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 curious ... July 7, 2006 03:25

potential flow vs. Euler flow

Can somebody explain the following to me:

Assume you have an airfoil (e.g. a NACA0012) at an angle of attack - let's say 5deg. If you want to describe this flow, using potential flow, circulation is needed in order to get the rear stagnation point onto the trailing edge of the airfoil. If you use an Euler code to simulate the same problem, you don't need to do anything - the code will never put the rear stagnation point anywhere else but in the trailing edge (provided there is enough resolution to see the stagnation area) If the grid is coarse you will not see the rear stagnation point, but still the flow will follow the shape of the airfoil and leave at the trailing edge. This will even happen if you fix the farfield conditions to parallel flow.

Why?

 mar July 7, 2006 04:51

Re: potential flow vs. Euler flow

The answer is in the wall boundary condition not only in the physical model you are using! The trailing edge is singular and so an appropriate boundary condition is needed to overcome this singularity (the kutta condition). Coherently to the kutta condition vorticity is generated at the trailing edge. The potentail flow is irrotational and so this "vorticity" must be introduced to correctly find the rear stagnation point. The Euler system of equation is rotational and so it can manage "automatically" this vorticity!

 Praveen. C July 7, 2006 05:25

Re: potential flow vs. Euler flow

 curious ... July 7, 2006 06:13

Re: potential flow vs. Euler flow

What happens if you run the same problem through an incompressible Euler solver?

 Tom July 7, 2006 06:13

Re: potential flow vs. Euler flow

>> The Euler system of equation is rotational and so it can manage "automatically" this vorticity!

Not quite - if the flow is irrotational at t=0 it will be irrotational for all t>0. The circulation is being induced by the numerical dissipation of the scheme (and possibly the application of the wall boundary conditions/unsteady Kutta condition).

 Gerrit July 7, 2006 07:58

Re: potential flow vs. Euler flow

I think they are mixing things here.

Mar was right in his first answer.

In order to get from the Full Navier-Stokes equations to the Euler equations, you only need to take out the viscous effects (set mu=0 in Navier Stokes), but it has nothing to do with the boundary conditions, those are flow far away is parrallel and flow on the airfoil is tangent to the surface (in both cases the same)

However, in order to get potential flow, the total rotation in the flow is supposed to be unchanged in time (DGamma/dt=0, which is not the same as Gamma=0), I think it doesn't necessarily have to be zero (I remember that I have seen publications with potential flow with rotation) BUT I'm not completely sure about that. Anyway that doesn't matter here, if the rotaion is zero at the moment that the flow starts to accelerate (Gamma=0 at t=0), it wil be zero in total if the same particles are considered when the steady flow is reached (Gamma=0 at t=t1>0, since dGamma/dt=0), doing so the potential flow equations are valid.

The potential flow solution you get with Gamma=0 and the flow not leaving from the Trailing Edge, is a perfect potential solution, it's mathematically as valuable as the solution we are looking for.

As in reality viscous effects (read rotational flow in the boundary layer) causes the flow to leave at the trailing edge, a rotation Gamma around the airfoil is added to the flow until this is the case. This is the Kutta trick that is applied, to get exactly that potential solution out of the many potential solutions (with different Gamma's around the airfoil) that approximates best the real viscous rotational flow solution. So your foreknowledge of some physical phenomena as seen in tunnels helps you to get the solution.

But then we have rotational flow? No, we don't because far far away behind the airfoil there is an equal and opposite rotation that cancels out the rotation around the airfoil, so considering the same particles as from the moment the flow started to accelerate the total rotation is still zero. This rotation far behind the airfoil came into being when the airfoil started to acelerate and the flow was able (for a short moment) to go around the trailing edge from pressure side to suction side, really!! This was beautifully demonstrated by Prandtl by means of windtunnel measurement pictures in the 20's

 curious ... July 7, 2006 08:37

Re: potential flow vs. Euler flow

and in 3D these two vortices are linked by the vortices starting from the wing tips, forming a closed loop.

Unfortunately our computational domain has only the first vortex and somehow the solver needs to create that one out of nothing.

What would you say is the difference between potential flow and incompressible "Euler flow" with parallel flow as a farfield condition?

 Gerrit July 10, 2006 04:17

Re: potential flow vs. Euler flow

Eeehm, I don't know whether there really IS a difference.

If the Euler flow happens to be incompressible, the only thing that changes is the numerical scheme that you use for solving it. As far as I know you could use artificial compressibility or solve the Poisson equation in stead (sure there will be more ways to Rome).

I don't see why this would change the fundamental physical difference between Euler flow and potential flow itself.

I am starting to wonder myself now, how this acceleration starting vortex comes into being. It's an inviscid effect, but viscosity makes sure that it will leave from the trailing edge after a very short while. ????

