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Pressure Problem (math error I think)

  1. Nov 9, 2006 #1
    Hi everyone,

    My question is as follows:

    Consider a point in the ocean about 9.60 km deep. The pressure at that point is huge, about 9.89 x 10^7 N/m^2.
    (a) Calculate the change in volume of 1.00 m^3 of water carried from the surface to this point in the ocean.


    I set up an equation:
    change in volume = [(change in pressure)(inital volume)] / Buoyance Force, which is [-(9.89 x 10^7 N/m^2)(1 m^3)] / (2.9 x 10^9 N), which finally equals -.034 m^3 for the change in volume, but that's wrong. Note: My TA gave me the value for the buoyance force.

    Any suggestions would be greatly appreciated. Thank you! :smile:
     
  2. jcsd
  3. Nov 9, 2006 #2

    OlderDan

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    Liquids are usually considered to be incompressible fluids, but in fact they are somewhat compressible. Under huge pressure the water mollecules will get closer together. You need to find some data on the compressibility of water. You might find this in terms of density of water as a function of pressure.

    If you know the buoyancy at that depth, you could find the density from that.
     
    Last edited: Nov 9, 2006
  4. Nov 9, 2006 #3
    What would I need the density for?
     
  5. Nov 9, 2006 #4

    OlderDan

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    Density is mass per unit volume. You are taking some mass of water that had a volume of 1m^3 at the water surface to a great depth. The mass is not going to change, but the volume is. If you know the new density you can calculate the new volume.
     
  6. Nov 10, 2006 #5
    I don't know how to find the density at this depth. This is an intro physics problem, so are there any "assumptions" I can make?
     
  7. Nov 10, 2006 #6

    OlderDan

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    At an introductory level, you can assume it doesn't change at all. I tried to find some information about this on the internet, and the several links I followed all made reference to the UNESCO International Equation of State (1980) for sea water, but nobody shows you the equation. Here is a place that programmed a calculator based on the equation

    http://fermi.jhuapl.edu/denscalc.html

    and here is a place that gives a graph with an explanation that says the change in density is primarily due to changes in temperature and salinity

    http://www.windows.ucar.edu/tour/link=/earth/Water/density.html&edu=high

    Salinity does not apply in this case because of the wording of the problem. The only place I found that says there is a simple connection between density and pressure (and temperature) is here

    http://www.lsbu.ac.uk/water/strange.html

    Unless you have been given some information about the compressibility of water that is assumed to apply over a great range of pressures, assume it does not compress.
     
  8. Nov 10, 2006 #7

    FredGarvin

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    Since this is an intro class, this HAS to be a simple [tex]P=\rho g h[/tex] problem. You know the pressure at and the depth (as well as g). Assume the density at the surface is [tex]\rho = 1000 \frac{kg}{m^3}[/tex]
     
  9. Nov 10, 2006 #8

    OlderDan

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    [tex]P=\rho g h[/tex] assumes constant density at all depths. How is this going to give you a change in density?
     
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