[jinny]

Dear all

- Euler simulation
- Test model :NACA0012 airfoil (symmetric airfoil)
- Mach number : 0.1 (incompressible flow)
- Angle of attack : 0 deg.

As above condition, there is no lift and drag
As you known, D'Alembert proved that for incompressible flow and inviscid potential flow - the drag force is zero on a body moving with constant velocity relative to the fluid

- Euler simulation
- Test model :NACA0012 airfoil (symmetric airfoil)
- Mach number : 0.1 (incompressible flow)
- Angle of attack : 2 deg.

However, in case of angle of attack of 2 deg. the drag force occur !
I don't know where the draq come from ?

please explain the drag force in inviscid flow (Euler simulation)

Thanks in advance
Jinny

[FMDenaro]

this is a classical issue in an aerodynamic course, generally the topic is illustrated by the flow problem of the lift generated by a 2D cylinder.
The key is in the circuitation generated by the position of the rear stagnation point (Kutta condition says it at trailing edge). Any textbook of basic fluid/aerodynamic can help you...
just start see in...wikepedia
http://en.wikipedia.org/wiki/Lift_%28force%29

[Martin Hegedus]

Artificial/numerical dissipation and numerical error. A high order code with very little numerical dissipation should produce zero drag.

[FMDenaro]

Originally Posted by Martin Hegedus

Artificial/numerical dissipation and numerical error. A high order code with very little numerical dissipation should produce zero drag.

As you said, the answer is that the code "implicitly" fix the Kutta condition ...it is as a check to compute the integral of the tangential velocity along the airfoil

[jinny]

Dear Martin Hegedus

Thnak you for your reply !

You means that the drag force generated in Euler simulation is numerical error ?
If I used high-oder scheme, such as DGM, the drag force does not occur ?
Um....Could you explain why the drag force does not occur in inviscid flow (Not potential flow)

Thanks in advance
jinny

[jinny]

Dear FMDenaro

Thanks, FMDerao
As you said, I have to check the surface integral routine which calculate a normal force to the aifoil surface

Thank you, have a nice day
Jinny

[Martin Hegedus]

Jinny,

Inviscid, incompressible, steady state flow is irrotational (due to vorticity transport equation), therefore inviscid incompressible steady state Euler is potential flow. Once the bound vortex reaches steady state, vorticity is not shed into the flow field, and then drag goes to zero. However, numerical error causes the bound vortex to decay and therefore vorticity needs to be constantly shed into the field to maintain it (and the implicit kutta condition). If there is no error, then the bound vortex does not decay and eventually all the vorticity in the field is transported away.

On this page is a grid refinement study of the NACA 0012 airfoil at M=0.5 and alpha=1.25. http://www.hegedusaero.com/examples/.../Vassberg.html. You can see that drag drops as the grid is refined. For the case of FLO82-HCUSP, drag basically approaches zero. AO_CFD has a more challenging time probably because of higher artificial dissipation. I assume the same is true for OVERFLOW.