1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fluid Dynamics: Static pressure in compressible Liquids

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data

    "Derive a relation between the static pressure P at a point and its depth y from the free surface of the liquid. Given the surface density of liquid is ρο, and compressibility of the liquid is k."

    2. Relevant equations

    ##ρ(P) = ρ_οe^{kP}##
    ##dP = ρg dy##

    3. The attempt at a solution

    I have already found that:
    ##ρ(P) = ρ_οe^{kP}##

    Then, from the relation:
    ##dP = ρg dy##

    and using ##ρ(P) = ρ_οe^{kP}##, we get:

    ##e^{-kP} dP = ρ_ο g dy##

    Integrating L.H.S. from 0 to P and R.H.S. from 0 to y, we get:

    $$P = \frac {-1}{k} \ln(1 - kρ_ο gy)$$

    Are my limits correct? Is the answer correct? I ask this because the answer looks like it could easily become undefined (if ##kρ_ο gy## became greater than 1). If it is wrong, then why? If it is correct, then how does it not become undefined?
     
    Last edited: Oct 9, 2015
  2. jcsd
  3. Oct 9, 2015 #2
    The answer is correct. In practice, you would never encounter values of y huge enough for the expression to become undefined. Imagine how much pressure it would take to compress liquid water to half its volume (say). Also, in reality, k varies gradually with P, but the starting equation is a good approximation over a substantial range of pressures.

    Chet
     
  4. Nov 24, 2015 #3
    Thank you very much! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Fluid Dynamics: Static pressure in compressible Liquids
Loading...