# External Flow Around A Body

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 June 2, 2007, 11:03 External Flow Around A Body #1 S. Guest   Posts: n/a Sponsored Links I am working on prediction of flow around a simple car body. I have created a box and placed the car body within this box and applied a sizing function between the car faces and the box faces. As I see it there are two options for the outer boundary conditions (i.e the side walls of the box)...pressure outlet or symmetry. When I use the symmetry boundary condition the computation quickly converges, which makes sense. When I use pressure, the residuals do not steadily decrease and convergence is not as good. I also receive a notification that the turbulent viscosity ratio is limited to a value when I run with the pressure boundary condition. I feel as though the pressure B.C. is the most appropriate in this case... is this correct? What can I do to improve convergence when using the pressure B.C.? S.

 June 3, 2007, 22:25 Re: External Flow Around A Body #2 AJ Guest   Posts: n/a Are you trying to model car in a wind tunnel where air is forced on one side and comes out on other side.... 1) There should velocity input for one of the box faces 2) There should be an outlet or pressure boundary condition for flow outlet 3) If you have half car model then you have to model symmetry boundary to represent the half model... 4) Other faces of the box would be wall ( although there will be wall effect), but these walls would generally be away from boundary layer around the car... So I don;t understand where and why are you planning to use pressure outlet.... explain.. Happy computing... AJ

 June 5, 2007, 19:14 Re: External Flow Around A Body #3 Carlos Guest   Posts: n/a Hi S, I'm doing a PhD in vehicle aerodynamics and I have a suggestion. People do use symmetry BC's for the sides and top of the domain but I use a frictionless wall instead. Providing that the blockage caused by the car is small enough and the boundaries are far enough away from the car, simply use walls to represent the sides and roof of the domain. However, to make the walls frictionless, you do not want the no slip condition, click on specified shear in the BC panel and leave the shear stress as zero. Hope this helps. Carlos.

 June 7, 2007, 18:57 Re: External Flow Around A Body #4 Prasad Dudhgaonkar Guest   Posts: n/a As AJ has mentioned, the upstream face should have velocity BC (speed of the car, and also it's a measured parameter, well... in most cases known to me). How is idea of using pressure outlet with atmospheric pressure (zero gauge pressure) and back-flow from neighboring cell for top and side surfaces which are supposed to be sufficiently away to avoid wall effect on the flow and also to the downstream surface?

 June 8, 2007, 08:24 Re: External Flow Around A Body #5 Carlos Guest   Posts: n/a To me it makes no sense at all to use pressure outlets for the sides of the domain. If you did that then air would enter through the inlet and immediately start to exit through the sides of the domain as well at the outlet. So I can't see how that will work at all. I have tried using the far field boundary condition before, this should work but you need to know a fair bit about the atmospheric conditions to apply it. I would recommend a frictionless wall far away from the vehicle or simply a series of symmetry planes. Carlos.

 June 9, 2007, 14:46 Re: External Flow Around A Body #6 Razvan Guest   Posts: n/a Dear Carlos, I'm really curious why you use frictionless walls instead of symmetry condition? Numerically speaking, as far as I know, a frictionless adiabatic wall is absolutely equivalent to a symmetry condition, so why even bother? All the best, Razvan

 June 9, 2007, 15:35 Re: External Flow Around A Body #7 Carlos Guest   Posts: n/a Dear Razvan, I had no idea they were the same! Intuitively a symmetry boundary condition would be used to get away with modeling half of a symmetrical solution domain by reflecting data points about the plane. It follows that why on earth would you use a symmetry condition on BOTH sides of a domain!?! It would be rather like having two mirrors facing each other and reflecting the results to infinity! Obviously it does work, hence the widespread use but to me a frictionless wall, weather the same as a symmetry condition or not, make more sense by its very nature. Furthermore, I don't use a rectangular cross-section in my domain for vehicle aerodynamics. It is semi-elliptic with a horizontal radius about 20% greater than the vertical radius. This allows greater freedom when looking at yaw, and my work involves vehicles typically 10m in length so it is necessary for me. I couldn't possibly imagine why a symmetry condition would work on a curved surface! Any comments would be welcome... Carlos.

 June 10, 2007, 04:05 Re: External Flow Around A Body #8 Razvan Guest   Posts: n/a Dear Carlos, I see that you missed the very most important idea of my intervention: I was speaking from the numerical POV, i.e. the numerical formulation/implementation of the BC. The wall BC is NO MORE than a symmetry-type condition with the no-slip effect added. Once you eliminate that, the frictionless wall becomes a symmetry BC, from mathematical POV. So you see, due to their mathematical equivalence, the two BC (the frictionless wall and the symmetry), MUST behave identically in a numerical model. You "see" the symmetry condition like a "mirror". That is true, but you think that walls are not reflecting waves, just as symmetries do? Even no-slip walls reflect waves very well, the only difference being the fact that they absorb some of the energy of the wave. Imagine for example the gravitational waves on water' surface: if you drop a stone in the middle of a bathtub, won't the generated waves reflect from the tub's walls again and again? Of course that this does not continue to infinity, because of the friction forces between water molecules. But in a frictionless numerical domain, this phenomena should continue forever, don't you agree? Unfortunately not even in a frictionless numerical model this is not true, the numerical diffision an dissipation will act as an energy absorbant and will eventually dump the waves. That is also the very explanation why we can use a symmetry condition if we place it far enough from the interest point of the flow domain. The numerical waves generated by the body will be dumped so much by the numerical disspation (discretisation methods on one side and the growth ratio, skewness and misalignment of the grid cells on the other), that the reflected waves will never get to spoil the results in the vicinity of the body. About the shape of the flow domain: as long as it is a sweeped-type shape, it really doesn't matter what cross section you use! It could be any shape, as long as the minimum-distance-from-body condition is fulfilled along the exterior boundary. And you can assign to it the symmetry BC or the frictionless wall BC, with identical final results. All the best, Razvan

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