# Rotating Frames

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 May 12, 1999, 04:08 Rotating Frames #1 Tom Guest   Posts: n/a Sponsored Links Hy my friends, I try to simulate a centifugal pump, but I do not understand the system of rotating frames. My problem: If I define one rotating zone with a diameter that is larger than the outlet-diameter of the impeller (I extend the mesh at the outlet of the impeller as suggested below and calculate only one channel (periodic))- what happens with the pressure in this extended area? Does it increase because of the additional centrifugal and coriolis forces in the Navier-Stokes-Eq. in the rotating zone ? I think, energy must increase, because the fluid has a radial velocity and a radial force (centrifugal force) also exists, and F_rad x dr is energy. Are there really additional components (centrifugal and coriolis) in the equations of rotating zones ? I ask, because I read here below, that you get good results with such a calculation configuration. thanks for your help tom

 May 12, 1999, 11:53 Re: Rotating Frames #2 John C. Chien Guest   Posts: n/a (1). To define a proper computational domain itself is an art work. (2). When in doubt, try to include the whole device in the computational domain, that is to include everything. (3). Sometimes, it is not possible to compute only portion of the flow field because of the lack of information at the artificial boundary locations. (4). If you insist on doing so, you can still solve only portion of the flow field,using the measured data at the specified boundary location.

 May 12, 1999, 15:02 Re: Rotating Frames #3 Tom Guest   Posts: n/a Hy, here are some additional informations. I use FLUENT 5 and the modell MRF (multiple reference frame) with only one (rotating) fluid zone. Does anyone know exact, how this model works ? sorry, that I have forgotten this informations before. tom

 May 12, 1999, 15:34 Re: Rotating Frames #4 R.Kurz Guest   Posts: n/a I think the problem is how you define your boundary conditions. If you define your problem in a rotating frame, then outside of the impeller you need to apply the boundadry conditions which may be constant in the absolute reference frame either as time averaged conditions or fluctuating boundary conditions in the rotating frame. The total energy outside of the impeller can not increase , because the only place where mechanical energy can be imparted to the fluid is in the impeller. I'm specifivcally writing total energy (or total pressure , if you will), because the static pressure may increase depending on the diffuser geometry.

 May 12, 1999, 17:37 Re: Rotating Frames #5 Tom Guest   Posts: n/a hy, my bc are: hub and shroud and the blades: walls, moving with the rotational speed (relative velocity=0), and in the extended area at the hub and shroud-side: walls with absolute-velocity=0. In cicumferential direction I use periodic bc., at the radial end of the extended region I use pressure outlet. in axial direction I have a velocity inlet. I also think, that energy (or ptot) can only increase for dd_impeller. but this seems not to be so. the results seem to be ok. but why ?????? I do not understand this ... Tom

 May 12, 1999, 17:40 Re: Rotating Frames #6 Tom Guest   Posts: n/a hy, my bc are: hub and shroud and the blades: walls, moving with the rotational speed (relative velocity=0), and in the extended area at the hub and shroud-side: walls with absolute-velocity=0. In cicumferential direction I use periodic bc., at the radial end of the extended region I use pressure outlet. in axial direction I have a velocity inlet. I also think, that energy (or ptot) can only increase for a diameter that is smaller than diameter_impeller. but why does energy increase ? because of the additional components (coriolis-centrifugal) in the N.S.-equations or because of what ? and my problem is still: if energy increases because of these components, then (I think) energy will also increase in the (radial) extended area for d>d_impeller. but this seems not to be so. the results seem to be ok. but why ?????? I do not understand this ... Tom

 May 13, 1999, 02:18 Re: Rotating Frames #7 Jyi-tyan Yeh Guest   Posts: n/a The total (actual) body force is the summation of conventive force and the additional centrifugal and coriolis force in rotating frame. Check the total body force inside the impeller zone and in the extended zone.

 May 13, 1999, 09:20 Re: Rotating Frames #8 Jordi Pallares Guest   Posts: n/a Hi Tom, You should include centripetal and Coriolis effects in the governing equations and boundary conditions (in one zone) ONLY if you want a rotating frame of reference in this zone. As I understood, your choice is to have one rotating zone (with a rotating frame of reference) and one non-rotating zone (non-rotating frame of reference) sharing a boundary. Then boundary conditions (BC) should be imposed according to this. BC for the rotating zone have to be imposed respect to the rotating frame of reference but, in the same boundary, BC for the non-rotating zone have to be specified respect to the non-rotating frame. see for example: AUTHOR : Vanyo, James P. TITLE : Rotating fluids in engineering and science PUBLISHED : Boston: Butterworth-Heinemann, cop. 1993 PAGES : x, 429 s. ISBN : 0-7506-9261-8 Hope it helps

 May 13, 1999, 10:05 Re: Rotating Frames #9 John C. Chien Guest   Posts: n/a (1). When you run into problems, try to isolate them first. (2). For the proper setup of the problem, you need to follow the sample cases and talk to the vendor's support engineer. I have not started running the new version yet,so I can't answer your question directly. I had used the old version, and I think it is important to talk to the support engineer and follow the setup setp-by-step (don't use your intuition.). (3). In the rotating frame of reference, you are fixed on the rotating blade. This is the only convenient way to solve the flow in the blade passage, and you can apply the wall boundary condition in the regular way.( how the code handle this is another story. I am not talking about the code here) (4). Everything you compute in the rotating frame uses the relative velocity field ( the velocity field on the rotating frame.). In this area, you really need to have a clear mind.(you don't care how the code handle this.) If you don't have a clear picture, try to read some text books on turbomachinery first to refresh your memory, because even an expert needs this from time to time. If you compute the total pressure using the relative velocity from the calculation, it is call the relative total pressure. When you compute the total pressure based on the absolute velocity ( combine the relative velocity with the rotating speed of the moving frame ),it is call the absolute total pressure. Refer to the textbook when you try to use it. It can be very confusing even for the expert. (5). That's all you need. Don't try to understand the terms created in the equations of rotating frame. ( assuming that the code you use can handle these terms properly and automatically as it should be.). (6). General CFD codes are designed to handle a wide range of problems, they are not designed to solve the problems in turbomachinery only. So, you must double check the numbers you compute. (even the graphic programs don't care whether it is relative or absolute. So, it is your responsibility to make sure that the number you are sending to the post processor is correct in the first place. The code can not make the decision for you. It is called " garbage-in-and-garbage-out".) (7). Everything will be all right, if you have a clear mind about the relative velocity field and about the exact steps required to run problems in rotating frame. (8). That is the reason why I always say the code alone is useless. (even if one has obtained results, he is not sure whether the results are right or not, he doesn't even know whether his inputs are correct or not. For meself,I always use hand calculation to double check the computed results. Try this, and you will like it.)

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