# I Potentiel of a source in a corner (Fluid Mechanics)

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1. Dec 7, 2016

### Nico045

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: Dec 7, 2016
2. Dec 12, 2016