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Consider the following system of interconnected tanks:

system of coupled tanks

The two tanks have cross-section areas $A_1$ and $A_2$ respectively and the levels of the liquid are denoted by $h_1$ and $h_2$ respectively.

The liquid can run from one vessel into the other through a tube of cross-section area $a$, which is significantly smaller than $A_1$ and $A_2$.

Let us denote the density of the liquid by $\rho$ and the gravitational acceleration by $g$.

I need to derive a dynamical model of this system that describes the evolution of $h_1$ and $h_2$ and I see in various publications (example) that

$$ F_1 = \rho a \sqrt{2g(h_1-h_2)}.\tag{*} $$

I would like to use Bernoulli's principle to derive this formula. I will assume that the head loss in the pipe due to friction is negligible.

From Bernoulli's principle between a point $x$ on the surface of the first tank and a point $y$ at its exit

bernoulli in first tank

we have

$$ P_{atm} + \rho g h_1 = P_y + \frac{1}{2}\rho v_y^2\tag{1a} $$

Similarly, in Bernoulli's principle in the second tank gives

bernoulli in second tank

$$ P_{atm} + \rho g h_2 = P_{y'} - \frac{1}{2}\rho v_{y'}^2\label{1b}\tag{1b} $$

I am not sure about Equation \eqref{1b} - I used the negative sign because I used the convention that a positive $v_y$ means that the liquid flows from the first tank into the second one, so the second tank gains kinetic energy.

Question 1. Can you please clarify whether this is correct?

From point $y$ to $y'$, Bernoulli's equation gives

$$ P_{y} + \frac{1}{2}\rho v_{y}^2 = P_{y'} + \frac{1}{2}\rho v_{y'}^2\tag{1c} $$

Because of the mass balance equation between the two ends of the tube (assuming incompressible flow), it is $v_y = v_{y'}$.

The mass balance between the two tanks is

$$ \rho A_1 \dot{h}_1 + \rho A_2 \dot{h}_2 = 0\tag{2}. $$

Lastly, we know that

$$ F_1 = \rho a v_y, $$

so it suffices to show that $v_y = \sqrt{2g(h_1-h_2)}$.

Question 2. I have tried to combine Equations (1a-1c) and (2) to derive Equation (*), but did not make it.

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  • $\begingroup$ It's correct but needlessly long-winded. Simply use Torricelli's Law but slightly modified and immediately get $v = \sqrt{2g(h_1-h_2)}$ for the flow speed between the tanks. This is essentially a conversion of potential-energy-to-kinetic-energy-problem. $\endgroup$ – Gert Apr 24 '19 at 22:18
  • $\begingroup$ @Gert thank you for your comment. It sounds reasonable, but could you post a complete answer? By the way, I am also concerned about the fact that Bernoulli's principle assumes that the flow is steady, but this one is not (the speed changes with time as the levels change). $\endgroup$ – Pantelis Sopasakis Apr 24 '19 at 22:39
  • $\begingroup$ "the speed changes with time as the levels change" Ah, I was wondering about that. In that case the system behaves like an oscillator. Would you like the solution to that? $\endgroup$ – Gert Apr 24 '19 at 22:50
  • $\begingroup$ @Gert Yes, I'd like to see the solution. I suspect that if the tanks are large, the deceleration of the flow will be negligible, so we'll be able to use Toricelli's law and the level dynamics will be described by the ODE $A_1\dot{h}_1 = -a\sqrt{2g(h_1-h_2)}$ and $A_2\dot{h}_2 = a\sqrt{2g(h_1-h_2)}$. $\endgroup$ – Pantelis Sopasakis Apr 24 '19 at 22:56
  • $\begingroup$ It's a bit late here now to formulate an answer immediately. For now just look at this very similar problem: physics.stackexchange.com/questions/222179/… $\endgroup$ – Gert Apr 24 '19 at 23:01
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the speed changes with time as the levels change

This is a crucial bit of information because it may be tempting to believe the flow will stop when the tank levels become equal but due to inertia this is not the case. Instead the system will enter a simple harmonic oscillation.

