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Old   May 29, 2001, 11:04
Default Euler + separation again
  #1
Oliver
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First I'm sorry to start that topic again. I have searched the forum for this topic and I thought I know the answer but everybody at my work tells me I am wrong. I am doing Euler calculations (4th order TVD Jameson) of a turbine row in an off-design point and I get separation and/ or voticities. So, I say the solution is wrong and has nothing to do with reality or the inviscid model. They say it's ok because it looks like reality. They say it is suffient to have a proper pressure gradient for separation and wall friction is not needed. One argument is that there is a separation at the trailing edge because of the thickness of the trailing edge and that the stagnation point is not defined by Kutta condition but "found" by the program itself. They say if I am right the stagnation point should be on the suction side. But it is on the trailing edge and there are vorticities because of the roudness of the trailing edge and so they are right (they say). They say it is possible to have vorticity because of the radial equilibrium. They say Re is infinite, does it make sence to argue with Re, if viscosity is zero? They say that I mix potential flow with Euler, and only in potential flow there should be no separation. I know this sound very confused, but I am totally confused! So what is the right answer and do you know any good arguments to proof this? Thanks in advance
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Old   May 29, 2001, 17:29
Default Re: Euler + separation again
  #2
phuang
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IF you run an Euler code over a step, you will have separation too.
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Old   May 29, 2001, 19:46
Default Re: Euler + separation again
  #3
Adrin Gharakhani
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>IF you run an Euler code over a step, you will have separation too.

Separation, yes - maybe, but flow reversal (recirculation), absolutely not!

As for Euler vs. potential flow; it is true that the two are different; however, a potential flow does satisfy the Euler equations (the converse is not true). You can modify the potential flow by putting vorticity sources and tracking their motion. The main issue is (here I'm assuming incompressible flow and this is not true in compressible flow), where does vorticity come from. The answer is from the solid boundaries and it is created due to viscous effects. Period. So, if there is no viscosity at the wall, no matter how infinitesimally small, there will be no vorticity generation, and if there is no vorticity you have potential flow, period. This is the physics of incompressible flow. For compressible flow, of course you have baroclinicity as a vorticity generation source.

So, again, in incompressible flow you can generate vorticity only at the wall and due to viscous effects. If you are actually seeing vorticity develop in a so called Euler simulation, have you considered the possibility that you have numerical diffusion, which is perhaps sufficient to balance out your pressure gradients and lead to flow separation and reversal?

Adrin Gharakhani
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Old   May 29, 2001, 20:01
Default Re: Euler + separation again
  #4
John C. Chien
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(1). My suggestion would be: (a). try to make your solution mesh independent. In this way, at least you are honest in getting the cfd solution. (b). don't try to convince others in terms of whether the solution is useful or not. Just present the solution, and let them to make their own decisions. (c). If you are still using inviscid equations, it is a good time to look at the possibility of using Navier-Stokes approaches, if feasible in your case. (2). CFD is numerical analysis and mathematical modeling in fluid mechanics. It does not mean that "numerical analysis" must use only the Navier-Stokes solutions. It is all right to use potential flow equation, Euler equation or parabolized equation. (3). In the research field, I think, we are well into the Low Reynolds number turbulence model and Navier-stokes solutions. I know that in turbomachinery industries, people are still using inviscid Euler equation and code, mainly becuase there are codes available at low cost, and Euler codes are relatively fast to produce solutions. (4). So, instead of focusing on the Euler solution, you can move on to a level higher and make your own comparison between solutions from inviscid and viscous equations.
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Old   May 30, 2001, 09:56
Default Re: Euler + separation again
  #5
Bernard Parent
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There were good comments made by Adrin and John, but I will add my own views on the subject.

(1) As pointed out by Adrin, for incompressible flow, you need viscosity to create vorticity. But I will add that the same is true for compressible flow. It can be shown through a modified form of Crocco's theorem that vorticity can only originate when a gradient of entropy is not aligned with a gradient of temperature. In other words, if the entire flowfield shares the same entropy level, the vorticity will always remain zero. For an entropy gradient to occur, an irreversible phenomenon must also take place, and irreversible phenomena are necessarily linked to viscous effects (a shock is a viscous phenomenon, although the states before and after the shock can be predicted through the integral form of the inviscid equations, which leads to some confusion).

(2) A vortex does not necessarily show high levels of vorticity (rotational versus irrotational vortices). Take the flow part of a cyclone for example: the vorticity will be very high near the inner core, but will vanish at a certain radius, at which point the flow is irrotational, although the flow follows a circular path around the vortex core. It is possible to have vortices while maintaining a zero-level of vorticity: water going down your bathtub is a good example of an irrotational vortex.

