# A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time T. How long will it take to empty the tank if it is half filled?

## I have to five the answer in terms of T.

Dec 21, 2016

Let the actual Height of the tank be $H$ and its uniform cross sectional area be $A$.

Now let us consider that at certain moment the height of water level be $h$ and the velocity of water emerging through the orifice of cross sectional area $a$ at the bottom of the tank be $v$.As the surface of water and the orifice are in open atmosphere, then by Bernoulli's theorem we have

$v = \sqrt{2 g h}$

Let $\mathrm{dh}$ represents decrease in water level during infinitesimally small time interval $\mathrm{dt}$ when the water level is at height $h$. So the rate of decrease in volume of water will be $- A \frac{\mathrm{dh}}{\mathrm{dt}}$.

Again the rate of flow of water at this moment through the orifice is given by $v \times a = a \sqrt{2 g h}$.

These two rate must be same by the principle of cotinuity.

Hence $a \sqrt{2 g h} = - A \frac{\mathrm{dh}}{\mathrm{dt}}$

$\implies \mathrm{dt} = - \frac{A}{a \sqrt{2 g}} \cdot {h}^{- \frac{1}{2}} \mathrm{dh}$

If the tank is filled to the brim then height of water level will be $H$ and time required to empty the tank T can be obtained by integrating the above relation.

$T = {\int}_{0}^{T} \mathrm{dt} = - \frac{A}{a \sqrt{2 g}} \cdot {\int}_{H}^{0} {h}^{- \frac{1}{2}} \mathrm{dh}$

$T = - \frac{A}{a \sqrt{2 g}} {\left[{h}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right]}_{H}^{0}$

$T = \left(\frac{A}{a} \cdot \sqrt{\frac{2}{g}}\right) {H}^{\frac{1}{2}}$

If the tank be half filled with water and the time to empty it be $T '$ then

$T ' = \left(\frac{A}{a} \cdot \sqrt{\frac{2}{g}}\right) {\left(\frac{H}{2}\right)}^{\frac{1}{2}}$

So $\frac{T '}{T} = \frac{1}{\sqrt{2}}$

$T ' = \frac{T}{\sqrt{2}}$

Dec 26, 2016

Bernoulli's theorem states that the total mechanical energy of the flowing fluid, comprising the energy associated with fluid pressure due to the gravitational potential energy associated with height $h$, and the kinetic energy of fluid motion associated with velocity $v$, remains constant.

In the instant case we need to apply Torricelli's law, which is a special case of Bernoulli's theorem. For a water particle of mass $m$ at height $h$, when it reaches bottom of the tank all the potential energy has been converted to its kinetic energy. We have
$m g h = \frac{1}{2} m {v}^{2}$
$\implies v = \sqrt{2 g h}$ ....(1)

Assuming that when tank is completely empty the velocity of water coming out of hole is $0$.
The average velocity of the emerging water for tank full up to height $h$ is
${v}_{\text{full}} = \frac{\sqrt{2 g h}}{2}$ .....(2)
Similarly, for the tank when half full, the average velocity is given by
${v}_{\text{half}} = \frac{\sqrt{2 g \frac{h}{2}}}{2}$ ....(3)

Let ${T}_{\text{full" and T_"half}}$ be the times taken to completely empty the tank in respective cases. Also let $a$ is area the orifice.

For case of full tank we have
Volume discharged$= {T}_{\text{full"xxv_"full}} \times a$ .......(4)

Similarly in case of half full tank we have
Volume discharged$= {T}_{\text{half"xxv_"half}} \times a$ ........(5)

We know that Volume discharged in case of equation (4) is twice the volume discharged in case of equation (5)

Hence we have the equality from (4) and (5)
T_"full"xxv_"full"xxa=2(T_"half"xxv_"half"xxa)

Using (2) and (3) and solving for ${T}_{\text{half}}$, we get
${T}_{\text{half"=1/2xxT_"full"xxv_"full"/v_"half}}$
$\implies {T}_{\text{half"=1/2xxT_"full}} \times \frac{\frac{\sqrt{2 g h}}{2}}{\frac{\sqrt{2 g \frac{h}{2}}}{2}}$
$\implies {T}_{\text{half"=1/2xxT_"full}} \times \frac{\sqrt{h}}{\sqrt{\frac{h}{2}}}$
$\implies {T}_{\text{half"=1/sqrt2T_"full}}$