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May 3, 2002, 04:46 
Lam Bremhorst

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I have written a 3d model using K epsilon model. I have used LaunderSpalding version successfully but I need to use a low Reynolds number model like Lam Bremhorst. The problem is, it doesnt converge with Lam Bremhorst while it converge with LaunderSpalding. It seems that LamBremhorst converges only in small domain, not big one. Can anybody help me about this problem? I have checked following boundary conditions. d(epsilon)/dy=0 , d(k)/dy=0 at wall also another boundary condition, epsilon=d2(k)/dy2*eta=2k/y*eta , d(k)/dy=0 , thank you


May 3, 2002, 05:49 
Re: Lam Bremhorst

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I read somewhere that LB convergence may be a problem with standard upwind. you can try with higher order UW.


May 3, 2002, 07:21 
Re: Lam Bremhorst

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In the original paper, k is equal to zero at the wall, not dk/dy.
Sylvain 

May 3, 2002, 08:34 
Re: Lam Bremhorst

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Hi,
I haven't had any problems with Lam Bremhorst KEpsilon stability  I'm not sure your wall boundary conditions are entirely correct. However I think your stability problem is more likely due to the relaxation you are using  Did you use the same relaxation for both domains ??? I would suugest that you use a false time step = ~(average time take for fluid to flow though a cell/3)  are you using false time step? 

May 3, 2002, 08:37 
Re: Lam Bremhorst

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May 3, 2002, 12:32 
Re: Lam Bremhorst

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In original paper in page 457, it is written that k=d(k)/dy=0. Only d(k)/dy=0 is applicable to the model and I cannt apply two boundary condition to the k equations at the same time and also k=0 on the wall in staggered mesh is not possible to apply.Thank you


May 3, 2002, 12:40 
Re: Lam Bremhorst

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Have you used an commercial package or you have written a 3d by yourself? If you have written by yourself what do you think about Marjori's message that I should use a high order scheme instead of upwind because I have used upwind. Also I dont think time step is problem, because with Launder Spalding I dont have problem and with Lam Bremhorst I dont see high osilations before crash. The domain which I want to solve has very low Reynolds number which laminar model gives good results(similar to experimental results), but I want to get even better results with turbulence. but by introducing turbulence , it seems that the results become worse.thank you


May 3, 2002, 12:53 
Re: Lam Bremhorst

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Is turbulence intensity very important in the inlet? when I choose small turbulence intensity( about 1%) , I have to decrease time step a lot. I want to use low turbulence intensity, to get lower turbulence in the domain. How much turbulence intensity can change the results? is it very effective?, thank you


May 7, 2002, 05:24 
Re: Lam Bremhorst

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It depends the way you do it. In RANS equation, turbulence acts on the flow throught the turbulence viscosity (nu_t = c_nu*k^2/eps = c_nu^(1/4)*k^(1/2)*l  where l is the turbulence lenght scale).
So if you decrease the turbulence intensity (sqrt(k)/U) without changing the value of eps or l, you decrease the value of the turbulent viscosity. Since the turbulent viscosity plays the same role than the dymamic viscosity in stabilizing the numerical scheme, this could explain why you have to decrease the time step. For the other part of your question, turbulence may change the result, it depends on what you are looking for. For the same value of eps, a greater turbulence intensity increase the turbulent viscosity and then act like diminishing the Reynolds number. On the other side, big velocity gradients increase turbulence intensity so that the inlet boundary condition becomes "invisible". To conclude, if the phenomenum you are looking for depends on the Reynolds number and presents very small velocity gradient, then variation of the inlet turbulence intensity may change the result. Concerning my other post and your answer, I think that dk/dy=0 at the wall is a physical nonsens. The wall vicinity is a region where turbulence is create (see the case of a boudary layer). A zero flux in this region can not modelize the real world. At the opposite, in the log law region, it is well known that dk/dy is equal to zero. Hope that helps, Sylvain 

May 7, 2002, 10:26 
Re: Lam Bremhorst

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Thank you for your reply.If I want to use LamBremhorst model, should I have a fine mesh near the wall that goes even to the linear sub layer?, I ask this question,because by using this model all f1 and f2 and fm coefficients are always equal unity and show that this model is equal to Launder Spalding. The domain which I want to model has low Reynolds number (about VR/eta=800 to 8000). from LamBremhorst model I understand that f2 is only important when the turbulence viscosity is very very low and about 10^7 which seems totaly unreasonable. I see values of turbulence viscosity in the order of about 10^4 during calculations which is very higher than 10^7. thank you