(3) With the compressible CFD methods I know of, it is impossible to obtain the exact solution to the Euler equations even for an infinitely refined grid: artificial dissipation is always present in the convective flux discretization scheme, and never vanishes. When tackling the Navier-Stokes equations, the physical diffusion overtakes the artificial dissipation when the mesh is refined sufficiently and hence it is possible to obtain the exact solution. The same is not true for the inviscid equations, and some errors will always be present in your solution. This does not mean that a grid refinement study is futile, but it should be taken with a grain of salt. As the grid is refined, you will probably notice the vortices to diminish in size in certain regions, but some fictitious gradients (at the surfaces, for example) might always be present, and might even be amplified with a refined grid..

To conclude, I would recomment not solving inviscid flow with CFD, but rather tackle directly the full NS, and perform a capable grid convergence study.

good luck.

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Old   May 30, 2001, 11:08
Default Re: Euler + separation again
  #6
kalyan
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Oliver,

I have a question regarding the need for entropy gradients to generate vorticity. If you use incompressible, variable density equations and equilibrium chemistry to model flames, is there no vorticity generation no matter what the initial shape of the flame is. In this case, the entropy gradients would be zero becauase of the equilibrium chemistry.

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Old   May 30, 2001, 17:27
Default Re: Euler + separation again
  #7
Peter
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Some codes imposed the Kutta condition at the trailing edge so that it is not possible to flow have flow coming from pressure to suction side. If your code "find" the stagnation point by itself, some recirculation it is possible.

If you are using a viscous model, the radial equilibrium can explain some vorticity and radial flows that occur near the wall surfaces (due to increment of the flow absolute velocity near the wall for rotating rows, the flow get extra radial velocity directed to the hub in turbines). But if you are running Euler, and extrictly Euler with non zero velocity on the wall, there is no increase in tangential velocity and therefore radial equilibrium cannot explain the vorticity.

Are you sure you are running a strictly Euler code???. In industry it is very common to use "zonal techniques". They compute with the inviscid model far from the wall, together they compute the integral boundary layer equations in the proximity of solid surfaces. They get close to reality solutions with very reduced computational time (and getting even boundary layer displacement thickness). If your code is one of this type, you can get separation without any problem.
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Old   June 1, 2001, 19:18
Default Re: Euler + separation again
  #8
clifford bradford
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The situation you posed can occur in inviscd compressible flow if there is a curved shockwave. This can be shown by Crocco's equation. I think also that certain types of body forces can cause vorticity in inviscid flow (which can also be shown by Crocco's equation).

Ah that Crocco's equation. Unfortunately I don't have my notes with "the nothing is assumed" derivation from grad school.
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Old   June 1, 2001, 20:31
Default Re: Euler + separation again
  #9
Bernard Parent
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But this assumes that a shock is a solution to the Euler equations. I won't deny that a curved shock wave induces vorticity. But a curved shock wave (and a straight one for that matter) is a viscous phenomenon, and is not more a solution to the differential Euler equations than a boundary layer or a shear layer. What causes the confusion is the fact that the properties after the shock can be obtained from the Rankine/Hugoniot approach, a technique which bypasses the viscous terms by integrating a control volume in which the shock is located. In that sense, a shockwave could be thought of as being a solution to the integral form of the Euler equations (but definitely not to the differential form). In short, while the properties after the shock can be estimated in the inviscid world, its solution can only be obtained using the full equations of motion including all viscous effects.
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Old   June 4, 2001, 10:35
Default Re: Euler + separation again
  #10
kalyan
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Is there any vorticity generation in an expansion fan created by a convex corner in supersonic flow. The entropy is uniform in this case.
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Old   June 4, 2001, 11:21
Default Re: Euler + separation again
  #11
andy
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Consider the flow over a backward facing step with a uniform inlet condtion.

(1) The Euler solution is for everything to keep going in a straight line (so long as the boundary condition treatment does not interfere). A line of infinite vorticity then exists separating the jet and the stationary region behind the step. Note that vorticity is present at the boundary and needs to be admitted by the boundary conditions.

(2) Potential flow does not allow vorticity and so the flow has to spread around the step giving an unreasonable solution (for high Reynolds Numbers) that looks something like creeping flow.

Of course, in reality your solution for the flow over an aerofoil will contain some numerical diffusion and this will entrain mass into the boundary and free shear layers leading to some recirculation. The reversed flow is predominantly a numerical artifact but the separation is not. Does that make it a draw?

