How to find area of a single mesh element

Dyls · January 30, 2014, 11:43

Hi

I have a temperature field along a x-y plane but I wish to multiply the corresponding temperature of a node with the area that it occupies.

For example, in my derived plane there exists many different sized elements of different shapes (because I am using a polyhedral mesh and cutting through a section). Is there a way to find what each corresponding nodes' areas are and then multiply by it's own temperature?

This would then be calculating a quantity with units of degree multiplied by area.

Thanks so much for your help!

Dylan

ping · January 31, 2014, 05:07

field function area is available on boundaries but not derived parts like a section plane. so you could place an interface between 2 regions at that location and you will be fine - create a field function $Temperature * mag($$Area) to display or report on.
an approximation might be possible using volume rather than area.

January 30, 2014, 11:43	How to find area of a single mesh element	#1
Dyls New Member Dylan Join Date: Mar 2012 Posts: 25 Rep Power: 14	Hi I have a temperature field along a x-y plane but I wish to multiply the corresponding temperature of a node with the area that it occupies. For example, in my derived plane there exists many different sized elements of different shapes (because I am using a polyhedral mesh and cutting through a section). Is there a way to find what each corresponding nodes' areas are and then multiply by it's own temperature? This would then be calculating a quantity with units of degree multiplied by area. Thanks so much for your help! Dylan

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January 31, 2014, 05:07		#2
ping Senior Member Ping Join Date: Mar 2009 Posts: 556 Rep Power: 20	field function area is available on boundaries but not derived parts like a section plane. so you could place an interface between 2 regions at that location and you will be fine - create a field function $Temperature * mag($$Area) to display or report on. an approximation might be possible using volume rather than area.