Chapter 14, Problem 44P

To determine

The expression for the variation of water height with respect to time.

Explanation of Solution

Given:

Height of the cone is H.

Base radius of the cone is R.

The cross section area of the hole is $A_h$.

The discharge coefficient is $C_d$.

Velocity of water is V.

Calculation:

From the law of conservation of mass,

  $\begin{array}{l}\frac{d}{dt}{\displaystyle \underset{CV}{\int }dV}+{\dot{V}}_{out}-{\dot{V}}_{in}=0\\ \frac{d{V}_{CV}}{dt}+0-{\dot{V}}_{in}=0\\ \frac{d{V}_{CV}}{dt}={\dot{V}}_{in}\\ \frac{d{V}_{CV}}{dt}=C_d{A}_{h}v\end{array}$        (I)

The volume of water at any time is given by the relation,

  $V_{CV}=\frac{\pi h}{3}\left(R^2+rR+r^2\right)$        (II)

The cross section of the cone is shown below.

From the diagram,

  $\begin{array}{c}r=\frac{R}{H}\left(H-h\right)\\ =R-\frac{R}{H}h\end{array}$        (III)

Substitute, Equation (III) in Equation (II).

  $\begin{array}{c}{V}_{CV}=\frac{\pi h}{3}\left[R^2+\left(R-\frac{R}{H}h\right)R+{\left(R-\frac{R}{H}h\right)}^2\right]\\ =\frac{\pi h{R}^2}{3}\left[1+\left(\frac{H-h}{H}\right)+{\left(\frac{H-h}{H}\right)}^2\right]\\ =\frac{\pi h{R}^2}{3H^2}\left[H^2+\left(H^2-hH\right)+\left(H^2+h^2-2hH\right)\right]\\ =\frac{\pi h{R}^2}{3H^2}\left(3H^2-3hH+h^2\right)\\ =\frac{\pi R^2}{3H^2}\left(3h{H}^2-3h^{2}H+h^3\right)\end{array}$        (IV)

Substitute Equation (IV) in Equation (I).

  $\begin{array}{l}\frac{d}{dt}\left[\frac{\pi R^2}{3H^2}\left(3h{H}^2-3h^{2}H+h^3\right)\right]=C_d{A}_{h}v\\ \frac{\pi R^2}{3H^2}\left(3H^2\frac{dh}{dt}-6hH\frac{dh}{dt}+3h^2\frac{dh}{dt}\right)=C_d{A}_{h}v\\ \frac{\pi R^2}{H^2}\left(H^2-2hH+h^2\right)\frac{dh}{dt}=C_d{A}_{h}v\\ \frac{\pi R^2}{H^2}{\left(H-h\right)}^2\frac{dh}{dt}=C_d{A}_{h}v\\ {\left(H-h\right)}^{2}dh=\frac{C_d{A}_{h}v{H}^2}{\pi R^2}dt\end{array}$

Integrating the above equation,

  $\begin{array}{l}{\displaystyle \int {\left(H-h\right)}^{2}dh}={\displaystyle \int \frac{C_d{A}_{h}v{H}^2}{\pi R^2}dt}\\ \frac{-1}{3}{\left(H-h\right)}^3=\frac{C_d{A}_{h}v{H}^2}{\pi R^2}t+C\end{array}$        (V)

Substituting, $h\left(0\right)=0$ in the above equation.

  $\begin{array}{l}\frac{-1}{3}{\left(H-0\right)}^3=\frac{C_d{A}_{h}v{H}^2}{\pi R^2}\left(0\right)+C\\ C=\frac{-H^3}{3}\end{array}$

Rewrite Equation (V).

  $\begin{array}{l}\frac{-1}{3}{\left(H-h\right)}^3=\frac{C_d{A}_{h}v{H}^2}{\pi R^2}t-\frac{H^3}{3}\\ {\left(H-h\right)}^3=H^3-\frac{3C_d{A}_{h}v{H}^2}{\pi R^2}t\\ H-h=\sqrt[3]{H^3-\frac{3C_d{A}_{h}v{H}^2}{\pi R^2}t}\\ h=H-\sqrt[3]{H^3-\frac{3C_d{A}_{h}v{H}^2}{\pi R^2}t}\end{array}$

Thus, the relation for variation of water height with time is $h=H-\sqrt[3]{H^3-\frac{3C_d{A}_{h}v{H}^2}{\pi R^2}t}$.