# Boundary conditions using SIMPLER algorithm

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 October 19, 2010, 09:20 Boundary conditions using SIMPLER algorithm #1 Senior Member   Join Date: Apr 2009 Posts: 118 Rep Power: 10 Sponsored Links I have seen a number of forum posts on this topic but I'm still struggling with my boundary conditions. I'm running a 1-D spherical calculation (only radial dependence considered). I decided to have velocity boundary conditions at the inlet and outlet. So the BCs I have are: inlet: outlet: which is zero gradient BC. I understand that for subsonic simulations you either have to specify velocity or pressure BCs. I have a staggered grid where the velocities are staggered forward. My discretised equation for the pressure looks of the form I used the discretised equation to calculate the pressure values at the boundary. At the inlet since I have therefore, I do not have in the discretised pressure equation and for the outlet I do not have . Is this correct?

 October 19, 2010, 11:01 #2 New Member   Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 8 How can you have 0 velocity in and non-zero velocity out, unless of course you are generating matter inside your volume?

 October 20, 2010, 05:54 #3 Senior Member   Join Date: Apr 2009 Posts: 118 Rep Power: 10 I'm studying combustion where the zero-upstream velocity is a far-field velocity and there is a flame in the middle of the domain and the boundary condition at the outlet is a zero-gradient boundary condition but the flow reaches about 60 m/s at this outlet. I was at first baffled by your question but I think the combustion, creates density difference which creates mass.

 October 20, 2010, 07:54 #4 New Member   Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 8 Could you maybe sketch your geometry? I understood it as: Flame (r=0)-----inlet (r=r_i)-----------outlet (r=r_o)----> radial coordinate Please bear with me if that is not how it looked :-)

 October 20, 2010, 08:46 #5 Senior Member   Join Date: Apr 2009 Posts: 118 Rep Power: 10 It's actually, Outlet (r=0) ----------------------------------------Flame------------------Inet (r=R) I use a normalised temperature to define the flame location. I study spherical explosions where the flame starts at the centre and moves radially outward. Therefore at the centre (r=0) is the highest temperature (burnt side) where the normalised temperature C=1. I end simulations before the flame reaches R (end of the domain). Since there is no flame here, the normalised temperature C=0 at the outlet. So the profile of normalised temperature (C) looks like a backward ramp function. So the flame actually moves from left to right during my calculations. However, relative to the flame is the flow velocity, so for a particle fixed on the flame the flow direction will be from right-to-left. That's why I defined my inlet to be at r=R, and outlet to be at r=0.

 October 20, 2010, 09:36 #6 New Member   Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 8 Ok, I think I get it finally. These are suggestions, I am by no means a CFD expert ... With regards to the pressure: for symmetry you should have a zero gradient at r=0, so then P(N)=P(N-1), where N is the outlet node and N-1 is the node right next to it (here to the EAST). You could do that by setting , , and in the pressure-equation written above. At the inlet boundary (r=R), I would specify that the pressure on the EAST point would be equal to the ambient pressure P0. You could do by lumping into the source term (b). In b, I think the velocities at the faces are also given, so there you would have to specify that the velocity at east face is 0. Let me hear what you think. Regards

 October 20, 2010, 16:10 #8 Senior Member   Join Date: Apr 2009 Posts: 118 Rep Power: 10 Actually I realise that I don't include the centre point of the sphere, r=0 because it creates a singularity. Instead I start it at r=a, some finite value. So I'm not sure if I need to have symmetry BCs in this case.

 October 21, 2010, 10:18 #9 New Member   Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 8 If your centre point for the pressure discretization is at 0, then wouldn't the control volume at that point would be a sphere, rather than a spherical shell? Then you only have flow out of there, right?. I think this should change your discretized equations there. With regards to the inlet, you could move the grid for the pressure so it does not end at r=R, but at . Then as you are staggering the velocity grid forward your last velocity node will be at r=R, right? In 'b' in the pressure-equation you could set u_e = 0, while keeping u_w from your solution to the momentum equation. Also you set a_E = 0 here, as you already wrote. Again, I'm by no means sure of this.. I don't remember seeing an example in the textbooks I've been reading except for Patankar, regarding discretization in spherical coordinates and that wasn't for the NS equations). Hopefully someone more qualified will answer you soon

 October 25, 2010, 18:27 #10 New Member   alex Join Date: Oct 2010 Posts: 4 Rep Power: 8 Hello, I am a mechanical engineering.I need to use a subroutine ables to solve (in unsteady condition) 3-D fluid flow (the pressure and velocity and temperature) fields in cylindrical coordinate .Who Can send me the source code for SIMPLE / SIMPLER scheme (preferably on fortran) so that I can get & proceed from it. please send me through my email, ahazbavi@ymail.com.

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