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raj calay April 20, 1999 10:03

CFD of turbulent flows
Please help with my understanding of this aspect of CFD. Flow becomes turbulent at some Re number. If we donot use any turbulent model and just solve N-S =ns, or using a laminar option. I know that the solution is not a physically true one. But is there any numerical difficulty also in obtaining a laminar solution for a flow which is in fact a turbulent. In other words does the equations (or computer) know that this is a turbulent flow and refuse to provide an answer or is it possible to get a converged solution.

John C. Chien April 20, 1999 11:09

Re: CFD of turbulent flows
In some cases, if you use a constant eddy viscosity model in the computation, the equations will be the same as the laminar flow but the solution is turbulent flow. So, in the turbulent flow computation, we are talking about " modeling", that is you can even model it as if it is laminar. In this case, the flow is turbulent, but the eddy viscosity is constant. In most cases, the eddy viscosity is not constant in the flow field. In general it is variable viscosity flow.

raj calay April 20, 1999 11:49

Re: CFD of turbulent flows
Well I get your point. But aren't we using 'a model' (eddy-viscosity) to solve a flow condition. My question is more about the mathematical or numerical difficulties that may occur if we solve a turbulent flow condition without any turbulent model. Do we get a solution or not. It may not be physically correct one. I ask this question because someone is finding problems with convergence. As a first guess he is solving it without a turbulent model although flow is turbulent. One of our friend says that he will get a solution and he has to use turbulent option. How the two things are connected. A Solution for a set of N-S equations and turbulent model.

John C. Chien April 20, 1999 12:11

Re: CFD of turbulent flows
(1). There is no mathematical difficulties is using the basic transient Navier-Stokes equations because even the Reynolds averaged equations are derived from these equations. (2). So, the problem is numerical difficulties, namely, the spatial resolution and the long-time accuracy. Direct simulation is a transient solution, so there is no steady-state solution. (3). Direct simulation is a research field which requires a lot of computer resources. I don't know whether the solution obtained is real or not, because I am not involved in this particular field. But some results are being used to guide the turbulence modeling.

David Creech April 20, 1999 13:05

Re: CFD of turbulent flows

Yes, you will have convergence problems when you try to solve turbulent flows with N-S equations on a coarse grid.

The turbulence model effectively adds viscosity, making the solver more robust and avoiding divergence.

David CFX User Subroutine Archive

Patrick Godon April 20, 1999 13:14

Re: CFD of turbulent flows
Hi there. The Navier-Stokes equations (i.e. density eq. and 3 velocity or momentum eqs., including a term for the dissipation of momentum due to the viscosity) are non-linear equations and depending on the conditions (initial, boundary, forcing, etc...) these equations can lead to chaos. In fuild dynamic this chaos is called turbulence. This means that (again under certain circumstances) if you perturb the laminar flow, the perturbation can grow (linear analysis). The simulations reveal that usually the grow of the perturbation saturates itself to some limit. Though the flow is turbulent (or chaotic if you prefer) there are some quantities which are perticular to the turbulent flow: the Kolmogov Spectrum, vortices in two-dimensional flow (inverse energy cascade), and so on. So in general, the answer is YES, the equations know very well that the flow has to be turbulent. As to the numerical issue, assume that you are solving for a turbulent flow (that is to say equations whose solution is unstable to a perturbation that leads to turbulence) using just a hydro code that solve the Navier-Stokes equations. This solver has on one side a physical viscosity which appears in the momentum equations (a second order in space diffusive term that has been introduced explicitly) and it has on the other side some numerical diffusion due to (for example) the fact that you are using a simple low order scheme to represent the derivatives. In addition, there might be in the code an artificial viscosity, which has been introduced into the code (for example) to smooth out strong shocks (in order to avoid two-point oscillations due to the shock). So in short the code has some numerical as well as viscous dissipation (the numerical one also depends on the resolution - number of grid points). Now when the flow is perturbed, there are two 'tendencies' in the flow. On one hand the perturbation wants to grow at a given 'growth' rate (which depends on the modes, high-order modes are growing faster than low-order modes for example) and on the other hand the diffusion in the code is trying to damp (reduce) the grow of the unstable modes (usually the high order modes are damped first). Whether the flow will eventually (in the simulations) be unstable depends on this competition between the natural growth of the instability and its damping. If the inverse Reynolds number (or just the numerical diffusion of the code) is too large then the flow is stable. Otherwise it is unstable. Now the main issue is how the code will be able to handle this instability. If the code is able to handle this, then YES for sure with a usual Navier-Stokes code you can solve for a turbulent flow. In this case usually the dissipative process is too strong and the code cannot resolve the fine structure of the flow (the high-order unstable mode are damped first by the diffusive process). In order to remediate to this problem people have written code especially to solve for turbulent flow (e.g. high-order spectral methods with little dissipation are used together with an 'hyper'viscosity). The other limit is when the code is unable to handle the turbulence because it has too little dissipation. Then the high-order modes grow too fast and what you basically get is some-kind of two-points oscillations in the solutions (here you can say that the code does not converge). By the way there is also a very important point here: a code that has too little dissipation can 'a priori' look like it is unstable and turbulent, because the high order modes are growing due to numerical (truncation) errors (and so on..). So if you obtain a turbulence with a 'regular' Navier-Stokes code it would be a good idea to check the Energy Spectrum of the flow to see how you handle the high order modes, etc.. I hope this helps somehow. PG.

