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Bernoulli equation problem

  1. Mar 21, 2016 #1
    Hi everybody! I'm training for an upcoming exam, and I'd like to know if I correctly use the Bernoulli equation.

    1. The problem statement, all variables and given/known data

    The problem is kind of difficult to describe without a drawing, please check out the attached jpg to see the situation and the known data.
    The questions are:
    a) How much time t is needed for a volume V = 1L of fluid to cross the hopper?
    b) What is the diameter of d2?

    2. Relevant equations

    Bernoulli equation, equation of continuity, formulas to calculate the flux of a fluid

    3. The attempt at a solution

    a) I use the Bernoulli equation at points (1) and (0):

    P1 + ρgh1 + ½ρv12 = P0 + ρgh0 + ½ρv02

    Now I think that P1 = 0 (if we take the gauge pressure), ½ρv12 ≈ 0 (because the diameter at point (1) is much larger than at point (0) (equation of continuity)) and ρgh0 = 0 (because I choose h0 = 0).

    What I am not sure of is that I assume P0 = 0 because the fluid is not being stopped at that point, so it should have no pressure. Is that correct? If so, here is where that takes me:

    ρgh1 = ½ρv02
    v0 = √(2gh1)

    We can then use the flux to determine the time it takes for a litre of fluid to cross the hopper:

    R = V/t = A0 ⋅ v0
    ⇒ t = V/(¼ ⋅ π ⋅ d02 ⋅ √(2⋅g⋅h1)) = 23.5 s

    b) I use the Bernoulli equation again at points (0) and (2).

    P0 + ρgh0 + ½ρv02 = P2 + ρgh2 + ½ρv22

    I assume P0 = 0 and P2 = 0 for the same reason as before (is that right?) and ρgh2 = 0 (because I now choose h2 = 0)

    ⇒ v2 = √(v02 + 2⋅g⋅h0)

    The equation of continuity says A0 ⋅ v0 = A1 ⋅ v1
    ⇒ A2 = A0 ⋅ v0/v2
    ⇒ r = 1.5 mm
    d2 = 3 mm

    My results are plausible, but I am unsure about how to deal with the pressure when there is a flow. Is there no pressure on the right side of the equation when there is a flow? Is that always the case?

    Thank you very much in advance for your answers.

    Attached Files:

  2. jcsd
  3. Mar 21, 2016 #2
    you didn't have to assume Po = 0 because it is the fact.
  4. Mar 21, 2016 #3
    Thank you very much for your answer. I'm still unclear about how is that a fact: does pressure only apply in a "closed" container? Anytime there is a hole or some sort of opening at the bottom of the container (or at a point we are interested in), there is never pressure because the water flows?

  5. Mar 21, 2016 #4
    In your case, the water flows but it is not at steady state. Static pressure only exists when liquid is steady. Assume that i haven't talked about this, only base on logic, how do you manage to solve this because Po = P1 + pGH1 => V2 = 0.
  6. Mar 22, 2016 #5


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    No, you should not assume P0 is 0 because it won't be!
    The state only changes relatively slowly (since d1>>d0). That means acceleration is negligible, so the usual equations can be used to find the velocity as a function of head height. Only then do we need to consider that head height, and therefore v, change over time.

    @JulienB, allow P0 to be nonzero and consider the flow in the narrow section.
  7. Mar 22, 2016 #6
    Oh, you win, i have made a huge mistake
  8. Mar 22, 2016 #7
    JulienB, i think that i have found the problem and here it is : P1 + pGH1 = P0 + 0,5pV^2 ( P1 = P0 = P air ) => pGH1 = 0,5pV^2 ( it still the same but P1 = P0 = P air)
    I forgot about air pressure
    I hope that it is right
  9. Mar 22, 2016 #8


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    No, it is not to do with air pressure. P0 > P1. Why do you think they should be equal?
  10. Mar 23, 2016 #9
    it always exists air pressure and i knew that Po = P1 + pGH but if i had really done it then the formula would have been like this: Po = Po + 0,5pV^2 then V = 0 , it is so confusing, huh.
  11. Mar 24, 2016 #10
    Hi everyone and thank you for your answers. It is a bit confusing though :)

    @haruspex I don't get why you're stating that P1 ≠ 0. If we consider gauge pressure (that is, ignoring air pressure), what other factors may cause a pressure at P1? Maybe I am wrong, but if so I would like to understand why.

