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Pressure drop in laminar flow

  1. Feb 3, 2016 #1
    1. The problem statement, all variables and given/known data
    as we all know , D=4R , D=hydraulic diameter , in the equation of P= (8μLV_avg) / (R^2) , i would get (128μLV_avg)/ (D^2) , am i right ? why the author gave (32μLV_avg)/ (D^2) ?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Feb 3, 2016 #2

    SteamKing

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    It depends on the shape of the duct or tube you are using.

    The hydraulic diameter is defined as DH = 4A/P, where A is the cross sectional area of the duct or pipe and P is the wetted perimeter..

    For a circular pipe, DH = D (internal) of the pipe, or ##D_H = \frac{4 ⋅ π ⋅ \frac{D^2}{4}}{π ⋅ D} = D##

    This article shows the hydraulic diameters for ducts of different shapes:

    https://en.wikipedia.org/wiki/Hydraulic_diameter
     
  4. Feb 3, 2016 #3
    yes , i know that . the object given is a long circular tube , so it is D=4R , right ? the notes is wrong ?
     
  5. Feb 3, 2016 #4

    haruspex

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    No, it's 2R as usual.
     
  6. Feb 3, 2016 #5
    why the note gave 4R ?
     

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  7. Feb 3, 2016 #6

    SteamKing

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    They used the wrong formula for calculating hydraulic diameter to start with

    ##D_H = \frac{4A}{P}##

    The rest is shoddy proof-reading and failing to catch the error.
     
  8. Feb 3, 2016 #7
    Some authors define the hydraulic diameter as A/P. However, I have never liked this, and I always used 4A/P. But, if you are reading a text, you need to make sure which definition they are using.
     
  9. Feb 3, 2016 #8

    haruspex

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    I believe you are confusing radius with hydraulic radius.
    The link from your original post mentioned R and D, with the implied relationship D=2R.
    Your link in post #5 does not mention R, but Rh. This is the hydraulic radius, which is meaningful in any cross section, and does not equate to the actual radius in the case of a cylindrical pipe. Indeed, the way it is defined at https://en.m.wikipedia.org/wiki/Manning_formula would make it R/2.

    Edit: to clarify, Rh=R/2 for a filled cylindrical pipe, and for a half-filled horizontal pipe.
     
    Last edited: Feb 4, 2016
  10. Feb 3, 2016 #9
    Oops. In my last post, I meant to say hydraulic radius, not hydraulic diameter. Thanks for catching this.
     
  11. Feb 4, 2016 #10
    so , D=4Rh or D=2Rh ?
     
  12. Feb 4, 2016 #11

    haruspex

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    D is the pipe diameter and is therefore twice the radius, D=2R.
    Rh is defined as the cross-sectional area of flow, A, divided by the wetted perimeter, P. For a filled cylindrical pipe radius R, A=πR2, P=2πR, so Rh=R/2=D/4.
    A half-filled horizontal cylindrical pipe has half the wetted perimeter and half the flow cross-section, so the same Rh.
    In between the two, Rh reaches a maximum somewhere.
     
  13. Feb 4, 2016 #12
    i have an example from my books ,

    it stated that Dh= 4A / P , why not Dh= 2A / P ? since 2 radius = diameter ?
     
  14. Feb 4, 2016 #13

    haruspex

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    I don't know why hydraulic diameter and hydraulic radius are defined in such ways that the ratio is 4:1 instead of 2:1. My guess is that it is an accident of history and the two definitions (in terms of A and P) were made independently.
     
  15. Feb 4, 2016 #14
    so , just follow D_h = 4A/ P ? R_h = A/ P ?
     
  16. Feb 4, 2016 #15

    haruspex

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    Yes. I suspect that you will not usually find both Rh and Dh used in the same text/problem. Authors will tend to adhere to one or the other. I could be wrong, though.
     
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