# Axisymmetry - heeelp!!!!

 Register Blogs Members List Search Today's Posts Mark Forums Read

 November 7, 2005, 05:11 Axisymmetry - heeelp!!!! #1 queram Guest   Posts: n/a guys, I have a FVM solver (MacCormack method) for 2D planar flows (euler equations). Now I'm modifying it to 2D axisymmetric. I've fixed boundary conditions (solid wall (no-slip), inlet, outlet, centerline), developed appropriate mesher etc. The code works nicely, even the results look OK. But: the equations in the solver remained the same as for 2D planar case. Is it a big mistake? Most of the books I went through say nothing on this, just few (ferziger-peric among them) say "the same code can be used for 2D planar and 2D axisymmetric cases, just to set x=z and y=r". Looking at the equations: 2D planar: dW/dt+dF/dx+dG/dy=0, where W=(rho, rho*u, rho*v, rho*e) F=(rho*u, rho*u^2+p, rho*u*v, rho*u*h) G=(rho*v, rho*u*v, rho*v^2+p, rho*v*h) and 2D axisym: dW/dt+dF/dz+dG/dr+1/r*R=0, where W=(rho, rho*u, rho*v, rho*e) F=(rho*u, rho*u^2+p, rho*u*v, rho*u*h) G=(rho*v, rho*u*v, rho*v^2+p, rho*v*h) R=(rho*v, rho*u*v, rho*v^2, rho*v*h) z=x, r=y, u is in x(z) direction, v is in y(r) direction; theta=0. obviously, 2D axisym case has the R term extra. If I don't use I'll (what?) ? If I use it, any idea how to dicretize it?

 November 9, 2005, 01:31 Re: Axisymmetry - heeelp!!!! #2 Mani Guest   Posts: n/a 2D flow is not the same as axisymmetric flow. Just think of the supersonic wedge versus the supersonic cone. The shock strength in front of the wedge (2D) is very different from that in front of a cone (3D) at the same freestream conditions and the same wedge/cone angle. These two flows are very different. Sure, you have to include the extra term to make it axisymmetric.

 November 9, 2005, 11:53 Re: Axisymmetry - heeelp!!!! #3 Sonny Guest   Posts: n/a You'll need to include the R term for axisymmetry. It doesn't need to be discretized so you can treat it as a source term. Problems to arise however if you have grid nodes on the symmetry axis due to the 1/r term where on the axis r = 0. My suggestion is to extrapolate the source term from the interior grid points for this case.

 November 9, 2005, 13:55 Re: Axisymmetry - heeelp!!!! #4 Mani Guest   Posts: n/a There is no real singularity, because v=0 at the centerline, so there are ways to deal with the centerline explicitly, by setting the source term to zero due to symmetry. Alternatively, the source term can be defined at the cell center, avoiding the r=0 boundary altogether.

 November 10, 2005, 05:05 Re: Axisymmetry - heeelp!!!! #5 queram Guest   Posts: n/a all right guys, I think I got it. R in axisym eq is actually the source term, the same thing than C in quasi-1D euler eq. R has to do with area variation, it simply needs to be taken into account, otherwise one would loose correct info on mass flow rate. as mani said, can be associated with cell centre (just as W is not associated with any point, it may represent an average state over the cell) thus avoiding r=0 at centerline. well and the cell centre should be, imho, the centre of gravity, right? then, it may be a bit challenging to evaluate its coordinates (r) correctly for nonunifrm quadrilateral cells (all cells are quadrilateral but their corners need not to have the same x- & r-coordinate)... can the effort of determining the coordinates of cells' centre of gravity be compensated by accuracy gain? compared to much simpler case when the cell centre is taken as diagonals intersection point.

 November 10, 2005, 15:06 Re: Axisymmetry - heeelp!!!! #6 Mani Guest   Posts: n/a >well and the cell centre should be, imho, the centre of gravity, right? Not necessarily. Gravity has nothing to do with it. Essentially, you are replacing a volume integral by some representative center value times the volume. If I remember right, this is referred to as the "midpoint rule" in calculus. The question is, is there any location inside the volume, where you actually observe that exact representative value? You cannot answer this without knowing the distribution of the integrand over the entire volume. It doesn't have to be the center of gravity (most likely not). In that sense it is not important to spend a lot of computational effort calculating a specific location. The geometric average of the surrounding vertices should be reasonable enough.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Eduardo FLUENT 0 August 4, 2008 09:23 Eduardo Main CFD Forum 0 August 4, 2008 09:18 Gianfranco FLUENT 0 June 3, 2008 17:35 aPpA FLUENT 4 March 22, 2007 09:33 Cote CFX 7 January 12, 2007 11:41

All times are GMT -4. The time now is 08:26.