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\begin{document}

\newpage
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\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\BUILDING WATER CLOCKS}





\markboth{Building  Water Clocks}{Building  Water Clocks}






\author{
\begin{minipage}[t]{9 true cm}
Sania Qureshi\\
Mehran University of Engineering and Technology\\
Jamshoro, Sindh PAKISTAN
\teacher{\\ sania.qureshi@faculty.muet.edu.pk}
\end{minipage} \ \
\begin{minipage}[t]{5 true cm}
Brian Winkel\\Director
SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
\end{minipage}
}
\makeStitlePDFLaTex



\begin{abstract}
We apply Torricelli's Law to the task of building a water clock in which the height of the water in a container falls at a constant rate when the   container has  a hole in the bottom to let the water flow out.  First, we review the principles and derivation (in Appendix) of the applicable physics in Torricelli's Law.  Second, we determine the shape of the container, given a constant spigot size for the exit hole, so that the water height falls at a constant rate. Third, for several shapes we determine spigot size as a function of time so that the water height falls at a constant rate.  Finally, we ask for a given spigot size function of time what must the shape of the container  be in order that  the height of the water in a container falls at a constant rate  
\end{abstract}


\section*{SCENARIO DESCRIPTION}

 
\section*{Modeling Activities}

\begin{enumerate}
  \item Given right circular CYLINDER with fixed spigot hole for water exit at the bottom of fixed size.  What shape container would we have to offer so that the water falls at a constant rate, meaning for a right circular cylinder what cross-sectional area $A(h)$ at height $h$ would we have to offer so that the height of the water falls at a constant rate? 
 \item We consider the problem of determining the area of the opening or spigot hole, $s(t)$,  at the bottom of a right circular cylinder of water of constant cross sectional area $A(h(t)) = K$ so that the water falls at a constant rate, i.e. $h'(t) = c$ ($c < 0$), where $h(t)$ is the height of the water at time $t$ and $A(h(t))$ is that cross sectional area of the cylinder at height $h(t)$.
 \item We consider the problem of determining the area of the opening or spigot hole, $s(t)$,  at the bottom of an inverted right circular cone of water of where the cone has a base radius of $R$ and height of $H$ so that the water falls at a constant rate, i.e. $h'(t) = c$ ($c < 0$), where $h(t)$ is the height of the water at time $t$ and $A(h(t))$ is that cross sectional area of the cone at height $h(t)$.
 \item We consider the problem of determining the area of the opening or spigot hole, $s(t)$, of the spigot at the bottom of an inverted hemisphere of water of where the hemisphere has a  radius of $R$ and height of $H$ so that the water falls at a constant rate, i.e. $h'(t) = c$ ($c < 0$), where $h(t)$ is the height of the water at time $t$ and $A(h(t))$ is that cross sectional area of the hemisphere at height $h(t)$.
\end{enumerate}


\section*{APPENDIX}


\subsection*{Derivation of Torricelli's Law from First Principles}


We reproduce here the derivation found in \cite{Winkel2015Torr}. We examine a model based on first principles, not just on empirical fit.  Given a column of water  in a cylinder with a tiny hole at the bottom of the cylinder through which the water exits, how long does it take to empty the cylinder down to the level of the hole? More specifically, if we are given the cross sectional area of the cylinder as a function of height and the area of the tiny hole at the bottom of the cylinder can we model the outflow of the water from the cylinder.

Now for some basic physics background.  We first consider a law of physics which can help us.  The Law of Conservation of Energy says that the sum of the potential energy and the kinetic energy of a particle of mass $m$ is constant, i.e., is conserved. So if we consider a particle of water of mass $m$ initially atop  a cylinder of water, some $h$  meters above a small, sharp-edged opening in the side wall of the cylinder through which the water can exit the cylinder, this mass of  water has initial potential energy $m*g*h$ ($PE_i$), where $g$ is the acceleration due to gravity, and initial kinetic energy  $\frac{1}{2}m v_i^2$ ($KE_i$), where $v_i$ is the velocity of the mass initially.  This gives us a total energy of $TE_i$, initially, when the mass of water is on the top of the cylinder of water:
\begin{equation*}
 TE_i = KE_i + PE_i = \frac{1}{2} m v_i^2 + m g h\, . 
\end{equation*} 

When this mass of water reaches the opening it has height 0 meters above the opening and a final velocity of $v_f$. Hence, the total energy at the final time ($TE_f$) the mass reaches the opening is 
\begin{equation*}
 TE_f = KE_f + PE_f = \frac{1}{2} m v_f^2 + m g*0 = \frac{1}{2} m v_f^2  \, . 
\end{equation*}
Now, by the Law Conservation of Energy we know $TE_i = TE_f$ and thus we have  
\begin{equation*}
\frac{1}{2} m v_i^2 + m g h = \frac{1}{2} m v_f^2 \, ,
\end{equation*}
which when one solves for $v_f$ yields $v_f = \sqrt{2 g h + v_i^2}$. In particular, if our element of mass of water was at rest initially, i.e. $v_i$ = 0 we have the classical Torricelli's Law

\begin{equation*}
v_f = \sqrt{2 g h} \, ,
\end{equation*}
where $v_f$ is the speed of the element of mass of water as it exits the opening in the cylinder. In particular we have the following differential equation for $v_f = h'(t)$:

\begin{equation}\label{eq:Torricelli}
h'(t) = \sqrt{2 g h(t)} \, ,
\end{equation}
where $h(t)$ is the height of the element of mass of water.

\begin{itemize}
\item [(2)] 
Solve this differential equation if we know the height of the column of water, initially, i.e., $h(0) = h_0$.
\end{itemize}




\subsection*{Modeling the outflow from a cylinder of water}

We offer some activities using Torricelli's Law as a practice for the activities of this modeling scenario.  