Rasputin April 21, 1999 03:19

Re: CFD of turbulent flows
Great answer Patrick, I always thought that the equations 'knew' when to be turbulent. A non-linear system tends to become chaotic when some 'forcing' factor goes above a critical threshold. Are you saying that the BCs in the CFD model are the forcing factor and that the systems of equations becomes unstable when, say, the inlet velocity/momentum becomes too high?


abdul aziz jaafar April 21, 1999 05:47

Re: CFD of turbulent flows
There is a paper of Morse A.P, for example, that support John opinion

A. P. Morse , 1991 Assessment of laminar turbulent transition in closed disc geometries J. Turbomachinery vol 113 pp 113-138


raj calay April 21, 1999 06:19

Re: CFD of turbulent flows
Thanks Patric for a great reply. Do you mean that if we try to solve steady state form of N-S equations (d/dt terms are zero) for a condition which we know fall in the turbulence regime (i.e. high Re)we will not get a converged solution. Will the solution oscillate or diverge?

abdul aziz jaafar April 21, 1999 07:30

Re: CFD of turbulent flows
Dear Raj

For the flow on rotating disc for example, at certain speed condition, the flow will be laminar at lower radius and turbulent at larger radius. I tried laminar computation using our group spesific code and commercial code, the solution is converged( satisfied the conservation mass, and momentum ) but when comparing with the existing experimental data, agreement is better with turbulent flow computation

For some rotating disc systems, where the flow could be unsteady, solving the steady state equation could end with solution residual oscillating as interation number increasing and this could give different result for each iteration number.

See for example (Wilson, M., Arnold, P.D., Lewis, T.W., Mirzaee, I, Rees D.A.S. and Owen, J.M., 1997,Stability of flow and heat transfer in a rotating cavity with a stationary outer casing, Eurotherm 55 - Heat Transfer in Single Phase Flows 5, IMechE HQ, Santorini, Greece.)


andy April 21, 1999 08:26

Re: CFD of turbulent flows
As is often the case with CFD, it depends on how the physics is represented in the simulation. For example, for no explicit turbulence model and a high Reynolds number:

(1) First order upwind differencing for the convection terms. This is about the most robust scheme you can use and if you are having problems starting up a turbulent calculation using this to get a reasonable initial flow field can be helpful. It may be a bit slow to converge if the flow is strongly aligned with the grid. However, the solution will generally be a poor representation of the high Reynolds number laminar solution.

(2) High order differencing for the convection terms. A steady state laminar solution is almost impossible to obtain with most numerical schemes and is physically unreasonable anyway. However (subject to appropriate boundary treatment and some numerical considerations) a good representation of the unsteady turbulent motion can be obtained.

