# D'Alembert paradox + kutta condition

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 January 4, 2006, 02:31 D'Alembert paradox + kutta condition #1 snegan Guest   Posts: n/a hi all, I am presenting 2 questions to the CFD commmunity 1. According to the D'Alembert paradox, for an inviscid incompressible flow, the drag for a body is zero. But through experiments this is a non zero quantity. Suppose if one is doing a numerical simulation of this what will cause the non-zero drag in the compuational methods. Is it a numerical dissipation associated with the scheme is responsible? pls do clarify me in this regard. 2. How will u incorporate the kutta condition at the trailing edge for the difference scheme, Finite volume scheme and the finite element scheme? -snegan

 January 4, 2006, 04:24 Re: D'Alembert paradox + kutta condition #2 Tian_FB Guest   Posts: n/a i will say some thing about first q, the drag for a body is zero in the direction of incoming flow velocity vector.so,does the drag in your simulation parallel to the velocity direction? if the non-zero drag in the compuational methods is very small,it is,maybe, a numerical dissipation associated with the scheme is responsible.

 January 5, 2006, 07:02 Re: D'Alembert paradox + kutta condition #3 Biga Guest   Posts: n/a Hi, 1. This paradox is only valid for 2-D cases. Remember that, even for an inviscid flow, if you got lift, you'll certainly get due-to-lift drag (induced drag). Therefore, if your simulating a 3-D INVISCID flow, it can be that you get drag. However, if your case is 2D: a) how are you computing the drag? Is it an integration of pressure forces along your profile? This may certainly generate some numerical errors. b) as Tain properly mentioned, is it along the flow direction? c) what's the order of magnitude of your resulting drag? d) artificial dissipation could also be blamed for an additional quantifyable drag, if it's strongly modifying the entropy field of your simulation. This can also be easily checked. 2. Just a curiosity: why do you need to set the Kutta condition in a finite volume code? Are you solving full potential flow with finite volume? As far as I know, in any case, you don't need to specifically modify your trailing edge to set the Kutta condition in a full potential flow (which uses a grid to solve the flow, differently from potential panel methods, which are surface methods, and won't use a grid for the flow). All you need to do is to set the circulation you'd get (adaptably along the iterations) in the lifting case to your farfield condition. This would properly generate your Kutta condition. Regards, Biga

 January 6, 2006, 11:45 Re: D'Alembert paradox + kutta condition #4 AnotherCFDUser Guest   Posts: n/a Although a Kutta condition is required for inviscid simulations most people choose not to apply this condition, instead relying on the inherent numerical dissipation of the scheme (from truncation errors) or added artificial viscosity. This means that they are relying on the fact that the discrete problem we are actually solving is related to an advection-diffusion equation rather than the advection problem described by the continuum governing equations. As we are now dealing with a 'viscous' problem there is now no paradox - although there will be drag. Thus the solution for the drag forces should tend towards zero asymptotically as the mesh is refined (a good verification test for a CFD code). Alarmingly a number of well known commercial CFD codes appear to asymptote to a small but non-zero value for drag! If we are dealing with a potential (irrotational) flow then even in 3D there should be no drag force. Although the governing equations convect vorticity, there is no mechanism for its generation.

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