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Old   September 18, 2002, 09:45
Default separated Euler solution
  #1
gecko
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Hello, is it allowed to have a separation in an Euler solution?

Thank you, gecko
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Old   September 20, 2002, 04:47
Default Re: separated Euler solution
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Tom
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This really depends upon what you mean by a solution (i.e. strong or weak)!

For incompressible flow with the velocity and pressure continuous (differentiable) functions the answer is NO. If you relax the differentiability constraint, so that you are working with the integral form of the equations instead of the point form (Euler), the answer is YES.

The basic point is that in order to have separation you must have a free-streamline (Kirchoff) and the derivatives, such as pressure, across the free-streamline are discontinuous; i.e. you have two different flow regimes and you cannot apply enough boundary conditions on the free-streamline to ensure that all derivatives exist).

A good introduction to separated flow is the book

"Asymptotic theory of separated flow" by Sychev et al. (Cambridge univerity press)
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Old   September 20, 2002, 09:29
Default Re: separated Euler solution
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Axel Rohde
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See earlier discussion on this topic:

www.cfd-online.com/Forum/main_archive_2001.cgi?read=19858
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Old   September 23, 2002, 07:12
Default Re: separated Euler solution
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Tom
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I don't really see the relevance of the previous discussion that you've added as a link - from what I can tell the previous discussion was more about numerical solutions to the Euler equations whereas my reply to the above question is basically a mathematical statement about the nature of the solution to the equations; i.e. the pointwise Euler equations do not permit separation since these equations are derived on the assumption that the spatial derivatives are continuous.

Separated inviscid flow solutions, such as Prandtl-Batchelor flows and the Sadovskii vortices, are always discontinuous across the separating streamline and so are not (strong) solutions to the Euler equations.

Numerical solutions of the Euler equations do not always represent approximations to the continuous equations just as the infinite Reynolds number limit of the Navier-Stokes solution is not necessarily equal to the inviscid solution.
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Old   October 24, 2002, 10:41
Default Re: separated Euler solution
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Kaveh Ghayour
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Dear Prof. Rohde,

The dissipation inherent in the low order discretization of the Euler equations is not the only culprit in the observation of boundary layer type behaviour near solid surfaces. I think the other factor is the implementation of the wall boundary conditions for the Euler equations. Charateristic theory tells us that for a slip solid wall, one piece of information has to be imposed and three other (for 2D) has to be computed somehow from the internal solution. Now, the question is which quantities shouls one choose to extrapolate from the inner domain to the solid boundary? For example, one may decide to extrapolate density, tangential velocity, and temperature (or total energy) ( this is the most common pratice I have seen in FV codes, while there are some codes out there that do it more elegantly by discretizing the characteristic equations at the boundary...). Eulerian flow does not allow for vorticity production at the wall but does such an implementation of wall BC's guarantee that?

Best regards,

Kaveh Ghayour.

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Old   October 24, 2002, 11:47
Default Re: separated Euler solution
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Tom
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In my comment I was lumping numerical dissipation issues with the obvious problems of extrapolation at the boundary - they are both issues which can stop the numerics converging to a true (properly posed) solution of the continuous equations while appearing to give a converged grid independant solution.

One of the problems with numerical solutions to the Euler equations is that many people use the same finite difference grid for the Euler equations as they do for the Navier-Stokes. Now since the NS equations are second order in space it is quite natural to use, for example in 2d, a 9-point stencil for the first order derivatives. The Euler equations on the otherhand are only first order and so it is more natural to use a 4-point stencil. This smaller stencil reflects the fact that the Euler equations have fewer boundary conditions. By using the NS's 9-point stencil for the Euler equations it becomes necessary, as you've quite rightly pointed, to extrapolate extra boundary condtions.

I'm pretty much certain that no extrapolated boundary conditions will work all the time and, for stably stratified flow for example, it's not even obvious what the extrapolated boundary conditions should be!

Tom (Not Prof. Rohde)
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Old   October 24, 2002, 12:21
Default Re: separated Euler solution
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Kaveh Ghayour
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Hi Tom,

Thanks for your reponse to my post.

No. of boundary conditions at the boundary doesn't have anything to do with the order of discretization. At a solid wall, physics of the hyperbolic problem tells us that only one quantity should be imposed and three ( four for 3D) other have to be determined from the inner solution. The number of BC's comes from the characteristic analysis of the linearized flow equations and should be respected by the discretization. You can use biased differences as you get close to the wall without introducing any ghost points for that matter. But, eventually, the question is how to populate the vector (\rho, \rho u_1, \rho u_2, \rho E) at the boundary.