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Old   June 4, 2001, 12:26
Default Re: Euler + separation again
  #12
Bernard Parent
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vorticity can only be generated through a gradient of entropy..
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Old   June 4, 2001, 13:42
Default Re: Euler + separation again
  #13
Adrin Gharakhani
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> (1) The Euler solution is for everything to keep going in a straight line (so long as the boundary condition treatment does not interfere). A line of infinite vorticity then exists separating the jet and the stationary region behind the step. Note that vorticity is present at the boundary and needs to be admitted by the boundary conditions.

This is pretty much what I implied in my earlier response: you may get "separation" at the edge of the step and thus develop a shear layer in the expanded region, BUT you will NOT see a recirculation bubble forming at the corner. For this to occur you will need no-slip boundary condition (and thus viscosity) at the walls.

Adrin Gharakhani
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Old   June 16, 2001, 05:43
Default Re: Euler + separation again
  #14
clifford bradford
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I get your point. I have two questions/comments:

(1) do you regard the weak solution (to the integral equation) to be somehow inferior to the the differential solution? personally I do not since we have more mathematical knowledge of the former than the latter. Moreover the former is more general because it's derivation requires no assumptions about the continuity of the flow field. I realise that shock wave is the Euler equation's weakness and we ought not to give too much importance to it's nature as practical fluid dynamicists (engineers).

(2) if I were to predict a curved shock using the method of characteristics in 2d (which would allow me to avoid the pesky viscous effects due to to artificial viscosity) would I still not see the production of vorticity?

In any case his issue is probably due to not having a kutta condition
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Old   June 16, 2001, 12:25
Default Re: Euler + separation again
  #15
John C. Chien
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(1). I think, CFD is "Numerical Analysis and mathematical Modelling in Fluid Mechanics". (2). And understand the physics of Fluid Mechanics is only the first step. (3). This is the reason why CFD requires the minimum training of a PhD. (in his specialized field) (4). Euler equation is inviscid, non-linear equation. Which can be subsonic, transonic or supersonic. (5). Shock relation is a formula which links the two states based on the control volume(integral) equations. In reality, it can only be used for flow going from supersonic to subsonic flow across a shock. It can not be used for flow from a subsonic state to a supersonic state, even though the formula can still give such solution. (6). This is because something is not real in this relation (shock relation). The physics of fluid is missing. The shock is viscous (that is, you can compute the shock structure by solving the Navier-Stokes equation).
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Old   June 18, 2001, 11:12
Default Re: Euler + separation again
  #16
kalyan
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There are inviscid solutions and there are viscous solutions in the limit of vanishing viscosity. Both need not coincide. Look at the discussion on "entropy solutions" in the book on level sets by Sethian.

Can you give me an example of a flow with a curved shock and no subsonic pockets. If you have subsonic pockets anywhere in the flow (like in the flow with a bow-shock), you can not use MOC.
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Old   June 18, 2001, 14:45
Default Re: Euler + separation again
  #17
Adrin Gharakhani
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I was present at the 15th AIAA CFD conference last week, and heard quite a number of presentations on the new fad "vorticity capturing" method, originally developed by Steinhoff (sp?) about a decade ago. The idea of the approach is interesting; it is a method for adding an "ad hoc" vorticity source term to negate the excessive dissipative effects of the commonly used finite volume methods.

A few nice examples were given. For example, using the standard method, one sees a fairly diffuse vorticity field (all over the place) when simulating flow over a ship deck. But with vorticity capturing added, one sees a very concentrated vortex filament near the edge of the deck, whick persists for quite a long time (in agreement with experimental observations).

OK, so what has this got to do with the current discussion?

In one of the presetations, results were shown using vorticity capturing and INVISCID flow over a backward facing step (and even flow over a cylinder), where the nice recirculation bubble was captured even though this was inviscid and the boundary condition, believe it or not was free slip! So, the "hope" is that they have a method (in the near future) for turbulent wall boundary conditions that is based on vorticity capturing and no "ad hoc" models. Of course, the method has an ad hoc parameter, which when you play with it you can get solutions from fully laminar all the way to infinite Reynolds number and even potential flow (how is this for ad hoc?)

My disappointment with this presentation was that a fairly well-known professor would stand up and claim (without expecting a challenge from the audience) that he can simulate physically viscous effects (no-slip boundary condition) using an inviscid method.

My comment/question at the end was similar to what I already posted earlier on this thread "The claim is that you are solving an inviscid problem, but no matter what you do and how fine a grid you have, you will always have residual amounts of viscosity in the calculaltions; enough for the solution near the wall to act as a viscous solution - perhaps at infinite Reynolds number - do you think it is fair to claim that you capture recirculation with an inviscid code and free slip BC?" His answer was "You are correct!" If I'm correct, then his claim is false, and that false claim is published in papers. Then people who believe that material published in respected journals by respected people must be correct/true or it won't be published, base their entire assumptions for future flow analyses on these false claims...

