# two-fluid model

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 July 25, 2005, 07:17 two-fluid model #1 enis Guest   Posts: n/a Hi everybody, I want to learn that what type of boundary condition should I use in simulating 3-D cylindrical flow at the beginning (0 degree) and at the end (360 degree) of the angular direction? thanks, enis.

 July 25, 2005, 08:37 Re: two-fluid model #2 Jim_Park Guest   Posts: n/a To start, the velocities and the first derivatives must be continuous across the boundary. If you're doing heat transfer, the temperature and its gradient (heat flux) must be continuous across the boundary. The pressure gradient must be set to insure mass flow across the boundary from the left and the right. There may be more. The details of how you do this depend on your algorithm, both approximation method (FDM, FVM, FEM, etc) and iteration details. I expect someone can recommend a good recent book to describe the latest.

 July 25, 2005, 08:41 Re: two-fluid model #3 Jim_Park Guest   Posts: n/a Sorry! I wrote answer before reading the title of your post. You didn't say if you have a mixture of two fluids (two gases for example) or two distinct fluids (oil on top of water for example). The details would certainly be different. In either case, the important consideration is that the fluxes and gradients across the (angle = 0) boundary must match the fluxes and gradients across the (angle = 360 degree) boundary.

 July 25, 2005, 09:27 Re: two-fluid model #4 enis Guest   Posts: n/a thanks for your reply dear Park, let me give you some more details I m trying to simulate a 3-D horizontal two-phase flow in a cylinder. it has a radial inlet and a radial outlet boundary. both phase are assumed to be incompressible and no energy equation is considered. I'm using IMF(implicit multifield model) model of Harlow and Amsden. but the main problem for me, at least for now, is seemed to impose such an angular boundary condition at 0 and 360 degrees numerically.(Boundary condition implementations are done by using ghost cells). As far as i understand from your message you recommend me to use, for example; if we set k=1 for the cell at 0 degree, k=kmax for the cell at 360 degrees, after starting to solve governing equations from k=1, when we reached k=kmax we need to impose W_{i,j,kmax}=W_{i,i,1} (W is angular velocity component) P_{i,j,kmax}=P_{i,j,1} what can we say about void fraction? thank you very much again.

 July 25, 2005, 17:55 Re: two-fluid model #5 Jim_Park Guest   Posts: n/a If you have a void fraction, does that mean you're talking about a free surface in your cylinder? Or did you mean a volume fraction? And you're doing a problem where the cylinder is full - no free surface? I'm not familiar with that Harlow and Amsden model. A sketch would make it easy to see how Tony Amsden usually handled this kind of thing. [I tried to do a sketch, but the display scrambles the picture.] If you have nc real cells, number them from 1 to nc + 2. Numbers 1 and nc+2 are ghost cells. The contents of 2 through nc + 1 are for real fluid. After each time step (or iteration), values from nc+1 are put into 1 & values from 2 are put into nc + 2 This closes the loop so that the fluid that is physically connected is also connected in your code. Hope I haven't forgotten something!

 July 26, 2005, 03:31 Re: two-fluid model #6 enis Guest   Posts: n/a Dear Park, This was exactly what i need. thank you very very much. As far as i understand this type of boundary condition is called as "periodic boundary condition" in the literature, am i right? P.s.:I meant actually volume fraction sorry. so the cylinder is considered to be full of water and air.

 July 26, 2005, 08:49 Re: two-fluid model #7 Jim_Park Guest   Posts: n/a Yes. It is a periodic boundary condition.

 July 27, 2005, 15:35 Re: two-fluid model #8 enis Guest   Posts: n/a i applied your recommendations about angular directions to my program. then i want to ask about another thing which is to specify radial boundary conditions in the centerline of the cylinder. I used free-slip condition for r=0. That is, radial velocity (U) is set to zero, while for the other velocities dV/dr=0 and dW/dr=0. pressure and volume fractions are extrapolated from the adjacent inner cell. Do you think there is any problem? also what shoul i use for the value of r at r=0.? so small value like 0.00001 or a negative one like -dr/2. thanks a lot.

 July 27, 2005, 17:22 Re: two-fluid model #9 Jim_Park Guest   Posts: n/a I need to ask some more questions. 1. Are you using a Harlow and Amsden code, or have you written your own, or are you using a commercial code? 2. Is your problem formulated as using finite difference, finite volume, or finite element - or something completely different? 3. Is your mesh laid out with the variables staggered or colocated? 4. Are the air and water mixed thoroughly or are they two distinct phases? Boundary conditions along an axis in cylindrical coordinates are difficult if the flow is not axi-symmetric. 5. It sounds from one of your first messages that your flow is not axisymmetric. Is that correct?

 July 28, 2005, 01:58 Re: two-fluid model #10 enis Guest   Posts: n/a 1. no i have written my own code which uses their solution algorithm. 2. it is a finite difference one. 3.i have been using staggered grid. velocities are defined in the cell-edges, volume fractions and pressure are defined in the cell-centers. 4.they are assumed to mixed. im not interested in the interface between them. 5.yes it is not-axisymmetric.

 July 28, 2005, 11:23 Re: two-fluid model #11 Jim_Park Guest   Posts: n/a Two things to 'discuss' here. First, the passing of the values for the axial velocity ('w'?) needs to be precise. My notation is that the radial momentum and continuity equations are axially at cell centers - the space goes from cell 2 to cell nc+1. The value from the right face of cell number 2 needs to go to the right face of cell nc+2 AND the value from the left face of cell nc+1 needs to go to the left face of cell 1. You'll solve the axial momentum equation for EITHER the right face of cell nc+1 OR the left face of cell number 2 - but not both. Then pass the value of w for the face you've solved to the other. Second, what to do about 1/r as r -> 0 on the axis? I don't KNOW what will work but have a suggestion for you to try. It might just blow up in your face, but I hope not: Only the radial momentum equation is evaluated on the axis, and then only for non-axisymmetric flows. [Of course you need BC's for the other equations on the boundary.] Suggestion [worth exactly what you're paying for it! : )]: Multiply your differential equation (not the difference equation!!) for u (radial velocity) by r and difference that. The 1/r terms are eliminated, leaving a soluble equation for r on the axis. I don't KNOW that this will work. Perhaps someone else has tried it and KNOWs it won't. Good Luck - and let us know how it comes out.

 July 28, 2005, 12:38 I Was Wrong! #12 Jim_Park Guest   Posts: n/a My suggestion or answer for the 1/r as r -> 0 problem is off base, or incomplete, or perhaps just wrong. Multiplying through by r might (or might NOT) help, but the big problem is that the regular mesh sort of collapses. I don't know how to handle that and I don't know how it's been handled in the past. Sorry to waste your time.

 July 29, 2005, 12:32 Re: I Was Wrong! #13 Angen Guest   Posts: n/a Jim, Congratulations. I admire your courage. You cannot see frequently this kind of answers these days. Angen

 August 1, 2005, 13:19 Re: two-fluid model #14 MT Guest   Posts: n/a Hi, Instead of solving for radial velocity (u), solving for (r*u) has been done....and is shown to produce good results....I'm just recalling from my memory and can't give you any reference right now...I guess it was a Journal of Computational physics paper. Just wanted to let you know....Jim isn't completely wrong... cheers MT

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