# Potential flow around a non-symmetric obstacle using stream functions

I've seen that there is a way to use the finite differences method, on a cartesian orthogonal grid, to perform calculations on potential flow about an obstacle without using the Neumann conditions, but only Dirichlet conditions. It is done using the stream function, i tested it and it seem to work. But this method has a big drawback: it can be used only with obstacles that are symmetric along an horizontal axis (the direction of the undisturbed stream). The method is based on the fact that there is a horizontal streamline that ideally follows all the boundary of the obstacle (so you can put a constant value of the stream function on it). Do you now any workaround or method to use only Dirichlet conditions, but on an arbitrary shaped obstacle (a wing, for example) ? Thank you in advance.

Before, someone advised me to use panel methods, coordinate transformations, etc. It would be good, but i'm looking for something simpler, if possible. It's just for demonstration of the finite differences method, it doesn't have to be particularly efficient of fast. Thank you.

Edit:

As GradGuy indicated, the stream function is constant on the boundary also when the obstacle is asymmetric; but I didn't say it was not. The method I used was this: I took a rectangular box, then I put two arbitrary values of the stream functions (but the same in all the side) in the upper and lower side; then I linearly varied it along the lateral sides (so there is continuity with the potential on the upper and lower sides). In this way I define the undisturbed flow far from the obstacle. Then I placed the symmetric obstacle right in the middle of the box, so I knew for sure there was horizontal streamline that divided the box in two and followed the obstacle boundary, so its potential was exactly the average value between the upper and lower side. I don't' know how to do same thing with an asymmetric obstacle... Any help would be greatly appreciated.

Hello, I tried to post a new question in which I posted the same requests as this question, but in a more clear way. It seems that question has been closed because it's a duplicate of this one. They complained also because I opened many accounts, I'm sorry for that but there was a problem with my browser that didn't allow me to do a password protected account (it crashed on the account creation page), so I can only open guest accounts. They advised me to edit that question instead. So, I appreciate your efforts but the answers I received till now didn't solve my problem. This question is to be considered still open... It may also be that what I'm asking is impossible...

• I'm not convinced what you say is true. I think the constant value simply from the condition that the normal velocity is zero on the boundary. – mmirzadeh May 29 '13 at 0:49

What you mention about the value of the stream function not being constant for non-symmetric boundaries is not correct. Indeed for any boundary for which the normal component of velocity is zero, the stream function should be constant on the boundary.

You can formally show this property as follows. In 2D, a stream function $\psi$ is defined such that the velocity vector can be written as:

$$\vec{v} = \hat{\mathbf{k}} \times \nabla \psi$$

where $\hat{\mathbf{k}}$ is the unit vector in the $z$ direction. Thus the normal component of velocity is given by:

$$v_n = \hat{\mathbf{n}} \cdot \left(\hat{\mathbf{k}} \times \nabla \psi\right) = -\hat{\mathbf{k}} \cdot \left(\hat{\mathbf{n}} \times \nabla \psi\right)$$

where $\hat{\mathbf{n}}$ is the unit vector normal to the boundary. For $v_n$ to vanish on the boundary, it is necessary to have $\hat{\mathbf{n}} \times \nabla \psi = 0$ on the boundary. This last requirement simply states that $\nabla \psi$ is parallel to the normal unit vector on the boundary. In other words, one must have:

$$\nabla_t \psi = 0$$

where $\hat{\mathbf{t}}$ denotes the tangential unit vector on the boundary. The only way this can be achieved for every point on the boundary is to have $\psi = \text{const.}$ on the boundary.