So, I still maintain, the only reason you are getting viscous effect with an "inviscid" code is because there is no way you can actually solve the inviscid flow problem. If you actually could do that, your solution would blow up because the Euler equations exist for small times (based on what I have read of theories on the Euler equations). You need that minute amount of viscosity to stablize the solution. And that is sufficient for separation, reverse flow, etc. to occur.

Adrin Gharakhani
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Old   June 18, 2001, 15:57
Default Re: Euler + separation again
  #18
John C. Chien
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(1). Some people think that one can add some viscosity to the inviscid equation to make it viscous. (2). Then they have the viscous solutions. (3). If the numerical method can include the viscous effect in the solution, then by adjusting the parameter they can obtain the good solution. (4). The thinking is the same as running a cfd code, if you have a code, and can run the code, then you should have a solution. (5). The fact is: companies are disappearing from the surface of the earth through this process. (6). I heard that long time ago, there were only a dozen people who really understand the general theory of relativity by Einstein. It is impossible to change a person, so it will follow the initial value problem. (I mean, the initial condition of a person was set long time ago. It will follow the predetermined or preprogrammed course.) (7). That's why Charlie Brown used to say "Good Grief!". (8). Anyway, for a flower to look beautiful, there must be a lot of leaves.
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Old   June 18, 2001, 16:51
Default Re: Euler + separation again
  #19
kalyan
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I am a bit surprised by some one who uses Lagrangian vortex methods claiming that Euler equations can not be solved.

I am not sure if your posting was in reply to mine. All that I said was the solution in the limit of vanishing viscosity need not match the inviscid solution. This I believe is true for fluid mechanics as well as other hyperbolic systems (that become parabolic upon adding the viscosity terms). Also, the additional set of boundary conditions (no-slip) need for N-S are not the reason why the viscous solution never reaches the inviscid solution in the limit. The difference also exists even if the flow is in unbounded/periodic domains where both (viscous and inviscid) equations have same boundary conditions.

The finite-time blow up of the Euler equations that you are refering to occurs in only finite domains due to the tendency towards equipartitioning. This does not happen on unbounded domains.

Could you provide the details of the paper that were referring to. This seems like something to be cautious about like you said.
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Old   June 18, 2001, 17:36
Default Re: Euler + separation again
  #20
Adrin Gharakhani
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> I am a bit surprised by some one who uses Lagrangian vortex methods claiming that Euler equations can not be solved

Nothing surprising here. Lagrangian vortex methods can do a _better_ job for solving the "Euler" equations, but better is relative. The method has significantly less numerical diffusion, but it is not 100% free of it; no finite-precision method can be free of numerical diffusion. The issue is whether it is so diffusive, like finite-volume methods, that you essentially solve laminar flow problems (unless you have significant grid density and/or apply grid adaptivity) or it has acceptable level of numerical diffusion so that you can still capture the global details of the vorticity field.

I have noticed that the finite volume people are finally coming around; numerical diffusion and the vorticity field capturing are inter-related (I heard a very interesting talk by Prof. Roe at the 15th CFD conference in this regard and am looking forward to reading about his vorticity capturing/confinement schemes in the future) ... but I digress (a little)

> Also, the additional set of boundary conditions (no-slip) need for N-S are not the reason why the viscous solution never reaches the inviscid solution in the limit.

Agree. And I don't think that is what I was claiming. I was specifically talking about separation, recirculation zones, etc., which I insist are due to viscous effects (in terms of their inception - once the flow is separated the process can be and it quite often is an "inviscid" phenomenon) But, you cannot use an inviscid code with inviscid boundary conditions to separate the flow. I am familiar with the ad hoc vorticity injection techniques at the sharp edges - the assumption here is that the diffusion of vorticity from the wall to the fluid domain has already taken place and we take care of the rest using inviscid codes. This does not contradict my prior statements. The issue is whether you can use the same approach to solve separating flow problems for geometries with smooth boundaries (like circles). The answer is no - you need to know exactly where vorticity has to be shed from. (I hope I'm making sense in trying to link the issue of BC and the viscous solutions obtained by inviscid codes)

> The finite-time blow up of the Euler equations that you are refering to occurs in only finite domains due to the tendency towards equipartitioning. This does not happen on unbounded domains.

I'm not sure about this. Take a simple vortex ring in free space, and perturb it in its most unstable mode and let it go inviscidly. What happens to the ring?

> Could you provide the details of the paper that were referring to. This seems like something to be cautious about like you said.

I'll search my notes from way back when (it will take some time since I've got heavy deadlines to meet, but I'll look for them for you)

(Thanks for the challenging questions)

Adrin Gharakhani
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