Dear Alex,
please find my response below
------------------------------------------------------------------------
Date: Wed, 25 Feb 2015 08:58:23 +0100
From: [email protected]
To: [email protected];
Subject: Re: Dumux Digest, Vol 49, Issue 10
Dear Sandeep,
I have a 3d, cube domain of 1mx1mx1m, gridded with 100 cells in
each direction.
I tried to setup a Dirichlet boundary on a circular area of the XY
face, The circle centre has X and Y valuesof 0.5
and radius of 0.5m. (Below is the code for boundary setup).
So, at the cells, where the circumference of the circle passes,
there might be two types of boundary conditionat a single cell.
I think this may be the problem ?
I am not quite sure I understand your code corrrectly. But I am
wondering why you are using for-loops (comparing int to Scalar)
within your boundaryTypesAtPos() function. If you want to assign a
Dirchilet boundary condition to a circular area within one of your
cube faces it should be sufficient to have a condition to see whether
you are on the correct cube face (as you have done correctly) and
another condition to make sure your integration point (ip, the point
at globalPos) is within your circle.
I used the for loopsto specify the coordinate
(X,Y) values for the center of the circle. With the condition
"if (((globalPos[0]*globalPos[0])+
(globalPos[1]*globalPos[1]))==(r_*r_))" i tried to ensure that my
integration
point (ip, the point at globalPos) is within
a circle. Without the for loops, the circle can be inscribed anywhere
on the XY face,
wherever the condition fulfils.
The way you assign it you would only allow the ips directly on your
circumference to be Dirichlet and as far as I understand you want the
whole circle area to be a Dirichlet BC? Probably at the moment none of
your boundary faces are assigned with a Dirichlet BC.
I have also tried with"if (((globalPos[0]*globalPos[0])+
(globalPos[1]*globalPos[1]))<=(r_*r_))" but the result is same. Yes,
you are right
i want the whole circular area as my boundary and i think even if, my
condition is
"if (((globalPos[0]*globalPos[0])+
(globalPos[1]*globalPos[1]))==(r_*r_))" then it applies to the
circumferences (as you said) of all
circles that can exist within my circle of given radius (concentric
circles), thus applies to the area within the circle.
(Please correct me if wrong)
For some reason, i am not able to download the file from listserver,
Would you please, send it again and cc to my email address.
I put you on cc.
Thanks, i would see if this can be used to create a complete
cylindrical domain rather than a piece of cake.
Finally, I need to setup a time dependent boundary, where the
Dirichlet boundary will
move from a sector, a piece of cake to the next piece, along a
circle, depending on the time.
So, i would need to have a circular domain as well ? Or can it be
done in a cube grid ?
If you assign this kind of a boundary condition to a cube grid you
will make some error depending on your refinement.
So, it might be better to have a cylindrical domain ?
Best regards
Sandeep
On 02/24/2015 01:13 PM, Sandeep Chakraborty wrote:
Dear Alex,
>
> > The convergence error was "residual_[10201][0] is not finite"
> > (attached is a screen shot).
> It is hard to tell what the problem could be. Maybe the boundary
> conditions are not set correctly.
>
> > So, is this due to describing a circular boundary over a
Cartesian grid ?
> Are you really having a circular domain 2d or 3d? Can you
describe that
in more detail?
I have a 3d, cube domain of 1mx1mx1m, gridded with 100 cells in
each direction.
I tried to setup a Dirichlet boundary on a circular area of the XY
face, The circle centre has X and Y valuesof 0.5
and radius of 0.5m. (Below is the code for boundary setup).
So, at the cells, where the circumference of the circle passes,
there might be two types of boundary conditionat a single cell.
I think this may be the problem ?
void boundaryTypesAtPos(BoundaryTypes &values,
const GlobalPosition &globalPos) const
{
values.setAllNeumann();
Scalar right = this->bBoxMax()[0];
Scalar top = this->bBoxMax()[1];
Scalar bottom = this->bBoxMax()[2];
Scalar front = this->bBoxMin()[2];
Scalar cx_=0.5;
Scalar cy_=0.5;
Scalar r_=0.5;
if (globalPos[2] < front +eps_)
{
for (int i = cx_ ; i <=r_+globalPos[0] ; i++)
{
for (int j = cy_ ; j<=r_+globalPos[1] ; j++)
if
(((globalPos[0]*globalPos[0])+(globalPos[1]*globalPos[1]))==(r_*r_))
values.setAllDirichlet();
}
}
#if !ISOTHERMAL
values.setDirichlet(temperatureIdx, energyEqIdx);
#endif
}
> > Can one describe polar coordinate (converted to x and y
values) on a
> > Cartesian grid in Dumux
> It has often been practiced to read in x,y corrdinates of a
radially
> symmetric grid (piece of cake). For the boundary conditions you
usually
> get a global position in x,y and can then transform these
coordinates
> into polar coordinates (r,alpha) to set the proper boundary
conditions.