I think that in case of an Euler solver, no matter whether it's incompressible or not, the opposite rotation is initiated when you start your time marching, even though the time marching in a steady flow solver is not physical at all. This equal and opposite vortex may then move in flow direction and disappear with time. But I'm not talking from experience now, more from imagination. Anyway to make this visible in the computer you should use an unsteady Euler solver to let the flow accelerate until the steady solution is reached, which maybe very time consuming. I can imagine that local time stepping with it's non physical solutions in the meantime makes it impossible to detect the start of the TE vortex.

Maybe there's someone with more experience out there who can give more clarity, although I fear that people who really bothered about understanding this died out :).

It's a very interesting question you're asking, I have been asking it myself as well. In fact it's more physics than CFD.

By the way, what you say about 3D is true indeed, I think that's the Helmholz theorem, saying that a vortex should always be a closed curve.

 Tom July 10, 2006 05:04

Re: potential flow vs. Euler flow

Basically the origin of the circulation lies within the viscous boundary-layer. If you consider, for a slender wing with an attached boundary layer, the steady state solution of the Navier-Stokes equations at successively higher Reynolds numbers then you may expect the flow everywhere outside of the boundary layer to tend towards a potential flow. The circulation is an artifact of this limiting process; i.e. the strength of the circulation is a reminant of the viscosity which does not vanish as the viscosity tends to zero (this is similar the the decay of a shock in Burgers equation which requires viscosity but results in a decay rate independent of the viscosity). Basically the Kutta condition is the requirement of a fully attached boundary layer (although the converse is not true).

The problem of circulation genration by the boundary layer is discussed in the book "Viscous Flow" by Ockendon & Ockendon. Also look at "Elementary fluid dynamics" by Acheson.

 Gerrit July 10, 2006 06:03

Re: potential flow vs. Euler flow

Does that mean, knowing the direction of the shedded circulation, that the boundary layer leaving the pressure side has more circulation CCW than the boundary layer leaving the suction side CW??? (Flying from right to left)

 Gerrit July 10, 2006 06:08

Re: potential flow vs. Euler flow

And if it has a viscous origin, how come that an Euler solver is able to solve this without imposing any Kutta condition like in potential methods???

Am I missing something?

 Tom July 10, 2006 06:24

Re: potential flow vs. Euler flow

Firsly you need to look up the definition of circulation since talking about the circulation on the upper and lower sides of the trailing edge is meaningless (circulation is the strength of the bound vortex).

The reason than an Euler solver can obtain the "correct" result is simply numerical dissipation - if the Kutta condition does not hold a large, infinite in the continuous case with a sharp trailing edge, velocity will be produced at the trailing edge which the advection scheme will have to dissipate. If you check your numerics you will find that the solutions during the course of the integration violate the Helmholtz theorems.

Check the books I referenced.

 Gerrit July 10, 2006 06:54

Re: potential flow vs. Euler flow

Quote: "Firsly you need to look up the definition of circulation since talking about the circulation on the upper and lower sides of the trailing edge is meaningless (circulation is the strength of the bound vortex)."

Comment: No, it is not meaningless at all. I know the definition, I think there's a miscommunication. I am NOT talking about the circulation in a closed curve around the airfoil. What I want to say is: If I make a closed curve INSIDE the boundary layer near the trailing edge on the suction side, there will be some Gamma<>0, if I do the same INSIDE the boundary layer on the pressure side there will again be some Gamma<>0, but opposite in sign. I am wondering whether we can conclude that the latter will be stronger in absolute value and if so why?

Quote: "The reason than an Euler solver can obtain the "correct" result is simply numerical dissipation - if the Kutta condition does not hold a large, infinite in the continuous case with a sharp trailing edge, velocity will be produced at the trailing edge which the advection scheme will have to dissipate. If you check your numerics you will find that the solutions during the course of the integration violate the Helmholtz theorems."

Comment: The fact that the Helmholz theorem does not hold during iteration, does not mean anything, as during iteration in usual steady flow solvers the solution is not physical by definition, it's just a bunch of numbers that starts to get physical after sufficient convergence, unless you use an unsteady flow solver of course.

I don't know enough about CFD to know whether the "correct" solution is because of numerical dissipation, but if so, how would it be in an imaginary tunnel if we suppose the existence of a inviscid fluid???

Quote: "Check the books I referenced."

Comment: I will

 Tom July 10, 2006 08:02

Re: potential flow vs. Euler flow

"Comment: No, it is not meaningless at all. I know the definition, I think there's a miscommunication. I am NOT talking about the circulation in a closed curve around the airfoil. What I want to say is: If I make a closed curve INSIDE the boundary layer near the trailing edge on the suction side, there will be some Gamma<>0, if I do the same INSIDE the boundary layer on the pressure side there will again be some Gamma<>0, but opposite in sign. I am wondering whether we can conclude that the latter will be stronger in absolute value and if so why?"