Firstly a bit of tedious algebra. Determine the equilibrium level $h_0$ (both tanks at same level), from total volume considerations.

$$h_1A_1+h_2A_2=h_0(A_1+A_2)$$ $$h_0=\frac{h_1A_1+h_2A_2}{A_1+A_2}$$ Now find the volume $V$ above the equilibrium level $h_0$ for any $h_1$,$h_2$: $$V=(h_1-h_0)A_1+(h_0-h_2)A_2$$ with: $$h_2=\frac{h_0(A_1+A_2)-h_1A_1}{A_2}$$ $$V=(h_1-h_0)A_1+(h_0-\frac{h_0(A_1+A_2)-h_1A_1}{A_2})A_2$$ $$V=(h_1-h_0)A_1+h_0A_2-h_0(A_1+A_2)+h_1A_1$$ $$V=2h_1A_1-2h_0A_1$$ Its mass $m$ is with density $\rho$: $$m=\rho(2h_1A_1-2h_0A_1)$$ So the net force acting on the total mass $M$ is: $$mg=\rho g(2h_1A_1-2h_0A_1)$$ The Equation of Motion, with $M$ the total mass, is: $$Ma+\rho g(2h_1A_1-2h_0A_1)=0$$ Where: $$a=\frac{\mathbf{d^2}h_1}{\mathbf{d t^2}}=\ddot{h_1}$$ Use the following substitution: $$y(t)=\rho g(2h_1A_1-2h_0A_1)$$ $$\dot{y}=2\rho gA_1\dot{h_1}$$ $$\ddot{y}=2\rho gA_1\ddot{h_1}$$ $$\ddot{y}+\frac{2\rho gA_1}{M}y=0$$ Set: $$\omega^2=\frac{2\rho gA_1}{M}$$ $$\Rightarrow y=y_0\cos(\omega t+\phi)$$ Where at $t=0$: $$y_0=\rho g(2h_1(0)A_1-2h_0A_1)$$ and if $\dot{y}(0)=0$, then $\phi=0$.

Note that this derivation only works for completely inviscid liquids. Where there's viscosity, there are friction losses and thus damping.

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  • $\begingroup$ Thank you for the answer. I wonder why the cross-section area of the hole does not appear in your solution? For example, if the second vessel were empty, we would have $F = \rho A_{\mathrm{hole}} \sqrt{2gh}$. It's also not clear to me how you used Newton's second law of motion. We have a system of changing mass, so there should be a term of the form $\dot{m}$. $\endgroup$ – Pantelis Sopasakis Apr 25 '19 at 17:29
  • $\begingroup$ Let me rephrase my questions: (i) Under what assumptions can we derive Equation (*) (if it's at all correct), (ii) Can we use Bernoulli's equations? (iii) Should we resort to Euler's equation for non-steady flows? $\endgroup$ – Pantelis Sopasakis Apr 25 '19 at 17:44
  • $\begingroup$ $F_1 = \rho a \sqrt{2g(h_1-h_2)}.\tag{*}$ is only true if $h_1-h_2=\text{Constant}$ OR if $h_2=0$. But it must be possible to develop a 'dynamic' version of it. In dynamic conditions, you need to apply Euler's version of Bernoulli. It also bothers me that in my derivation $a$ (the hole or pipe diameter) doesn't feature, I'm looking into that now. I will also clarify the EoM a little. $\endgroup$ – Gert Apr 25 '19 at 19:56
  • $\begingroup$ I'm not able to derive (*) using Bernoulli's principle (unless I am justified to assume that $P_{y'}\approx \rho g h_2 + P_{atm}$). By the way, I found this article scielo.br/… - the authors propose the use of Euler's equation and the mass balance equation for each tank. I feel however that they don't justify their assumptions much. $\endgroup$ – Pantelis Sopasakis Apr 25 '19 at 20:04
  • $\begingroup$ That's a great reference and confirms some of my derivation (although mine is simpler). They too find a SHO where $a$ doesn't feature. It doesn't feature because we assume a perfectly inviscid liquid ('ideal liquid'). Such a liquid experiences no resistance to flow at all: pipe diameter or length have no handle on it whatsoever. This is of course unrealistic. A real system would be damped due to viscous friction. If you're interested in 'real world' system of communicating vessels (the reference doesn't cover it either) then I suggest you pose another question, with specifics re. the pipe. $\endgroup$ – Gert Apr 25 '19 at 20:44

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