Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fully developed flow

  1. Oct 7, 2005 #1
    Hi all!
    What actually does a "fully developed flow" mean?
    Is it equivalent to saying the the velocity is uniform along the directon of flow?
     
  2. jcsd
  3. Oct 7, 2005 #2
    Not really (at least IMO).
    Rather, with "fully developed flow" you should think that viscous effects have spread throughout the fluid in the pipe, i.e, that the "inviscid core" has disappeared.

    Think of a jet of water entering a pipe.
    Close to the entrance region (on the inside of the pipe), significant viscous effects will be concentrated to a thin boundary layer attached to the pipe wall. The fluid in the middle is basically inviscid.
    As you progress further into the pipe, these boundary layers will increase in thickness until you reach a point where they merge, so that the whole fluid is significantly affected by viscosity. That is when "fully developed flow" has its onset.
     
  4. Oct 7, 2005 #3
    why does it turn into an inviscid fluid further long the pipe?
    does the no-slip condition increase with length of pipe?
     
  5. Oct 7, 2005 #4

    FredGarvin

    User Avatar
    Science Advisor

    It doesn't turn into an inviscid flow further down the pipe, like Arlindo mentioned. The boundary layer increases in size as you make your way down the length of the pipe and since the boundary layer is where viscous forces are dominant, viscous forces take over.

    Here's a graphic:
    http://www.coolingzone.com/Guest/News/NL_JAN_2001/Biber/channelFlow.jpg
     
    Last edited: Oct 7, 2005
  6. Oct 7, 2005 #5
    Hmm..where did I mention that the fluid turns inviscid? :confused:
     
  7. Oct 7, 2005 #6

    Astronuc

    User Avatar

    Staff: Mentor

    I think this comment is misundertood.

    In a wide open area, the fluid is all moving at the same velocity (speed) locally. One it enters a pipe, the fluid in contact with the pipe basically stops while the fluid away from the pipe surface continues to move. It takes some distance from the entrance of the pipe for the fluid in the center to begin to experience the shear forces from the fluid toward the wall.

    I suppose one would also wish to discuss laminar vs turbulent flow.
     
  8. Oct 7, 2005 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yes, I may have phrased myself rather unfortunately there..
     
  9. Oct 7, 2005 #8

    Astronuc

    User Avatar

    Staff: Mentor

    I guess it was your other personality - welcome back, arildno. :biggrin:
     
  10. Oct 7, 2005 #9

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    Some little collaboration here:

    In the example of Arildno about a jet of water entering in a pipe, there is a characteristic length of flow developing until the viscosity force has slown down the flow velocity in a value of the same order of the entrance velocity. We call such length the Hydrodynamic Entrance Length [tex]L_e[/tex], and it can be roughly estimated imposing the balance of the convective term and the viscous term:

    [tex]
    u\frac{\partial u}{\partial x}\sim \nu \frac{\partial^2 u}{\partial r^2}
    [/tex]

    if the pipe diameter is [tex]2a[/tex], then one can estimate these terms as:

    [tex] \frac{U^2}{L_e}\sim \nu \frac{U}{a^2}[/tex]

    So that [tex]L_e\sim a Re[/tex] where [tex] Re[/tex] is the Reynolds #. This scalement is used many times in fluid mechanics. Also, you can check that there is a Thermal Entrance Length, which caracterizes a fully developed flow regarding the temperature profile. This last length will be scaled with Prandtl # also. As you may see, as the Reynolds # increases, the needed lenght for having fully developed flow increases. That's because the flow has greater inertia and it takes more distance to brake it by means of viscous shear stress.
     
  11. Oct 8, 2005 #10

    FredGarvin

    User Avatar
    Science Advisor

    You didn't Arlindo, Shawnzyoo did in the question right after your post. I thought your post was right on.

    I should use quotes more in my posts. Sorry about the confusion.
     
  12. Oct 8, 2005 #11

    FredGarvin

    User Avatar
    Science Advisor

    Well said. It brings back memories of my senior fluids labs in school. I was trying to find on line, the characteristic pressure drops along the length of the pipe one sees in this situation for both the entrance loss and the inlet length drop. It also gives a very good indication of where fully developed flow begins. I'm still looking.

    Welcome back Clausius.
     
  13. Oct 8, 2005 #12
    hi all! Thanks for all replies.

    1. Why boundary layer will increase when the flow go along the pipe?

    2. When the viscous force take the dominance, what will happen?

    3. Actually I don't really catch the meanings and can't visualize what "inertia effect" and "viscous effects" are. Could anyone help explain in plain terms?

    3. I just read a book which says "the flow is fully developed, i.e. the velocity profile is constant along the direction on flow" is it true? And that's a consequence of the dominance of the viscous effect?

    4. I just read a question in which the following assumptions are made:
    "The open channle is wide and long such that the flow is fully developed"
    And then the solution says by this assumption, the velocity compoent v =0 and also w=o. Therefore this is just a 1-D flow.
    Can anyone explain me why?
    For my understanding, I know that since the channel is wide so the velocity along the z-direction =0. However, why the component of velocity perpendicular to the flow direction also equals zero? Is it because the flow is fully developed? But from the book's definition, fully developed just mean the velcoty profile is constant and this implies the velocity component v =0?
     
  14. Oct 8, 2005 #13

    Astronuc

    User Avatar

    Staff: Mentor

    As the fluid travels into the pipe, the effect of viscous forces are felt further into the fluid, from the outside, i.e. pipe/channel wall.

