# Torricelli's law

Not to be confused with Torricelli's equation or Torricelli point.

Torricelli's law, also known as Torricelli's theorem, is a theorem in fluid dynamics relating the speed of fluid flowing out of an opening to the height of fluid above the opening.

Torricelli's law states that the speed of efflux, v, of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h, i.e. $v = \sqrt{2gh}$, where g is the acceleration due to gravity (9.81 N/kg). This last expression comes from equating the kinetic energy gained, $\frac{1}{2}mv^2$, with the potential energy lost, mgh , and solving for v.

The law was discovered (though not in this form) by the Italian scientist Evangelista Torricelli, in 1643. It was later shown to be a particular case of Bernoulli's principle.

## Derivation

Bernoulli's principle states that:

${v^2 \over 2}+gz+{p\over\rho}=\text{constant}$

where v is fluid speed, g is the gravitational acceleration (9.81 m/s^2), z is the fluid's height above a reference point, p is pressure, and ρ is density. Define the opening to be at z=Ø. At the top of the tank, p is equal to the atmospheric pressure. v can be considered 0 because the fluid surface drops in height extremely slowly compared to the speed at which fluid exits the tank. At the opening, z=Ø and p is again atmospheric pressure. Eliminating the constant and solving gives:

$gz+{p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}$
$\Rightarrow v^2=2gz\,$
$\Rightarrow v=\sqrt{2gz}$

z is equivalent to the h in the first paragraph of this article, so:

$v=\sqrt{2gh}$

## Experimental evidence

Torricelli's law can be demonstrated in the spouting can experiment, which is designed to show that in a liquid with an open surface, pressure increases with depth. It consists of a tube with three separate holes and an open surface. The three holes are blocked, then the tube is filled with water. When it is full, the holes are unblocked. The lower a jet is on the tube, the more powerful it is. The fluid's exit velocity is greater further down the tube.[1]

Ignoring viscosity and other losses, if the nozzles point vertically upward then each jet will reach the height of the surface of the liquid in the container.

## Application for time to empty the container

Consider a container containing water to height h is being emptied through a tube freely. Let h be the height of water at any time. Let the velocity of efflux be $v= \sqrt{2gh}\ = {dx \over dt}$

Now, A dh= a dx where, A and a are the cross sections of container and tube respectively. dh is the height of liquid in container corresponding to dx in tube which decreases in same time dt.

so, :

${A .dh \over\ a .dt} = \sqrt {2gh}$

$\Rightarrow{A .dh \over\ a .\sqrt {2gh}} = dt$
$\Rightarrow\int\ {A .dh \over\ a .\sqrt {2gh}} = \int\ dt$
$\Rightarrow {A \sqrt {h}\ \over\ a \sqrt {2g}} = t$
$\Rightarrow t = {A \sqrt {h}\ \over\ a \sqrt {2g}}$
$\Rightarrow t = {A \over\ a \sqrt {2g}} (\sqrt {h1} - \sqrt {h2})$ is the time required to empty the water from height h1 to h2 in the container.