Fluids

Area-weighted average in axisymmetric problems

• aitor.amatriain
Subscriber

I am simulating an axisymmetric problem. I have defined an iso-surface of interest,

• Rob
Ansys Employee
In 2d we assume the domain is 1m thick (I advise against changing the reference depth) and in 2d-axi it's either PI or 2PI, but I can never remember which. As your surface is at different radial positions I'd check which it is using one of the other boundaries.
• aitor.amatriain
Subscriber
Hello Rob Thank you for your answer. I understand that the area of the surface is computed by means of the usual formula for revolution surface, but I still have a doubt in terms of the area of the cells.
In a 3D mesh, all cells in contact with a surface have a well defined area. However, in a 2D mesh and an additional assumption has to be done. How is the area of the cells in contact with the revolution surface computed? ANSYS Fluent considers that each of the cells covers all angles from 0 to 2*pi?
Thank you Aitor
• Rob
Ansys Employee
As you suspect we assume each cell is PI or 2PI radians (if you find the documentation post the link in here for future reference), so represents an annular volume. Remember your cell has a volume, and the surface (boundary) area is that of the sector. So the area may be PI r^2 (assuming it's circular) or PI d length in the event it's an annular surface.
• aitor.amatriain
Subscriber
Thank you As far as I am concerned, this issue related to the cell area and volume in 2D axisymmetric problems affects to all the Chapter 30
Chapter 30: Reporting Alphanumeric Data (ansys.com)
It would be interesting for other users to include a warning/note mentioning that cells are assumed to cover all the circumference.
• Rob
Ansys Employee
• aitor.amatriain
Subscriber
Yes! Thank you!
• DrAmine
Ansys Employee
Welcome!