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Posted:
1 decade ago
4 mag 2011, 11:32 GMT-4
Yeah,
You can define a surface and ask the solver to integrate over that specific surface.
Cheers
Yeah,
You can define a surface and ask the solver to integrate over that specific surface.
Cheers
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Posted:
1 decade ago
5 mag 2011, 01:42 GMT-4
You are right,
it is possible to define a surface being bounded by geometric entities.
What I want is to define a surface whose boundaries depend on the value of a variable which was solved for before.
Regards,
Marcel
You are right,
it is possible to define a surface being bounded by geometric entities.
What I want is to define a surface whose boundaries depend on the value of a variable which was solved for before.
Regards,
Marcel
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Posted:
1 decade ago
5 mag 2011, 04:23 GMT-4
Hi
You want to average the pressure of a part of a boundaray, which has a temperature higher than a reference temperature 'Tref'?
Not tested:
integrate the following expression over the boundary:
(T>=Tref)
result: length/area of the part of the boundary.
next: integrate the following expression over the boundary:
(T>=Tref)*p
divide the result of B by A. That's what you want?
best regards
Hi
You want to average the pressure of a part of a boundaray, which has a temperature higher than a reference temperature 'Tref'?
Not tested:
integrate the following expression over the boundary:
(T>=Tref)
result: length/area of the part of the boundary.
next: integrate the following expression over the boundary:
(T>=Tref)*p
divide the result of B by A. That's what you want?
best regards
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Posted:
1 decade ago
5 mag 2011, 05:01 GMT-4
Thanks for your answer.
With result A I get the area on the boundary where T>Tref.
Integrating the whole surface and just multiplying (T>Tref) simply uses a value of zero when T<Tref and the average value is far too low.
Regards,
Marcel
Thanks for your answer.
With result A I get the area on the boundary where T>Tref.
Integrating the whole surface and just multiplying (T>Tref) simply uses a value of zero when T
Ivar KJELBERG
COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted:
1 decade ago
5 mag 2011, 09:14 GMT-4
Hi
yes there is the fact that an average divides by the full area, so you need to do two separate integrations and make the division manually (or on the same line) and not use "average"
--
Good luck
Ivar
Hi
yes there is the fact that an average divides by the full area, so you need to do two separate integrations and make the division manually (or on the same line) and not use "average"
--
Good luck
Ivar