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Question: A rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tan...

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

Explanation

Solution

To solve this above question, we will apply Bernoulli’s theorem to determine how long it will take to empty the tank if it is half filled and we will integrate the obtained equation to get the required solution.

Formula used:
v=2ghv = \sqrt {2gh}
Where, gg is the acceleration due to gravity and hh is the height.

Complete step by step answer:
Let us assume the actual height of the tank is HH and its cross sectional area is AA. Now let us consider that at some moment of time the height of water level is hh and the velocity of water emerging through the cross section of the area is aa at the bottom of the tank be vv.As the surface of water and the orifice are in open atmosphere, then with Bernoulli's equation we can say that,
v=2ghv = \sqrt {2gh}

And now, let dhdh is the decrease in level of water during infinitesimally small time intervals dtdt. When the level of water is at the height of hh. So, we can say that the rate of decrease in the volume of water will be Adgdt - A\dfrac{{dg}}{{dt}}.And the rate of flow of water at this moment through the orifice is given by,
v×a=a2ghv \times a = a\sqrt {2gh}
Then we can say that with the principle of continuity the both be same i.e.,
a2gh=Adgdt dt=Adga2g×h12dh a\sqrt {2gh} = - A\dfrac{{dg}}{{dt}} \\\ \Rightarrow dt = - A\dfrac{{dg}}{{a\sqrt {2g} }} \times {h^{\dfrac{{ - 1}}{2}}}dh \\\
Now, we can integrate the above equation to get the time taken to empty the tank. If the tank is filled to the brim, then the height of the water level will be HH and the time required to empty the tank is TT .
i.e.,
T=0T0dt=Adga2g×H0h12dh T=Aa2g[h1212]H0 T=A2Ha2g T = \int\limits_0^{{T_0}} {dt} = - A\dfrac{{dg}}{{a\sqrt {2g} }} \times \int\limits_H^0 {{h^{\dfrac{{ - 1}}{2}}}dh} \\\ \Rightarrow T = \dfrac{{ - A}}{{a\sqrt {2g} }}\left[ {\dfrac{{{h^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right]_H^0 \\\ \Rightarrow T = \dfrac{{A\sqrt {2H} }}{{a\sqrt {2g} }} \\\
And according to the question if the tank is half filled with water and the time to empty be TT' then

\Rightarrow \dfrac{{T'}}{T} = \dfrac{1}{{\sqrt 2 }} \\\ \therefore T' = \dfrac{T}{{\sqrt 2 }} \\\ $$ **So, time taken to empty the tank if it is half filled is $$\dfrac{T}{{\sqrt 2 }}$$.** **Note:** The rule of conservation of energy is used to derive Bernoulli's equation. If you don't remember Bernoulli's efflux equation, you can apply the law of conservation of energy, which states that potential energy must be transferred to kinetic energy.