    In laminar flow, Re < ~2000, there will be a parabolic velocity profile. Turbulent flow has a different profile, but it is more or less similar the further downstream once the flow is 'fully developed'.

    Inertial referes to the kinetic energy and momentum of the flow, and the pressure parallel with the flow. The viscous effects refers to the friction and shear forces on the flow.

    For incompressible flow. It has to do with conservation of mass (continuity equation), not viscous effect.

    I'm not sure about v=0, w=0 unless that refers to y, z directions, and u refers to velocity in x. The fluid is contrained by the channel/pipe in one or 2 dimensions. Wide and long could be a river, for example, so there is a characteristic velocity profile which is a function of depth only (ignoring the channel profile and banks). But basically, for a very wide channel, the velocity profile is a function of depth only (hence 1-D), and the profile is much the same far downstream, assuming the no change in the elevation or the channel geometry, i.e. uniform depth and channel profile, across and along the channel.
     
  15. Oct 9, 2005 #14
    Um...I am really such a dummy that I cant fully understand you guys words.

    Can I say that a flow is fully developed when the velocity field does not change in the flow direction, so the velocity vector is independent of the coordinate along that direction?

    Actually all my questions came from a questoin as follow:
    Consider there is a flow of water down on an inclinded wedge.
    The wedge is wide, long and the flow is fully developed. Defining the x-axis along the inclinded surface of the wedge and also y, z axis a shown in the figure, and u being the velocity along x-direction, v for y-dir and w for z-dir. After assuming the wedge is wide, long and fully developed, the solution says that the velocity field is simply u=u(y).

    I can't exactly understand how the relation is obtained from the assumptions"long, wide and fuly developed".

    Below is my thought:
    In general, Velocity field = u(x,y,z,t)i+v(x,y,x,t)j+w(x,y,z,t)k
    By the physical boundaries of the wedge and just some intuitive feelings, the component v and w are discarded, remaining
    velocity field =u(x,y,z,t)i
    Long wedge=?Fully developed => discard "t"?
    velocity field =u(x,y,z)
    Wide wedge, so end effects along z-direction is insignificant, so velocity should be independent of "z"...
    velcoty field=u(x,y)
    Since long wedge, so end effects along x-direction is insignificant, so velocuty should be independent of "x" also?
    finally velocity field = u(y) ??!!

    I am sure that my arguments are invalid somewhere....
    Can you guys kindly correct me and tell me how to deal with the assumptions?
     
  16. Oct 9, 2005 #15
    The diagram:
     

    Attached Files:

    • h.JPG
      h.JPG
      File size:
      3 KB
      Views:
      295
  17. Oct 11, 2005 #16

    Clausius2

    User Avatar
    Science Advisor
    Gold Member

    OOPs Fred, sorry for not having too much time for thinking about pressure. Next time I'll try it. Now this homeworks are driving me crazy....

    Hanson, IS ACTING THE GRAVITY IN YOUR PROBLEM???

    In open channel flow, with an SLIGHTLY inclined channel, one says:

    i) Steady flow: none variable is dependant on time

    ii) Wide of the channel large enough to neglect z component variations.

    iii) Height of water small enough to neglect y component velocity. This can be demonstrated by means of the continuity equation: [tex]\partial u\partial x\sim \partial v\partial y[/tex], from where one can estimate [tex]V/U\sim O(H/L)[/tex] being H and L the characteristic lengths in y an x directions.

    iv)Length of the channel long enough to enhance the fully developed flow. Just in the beginning of the channel, the convective terms are predominant (i.e. the profile is almost plane except in a boundary layer of negligible thickness). As you go down the channel, the boundary layer has """ideally""" growth to thickness of the same order of H (don't forget we are assuming laminar flow everywhere).
     
  18. Mar 15, 2007 #17
    fully developed turbulent flow over plate

    Is it possible to have a “fully developed turbulent flow” over plate?

    I am interested is incompressible subsonic flow and I know that the fully developed turbulent flow, i.e., constant velocity profile can happen inside pipes or channels or between two parallel plates.

    Does the boundary layer of turbulent flow over plate reach a constant value?
    If so how are boundary layer displacement and boundary layer displacement thickness obtained?

    Thanks
     
  19. Mar 15, 2007 #18

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    What "fully developed" means here is:

    If you take two sections through the wedge at different points in the flow, the flow profile is the same at each section. Also if you look at the flow at one pont at different times, it will be the same.

    Most of the previous posts are about WHY the flow gets to be fully developed (or not) rather than what the consequences are.

    From the second part of my definition, the flow is independent of t.

    From the first definition, it follows that the v and w components must be zero everywhere. If there was a w velocity component to the left (say) at one point down the wedge, then "fully developed" means there would be the same velocity component left at EVERY point down the wedge, so all the water will finish up on one side, which is nonsense. The same applies to the v component.

    So the velocity components v and w are 0. and the velocity does not depend on t or on x.

    In other words, the general expression for the velocity is u(y,z).
     
  20. Mar 15, 2007 #19

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    No it doesn't reach a constant thickness. Look up the standard solution for flow over an infinite flat plate.

    At least, thats what the standard theory says. It's usually safe to bet that CFD theories don't fit ALL the experimental facts, though.
     
  21. Mar 15, 2007 #20
    I still don't know if it is possible to have a “fully developed turbulent flow” over plate?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fully developed flow
Loading...