Hi,

I am struggling with this problem. It will be appreciated is someone can help.

A cup, with opening on the top, is filled with 50% of liquid. This cup is glued to a turntable. CG of the cup is R meter away from the center of the turntable. The turntable starts to rotate at variable speed, that is omega(t). This problem will be solved in cartesian coordinates. So, for the wall, I will specify wall velocity as r(x,y) * omega(t)*[cos(theta), sin(theta), 0]. How about the opening on top? I do not think I should specify velocity for this face because air/liquid can go in/out of the domain due to rotation.

Can someone shed some light on this?

Thanks!

Pei-Ying

The top boundary is so called free surface boundary. For incompressible flow, you should specify two physically meaningful conditions:

1)Kinematic free surface condition. I.e. d(eta)/dt + u*d(eta)/dx+v*d(eta)/dy = w. where eta(x,y,t) is the surface elevation, (u,v) is the horizontal velocity, w is the vertical velocity. This should be satisfied at z= eta(x,y,t) which itself is unknown.

2)Dynamic free surface condition, i.e. pressure at the surface = 0 or constant atmospheric pressure.

If you are looking for steady state solution, it is relatively easy. If you are looking for unsteady solution, it is actually a very challenging problem. If you are using commercial software, make sure that software can deal with free surface problems.

Wen

If you have access to Fluent Software and Tutorials, this is one of their tutorials, as Wen suggested, you have to use a VOF method. Hope that will help

Hi,

Thanks for the reply!

I used Fluent for the past 6 years. I went through the VOF tutorial before. The problem that I am interested is somewhat different from the fluent tutorial. In fluent tutorial, the center axis of the bowl is the the same as the rotational axis. Ihe problem that I am interested, the cup sits "away" from the rotational axis.

Yes, VOF will be used to track the interface between liquid and air. What I am confuse is how the BC on the opening face should be specified (zero gradient for U?V?W? specify U?V?W? specify pressure?)?

Pei-Ying

The difference between your case and Fluent tutorial is the location of the rotation axis, not a big deal, in gambit define an axis then define it in fluent as the rotation axis, hope I am not misleading you. Pressure at the interface is atmospheric

You can't just specify u,v,w, because you don't know what they are on the surface, you can't also just specify the surface elevation eta, because it is unknown. You know pressure, but that's not enough.

Fluent should be able to deal with the problem but may not be trivial. Actually, if you are looking for unsteady solution, on the wall boundary you know the velocity and on the top you known pressure, and that's all you have. You may want to simplify your problem before you throw the equations into a commercial solver. For example, you can transfer your coordinates frame to be mounted on the center of cup, then you will have a so called inertial term due to the acceleration (both centripetal and angular acceleration) of the frame. This term can be seen as body force in addition to gravity. Since the spinning of the table is unstable, this extra body force is unsteady in time, that means the fluid in the cup is accelerated or decelerated or more precisely activated to move. For example, if you are standing on the sun, and you want to simulate ocean waves/tides on the earth, it will be not easy to do using fluent. But when you disregard the sun, yet only sit on the earth, with some specified rotation speed of the earth, you get the problem solvable.

If you use VOF, it should be alright to only specify the free surface pressure,

wen

Dear Pei-Ying,

You mentioned that the centre axis of the bowl is not the same as the rotational axis.

Question: Would the cup not then impose different forces/accelerations on the liquid (at cup/liquid interface), than for the first case? How would you handle this, if at all?

diaw...

Hi, Ahmed,

Thanks for the reply. Actually, I no longer use Fluent. But, my confusion is related to setting up the BC of the top opening face, not meshing in gambit.

Pei-Ying

Hi, Wen,

I was thinking about transforming the problem from moving wall to adding a time varying body force into momentum equation like you suggested. But, are these two problems the same? For solid objects, I believe that this is OK, but, I am not sure that this transformation is the correct with liquid/air binside a cup ecause lquid/air moves (and may flow out or outside air flows in). I have to think about this more.

Pei-Ying

The extra body force is only related to the acceleration of the frame, it can't tell whether there is fluids inside the cup or not. please look for N-S equations in a rotating frame for more reference. I'm sure you can do it.

Wen