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How can one give bounbary conditions other than symmetry boundary condition on the same egde?
Posted 2010年6月18日 GMT-4 08:17 9 Replies
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I would like to solve a multiphisics problem which consists of navier stokes and heat equation.
I use a 2D- axysimmetrical domain. On the symmetry axis I would like to give the boundary conditions written below.
du/dr=0, v=0 and dT/dr , theses derivatives are partial derivatives, u and v are velocities in the r and z directions. T is temperature..
But I could not mange to give these boundary conditions. I used comsol 3.5a and I could only give the axial symmetry boundary condition r=0
How can I give the mentioned boundary conditions.
I use a 2D- axysimmetrical domain. On the symmetry axis I would like to give the boundary conditions written below.
du/dr=0, v=0 and dT/dr , theses derivatives are partial derivatives, u and v are velocities in the r and z directions. T is temperature..
But I could not mange to give these boundary conditions. I used comsol 3.5a and I could only give the axial symmetry boundary condition r=0
How can I give the mentioned boundary conditions.
9 Replies Last Post 2017年8月13日 GMT-4 21:48
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Posted:
1 decade ago
Hi
first of all carefull with the naming, COMSOL uses quite some letter internally, such as r,s,t,u,v,w,x,y,z ... with specific meanings.
Then on a symmetry axis, you are limited by what you can define, othervise how do you obtain "symmetry"
think it over, good luck
Ivar
first of all carefull with the naming, COMSOL uses quite some letter internally, such as r,s,t,u,v,w,x,y,z ... with specific meanings.
Then on a symmetry axis, you are limited by what you can define, othervise how do you obtain "symmetry"
think it over, good luck
Ivar
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Posted:
1 decade ago
Hi Atilla,
I don't think it makes any physical sense to define boundary conditions on a symmetry axis. The symmetry axis is just a geometrical object that defines how to rotate the 2D cross section into 3D to make a real world object of it.
The symmetry axis is actually not a boundary.
Best regards
Edgar
I don't think it makes any physical sense to define boundary conditions on a symmetry axis. The symmetry axis is just a geometrical object that defines how to rotate the 2D cross section into 3D to make a real world object of it.
The symmetry axis is actually not a boundary.
Best regards
Edgar
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Posted:
1 decade ago
Hi
Well I do not 100% agree it's a type of boindary, as COMSOL too adds some conditions, such as i.e. in structural u=0 and test(-u)=0 which means that (lateral) displacements are zero on axis (for me this is the definition of an "axis"), and gradients are normal I believe, would have to think it over a bit furter for tha latter item ;)
Good luck
Ivar
Well I do not 100% agree it's a type of boindary, as COMSOL too adds some conditions, such as i.e. in structural u=0 and test(-u)=0 which means that (lateral) displacements are zero on axis (for me this is the definition of an "axis"), and gradients are normal I believe, would have to think it over a bit furter for tha latter item ;)
Good luck
Ivar
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Posted:
1 decade ago
Dear Mr Kjelberg and Kaiser,
Thank you for your replies, it is now clear to me that there won't be any boundary conditions on the symmetry axis.
But I have other questions to ask, why can't I get convergengence while solving a multiphysics problem including navier stokes and heat transfer. Where should I look and check in order to achive convergence.
I thought that the first place to check is the boundary conditions , am I right?
Thank you for your replies, it is now clear to me that there won't be any boundary conditions on the symmetry axis.
But I have other questions to ask, why can't I get convergengence while solving a multiphysics problem including navier stokes and heat transfer. Where should I look and check in order to achive convergence.
I thought that the first place to check is the boundary conditions , am I right?
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Posted:
1 decade ago
Hi
in V4 its not much different than in V3.5, the only thing is that the fileds to control the solver and rthe physics are placed differently and some has changed name.
for me when you do multiphyiscs, first one should ensure that each multiphysics solves by itself, then to couple them. This requires often to enable/disable some BCs or tree entries.
Further, when mixing physics, one must call different solers, which can be done in several ways (I have learned recentls).
