As to the title, I have computed a gas-liquid two-phase flow by "laminar-flow,level-set ". I wish to get the velocity of the head of the bubble vs. time, i.e., the velocity of the interface(phils=0.5). How could i get this velocity? Thanks a lot.
You may test some possibilities:
1. Determine the location of the interface and divide it with time.
2. Compute the velocity surface average on the interface
Usually, a smooth Heaviside function (phi) is used to mark different regions. You can then mark the interface region using phi*(1-phi).
I usually use the hyperbolic tangent function from the phase-field model:
phi=0.5*(1+tanh(-2*distance_function/interface_width))
and use weight = phi*(1-phi) in calculating an integral value in the interface region.
Yos Sitompul thank you. Will its integration give me the interface velocity or shape?. Actually, I need the interface or drop velocity as time progresses.
In two-phase fluid flow, the velocity at the interface can be found by equating the individual phase velocity and stress because they both remain continuous. Kindly follow the steps given below:
1. Write the constitutive equations for each phase and solve them to find the general solution in their respective domain.
2. The particular solution will depend on 3 conditions:
(i) wall boundary conditions (slip or no slip)
(ii) continuity of individual phase velocity at the interface and
(iii) continuity of the stress at the interface.
I hope it will help you.
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