My intuition says that if your velocity (streamfunction gradient) is constant that vorticity should be zero.

However, I think the typical procedure is that you will likely avoid specifying vorticity at the boundaries directly and its implementation is somewhat algorithm dependent. The problem is essentially fourth-order for streamfunction so you will be able to specify two boundary conditions for the streamfunction and can simply link the vorticity to the second derivative of streamfunction. Not sure if that helps, but hopefully it points you in a better direction. 

The book "Spectral Methods for Incompressible Viscous Flow" by Peyret has a good discussion on this.

Given normal boundary condition (dPsi/dn) = 1 on the moving boundary, this does not mean that vorticity is zero on the boundary. However, the approximation on the moving boundary depends method used for finding solution. One has to use the equation          -w = Laplace(psi) to get the value of vorticity w on the boundary. One can use simple first order formula  ω1 = 2[ψ2 – h (ψy)1 ]/ (h*h) at the boundary, where 2 is interior point and 1 is boundary point at the moving boundary. Only if you iterate and find solutions then under relaxation is to be used as suggested by Rigal.

I used finite difference method. Indeed, there is a formulation, where the author specified zero at the entrance cavity for vorticity. I did it, but the results was incorrect too. The finite difference schemes, which i used are correct, because i've already test them on another problems.  The link between vorticity and the second derivative of streamfunction was described in Rouch. 

Thans's a lot for your answers

done.

Hi Alexander, please could you explain what you mean if talking about vorticity? In a turbulent flow there are two options: either (a) r.m.s. vorticity of the turbulence ("turbulent vorticity") or (b) the vorticity of the mean flow. In case (b) you may look at the mean-flow conditions in terms of r.m.s. velocity and pressure and simply derive therefrom the corresponding boundary condition for the r.m.s. vorticity. If you talk about turbulent vorticity I would refer to a paper which I attach. Best regards, Helmut

Hai Alexandar,

Vorticity is not zero from any solid walls for viscous flow. You need to calculate based on its definition. For instance from moving wall it will be = - {1-u_(wall-1) } / delta y; 1 stands for dimensionless moving plate velocity.

You can refer our paper for more details.

Das, M.K. and Kanna, P. R., Application of an ADI Scheme for Steady and Periodic Solutions in a Lid-Driven Cavity Problem, International Journal of Numerical Methods for Heat & Fluid Flow, Vol. 17, Issue 8, pp. 799-822, 2007

best regards,

Dr PRK

To Helmut Ziegfeld Baumert

2D laminar flows are described by Navier-Stokes and continuity equations in variables u,v,p, but also you can add stream function and vorticity and get vorticity transfer and Poisson equations. I attached my article(1), where i used this method in modelling natural convection in closed rectangular area.

To P. Rajesh Kanna

You misunderstood me a little bit. The entrance cavity is not a solid moving wall. It is the space, where flow is coming to the solution domain. In case of solid moving wall i can simple specify Psi = 0, dPsi/dn=1 and calculate boundary vorticity value by Thom's formula. I've solved a test problem using this method

The problem is that vorticity is not, in this case, a natural condition!.

You must use an approximation using the natural condition (velocity, stream function), but the difference approximation must be of the same order than the schemes used for vorticity equation.

"Given normal boundary condition (dPsi/dn) = 1 on the moving boundary, this does not mean that vorticity is zero on the boundary. However, the approximation on the moving boundary depends method used for finding solution. One has to use the equation -w = Laplace(psi) to get the value of vorticity w on the boundary. One can use simple first order formula ω1 = 2[ψ2 – h (ψy)1 ]/ (h*h) at the boundary, where 2 is interior point and 1 is boundary point at the moving boundary. Only if you iterate and find solutions then under relaxation is to be used as suggested by Rigal."

Dear Alexander, I didn't realize that you are working on the classical laminar problem. In this case you may find the solution for your problem in classical textbooks. I recommend to have a look in the many monographs and papers of A. A. Samarskij  (all in Russian language) and his students - a Russian classic in difference schemes. I guess there you may find all you need.

Thank's a lot, but i've already done it. 

Dear colleague,                                                                                                                       Since in general the laplacian of \psi (stream function) is poorly approximated by numerical schemes, rigorously the right way to deteremine \omega (vorticity) Dirichlet boundary conditions is to exploit an overabundant boundary condition, say d\psi/dn given, by solving a boundary problem for the trace of  \omega on the boundary \Gamma of the flow domain \Omega. For a stationary flow the variational formulation for this problem is: 

\int_{\Omega} (\xi +  \omega_0) \chi dx + \oint_{\Gamma} d\psi/dn \xi ds = 0,

for all harmonic \chi, \xi being the harmonic function whose trace on \Gamma coincides with the one of \omega. \omega_0 is defined as \omega - \xi, and has thus a zero trace on \Gamma. It can be determined from the momentum equation, by plugging given values of \psi and \xi in an iterative Navier-Stokes solver. For further details, please see my ZAMM papers published in 1994 and 1995. For transient problems you may come back to me if you wish. Good luck,                                  Vitoriano Ruas

Hi Alexendre

Sometimes, general finite difference schemes doesn't give good results for stiff problems. Instead of changing the boundary condition, try to change the numerical technique or use Shishkin meshes in your finite difference scheme. You will definitely get correct results.

Hi all

I have a solved domain (2D flow around a circular cylinder) with U field. Now I want to calculate stream function with this formulation:

omega = dv/dx - du/dy

solve (laplacian (sai) = - omega)

But at last the results are wrong!

The BC for sai are:

for top and bottom sai is calculated from the volumetric flux (Q = A.U_in) that is: sai_top - sai_bottom = U*delta(y) = 1*20 = 20, so:

top boundary: sai = 10

bottom boundary: sai = -10

cylinder surface boundary: sai = 0

inlet and outlet boundaries: d(sai)/dn = 0

and for omega I don't have BC, because it will be calculated from velocity field. However it was very strange that omega is in order of 1 on cylinder surface and in order of 0.01 on inlet and outlet and in order of 0.001 on top and bottom boundaries.

Could you please help me in calculating that?

Are the boundary C. correct?