Question

A woman is draining her fish tank by siphoning the water into an outdoor drain as shown in Figure P14.82. The rectangular tank has footprint area A and depth h. The drain is located a distance d below the surface of the water in the tank, where $D>>h$. The cross-sectional area of the siphon tube is$A\text{'}$. Model the water as flowing without friction. Show that the time interval required to empty the tank is given by

$\Delta t=\frac{Ah}{A\text{'}\sqrt{2gd}}$

Short answer

It is proved that the time interval required to empty the tank is given by$\Delta t=\frac{Ah}{A\text{'}\sqrt{2gd}}$

Step-by-step solution

Bernoulli’s equation:

The sum of pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume have the same value for an ideal fluid at all points along the streamline. This result is summarized in Bernoulli's equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Here, $P$is the pressure, $\rho$is the density, $g$is the gravity, and $y$is the distance.

The time interval required to empty the tank:

Area of the cross section of the tank is$A$ .

Area of cross section of the tube is $A\text{'}$.

Depth of tank is $h$.

Height from Bernoulli’s principle is,

$\frac{1}{2}\rho v^2=\rho gd$ ….. (1)

Here, $v$is the speed with which the water flows out of the pipe.

But as $d>>h$, equation (1) become.

From the continuity equation, if $\Delta t$be the time required t empty the tank of volume $V$, then

The volume flow rate is

$$\begin{gathered} \frac{V}{\Delta t}=\frac{Ah}{\Delta t} \\ =vA\text{'} \end{gathered}$$

Here, the volume $V=Ah$. Therefore, the above equation for time is,

$$\begin{gathered} \Delta t=\frac{Ah}{vA\text{'}} \\ =\frac{Ah}{A\text{'}\sqrt{2gd}} \end{gathered}$$

Hence, the time required to the empty tank is $\Delta t=\frac{Ah}{A\text{'}\sqrt{2gd}}$.