1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Fluid Mechanics: Compressible Fluids, density function

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data
    "Suppose that a liquid has an appreciable compressibility. Its density therefore varies with depth and pressure. The density at the surface is ρ0
    (a) Show that the density varies with pressure according to ρ=ρ0ekp where P is gauge pressure at any depth and k is the compressibility, a constant.
    (b) Find P as a function of depth y.

    2. Relevant equations
    ρ = m/V (Density definition)
    k = (1/V)(ΔV/ΔP) (Compressiblity definition)
    dP/dy = ρg (Pascal's Law)
    My professor also hinted that I should replace any deltas with differentials, that I need to eliminate V, and to find dρ/dV (which is -m/V2
    3. The attempt at a solution
    From the second equation, I multiplied both sides by dP (after changing ΔV to dV and ΔP to dP) to obtain the following:
    k dP = (1/V) dV
    I then integrated both sides:
    ∫k dP (from P0 to P) = ∫(1/V) dV (from 0 to V)
    ekP(gauge) = V
    I get the feeling that I'm going in the correct direction, but haven't found a viable solution from this point. I fooled around with the density definition and Pascal's Law to no avail. Any assistance would be much appreciated.
  2. jcsd
  3. Feb 18, 2013 #2
    Good start!!! Your integration limits on V are incorrect. They should be from V0 to V. Also, for a given amount of mass the product of density and volume is constant.
  4. Feb 18, 2013 #3
    When I intergrate on those limits I get kP(gauge)=ln(V)-ln(V0), which seems to be worse than before.
    Also, if m = ρV is constant, how does that work with ρ and V both being varying functions?
  5. Feb 18, 2013 #4
    The compressibility condition should have a minus sign (the volume decreases with increasing pressure):

    k dP = -(1/V) dV

  6. Feb 18, 2013 #5
    I understand where the v/v0 came from (the ln) but how did you derive p0/p? Did it come from another integral somewhere?
    EDIT: nvm, I understand now. Thanks a bunch!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook