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Chetan Kadakia April 3, 2001 22:03

Turbine Flow - how many cells
 
Hi I am modeling turbine flow with various turbulence models. I have a Pentium 3 900MHz computer. How many cells should I be able to allow in my mesh? How long should it take to converge? How precise should I have my convergence criterion?


Scott W April 4, 2001 12:55

Re: Turbine Flow - how many cells
 
You should keep adding more cells until the answer you want does not change anymore. For example, if 100,000 cells gives a significantly different answer than 200,000 cells, then the 100,000 cell model is junk. However, that doesn't prove that 200,000 cells is enough. So try 300,000 cells, and repeat until the answer you want doesn't change. These numbers were just picked out of my head, so they don't apply to any specific problem. For your specific problem the answer depends on 1) how much time do you have, 2) how much memory does your computer have, and 3) how accurate do you need to be?

Suppose you need an answer in 10 minutes, then you cannot possibly use 1 million cells with a single processor pentium 3. If you have 10 weeks, then you can use 1 million cells (assuming your computer has enough memory).

Other factors: time dependent vs time independent, number of governing equations solved, are you properties constant or functions, etc. Without this knowledge no one can answer your first two questions.

I can answer your final question. Determine exactly what you want to answer. Then keep iterating (who cares what the residuals or convergence criterion say) until your answer doesn't change with more iterations. When your desired answer is unchanged, you can stop even if other things are not fully converged.

Chetan Kadakia April 4, 2001 15:18

Re: Turbine Flow - how many cells
 
Lets say I have 12 hours and 256MB of RAM on a 900MHz processor with a good initial conditions. What do you think?


Scott W April 4, 2001 15:28

Re: Turbine Flow - how many cells
 
For 3D flow I'd start with 200,000 cells (since that would take roughly 200MB of RAM, the remaining memory can be used by your operating system). You really need more memory for good 3D simulations. For 2D flows it may be fine to start with 20,000 cells. But please follow my test and try two different cell sizes, and then determine the number of cells you need. It will vary with each simulation.

Chetan Kadakia April 4, 2001 15:51

Re: Turbine Flow - how many cells
 
I will try starting with 100,000 cells. Obtain an inviscid solution. Adapt the region of high gradients and boundary layer and obtain a laminar solution. Then I will refine the grid some more and obtain a turbulent solution with the standard k-epsilon solution. All of these runs will be steady such that I can provide an initial flow field for my unsteady models. After all the refinements, I should have 200,000 cells. I will then refine for a grid independent solution for the unsteady solutions. Until I can find grid independence. I may have to get into parallel processing. What do you think?


John C. Chien April 4, 2001 17:15

Re: Turbine Flow - how many cells
 
(1). Even if you are using a custom made code for 3-D turbine passage flow (Navier-Stokes), to get a converged solution for compressible flow in 12 hours is somewhat tight for routine calculations, using a workstation. (2). To get a somewhat accurate result, you need to push the mesh size over 350,000 to one mega cells. (3). And even after that, we know that the solution will be incorrect in the loss prediction, unless you are an expert in turbulence modeling using your own code.

Chetan Kadakia April 4, 2001 19:04

Re: Turbine Flow - how many cells
 
Perhaps I can parallel several machines together and get a million cells plus over several days. I can use LES, RSM or k-epsilon models. Do you think I may get a somewhat close solution then?

John C. Chien April 4, 2001 19:39

Re: Turbine Flow - how many cells
 
(1). I don't think so. (2). But from a company's point of view, whether it is a large company, a school or a government lab, going parallel is a logical execuse. (3). It is not difficult to do CFD research. You don't need fancy mesh generation or colorful post-processing. (4). But you need good test data and the flexibility to do analysis (write your own code so that you can easily change your model), if you are serious about getting good results. It is possible, but not through the commercial CFD codes. (5). The commercial cfd codes are designed to be general, therefore the codes can not be changed easily for the particular problem to be solved. (6). So, if you are serious about getting good cfd solutions (nobody is interested in this subject right now, anywhere), here is my suggestion: (a). get a good set of data for your problem, (b). write a code to solve Navier-Stokes equations with a two-equation turbulence model (Reynolds average equation) based on the technology of 80's, (c). create algebraic multi-block mesh for your problem so that you can adjust the mesh easily,(algebraic means analytical, and you can code it easily) (d). do a parametric study, adjust the turbulence model and check the results against the test data. (7). This is a sure way of getting good results for your problem. Using this approach, the machine you have right now should be adequate. But the Window operating system alone could take a very large chunk of your RAM away. In terms of the RAM memory, you should try to increase it, the more the better.(it's rather cheap these days) And remember to get a UPS for your computer.(so you don't run into power blackout. Save your results often also is a good practice.)

Scott W April 5, 2001 11:40

Re: Turbine Flow - how many cells
 
Sounds good, but with all those steps you may need more than 12 hours. In my opinion, I'd skip the grid adaption between the inviscid and laminar solutions. It may help you, but it might just give you a very small mesh in an area which isn't important when you switch turbulence models (thus leading to extra computer time spent in an unimportant area).

Instead, you might try this solution: run the problem once with full turbulence but with a very rough grid. Those results will be quick and probably not too accurate since it will not be grid-independent. However, it will show where the largest gradients are, and thus when you make a real mesh, you can mesh that region properly without the need for much grid adaption later.

I do like your idea of increasing the complexity and using the results from previous models as initial conditions for the more complex model (it certainly helps to converge quickly).

For parallel calculations, I've read on this board that your main computer requires much more memory (although I can't tell you if this is true). Thus if your main computer has 256 MB, the rest of the computers may be able to only use roughly 128 MB. So each additional computer can't help as much as you might want. Of course, if you are able to upgrade to 512 MB on your main computer, you will be much better off for 3D calculations.

As for John's comments, neither he nor I know what your goals are. If you just want to compare common turbulent models, then you have no need to program your own since the standard models are already in your commercial code. If you want one heck of a good solution, then read the research in books and journals, to identify problems with the standard models. Then spend months/years to determine a correction (or even a better model), and then program the model yourself. I don't think you can do that in 12 hours :)

Good luck.

Scott W April 5, 2001 11:44

Re: Turbine Flow - how many cells
 
I forgot to add: I fully agree with John on his points (6) and (7) on his 5:39 pm 4 Apr 2001 response. You need good experimental data if you want to have confidence in your CFD model.

Xiao Hu April 13, 2001 16:37

Re: Turbine Flow - how many cells
 
The cell numbers depends on which turbulent model and wall function you use. If you use standard wall function or non-equilibrium wall function, check your y+ value at the first grid and make sure it is between 50 to 500. If you chose two-layer zone wall treatment, then make sure that y+ is approx. 1.


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