Solve axisymmetric by using rectangular grid
Dear sir,
I have read from one paper that the author write the Fortran code to solve NavierStokes equations in rectangular grid. Anyway he can use his program to solve the problem with axisymmetric domain shape by constructing* pie shape and doing something that I don't understand. Do anybody have some idea about this? Please give me some explaination. *** Thank you very much sir. Best regards, Atit Koonsrisuk 
Re: Solve axisymmetric by using rectangular grid
If you are solving an axisymmetric problem (only the problem details can determine if it is axisymmetric or not), then you can solve the problem on a 2D grid (rectangular cells is only one of many possible cell types) by prescribing the boundary that is along the actual axis of symmetry as being a symmetry boundary.
The problem can also be solved as a 3D pie wedge with the two sides of the pie being prescribed as either symmetric or periodic. 
Re: Solve axisymmetric by using rectangular grid
Yes,
In cylindrical coordinates for example, (r, z, phi), if you assume axisymmetry around the z axis, then there is no dependance on phi. In this case you can solve the problem in the (r,z) plane using cartesian coordinates, that's fine. Patrick Godon 
Re: Solve axisymmetric by using rectangular grid
Patrick,
Your previous message is not very clear. It is possible to solve an axisymmetric flow problem on a grid that is defined in the cartesian plane i.e. consisting of only (r,z). However you must still use the cylindrical equations of motion. The continuity equation has an 'r' in it which is not present in the cartesian equation of continuity. Furthermore, if the flow has an azimuthal component of velocity the momentum equations have a couple of extra terms to account for so called 'apparent forces' Regards, George 
Re: Solve axisymmetric by using rectangular grid
Well,
I wrote you solve the problem in the (r,z) plane using cartesian coordinates. Solving the problem in the (r,z) plane implies that you write the equations in the cylindrical coordinates form. Then using cartesain coordinates means that you can replace r by x and z by y, after you wrote down the equations, and use a (x,y) grid. Sorry, if I was not clear, I though it was obvious. 
Re: Solve axisymmetric by using rectangular grid
Your note is correct as far as it goes, but I think it's more than a couple of terms if "the flow has an azimuthal component of velocity".
A third momentum equation is required. 
Re: Solve axisymmetric by using rectangular grid
Dear sir, I would like to give you more information. The program, that I read, use equation in rectangular coordinate, and use rectangular grid. After I read all kind explaination, I conclude that first we split the cylinder to be pie shape and then set the boundary conditions on both faces of the pie shape to be the same values as we call periodic boundary condition. Do I understand correctly? The next question is that why can that program use the equation in rectangular coordinate to solve this cylindical system? Thank you very much for all kind explaination.
Best regards. Atit Koonsrisuk 
Re: Solve axisymmetric by using rectangular grid
(1). It is normally done this way, that is solve the equation in Cartesian coordinates. (2). In CAD, you also do the same thing, use Cartesian coordinates for all geometric shapes. (3). Most commercial codes use Cartesian coordiantes. There is no need to cast the equation in Cylindrical coordinates, just because there is a pipe in the flow field. Using cylindrical coordinates sometimes can simplify the boundary conditions, but not always. (4). Even if you are using the coordinate transformation approach in finite difference approach, the base coordinate system in most cases is Cartesian coordinates. (you could use Cylindrical coordinates as the base, but it is not as general to handle noncylindrical geometry)

Re: Solve axisymmetric by using rectangular grid
Hi Atit.
From your question, when you solve axisymmetric ploblem that means d( any thing)/d Phi is zero. So The continuity eq. is d(den*U)/dX + d(r*den*V)/dr = 0 Radial Momentum is d(den*U*V)/dX + d(r*den*V*V)/dr = dP/Dr + d(vis*dV/dX)/dX + d(r*vis*dV/dr)/dr 2*vis*V/r**2 and Axial Momentum is change in the same way. Then, when you write your code you just set r equal to 1 when you solve Cartisian coordinate and use the value of r when you solve axisymmetric problem. Hope this help. Suthichock Nunthasookkasame P.S. Are you study in Thailand ? 
Re: Solve axisymmetric by using rectangular grid
Everything previously said is correct, but I saw many times something resembling the procedure you described in a 2D approach to solve an axisymmetric problem. So I have a question for you is it your code solving 2D NavierStokes equations or not ? If not forget this post and follow previous advices. Bye

Re: Solve axisymmetric by using rectangular grid
Dear sir, Actaully the mentioned code is not mine. I read from the comment of the code that this code use a rectangular grid and a rectangular equations. It is the genaral code; because it can solve one, two, or three dimensional flow and can solve the axisymmetric problem with the same grid and the same equations. I think I understand why we can solve one, two, and three dimensional flow by the same code. However I do not understand why this code can solve axisymmetric problem without changing the equation. So it must be empasized that this code use the same equations for solving a rectangular or an axisymmtric problem.
Thank you very much sir. Best regards, Atit Koonsrisuk. 
Re: Solve axisymmetric by using rectangular grid
Usually this kind of procedure it is used to fast implement the axisymmetric case, generally it works good enough. In general it is necessary to adjust a little bit the evaluation of lateral fluxes; but I can't say ,without take a look at the code, if this is the case. Bye

Re: Solve axisymmetric by using rectangular grid
Hi Atit,
You can find good explanation of your question in following reference : Computational Methods for Fluid Dynamics by J. H. Ferziger & M. Peric (2nd ed. 1999 published by Springer) section 8.9 at page 244  246 One thing I am not clear here is that why back and front surfaces are also taken into account when integrating the radial component of pressure forces. Can anybody explain this in further details or give any reference that explainsit in more details ? Thanks. Suman. 
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