# [Q] Boundary conditions in Turbomachinery

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 July 21, 1999, 11:33 [Q] Boundary conditions in Turbomachinery #1 ghlee Guest   Posts: n/a Dear friends ! I'm calculating centrifugal compressor in steady state. Unshrouded impeller & vainless diffuser is used. Pressure ratio is nearly 1.0, thus incompressible flow assumption is made. I used rotating coordinate system, relative velocity components, standard k-epsilon turbulence model. In this situation, addtional force terms(Coriolis force & Centrifugal force) is included in source term. In general, boundary conditon for turbomachinery has four types. (1) Inflow (2) Outflow (3) Wall (4) Periodic My questions are below. (1) We put upstream & downstream region from impeller in order to give boundary conditions easily. These regions are stationary (i.e. not rotating), thus can I omit additional force term due to rotation? (2) Because the coordinate system is fixed to impeller, shroud casing is rotating in oppose direction, and it has circumferential velocity. We treat the shroud casing as wall in using wall function. In this case, is there anything to change in wall function? Though many questions are remain, I would end question. Thank for your reading...

 July 21, 1999, 16:27 Re: [Q] Boundary conditions in Turbomachinery #2 clifford bradford Guest   Posts: n/a i can answer question 1. stationary reqions don't need body force term. however you should not make the incompressible flow assumption. the flow is compressible (your pressure ratios is 1!!).

 July 22, 1999, 03:04 Re: [Q] Boundary conditions in Turbomachinery #3 John C. Chien Guest   Posts: n/a (1). Look like that you are working on a centrifugal fan instead of compressor. Well, that is not important. (2). You really have to be very careful here. (3). First of all, the appearance of the governing equations depends on the coordinate system used, whether it is stationary (fixed, non-moving) Cartesian, stationary cylindrical, or moving (rotating) cylindrical coordinates. You do not add or substract terms from the equations! They are there because of the coordinate system you selected in the first place. (4). Once you have the coordinate system selected, the boundary conditions must be specified on that coordinate system as well. (5). If the wall was stationary in the stationary coordinate system originally, then it will become a moving wall (rotating wall) in the rotating coordiante system. If the rotating coordinate system is moving clockwise, then the wall will rotate counter-clockwise. And the inlet condition in the new rotating coordinate system will likewise pick up this counter-rotating velocity component. That is all you need. (6). For steady-state stage calculation, the situation is quite different. Here, you are trying to joint the two separate problems ( one is formulated on rotating coordiante system and the other is formulated on stationalry coordinate system) into one combined problem through the use of "mixing plane boundary conditions". In principle, they are still two separate problems, but now they exist in one program and the solution is obtained through this mixing plene boundary condition iteratively. (7). For the real transient flow problems, the formulation is still on one single coordinate system , say stationary coordiante system, but the boundary conditions on the balde will become moving boundary. In this case, the steady-state mixing plane boundary between two computational zones is no longer used. (8). So, make sure that you know what kind of cases you are dealing with before setting the boundary conditions. The guideline is When one coordiante system is used, there is only one probem. If you have to use both stationary and moving coordinate systems, then there are two separate problems. These two problems are separated by the mixing plane boundary conditions. (9). It can be very confusing if you are not careful.

 July 23, 1999, 17:05 Re: [Q] Boundary conditions in Turbomachinery #4 Tom Guest   Posts: n/a 1. it is not a must to use a mixing-plane between a stationary and a moving zone. often it is possible to calculate in "one" piece, but then you have to pay attention, that the flow in the interface is mixed out. 2. the use of different frames is complicated and confusing. often there are tutorial-manuals with an example of this (in fluent it is) 3. don't think too much about the problem in which direction which wall in which coordinate-system moves. think a little bit and then try a calculation in the optimal point of the impeller. you see then very quick (if the impeller is not completely wrong constructed) if the flow enters the channel correct. always check after the calculations the wall velocities. they are easy to control: v-tan=omega*r, so you see, if the direction and the magnitude of the different wolls are ok. 4. you can use the wall-function

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