[Siba11]

I am trying to simulate the rise of a bubble in a tube open from two ends. I need to evaluate the area of the geometry at the beginning and end of the simulation. Despite searching about the methods to do that I could not find it. Can anyone please guide me on how to do that.

Thanks

[vinerm]

What do you mean by geometry? Do you mean the bubble shape? Apart from that, the only geometry is of the duct for which you have supplied the mesh. And for that, you obviously have everything available.

[Siba11]

Yes, I need to find the area of the bubble at the end of the simulation.
Now, here I there are two cases one in which only the shape of the bubble is deformed and the bubble remains intact. In the other case , the bubble breaks up . Can you please guide me on how to find the area of the bubble in both the cases.

Thanks.

[vinerm]

To determine the area, you need to identify the interface. In multiphase flows, the interface is defined in terms of volume fraction of the phases. So, create an iso-surface with volume fraction of either of the phases using a value of 0.5. This surface defines the shape of the bubble or drop. Just reports its area.

[Siba11]

Thanks for the message.
I have tried according to your suggestion. But the answer does not seem convincing. If you could just check the screenshot for the parameters that I have attached I shall be grateful.
Thanks in advance.

[vinerm]

Yes, the approach is correct but do not select anything under From Surface or From Zones. In your case, selecting the zone under From Zones will not make a difference since there is only one zone. Do note that the you are reporting is not for bubble since you appear to be using a 2D planar approach. So, it is not a bubble but a cylinder with 1 m height. For bubble, use 2D axisymmetric approach.

[Siba11]

Thanks for the advice.
I tried to find out the area of the patched bubble before the 2D axisymmetric simulation. As per it, the area should be pi*r^2 but the result came out to be different. Can you please guide me where could I have gone wrong. The radius of the circle is 0.05m. I have captured the screen. If you could take a look it will be helpful.
Thanks.

[vinerm]

returns area of projection and not the area of the surface. You do not need to use any formula. Just create an iso-surface with volume fraction of 0.5 and report its area. It should be equal to , surface area of sphere.