Fluid mechanics: Separation surface of a source in a parallel flow (3D)

  • Thread starter steem84
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  • #1
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Hello,

I have the following problem with respect to fluid mechanics:

A source of strength Q is in a 3D space and subjected to a parallel flow U along the x-axis. The position of the source is at (xq,yq,zq). This will lead to the following velocity potential in cartesian and cylindrical coordinates

figure 1

With the velocity in the x-direction determined to be

figure 2

To calculate the surface rho(x) which separates the fluid (coming from the source) from the fluid, (coming from the parallel flow), one can use the law of volume conservation (because of constant density). In that body which is described by that surface, the flow through any plane x=constant should equal to Q

figure 3

I know this is the correct way to calculate the separation surface, but from this point I can not go further: how do I solve the integral? Or can I use some trick?

Btw: sorry for the clumsy format, but I can’t LateX

Thanks!

Steven

If this is the wrong sub-forum, please let me know
 

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Answers and Replies

  • #2
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Figure 3 shows an integration with respect to rho and theta then shows an integral with respect to rho after integrating with with respect to theta. This second expression should not have a theta term, namely cos(theta - thetaq). This is a single variable integral with two independent variables. Evaluate the first integral for theta taking into account the cosine term.
 
  • #3
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yes, that does make sense. Thank you
 

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