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Homework Help: Bernoulli's equation and water exit speed from tank opening

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Use Bernoulli’s equation to calculate how fast the water emerges from the open tap (at position 2) in the figure(a). You may assume that the water at position 1 moves negligibly slowly
    (b) The tap is rotated to create a fountain as shown in (b) Calculate the maximum height h that
    the water could reach
    2. Relevant equations
    Bernoulli's equation
    3. The attempt at a solution
    Im looking to answer question b, I'm pretty sure I've done part A right.
    So for B what I did was since I got v from part a which was 3.96, hence I did follow on the part from part a, that is= rho h1= 1/2 v^2 changing to solve for h, then I got h as 1.25 meters. Idk If i did part b right.
     

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  3. Sep 26, 2016 #2

    Bystander

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    Looks good based on what I can see ... I'll give you a conditional thumb's up. Haven't actually "crunched" the square root.
     
  4. Sep 26, 2016 #3

    gneill

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    The image shows the tank's total height to be 80 cm. That's 0.80 m. How can the water from the "fountain" exceed that height? Sounds like a magnificent start for an over unity machine o_O
     
  5. Sep 26, 2016 #4

    Bystander

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    Google "hydraulic ram water pump."
     
  6. Sep 26, 2016 #5

    gneill

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    The column of water above the tap inside the tank is not literally falling (except perhaps minutely due to the overall water level falling). So it doesn't carry momentum before suddenly hitting the exit port.The water is forced out by ambient pressure.

    If this worked and the height of the jet of water ended up above the level of the tank, one could direct it to fall back into the tank. Perpetual motion would ensue, and you could extract energy by having it fall through a waterwheel on its way to the tank.
     
  7. Sep 26, 2016 #6

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    some of
    You are totally annihilating the water at the apex of its trajectory.
     
  8. Sep 26, 2016 #7
    I concur with gneill. Bernoulli will predict the height will be the same 80 cm as in the tank.
     
  9. Sep 26, 2016 #8

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    Map the stream lines; I took the intuitive route initially, also.
     
  10. Sep 26, 2016 #9
    It's not intuitive. It's just straight application of the Bernoulli equation. It even gives the same result in the second part as just taking an initial upward velocity of ##v_0##, and applying ##v_f^2 = v_0^2-2gz##.
     
  11. Sep 26, 2016 #10

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    Are you annihilating the water at the apex?
     
  12. Sep 26, 2016 #11
    I assume that, by the apex, you mean the high point of the water spout. If that's the case, the answer is "no." Approaching this point, the streamlines diverge and then turn downward again (a little like an umbrella).
     
  13. Sep 26, 2016 #12

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    Think about that.
     
  14. Sep 26, 2016 #13
    I guess we have a difference between experts here, and we are going to have to agree to disagree (unless someone else like @boneh3ad wants to chime in).
     
  15. Sep 26, 2016 #14
    211000 resposes...which one do you recommend? for this post
     
  16. Sep 26, 2016 #15

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    Alternatively, Bernoulli isn't necessary --- one can simply look at the the difference in head heights, and declare that the static maximum difference (zero) is the maximum possible difference.
     
  17. Sep 26, 2016 #16

    gneill

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    There's no reason why the nozzle of the outlet couldn't be angled a tad to allow the stream to form a parabolic arc that ended up over the tank. The tank height is only 80 cm above the nozzle, and some have predicted that the stream could reach 1.25 meters in height if directed straight up. Then:

    upload_2016-9-26_15-38-0.png

    I cannot justify this energetically. It would be akin to a ball bouncing higher than it falls.
     
    Last edited: Sep 26, 2016
  18. Sep 26, 2016 #17

    'experts' ?? how many? between implies 2
     
  19. Sep 26, 2016 #18

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    I've not found the one that raised my eyebrows twenty years ago, an open irrigation ditch, running downhill, and "magically" lifting two or three per cent of the flow to a height one or two feet greater than the source.
     
  20. Sep 26, 2016 #19
    Ok thankyou
     
  21. Sep 26, 2016 #20

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    ... and the lateral motion comes from where?
    What about part of the ball? Have you never "splashed" a liquid?
     
  22. Sep 26, 2016 #21

    gneill

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    It's just projectile motion. A bit of the vy is "stolen" to allow some vx. Apparently we have height to spare!
    Sure. But the whole volume of the tank is not "hitting" the tap opening to impart momentum in a collision with the water there. For a large tank the water is practically stagnant. A hydraulic ram takes advantage of the energy in a large quantity of moving water and sets up a "collision" that imparts momentum to a smaller quantity of water. Like a large mass hitting a small mass. But you need to have the momentum in the large mass of water prior to the "collision".

    I maintain that if this phenomena were true for this non-flowing tank then we would have perpetual motion machines and over unity power generation as the tank water continuously cycled around the loop. The best you could hope for is to have the jet just reach the height of the surface water of the tank.
     
  23. Sep 26, 2016 #22
    Hi i got 0.8 meters, is it right?
     
  24. Sep 26, 2016 #23

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    You're going to have to give us the entire problem statement.
     
  25. Sep 26, 2016 #24
    that's in page 1
     
  26. Sep 26, 2016 #25

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    Orifice/tap dimensions/configurations?
     
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