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Measuring surface waves using the a pressure sensor?

  1. Oct 26, 2015 #1
    Not sure if this is an engineering or physics question, but here it goes:

    I'm trying to wrap my head around the pressure field caused by waves. I'll recap to so anyone can check if I have made any incorrect assumptions:

    If we ignore atmospheric pressure, the pressure in the water has a hydrostatic part -ρgz (z<0 if under water) and a dynamic part δΦ/δt, where Φ is the velocity potential. The kinematic part 0.5ρv2 is approximated away.

    The total pressure is therefore P=-ρgz+δΦ/δt. This assumes a constant level for the surface. To solve the free surface we require that they cancel at the surface, essentially solving the free surface as η=(δΦ/δt)/(ρg), which will be a sine wave.

    By solving Φ we also know that the dynamic pressure δΦ/δt is attenuated at a rate K=cosh(z(k+h))/cosh(kh), where h is the total water depth, k=2π/λ is the wave number and h<z<0 is the depth we are calculating the pressure.

    So here is the thing I am having problem with: The hydrostatic pressure is now fluctuating because of the elevation changes in the free surface. At the surface these fluctuations are equal to the fluctuations in the dynamic pressure (caused by the variations in the horizontal velocity, right?). If we now measure the pressure at e.g. 20 m depth, we will measure a fluctuating hydrostatic pressure and an attenuated dynamic pressure. The fluctuations in the dynamic pressure will be smaller than the fluctuations in the hydrostatic pressure (or equal in shallow water). I have understood that when wanting to measure waves the pressure readings are compensated by the factor K=cosh(z(k+h))/cosh(kh) to "move back" to the surface and solve the surface. However, isn't a major part of both the measured absolute pressure and the measured fluctuations from the hydrostatic part, which is not attenuated no matter how deep we go?

    E.g the total measured pressure is the fluctuating hydrostatic part Ph=-ρgz+ρgη and the dynamic part which is equal to the fluctuations in the hydrostatic part, but attenuated Pd=Kρgη. Total pressure is therefore P=Ph+Pd=(K+1)ρgη-zρg. Do we just enter K and solve for η, or is there any way to measure just the dynamic part Pd at any depth by measuring the varying horizontal flow? If we have to measure the total pressure, why can't we just solve η from the fluctuations in the hydrostatic part even if we would be deep enough that K~0 (i.e. over λ/s deep)? Sure they are small in comparison to the total pressure, but the dynamic part is even smaller?

    What am I missing here? Can anyone please help?
     
  2. jcsd
  3. Oct 26, 2015 #2

    Nidum

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  4. Oct 26, 2015 #3
    The theory I outlined is the linear theory, not the Stokes' theory. Also, the only time pressure is mentioned in the article is when the dynamic boundary condition is discussed. Since it is similar as in the linear theory, it does not deepen my understanding. If I missed something, could you point out which part of the article sheds light on my problem?
     
  5. Oct 26, 2015 #4

    Nidum

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    Input terminated .
     
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