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Cut Plane 3d not through whole geometry

Posted Jun 7, 2016, 8:22 a.m. EDT Geometry, Results & Visualization Version 4.4 4 Replies

Alexander Rave
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Hello,

I am simulating a fluid flow with heat transfer in a bent rectangular microchannel. For postprocessing, I defined several cut planes where I do integrations on (derived values). It all worked well as long as I had only one arc. Now I combined two identical arcs. Unfortunately, some of the cut planes cut two or three times through my channel, making it impossible to do evaluations on it (see attached picture ). So here are my questions:

Is there a way to tell COMSOL to stop cutting through the geometry at a certain (for example) x-value?

Or is it possible to do the surface integration only on a part of the whole cut plane?

I am looking forward for some ideas.

Kind regards,
Alexander

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4 Replies Last Post Jun 7, 2016, 9:43 a.m. EDT
Jeff Hiller COMSOL Employee

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Posted: 9 years ago Jun 7, 2016, 8:52 a.m. EDT
Hello Alexander,
Cut Planes are planes in the mathematical sense, hence infinite.
The trick is to include a boolean in the expression your integrand.
Say you want to intergrate "u" in the part of the cur plane that has coordinates x>15e-3 (i.e. the rightmost of the three intersections between your geometry and the cut plane in your screenshot) the you can ask COMSOL to integrate "u*((x>15e-3)".
Best,
Jeff
Hello Alexander, Cut Planes are planes in the mathematical sense, hence infinite. The trick is to include a boolean in the expression your integrand. Say you want to intergrate "u" in the part of the cur plane that has coordinates x>15e-3 (i.e. the rightmost of the three intersections between your geometry and the cut plane in your screenshot) the you can ask COMSOL to integrate "u*((x>15e-3)". Best, Jeff
Alexander Rave
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Posted: 9 years ago Jun 7, 2016, 9:19 a.m. EDT
Hello Jeff,

thank your very much for your quick response. It seems that this works.

I have one further question: When I am doing not an integration, but an averaging, will I get the right result? Because when COMSOL takes just the integrand as zero, but continues to integrate over the full cut plane, than I will get the wrong average value because the surface is to large.

But if COMSOL does not sum up the surface elements where the boolean value is false, then this should work, too.

Best wishes from Hamburg, Germany,
Alexander

Hello Jeff, thank your very much for your quick response. It seems that this works. I have one further question: When I am doing not an integration, but an averaging, will I get the right result? Because when COMSOL takes just the integrand as zero, but continues to integrate over the full cut plane, than I will get the wrong average value because the surface is to large. But if COMSOL does not sum up the surface elements where the boolean value is false, then this should work, too. Best wishes from Hamburg, Germany, Alexander
Jeff Hiller COMSOL Employee

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Posted: 9 years ago Jun 7, 2016, 9:24 a.m. EDT
You can use the boolean trick above also to compute the surface area of the rightmost cross-section: integrate "x>15e-3". Then you get the average by dividing the integral of "u*(x>15e-3)" by the integral of "x>15e-3" .
Jeff
You can use the boolean trick above also to compute the surface area of the rightmost cross-section: integrate "x>15e-3". Then you get the average by dividing the integral of "u*(x>15e-3)" by the integral of "x>15e-3" . Jeff
Alexander Rave
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Posted: 9 years ago Jun 7, 2016, 9:43 a.m. EDT
Thank you again.

Yes, I have to do it as your suggested. Just applying the averaging function with the "boolean trick" yields the wrong value because obviously Comsol devides the right integrated value by the total and not the "boolean" surface.

That's everything I wanted to find out :-)

Kind regards,
Alexander
Thank you again. Yes, I have to do it as your suggested. Just applying the averaging function with the "boolean trick" yields the wrong value because obviously Comsol devides the right integrated value by the total and not the "boolean" surface. That's everything I wanted to find out :-) Kind regards, Alexander

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