Emptying of a spherical tank

Take 20 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

A spherical tank, of radius \(R = 40\; cm\), is initially half filled with water density \(\rho=10^3\;kg.m^{-3}\).

Atmospheric pressure \( P_0\) prevails above the free surface of the water through an opening at the top \(S\) of the tank.

One opens, at \(t = 0\), a circular hole \(A\) of a small cross section \(s = 1 \;cm^2\) at the tank bottom.

Question

Establish the differential equation in \(z_s(t)\), if \(z_s(t)\) is the water level in the reservoir counted from \(A\) at time \(t\).

Solution

Neglecting the velocity of the free surface of the water, the Bernoulli theorem between surface and the output \(A\) gives :

\(P_0 + \mu gz_S = P_0 + \frac{1}{2}\mu v_A^2\)

Where :

\(v_A = \sqrt {2gz_S }\)

Torricelli's formula is found.

Since water is incompressible, the volume flow rate is conserved :

\(sv_A = - \pi r^2 \frac{{dz_S }}{{dt}}\)

Or (see figure) :

\(r^2 = R^2 - (R - z_S )^2 = z_S (2R - z_S )\)

Either after the separation of the variables :

\((2R - z_S )\sqrt {z_S } \;dz_S = - \frac{{s\sqrt {2g} }}{\pi }\;dt\)

Question

Literally express and calculate the duration \(T_S\) of emptying of this tank.

Solution

Emptying duration \(T_S\) is :

\(T_S = - \frac{\pi }{{s\sqrt {2g} }}\int_R^0 {(2Rz_S ^{1/2} - z_S ^{3/2} )dz_S }\)

Either :

\(T_S = \frac{{7\pi R^2 }}{{15s}}\sqrt {\frac{{2R}}{g}}\)

The numerical application gives \(11\) minutes and \(10\) seconds.