Well maybe you should ask yourself why you are bothered what the circulation is either side of the trailing edge - the Helmholtz theorems don't hold here since the flow is viscous. Furthermore, since you are only interested in the inviscid solution, the Kutta condition is nothing more than the requirement that large adverse pressure gradients don't occur at the trailing edge (e.g. dynamic stall). This by itself does not guarantee a consistency with the boundary layer - a solution not satisfying the Kutta condition will always separate the boundary-layer. A solution satisfying the Kutta condition however may still cause separation => a violation of the potential flow assumption. A similar type of argument yields Brillouin-Villiat condition for the Kirchoff freestreamline leaving a bluff body.

Why not write down the potential flow solution for flow past a Joukowski aerofoil satisfying the Kutta condition and then solve the corresponding boundary-layer equations to find out the answer to this.

"Comment: The fact that the Helmholz theorem does not hold during iteration, does not mean anything, as during iteration in usual steady flow solvers the solution is not physical by definition, it's just a bunch of numbers that starts to get physical after sufficient convergence, unless you use an unsteady flow solver of course."

It does matter for the simple reason, and why I placed quotes around correct, that the final steady-state solution is not necessarily the correct one. A simple example of this is Euler flow past a circular cylinder where the solution is symmetric for potential flow. Here it is not uncommon for the numerics to produce small separation bubbles at the rear stagnation point which does not correspond to a differentiable solution of the full Euler equations - and are actually not correct weak solutions either.

"I don't know enough about CFD to know whether the "correct" solution is because of numerical dissipation, but if so, how would it be in an imaginary tunnel if we suppose the existence of a inviscid fluid???"

It would look like the zero circulation solution. Have a look at "the album of fluid motion" by van Dyke - the section on Hele-Shaw cells.

 Gerrit July 10, 2006 08:53

Re: potential flow vs. Euler flow

Thanks a lot for your explanations, the whole thing has never been clear to me, so I tried to make people answer something more than just one sentence... ...and I didn't even ask the initial question :)

I was wondering about the Gamma in the boundary layers leaving the T.E. because you said that the circulation coming from the T.E. had a viscous origin. So I asked myself: couldn't it be just the sum of the two circulations found in both boundary layers just before mixing? Or am I making a conceptual mistake?

May I conclude that the strength of the circulation that's being shed from the trailing edge in unsteady panel methods, generating wake panels at each time step, is fully determined by the Kutta condition only?

 Tom July 10, 2006 09:54

Re: potential flow vs. Euler flow

"May I conclude that the strength of the circulation that's being shed from the trailing edge in unsteady panel methods, generating wake panels at each time step, is fully determined by the Kutta condition only?"

It's a long time since I looked at panel methods (and I've never used them myself) but as I recall if you only specify the condition for the Kutta condition (i.e. finite velocity at the trailing edge) then you obtain a solution equivalent to those obtained by classical conformal methods; i.e. no trailing vortex sheet. If you add in a trailing vortex sheet you need to include extra information about the evolution of the location of the sheet along with boundary conditions for the continuity of pressure and normal velocity. The unsteady evolution of the resulting sheet is however likely to be ill-posed (see for example the theory of the Birkhoff-Rott equation).

However ignoring awkward mathematical issues such as well-posedness there was a paper by Tuncer Cebeci in the early/mid eighties (?) (in JFM*) where this is explained (he also solves for the boundary layer as I recall).

Hope this helps - if your lucky an expert on panel methods may read this and explain it better and fill in the missing gaps.

(*) I think the reference for this is "Airfoils with separation and the resulting wakes" Cebeci et al (1986) JFM 163.

 Gerrit July 10, 2006 10:19

Re: potential flow vs. Euler flow

Thanks a lot for your comments on this. Lots of stuff left to read as well, pfew!

Maybe someone could comment also something on my circulation question inside both boundary layers, I wonder if what I said is plausible? It seems plausible to me, because in that way, the circulation coming from the T.E. for an accelerating airfoil depends on the history in pressure gradients on both sides, i.e. on the geometry as it should be. I imagine taking two closed curves from boundary layer edge to surface and back.

"If you ask a question, you're a fool for a moment. If you don't ask that question, you're a fool forever" Lao Tse :)

 Peter July 10, 2006 11:29

Re: potential flow vs. Euler flow

The two-dimensional Kutta condition can be used along the trailing edge in panel methods to fix the wake strength. Also the Kelvin condition dGamma/dt=0 can be used to calculate the change in the wake circulation. Wake shapes usually are assumed to flat but methods which give a correct "force-free" wake can be used..but are costly. These are physical considerations which are needed since without them the solution would be non-unique. See Katz and Plotkin's book "Low-Speed Aerodynamics" for more details on panel methods.

 Adrin Gharakhani July 10, 2006 16:50

Re: potential flow vs. Euler flow

> By the way, what you say about 3D is true indeed, I think that's the Helmholz theorem, saying that a vortex should always be a closed curve.

Not generally true!