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Francisco Saldarriaga August 10, 1999 20:06

Turbulent model in 2D calculations
 
Flow community: Question: What is the validity of using turbulent models in 2D calculations? I use the k-epsilon standard model of fluent with the law of the wall in a curvilinear profile which contains 2 or 3 stirring blades. Using the MRF (moving reference frame) model I obtain a general velocity vector diagram on a plane for my domain. This compares rather well with flow lines, I observe in a tank. In the model, I input a value for the angular velocity (omega). When I look at the equations of motion all the components involving the z derivative and values of w are equal to zero. How can I explain, the fact that I am solving for 2D but it seems that all the centrifugal term and the coriolis term seem to go to zero. I conclude the program takes a constant value for Z. Is this correct? Does any body know this in fluent?

John C. Chien August 10, 1999 23:10

Re: Turbulent model in 2D calculations
 
(1). Turbulence models are all based on the 2-D turbulent boundary layer flow over flat plate. That should answer your first question. (2).So, you have used a two-equation k-epsilon turbulence model with the standard law of the wall. That part is all right. The model has been used for 2-D and 3-D flows, so this part is all right. (3).But I am having difficulty in figuring out this " in a curvilinear profile which contains 2 or 3 stirring blades". There is definitely something missing here. ( lost between your keyboard and the cfd-online forum?) It is likely that you have two configurations, one with 2 stirring blades and the other has 3 stirring blades. It is safe to say that these stirring blades are 3-D blades.( unless otherwise specified.) I can only say that it must be a container or something like that, instead of a curvilinear profile (which does not make any sense at all.) (4). Since you are using "moving reference frame", those stirring blades must be rotating and are fixed on this rotating frame of reference. This is perfectly all right. It is convenient to use a rotating reference frame, although it is not absolutely necessary. ( just for convenience of handling of the blade boundary conditions). (5).At this point, you are assuming that the governing equations in the code have been written in term of this rotating coordinate system. (6).If that is the case, the resultant equations should contain the centrifugal terms. (7). In old days, as I remembered, there were 2-D code, 3-D code, and 3-D axisymmetric code with swirl for low speed flows. If the problem is 2-D, you just select the 2-D module at the begining when you start the code. In 2-D module, it is just plane 2-D, you should not have centrifugal terms in it. (8). In the axisymmetric with swirl case , it is really 3-D but written in cylindrical coordinate system. You should also find the centrifugal terms in it. (9). Otherwise, in 3-D module, it is written in Cartesian x,y,z coordinates with Ux,Uy,Ux velocity components. I am sure that is the way it is coded in the program because I had to create cylindrical model and expressed the coordinates in Cartesian x,y,z. So, in this coordinate system, there is no centrifugal term in it (explicitly!!). (10). Sound confusing? Well, don't use your common sense to guess at the inner structure of a black box code. (11). I could be wrong at assuming that the stirring blades are rotating around an axis, these could be just sliding in a linear fashion. Then the problem will be 2-D only. ( or 3-D cartesian.) In this 2-D case ( or 3-D), you should not see any centrifugal terms or effect at all. (12). So, You will be able to see the centrifugal terms only if the variables and equations are written in cylindrical coordinates. (13). It is a good idea to look at the governing equations written in Cartesian coordinates system and the cylindrical coordinate system, and check the difference in the two equations. (make sure that you are using a rotating periodic boundary condition, not a translating periodic boundary condition!!!). A good exercise for my brain anyway. Hope this will help.


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