    Thank you very much again.

  12. Mar 24, 2016 #11
    @haruspex What precisely confuses me, is that I have understood that for example in the case of a side hole on a recipient we should consider P = 0 at that exit. In th case of the present problem, the "hole" is located underneath the fluid. Is that why we should consider the pressure not being 0 for P0? It is kind of unlogical to me that the pressure of the water matters when it can move away, but if you say so I'm ready to believe :)
  13. Mar 24, 2016 #12


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    Consider the linear velocity of the water in the two sections of pipe. Are they the same? If not, which is greater, and what can account for the acceleration?
  14. Mar 24, 2016 #13
    @haruspex Well the equation of continuity (A1⋅v1 = A0⋅v0) states that the water flows faster by the smallest opening (point (0)). And I guess that that is happening because of a greater pressure at that point? Now the formulas I know for pressure are P = F/A or for an submerged object P = ρgh (probably irrelevant here). Is there another one I don't know, or am I missing something important? I can't say much about the force applying on the fluid at that point.
  15. Mar 24, 2016 #14


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    Exactly. So, as I suggested in post #5, apply the Bernoulli equation to the flow in the narrow section.
  16. Mar 24, 2016 #15
    As I see it, the pressure at point 2 is atmospheric and the pressure at the very top of the hopper is atmospheric. So, from Bernoulli, $$v_2=\sqrt{2g(h_1+h_2)}=\sqrt{2g(0.115+0.026)}=1.66m/s$$
    This means that the pressure immediately below point 0 (just below the convergence) is sub-atmospheric. The gauge pressure there would just be ##-\rho g h_2=-\rho g(0.026)##. It seems that there would be a jump change in pressure from just above point 0 to just below point 0, with the pressure above being higher than the pressure below. But, the question is, "how much higher?" It seems to me there is not enough information provided to solve this problem without knowing d2.

    Last edited: Mar 24, 2016
  17. Mar 24, 2016 #16
    @haruspex so I took P0 = -ρgh1 -a little out of despair, I must admit- and got that result:

    Attached Files:

  18. Mar 24, 2016 #17
    Oops the "2" in the last square root should obviously not be there, but the result was obtained without including it in the calculations. With that t I now get for next question d2 ≈ 2.5 mm.

    I'm very skeptical about my negative pressure though :/
    Last edited: Mar 24, 2016
  19. Mar 25, 2016 #18
    @Chestermiller Are you saying that the gauge pressure just below point (0) is -pgh2? I'm afraid I don't really understand why: I thought the pressure at point (0) would be depending on the height of the fluid above that point (h1) and fail to see the connection with h2.
  20. Mar 25, 2016 #19
    Here is a more fleshed out analysis of my discussion in post #15.

    Let z = h1= elevation of top surface of liquid in hopper
    Let z = 0+ = elevation of top of short diameter convergence
    Let z = 0- = elevation of bottom of short diameter convergence
    Let z = - h2 = exit elevation

    Bernoulli term at each location:

    ##z = h_1##: $$0 + \rho g h_1 + 0$$
    ##z=0^+##:$$p^++0+\frac{1}{2}\rho v_0^2$$
    ##z=0^-##:$$p^-+0+\frac{1}{2}\rho v_2^2$$
    ##z=-h_2##:$$0-\rho g h_2+\frac{1}{2}\rho v_2^2$$

    All the above Bernoulli terms must be equal to one another.

    In addition, the velocities in the two tubular sections must satisfy:

    These equations cannot be solved unless we know d2.
  21. Mar 25, 2016 #20
    See my post #19 for more details. If you can find something wrong with that, please point it out. Setting the Bernoulli terms at the lowest two elevations equal to one another clearly produces this result.
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