Solving Water Level Puzzle: Page 6 of 1902 Exam Solutions 2004

[jdstokes]

Please refer to page 6 of

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

I'm quoting from the solution guide:

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

$P_1 = P_A + \rho g y_1$ and $P_2 = P_A + \rho g y_2$

Hence

$\Delta h = y_1 - y_2$.

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

Thanks.

James

 

[FredGarvin]

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 $$y_1 = y_2$$ 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 $$P_1 = P_a + \rho g y_1$$. At point 2, you have atmospheric pressure in equillibrium with the fluid's static pressure at point 2 or $$P_2 = P_a + \rho g y_2$$.

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

I guess the best thing would be for you to post how you came up with your answer and we can go from there.

 

[jdstokes]

The diagram is mislabeled. The delta h in the diagram is $y_1 - y_2 + \frac{D_2 - D_1}{2}$. The question makes sense as long as you "assume" that they actually want $y_1 - y_2$. Quite a silly question.

 

[FredGarvin]

The diagram is mislabeled. The delta h in the diagram is $y_1 - y_2 + \frac{D_2 - D_1}{2}$. The question makes sense as long as you "assume" that they actually want $y_1 - y_2$. Quite a silly question.

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

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