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Density of water as a function of depth

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Estimate the density of the water 5.3 km deep in the sea. (bulk modulus for water is B=2.0 *10^9
    2. Relevant equations

    [tex]\Delta[/tex]V/V=([tex]-1/B[/tex])*[tex]\Delta[/tex]P

    dP/dY=-[tex]\rho[/tex]g

    3. The attempt at a solution

    I tried to use the equation:
    P=Po*e^(-[tex]\rho[/tex]o/Po)*gh
    But i kept getting ridiculously large answers that i dont think could possibly be true.

    I said that the initial pressure was 1atm or 1.013*10^5 Pa and that initial density was 1*10^3(kg/m^3).

    I know that once i get the pressure under water, I can find the change in volume divided by volume (delta V over V) and from there figure out how what the density of a theoretical block of water would be if submerged 5300 meters. Its just finding out the pressure that trips me up. Am i using the wrong constants or must i derive a different equation for pressure as a function of height?

    I tried to find volume as a function of pressure, but its finding the pressure with a variable density that keeps throwing me off. I'm not sure the third equation i used is correct for this kind of problem.


    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    as a sense check, first assume the density is constant, calculate the pressure at 5.3km (10m of water is ~1atm so will be around 530atm). now use this to estimate the density as the chaneg will probably be reasonably small... also gives you an idea of your error


    now for a more accurate assumption, set up your de as follows..
    [tex] \rho(V) = \frac{m}{V} [/tex]
    [tex] \frac{d}{dV} \rho(V) = -\frac{m}{V^2} [/tex]
    [tex] d \rho = -\frac{dV}{V} \rho [/tex]

    then from the bulk modulus
    [tex] \frac{dV}{V} = -\frac{dP}{B}[/tex]
    [tex] d \rho= \frac{dP}{B} \rho [/tex]
    or
    [tex]dP = B \frac{d \rho}{\rho} [/tex]


    then the pressure gradient with denisty you know
    [tex] \frac{d}{dy} P(y) = \rho(y).g [/tex]
    re-arranging
    [tex] dP(y) = \rho(y).g.dy [/tex]

    then combine the equations & integrate
     
    Last edited: Sep 27, 2009
  4. Sep 27, 2009 #3
    thank you for your help, and i certainly will look this this over to get a better understanding of the problem, but I actually made this problem more difficult than it had to be and i dont need differentials. thanks again though
     
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