# Karman vortex street: boundary conditions for velocity, pressure

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 September 12, 2013, 17:07 Karman vortex street: boundary conditions for velocity, pressure #1 New Member   Join Date: Sep 2013 Posts: 2 Rep Power: 0 Hi all, with incompressible Navier-Stokes, I'd like to build a PDE for Kármán's vortrex street and I wonder what consistent boundary conditions for the velocity u and the pressure p would be. While on the obstacle and the tube walls I'd employ no-slip-no-penetetration for u and free boundary conditions for p, i.e., u = 0, n.grad(p) = 0, (where n is the outer normal), the situation for the inlet and outlet are less clear to me. To force a inflow on the left, one could certainly use something along the lines of u = ((y-tube_lower)*(tube_upper-y)) (0.0 ) How about the pressure here, though? What about the outlet?

September 12, 2013, 17:21
#2
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Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
 Originally Posted by nschloe Hi all, with incompressible Navier-Stokes, I'd like to build a PDE for Kármán's vortrex street and I wonder what consistent boundary conditions for the velocity u and the pressure p would be. While on the obstacle and the tube walls I'd employ no-slip-no-penetetration for u and free boundary conditions for p, i.e., u = 0, n.p = 0, (where n is the outer normal), the situation for the inlet and outlet are less clear to me. To force a inflow on the left, one could certainly use something along the lines of u = ((y-tube_lower)*(tube_upper-y)) (0.0 ) How about the pressure here, though? What about the outlet?

For incompressible flows, the "pressure" is not a thermodinamic function but it enforces the divergence-free velocity field by means of the Poisson equation.
Said that, BC.s for velocity and pressure can not be prescribed independently. If you fix the velocity, the pressure BC for the Poisson equation is constrained by the congruent Neumann condition.

 September 12, 2013, 17:40 #3 New Member   Join Date: Sep 2013 Posts: 2 Rep Power: 0 That is my understanding as well. If I see things correctly, I can use n.grad(p)=0 everywhere if n.u is specified everywhere, and in this case I have to make sure that \int_{boundary} n.u = 0 to have a consistent PDE. -- That corresponds with the incompressibility assumption. In the case of Karman's vortex street, though, I don't want to prescribe n.u at the outlet. What is the consequence for the pressure boundary conditions? Will I have prescribe the pressure somewhere?

 September 13, 2013, 06:45 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 1,645 Rep Power: 23 outlet can be managed in terms of some free-stream condition, e.g. n.grad (u) = 0. That leads to change the source term and the coefficients of the pressure matrix. It is quite easy to find in literature papers about the outlet bcs.

September 15, 2013, 15:05
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Quote:
 Originally Posted by nschloe Hi all, I'd like to build a PDE for Kármán's vortrex street
I don't understand what you mean.
Navier-Stokes equations are the PDE which rule the Karman's vortex street as any flow!!

Quote:
 Originally Posted by nschloe and I wonder what consistent boundary conditions for the velocity u and the pressure p would be.
the best ones are Orlansky outlet boundary conditions:

du/dt + ubarre du/dx =0
dv/dt + ubarre dv/dx =0

dp/dx=0 if you use SIMPLE like algorithm or projection method for velocity-pressure coupling algorithm

ubarre is the mean longitudinal velocity in the domain
here d /dt or d/dx are the partial derivative.

 Tags boundary conditions, navier stokes, vortex street

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