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Question regarding Pipe flow

  1. Oct 26, 2007 #1
    This question has been bugging for a while now. I have been trying to solve it, but not getting anywhere. Hopefully, you guys can be of some help. Here's the question:

    For laminar fully developed pipe flow, show the average velocity V = 0.5u(max).

    Any help will be greatly appreciated! Thanks!
     
  2. jcsd
  3. Oct 26, 2007 #2

    FredGarvin

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    What does the velocity profile for fully developed laminar flow look like?
     
  4. Oct 28, 2007 #3
    laminar flow in a pipe will get its maximum velocity in the very center of the pipe. the profile is a parabola. U function of the radial position U=U(r); Radius = R; U(0)=Umax and U(R)=0

    u(r)=Umax*(1-r**2/R**2)

    Umoy is indeed equal two the half of Umax.

    do you need more info about how to get there?
     
  5. Oct 29, 2007 #4
    Here it is!!

    We wand to find average velocity. Now lets start from basics:- Average velocity means average velocity of all the particles in a unit cross-section of the tube. So consider a ring of radius r having a differential are equal to 2*Pi*r*dr. So multiplying this by velocity gives the sum of velocity of all the particles in that differential tube. So if we integrate from 0 to R, we get total velocity of all the particles. and hence dividing the expression by area gives us the average velocity which comes out top be equalt to 0.5 times the Vmax!!!!

    Regards,
    Vishal
     
  6. Feb 13, 2011 #5
    Hello,

    Could you please tell me how to get to the equation for a pipe flow.

    u(r)=Umax*(1-r**2/R**2)

    Thank you so much in advance.



     
  7. Feb 15, 2011 #6
    you "get it" by solving the Navier-Stokes equations with the appropriate boundary conditions and simplifications (i.e. incompressible, v . ∇ρ ≈ 0 )
     
  8. Jun 22, 2012 #7

    LFS

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    This is a chicken/egg question: parabolic or navier-stokes?

    I "read/know" that the velocity profile for laminar flow is parabolic with v=vmax as maximum and v=0 on the edges let us say D=2d. Mathematically I immediately say: the equation of such a parabola is v=vmax•(1-x²/d²) for xε[-d,d]. (This is just the equation of a parabola with max=vmax and roots x=-d and x=d.) No Navier-Stokes required.

    If I have a circular pipe, I change x=r, d=R (radius of pipe) and use polar integration to get
    Q=vmax•∫∫(1-r²/R²)rdrdθ on intervals rε[0,R] and θε[0,2π] to get Q=(vmax/2)•πR².

    Then vavg=Q/A=(vmax/2)•πR²/πR²=vmax/2 (and derive the formula asked at beginning).

    So - if I know laminar flow has a parabolic velocity profile, why use NS to find the equation with vmax? I can see using NS to find vmax, but not to get the equation with vmax in it.

    BTW: Anybody know of a link with (solved) laminar flow problems for rectangular ducts with a given vmax? Thanks!
     
  9. Jun 22, 2012 #8
    Integrate the velocity profile then divide by the length of the integration.
     
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