Question

Suppose water is leaking from a tank through a circular hole of area A_hat its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to$\mathit{c}{\mathit{A}}_{\mathbf{h}}\sqrt{\mathbf{2}\mathbf{g}\mathbf{h}}$, where$\mathit{c}\mathbf{(}\mathbf{0}\mathbf{<}\mathit{c}\mathbf{<}\mathbf{1}\mathbf{)}$is an empirical constant. Determine a differential equation for the height hof water at time tfor the cubical tank shown in Figure 1.3.11. The radius of the hole is 2 in., and$\mathit{g}\mathbf{=}\mathbf{32}\mathbf{\text{ft}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}$.

Short answer

The differential equation is $\frac{dh}{dt}=-\frac{c\pi }{450}\sqrt{h}$.

Step-by-step solution

Definition:

A differential equation is an equation with one or more derivatives of a function.

Assumption:

Suppose water is leaking from a tank through a circular hole of area A_h at its bottom as shown below.

When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to $c{A}_h\sqrt{2gh}$where $c(0<c<1)$is an empirical constant.

Compute area of hole:

Area of the circular hole is$A_h=\pi r^2$

Here r = 2 inches

$$\begin{gathered} =\frac{2}{12}\text{feet} \\ =\frac{1}{6} \end{gathered}$$

So area of the circular hole is,

$$\begin{gathered} A_h=\pi {\left(\frac{1}{6}\right)}^2 \\ =\frac{1}{36}\pi \end{gathered}$$

Find rate of change of volume:

From the figure volume of the tank is,

$$\begin{gathered} V=\left(10\right)\left(10\right)\left(h\right) \\ =100h \end{gathered}$$

Differentiating it with respect to t,

$\frac{dV}{dt}=100\frac{dh}{dt}\dots \dots \text{(i)}$

From the data,$\frac{dV}{dt}=-c{A}_k\sqrt{2gh}\dots \dots$(ii)

[Here -ve sign indicates water leaving the tank.]

Find differential equation:

From equations (i) and (ii),

$$\begin{gathered} 100\frac{dh}{dt}=-c{A}_h\sqrt{2gh} \\ 100\frac{dh}{dt}=-c\left(\frac{\pi }{36}\right)\sqrt{2gh}\left[\text{use}{A}_h=\frac{\pi }{36}\right] \\ \frac{dh}{dt}=\frac{-c\left(\pi \right)}{36\times 100}\sqrt{2\times 32\times h} \\ =\frac{-c\left(\pi \right)\left(8\right)}{36\times 100}\sqrt{h} \\ =\frac{-c\left(\pi \right)\left(2\right)}{9\times 100}\sqrt{h} \\ =\frac{-c\left(\pi \right)}{9\times 50}\sqrt{h} \\ =-\frac{c\pi }{450}\sqrt{h} \end{gathered}$$

Hence the required differential equation is, $\frac{dh}{dt}=-\frac{c\pi }{450}\sqrt{h}$.