|April 10, 2009, 08:18||
Formula One WHEEL
Join Date: Mar 2009
Posts: 8Rep Power: 9
I have just been told by, fluent help, that I have to place my rotating wheel slightly above the bottom of the wind tunnel, however this would not be a true study of the aerodynamics around it. However, I have seen people doing analyses with the bottom of the wheel slightly cut off/straight and they have placed it at the bottom of the wind tunnel. Does anyone know if this is possible and would I have to follow the same principles in setting the boundary conditions as the one with the wheel slightly above the bottom of the wind tunnel.
|April 11, 2009, 05:09||
Join Date: Mar 2009
Blog Entries: 14Rep Power: 18
I don't have any experience but i think that what they suggested is wrong.
I think that a much more realistic condition is to have a contact between the wheel and the ground, probably with the wheel somehow compressed so that you don't have a contact line but a contact surface (whose definition is probably guided by experience). Than i would set the following boundary conditions:
1) Ground and inflow: uniform velocity inlet
2) For the wheel you can define a local polar coordinate system with it's axis on the wheel's axis and set the velocity of the wheel based on the radial distance from the axis.
i.e. if the car speed is V and the distance of the wheel axis from the ground is h, you should set a boundary condition on the wheel surface that is U = V*r/h which is a tangential velocity based on the radial distance from the wheel axis.
Maybe this is not so much accurate near the contact surface between the wheel and the ground, where i expect to have a component of the wheel's velocity in radial direction but as a first attempt it should be enough (and you'll probably find out that it is also quite accurate).
A kind of problem that can rise is that near the contact surface some kind of singularity will rise because of the abrupt change of the velocity direction. However, i guess that you can easily find some paper which describes the proper approach for this kind of flow.
|Thread||Thread Starter||Forum||Replies||Last Post|
|how to define pressure drop formula in a pipe||Yessica Pomarinni||FLUENT||2||August 31, 2007 11:09|
|Output Pressure reults for rotating wheel. HELP||Krishan Parmar||Phoenics||0||July 2, 2007 14:11|
|Rotating bike wheel||E. Turner||FLUENT||3||October 1, 2006 00:44|
|Water wheel formula, please ?||Externet||Main CFD Forum||0||August 30, 2005 01:06|
|Benetton Formula 1||CD adapco Group Marketing||CD-adapco||13||February 7, 2002 10:33|