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Navier-stokes flow around a sphere

  1. Apr 28, 2014 #1

    joshmccraney

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    hi pf!

    basically, i am wondering how to find the velocity profile of slow flow around a sphere in terms of a stream function ##\psi = f(r,\theta)## where we are in spherical coordinates and ##\theta## is the angle with the ##z##-axis. (i think this is a classical problem).

    i understand the situation like this: if we are trying to satisfy continuity then we have ##\nabla \cdot \vec{V} = 0##. thus, since we can describe the flow in two parameters ##r, \theta##, we say ##\vec{V} = \nabla \times \psi \hat{\phi}## where ##\hat{\phi}## is the phi unit vector. doing this in spherical coordinates gives me a different solution than the proposed ##\vec{V} = -\frac{1}{r^2 sin \theta} \frac{\partial \psi}{\partial \theta}\hat{r} + \frac{1}{r sin \theta} \frac{\partial \psi}{\partial r} \hat{\theta}##

    can someone shed some light on this? thanks so much!
     
  2. jcsd
  3. Apr 28, 2014 #2
    This problem is solved in detail in Transport Phenomena, by Bird, Stewart, and Lightfoot. If you don't already have this book, Josh, you will find it highly worthwhile to get a copy.

    Chet
     
  4. Apr 28, 2014 #3

    joshmccraney

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    thanks for the fast response! i actually do have the book, but it is not showing a derivation of where they came up with the velocity profiles. they just present them in a table. i cross referenced this with other materials and the above approach seemed to work with rectangular coordinates, although i am having troubles with it in spherical coordinates.

    do you have any suggestions here?

    thanks!
     
  5. Apr 28, 2014 #4

    joshmccraney

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    nevermind, chet. after checking more sources it seems they are letting ##\vec{V} = \nabla \times \frac{\psi}{r \sin \theta}##. i'm assuming to make the algebra easier.

    thanks again!
     
  6. Apr 28, 2014 #5
    Chapter 4. Worked Example 4.2.1

    Chet
     
  7. Apr 28, 2014 #6

    joshmccraney

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    Yea they definitely use the velocity here as boundary condition but they don't explain rigorously where the b.c. Comes from. But it's ok another source had it in full.
     
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