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\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\Figuring Fluid Flow}


\markboth{Figuring Fluid Flow}{Figuring Fluid Flow}

\author{Brian Winkel, 
Director SIMIODE\\ 
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
}

\makeStitlePDFLaTex


\begin{abstract}
We propose three differential equations models for the height of a column of falling water as the water exits a small bore hole at the bottom of the cylinder and ask students to determine which model is the best of the three.
\end{abstract}


\section*{SCENARIO DESCRIPTION}

Evangelestia Torricelli (1608-1647) was an Italian physicist and mathematician who also served Galileo as a secretary. He investigated the concepts of atmospheric pressure and vacuums and he invented the barometer.

\begin{figure}[!h]
\begin{center}
\includegraphics[width=2.0in]{Torricelli.jpg}
\caption{Evangelestia Torricelli  (1608-1647) - Italian physicist and mathematician.}\label{fig:torricelli}
\end{center}


\end{figure}


Torricelli was interested in fluids and their flow rates. He studied the following situation. Consider a cylindrical container (e.g., tin can open at top and closed at bottom) containing water. A small hole has been placed in the side of the can and the can is resting at the edge of the surface of a platform (on a table) with the hole in the can over the edge of the platform.

Now the distance from the table to the hole in the can is $h_2$ cm. And the water level in the can is $h_1$ cm above the hole in the side of the can. We can assume the diameter of the hole is small (but real) compared to heights $h_1$ and $h_2$, e.g.,  $h_1 = 15$ cm and
$h_2= 10$ cm, while the diameter of the hole is only 0.3 cm.

Torricelli was interested in the rate at which the water would flow out of the hole both as a function of height and time. Equivalently he was interested in the change in volume of the water in the cylinder per unit time or the change in the height, $h$, (for a
fixed radius, $r$, cylinder) per unit time, i.e.~he was interested in $h'(t)$ where $h(t)$ is the heighit of the ater at time $t$.

Certainly one could reason that with more water in the cylinder, hence more weight on the water above the hole, there would be more force pressing down, thus causing more force on the water going out. This would mean there is more water per unit time exiting the hole when $h$ is big than when $h$ is small.

This suggests that when $h$ is big, $h'(t)$  ought to be big and when $h$ is small, $h'(t)$  ought
to be small. Indeed, when $h = 0$ we must have	$h'(t)= ?$ 

Could	$h'(t) = -\alpha h(t)$?  Could $h'(t) = -\alpha \sqrt{h(t)}$? Could $h'(t) = -\alpha h(t)^2$?  Could $h'(t) =  -\alpha h(t) + \beta$?
Why would you not expect the latter to be the general form?


There are a number of possibilities for	$h'(t)$ --  the understatement of the project! What 
other function of $h$ could we place on the right hand side of $h'(t) =\quad$? What characteristics would
this function have to possess? Think about the physical situation under study. Let us consider just two of these possibilities:

\begin{equation}\label{eq:first}
h'(t) = -\alpha h(t)\, ,
\end{equation}
\noindent and
\begin{equation}\label{eq:root}
h'(t) = -\alpha \sqrt{h(t)}\, ,
\end{equation}

We shall offer some data to test (\ref{eq:first}) and (\ref{eq:root}) .

We shall need a plastic jar (either translucent or transparent) of constant radius, marked with cm markings going vertically (0 being at the small exit hole in the side of the jar), a watch - accurate to the nearest second will do, some water, a collection container for when the water exits the jar, and some graph paper.

We shall fill the jar to a height of $h_1$ above the exit hole, uncover the hole, observe the height of the water at various intervals (try some to get a sense of how often to take data), and record that height. Now we can plot $h(t)$, the height of the water from the hole versus $t$ on the graph paper.

If it is not possible to build an apparatus and collect data then we provide historical data in Table~\ref{table:data}. 

\subsection*{Activity 1}
If $h'(t) = -\alpha h(t)$ (our first conjecture) then we can solve for $h(t)$ explicitly using $h(0) = h_0$ which we can set at the start of our experiment as in our data in  Table~\ref{eq:data} in the APPENDIX. 


Somehow we have to find a reasonable value of $\alpha$ which will permit the solution of our equation, $h'(t) = -\alpha h(t)$,   to fit the data. Do that now and plot the model (theoretical) equation over the data. How does it look?



\subsection*{Activity 2}
If $h'(t) = -\alpha \sqrt{h(t)}$ (our second conjecture) then we can solve for $h(t)$ explicitly using $h(0) = h_0$ which we can set at the start of our experiment as in our data in  Table~\ref{table:data} in the Appendix. 

Again, we have to  find a reasonable value of $\alpha $ which will permit the solution of our equation, $h'(t) = -\alpha \sqrt{h(t)}$ , to fit the data. Do that now and plot the model (theoretical) equation over the data. How does it look?


\subsection*{Activity 3}
If $h'(t) = -\alpha h(t)^2$ (our third conjecture) then we can solve for $h(t)$ explicitly using $h(0) = h_0$ which we can set at the start of our experiment as in our data in  Table~\ref{table:data} in the Appendix. 

Again, we have to  find a reasonable value of $\alpha $ which will permit the solution of our equation, $h'(t) = -\alpha h(t)^2$,  to fit the data. Do that now and plot the model (theoretical) equation over the data. How does it look?

\vspace*{.3 true cm}

Thus we have three conjectures, models if you will. One may appear ``better'' than the other. Which one appears better and why?


Write up your results so far, being sure to address the questions asked.

