# Torricelli's law

Torricelli's law describes the parting speed of a jet of water, based on the distance below the surface at which the jet starts, assuming no air resistance, viscosity, or other hindrance to the fluid flow. This diagram shows several such jets, vertically aligned, leaving the reservoir horizontally. In this case, the jets have an envelope (a concept also due to Torricelli) which is a line descending at 45° from the water's surface over the jets. Each jet reaches farther than any other jet at the point where it touches the envelope, which is at twice the depth of the jet's source. The depth at which two jets cross is the sum of their source depths. Every jet (even if not leaving horizontally) takes a parabolic path whose directrix is the surface of the water.

Torricelli's law, also known as Torricelli's theorem, is a theorem in fluid dynamics relating the speed of fluid flowing from an orifice to the height of fluid above the opening. The law states that the speed v of efflux 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. ${\displaystyle v={\sqrt {2gh}}}$, where g is the acceleration due to gravity (9.81 m/s2 near the surface of the Earth). This expression comes from equating the kinetic energy gained, ${\displaystyle {\frac {1}{2}}mv^{2}}$, 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

Under the assumptions of an incompressible fluid with negligible viscosity, Bernoulli's principle states that

${\displaystyle {\frac {v^{2}}{2}}+gy+{\frac {p}{\rho }}={\text{constant}},}$

where ${\displaystyle v}$ is fluid speed, ${\displaystyle g}$ is the acceleration due to gravity (about 9.81 m/s2 on the Earth's surface), ${\displaystyle y}$ is the height above some reference point, ${\displaystyle p}$ is the pressure, and ${\displaystyle \rho }$ is the density. Thus, for any two points in the liquid,

${\displaystyle {\frac {{v_{1}}^{2}}{2}}+g_{1}y_{1}+{\frac {p_{1}}{\rho _{1}}}={\frac {{v_{2}}^{2}}{2}}+g_{2}y_{2}+{\frac {p_{2}}{\rho _{2}}}.}$

The first point can be taken at the liquid's surface, and the second just outside the opening. Since the liquid is assumed to be incompressible, ${\displaystyle \rho _{1}}$ is equal to ${\displaystyle \rho _{2}}$; both can be represented by one symbol ${\displaystyle \rho }$. In addition, when the opening is very small relative to the horizontal cross-section of the container, the velocity of the surface is assumed to be negligible (${\displaystyle v_{1}=0}$). ${\displaystyle g}$ is assumed to be practically the same at both points, so ${\displaystyle g_{1}=g_{2}=g}$.

${\displaystyle gy_{1}+{\frac {p_{1}}{\rho }}={\frac {{v_{2}}^{2}}{2}}+gy_{2}+{\frac {p_{2}}{\rho }}}$
${\displaystyle \Rightarrow v_{2}={\sqrt {2g(y_{1}-y_{2})+2(p_{1}-p_{2})/\rho }}.}$

${\displaystyle y_{1}-y_{2}}$ is equal to the height ${\displaystyle h}$ of the liquid's surface over the opening. ${\displaystyle p_{1}}$ and ${\displaystyle p_{2}}$ are typically both atmospheric pressure, so ${\displaystyle p_{1}=p_{2}\Rightarrow p_{1}-p_{2}=0}$.

${\displaystyle v_{2}={\sqrt {2gh}}.}$

## 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 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.

## Coefficient of discharge

If one compares the theoretical predictions about the discharge process of a tank with a real measurement, very large differences can be found in some cases. In reality, the tank usually drains much more slowly. In order to obtain a better approximation to the volumetric flow rate actually measured, a discharge coefficient is used in practice:

${\displaystyle {\displaystyle {\dot {V}}_{\text{real}}=\mu \cdot {\dot {V}}_{\text{ideal}}}}$

The coefficient of discharge takes into account both the reduction of the discharge velocity due to the viscous behaviour of the liquid ("coefficient of velocity") and the reduction of the effective outflow cross-section due to the vena contracta ("coefficient of contraction"). For low viscosity liquids (such as water) flowing out of a round hole in a tank, the discharge coefficient is in the order of 0.65.[2] By using rounded pipe sockets, the coefficient of discharge can be increased to over 0.9. For rectangular openings, the discharge coefficient can be up to 0.67, depending on the height-width ratio.

## Applications

### Horizontal distance covered by the jet of liquid

If ${\displaystyle h}$ is height of the orifice above the ground and ${\displaystyle H}$ is height of the liquid column above the orifice, then the horizontal distance covered by the jet of liquid to reach the same level as the base of the liquid column can be easily derived. Since ${\displaystyle h}$ be the vertical height traveled by a particle of jet stream, we have from the laws of falling body

${\displaystyle h={\frac {1}{2}}gt^{2}\quad \Rightarrow \quad t={\sqrt {\frac {2h}{g}}},}$

where ${\displaystyle t}$ is the time taken by the jet particle to fall from the orifice to the ground. If the horizontal efflux velocity is ${\displaystyle v}$, then the horizontal distance traveled by the jet particle during the time duration ${\displaystyle t}$ is

${\displaystyle D=vt=v{\sqrt {\frac {2h}{g}}}.}$

Since the water level is ${\displaystyle H-h}$ above the orifice, the horizontal efflux velocity ${\displaystyle v={\sqrt {2g(H-h)}},}$ as given by Torricelli's law. Thus, we have from the two equations