We note that (\ref{eq:drain}) blow would be a typical differential equaiton for the height of the water in our container given the geometry of the situation described leading up to (\ref{eq:drain}).  For the activities in this modeling scenario the geometry and hence the terms of the appropriate differential equation  similar to (\ref{eq:drain}) would be different.

Suppose we have a cylinder of height 0.20 m and radius  0.05 m and it is filled with water. There is an  opening in the side of the cylinder at the bottom of the cylinder in the form of a sharp edged circle of radius $r = 0.01$ m. 
 
Empirically, it is known that while the area of the opening is $\pi r^2 = \pi (.01)^2$ m$^2$ the effective area for fluid flow through the opening is only a fraction of the area. That fraction, $\alpha$, is called the {\it discharge or contraction coefficient\/}. Let us assume $\alpha = 0.7$, which is a reasonable value from the literature for various opening shapes and sizes. 

We now build a differential equation to determine the height of the water in the cylinder as a function of time, $h(t)$ in meters -- in this case. To do this we need to use accounting!! This will prove to be a nice way to build models, namely compute some quantity in two different ways and set the two different computations equal to each other. If the computations involve derivatives we have built a differential equation!

So let us compute, in two different ways,  the rate at which the volume of water changes at time $t$. We shall do this for a general cross sectional area of our cylinder as a function of height, $A(h(t))$.  

\begin{itemize}
\item [(L)] At time $t$ a small element of height changed, $h'(t)$, over a cross sectional area $A(h(t))$. Hence we will have lost 
\begin{equation}\label{eq:slab}
A(h(t))*h'(t)
\end{equation}
 volume of water. In our case, for our circular cylinder of constant radius 0.05 m, we have the following computation:
\begin{equation}
A(h(t))*h'(t) = \pi (.05)^2 h'(t)
\end{equation}
\item [(R)]  Since the water is flowing out of the opening with a velocity (according to Torricelli's Law) of $\sqrt{2 g h(t)}$ m/s and our opening's cross sectional area is $a = \pi r^2 = \pi (.01)^2$ m$^2$ then we are losing water at a rate of 
\begin{equation}\label{eq:opening}
\alpha a \sqrt{2 g h(t)} = \alpha \pi r^2 \sqrt{2 g h(t)} = .7 \pi (.01)^2 \sqrt{2 g h(t)} \quad {\rm m}^3{\rm /s}\, ,
\end{equation} 
where we recall the effective area for fluid flow through the opening is only a fraction ($\alpha = .7$) of the area.  
\end{itemize}
Now if we equate the computations for the loss of water found in (\ref{eq:slab}) with that computed in (\ref{eq:opening}) we obtain the following differential equation (adding our initial height $h(0) = h_0 = 0.20$ m as an initial condition):
\begin{equation}\label{eq:changevol}
A(h(t))*h'(t)= \alpha\, a \sqrt{2 g h(t)}\, ,\quad h(0) = h_0
\end{equation}
in general, and in our case, using $g = 9.8$ m/s$^2$,
\begin{equation}\label{eq:drain}
\pi (.05)^2 h'(t)= .7 \pi (.01)^2 \sqrt{2 (9.8) h(t)}\, , \quad h(0) = 0.20\, .
\end{equation}

We repeat the activities from \cite{Winkel2015Torr} to give some idea of what can be accomplished with the resulting models.

\begin{itemize}
\item [(3)] Solve the differential equation~(\ref{eq:drain}) (be sure you pick the correct branch of the solution) and determine how long it takes to empty the cylinder of water. Plot the height $h(t)$ as a function of time, $t$, for the duration of the draining. 
\item [(4)] Suppose $\alpha = 0.6$ instead of 0.7, then how long would it  take to empty the cylinder of water?  Plot the height $h(t)$ as a function of time, $t$, for the duration of the draining. Compare this answer and the plot with those from (a). Explain why this is reasonable?
\item [(5)] Continuing with (3) determine the velocity of the surface of the water and plot it over the  duration of the draining. Explain why what you see is reasonable?
\end{itemize}

\teacher{
\section*{NOTES FOR TEACHERS}

This activity is a direct applications of Torricelli's Law as offered in \cite{Winkel2015Torr}. We offer a complete analysis in an accompanying notebook 1-28-T-Mma-WaterClocks-TeacherVersion.nb and a pdf version of this file for those who do not use Mathematica.





\begin{thebibliography}{99}




\bibitem{Winkel2015Torr} Winkel, B. 2015. 1-015-T-Torricelli.





\end{thebibliography}
}





\end{document}

\begin{figure}[h]
\centering
\includegraphics[width=6cm]{1-29-ConeCube.eps}


\parbox{9.7 true cm}{\caption{\label{fig:1-29-ConeCube}  Shown is a cone   of base radius $R$ and height $H$ sitting above a cube of side $s$ with the water running out of the cone into the cube, whereupon the water runs out of the cube.}}

\end{figure}