May 7, 2002, 10:55 
Re: Lam Bremhorst

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Hi,
just to answer a couple of points, I had no problems using standard upwind. I used K = 0.0 at the wall I used turbulence intensity of 1%  The value you choses depends on the case you are trying to simulate  If you pick a value that is too large then it tends to drop to the correct vlaues fairly rapidly downstream of the inlet. With all low reynolds number models it is adviable to get at least 1 grid point in the viscous sub layer (Yplus <25) Note for KEpsilon Standard high Reynolds number version keep the first cell out of the viscous sublayer!!! You suggest you have a turbul;ent viscosity of 1E4 and a laminar Visc of 1e7? Giving a ratio of 1000 times  this is a little high though not impossible  is 1e4 the max value you have in the domain ? I still think you are suffering a stability problem due to unsuitable relaxation  use false time step (false DT = average time for flow to pass through a cell/3) What type of relaxation are you using ?? good luck Tim :> 

May 8, 2002, 08:05 
Converged!

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Hi, Tim. Thank you for your reply which was clear and good. my model converged with following conditions, k=0 in boundary or k boundary= k inside. Also it was needed to give turbulence viscosity in boundary = inside turbulence viscosity. Now I have some graphs,but they are not similar to experimental model. infact laminar model gives better results than turbulent. Should I change coefficients f1,f2,fm in Lam Bremhorst model to get better results or they are universal. I ask this question because in LB paper, I read that they have found these functions by trial and error in modelling of flow in pipes. My modelling is in a open serpantine canal form. Also I dont know, how can you make new paragraph in your messeges?, Thank you


May 9, 2002, 11:51 
Re: Converged!

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Hi,
I wouldn't worry about the f1,f2,fm constants just yet. How does the Lam Bremhorst model compare with standard ke model ??? what Reynolds number in the canal you are modelling ? I think the main things to check are the level of convergence ( how many orders of magnitude do your residuals drop?) and grid dependancy ~( how many cells are you using and what happens when you increase the number?) Tim. 

May 9, 2002, 14:57 
Re: Converged!

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hi,Tim, thank you for your reply. Graphs of LB and standard kepsilon methods are very similar(velocities in the domain). Actually I am not very surprised by this, because f1,f2,fm are usually unity or very near to unity and therefore LB is equal to standard kepsilon. the only difference between them is in boundary condtion which in standard form considers loglaw profile(k=Ut^2/Cmu...). Reynolds number in the domain are very law about (800 to 8000). The convergence is good and ((eta(next step)eta(prev. step))/eta(prev. step)<0.001) in convergance. the grid size is good and gives good results(good agreement with experimental results) for laminar flow. I can send some information by email to you (if you like). I have given my email adress. If you send an email to me I will send you all informations(if you like). I dont understand f2 function in LB model. it should be unity even in the finest possible grid near wall, becasue if Re>2 , f2 will be unity. even in laminar flow we have Re about 400 or 500. I dont understand the usage of this function in any kind of flow. it is worthless. completely unreasonable. I also dont understand the physical meaning of turbulence viscosity outside doamin= turbulence viscosity inside for convergance. Thank you


January 27, 2011, 05:53 

#15 
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Robert Sawko
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Hello,
I just wish to add two comments. I also encountered problems with stability but they were alieviated by adjusting time step. Thanks. I'd like to ask about the criticism of dk/dy=0 as boundary condition. BC for turbulence quantities all come from the expansion of Taylor series of fluctuating velocities in the vicinity of the wall, application of wall BC and substituting back to expressions for k and epsilon. When you do it you will see that dk/dy = 0 and epsilon = nu d2k/dk2 at the wall. You will also get k=0. Honestly, I can't see why d(epsilon)/dy = 0. I understand the first statement as: k vanishes at the wall faster than linearly. Actually it vanishes as y^2. Does this give it more of a physical sense? Also the following paper Hrenya, C. M., E. J. Bolio, D. Chakrabarti and J. L. Sinclair, "Comparison of Low Reynolds Number ke Turbulence Models in Predicting Fully Developed Pipe Flow," Chemical Engineering Science, 50, 19231941 (1995). gives for LB model epsilon = nu d2k/dk2 dk/dy = 0 

January 28, 2011, 05:49 

#16 
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Robert Sawko
Join Date: Mar 2009
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I have just found in Wilcox "Turbulence modelling" (1993) a relevant passage. On page 141 the bc are given as:
k = 0 and epsilon = d2k/dn2 The only justification for these condition is asymptotic consistency with the behaviour at the wall. 

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