${\displaystyle D=2{\sqrt {h(H-h)}}.}$

The location of the orifice that yields the maximum horizontal range is obtained by differentiating the above equation for ${\displaystyle D}$ with respect to ${\displaystyle h}$, and solving ${\displaystyle dD/dh=0}$. Here we have

${\displaystyle {\frac {dD}{dh}}={\frac {H-2h}{\sqrt {h(H-h)}}}.}$

Solving ${\displaystyle dD/dh=0,}$ we obtain

${\displaystyle h^{*}={\frac {H}{2}},}$

and the maximum range

${\displaystyle D_{\max }=H.}$

### Clepsydra problem

An inflow clepsydra

A clepsydra is a clock that measures time by the flow of water. It consists of a pot with a small hole at the bottom through which the water can escape. The amount of escaping water gives the measure of time. As given by the Torricelli's law, the rate of efflux through the hole depends on the height of the water; and as the water level diminishes, the discharge is not uniform. A simple solution is to keep the height of the water constant. This can be attained by letting a constant stream of water flow into the vessel, the overflow of which is allowed to escape from the top, from another hole. Thus having a constant height, the discharging water from the bottom can be collected in another cylindrical vessel with uniform graduation to measure time. This is an inflow clepsydra.

Alternatively, by carefully selecting the shape of the vessel, the water level in the vessel can be made to decrease at constant rate. By measuring the level of water remaining in the vessel, the time can be measured with uniform graduation. This is an example of outflow clepsydra. Since the water outflow rate is higher when the water level is higher (due to more pressure), the fluid's volume should be more than a simple cylinder when the water level is high. That is, the radius should be larger when the water level is higher. Let the radius ${\displaystyle r}$ increase with the height of the water level ${\displaystyle h}$ above the exit hole of area ${\displaystyle a.}$ That is, ${\displaystyle r=f(h)}$. We want to find the radius such that the water level has a constant rate of decrease, i.e. ${\displaystyle dh/dt=c}$.

At a given water level ${\displaystyle h}$, the water surface area is ${\displaystyle A=\pi r^{2}}$. The instantaneous rate of change in water volume is

${\displaystyle {\frac {dV}{dt}}=A{\frac {dh}{dt}}=\pi r^{2}c.}$

From Torricelli's law, the rate of outflow is

${\displaystyle {\frac {dV}{dt}}=av=a{\sqrt {2gh}},}$

From these two equations,

{\displaystyle {\begin{aligned}a{\sqrt {2gh}}&=\pi r^{2}c\\\Rightarrow \quad h&={\frac {\pi ^{2}c^{2}}{2ga^{2}}}r^{4}.\end{aligned}}}

Thus, the radius of the container should change in proportion to the quartic root of its height, ${\displaystyle r\propto {\sqrt[{4}]{h}}.}$

### Time to empty a cylindrical vessel

Assuming that a vessel is cylindrical with fixed cross-sectional area ${\displaystyle A}$, with orifice of area ${\displaystyle a}$ at the bottom, then rate of change of water level height ${\displaystyle dh/dt}$ is not constant. From the previous relations, we have

${\displaystyle A{\frac {dh}{dt}}=a{\sqrt {2gh}}\quad \Rightarrow \quad A{\frac {dh}{\sqrt {h}}}=a{\sqrt {2g}}\;dt}$

Integrating both sides and re-arranging, we obtain

${\displaystyle T={\frac {A}{a}}{\sqrt {\frac {2h}{g}}},}$

where ${\displaystyle h}$ is the initial height of the water level and ${\displaystyle T}$ is the total time taken to drain all the water and hence empty the vessel.

This formula has several implications. If a tank with volume ${\displaystyle V}$ with cross section ${\displaystyle A}$ and height ${\displaystyle H}$, so that ${\displaystyle V=AH}$, is fully filled, then the time to drain all the water is

${\displaystyle T={\frac {V}{a}}{\sqrt {\frac {2}{gH}}}.}$

This implies that high tanks with same filling volume drains faster than wider ones.

Likewise, if the shape of the vessel of the outflow clepsydra cannot be modified according to the above specification, then we need to use non-uniform graduation to measure time. The above formula tells us that the time should be calibrated as square root of the discharged water height, ${\displaystyle T\propto {\sqrt {h}}.}$ More precisely,

${\displaystyle \Delta t={\frac {A}{a}}{\sqrt {\frac {2}{g}}}({\sqrt {h_{1}}}-{\sqrt {h_{2}}})}$

where ${\displaystyle \Delta t}$ is the time taken by the water level to fall from the height of ${\displaystyle h_{1}}$ to height of ${\displaystyle h_{2}}$.

Lastly, we can re-arrange the above equation to determine the height of the water level ${\displaystyle h(t)}$ as a function of time ${\displaystyle t}$ as

${\displaystyle h(t)=H\left(1-{\frac {t}{T}}\right)^{2},}$

where ${\displaystyle H}$ is the height of the container while ${\displaystyle T}$ is the discharge time as given above.

## References

1. ^
2. ^ tec-science (2019-11-21). "Discharge of liquids (Torricelli's law)". tec-science. Retrieved 2019-12-08.