Draining a Container

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Draining a Container

Question: Consider a container with area A_1  and height h . A small hole of A_2  is put into the bottom of the container. How much time t  does it take to drain the container? Assume A_2 \ll A_1 .

Container1We will show:

t = \frac{A_1}{A_2} \sqrt{\frac{2h}{g}} 

Applicable Ideas:

Continuity Equation

A_1 v_1 = A_2 v_2 

There is no accumulation in the amount of fluid anywhere in the container.

Bernoulli Equation

P_1 + \frac{1}{2} \rho {v_1}^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho {v_2}^2 + \rho g h_2 

The statement of conservation of energy for the fluid. Applying this to the problem, where P_o  is atmospheric pressure:

P_o + \frac{1}{2} \rho {v_1}^2 + \rho g h = P_o + \frac{1}{2} \rho {v_2}^2 

\frac{1}{2} \rho {v_1}^2 + \rho g h = \frac{1}{2} \rho {v_2}^2 

{v_1}^2 + 2gh = {v_2}^2 

v_1 = \frac{dh}{dt} 

Sidenote: If A_1 \gg A_2 , then v_1 \ll v_2  and we have v_2 = \sqrt{2gh} . This is Torricelli’s law where the speed of the fluid emerging from the bottom of the container is equal to the speed it would have attained falling through a distance h .

Here we can use that fact to write:

A_1 \frac{dh}{dt} = - A_2 \sqrt{2gh} 

The negative sign is introduced because the water is leaving the container.

\frac{dh}{dt} = - \frac{A_2}{A_1} \sqrt{2gh} 

\int^0_h \frac{1}{\sqrt{h}}\,dh = \sqrt{2g} \frac{A_2}{A_1} \int^t_0\,dt 

2\sqrt{h} = \sqrt{2g} \frac{A_2}{A_1} t 

t = \frac{A_1}{A_2} \sqrt{\frac{2h}{g}} 

Bonus Question: Consider a hemispherical container of radius R . A small hole of radius r = 1  is put into the bottom of the container. How much time t  does it take to drain the container? Assume r \ll R .

Container2We will show:

t = \frac{14}{15}\frac{R^{5/2}}{r^2\sqrt{2g}} 

r = 1 

t = \frac{14}{15}\frac{R^{5/2}}{\sqrt{2g}} 

We will apply the continuity equation and Bernoulli’s equation just like before.

Container3R(t)^2 = R^2 - (R - h)^2 = 2Rh - h^2 

A_1 = \pi R(t)^2 = \pi (2Rh - h^2) 

A_2 = \pi r^2 

From the Bernoulli equation:

{v_1}^2 + 2gR = {v_2}^2 

We will approximate {v_2}^2 = 2gR  because A_1 \gg A_2  holds most of the time.

\pi (2Rh - h^2) \frac{dh}{dt} = - \pi r^2 \sqrt{2gh} 

\frac{h^2 - 2Rh}{\sqrt{h}} dh = r^2 \sqrt{2g}\,dt 

\frac{1}{r^2\sqrt{2g}}\int^R_0 2Rh^{1/2} - h^{3/2}\,dh = \int^t_0 \,dt 

\left[\frac{4}{3}Rh^{3/2} - \frac{2}{5}h^{5/2}\right]^R_0 = \frac{4}{3}R^{5/2} - \frac{2}{5}R^{5/2} 

\frac{14}{15}R^{5/2} = t r^2 \sqrt{2g} 

t = \frac{14}{15}\frac{R^{5/2}}{r^2\sqrt{2g}} 

r = 1 

t = \frac{14}{15}\frac{R^{5/2}}{\sqrt{2g}}