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Bernoulli and darcy equation in split flow system

  1. Jun 5, 2010 #1
    Refering to the image uploaded can we say by applying bernouli equation and darcy equation the following :

    Pa/(guamma) + Va^2/2.g + Za - (head loss in pipe ab using darcy equation )-(head lossin pipe bc using darcy equation) = Pc/(guamma) + Vc^2/2.g + zc . ??

    and the same thing between point a and point d ?

    if we can do so , thn the head loss in pipe ab must be diferent thn the head loss in pipe bd with a value that is equal to the difference between the dinamic velocities energy in c and d ? am I right ???

    why we used to say that head loss are both equal in pipe bc and bd ???

    P.S : pc and pd are at 0 atmospheric pressure .
     

    Attached Files:

  2. jcsd
  3. Jun 5, 2010 #2
    Here is what I am completely sure of

    bernouli is not applicable in pipes because the flow is not potential. There is mean vorticity. The head loss in pipes comes from the friction on walls (what you call darcy law).

    Here is what I am not really sure of, I have to think more

    if we assume that the mean flow is steady, then in the vertical part of bd the friction is balanced by the gravity (no acceleration), and there is no loss. Then the horizontal part of bd is of the same length as bc, so I would expect no difference in the head loss.
     
  4. Jun 5, 2010 #3
    well zeebek thnks for ur response but that did not answer my question , I want to know if the equation I wrote is true or not .
     
  5. Jun 5, 2010 #4
    as I said bernoulli part is defintely wrong, it just does not work for such flow. For the rest - I will consult with my profs on monday. In principle I can measure what you are asking :)
     
  6. Jun 5, 2010 #5
    man bernouli equation is an essential law in fluid mechanics , how can u say it dosent work in pipes !
     
  7. Jun 5, 2010 #6
    Please, open the book and read under which assumption bernoulli is derived from the Navier-Stokes and where it is applicable. Bernoulli works for potentials flows only. It means no viscosity and vorticity. In pipes both are present.
     
  8. Jun 5, 2010 #7
  9. Jun 5, 2010 #8
    you are being theoretical , In real life that how engineers do their calculation , they assume incompressible flow ...
     
  10. Jun 5, 2010 #9

    Q_Goest

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    See equation #15 and #16 in the "Pipe-Flo Pro.PDF" file attached to the second post in this thread:
    https://www.physicsforums.com/showthread.php?t=234887
    The Darcy-Weisbach equation is added in to show frictional losses. This accounts for permenant pressure drop in a pipe.
     
  11. Jun 5, 2010 #10
    what about in split flow ? , can we have darcy-weisbach and bernoulli equation in abc ?
    can we consider abc as 1 streamlime that we can apply between any 2 points in this "abc streamline" the darcy-weisbach and bernoulli equation ??? .
    my question is very clear ^ the pdf u gave me dont answer my question which is about split flow .
     
  12. Jun 5, 2010 #11

    Q_Goest

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    The equations apply equally to both branches. Problem is you need to determine what is a known on your 2 branches. You may know flow rate, or you may know pressure at a point. What do you know about your system?

    For example: Is c and d at the same pressure? If so, the pressure at b is the same for both, and you can use the equation shown in that text to analyze flow rate. Typically, this is an iterative calculation that must be done over and over till you equalize the pressure differential in the two legs, and so that the sum of the two flows (bc and bd) add up to the flow rate ad.

    Computer programs are often made to solve these kinds of iterative problems.
     
  13. Jun 5, 2010 #12

    stewartcs

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    Uh, Bernoulli most certainly can be applied for the flow of fluid in pipes, one just needs to allow for the friction with a separate term, Darcy is the most common...

    CS
     
  14. Jun 5, 2010 #13

    stewartcs

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    When you have a fluid flow through a single pipe that then branches into two and then rejoins into one, the head loss in each branch must be the same since a fluid element will end up with the same energy level regardless of the branch that it takes.

    The key point here is to note that the fluid elements (through each branch after it splits) will recombine into a single pipe so regardless of the path they take they must have lost the same amount of energy since they both end up at the same energy level at the point where they recombine.

    CS
     
  15. Jun 6, 2010 #14
    If I have the height from the top of the tank in the figure I uploaded , the length of each section of the pipes the friction factor in each pipe the diameter of each pipe section cant we use bernouli eauation through abc and abd and calculate the pressure between a and c and and and d ?
     
  16. Jun 6, 2010 #15

    stewartcs

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    If you assume the energy level in each flow is the same at the exit (i.e. they come together to a single node again) then it's a simple solution and can be found in any fluid mechanics text and most likely on the web somewhere. In this case, the head loss in each branch will be the same (for the reasons I explained previously) and that key piece of information will allow you to solve the equations.

    Since there are only two pipes, you can solve the equations simultaneously instead of iteratively. If you have 3 or more branches, it will require iterations.

    CS
     
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