Question

I need examples of the law of torricelli (using differentials) of regular tanks of truncated pyramids with hexagonal, triangular, circular base

Answer 1

Torricelli’s Law

The Italian scientist Evangelista Torricelli investigating fluid flow experimentally found in 1643 that the velocity of fluid flowing out through a small hole at the bottom of an open tank (Figure 1) is given by the formula:

$v=\sqrt{2gh}$

where h is the height of fluid above the opening, g is the gravitational acceleration.

The same formula describes the velocity of a free solid particle falling from the height h in the Earth’s gravitational field in a vacuum. Actually this formula is not quite precise. In fact, the velocity of fluid depends on the shape and size of the opening, the fluid viscosity and flowing mode. Therefore, the Torricelli’s formula is often written with an additional factor φ:

$v= \varphi \sqrt{2gh}$

where the coefficient $\varphi$ is close to 1. The values of the coefficient $\varphi$ for openings of different shape and size are given in hydraulic handbooks.

Differential Equation of Fluid Flowing Out

We can derive the differential equation considering fluid balance in a vessel. Take as an example a cylindrical vessel with a broad base of radius R. Suppose that fluid flows out through a small opening of radius a at the bottom of the vessel (Figure 3).

Figure 3.

The fluid velocity is described by the Torricelli’s formula:

$v=\sqrt{2gz},$

where z is a the height of the fluid above the opening. Then the fluid flow is given by

$q=-\pi a^2 \sqrt{2gz}$

Here $\pi a^2$ corresponds to the area of the opening through which the fluid flows out, and the “minus” sign means that the height of the fluid decreases when it flows out of the reservoir.

The fluid balance equation in the reservoir is written as follows:

$\frac{\mathrm{dV} }{\mathrm{d} t}=q$

Since the volume change dV can be expressed as

$dV=S(z)dz,$

we have the following differential equation:

$\frac{S(z)dz}{dt}=q(z)$

Putting the function q(z) into this equation gives:

$\frac{S(z)dz}{dt}=-\pi a^2 \sqrt{2gz}.$

The cross section S(z) of the cylindrical vessel does not depend on the height z and is given by

$S(z)=\pi R^2,$

where R is the base radius of the cylinder. Then

$\pi R^2\frac{\mathrm{d}z }{\mathrm{d} t}=- \pi a^2 \sqrt{2gz}$

As a result, we obtain the separable equation:

$\frac{dz}{\sqrt{z}}=-\frac{a^2}{R^2}\sqrt{2g}dt$

Now we integrate this equation assuming that the initial height of the fluid is H and the fluid level decreases to 0 for the time T.

$\int_{0}^{H}\frac{dz}{\sqrt{z}}=-\int_{0}^{T}\frac{a^2}{R^2}\sqrt{2g}dt=2\left [ \left ( \sqrt{z} \right ) \right ]_H^0=-\frac{a^2}{R^2}\sqrt{2g}\left [ \left ( \sqrt{t} \right ) \right ]_0^T$

$\Rightarrow 2\sqrt{H}=\frac{a^2}{R^2}\sqrt{2g}T,\Rightarrow \sqrt{2H}=\frac{a^2}{R^2}\sqrt{g}T$

It follows from here that the time T is defined by the expression:

$T=\frac{R^2}{a^2} \sqrt{ \frac{2H}{g} }$

This is an example of circular base. More can be obtained by manipulating the base area based on the shape desired.

For triangular base: $S(z)=\frac{\sqrt{3}}{4}a^2$ where a is the side length

For hexagonal base: $S(z)=\frac{3\sqrt{3}}{2}a^2$ where a is the side length