# Complex boundary conditions??

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 April 21, 2000, 11:32 Complex boundary conditions?? #1 Moussa Diarra Guest   Posts: n/a Hi, Can anybody help me? i'am trying to simulate Air Flow between two independant rotating disks and i have a problem to define the boudary conditons (there is no inlet/outlet boundaries).The only condition i know is the static pressure (atmosphere 1.013e5 Pa) is present everywhere before the rotor disk starts. all ideas are welcome. Regards

 April 21, 2000, 19:02 Re: Complex boundary conditions?? #2 Bernard Parent Guest   Posts: n/a In 2D, just set the right BC equal to the left and vice-versa (the so-called periodic BC if I recall correctly). But the interesting thing about your problem is that you could solve it using a 1D algorithm, where the BC's would be wall on the left and wall on the right, no need to worry about the periodic BC. -bernard

 April 22, 2000, 13:41 Re: Complex boundary conditions?? #3 Jim Park Guest   Posts: n/a Do you have two disks with a common axis of rotation and different rotating speeds? If the answer to these questions is yes, is the space between the disks sealed so that no mass is added to or removed from that space after rotation begins? If this is the case, the problem appears to be at last two-dimensional, with radial variation in pressure to balance the centrifugal force created by the rotation and axial variation from the bottom plate to the top plate. In addition, the azimuthal momentum equation must also be solved (but also only in the radial and axial directions) to account for the different rotational speeds between these disks. So, leaving an compressibility and thermal energy changes aside, you need to solve three momentum equations and a continuity equation in two space dimensions. If the difference in rotational speeds is sufficient that turbulence develops, you have to deal with that. And all of this assumes that the mean flow is steady. So, for the incompressible, steady state, laminar case, I'd recommend the following bc's: On the disk surfaces, normal velocity is 0, tangential velocity matches the local disk velocity (include the rotation speeds here). At the inner radius (if > 0) and the outer radius, same specification. I think Bob Ribando, at the University of Virginia, did something along these lines in the early 80's. Good luck.

 April 22, 2000, 15:54 Re: Complex boundary conditions?? #4 Bernard Parent Guest   Posts: n/a A problem defined by (1) two disks with a common axis of rotation and different rotating speeds, and (2) the space between the disks sealed so that no mass is added to or removed from that space after rotation begins can be solved in two dimensions (writing the governing equations in r and theta) or in one dimension (writing the governing equations as a function of r only, since there will be no gradient at any time along theta). If it is solved in 2D, the results should collapse to the ones obtained by a one-dimensional formulation (that is, you should get the same properties along r, at any theta). -bernard

 April 22, 2000, 18:29 Re: Complex boundary conditions?? #5 Jim Park Guest   Posts: n/a It is true (as I mentioned) that there is no dependence of any flow property on angle if the disks are parallel. But there is dependence on the distance between the disks. Since the disks are rotating and the fluid sticks to the disks at the surfaces (the no slip condition), the fluid is rotating also. That is, the fluid has a tangential velocity. This is the key point. At any radius r, the fluid on the first disk has the (no slip) boundary condition v1(r) = r x om1, where om1 is the angular rotation rate. Similarly, on the second disk, v2(r) = r x om2. Between these points on the boundaries, v must vary smoothly. That is, if the distance between the disks is measured by z (the axial direction) and the surface velocity of the two disks is different, then v must depend on z. Clearly the tangential velocity component v is a function of axial position as well as radial position. That is, tangential velocity in the fluid = v(r, z), but not angle. In the radial momentum equation for a rotating fluid, the "centrifugal force", [v squared]/r comes into play. This is balanced by the radial variation in shear stress and by a radial pressure gradient. In general, since v is a function of elevation z so is the pressure between the disks. Thus the flow field between two rotating disks depends on the radial and axial location. It is not a function of r only. However, in a fluid bearing the distance between the disks is small (lubrication theory), and an analysis similar to boundary layer analysis shows that the axial pressure gradient is negligible. In that case the pressure is a function of radius only.

 April 22, 2000, 20:05 Re: Complex boundary conditions?? #6 John C. Chien Guest   Posts: n/a (1).It is a very simple problem, but you need to define the problem more clearly. (2). So, try to describe the problem as if you are going to run a real test in the lab. (3). A disk will have two sides, so, two disks will have four sides. And each will have its own shaft. It will be driven independently by separate motor. (4). If you are not going to have inlet and outlet, then you need to design a sealed casing?.......(5). And what about the size of the disks? the thicknesses?, the spacing? the size and shape of the casing?, the RPM? the fluid? (6). The ASME/Journal of Turbomachinery has many papers on verious kinds of rotating cavity flows, and it is a good place to start paper search. good luck.

 April 25, 2000, 05:37 Re: Complex boundary conditions?? #7 Moussa Diarra Guest   Posts: n/a Thank you very much for your help. Moussa

 April 25, 2000, 11:25 Re: Complex boundary conditions?? #8 Bernard Parent Guest   Posts: n/a I see what you mean, I was thinking of a different problem, where the flow is located between an outer and inner ring of different radii, not between two disks of equal radii. Sorry about that.. -bernard

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