During iterations running the turbulent model I am regularly getting the report turbulent viscosity limited to viscosity ratio of 1e5 in xxx cells. I cannot find in the Fluent manual what the viscosity ratio is. Is it simply the definition of viscosity ie ratio of shear stress to shear rate? What does this warning signify in the model, are there common causes which lead to this problem?

Hi.

It is a turbulent/laminar viscosity ratio, and you can set the limit in Fluent.

Have you scaled your model ? Sometimes that helps, if forgotten.

Christian

what do you mean by "scaled you model" ? Could you shed some more light on that?

Read Chapter 6 for info. on turb. viscosity ratio. "scaled your model" means the units in Gambit grid file should match with those to be modeled using Fluent. It can be done using foll. commands....

In Fluent Read the .msh file then, in Grid menu, click on "Scale" & select proper option of units i.e. meters or cm etc.

Sharad

I have scaled the model so I am assuming this is not the problem as my model is the physical size I expect it to be. Is the default setting of 1e5 something that is sensible to change so that Fluent isn't limiting the ratio, or is this usually an indication of another underlying problem in the model. Thanks for your help so far.

Phil

Hi.

As Sharad says there are some theory about the ratio in the Fluent book.

If you have a domain where you have very large and very small velocitys, then I would not be surprised if you get a high ratio. But I would not expect a high ratio in incompressible cases. I usually only have a high ratio in the beginning of the calc.

Christian

The viscosity ratio correlates with Reynolds number. Both are measures of "how turbulent is the flow ?". With large dimensions or velocities and a low viscosity fluid (e.g. water) it is quite expected to get ratios higher than 1e5 (I have seen >1e6 in a converged solution in thousands of cells at Re~1e7).

Joakim

i have the same question. in one of my case.the vilosity is high to 40(m/s), what shall i do with the viscosity ratio. it divergence quickly!

I also have the same problem in modelling of water flow. If i change the the default limit of viscosity ratio to higher value will that have significant effect on my result? How does it affect? Will that be more accurate than if i limit it to 1e+05?

In my point of view turbulent viscosity ratio isn't a physcical value of any kind for the fluid. This term arise from the treatment of turbulence in most turbulence model which are based on bousinesq hypothesis.

It only shows the adequacy of your turbulence model with your kind of flow. In most real flow, turbulent viscosity ratio may not exceed 500-1000.

If you have more than 1e5 this is the sign of a problem : poor grid discretisation, poor convergence, ill posed BCs, turbulence overproduction due too flow feature,...

In FLUENT this default value is set in order to avoid divergence in this kind of situation. To put it at higher value won't correct the problem.

At last, the turbulent viscosity is the viscosity used by the solver to calculate his momentum and heat conservation law. So if you have a value of 1e5 you are not dealing with water but with cement !

Best regards

Alain

Check your turbulence quantities at your inlet BC's. You can dial in a viscosity ratio if need be. If you are using values of k&e that are "out to lunch", you can get viscosity ratio errors.

Hi, My experience is that it is not uncommon to get the limitation of viscosity for the first few iterations in Fluent. However, I had a scenario when it converged nicely for 100 iter and then gradually got the viscosity limited in more and more cells and finally diverged. This turned out to originate from poor resolution. Two walls where fairly close and had too coarse surface mesh, giving only one or a few volume cells between them. After remeshing it converged very nice. So, my suggestion is to stop the simulation when it starts saying 'limiting viscosity ratio in ..' make isosurfaces of it and check your mesh in the region wiht high values of viscosity ratio.

Regards Anders

how can i know the viscosity ratio for special parts

This is simply not true! Viscosity ratio can be large. E.g. in mixing vessels the measured value for k near the tip of the impeller is about 0.1*Utip^2, where Utip is the speed of the tip of the impeller in m/s. Epsilon at the same location is about 3*k^1.5/D, where D is the radius of the impeller. As we all know the turbulent viscosity is µt=rho*0.09*k^2/epsilon. Using the above equation we get µt=rho*0.03*k^0.5*D=rho*0.009*Utip*D. From here we can see that µt/µ is proportional to Reynolds number (Re=rho*N*D^2/µ ~ rho*Utip*D/µ, where N is rps) and there is no limit to the viscosity ratio. For a typical case of rho=1000 (water), Utip=5 m/s and D=1m we get µt=1000*0.009*5*1=45 Pas and µt/µ=4.5E4 (µ=0.001, water). Typically the maximum value of µt is not near the tip of the impeller and Utip and D can be much larger (e.g. propellers of the ships, turbines in hydropower etc.). Similar calculations can be done for pipe flows, but I don't no the exact behaviour of epsilon in that case.

Joakim

Dear Joakim,

in certain point of view you are right.

I didn't want to discuss "what is the good value for turbulent viscosity" since it is often pointless. I just wanted to point out the lack of physical meaning of turbulent viscosity.

In standard k-eps, turbulent viscosity can have any value, and this is one of his weakness ; turbulent viscosity is high expecially in strongly anisotropic and/or non equilibrium flow like you can find in turbomachinery application, rotating flow, impinging jet,.... In this cases it is also well known that k-eps model would give poor accuracy.

regarding your demonstration, I don't fully agree with your formulas. I use these ones :

k = 2/3 (U I)^2

eps = Cmu^3/4 k^1.5/l

with I turbulent intensity which near 10 - 20% at a mixing vessel impeller tip and l turbulent scale.

This may give you a different value for mu_t depending on your turbulent scale (i don't think it can be equal to the radius of the impeller) and the value of C_mu.

The formulas I gave are based on _measured_ (LDA by e.g. Ranade & Joshi and many other groups) values. E.g. for 45 degree pitched blade turbine k is 0.06*Utip^2 at Re>~10000. For a Rushton it is higher. With your formula it would be much less.

Actually standard k-epsilon works very well in a baffled mixing vessel. The k-values are quite close to measured ones and the integral of epsilon is close to power of the impeller. The "better" models such as RNG k-e and RSM give considerably worse results (~ order of magnitude too low values for k).

It is true, that k-epsilon can overpredict µt, but µt/µ is not limited in real world.

Joakim