# Putting Boundary Conditions in ANSYS CFX 12.1

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 November 27, 2010, 18:37 Putting Boundary Conditions in ANSYS CFX 12.1 #1 New Member   Join Date: Nov 2010 Posts: 9 Rep Power: 6 Hey Guys!! I am doing 2-way coupled 'Fluid-Structure Interaction' simulation of blood flow in axisymmetric abdominal aortic aneurysm, using ANSYS 12.1 mechanical and Fluid Flow(CFX). I have the following questions :- 1) The boundary conditions at the inlet and outlet of the fluid domain are those of velocity and pressure respectively as follows: uz(r, t) = 1.22um(t)[1−(2r/d)]^(1/7), ur=0 at z = 0 σnn = ˆn · p(t)I · ˆn at z = Le I have the time variations of um(t) and p(t) waveforms. How do I put these boundary conditions in ansys Fluid Flow(CFX)?? 2) How do I put the boundary condition of symmetry about the centreline?? 3) The boundary conditions at the interface between solid and fluid domains include no-slip, displacement and traction boundary conditions. Do I need to put them explicitly or they are assumed on their own by ANSYS?? If I do need to put them, then how?? If anybody could help, it would be great!!

 January 3, 2015, 11:05 Putting Boundary Conditions in ANSYS CFX 15 #2 New Member   David Join Date: Aug 2013 Posts: 27 Rep Power: 3 Does anyone know how to apply zero traction boundary condition in ANSYS CFX? zero traction boundary condition definition (t is the stress vector): t=T.n=0 where T=-pI +2mu D D is the rate of deformation tensor and I is the unit tensor. Or can anyone tell me how to set P and dP/dx to be zero at the same time in ANSYS cfx?

 January 3, 2015, 23:44 #3 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,797 Rep Power: 84 CFX does not have this boundary condition. CFX can set a specified pressure, but the dP/dX is set by the convected fluid values, not an artificial imposed gradient condition. In my opinion the CFX approach is superior to the zero traction condition you describe. If you REALLY want to impose a zero traction boundary condition here is my recommendation: You may be able to do a user momentum source which adjusts the velocities to give p=0 and dP/dX=0. I am not sure you can do this on a boundary patch, (I suspect not), you probably need to do it on a volume source where the volume is adjacent to the boundary and a few nodes in. Note that having p=0 and dP/dX=0 at the boundary implies that not only the boundary node has zero pressure, but the next node into the domain also have zero pressure (to give zero gradient). So your boundary condition will require controlling the pressure at some internal nodes as well. This is why I suspect you will need to use a volume source term to do this.

 January 4, 2015, 11:58 #4 New Member   David Join Date: Aug 2013 Posts: 27 Rep Power: 3 Hi Glenn, Thank you for your useful information, you are right, It cannot be applied to a face which is normal to the flow direction. Therefore I created an adjacent volume to my outlet boundary condition. but how can I adjust velocities to give p=0 and dP/dX=0? I have not any data about my domain outlet and I found that U,V,W options in this volume sources tab is only for sources and relevant sink options.

 January 4, 2015, 17:47 #5 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,797 Rep Power: 84 It's just a hunch, I will let you do the detailed work Take the current pressure and using Bernoulli work out the velocity you need to have zero pressure. Apply that velocity as a source term. And do it over a volume so that inside the volume source region the gradients are zero. This approach is inferior to the outlet flow approach built into CFX. Why do you want to put an inferior boundary condition on your flow?

 January 5, 2015, 03:16 #6 New Member   David Join Date: Aug 2013 Posts: 27 Rep Power: 3 Yes,yes,yes, It is a great hunch The problem is with the radial displacement of a flexible tube. I read an excellent thesis which I sent it to you by private message, in the section 5.3 (page130) it compares various outlet boundary conditions for a case which is similar to mine, and in the section 5.3.3 it is mentioned that using a constant Pressure Outlet Boundary Condition will lead to smaller radial deformation. I'm simulating the blood flow in a geometry which is given in the article which I sent it to you by private message, and my radial deformation is smaller than the article's deformation that is given in the fig.3, I think that the problem is arising with the boundary condition. The article's outlet boundary condition is zero traction. Thank you for your help, Now I should try what you said.

 January 5, 2015, 05:27 #7 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,797 Rep Power: 84 Maybe I have not read the thesis in enough detail but I cannot see anywhere where it says the outlet boundary condition is p=dP/dX=0. In section 4.2.1 it describes the impedance boundary condition as a complex condition using the phase lag between pressure and flow waves. This sounds a lot more complex than you describe. But regardless of that, I would use CFX's built in outlet boundary models and show they are inadequate before doing a big heap of work implementing your own boundary condition. You may find they are not so bad after all. Have you shown the built in options are unacceptable? There are also non-documented options available such as the non-reflecting boundary condition which may also be useful. And of course there is the safe option: just extend the domain into where ever it is going, then reduce the importance of the outlet boundary. This may be expand it out into a big box or continue the tube on for longer. Then you do not need to do tricky boundary conditions as simple ones will be fine.

 January 5, 2015, 06:51 #8 New Member   David Join Date: Aug 2013 Posts: 27 Rep Power: 3 I found about P=dP/dx=0 for zero traction boundary condition in this link: https://www.researchgate.net/post/Wh...ndition_in_CFD You are right, Farfield boundary condition probably solves my problem. The problem with the boundary condition is now solved. Thank you Last edited by mrkmrk; January 5, 2015 at 09:41.

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