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Bernoulli Equation Simple derivation help

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Please help me as I am quite confused in Bernoulli's theorem derivation...In my text book,it is considered that Fluid Moves from a Greater height h1 to lower height h2,The pressure on upper end is positive,while at the lower end,it is negative,i.e against the motion of fluid,I want to know why it is negative at the lower end of pipe


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 5, 2013 #2

    TSny

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    Hi kashan123999. Welcome to PF!

    The pressure is not negative at h2 and hopefully your text is not really claiming that.

    The derivation of the Bernoulli equation includes determining the work done by the pressure forces at each end of the section of fluid.

    It should be clear that when the section of fluid moves a bit, the work done on the section of fluid by the pressure force at height h1 is positive while the work done by the pressure force at h2 is negative. The work is negative at h2 because the force created by the positive pressure P2 acts opposite to the displacement Δx.

    So at h2, work = F2 Δx cos(180o) = -F2Δx = -(P2A2)Δx = -P2A2Δx.

    So, P2 is not negative, but there is a negative sign in the work done by P2.
     
  4. Apr 5, 2013 #3
    Thank you very very much sir but still,Please tell me why the work done by p2 is negative...I mean at P1,the work is done on fluid by the fluid moving behind it which is intuitively positive,But what happens at P2,how come p2 has opposite direction???
     
  5. Apr 5, 2013 #4
    Thank you very very much sir but still,Please tell me why the work done by p2 is negative...I mean at P1,the work is done on fluid by the fluid moving behind it which is intuitively positive,But what happens at P2,how come p2 has opposite direction than that of displacement of fluid x??? which pressure does p2 represent?
     
    Last edited: Apr 5, 2013
  6. Apr 5, 2013 #5

    TSny

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    In the figure, the purple shade shows the section of fluid that you are interested in. The light blue shows the fluid behind and ahead of this section. At the lower right end of the purple section, the fluid ahead of that section is pushing back to the left on the purple section with a force F2. Since the fluid is displacing to the right, the work done by F2 is negative.

    [Edit: Pressure doesn't have a direction. But force created by pressure does have a direction. There is only one pressure at h2 and it's positive. The force which the purple fluid exerts on the blue fluid at h2 has a magnitude of P2A2 and acts to the right on the blue fluid. At the same time, the blue fluid exerts a force on the purple fluid of the same magnitude but toward the left (as shown in the diagram). This force acting to the left on the purple fluid does negative work on the purple fluid.]
     

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    Last edited: Apr 5, 2013
  7. Apr 7, 2013 #6
    Thank you very much sir,Now I got your point...but if we take a section of purple fluid on the upper portion,and after this section lies the blue fluid again,we can assume that the blue fluid is again exerting the force on purple fluid to the left...so work on the upper section can also become negative,how plausible is this assumption?
     

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  8. Apr 7, 2013 #7

    TSny

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    Yes, the force acting to the left will do negative work on that purple section. At the same time the force (F1) acting to the right on that section will do positive work. As long as that purple section remains on the same horizontal level with constant cross sectional area A, the negative and positive works cancel.
     
  9. Jan 6, 2015 #8
    Brilliant response. I was having a lot of trouble with this derivation, but it's clear now that the external work done at the second end of the tube is negative because the fluid at the external end is pushing against the flow of the fluid. By the dot product, the directions of these two vectors are separated by 180 degrees, and the work is therefore negative. For me, the key to understanding this was comprehending that the work is done by the external fluid on the surface area of the flowing fluid... Thank you, sir.
     
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