Boundary conditions using SIMPLER algorithm

lost.identity · October 19, 2010, 09:20

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?

johnhelt · October 19, 2010, 11:01

How can you have 0 velocity in and non-zero velocity out, unless of course you are generating matter inside your volume?

lost.identity · October 20, 2010, 05:54

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.

johnhelt · October 20, 2010, 07:54

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 :-)

lost.identity · October 20, 2010, 08:46

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.

johnhelt · October 20, 2010, 09:36

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

lost.identity · October 20, 2010, 13:00

Hi thanks very much for the quick reply.

I'm still confused. This is a lengthy reply but I'm trying to understand certain things between what I have now and what you've just told me.

So I'm using SIMPLER as I said before and I use a staggered grid.

What I have now is a zero-gradient BC at the outlet (r=0) for velocity which says

. Am I correct in thinking that if I have this velocity BC then I can't have a pressure BC at the same location for incompressible flow?

At this boundary I discretise the pressure equation to obtain (just neglecting a_w term):

and according to SIMPLER method the mass source term b looks like

$b = (\rho_P + \rho_P^0)r_P^2\frac{\Delta{r}}{\Delta{t}} - r_e^2(\rho\hat{u})_e + r_w^2(\rho\hat{u})_w$ ------------- (1)

where r_P, r_w and r_e are the radii at the relevant points along the grid (these terms only come about because of the spherical discretisaiton). Also my grid is staggered and I store the velocities at the velocity-cell faces. So with my current boundary conditions I set $u_w = \hat{u}_e$ , where the pseduo-velocities are only calculated for the internal nodes. The problem with this is that the

term doesn't go to zero at this location for spherical coordinates. This is ecause even though the velocities

and

are the same, the change in radii

and

creates a source. So even if I have neglected the

in the discretised equation for this boundary $p_1\neq{p_2}$ . Therefore, this violates symmetry. So am I right in thinking that rather than having zero-gradient with respect to velocity I should have zero-gradient with respect to pressure?

With regards to the inlet (r=R) BC, I have the discretised equation

by just neglecting a_E. A problem I have now is that I don't know what the cell face velocities should be at this location for a staggered grid to evaluate

according to Eq.(1). At the end I do not have

since it's that point is outside the grid, but I set

. However, my boundary condition for

. So despite the fact that there is a radius change the mass term b goes to zero here since the velocities are zero anyway. I'm really not sure about this bit.

lost.identity · October 20, 2010, 16:10

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.

johnhelt · October 21, 2010, 10:18

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 $R-1/2\Delta r$ . Then as you are staggering the velocity grid forward $1/2 \Delta r$ 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

ahazbavi · October 25, 2010, 18:27

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.

October 19, 2010, 09:20	Boundary conditions using SIMPLER algorithm	#1
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	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?

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October 19, 2010, 11:01		#2
johnhelt New Member Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 15	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
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	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
johnhelt New Member Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 15	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
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	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
johnhelt New Member Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 15	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, 13:00		#7
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	Hi thanks very much for the quick reply. I'm still confused. This is a lengthy reply but I'm trying to understand certain things between what I have now and what you've just told me. So I'm using SIMPLER as I said before and I use a staggered grid. What I have now is a zero-gradient BC at the outlet (r=0) for velocity which says . Am I correct in thinking that if I have this velocity BC then I can't have a pressure BC at the same location for incompressible flow? At this boundary I discretise the pressure equation to obtain (just neglecting a_w term): and according to SIMPLER method the mass source term b looks like $b = (\rho_P + \rho_P^0)r_P^2\frac{\Delta{r}}{\Delta{t}} - r_e^2(\rho\hat{u})_e + r_w^2(\rho\hat{u})_w$ ------------- (1) where r_P, r_w and r_e are the radii at the relevant points along the grid (these terms only come about because of the spherical discretisaiton). Also my grid is staggered and I store the velocities at the velocity-cell faces. So with my current boundary conditions I set $u_w = \hat{u}_e$ , where the pseduo-velocities are only calculated for the internal nodes. The problem with this is that the term doesn't go to zero at this location for spherical coordinates. This is ecause even though the velocities and are the same, the change in radii and creates a source. So even if I have neglected the in the discretised equation for this boundary $p_1\neq{p_2}$ . Therefore, this violates symmetry. So am I right in thinking that rather than having zero-gradient with respect to velocity I should have zero-gradient with respect to pressure? With regards to the inlet (r=R) BC, I have the discretised equation by just neglecting a_E. A problem I have now is that I don't know what the cell face velocities should be at this location for a staggered grid to evaluate according to Eq.(1). At the end I do not have since it's that point is outside the grid, but I set . However, my boundary condition for . So despite the fact that there is a radius change the mass term b goes to zero here since the velocities are zero anyway. I'm really not sure about this bit.

October 20, 2010, 16:10		#8
lost.identity Senior Member Join Date: Apr 2009 Posts: 118 Rep Power: 17	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
johnhelt New Member Tobias Elmøe Join Date: Oct 2010 Location: Denmark Posts: 9 Rep Power: 15	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 $R-1/2\Delta r$ . Then as you are staggering the velocity grid forward $1/2 \Delta r$ 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
ahazbavi New Member alex Join Date: Oct 2010 Posts: 4 Rep Power: 15	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.