Water in a tank being driven up and down

[member 428835]

Hi PF!

Suppose I have a square tank of water. If I drive the tank of water such that it's height as a function of time is $h(t) = \cos(t)$, ignoring gravity, is this the same thing as saying the tank is holding still but the pressure field is changing as $p(t)=\cos(t)$?

 

[BvU]

Pressure can not be negative

 

[A.T.]

Hi PF!

Suppose I have a square tank of water. If I drive the tank of water such that it's height as a function of time is $h(t) = \cos(t)$, ignoring gravity, is this the same thing as saying the tank is holding still but the pressure field is changing as $p(t)=\cos(t)$?

You mean replacing gravity with acceleration of the tank? How is your artificial gravity related to h?

 

[jbriggs444]

Hi PF!

Suppose I have a square tank of water. If I drive the tank of water such that it's height as a function of time is $h(t) = \cos(t)$, ignoring gravity, is this the same thing as saying the tank is holding still but the pressure field is changing as $p(t)=\cos(t)$?

As I understand it, you want to set $g=1$ and $\rho=1$ so that $p(h)=\rho g h = h$

Then you want to move the tank up and down so that $h(t) = \cos t$.

And then you want to slam these two equations together so that $p(h(t)) = h(t) = \cos t$

No. That makes no sense. For one thing, it is the fallacy of equivocation. You use the same variable names with different meanings in different equations and then use those equations together. In particular, the equation: $p(h) = \rho g h$ assumes that h is a position within a column of water at equilibrium. But the equation: $h(t) = \cos t$ assumes that h is the position of a small isolated tank.

 

[BvU]

drive the tank of water

Can you explain what you mean ? Driving with a car or an elevator -- or filling and emptying ?

 

[member 428835]

Can you explain what you mean ? Driving with a car or an elevator -- or filling and emptying ?

By driving I mean moving the tank up and down. Like if the tank sits on a shaker table and the table height changes as $\cos(t)$. An example of the table I'm referring to is here.

Case 1) Tank is accelerated up and down on a shaker table with frequency $\cos(t)$
Case 2) Tank is resting on a table with some applied pressure field (time-changing gravity) $g(t)$.

What should $g(t)$ be to generate the same accelerations as the shaker table?

Was not referring to what jbriggs444 is saying, though I see your point.

 

[BvU]

ignoring gravity

Is a messy idea. Better have gravity on.
And if you drive your support up and down such that the floor of the tank goes up like $z=A\cos(t)$, it would be the same as if $g$ varied like $g = g_0-A\cos t$.

So you can still use $\Delta p = \rho g \Delta h$

(with reasonable bounds for $A$)

 

[BvU]

An example of the table I'm referring to is here

The frequency is a bit higher there ! And I wonder if the plate is remaining flat.

 

[member 428835]

And if you drive your support up and down such that the floor of the tank goes up like $z=A\cos(t)$, it would be the same as if $g$ varied like $g = g_0-A\cos t$.

Okay, so I assume you took the derivative of the driving frequency twice w.r.t. time? In other words, if the table's driving frequency is $\Omega$ such that the table height is $h=A\cos(\Omega t)$, then a static tank undergoing the same accelerations of a tank on the shaker table would experience a gravitational acceleration $g = g_0 + (A\cos (\Omega t))'' = g_0 - A\Omega^2\cos(\Omega t)$? Then if $g_0 = -9.8$ we conclude acceleration of a still water tank would be $-9.8 - A\Omega^2\cos(\Omega t)$. Right?

The frequency is a bit higher there ! And I wonder if the plate is remaining flat.

Yea, it's not perfect, but I was trying to convey the idea.

 

[BvU]

Right?

That's the acceleration of the container, yes. [edit] correction -- see below
I expect that a duck on the surface experiences a reduction in perceived gravity when $\cos = +1$ and an increase when the supporting table is at its lowest point.

[edit]
I found this haphazard determination of the sign unsatisfactory.

The acceleration of the container is $-a\cos t$ -- pointing downward at the highest point.
So: no, not right.

 

[member 428835]

That's the acceleration of the container, yes.
I expect that a duck on the surface experiences a reduction in perceived gravity when $\cos = +1$ and an increase when the supporting table is at its lowest point.

Riiiiiiiiiiight!

Okay, so really what I've suggested is adding a body force to the Navier-Stokes equal to $-\rho A \Omega^2 \cos(\Omega t)$ where $\rho$ is water density. Now this is not perfect (for reason you pointed out), but wouldn't this do a "decent" job at predicting the motion of the water within the tank when the tank is on a shaker table (elevator) who's height changes according to $h = A\cos(\Omega t)$?

 

[BvU]

See my edit above.

but wouldn't this do a "decent" job at predicting the motion of the water within the tank when the tank is on a shaker table (elevator) who's height changes according to $h = A\cos(\Omega t)$?

From Wikipedia:

I therefore expect wou want to add an inertial acceleration ${\bf +} A\cos t$

 

[member 428835]

See my edit above.From Wikipedia:
View attachment 263086

I therefore expect wou want to add an inertial acceleration ${\bf +} A\cos t$

Got it; thanks!

Let me elaborate a little bit more: I'm running a CFD analysis of liquid on a shaker table. Rather than having the table move up and down (expensive) I want to have a time-varying body force that emulates the tank going up and down. I'm reading from a paper and it seems what they are doing is saying there is a time-dependent pressure field $p(t)=A\cos(\Omega t)$. When they apply the linearized Bernoulli equation to the liquid in the tank, they get the expression $p = \rho \Psi_t + A\cos(\Omega t)$ where $\Psi$ is the liquid velocity potential.

