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Posted:
1 decade ago
Mar 25, 2013, 9:45 a.m. EDT
as the cross section is constant we have constant current through each line, so I was thinking to use point evaluation but I need help how to do it.
Thanks in advance
as the cross section is constant we have constant current through each line, so I was thinking to use point evaluation but I need help how to do it.
Thanks in advance
Ivar KJELBERG
COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted:
1 decade ago
Mar 25, 2013, 1:00 p.m. EDT
Hi
you have also the "arrow lines" you could try
--
Good luck
Ivar
Hi
you have also the "arrow lines" you could try
--
Good luck
Ivar
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Posted:
1 decade ago
Mar 26, 2013, 6:47 a.m. EDT
Dear Ivar,
I really appreciate your help in this forum. It already has helped me a lot when I am reading other posts with your commetns on them.
I used arrow lines. it was good for showing the current direction. however I want to do something similar to line integration that is used to get the total heat flux in 2D, for example here:
www.youtube.com/watch?v=N784M5Mkp0o
I want to do the same, how ever on the boundary points to get the total current.
Thanks again
Ehsan
Dear Ivar,
I really appreciate your help in this forum. It already has helped me a lot when I am reading other posts with your commetns on them.
I used arrow lines. it was good for showing the current direction. however I want to do something similar to line integration that is used to get the total heat flux in 2D, for example here:
http://www.youtube.com/watch?v=N784M5Mkp0o
I want to do the same, how ever on the boundary points to get the total current.
Thanks again
Ehsan
Sven Friedel
COMSOL Employee
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Posted:
1 decade ago
Mar 26, 2013, 6:47 a.m. EDT
Dear Ehsan,
what do you mean by "derive it"? Would you like to export the data as txt file, display the values graphically ... ?
Please note that tangential derivatives are not defined in points but only on edges, because points have no tangent.
There is however a way to derive the current in a point "on the side of a neighbroung edge".
Take for instance in the model I sent earlier and attach here again point 7. It is connected to the edges 10, 12 and 7.
If you want to know the current in point 7 on the side of edge 12, you can access that by using a point evaluation for the expression
side(12,sqrt(uTx^2+uTy^2+uTz^2))*(rho/A)
I have addeed that to the file attached. Hope that helps.
Sven Friedel
Dear Ehsan,
what do you mean by "derive it"? Would you like to export the data as txt file, display the values graphically ... ?
Please note that tangential derivatives are not defined in points but only on edges, because points have no tangent.
There is however a way to derive the current in a point "on the side of a neighbroung edge".
Take for instance in the model I sent earlier and attach here again point 7. It is connected to the edges 10, 12 and 7.
If you want to know the current in point 7 on the side of edge 12, you can access that by using a point evaluation for the expression
side(12,sqrt(uTx^2+uTy^2+uTz^2))*(rho/A)
I have addeed that to the file attached. Hope that helps.
Sven Friedel
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Posted:
1 decade ago
Mar 26, 2013, 7:10 a.m. EDT
Dear Sven, Hello again,
I need to get the total current in the structure.
I can do it by exporting the data of the arrow lines plot (for the edges connected to the boundary points) but I have to choose them manually, which is not actually practical when I go to my bigger models.
So the best choice(in my opinion) is to choose the same nodes of the constraint(u-1 or u-0) and find the total current. However I'm not sure if this is the right way.
Thanks again for your help,
Ehsan
Dear Sven, Hello again,
I need to get the total current in the structure.
I can do it by exporting the data of the arrow lines plot (for the edges connected to the boundary points) but I have to choose them manually, which is not actually practical when I go to my bigger models.
So the best choice(in my opinion) is to choose the same nodes of the constraint(u-1 or u-0) and find the total current. However I'm not sure if this is the right way.
Thanks again for your help,
Ehsan
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Posted:
1 decade ago
Mar 26, 2013, 7:15 a.m. EDT
Sorry I didnt see your message completely. Yes this is exactly what I want. is there a way to find the current in all the edges(or any) connected to point 7?
so I can use it automatically for a set of points.
By the way, I have searched a lot to find a list of these kind of expressions. Are they listed somewhere in the documentations.
Thanks a lot
Sorry I didnt see your message completely. Yes this is exactly what I want. is there a way to find the current in all the edges(or any) connected to point 7?
so I can use it automatically for a set of points.
By the way, I have searched a lot to find a list of these kind of expressions. Are they listed somewhere in the documentations.
Thanks a lot
Ivar KJELBERG
COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted:
1 decade ago
Mar 26, 2013, 1:00 p.m. EDT
Hi
Sven
Where is "side()" documented ? I have noticed that there are quite some goodies under the hood not really documented yet ?
__
Having fun Comsoling, mostly ...
Ivar
Hi
Sven
Where is "side()" documented ? I have noticed that there are quite some goodies under the hood not really documented yet ?
__
Having fun Comsoling, mostly ...
Ivar
Sven Friedel
COMSOL Employee
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Posted:
1 decade ago
Mar 27, 2013, 1:02 p.m. EDT
Hi,
side is one of my favourites recently. It will be in the doc 4.3b shortly, I heard.
Sven
Hi,
side is one of my favourites recently. It will be in the doc 4.3b shortly, I heard.
Sven
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Posted:
1 decade ago
Mar 27, 2013, 1:55 p.m. EDT
Hi,
I tried to invent somethings like side(:,sqrt(uTx^2+uTy^2+uTz^2))*(rho/A) to get the current in all the connected edges but they don't work(obviously).
However thanks for your help.
Ehsan
Hi,
I tried to invent somethings like side(:,sqrt(uTx^2+uTy^2+uTz^2))*(rho/A) to get the current in all the connected edges but they don't work(obviously).
However thanks for your help.
Ehsan