If you want a steady-state solution for a high Reynolds number problem then using a turbulence model will introduce (modelled) Reynolds stresses for the mean flow to work against (work with in only a few regions for the more sophisticated models). This extra dissipation will steady down the predicted flow and, more often than not, lead to a steady solution but it is problem dependent. For example, it is very difficult to obtain a steady solution for the flow over a cylinder by inhibitting the vortex shedding.

Patrick Godon April 21, 1999 08:27

Re: CFD of turbulent flows
I am looking at the problem in general, not at a specific problem. A system can be unstable locally or globally. Locally means that it is unstable anywhere in the flow when you make a local approximation. Globally means that the flow is unstable when you consider the whole system, and this includes the boundary conditions. Some non-axisymmetric centrifugal instabilities appear in some two-dimensional flow ONLY when one of the boundary is reflective and the unstable modes grow because waves can be trapped between the reflective boundary and an evanescent region in the flow where waves can also be trapped. This is why I mentioned the boundary conditions as an additional factor. In the particular case that I know it is not when the velocity at the boundaries is too high but just when the boundary is reflective. AN expamle of such an instability is for example given in Tomasini, Dolez, Leorat, J.Fluid Mech., 306, 59 (1996). Another example (and there it is called the Papaloizou-Pringle instability) is given in Glatzel 1989, J.Fluid Mech., 202, 515. Gat and Livio, (1992, the astrophysical journal, 396, 542) showed that it is due to a reflective boundary conditions. PG.

Patrick Godon April 21, 1999 08:40

Re: CFD of turbulent flows
I must admit that I never tried to solve for steady state flows, but I always solved for time dependent solutions. Also the linear analysis (i.e. the perturbation analysis) is always carried out in the time dependent equations, since you want to know if the perturbation grows with time. Now if you assume from the start d/dt=0 and try to solve the system of equations, my guess is that it is not consistent with the solution and therefore I would expect that your scheme will not converge, but most probably diverge. Since I have not much experience with steady state equations (d/dt=0 from the start) I cannot say too much about it. The insconsistency comes from the fact that you try to put zero where it should not be (and in all the equations at the same time), and there is no solution for such a new system of equations, this is why I guess it would diverge. If you were solving the time dependent equations, you would get a laminar flow that would eventually get unsable physically and then numerically (divergent with time). PG.

raj calay April 21, 1999 10:16

Re: CFD of turbulent flows
It is becoming really an interesting discussion. I understand turbulent flows are not steady i.e. But even when we model as transient we truncate our simulation at certain time T. Correct me if I am wrong I have dealt because mostly with diffusion type problems, if the BC's donot change with time and then system will eventually asymptote to a steady state solution. Theoretically we should be able to get a similar situation statring off with steady-state form of equations (i.e. when T approaches infinity d/dt becomes zero. we can set this term as zero at the start). With diffusion type problems it works. Unless we are interested in how a system develops w.r.t time steady state approach works. I think same should work with convection type problems.

raj calay April 21, 1999 10:27

Re: CFD of turbulent flows
It is becoming really an interesting discussion. I understand turbulent flows are not steady i.e. But even when we model as transient we truncate our simulation at certain time T. Correct me if I am wrong I have dealt because mostly with diffusion type problems, if the BC's donot change with time and then system will eventually asymptote to a steady state solution. Theoretically we should be able to get a similar situation statring off with steady-state form of equations (i.e. when T approaches infinity d/dt becomes zero. we can set this term as zero at the start). With diffusion type problems it works. Unless we are interested in how a system develops w.r.t time steady state approach works. I think same should work with convection type problems.

With flow problem as Andy mentioned about vortex shedding which is a transient phenomenon. Then the transient solution represent snap-shots of flow at different times that is equal to the chosen time step dT. I insist that if your BC are not changing with time you should reach a steady state (numerically). OR if you have a time-dependent BC's then solution is dictated by the time function of the boundary conditions. e.g. Periodic boundary condition is bound to give a periodic solution.

In reality we don't achieve steady-state because boundary conditions change with time.