You have the obvious solver sequence, BUT ALOS the physics Advanced tan, to replace the "automatic by a specific solver method + additional inputs such as sweep ranges etc, when applicable.
see also www.comsol.eu/community/forums/general/thread/6332/#p17416
Have fun Comsoling
Ivar
in V4 its not much different than in V3.5, the only thing is that the fileds to control the solver and rthe physics are placed differently and some has changed name.
for me when you do multiphyiscs, first one should ensure that each multiphysics solves by itself, then to couple them. This requires often to enable/disable some BCs or tree entries.
Further, when mixing physics, one must call different solers, which can be done in several ways (I have learned recentls).
You have the obvious solver sequence, BUT ALOS the physics Advanced tan, to replace the "automatic by a specific solver method + additional inputs such as sweep ranges etc, when applicable.
see also www.comsol.eu/community/forums/general/thread/6332/#p17416
Have fun Comsoling
Ivar
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Posted:
1 decade ago
I learnt from the user guide that another way of reaching convergence to give some initial values from the subdomain setting even though you are dealing with stationary problem.
The user guide says:
The Navier-Stokes equations are nonlinear and therefore require an educated guess as an initial solution to the nonlinear solver. A good initial value is necessary for the model to converge.
Have a nice weekend.
atilla
The user guide says:
The Navier-Stokes equations are nonlinear and therefore require an educated guess as an initial solution to the nonlinear solver. A good initial value is necessary for the model to converge.
Have a nice weekend.
atilla
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Posted:
1 decade ago
Hi
indeed often its usefull to run a static analysis on as many variables as possible to get a good starting point, it often helps for the convergence.
Have fun comsoling
Ivar
indeed often its usefull to run a static analysis on as many variables as possible to get a good starting point, it often helps for the convergence.
Have fun comsoling
Ivar
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Posted:
7 years ago
Hi, Edgar!
I don't agree that the axis is not a boundary. Maybe from the 3D view the axis is inside the body and doesn't appear as a boundary. But physically for the system to be cylindrically symmetrical there are constraints on physical quantities, e.g., the r-component of a vector should be zero on the axis. Mathematically it is even more clear it is a boundary. For the unique determination of the solution of the PDE boundary conditions are needed.
However, I'm not sure how COMSOL implements it behind the scene. In many cases I see no expression under the node axial symmetry, which does seem to enforce no condition. This couldn't work universally, as the situation is very different for m = 0 and m != 0.
Also from my experience sometimes a reasonable solution can only be obtained if I set additional constraints on the axis.
Cheers!
--
Pu, ZHANG
DTU Fotonik
I don't agree that the axis is not a boundary. Maybe from the 3D view the axis is inside the body and doesn't appear as a boundary. But physically for the system to be cylindrically symmetrical there are constraints on physical quantities, e.g., the r-component of a vector should be zero on the axis. Mathematically it is even more clear it is a boundary. For the unique determination of the solution of the PDE boundary conditions are needed.
However, I'm not sure how COMSOL implements it behind the scene. In many cases I see no expression under the node axial symmetry, which does seem to enforce no condition. This couldn't work universally, as the situation is very different for m = 0 and m != 0.
Also from my experience sometimes a reasonable solution can only be obtained if I set additional constraints on the axis.
Cheers!
Hi Atilla,
I don't think it makes any physical sense to define boundary conditions on a symmetry axis. The symmetry axis is just a geometrical object that defines how to rotate the 2D cross section into 3D to make a real world object of it.
The symmetry axis is actually not a boundary.
Best regards
Edgar
--
Pu, ZHANG
DTU Fotonik
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Posted:
7 years ago
Hi, Ivar!
I have the same feeling that certain boundary conditions need to be enforced on the axis. But there's essentially no related information in COMSOL documentation. It would be great if you can share your opinion on the boundary condition on the axis.
Thanks!
--
Pu, ZHANG
HUST
I have the same feeling that certain boundary conditions need to be enforced on the axis. But there's essentially no related information in COMSOL documentation. It would be great if you can share your opinion on the boundary condition on the axis.
Thanks!
Hi
Well I do not 100% agree it's a type of boindary, as COMSOL too adds some conditions, such as i.e. in structural u=0 and test(-u)=0 which means that (lateral) displacements are zero on axis (for me this is the definition of an "axis"), and gradients are normal I believe, would have to think it over a bit furter for tha latter item ;)
Good luck
Ivar
--
Pu, ZHANG
HUST