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Fluid Mechanics - Equations of motion circular flows

  1. Jan 18, 2005 #1
    Departing from the Navier-Stokes equations for an incompressible flow (with [itex]\rho[/itex] and [itex]\mu=\rho \nu[/itex] constant):

    [tex] \frac{Du_i}{Dt}=\frac{\partial u_i}{\partial t}+u_j \frac{\partial u_i}{\partial x_j} = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} +\nu \frac{\partial^2 u_i}{\partial x_j^2}[/tex]

    my book says it follows for a circular flow

    [tex]\frac{\rho u_{\theta}^2}{r} = \frac{\partial p}{\partial r}[/tex]
    [tex]\frac{u_{\theta}}{\partial t} = \nu [\frac{\partial^2 u_{\theta}}{\partial r^2} + \frac{1}{r} \frac{\partial u_{\theta}}{\partial r}-\frac{u_{\theta}}{r^2}] [/tex]

    I can understand part of it. For a circular flow [itex]u_r=0[/tex] So the incompressibility yields that [itex] u_{\theta}[/itex] is not a function of theta. So the nonlinear term in the material derivative vanishes. But where does the last term [itex]-\frac{u_{theta}}{r^2} [/itex]
    come from? And the entire first equation?!
    Last edited by a moderator: Jan 18, 2005
  2. jcsd
  3. Jan 18, 2005 #2


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    How about
    [tex] a_{cp}=\frac{\omega^{2}}{r} [/tex]


    After all,it is a circular motion of a fluid and the NS equations are the Newtons equations for a newtonian fluid,right??

    As for the second,think about,make some calculations.Express that laplaceian in polar coordinates and find out where it comed from.i ain't gonna do those calculations for u.

  4. Jan 18, 2005 #3


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    1. By continuity, you must show that a 2-D circular motion for an incompressible fluid is only possible if the velocity field is independent of the angular coordinate;
    2.Write N-S as follows:

    Calculate the various terms this implies!
    3) Show that the pressure must be independent of angle, since the pressure must be a continuous function.

    "So the nonlinear term in the material derivative vanishes."
    This is TOTALLY WRONG!!
    You have curved motion.
    Last edited: Jan 18, 2005
  5. Jan 18, 2005 #4
    But [itex]u_r[/itex] is zero for a circular flow so [itex]\vec{u}=(u_r,u_\theta)=(0,u_\theta)[/itex]. The nonlinear term in the material derivative is [itex](\vec{u} \cdot \nabla )\vec{u}=u_\theta \frac{\partial u_\theta}{r\partial \theta}[/itex]. And the continuity equation says: [itex]\nabla \cdot \vec{u}= \frac{\partial u_\theta}{r \partial \theta}=0[/itex]. So the nonlinear term vanishes, right?
  6. Jan 18, 2005 #5


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    1. [tex]\vec{v}\cdot\nabla=\frac{u_{\theta}}{r}\frac{\partial}{\partial\theta}[/tex]
    This is the centripetal acceleration term.
    Last edited: Jan 18, 2005
  7. Jan 18, 2005 #6
    Aha that explains a lot. Those darm changing unit vectors... But anyway, thanks a lot. The origin of the first equation is clear now, it is the r-component of N-S. As for the second equation; using the expression for the laplacian:

    [tex]\nabla^{2 }=\frac{\partial^{2}}{\partial{r}^{2}}+\frac{1}{r} \frac{\partial}{\partial{r}}+\frac{1}{r^{2}}\frac{ \partial^{2}}{\partial\theta^{2}}[/tex]

    I get for the theta component:

    [tex]\frac{\partial^2 u_{\theta}}{\partial r^2} + \frac{1}{r} \frac{\partial u_{\theta}}{\partial r}+\frac{\partial ^2 u_{\theta}}{r^2 \partial \theta ^2}[/tex] Where the last term vanishes because [itex]u_{\theta}[/itex] is not a function of theta. So is the last term I mentioned in my first post a wrong?
  8. Jan 18, 2005 #7


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    Again, the darn changing unit vectors..:wink: :

    All that remains, is to show that the pressure must be independent of the angle.
  9. Jan 19, 2005 #8
    Ouch, thats painful :blushing: I hope I learned my lesson now... And about the independence of angle of the pressure. Isn't this obvious by symmetry? As the velocity profile does not depend on angle, there is no way to distinguish one direction form the other. The physical conditions are independent of the angle so the pressure must be as well?
  10. Jan 19, 2005 #9


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    It is very easy to prove that independence:
    Look at the component equation in the angular direction.
    Transfer the viscous term onto the other side; we gain therefore:
    where G is some function solely of r and t, since the velocity is so (proven through continuity equation).

    But this means, that the pressure must be, through integration:

    But, the polar coordinate points [tex](r,0)[/tex] and [tex](r,2\pi)[/tex] is the SAME point; hence, the pressure function must prescribe the same pressure value there.
    (Alternatively, if you regard the angle to lie in the half-open interval [tex][0,2\pi)[/tex], the limiting value of the pressure when the angle tends towards [tex]2\pi[/tex] must be p(r,0,t); i.e, continuity of pressure)

    We therefore must have:
    Hence, p=K(r,t)=p(r,t), that is, independent of the angle.
    Last edited: Jan 19, 2005
  11. Jan 19, 2005 #10
    nice, very elegant. Thank you very much, you've been very helpful.
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