I have tried a subcritical Ma=0.35 flow over a cylinder with fourth and sixth order central differences augmented with a sixth order artificial dissipation ( ~d^6u/dx^6\times \dx^6). If you plot vorticity , you can clearly see that it is basically zero everywhere except at the TE where a blob of high intensity vorticity is present. By refining the mesh, the blob gets smaller but intensifies greatly. It looks as if we approach the Eulerian flow as the limiting case of a viscous flow with \mu -> 0 rather than a potential flow that can advect vorticity (but still cannot produce it except at a flow discontinuity)

To get back to your note on the order of finite differences, I have also tried solving this problem with the discontinuous Galerkin method (DG) with polynomial orders from 0 to 8 over each element, and exactly the same thing can be observed regardless of the order of polynomial approximation. As a side note, in the DG method, the boundary conditions are applied in a weak manner in contrast with FD where it is usually done in the hard way.

Dissipation is definitely a culprit but considering the fact that the second and higher derivatives of the velocity field are not usually very large close to the wall for an inviscid flow which can easily slide past it, I wonder how big the effect of dissipation is in the production of vorticity that I see at the TE.
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Old   October 25, 2002, 04:56
Default Re: separated Euler solution
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Tom
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Kaveh,

I think you've partly missed my point which was that the order of the equation determines the number of boundary conditions that can by applied and the discretized version should respect this.

In my example of 2D flow if you use second order finite differences with a 4-point stencil and a suitably staggered grid boundary problems should be reduced; i.e. ghost points do not need to be introduced and the same finite difference representation can be used everywhere in the domain. If on the other hand you use a 9-point stencil the finite difference equations become higher order than the underlying continuous pde and hence the problems at the boundary- this is effectively why the leapfrog scheme has a computational mode but Crank-Nicolson does not.

What extra boundary conditions are you placing on your equations due to the artificial diffusion? If you're adding a grad^6 term you need extra-conditions at the wall that cannot be determined by the internal consistency of the solution!

Hope this better explains my point,

Tom.

P.S. you don't need to linearize the Euler equations to do a characteristic analysis to determine the correct number of bcs since the equations are quasi-linear hyperbolic and there's quite a bit known about such equations. Actually, the number of bcs required for a pde is usually deduced from the Cauchy-Kowalewsky theorem and not through the use of characteristics.

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Old   October 25, 2002, 10:01
Default Re: separated Euler solution
  #9
Kaveh Ghayour
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Tom,

There are a few points in your comments that I disagree with:

1) In the derivation of the characteristics of the Euler equations, linearization is indeed done in a very subtle way! Characteristic analysis of a system of nonlinear or quasi-linear (e.g Euler equations) ODE's or PDE's, requires linearization of the nonlinear ot quasi-linear terms. To derive the characteristic directions and compatibility relations, one inserts a wave-like solution in the Euler equations, linearizes the fluxes by writing ( dF/dx = A (dU/dx) ) the jacobians of fluxes are then assumed to be 'FROZEN' (because they are functions of the flow variables too!). Otherwise, you cannot arrive at the well-known eigenvalue problem that is solved to find the characteristic directions, ...

Along characteristic directions a nonlinear ODE becomes a quasi-linear ODE, a quasi linear ODE becomes a linear ODE, and a linear ODE becomes a algebraic system. That's the whole attraction of characteristic method and that is why so much work was done with this method in 50's and 60's.

2) With regard to your question about the treatment of artificial dissipation at the wall: if you are solving the conservative form of the equations, it is preferrable NOT to add any dissipation (nonphysical flux) 'at' the wall so that boundary garbage does not change the balance of conserved quantities!

3) I never knew that anything can prevent me from using a richer function space to obtain better numerical solutions to any PDE! What do you think about high order continuous finite element or discontinuous galerkin research efforts where piles of results are shown for the solution of a (first order) linear advection equation with very high order elements? Don't you think that if it can be done in one numerical scheme, one should be able to mimick it (to a great extent) in other discretizations.

With my best regards,

Kaveh Ghayour.

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Old   October 25, 2002, 14:40
Default Re: separated Euler solution
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Tom
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Kaveh,

Strictly speaking you are not linearizing when introduce characteristic directions - linearization would imply that the results would not yield exact solutions to the pdes which they do! A characteristic is a curve in space-time along which the pde becomes degenerate and reduces to an ode. It's this degenaracy condition which yeilds the "well-known eigenvalue problem" along with the other well known result that the initial data must not be prescribed along a characteristic. See, for example, the book (Ch.2)

Partail differential equations by P.R. Garabedian

for an explaination of this geometric argument. (I actually do know the literature on pdes! I've had a Ph.d in Appliend Mathematics for over 10 years).

For your other point. As long as the solution you are seeking lies within your "richer" function space there should be no problem. All I'm saying is that it is quite common for people to use schemes derived for the NS equations to solve the Euler equations and then have problems with the bcs whereas a scheme derived solely to solve the Euler equations will not have this problem. Conversely consistent scheme for the Euler equations will have problems incorporating the diffusion terms of the NS equations. (The Euler equations are hyperbolic implying real characteristics while the NS equations are parabolic so that the only real characteristic is time).

Tom.
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