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Velocity as a function of area in pipe flow

  1. Apr 7, 2016 #1
    There is an equation which perplexes me and it is about calculation of average velocity in a pipe but over area.


    In the image, the velocity is already function of diameter, i.e, u=u(r) so how can we think velocity as a function of area?

    Source: Fluid Mechanics, Fundamentals and Applications by Çengel/Cimbala

    Thank you.
     

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    Last edited: Apr 7, 2016
  2. jcsd
  3. Apr 7, 2016 #2

    boneh3ad

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    You've sort of answered your own question. It's the difference between the velocity at a point, which is a function of ##r##, and the average velocity across the cross section, which averages that ##r##-dependence away and depends on ##A##.
     
  4. Apr 7, 2016 #3
    u(r) is the local axial velocity at radial coordinate r. The axial velocity is maximum at the center of the tube u(0), and is zero at the wall, u(R) = 0. If you want to get the average axial velocity, you calculate the volumetric flow rate and divide by the total cross sectional area of the tube.
     
  5. Apr 7, 2016 #4
    Can thinking of u(Area) instead of U(D) be the same? Because Area is function of diamter as well. This looks like a differential equation problem and while integrating over area it becomes integration over diameter because dA=2*pi*r*dr

    Thank you.
     
  6. Apr 7, 2016 #5

    boneh3ad

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    In general, the velocity varies over the range of ##r##. Doing the integration so that it is ##u_{avg}(A)## loses all that information about the ##r## dependence. You could certainly cast it in terms of ##u_{avg}(D)## as long as you are just referring to ##D## as the inner diameter of the pipe and not some kind of surrogate for ##r##, but you couldn't get ##u(D)## just like you couldn't get ##u(A)##. It is less intuitive though, since using ##A## comes from conservation of mass.
     
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