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

Click For Summary
SUMMARY

The discussion focuses on calculating the separation surface of a source in a parallel flow within fluid mechanics, specifically when a source of strength Q is positioned in 3D space and subjected to a parallel flow U along the x-axis. The velocity potential is expressed in both Cartesian and cylindrical coordinates, with the velocity in the x-direction derived accordingly. To determine the separation surface ρ(x), the law of volume conservation is applied, ensuring that the flow through any plane x=constant equals Q. The main challenge discussed is solving the integral related to this calculation, particularly addressing the integration with respect to the variables ρ and θ.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly flow dynamics.
  • Familiarity with velocity potential in Cartesian and cylindrical coordinates.
  • Knowledge of the law of volume conservation in fluid systems.
  • Proficiency in integral calculus, especially with multiple variables.
NEXT STEPS
  • Study the derivation of velocity potentials in fluid mechanics.
  • Learn about the application of the law of volume conservation in fluid flow problems.
  • Explore techniques for solving multiple variable integrals in calculus.
  • Investigate the implications of source strength Q in fluid dynamics scenarios.
USEFUL FOR

Fluid mechanics students, researchers in fluid dynamics, and engineers working on flow analysis and separation surface calculations will benefit from this discussion.

steem84
Messages
13
Reaction score
0
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
 

Attachments

  • pf1.PNG
    pf1.PNG
    2.9 KB · Views: 486
  • pf2.PNG
    pf2.PNG
    1.6 KB · Views: 504
  • pf3.PNG
    pf3.PNG
    5.3 KB · Views: 499
Physics news on Phys.org
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.
 
yes, that does make sense. Thank you
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K