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Homework Help: Venturi Meter

  1. Nov 12, 2005 #1
    Please refer to page 6 of

    http://www.physics.usyd.edu.au/ugrad/jphys/jphys_webct/jp_exams/1902_exam_2004.pdf [Broken]

    I'm quoting from the solution guide:

    http://www.physics.usyd.edu.au/ugrad/jphys/jphys_webct/jp_exams/1902_exam_solutions_2004.pdf [Broken]

    [itex]P_1 = P_A + \rho g y_1 [/itex] and [itex]P_2 = P_A + \rho g y_2 [/itex]

    Hence

    [itex]\Delta h = y_1 - y_2[/itex].

    Is it just me or does this last step total nonsense? AIUI, [itex]y_1[/itex] and [itex]y_2[/itex] refer to the position of the water levels measured with respect to two different coordinate systems. So how is it justified to say [itex]\Delta h = y_1 - y_2[/itex]? I drew a diagram and calculated the vertical separation between the water levels to be [itex]y_1 - y_2 + \frac{D_2 - D_1}{2}[/itex]. Could someone please point out if I am missing something obvious.

    Thanks.

    James
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 13, 2005 #2

    FredGarvin

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    You're only asked to calculate the height differences on the columns. That's pretty much just a fluid statics problem. The pressures are all you care about. When you ask how is it justified to say [tex]y_1 = y_2[/tex] just take a look at the fluid static FBD:

    At column number 1, you have atmospheric pressure in equillibrium with the fluid static pressure at point one, or [tex]P_1 = P_a + \rho g y_1[/tex]. At point 2, you have atmospheric pressure in equillibrium with the fluid's static pressure at point 2 or [tex]P_2 = P_a + \rho g y_2[/tex].

    Since it is assumed incompressible and no local changes in g, then that means that the only thing that can change as [tex]P_1[/tex] amd [tex]P_2[/tex] change is [tex]y[/tex].

    I guess the best thing would be for you to post how you came up with your answer and we can go from there.
     
    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3
    The diagram is mislabeled. The delta h in the diagram is [itex]y_1 - y_2 + \frac{D_2 - D_1}{2}[/itex]. The question makes sense as long as you "assume" that they actually want [itex]y_1 - y_2[/itex]. Quite a silly question.
     
  5. Nov 14, 2005 #4

    FredGarvin

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    You've lost me on that one. The [tex]\Delta h[/tex] is the pressure drop across the venturi. [tex]\Delta h[/tex] has to equal [tex]y_1 - y_2[/tex]. How can they be different values? All that is done is to take the relationship derived for [tex]P_1 - P_2 = \frac{1}{2}\rho \Delta V^2[/tex] and replace the velocity terms with [tex]V = \frac{R}{A}[/tex]

    Show how you arrived at your conclusion. That would help.
     
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