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Fluid Mechanics problem - cylinder over a orifice

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    The horizontal bottom of a wide vessel with an ideal fluid has a round orifice of radius ##R_1## over which a round closed cylinder is mounted, whose radius ##R_2>R_1##. The clearance between the cylinder and the bottom of vessel is very small, the fluid density is ##\rho##. Find the static pressure of fluid in the clearance as a function of the distance r from the axis of the orifice (and the cylinder), if the height of the fluid is equal to ##h##.


    2. Relevant equations



    3. The attempt at a solution
    I think I have to apply the Bernoulli's equation but I need two points to apply the Bernoulli equation. How do I find those points? :confused:

    I understand I should show some more attempt but I really can't get the intuition behind these fluid mechanics problem so I am unable to make any attempt.

    Any help is appreciated. Thanks!
     

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  2. jcsd
  3. Nov 17, 2013 #2

    TSny

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    You might start with two points where you know some information. For example, what can you learn by picking points 1 and 2 as shown?
     

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  4. Nov 17, 2013 #3

    tiny-tim

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    Hi Pranav-Arora! :smile:
    No, you need a streamline

    Bernoulli's equation is always and only valid along a streamline.​

    Once you've drawn a streamline, it should be obvious what to do next. :wink:
     
  5. Nov 18, 2013 #4
    Hi TSny and tiny-tim! :smile:

    I use the streamline shown by TSny and the reference line is a line passing through 2.

    From Bernoulli's equation,
    $$P_1+\rho gh+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$$

    I am not sure but is ##P_1=P_2## as both points are open to atmosphere or medium outside? Also, how do I relate ##v_1## and ##v_2##? :confused:
     
  6. Nov 18, 2013 #5

    tiny-tim

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    Hi Pranav-Arora! :smile:
    Draw circles round the hole …

    since the density is constant, isn't it obvious from the geometry how the speed must increase as the circles get smaller? :wink:
     
  7. Nov 18, 2013 #6
    Yes, I do understand that speed increases as the hole size decreases. Okay, so let the area of hole in clearance be s and that of wide vessel be S. From equation of continuity, ##Sv_1=sv_2## but what now? Do I approximate ##v_1## equal to zero? :confused:
     
  8. Nov 18, 2013 #7

    tiny-tim

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    the pressure (under the cylinder) is a function of r

    the pressure in the outlet should be the same as if the cylinder wasn't there :wink:
     
  9. Nov 18, 2013 #8
    I don't think I get this, can you please elaborate a bit more? :redface:
     
  10. Nov 18, 2013 #9

    tiny-tim

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    the gauge pressure for an outlet (to the atmosphere) is zero, so that gives you v in the outlet :wink:
     
    Last edited: Nov 18, 2013
  11. Nov 18, 2013 #10
    So that means ##v_2=\sqrt{2gh}##? But the question asks something else....
     
  12. Nov 18, 2013 #11

    tiny-tim

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    yes, it asks for P as a function of r, as r moves away from the rim of the hole
     
  13. Nov 18, 2013 #12
    Hehe tiny-tim. :P

    But I don't get why you asked me to find the speed, I am honestly not sure what to do with it. How do I find the pressure as a function of r? :confused:
     
  14. Nov 18, 2013 #13

    tiny-tim

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    Bernoulli's equation gives you the pressure if you know the speed. :wink:
     
  15. Nov 18, 2013 #14
    The fluid is going to be entering the hole horizontally (i.e., radially), so the volumetric throughput rate entering the hole is going to be [itex]2\pi R_1t\sqrt{2gh}[/itex], where t is the clearance. Once you know the volumetric throughput rate, you can get the fluid velocity at any radial location r.
     
  16. Nov 18, 2013 #15
    Hi Chestermiller!

    Okay, I see that you find the rate of flow of liquid but I still don't get how to find the pressure as a function of r.

    Do I have to apply Bernoulli at 2 and at some distance r from the axis?
     
  17. Nov 18, 2013 #16
    Yes. You have the pressure and velocity at r = R1, and you have the velocity at arbitrary r, so you use the Bernoulli equation between these two locations to get the pressure at r.
     
  18. Nov 18, 2013 #17
    Thanks Chestermiller and tiny-tim! :smile:

    Applying Bernoulli at 2 and at a distance r,
    $$P_o+\frac{1}{2}\rho(\sqrt{2gh})^2=P(r)+\frac{1}{2}\rho v^2$$
    From equation of continuity, I have
    $$A(r)v=2\pi R_1t\sqrt{2gh} \Rightarrow 2\pi rtv=2\pi R_1t\sqrt{2gh} \Rightarrow v=\frac{R_1\sqrt{2gh}}{r}$$

    Substituting in Bernoulli and solving for P(r),
    $$\boxed{P(r)=P_o+\rho gh\left(1-\frac{R_1^2}{r^2}\right)}$$
    (##P_o## is the atmospheric pressure)
     
  19. Nov 18, 2013 #18

    tiny-tim

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    yes, that looks right :smile:
     
  20. Nov 19, 2013 #19
    Thanks tiny-tim! :)

    I compared my answer with the answer given in book. The book shows the condition that ##R_1<r<R_2##. I don't get this. Why not ##R_1 \leq r \leq R_2##? The P(r) I have found is valid for ##R_1##, I am not sure about ##R_2## though.
     
  21. Nov 19, 2013 #20

    tiny-tim

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    i agree with you …

    it seems obviously valid for R1, but "at" (or even near) R2 there's likely to be some "edge effect" (related to the movement not being entirely horizontal)

    all this is academic … for an "ideal" fluid and "ideal" cylinder, your limits are just as good as the book's! :smile:
     
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