> There are no specific functions in dumux for this
transformation, but
> you could easily implement them into your problem file.
>
> > if there is polar grid creator in Dumux ?
> There is a Matlab script you could use for setting up 3d radially
> symmetric grids (piece of cake) in cartesian coordinates which
creates a
> dgf file (see attached files).
For some reason, i am not able to download the file from listserver,
Would you please, send it again and cc to my email address.
Finally, I need to setup a time dependent boundary, where the
Dirichlet boundary will
move from a sector, a piece of cake to the next piece, along a
circle, depending on the time.
So, i would need to have a circular domain as well ? Or can it be
done in a cube grid ?
Thanks and Best Regards
Sandeep
> From: [email protected]
<mailto:[email protected]>
> Subject: Dumux Digest, Vol 49, Issue 10
> To: [email protected]
<mailto:[email protected]>
> Date: Tue, 24 Feb 2015 12:00:07 +0100
>
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> Today's Topics:
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> 1. Re: Describing radial boundary on a Cartesian grid
> (Alexander Kissinger)
>
>
>
----------------------------------------------------------------------
>
> Message: 1
> Date: Mon, 23 Feb 2015 15:22:01 +0100
> From: Alexander Kissinger
<[email protected]>
<mailto:[email protected]>
> To: DuMuX User Forum <[email protected]>
<mailto:[email protected]>
> Subject: Re: [DuMuX] Describing radial boundary on a Cartesian grid
> Message-ID: <[email protected]>
<mailto:[email protected]>
> Content-Type: text/plain; charset="windows-1252"; Format="flowed"
>
> Dear Sandeep,
>
> > The convergence error was "residual_[10201][0] is not finite"
> > (attached is a screen shot).
> It is hard to tell what the problem could be. Maybe the boundary
> conditions are not set correctly.
>
> > So, is this due to describing a circular boundary over a
Cartesian grid ?
> Are you really having a circular domain 2d or 3d? Can you
describe that
> in more detail?
>
> > Can one describe polar coordinate (converted to x and y
values) on a
> > Cartesian grid in Dumux
> It has often been practiced to read in x,y corrdinates of a
radially
> symmetric grid (piece of cake). For the boundary conditions you
usually
> get a global position in x,y and can then transform these
coordinates
> into polar coordinates (r,alpha) to set the proper boundary
conditions.
> There are no specific functions in dumux for this
transformation, but
> you could easily implement them into your problem file.
>
> > if there is polar grid creator in Dumux ?
> There is a Matlab script you could use for setting up 3d radially
> symmetric grids (piece of cake) in cartesian coordinates which
creates a
> dgf file (see attached files).
>
> best regards
> Alex
>
>
> On 02/23/2015 12:55 PM, Sandeep Chakraborty wrote:
> > Dear All,
> >
> > Greetings to everyone.
> > I was trying to describe a circular boundary over a fine
Cartesian
> > grid, in the 2p2cni problem.
> > The problem didn't converge and more interestingly with different
> > timesteps the convergence relative error was always same.
> > The convergence error was "residual_[10201][0] is not finite"
> > (attached is a screen shot).
> >
> > So, is this due to describing a circular boundary over a
Cartesian grid ?
> > Can one describe polar coordinate (converted to x and y
values) on a
> > Cartesian grid in Dumux and if there is polar grid creator in
Dumux ?
> >
> > I appreciate your advice in advance.
> > Thanks and Best regards
> > Sandeep
> >
> >
> > _______________________________________________
> > Dumux mailing list
> > [email protected]
<mailto:[email protected]>
> > https://listserv.uni-stuttgart.de/mailman/listinfo/dumux
>
>
> --
> Alexander Kissinger
> Institut f?r Wasser- und Umweltsystemmodellierung
> Lehrstuhl f?r Hydromechanik und Hydrosystemmodellierung
> Pfaffenwaldring 61
> D-70569 Stuttgart
>
> Telefon: +49 (0) 711 685-64729
> E-Mail: [email protected]
<mailto:[email protected]>
>
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--
Alexander Kissinger
Institut für Wasser- und Umweltsystemmodellierung
Lehrstuhl für Hydromechanik und Hydrosystemmodellierung
Pfaffenwaldring 61
D-70569 Stuttgart
Telefon: +49 (0) 711 685-64729
E-Mail:[email protected]
<mailto:[email protected]>
