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Difference in height in a U-shaped tube

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    The horizontal pipe, shown in the figure, has a cross-sectional area of [itex]40,0 cm^{2}[/itex] at the wider portions and [itex]10,0 cm^{2}[/itex] at the constriction. Water is flowing in the pipe, and the discharge from the pipe is [itex]6,00*10^{-3}\frac{m^{3}}{s} (6,00\frac{L}{s})[/itex]. The density of mercury is [itex]\rho_{Hg}=13,6*10^{3}\frac{kg}{m^{3}}[/itex] and the density of water is [itex]\rho_{w}=1,00*10^{3}\frac{kg}{m^{3}}[/itex].

    What is the difference in height between the mercury columns in the U-shaped tube?

    YF-14-45.jpg

    2. Relevant equations

    I've been using Bernoulli's Principle.

    [itex]P_{1}+\rho*g*H_{1}=P_{2}+\rho*g*H_{2}[/itex]

    3. The attempt at a solution

    When I use the formula above I get 0.127 m which is incorrect. I also found this problem on the internet where they get the same answer so I was pretty sure but apparently it's wrong. :(
     
    Last edited: Nov 8, 2011
  2. jcsd
  3. Nov 8, 2011 #2
    Where is your picture? I suspect one end of the manometer attached to the smaller diameter while the other is sampling the larger diameter. If so there is a difference in velocity so the pressure must change. You'll probably need some V^2 terms in your equation to account for velocity differences. We need a picture of your problem.
     
  4. Nov 8, 2011 #3

    SteamKing

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    You have the density of water and the density of mercury as roughly equal.
     
  5. Nov 8, 2011 #4
    Sorry! I've put in the picture and the correct density of water. :)
     
  6. Nov 8, 2011 #5
    OK I see your picture now. Don't know where it was before because it was not showing up on my screen. Because the velocity changes you must have a V^2 on each side of the equation. You compute the velocities from the cross sectional areas and flow rate.
     
  7. Nov 8, 2011 #6
    Should I use [itex]A_{1}v_{1}=A_{2}v_{2}[/itex]? A and v are the areas and velocities. I can use [itex]A_{1}[/itex] and [itex]v_{1}[/itex] for the wide part... then I can find [itex]v_{2}[/itex]. Does that help?
     
  8. Nov 8, 2011 #7
    Okay my calculations are [itex]A_{1}v_{1}=40 cm^{2}*6,00\frac{L}{s}=240\frac{L*cm^{2}}{s}[/itex]. Then I get [itex]A_{2}v_{2}=10 cm^{2}*v_{2}=240\frac{L*cm^{2}}{s}[/itex] so [itex]v_{2}=24\frac{L}{s}[/itex].
     
  9. Nov 8, 2011 #8
    Your units are liters per second. That is not the unit of velocity. You need meters/second. V1 is not 6 liters/second.

    Always check to ensure your units describe the variable you seek.
     
    Last edited: Nov 8, 2011
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