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Bulk Modulus Application

  1. Jan 21, 2016 #1
    1. The problem statement, all variables and given/known data
    See image.

    2. Relevant equations
    pressure = density*gravity*depth

    3. The attempt at a solution
    The pressure at 5000 ft according to the book is 322,000 psf. This makes sense because density*gravity*depth = 2*32.2*5000 = 322,000 psf. How do I apply the bulk modulus equation to find the pressure at 5000 ft factoring a variable density with depth (The book says its 323,200 psf which makes sense because density increases with depth, although very slightly)?
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2016 #2
    What is the hydrostatic equation, expressed in terms of the derivative of pressure with respect to depth?
     
  4. Jan 21, 2016 #3
    That would be just [density*gravity]. So how do I apply that?
     
  5. Jan 21, 2016 #4
    Right. $$\frac{dp}{dz}=\rho g$$
    Now, from the bulk modulus equation, if the density approaches ##\rho_0## at low pressures, what is the density at pressure p?
     
  6. Jan 21, 2016 #5
    That's where I have trouble. I always end up with just [density*gravity*depth]. Is it just (1/g)*(dp/dz) ?
     
  7. Jan 21, 2016 #6
    The relationship between density and pressure does not involve z. It's strictly a physical property relationship (sort of like the ideal gas law, except for a liquid).

    Chet
     
  8. Jan 21, 2016 #7
    So, solving for density using the Bulk Modulus equation:

    ρ = B*(Δρ/p)
     
  9. Jan 21, 2016 #8
    Actually, the equation is $$\frac{1}{\rho}\frac{d\rho}{dp}=\frac{d(\ln \rho)}{dp}=\frac{1}{B}$$ What do you get if you integrate that, subject to the initial condition ##\rho=\rho_0## at p --> 0?
     
  10. Jan 21, 2016 #9
    I get:

    (1/B)*p = ln(ρ/ρ0) when factoring in the initial conditions.
     
  11. Jan 21, 2016 #10
    Good. Now solve for ##\rho## in terms of p. What do you get?
     
  12. Jan 22, 2016 #11
    ρ = ρ0ep/B
     
  13. Jan 22, 2016 #12
    OK. Now substitute that into the hydrostatic equation in post #4. What do you get? Can you integrate that from z =0?
     
  14. Jan 22, 2016 #13
    (dp/dz) = ρ0gep/B

    How do I rearrange that to integrate from z=0 to h?
     
  15. Jan 22, 2016 #14
    Cmon man.

    $$e^{-\frac{p}{B}}dp=\rho_0 g dz$$
     
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