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Fluid physics problem

  1. Nov 25, 2009 #1
    Im trying to complete homework for my physics class but i can't figure this problem out. here's the problem:

    The axis of a cylindrical container is vertical. The container is filled with equal masses of water and oil. The oil floats on top of the water, and the open surface of the oil is at a height h above the bottom of the container. What is the height, h, if the pressure at the bottom of the water is 9.6 kPa greater than the pressure at the top of the oil? Assume the oil density is 875 kg/m3.

    density of water i assume to be 1000 kg/m3. The help for this problem said to assume that the masses of the two liquids are the same mass, and use this obtain a relationship between the two fluids for a height. So F = mg = p(density)*A*dh*g

    p(water)*dh*g = p(oil)*dh*g area(A) would cancel because it is the same

    then i figured we would have to use the equation
    P(pressure)=P0(pressure at the top) + p*g*(dh)

    From there im assuming that P-P0 = 9.6kPa

    so 9.6kPa = p(water)*dh*g + p(oil)*dh*g

    then converted 9.6kPa to 9600 Pa, and solved for dh, but im getting the wrong answer. Can anyone help me out, show me what im doing wrong. Any help would be greatly appreciated.
     
  2. jcsd
  3. Nov 25, 2009 #2

    Doc Al

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    Staff: Mentor

    You're on the right track. The height of the water column is different than the height of the oil column, so use different symbols for each. You can't use 'dh' to represent both heights.

    Try h(total) = h(water) + h(oil).
    Rewrite this equation, using h(water) and h(oil) instead of 'dh'.
    Same issue here.
     
  4. Nov 25, 2009 #3
    instead of using dh
    use total height ...let say
    H(w) for total height of water
    and H(o) for total height of oil

    and then as Doc wrote...

    ro(water)*H(w)=ro(oil)*H(o)=m/A......ro-> density......
    verify by urslf....
    and second equation will again be in terms of H(w) and H(o).....
    u will get the answer.....
     
  5. Nov 25, 2009 #4
    ok thanks that makes sense. So correct me if i'm wrong, my second equation

    9.6kPa = p(water)*dh*g + p(oil)*dh*g

    is not correct. It should look like this:

    9600Pa = p(w)*g*H(w) and 9600Pa = p(o)*g*H(o)

    so H(w) = 9600Pa/[p(w)*g] and H(o) = 9600Pa/[p(o)*g]

    so then H(w) + H(o) = H(total)... which is the answer im looking for. Is that right? I'm getting down to the last couple of allowable attempts so i want to make sure I am correct before submitting an answer.
     
  6. Nov 25, 2009 #5
    no it shud be
    9600Pa = p(w)*g*H(w) + p(o)*g*H(o) ....bcoz 9.6kPa is total pressure drop....

    and then from 1st eq, replace any one variable in terms of other....and put in this 2nd equation...
    u will get the answer...
     
  7. Nov 25, 2009 #6

    Doc Al

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    Staff: Mentor

    Actually, it's close to being correct. You just need to replace the 'dh' with the correct heights as I stated earlier.

    That's not correct. The 9600 Pa is the pressure difference for both the water and oil combined, not each separately.
     
  8. Nov 25, 2009 #7
    ok gotya, that makes sense. So the equation should look like this:

    9600Pa = p(w)*g*H(w) + p(o)*g*H(o)

    Then i would substitute H(w) and H(o) from p(water)*dh*g = p(oil)*dh*g into the above equation. So i would get two equations:

    9600Pa = pw*H(w)*g + po*[(pw*H(w))/po]*g "po"'s cancel and you get

    9600Pa = 2*[pw*H(w)*g] then solve for H(w)

    then do the the same and solve for H(o). Then add H(w) + H(o) to get H(total)

    is that correct or am i still missing something?
     
  9. Nov 25, 2009 #8
    yup now on right track :)
     
  10. Nov 25, 2009 #9
    I decided to go ahead and try it and it worked. Thank you for all your help i greatly appreciate it =)
     
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