John C. Chien April 21, 1999 13:49

Re: CFD of turbulent flows
(1). The steady-state solution of the steady-state equations is always possible. Even for the high Reynolds number flows, the equations can be reduced to Euler equation or potential flow equation. Mathematically, the steady-state solution can exist at any Reynolds number ( if you also consider Euler equation as the Reynolds number reaches infinity.) (2). But, in reality, it is hard to position an egg in its upright position. It is generally called the stability problem. (3). One way to obtain the steady-state solution for flow over a cylinder is to use symmetry condition. In this way, you forced the solution symmetric. But, still there is no guarantee that you can easily obtain the steady-state solution because of the iterative nature of the numerical solution procedure. (4). The other example is the fully developed laminar flow in a tube. The exact analytical solution exists, but at high Reynolds numbers the flow becomes turbulent. ( I am not sure whether it is easier to keep the flow laminar in outer space or not.) (5). I think, some people were able to avoid this stability problem by simply solving the Euler equations.

Duane Baker April 21, 1999 22:19

Re: CFD of turbulent flows
Hi Raj,

This certainly has become a very interesting discussion especially for those who are slanted toward non-linear dynamics.

FOR NONLINEAR PROBLEMS: unsteady solutions can result from steady boundary conditons. Similarly, assymetric solutions in space can result from symmetric domains and boundary conditons. This is why imposition of a symmetry boundary condition reduces a spacial mode when the domain is symetric! This all comes under the topic of symmetry breaking. A good intro reference is Seydel's (sp?) "From Equilibrium to Chaos"

In the instance of CFD modelling of flows the transient laminar flow will capture the scales of unstedyness that it can based on the time and spacial resolution.

1. IF THE COMPUTATIONAL SCALES ARE FINE ENOUGH then you have a DNS solution and is belived to be correct!

2. IF THE COMPUTATIONAL SCALES ARE NOT FINE ENOUGH you have a truncated resolution and is wrong unless some model for the unresolvable terms is given. This is done in DNS. (This is a general pronciple in all of continuum problems that information is lost in an averaging process and must be supplied in some form of a model eg. integral methods, porous media, continuum approach to molecular scale processes, turbulence, etc.)

3. If you use the RANS eqns then the modeled turbulence scales appear as a diffusive flux of momentum which is typically two orders or more greater than molecular diffusive fluxes. The result is the time scale of any resulting unsteadyness is also drastically increased and in most cases a steady solution may be obtained by solving the insteady RANS equations.

Regards........................................... Duane

rick April 21, 1999 23:03

Re: CFD of turbulent flows
It is really an interesting topic. I absolutely agree with Patrick's opinon, the equations know the flow should be a turbulent one. with the large Re. number, the effects of the non-linear convective terms become great. regardless of the reality or the physical perturbation, the error in the numerical process is enough to cause the unstability or the divergence of the solution. that is to say enev with the symmetric boundary condition, with the infinity time, we can not get a stable unique solution under large Re. someone has numerically simulated a flow in an enlargement tunnal, the boundary condition and the geometry was surely symmetric,( sorry I can not find the paper on my hand), when the Re was about 400, the flow will be a unstability.

raj calay April 22, 1999 04:41

Re: CFD of turbulent flows
I agree. At high Re i.e. the non-linearity increases which induces numerical instabilities. The solution may oscillate or diverge depending upon the scheme used to dampen the instabilities. But I think John is correct in saying that it is possible to obtain mathematical solution of steady-state equations. Problems arise with numerics which also include machnine tolerance etc. I am puzzled by how the numerical instabilities or unsteadiness is related to the physical instabilities or turbulence. I would be nice if you let me have the details of the reference you mentioned.

rick April 22, 1999 05:37

Re: CFD of turbulent flows
I dont think the numerical instability and the physical instability are the same thing. i just say one could get asymmetric result under symmetric structure. ok, i will try to find that paper. another example, with the CFD of the senonsis artery, which is a axial-symmetric structure, around 1000 Re, someone got a non-symmetric result. i will try to find the details.

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