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Bernoulli problem

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I'm back with another Bernoulli problem :)

    A reservoir with constant pressure P = 105 Pa releases through a tube water (density ρ = 1 g/cm3) with the velocity v = 4 m⋅s-1. That tube is then connected to another perpendicular tube (area A = 0.1 m2, height h = 2m) in which the water doesn't move and holds a mass m covering the whole area.
    What is the mass m?

    (sorry for the poor problem statement, it's hard to translate from German to English since neither of those languages are my mother language :) see attached picture for a better comprehension).

    2. Relevant equations

    Bernoulli equation, pressure, forces.
    I also immediately convert ρ into ρ = 1000 Kg/m3.

    3. The attempt at a solution

    Okay I'm posting that problem here because I feel like I still get confused about how to deal with some inputs in the Bernoulli equation. Here is what I've done:

    Since the mass is not moving, I believe ΣF = 0, which means that the force of the water on the mass (FU) is equal to the force of gravity of the mass on the water:

    ΣF = 0 ⇔ mg = FU

    Next step I want to know the pressure (P1) at the point where the water and the mass meet, and I am a little unsure about my formula:

    P1 = FU/A = mg/a

    Is that correct? I am often unsure what force should be placed in that equation...However I then attempted to substitute that pressure into my Bernoulli equation:

    P + ρgh0 + ½ρv2 = P1 + ρgh + ½ρv12

    (ρgh0 = 0 because h0 = 0, ½ρv12 = 0 since the water is not moving)

    ⇔ P + ½ρv2 = mg/A + ρgh
    ⇔ m = (A/g)(P + ½ρv2 - ρgh)
    m = 880 Kg

    What do you guys think?

    Thank you very much in advance for your answers, I appreciate your help.


    Julien.
     

    Attached Files:

  2. jcsd
  3. Mar 26, 2016 #2
    The pressure in the tank is P (and the velocity inside the tank is zero), but the pressure within the tube is ##P-\frac{1}{2}\rho v^2##. This is also the pressure underneath the vertical pipe. The reason that pressure in the tube is less than P is that part of the pressure was used to accelerate the fluid to the tube velocity. Use Bernoulli to determine the pressure directly underneath the vertical pipe, and you will see this. So, in your final equations, there should be a minus sign in front of the ##\frac{1}{2}\rho v^2##, rather than a plus sign. Everything else is done correctly.

    Chet
     
  4. Mar 26, 2016 #3
    @Chestermiller Hi Chet, and thank you for your answer! Indeed I see what you mean with the pressure within the tube, I didn't reflect upon the fact that the water has no velocity inside the tank. But I fail to obtain P - ½ρv2 on the left side of my Bernoulli equation, since the - ½ρv2 cancels with the + ½ρv2 (which also kind of makes sense). I'm sure I am missing something, because the way I've done it makes the information about the velocity in the tube irrelevant.

    I get m = 800 kg now. I've attached my equations with the post.

    Julien.
     

    Attached Files:

  5. Mar 26, 2016 #4
    Bernoulli can't be applied to points 1 and 2 because point 2 is not on a streamline. Between points 1 and 2, you just apply the hydrostatic equilibrium equation. If you didn't know the lower pipe was there (say it was covered), you would conclude that the material in the vertical column would have to satisfy hydrostatic equilibrium.
     
  6. Mar 26, 2016 #5
    @Chestermiller Thank you, that was a very clear explanation. I now see how to obtain the - ½⋅ρ⋅v2, but I do not have the - ρ⋅g⋅h anymore, which I guess should be there.
     

    Attached Files:

  7. Mar 26, 2016 #6
    (Also, I assumed we are dealing with gauge pressure throughout the whole problem - is that a correct assumption?)
     
  8. Mar 26, 2016 #7
    Yes.
     
  9. Mar 27, 2016 #8
    @Chestermiller Could it be that the -ρgh is the pressure at the top (point (2))?

    ΣF = 0 = P1⋅A - P2⋅A - m⋅g
    ⇔ m⋅g = A⋅(P0 - ½⋅ρ⋅v2 - ρ⋅g⋅h)
    m = 720 kg
     
  10. Mar 27, 2016 #9
    Your force balance on the piston showed that the pressure at point 2 is mg/A. However, the equations you have written here are correct.
     
  11. Mar 27, 2016 #10
    @Chestermiller I think I get it now, thank you once again for your very valuable help!

    I attach to this post my final complete answer for other people searching infos on the topic.
     

    Attached Files:

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