Axisymetric cone flare

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

 December 17, 1999, 07:59 Axisymetric cone flare #1 omar Guest   Posts: n/a I am using a 3D Finite Volume solver to compute the flow around a cone flare. The flow is hypersonic so I use a shock capturing method because of my very fine mesh. The problem is that, instead of doing a 3D computation, I have a 2D mesh (because the problem is axisymetric) to reduce the cost of the computation. Indeed, I have to impose a boundary condition at the symetry line and two periodic boundary conditions at each face. It seems that my boundary condition formulation (imposing at the ghost cells below the symetry line the same velocity as in the flow in the x-direction, a velocity in the y-direction and z-direction in an opposite direction, and the same pressure) doesn't work at all. I have very bad results and the "leading edge" is not at the symetry line. Do you have any idea whether it is coming from the nature of the problem (hypersonic flow with strong shock)? How to impose the bc's at the symetry line. Thank you very much for your attention, Omar S.

 December 17, 1999, 10:56 Re: Axisymetric cone flare #2 Patrick Godon Guest   Posts: n/a If you have a conical shape, then you have axi-symmetry: rotational symmetry around one of the axis. Then you are left to solve the problem in two dimensions, however, because of the symmetry you need to solve only for half of the (2D) computational domain (above and below the rotational axis the flow is symmetric). On the rotational axis you need to have the velocity normal to the rotational axis set to zero. The velocity parallel to the rotational axis needs to be symmetrical, which means that its DERIVATIVES needs to be set to zero. THis DOES NOT especially means that you have to set the ghost points to the same value, but that when you write down the derivative of the parallel velocity (at the point on the axis of rotation) its actual value must be zero. And this depends on your numerical scheme. In some numerical scheme seting the derivative to zero or seting the ghost points to the same value as the computational points is equivalent, but not always. So you need to write down the derivative in the difference form (the same way you solve the equations) and set it to zero. THIS will give you the value of the velocity at the ghost points and/or on the axis. The third velocity is the velocity rotating around the axis of rotation, and on the axis it must be zero. For the pressure, you need also to set its derivative (normal to the rotational axis) set to zero, and then obtain from there the value of the pressure on the axis or at the ghost points. You need FIRST, however, to solve for the problem in axi-symmetric coordinates (cylindrical coodrinates) and not CARTESIAN coordinates. You need (r,phi,z). Where, the z-axis is the axis of rotational symmetry. Then you will have on the z-axis: dVz/dz=0 (symmetry of Vz) Vr=0 (no flow accros the z-axis) Vphi=0 (the radius of rotation is zero, therefore Vphi=0) and in addition Vphi is antisymmetric on each side of the z-axis: Vphi on one side = -Vphi on the other side dP/dr=0 (symmetry of P) So the first thing you need to do is write and solve the equations in a cylindrical coordinate system, using the above bondary condtions. You will also need to specify boundary coditions on the surface of the cone (I guess no-slip boundary conditions). There is I guess one boundary with an inflowing flow and the other with outflowing and the last boundary is probably an open one. Cheers, Patrick

 December 17, 1999, 13:35 Re: Axisymetric cone flare #3 John C. Chien Guest   Posts: n/a (1). If the code is a 3-D code, then the mesh also has to be a 3-D mesh. (2). If the geometry is axi-symmetric, then you can first create a 2-D mesh on one symmetric plane (r-z plane) and then rotate it to create meshes in the theta direction. This will give you the complete 3-D mesh. (3). If you assume the flow is also axisymmetric, then you can solve only a small sector in the theta direction, instead of the full 360 degree range. But you still need to have several mesh points in the theta direction to cover the symmetry plane boundary conditions in the theta direction. (at theta_zero and theta_max) (4). The extra grid points (or cells) outside the symmetry planes must also be generated in the same way as the interior points. (5). Along the axis, you can set the radial velocity to zero (you will have to do some calculation to convert it into 3-D x,y,z-componets. And the the theta-direction velocity is also zero. the combined effect is that the x-,y-components of the velocity are zero along the axis. The only non-zero component is the Uz axial velocity. (even at the axis, you still have several mesh points in the theta direction. So, there will be several Uz at the axis.) (6). A better way to see it is to start the mesh or the computational domain at a very small radius from the axis, so that you can still have non-zero cell near the axis. The symmetric condition in the r-direction remains the same for each theta plane. (7). Basically, you have to deal with a 3-D fan shaped mesh, (looking down stream along the axis.) and you still need several mesh points (cells) in the theta direction. ( the small needle size cylinder along the axis is not going to cause any problem ).(the other option is to use an axi-symmetric version of the code. even in that case, it is a good idea to avoid the axis region first.)

 December 17, 1999, 15:00 Re: Axisymetric cone flare #4 omar Guest   Posts: n/a Thanks for getting back to me so quickly, Your advices are very helpful, hear from you soon, OMAR S.

 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 launchpad FLUENT 0 June 30, 2011 03:22 JaC FLUENT 0 January 5, 2010 10:20 JaC FLUENT 1 December 3, 2009 04:03 beginner Main CFD Forum 0 March 19, 2003 12:49 Tylor Xie Main CFD Forum 0 June 9, 1999 07:33

All times are GMT -4. The time now is 08:40.