I'm not sure how to implement this in a CFD package, but assumed it would be something similar to generating a body-force term proportional to $A\cos(\Omega t)$. What do you think?

 

[BvU]

Pinging @Chestermiller -- I am a CFD dimwit; perhaps he knows (or knows someone...)

 

[member 428835]

Pinging @Chestermiller -- I am a CFD dimwit; perhaps he knows (or knows someone...)

Chet can model anything. I'm sure he knows what the implications are.

 

[A.T.]

I'm not sure how to implement this in a CFD package, but assumed it would be something similar to generating a body-force term proportional to $A\cos(\Omega t)$. What do you think?

Yes, the inertial force (-m*a_frame) that you get in the accelerating rest-frame of the tank is equivalent to gravity, so you can model it just like you model gravity (but time dependent).

 

[Chestermiller]

If the upward motion of the plate is $$h=A\cos{\omega t}$$, then the upward acceleration of all the fluid is $$-A\omega^2\cos{\omega t}$$. So, from the moving frame of reference, the equivalent upward gravity would be $$-g+A\omega^2\cos{\omega t}$$ and the equivalent downward gravity would be $$g-A\omega^2\cos{\omega t}$$

 

[member 428835]

If the upward motion of the plate is $$h=A\cos{\omega t}$$, then the upward acceleration of all the fluid is $$-A\omega^2\cos{\omega t}$$. So, from the moving frame of reference, the equivalent upward gravity would be $$-g+A\omega^2\cos{\omega t}$$ and the equivalent downward gravity would be $$g-A\omega^2\cos{\omega t}$$

Thanks Chet!

Now suppose I want to do a frequency sweep/scan to detect resonance frequencies (analogy: forced harmonic ocillator). In practice this means turning a control dial that changes the frequency of the driver. Assuming we turn the control dial steadily, doesn't this imply a height equal to $h(t) = A \cos((a + bt)t)$ where $A$ is amplitude, $a$ is the starting frequency, and $b$ is the speed at which we steadily turn the knob?

 

[Chestermiller]

Thanks Chet!

Now suppose I want to do a frequency sweep/scan to detect resonance frequencies (analogy: forced harmonic ocillator). In practice this means turning a control dial that changes the frequency of the driver. Assuming we turn the control dial steadily, doesn't this imply a height equal to $h(t) = A \cos((a + bt)t)$ where $A$ is amplitude, $a$ is the starting frequency, and $b$ is the speed at which we steadily turn the knob?

Don't you think that, if you are looking at things like that, you need to include liquid compressibility?

 

[member 428835]

Don't you think that, if you are looking at things like that, you need to include liquid compressibility?

Experiments and theory suggest no. See this paper for a quick reference.

But what do you think about the $h=A((a+bt)t)$ approach for modeling the turning dial?

 

[Chestermiller]

Experiments and theory suggest no. See this paper for a quick reference.

But what do you think about the $h=A((a+bt)t)$ approach for modeling the turning dial?

I think that you would then have to differentiate that twice with respect to time to get the instantaneous acceleration.

 

[member 428835]

I think that you would then have to differentiate that twice with respect to time to get the instantaneous acceleration.

Agreed, but what do you think about this as a model for the frequency sweep?

 

[Chestermiller]

Agreed, but what do you think about this as a model for the frequency sweep?

I don't understand what a model of the frequency sweep means.

 

[member 428835]

I don't understand what a model of the frequency sweep means.

The height of the shaker table that the water rests on is $h(t) = A \cos(\Omega t)$. There is a dial I can turn to change the frequency $\Omega$. During a frequency sweep, to detect resonant frequencies we start at some frequency $\Omega_1$ and turn the nob, which increases $\Omega_1$ to a higher frequency. Assuming this increase occurs linearly, the frequency as a function of time is $\Omega(t) = \Omega_1+bt$ where $b$ is a constant related to how fast I turn the dial.

For me, this implies the height of the table as I turn the dial is then $h(t) = A \cos((\Omega_1+bt)t)$. Do you agree?

 

[Chestermiller]

The height of the shaker table that the water rests on is $h(t) = A \cos(\Omega t)$. There is a dial I can turn to change the frequency $\Omega$. During a frequency sweep, to detect resonant frequencies we start at some frequency $\Omega_1$ and turn the nob, which increases $\Omega_1$ to a higher frequency. Assuming this increase occurs linearly, the frequency as a function of time is $\Omega(t) = \Omega_1+bt$ where $b$ is a constant related to how fast I turn the dial.

For me, this implies the height of the table as I turn the dial is then $h(t) = A \cos((\Omega_1+bt)t)$. Do you agree?

I think you are going to want the frequency to be changing a small amount over each cycle.

 

[member 428835]

I think you are going to want the frequency to be changing a small amount over each cycle.

So really small $b$?

 

[Chestermiller]

So really small $b$?

small $b/\Omega_1$. Make a graph and see what it looks like.

 

[A.T.]

Now suppose I want to do a frequency sweep/scan to detect resonance frequencies

What is resonating here? Is the tank not filled and sloshing around? Or are you talking about pressure waves within the fluid resonating?

 

[BvU]

Now suppose I want to do a frequency sweep/scan to detect resonance frequencies

Along the same lines: why on Earth force a tank up and down if you can reasonably expect resonances mainly in perpendicular directions -- and they can be excited from the sides much more reproducibly ?

 

[member 428835]

What is resonating here? Is the tank not filled and sloshing around? Or are you talking about pressure waves within the fluid resonating?

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 $\Omega_2$).

Along the same lines: why on Earth force a tank up and down if you can reasonably expect resonances mainly in perpendicular directions -- and they can be excited from the sides much more reproducibly ?

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.

 

[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).