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Bernoulli's Equation Applied to Two Free Surfaces

  1. Sep 12, 2012 #1
    Consider two cylindrical containers of the same diameter. Fill them with a liquid to different levels. We all know that the liquid can be "siphoned" from one container to the other using a thin hose, which we will consider to be thin and of uniform diameter.

    Bernoulli's Equation can be used to derive the flow rate (which is essentially the same as Torricelli's Theorem).

    However, consider applying Bernoulli's Equation to the free surface of each container.

    P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2

    Because the cylinders have the same diameter, v1 = v2.

    Because a free surface is at atmospheric pressure, P1 = P2.

    Bernoulli's Equation then reduces to h1 = h2, which makes no sense.

    Does anyone have any insight to why Bernoulli's Equation does not work in this case?
  2. jcsd
  3. Sep 12, 2012 #2
    It would help if you drew a diagram and identified your symbols.

    The three points of interest in the flow are

    1) The free surface in the first vessel. Here the velocity is zero. Call it A

    2) The knuckle of the connecting pipe. Call it B This can be calculated from Bernoulli.

    3) The discharge point into the second vessel. Call it C The velocity here is the same as at B
  4. Sep 12, 2012 #3
    I understand that, but my question is why Bernoulli's Equation give nonsensical results when applied to the liquid surface on each side.
  5. Sep 12, 2012 #4

    What is v1 and v2? Or rather where are they taken?
  6. Sep 12, 2012 #5
    In the diagram attached, Bernoulli's Equation applied to points 1 and 2 gives h = 0, which makes no sense.

    Attached Files:

    • tmp2.png
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  7. Sep 12, 2012 #6
    That is not a syphon, that is two vessels connected by a pipe at the bottom.

    Look at my diagram and think about the value of velocity at A

    Attached Files:

  8. Sep 12, 2012 #7
    I'm familiar with your analysis. My question is why does applying Bernoulli's Equation at points 1 and 2 in my diagram give h = 0. Applying Bernoulli's Equation in this way yields the same result whether it's a siphon, or if the vessels are connected by a pipe at the bottom.
  9. Sep 12, 2012 #8
    Why do you think your v1 = your v2 ?
  10. Sep 12, 2012 #9
    Because the cylinder's have the same cross sectional area, as I said in the original post.
  11. Sep 12, 2012 #10
    I know you said that but you are not correctly applying Bernoulli's equation.

    You will understand much more quickly if you stop accusing the equation of being wrong and look at your own work.

    At the start of the process (vessel 1 full, vessel 2 empty) v1 = 0, v2 can be calculated by Bernoulli and h1 is not equal to h2.

    Since the vessels are of limited capacity the velocity changes through the discharge.

    At the end of the process v1 = v2 = 0 and h1 = h2 as you have already observed.
  12. Sep 12, 2012 #11
    OK. Why can't we start the process with initial conditions as they are in my diagram, and why does the application of Bernoulli's Equation to points 1 and 2 yield the non-sensical result that h = 0 in that case? I'm not accusing Bernoulli's Equation of being wrong, I'm simply trying to understand its range of application.
  13. Sep 12, 2012 #12
    Of course you can, but you don't have the information to fill in the Bernoulli equation part of the way through the process so you cannot solve it.

    What you have written down is not correct for your diagram.
  14. Sep 12, 2012 #13
    OK, let's work through this:

    v1 = v2 because the cylinders are identical.

    P1 = P2 because both points are at the atmosphere.

    So we get h = 0.

    What did I do wrong?
  15. Sep 12, 2012 #14
    You have shown a picture.

    Is this your initial state, ie is there initially some water in each vessel and a tap in your pipe?

    So do you start the process by opening the tap?
  16. Sep 12, 2012 #15

    Sure. The initial state is as shown, and the process starts by opening a tap.
  17. Sep 12, 2012 #16
    So at the instant of opening the tap, can you describe the situation, that is put values into Bernoulli's equation?
  18. Sep 12, 2012 #17
    It doesn't matter how the cylinders are connected or whether or not there are valves. We can start with one empty and one full, and then apply Bernoulli's equation at the time the situation is as depicted in my diagram.
  19. Sep 12, 2012 #18
    If you know the answers, why are you asking?

    Of course it matters or I wouldn't have introduced a valve.

    At the instant of opening the valve the velocity at both ends of the tube is zero.

    As a result, Bernoulli will give you the result that the difference in fluid levels equals the static pressure difference between the fluids in each vessel.

    You have three variables to insert into each side of Bernoulli. You can reduce this to two (or even one) if you can make one (or two) of the height, static pressure or velocity equal zero, or some other known value such as atmospheric pressure.

    The problem arises because as soon as flow starts the static pressure drops and you therefore don't know either.

    You keep repeating that the velocities are equal because the vessels are the same size. The velocity is quite independent of the vessel size.
  20. Sep 12, 2012 #19
    Points 1 and 2 are at the surface in each container. If the containers have the same diameter, then v1 = v2.
  21. Sep 12, 2012 #20
    To solve this you obtain an expression for v using Bernoulli.

    If h is the head difference between the higher free surface and the lower free surface, producing the flow then by Bernoulli

    h = v2/2g

    You can estimate the time to equalise levels by integrating the equation

    -Adh = Qdt.

    Where A is the area of the upper free surface and t is time

    You can obtain an expression for Q in terms of t from the velocity and connecting pipe diameter.

    To complete the task you may need to consider the reverse pressure build up in the second vessel, due to increasing head
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