Potentiel of a source in a corner (Fluid Mechanics)

  • #1
10
0
Hi, I made some calculations abiout this problem biut my results are a bit complicated :

We consider two walls in x=0 and y=0, in (a,bi) , a>0, bi>0 ,we have a source in this corner and I need to find the potential biy using the method of image. After this I need to find v, the pression and the forces applied on the walls.

So from what I know, I can add 3 images of the source in (a,-bi) ; (-a,-bi) ; (-a,bi)

Then I would have
W(z) = m/2π * (ln(z-a-bi) + ln(z-a+bi) + ln(z+a-bi) + ln(z+a+bi) )

(conjugate)v=dW(z)/dz = m/2π * (1/(z-a-bi) + 1/(z-a+bi) + 1/(z+a-bi) + 1/(z+a+bi) )

However to find the pression, from bernoulli's equation, I need to calculate v2 = conjugate(v)*v and it gives me something like :

v2= (m/2π)2 * (1/(x+yi-a-bi) + 1/(x+yi-a+bi) + 1/(x+yi+a-bi) + 1/(x+yi+a+bi) ) *(1/(x-yi-a-bi) + 1/(x-yi-a+bi) + 1/(x-yi+a-bi) + 1/(x-yi+a+bi) )

which is very complicated since I have to integrate this expression to get the forces.

Have I done something wrong in the beginning ?
 
Last edited:

Answers and Replies

  • #2
18,633
8,579
Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
 

Related Threads on Potentiel of a source in a corner (Fluid Mechanics)

  • Last Post
Replies
2
Views
1K
Replies
7
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
1K
Top