# Euler + separation again

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 May 29, 2001, 11:04 Euler + separation again #1 Oliver Guest   Posts: n/a 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

 May 29, 2001, 17:29 Re: Euler + separation again #2 phuang Guest   Posts: n/a IF you run an Euler code over a step, you will have separation too.

 May 29, 2001, 19:46 Re: Euler + separation again #3 Adrin Gharakhani Guest   Posts: n/a >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

 May 29, 2001, 20:01 Re: Euler + separation again #4 John C. Chien Guest   Posts: n/a (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.

 May 30, 2001, 11:08 Re: Euler + separation again #6 kalyan Guest   Posts: n/a 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.

 May 30, 2001, 17:27 Re: Euler + separation again #7 Peter Guest   Posts: n/a 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.

 June 1, 2001, 19:18 Re: Euler + separation again #8 clifford bradford Guest   Posts: n/a 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.

 June 1, 2001, 20:31 Re: Euler + separation again #9 Bernard Parent Guest   Posts: n/a 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.

 June 4, 2001, 10:35 Re: Euler + separation again #10 kalyan Guest   Posts: n/a Is there any vorticity generation in an expansion fan created by a convex corner in supersonic flow. The entropy is uniform in this case.

 June 4, 2001, 11:21 Re: Euler + separation again #11 andy Guest   Posts: n/a 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?

 June 4, 2001, 12:26 Re: Euler + separation again #12 Bernard Parent Guest   Posts: n/a vorticity can only be generated through a gradient of entropy..

 June 4, 2001, 13:42 Re: Euler + separation again #13 Adrin Gharakhani Guest   Posts: n/a > (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

 June 16, 2001, 05:43 Re: Euler + separation again #14 clifford bradford Guest   Posts: n/a 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

 June 16, 2001, 12:25 Re: Euler + separation again #15 John C. Chien Guest   Posts: n/a (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).

 June 18, 2001, 11:12 Re: Euler + separation again #16 kalyan Guest   Posts: n/a 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.

 June 18, 2001, 15:57 Re: Euler + separation again #18 John C. Chien Guest   Posts: n/a (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.

 June 18, 2001, 16:51 Re: Euler + separation again #19 kalyan Guest   Posts: n/a 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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