 \section*{APPENDIX}

We enter data collected in a classroom laboratory by students in 1989. The data consists of the height of water, $h(t)$,  in cm in a graduated cylinder at time $t$ in s. The data was collected  by reading the times on the clock as the water passed through the given height. Notice the data is presented backwards ferom when the cylinder is empty ($t = 14$ s)  until it is full ($t = 0$ s).

\begin{table}[h]
\begin{center}

\begin{tabular}{|c|r|r|r|r|r|r|r|r|r|}\hline \hline 
Height - cm& 0 & 36.5 & 72.7 & 111.6 & 154.5&  195.8 &  241.1& 285.6\\ \hline 
Time - s& 14 &  13 &  12 &  11 &  10 &  9 &  8 & 7\\ \hline  \hline \hline
 Height - cm & 337.4 & 394.4 &  455.3 & 512.9 & 586.1 & 661.3 & 756.8&\\ \hline
 Time - s& 6 &  5 &  4 &  3 &  2 &  1 &  0&\\ \hline \hline 
\end{tabular}

\parbox{13 true cm}{\caption{\label{table:data} Data collected on height in cm vs. time in s for water in graduated cyilinder.}}
\end{center}
\end{table}



\section*{ACKNOWLEDGMENTS}

This material was developed with the help of students from the Integrated First-Year Curriculum in Science, Engineering, and Mathematics, at Rose-Hulman Institute of Technology, Terre Haute IN USA, in 1989. We thank the following students: Matthew Drew, Greg Gass,  Garth Grimm, and David Smith.




\teacher{
 
\section*{NOTES FOR TEACHERS}


We see that we have three models and one may be best at predicting the height 
$h(t)$ of the water in our cylinder. All models were built upon assumptions
about  $h'(t)$ based on some reasoning with regard to force due to the weight of water, namely the higher the column of water the more weight of water there is and hence as height, $h$, increases then $h'(t)$ should increase, as all three candidates we offered do. However, none of these models are based on physical principles and so we address this here with a derivation of Torricelli's Law.  This Modeling Scenario is fine for testing out which of three models, indeed, reasonable models, is best, but it is not inclusive, e.g., it does not go after a general model $h'(t) = -\alpha h(t)^r$. Thus we could determine BOTH parameters $\alpha$ and $r$ for a best fit model, but, instead, we shall derive a model from physical principles and we find from this that our second model, $h'(t) = -\alpha \sqrt{h}$, is grounded in physical principles and hence is the right model, even though the data in our Mathematica notebook 1-57-T-Mma-FiguringFluidFlow-TeacherVersion (with an accompanying pdf version for those who do not use Mathematica) and our second model are not a perfect match. 

Let us now view this situation in light of the Principle of Conservation of Energy. In this application we can say that the total energy (kinetic + potential) at the surface of the water, height $h_1$ from the exit hole to the surface of the water, is the same as the total energy (kinetic + potential) at the hole, height $h_2$ from the table top to the exit hole. Thus

\begin{equation}\label{eq:basic}
KE_S+PE_S=KE_H+PE_H\, ,
\end{equation}
where $KE_S$ and $PE_S$ are the kinetic energy and potential energy, respectively, of the water at the surface and $KE_H$ and $PE_H$ are the kinetic energy and potential energy, respectively, of the water at the height of the exit hole.  

Knowing that potential energy of a mass $m$ at height $h$ from a reference point is $m g h$ where $g$ is the acceleration due to gravity and the kinetic energy of the mass $m$ with velocity $v$ is $\frac{1}{2} m v^2$ we can compute the terms of (\ref{eq:basic}):

\begin{equation}\label{eq:fill}
mg(h_1 + h_2) + \frac{1}{2}mv_1^2 = mgh_2 +  \frac{1}{2}mv_2^2\, ,
\end{equation}
where $v_1$ is the velocity of the water on the surface and $v_2$ is the velocity of the water at the exit hole. 

Now, if we presume that $h_2 = 0$, that is the height of the exit hole is our ground, and $v_2 = v$,  and if we follow a mass of water from the surface at height $h_1 = h$ with initial velocity $v_1 = 0$ then (\ref{eq:fill}) becomes:
\begin{equation}\label{eq:fillFin}
m g h =   \frac{1}{2}mv^2\, .
\end{equation}
From 
(\ref{eq:fillFin}) we can solve for $v$ to obtain
\begin{equation}\label{eq:vel}
v = \sqrt{2 g h}\, .
\end{equation}
Equation (\ref{eq:vel}) means the velocity at the exit hole of the water is proportional to the square root of height, not just to height, and not to height squared.


Thus we see that if we multiply, $v$ (in cm/sec), the velocity of the water at the exit hole, by the cross sectional area of the hole, $A$  (in cm$^2$), we shall find the rate at which the volume, $V$, is changing, i.e.

\begin{equation}\label{eq:colChange}
V'(t) = -A \sqrt{2 g h}\, ,
\end{equation}
   

\begin{equation}\label{eq:cyl}
\pi r^2 h'(t) = V'(t) = -A \sqrt{2 g h}\, .
\end{equation}

Now if in (\ref{eq:cyl}) we set $\alpha =\frac{A}{\pi r^2} \sqrt{2g}$, then our final equation for the rate at which the height, $h(t)$ of our column of water changes becomes,

\begin{equation}\label{eq:Torr}
h'(t)  = -\alpha \sqrt{h(t)}\, .
\end{equation}
Equation  (\ref{eq:Torr}) is one form of what is known as Torricelli's Law.
 
 
Thus we have a derivation of a differential equation which attempts to describe the rate at which the water would flow out of an exit hole in the bottom of a cylinder as a function of height.

This appeal to the Law of Conversation of Energy is a rational approach to deriving a model we only surmised in our previous discussion. This, of course, differs from our somewhat informed  conjecture of the three models:  $h'(t) = -\alpha h(t)$, $h'(t) = -\alpha \sqrt{h(t)}$, and  $h'(t) = -\alpha h(t)^2$ .
}











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