Water in a tank being driven up and down

[Chestermiller]

So the liquid is not displacing up and down like a rigid body?

 

[member 428835]

So the liquid is not displacing up and down like a rigid body?

The tank that holds the liquid is a rigid body. The liquid inside is not. Think the approximation was initially done here.

 

[BvU]

Funny you say this, because we're actually sending a rocket up to the ISS this September (so perhaps not on earth). And we go up and down because the experiments are controlled this way (I don't do the experiments, just the math mentioned above). The theory suggests looking at oscillations normal to the equilibrium configuration (zero-g, so perfect curvature arc). Vibrating up and down isn't perfect, but the experiments agree very well with theory.

Should be interesting, these experiments. But with zero g you don't even have an equilibrium liquid surface.
I take it no XL heavy beam as described in the link is going up; what is the scale of the experiment ?

Anything else you forgot to mention ? Any preliminary practical tests in zero g planes or at NASA 0 g ?

 

[A.T.]

The liquid pressure waves, specifically capillary pressure. There's an analogy that can be drawn from the linearized Navier-Stokes equations to a damped harmonic oscillator. The eigenvalues of this equation are the fundamental frequencies of the liquid, call the first $\Omega_1$. When we turn the shaker table (think elevator) on at a particular frequency ($h=A\cos (\Omega t)$) we get minimal response (can't see the drop of water do anything). However, as we sweep through a range of frequencies, once $\Omega=\Omega_1$ we see a large disturbance in the shape of the predicted mode. As the frequency sweep exceeds $\Omega_1$, the disturbances die down again (until $\Omega_2$).

What happens if you set the frequency directly to $\Omega=\Omega_1$ ? Is the "sweep" somehow relevant here?

And what do you mean by "drop of water". I thought it's a tank with water. Is the tank filled completely and closed, or does the water have have a free surface?

 

[member 428835]

Should be interesting, these experiments. But with zero g you don't even have an equilibrium liquid surface.

The equilibrium in zero g is a circular arc (in 2D). So taking normal disturbances from that.

I take it no XL heavy beam as described in the link is going up; what is the scale of the experiment ?

Yea, definitely not a heavy beam. The scale is several cm footprint drops, so pretty big.

Anything else you forgot to mention ? Any preliminary practical tests in zero g planes or at NASA 0 g ?

I don't think so. Again, I pretty much just do the math, but some modeling enters and I like to double check with you all on here.

Should say we have done lots of terrestrial work below the capillary length scale. Have also worked with a university using their drop tower. But no aircraft tests (those would be fun I think though)!

What happens if you set the frequency directly to $\Omega=\Omega_1$ ? Is the "sweep" somehow relevant here?

Yea, my first thought was to do this. But numerics and experiments don't always agree with theory, so while we should see resonance, perhaps not. Even a 5% error could cause lots of confusion without a sweep.

And what do you mean by "drop of water". I thought it's a tank with water. Is the tank filled completely and closed, or does the water have have a free surface?

Sorry, I mixing and matching terms here (mainly because I was only interested in the shaker table up and down).

Some of our tests do drops of water on the shaker table. Something I'm looking at is liquid in channels (tanks), which are on shaker tables. And yea, they have a free surface (not closed).