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To familiarize myself with concepts from my system modeling class, I have posed myself the following problem, which I have been struggling with for a good while now. I am an absolute physics beginner, so it is likely I fundamentally misunderstood some things about pressure etc.


The Problem

There are two water tanks with cross-sectional areas $A_1$ and $A_2$, and water level heights $h_1(t)$ and $h_2(t)$ respectively. They are connected by a duct of cross-sectional area $A_d$ and length $l$. Density of water is $\rho$.

Given initial water heights $h_1(0)$ and $h_2(0)$, how can the velocity of the water $v_d$ flowing through the duct be modeled?

Sketch


My Attempt

I assume that atmospheric pressure is zero and the water velocity at the top of the tanks can be neglected as the duct should be much smaller than the tanks and thus water flows there much faster. By Bernoulli's principle, I think that the pressure at the surface of tank 1 and the pressure right before the duct ($p_1$) should be related as follows:

$\underbrace{\frac{1}{2} 0^2 + \rho g h_1(t) + 0}_\text{top of tank 1} = \underbrace{\frac{1}{2}v_d^2 + 0 + p_1}_\text{point p_1}$

Leading to $p_1 = \rho g h_1(t) - \frac{1}{2}v_d^2$. Similarly, for the pressure right at the entrance of the duct at tank 2: $p_2 = \rho g h_2(t) - \frac{1}{2}v_d^2$

Now that I have these two pressures, I tried to model the force acting on the mass of water inside the duct (which is $\rho A_d l$) as follows:

$F = A_d p_1 - A_d p_2 \\ \rho A_d l \cdot \frac{d}{dt} v = A_d ( p_1 - p_2 ) \\ \frac{d}{dt} v = \frac{p_1 - p_2}{\rho l} = \frac{1}{\rho l} \cdot \left( \rho g h_1(t) - \frac{1}{2} v_d^2 - \rho g h_2(t) + \frac{1}{2}v_d^2 \right) = \frac{1}{l} g \left( h_1(t) - h_2(t) \right)$

I modeled the height of the first water tank based on the volume of water in it ($V_1(t)$) as follows:

$\frac{d}{dt}V_1(t) = - v_d(t) \cdot A_d \\ h_1(t) = \frac{V_1(t)}{A_1} \\ \frac{d}{dt}V_2(t) = v_d(t) \cdot A_d \\ h_2(t) = \frac{V_2(t)}{A_2}$

When I run this model in Simulink, I the height of the two water tanks keeps fluctuating like a pendulum, instead of reaching the expected equilibrium where the water heights are the same.


Questions

  1. Is my assumption correct that this system should arrive at an equilibrium after a while such that $h_1(t) = h_2(t)$?
  2. How would I correctly model the velocity inside the duct neglecting friction?
  3. How would I correctly model the velocity inside the duct with friction?
  4. What is wrong with my modeling attempt? Are the calculated pressures $p_1$ and $p_2$ wrong? Can my attempt be corrected or is it fundamentally wrong?
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    $\begingroup$ Without including viscous friction, the predicted oscillation will continue forever, so there is no equilibrium. To model it with viscous friction, the focus would be the duct. You would have to determine the pressure drop/flow rate relationship for the duct. $\endgroup$ – Chet Miller Oct 31 '18 at 16:19
  • $\begingroup$ Thanks for the perspective. As a physics beginner I'm missing this kind of intuition and thought the equilibrium would be reached even if there is no friction. Why exactly does the oscillation continue forever? $\endgroup$ – anroesti Oct 31 '18 at 16:23
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    $\begingroup$ The oscillation continues forever because there is no friction to dissipate the combination of kinetic and potential energy. $\endgroup$ – Chet Miller Oct 31 '18 at 16:26
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    $\begingroup$ What you have is a combination of two limiting cases. If the tube is very short/wide, flow rate squared dominates, due to Bernoulli. If the tube is very long/narrow, flow rate to the first power dominates, due to viscosity. $\endgroup$ – Mike Dunlavey Nov 1 '18 at 20:31
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An approach that gives a qualitative feel for this problem is as follows. It is well known that $\Delta P = kF^2$, where P is pressure, F is flow, and k is a constant that depends on Reynold's number, duct cross-sectional area, duct roughness, duct length, etc. Thus, k is essentially a "fudge factor" that greatly simplifies the problem.

For the $\Delta P$ term, determine the pressure difference between the two tanks by using the equation $\Delta P = \rho g (h1 - h2)$, where $\rho$ is the density of the fluid that you are using.

Note that flow can be solved directly by using a bit of algebra. Also note that as the height difference between the two tanks decreases, the flow decreases.

Regarding the constant k, if you know one flow rate at given liquid heights, you can calculate it directly. If you don't know this information, a rather tedious and detailed calculation will be needed to calculate the required value. Rather than do that, I would recommend running an experiment where you record liquid heights vs. time. Then, develop a digital model of the process to calculate flow rate vs. time, and manipulate k until your simulation matches your data.

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  • $\begingroup$ Does your first equation $\Delta P = kF^2$ have a name? $\endgroup$ – anroesti Oct 31 '18 at 16:30
  • $\begingroup$ @anroesti, no. That "kluge" is admittedly a bastardization of the Bernoulli equation in order to get a very simple model, and thus obtain a better intuitive understanding of what is happening. I find that such simplifications are helpful when one wants a "quick and dirty" answer to what would otherwise be a very tedious and drawn-out problem. $\endgroup$ – David White Oct 31 '18 at 16:33
  • $\begingroup$ This approach, of course, includes viscous friction. $\endgroup$ – Chet Miller Nov 1 '18 at 16:09
  • $\begingroup$ @ChesterMiller, indeed it does. The fudge factor "k" takes care of a "multitude of sins", including changes in duct area, unknown relative roughness of the duct, viscous friction, etc. However, I doubt that pressure drop is quadratic for laminar flow, so there are limitations to this approach. $\endgroup$ – David White Nov 1 '18 at 16:19
  • $\begingroup$ Right. I also feel that the inertia (ma) of the fluid in the duct should be included. $\endgroup$ – Chet Miller Nov 1 '18 at 19:33
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This is for @David White.

What I have in mind is $$F=-A_1\frac{dh_1}{dt}=A_2\frac{dh_2}{dt}$$

For the tube: $$\rho l\frac{dF}{dt}=\rho g A(h_1-h_2)-\frac{1}{2}\rho \left(\frac{F}{A}\right)^2f\left(\frac{4l}{D}\right)A$$where f is the friction factor, A is the cross sectional area of the tube, and D is the tube diameter. The left hand side is the mass times acceleration of the fluid in the tube. The first term on the right hand side is the pressure driving force. The 2nd term on the right hand side is the drag force. This equation is the same as the one you suggested, except it includes the inertia term on the left hand side.

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