Learning Forum

[AbdulZahir]

I need to calculate the liquid holdup in a 2D geometry. In 2D geometry the liquid holdup is the ratio of area of liquid/area of flow domain. For that purpose I need to find the area covered by the liquid in the flow domain.

The actual geometry and the contours of my case are attached for the better understanding of my geometry.

Your kind guidance in this regard will be highly appreciated.

[Rob]

Use a volume integral: it's a volume in 2d. Surface integrals will also work as it's also (sort of) a surface. Just report the volume of each phase, or volume total & of which ever phase is of interest.

[AbdulZahir]

Thank you for your prompt response sir.
Yes volume in flow domain can also be used for this purpose but the problem is i am dealing with a 2D geometry (like a computer disk). We can calculate the area of a sphere (V=1.33*3.14*r3) but cannot calculate the volume of a circle as my geometry is 2D. Therefore, i cannot use volume of liquid for liquid holdup calculation. For L.H i need area of liquid instead.

[Amine Ben Hadj Ali]

Surface integral or Volume integral with default depth of the Volume fraction to get the wetted area.
Probably you are using planar 2d and your run is wrong as you rather require axis symmetric or 3d.