# Torricelli’s theorem: what it consists of, formulas and exercises

The **Torricelli theorem** or **Torricelli** principle states that the rate of liquid leaving the hole in the wall of a tank or vessel is identical to that which acquires an object and freely falls from a height equal to the free liquid surface into the hole.

The theorem is illustrated in the following figure:

Due to Torricelli’s theorem, we can state that the velocity of the liquid that leaves a hole at a height h below the free surface of the liquid is given by the following formula:

Where g is the acceleration due to gravity and h is the height of the hole to the free surface of the liquid.

Evangelista Torricelli was a physicist and mathematician born in the city of Faenza, Italy, in 1608. Torricelli is credited with having invented the mercury barometer and, in recognition, there is a pressure unit called “torr”, equivalent to one millimeter of mercury ( mm Hg).

**Proof of theorem**

In Torricelli’s theorem and in the formula that gives the velocity, he assumes that losses due to viscosity are negligible, just as in free fall, friction due to the air surrounding the falling object is negligible.

The above assumption is reasonable in most cases and also implies the conservation of mechanical energy.

To prove the theorem, we will first find the velocity formula for an object that is released with zero initial velocity from the same height as the surface of the liquid in the reservoir.

The principle of energy conservation will be applied to obtain the speed of the falling object only when a height *h* equal to that of the hole until the free surface has descended.

As there are no friction losses, it is valid to apply the principle of mechanical energy conservation. Suppose the falling object has mass m and height h is measured from the exit level of the liquid.

**falling object**

When the object is released from a height equal to the free surface of the liquid, its energy is just a gravitational potential because its velocity is zero and therefore its kinetic energy is zero. The potential energy Ep is given by:

Ep = mgh

When it passes in front of the hole, its height is zero, so the potential energy is zero, so it only has kinetic energy Ec given by:

Ec = ½ mv ^{2}

How energy is conserved Ep = Ec from what is obtained:

½ mv ^{2} = mgh

Clearing the velocity *v* , we obtain the Torricelli formula:

**liquid coming out of the hole**

Next, we will find the exit velocity of the liquid through the orifice, in order to demonstrate that it coincides with the one calculated only for a free-falling object.

For this, we will base ourselves on Bernoulli’s principle, which is nothing more than the conservation of energy applied to fluids.

Bernoulli’s principle is formulated like this:

The interpretation of this formula is as follows:

- The first term represents the kinetic energy of the fluid per unit volume
- The second represents the work done by pressure per unit of cross-sectional area.
- The third represents the gravitational potential energy per unit of fluid volume.

As we start from the premise that it is an ideal fluid, in non-turbulent conditions and with relatively low velocities, it is pertinent to state that the mechanical energy per unit volume in the fluid is constant in all its regions or cross sections.

In this formula, *V* is the fluid velocity, *ρ* the fluid density, *P* the pressure and *z* the vertical position.

The figure below shows the Torricelli formula based on the Bernoulli principle.

We apply Bernoulli’s formula on the free surface of the liquid that we denote by (1) and on the exit hole that we denote by (2). The zero height level was chosen flush with the exit hole.

Under the premise that the cross section in (1) is much larger than in (2), we can then assume that the rate of descent of the liquid in (1) is practically negligible.

This is why V _{1} = 0 was set, the pressure to which the liquid is subjected (1) is atmospheric pressure and the measured height from the orifice is *h* .

For the outlet section (2), we assume that the outlet velocity is v, the pressure to which the liquid is subjected to the outlet is also atmospheric pressure, and the outlet height is zero.

Values corresponding to sections (1) and (2) in Bernoulli’s formula are substituted and equalized. The equality holds because we assume that the fluid is ideal and there are no viscous friction losses. Once all terms are simplified, the velocity at the exit orifice is obtained.

The box above shows that the result obtained is the same as an object in free fall,

**solved exercises**

**Exercise 1**

**I** ) The small outlet pipe of a water tank is 3 m below the water surface. Calculate the leaving water velocity.

**Solution:**

The following figure shows how the Torricelli formula is applied in this case.

**Exercise 2**

**II** ) Assuming that the outflow pipe of the tank from the previous exercise has a diameter of 1 cm, calculate the outflow water flow.

**Solution:**

Flow rate is the volume of liquid exiting per unit of time and is calculated simply by multiplying the exit orifice area by the exit velocity.

The following figure shows the calculation details.

**Exercise 3**

**III** ) Determine at what height is the free surface of water in a container, if known

that in a hole in the bottom of the container, the water comes out at 10 m/s.

**Solution:**

Even when the hole is at the bottom of the container, the Torricelli formula can still be applied.

The following figure shows the details of the calculations.