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Venturi Meter

  • Thread starter jdstokes
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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
 
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FredGarvin

Science Advisor
5,050
6
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:
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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.
 

FredGarvin

Science Advisor
5,050
6
jdstokes said:
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.
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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