**what is the** theorem** from Torricelli?**

He **Torricelli’s Theorem,** or Torricelli’s principle, states that the speed of the liquid that comes out through the hole in the wall of a tank or container is identical to that acquired by an object that is dropped freely from a height equal to that of the free surface of the liquid. up to the hole. The theorem is illustrated in the image above.

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

where g is the acceleration of gravity and h is the height from the hole to the free surface of the liquid.

Evangelista Torricelli was a physicist and mathematician born in Faenza, Italy, in 1608. Torricelli is credited with inventing the mercury barometer, and in recognition there is a unit of pressure called a «torr,» equal to one millimeter of mercury (mm of Hg).

**proof of the theorem**

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

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

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

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

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

**falling object**

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

Ep = mgh

As it passes the hole, its height is zero, so the potential energy is zero, so it only has kinetic energy Ec given by:

Ec = ½ m v2

Since energy is conserved Ep = Ec of what you get:

½ m v2 = mgh

clearing speed *v* then Torricelli’s formula is obtained:

**fluid coming out of the hole**

Next, we will find the exit speed of the liquid through the hole, in order to show that it agrees with the one just calculated for a freely 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 as follows:

The interpretation of this formula is as follows:

The first term represents the **Kinetic energy** of fluid per unit volume.

The second represents the **work done by pressure** per unit cross-sectional area.

The third represents the **gravitational potential energy** per unit volume of fluid.

Since we start from the premise that we are dealing with an ideal fluid, in non-turbulent conditions, with relatively low speeds, then it is pertinent to state that the mechanical energy per unit volume in the fluid is constant in all regions or cross sections thereof.

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

The image below shows Torricelli’s formula based on Bernoulli’s principle.

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

For this reason V1=0 has been placed, the pressure to which the liquid is subjected in (1) is atmospheric pressure and the height measured from the hole is *h*.

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

Substitute the values corresponding to sections (1) and (2) into Bernoulli’s formula and equalize. The equality is valid because we assume that the fluid is ideal and there are no losses due to viscous friction. Once all the terms have been simplified, the velocity at the exit orifice is obtained.

The box above shows that the result obtained is the same as that of a freely falling object,

with which the Torricelli principle is demonstrated.

**Solved problems of Torricelli’s theorem**

**Exercise 1**

**Yo**) The small outlet tube of a water tank is 3 m below the surface of the water. Calculate the outflow velocity of the water.

**Solution:**

The following figure shows how Torricelli’s formula applies to this case.

**Exercise 2**

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

**Solution:**

The flow rate is the volume of liquid that exits per unit of time, and is calculated simply by multiplying the outlet orifice area by the outlet velocity.

The following figure shows the details of the calculation.

**Exercise 3**

**II**) Determine how high is the free surface of the water in a container, if it is known Water flows out of a hole in the bottom of the container at 10 m/s.

**Solution:**

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

The following figure shows the detail of the calculations.

**References**

Torricelli’s theorem. Retrieved from es.wikipedia.org.

Hewitt, P. Conceptual Physical Science. Fifth edition.

Young, Hugh. Sears-Zemansky’s University Physics with Modern Physics. 14th Ed. Pearson.