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I Laminar and steady flow(Distinguish)

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  1. Apr 16, 2017 #1
    While deriving Poiseuille's law (the relation between flow rate and pressure gradient for fluid flow in a rigid cylindrical tube under a pressure gradient) , we make an assumption that flow is both laminar and steady.Why we need the flow to be laminar. Is it not enough to consider only steady flow?
    Is there any connection between these two terms? Is steady flow always laminar ?

    I am confused with these terms.
     
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  3. Apr 16, 2017 #2

    anorlunda

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    :welcome:

    The image shows the difference between laminar and turbulent flows. Both can be consiered steady in terms of volume of fluid per second. But laminar is obviously easier to analyze.

    laminar_turbulent_flow.gif
     
  4. Apr 16, 2017 #3

    boneh3ad

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    If you start from the beginning of the derivation of steady, incompressible flow in a pipe without assuming anything about laminar versus turbulent flow, you can come up with the relationship
    [tex]\tau_{rz} = \dfrac{r}{2}\left(\dfrac{\partial p}{\partial z}\right).[/tex]
    In other words, the pressure gradient in pipe flow is exactly balanced by the shear stress so that you get a steady flow. If you assume the flow is truly steady and therefore laminar, then you just use the relationship for shear stress in a Newtonian fluid to come up with Poiseuille's law:
    [tex]\tau_{rz} = \dfrac{r}{2}\left(\dfrac{\partial p}{\partial z}\right) = \mu \dfrac{\partial v_z}{\partial r}.[/tex]

    Now, you have to be a little careful here when discussing steady flows. All real fluid flows have some small fluctuating component. One common way to treat this is to recognize that you can break all of your velocity components into a time averaged and a fluctuating component (a process called Reynolds averaging). For example, the ##v_z## component would like like this,
    [tex]v_z(r,\theta,z,t) = \bar{v_z}(r,\theta,z) + v_z^{\prime}(r,\theta,z,t).[/tex]
    This works as long as the time average of ##v_z##, or ##\bar{v_z}##, is steady over some length of time that you care about. Then, even though the velocities are technically fluctuating, their mean quantities are still steady and you can treat the flow as steady for all intents and purposes. However, by performing this Reynolds averaging, you've also modified your basic Navier-Stokes equations. For a Newtonian fluid, the shear stress tensor now has both viscous and turbulent components and looks something more like this
    [tex]\tau_{rz} = \mu\dfrac{\partial \bar{v_z}}{\partial r} - \rho\overline{v_z^{\prime}v_r^{\prime}}.[/tex]
    So basically you still have your Newtonian viscous component in the total shear stress, but you now have this new inertial term that shows up that represents a time average of a pair of velocity fluctuation terms. These are typically called Reynolds stresses. This means the solution to the differential equation we started with is now different and is no longer parabolic. It is no longer Poiseuille's law.

    Now, I mentioned that no real flow, even a laminar one, is truly without fluctuations. It just turns out that, for a typical laminar flow, the fluctuations are so tiny that the Reynolds stress term is effectively zero and the steady assumption works without any additional stress term. For a turbulent flow, with its characteristic large fluctuations, this is no longer true.
     
    Last edited: Apr 17, 2017
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