). In that case, you get nonlinear effects, cavitation and quite possibly destruction of your hydraulics. $\endgroup$ – orion Nov 13 '14 at 10:16
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The maximum rate at which the pressure "information" can be transferred in the pipe is the speed of sound for the liquid.

This can be calculated using the Newton-Laplace equation:

$$ a = \sqrt{\frac{K}{\rho}} $$

where,

  • a = speed of sound
  • K = bulk modulus of elasticity
  • $ \rho $ = density

So the time for the pressure pulse to reach the other end of the pipe is:

$$ t = L/a $$

where,

  • t = time for pressure to reach other end
  • L = length of pipe line

If the pipe is not perfectly rigid you can apply a factor to account for the wall thickness and the anchor points of the pipe.

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The disturbance caused by the moving piston moves at the speed of a compression wave in the fluid. Since sound is a compression wave, the time it takes for the piston B to start moving is equal to the time it takes for sound to get there.

Speed of sound is given by

$$ c = \sqrt{\frac{\partial p}{\partial \rho}} $$

Since an incompressible material can, by definition, not change in density, the speed of sound in such a material is infinity. The disturbance will therefore reach the other end instantly in an incompressible material.

Even in this case, the signal will travel only at sonic speed, and not more, so it will not break the sound barrier.

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    $\begingroup$ Note that the maximum speed information can propagate is $c=2.99\times10^5$ km/s, even if $c_s=\infty$--though, of course, $v<c$ for objects with non-zero mass. $\endgroup$ – Kyle Kanos Oct 25 '14 at 14:49
  • $\begingroup$ Incompressible is a relative term. If you move at normal human speeds (i.e., not flying in a jet or plane), nearly everything you do can be described as having had interacted with an incompressible fluid. Meaning, if you go running, the speed at which you run is so much smaller than the speed of sound in air that you can treat the air as being incompressible and then approximate the fluid equations of motion. The results will be close enough to the "exact" results for most intents and purposes. $\endgroup$ – honeste_vivere Oct 25 '14 at 15:41
  • $\begingroup$ In (fresh) water, the speed of sound is >1400 m/s (at or near surface), which is significantly higher than in air. At much greater depths (e.g., ocean trenches), I imagine that speed increases. In any case, 1400 m/s is as close to instantaneous as anyone will ever need for most situations using hydraulics. $\endgroup$ – honeste_vivere Oct 25 '14 at 15:44
  • $\begingroup$ @honeste_vivere my answer addresses 'perfectly' incompressible fluids, as asked in the question. $\endgroup$ – Pranav Hosangadi Oct 25 '14 at 16:15
  • $\begingroup$ @PranavHosangadi - I know, but I was trying to point out that such things (most likely) do not exist. Therefore, we should concern ourselves with relative terms and approximations. Dividing by zero (which would be necessary for perfectly incompressible fluids) is bad even for physicists (sorry, my poor attempt at humor). $\endgroup$ – honeste_vivere Oct 28 '14 at 13:04

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