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Homework Help: Flow velocity

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data
    hello everybody,

    I am trying to obtain the flow velocity in two dimensions [tex] u(x,y) [/tex] for the case of a flow past a circle. The equations to solve are:

    [tex] \vec{\nabla} p = \mu \vec{\nabla}^2 u [/tex]

    [tex] \vec{\nabla} \cdot \vec{u} = 0 [/tex]

    I am very blocked at the moment and I don't become any idea at which can I start to calculate [tex] u(x,y) [/tex]. I want to solve it, but could some one give some sugerence to start to solve alone.

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 19, 2009 #2
    Do you have access to the book "Transport Phenomena" by Bird, Stewart, and Lightfoot?
  4. Aug 19, 2009 #3
    First list you assumptions/postulates. This will allow you to stary simplfying the equations.
  5. Aug 19, 2009 #4
    you can write down the stream function first since the flow passes a known circle; then, it wont be difficult.
  6. Aug 21, 2009 #5
    this reminds me of the heatring problem that fourier studied
  7. Aug 24, 2009 #6
    My attempt following the case of 3 dimensions that I found in the book Fluid Mechanics by Kundu.

    First of all, I have to obtain the stream function, for that we now

    [tex] \vec{u} = \nabla \times \phi [/tex]

    in cartesians coordinates:
    [tex] u_x = \frac{\partial \phi}{\partial y}[/tex]
    [tex] u_y = -\frac{\partial \phi}{\partial x}[/tex]

    The Navier - Stokes equation can be written as a function of the vorticity:
    [tex] \vec{\nabla} p = \mu \vec{\nabla} \times (\vec{\nabla} \times \vec{u}) = \mu \vec{\nabla} \times \vec{w} [/tex]

    with [tex]\vec{w} [/tex] the vorticity:

    [tex]\vec{w} = -\frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} = \vec{\nabla}^2 \phi [/tex]

    now I insert this result in the momentum equation and

    [tex] \vec{\nabla} \times \vec{w} = \frac{\partial}{\partial y} \left( \frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2}\right) e_x - \frac{\partial}{\partial x} \left( \frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2}\right) e_y = \left( \frac{\partial^3 \phi}{\partial x^2 \partial y}+ \frac{\partial^3 \phi}{\partial y^3}\right) e_x - \left( \frac{\partial^3 \phi}{\partial x^3}+ \frac{\partial^3 \phi}{\partial y^2 \partial x}\right) e_y[/tex]

    [tex] \vec{\nabla} p = \mu \vec{\nabla} \times \vec{w} = \mu \left[ \left( \frac{\partial^3 \phi}{\partial x^2 \partial y}+ \frac{\partial^3 \phi}{\partial y^3}\right) e_x - \left( \frac{\partial^3 \phi}{\partial x^3}+ \frac{\partial^3 \phi}{\partial y^2 \partial x}\right) e_y \right] [/tex]

    Equating terms for [tex]e_x[/tex] and [tex]e_y[/tex]:
    [tex]\frac{\partial p}{\partial x} = \mu \left( \frac{\partial^3 \phi}{\partial x^2 \partial y}+ \frac{\partial^3 \phi}{\partial y^3}\right) [/tex]

    [tex]\frac{\partial p}{\partial y} = \mu \left( \frac{\partial^3 \phi}{\partial x^3}+ \frac{\partial^3 \phi}{\partial y^2 \partial x}\right) [/tex]

    How can I resolve this two equations?

    The next step is to fix the boundary conditions. In the circle surface it should be no slip [tex] \frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial y} =0 [/tex]

    and far of the circle is the boundary condition like this [tex]u_x = u_0 x[/tex] and [tex]u_y = u_0 y[/tex]
  8. Aug 24, 2009 #7


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    Homework Helper
    Gold Member

    You haven't stated whether you are trying to model a compressible or an incompressible flow yet (or any other assumptions you are making about the fluid flow!). For the latter case, what is [itex]\mathbf{\nabla}\times\textbf{u}[/itex]?
  9. Aug 24, 2009 #8
    hi, rb

    from the second, third equations and the last two equations in your last line, exact stream function can be found. when you get the stream equation, you can put it into rest equations.
  10. Aug 25, 2009 #9
    Sorry, I forgot to said that I am considering a incompresible and viscous flow with very small Reynolds numbers.

    [itex] \mathbf{\nabla}\times\textbf{u} [/itex] is the curl for u, that you can become from the properties of the laplacian operator:
    [itex]\mathbf{\nabla}^2 \mathbf{u} = \mathbf{\nabla} ( \mathbf{\nabla} \mathbf{u})- \mathbf{\nabla} \times ( \mathbf{\nabla} \times \mathbf{u}) [/itex]

    Second, I am not sure if my boundary conditions far from the circle are correct or not.
  11. Aug 25, 2009 #10
    This problem is laid out in great detail in the book "Transport Phenomena" by Bird, Stewart, and Lightfoot.
  12. Aug 25, 2009 #11


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    Science Advisor

    I've cited this paper many of times. Taken from NASA Report No. 496, "General Theory of Aerodynamic Instability and the Mechanism of Flutter" by Theodorsen. (Available on the NASA technical report server).

    \\Start quoted
    Let us temporarily represent the wing by a circle. The potential of a source [tex]\epsilon[/tex] at the origin is given by:
    \phi = \frac{\epsilon}{4\pi}\log(x^2 + y^2)
    For a source [tex]\epsilon[/tex] at [tex](x_1,y_1)[/tex] on the circle:
    \phi = \frac{\epsilon}{4\pi}\log[(x-x_1)^2 + (y-y_1)^2)]
    Putting a double source [tex]2\epsilon[/tex] at [tex](x_1,y_1)[/tex] and a double negative source [tex]-2\epsilon[/tex] at [tex](x_1,-y_1)[/tex] we obtain for the flow around a circle:
    \phi = \frac{\epsilon}{4\pi}\log\frac{(x-x_1)^2 + (y-y_1)^2)}{x-x_1)^2 + (y+y_1)^2)}
    The function [tex]\phi[/tex] on the circle gives directly the surface potential of a straight line pq, the projection of the circle on the horizontal diameter. In this case, [tex]y=\sqrt{1-x^2}[/tex] and [tex]\phi[/tex] is a function of x only.
    \\end quote

    The paper goes on to talk about pitching and plunging an actual airfoil shape, and the math involved is....fun. Either way, the link to the paper is (helpful for the figure)

    Good luck!
  13. Aug 26, 2009 #12
    In this book, they threat the problem in spherical coordinates (r, [tex]\vartheta[/tex], [tex]\phi[/tex]) or cartesian coordinates (x,y,z) and the problem proposed here is to resolve in two dimensions only, polar coordinates (only r, [tex]\vartheta[/tex]) or cartesian coordinates (x,y).
  14. Aug 26, 2009 #13
    In the Transport Phenomena book this problem is outlined as an example of how to solve the flow of an inviscid fluid (one with negligible viscosity) through the use of a velocity potential. The problem is not solved in spherical cooridinates. (Why whould you use spherical coordinates for a cylindrical problem?)

    Anyways, I am not going to post what the book has done but if you want to view it yourself then pick up a copy of the 2nd edition printed in 2007 and read section 4.3.

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