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Simple fluid dynamics

  1. Sep 20, 2005 #1
    for some reason my brain is dead on this one:

    diagram: http://img.photobucket.com/albums/v11/biggm/diagram.jpg

    here is a u-tube of uniform cross sectional area, partially filled with Hg. water is poured into both sides and the equilibrium is as shown. the value of h2=1.00cm. I am to calculate the value of h1.

    I know it has to do with P= Po + [tex]\rho[/tex]gh but am not sure how to apply it correctly.

    thanks
     
  2. jcsd
  3. Sep 20, 2005 #2

    LeonhardEuler

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    Consider what force must be pushing down on the mercury in the left of the column to hold the mercury and water up on the right.
     
  4. Sep 20, 2005 #3
    the pressure at said point would just be [tex]\rho[/tex]gh where
    h=h1 + h2 + height of the right column of water

    this doesnt really allow me to solve this though. unless i am thinking wrongly still
     
  5. Sep 20, 2005 #4

    Pyrrhus

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    Is the tube open to the atmosphere?
     
  6. Sep 20, 2005 #5
    yes its open to the atmosphere so both sides experience the same pressure.
     
  7. Sep 20, 2005 #6

    LeonhardEuler

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    This is true, but what I mean is that the pressure at the mercury surface on the left side must be equal to the pressure at the same depth on the right because within the mercury the pressure is varying by the same amount for a given change in depth. So what you just said is one side of the equation. The other side is the pressure at the same depth on the right side. (Note that this only works because up to that point we are talking about pure mercury. Once you throw water in, then the pressure at equal hieghts no longer has to be the same because the rate of change of pressure over hieght is proportional to density.)
     
  8. Sep 20, 2005 #7

    Pyrrhus

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    Yes like Leonhard said, You must make the equation for the pressure of mercury on the left side, and make a equation for pressure of mercury on the right side at the same height (static fluids), because it must be the same pressure, Remember Pascal's Paradox. The answer will be about 12.6 cm
     
  9. Sep 20, 2005 #8
    the pressure on the mercury at the water-mercury interface on the left is given by what i said. but on the right side at this same depth is mercury and water so therefore i have to add together the pressures due to the depths of water and mercury on the right? this does not make sense since i am not given the density of mercury...
     
  10. Sep 20, 2005 #9

    LeonhardEuler

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    You definitly need the density of mercury to solve this problem because the answer will change depending on it. If the density of mercury were the same as water, then h1 would necessarily be zero as long as the pressures on boths sides of the tube are the same.
     
  11. Sep 20, 2005 #10
    ok so ill find the density of mercury and my equation should be the following, correct?:

    letting L=length of column of water on right,

    pwater*g*(h1+h2+L)=pwater*g*L + pmerc*g*h2

    where pwater and pmerc are the densities of water and Hg respectively.

    is this correct? i think there should be an L in the last term of the equation.
    thanks for your help
     
  12. Sep 20, 2005 #11

    LeonhardEuler

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    Looks right to me.
     
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