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Minor loss in pipe formula

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    what is the meaning of number of diameter N ? why the author make L = ND ? and then divide it by D ?

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 25, 2016 #2

    BvU

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    This is a practical way to do it: you get an equivalent length for fittings, elbows etc. that you can add to the sum of lengths of straight sections and use in friction factor formulas (e.g. Darcy).

    Head loss is a function of ##L\over D## .

    Didactically the sheet you show is indeed rather ready for improvement. I find it confusing.
     
  4. Apr 25, 2016 #3
    can you explain about what is ND / D ? i am confused
     
  5. Apr 25, 2016 #4

    BvU

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    ND is L so ND / D is L/D. That is the factor that appears in the friction factor equations such as Darcy and Fanning (*). The approach exploits the observed similarity in flow properties between a flow in a pipe of 100 m and 1 m diameter and a flow in a pipe of 10 m with a diameter of 10 cm.

    (*)
    And I would almost wish one of the two never existed :smile:. Now you have to be really careful if you divide 16 or 64 by Re for laminar flow.....
     
  6. Apr 25, 2016 #5
    why L = ND ? i dont understand it
     
  7. Apr 25, 2016 #6

    BvU

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    Length of the pipe expressed in number of diameters. Nicely dimensionless. What can I say ?
     
  8. Apr 25, 2016 #7
    what does it mean by number of diameter ?
     
  9. Apr 25, 2016 #8

    BvU

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    The two pipes in #4 have the same ##{L\over D} = 10## so they will show the same pressure drop for a given fluid with widely different volume flows (factor 100) but the same flow velocity.
     
  10. Apr 25, 2016 #9
    ys , they have L/ D of factor 100 ,why they will have the same pressure drop ?
     
  11. Apr 25, 2016 #10

    BvU

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    Allright, L/D = 100 o:) .
    That's what has been observed to be the case :smile: .
    'Apparently' ##\Delta p## is a function of L/D, something that probably also comes out of similarity considerations.

    Something with ##{\rm Re} = {\rho v D\over \mu}##
     
  12. Apr 25, 2016 #11
    ok , how does the case that you mentioned relate yo number of diameter ?
     
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