# Water in a tank being driven up and down

• I
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## Main Question or Discussion Point

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 $p(t)=\cos(t)$?

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BvU
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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
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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.

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BvU
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drive the tank of water
Can you explain what you mean ? Driving with a car or an elevator -- or filling and emptying ?

jbriggs444
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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.

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BvU
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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 $A$)

joshmccraney
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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.

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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
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Right?
That's the acceleration of the container, yes.  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.

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.

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joshmccraney
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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
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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 ${\bf +} A\cos t$

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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?

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

joshmccraney
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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*aframe) 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).

joshmccraney
Chestermiller
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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 $$g-A\omega^2\cos{\omega t}$$

joshmccraney and BvU
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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
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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?

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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
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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.

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

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

Gold Member
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
Mentor
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.