# Torricelli’s Theorem: What It Consists Of, Formulas And Exercises

The theorem Torricelli  or principle Torricelli states that the rate of the liquid exiting the orifice in the wall of a tank or container, is identical to that acquires an object is dropped freely from a height equal to the surface free of liquid to the hole.

The theorem is illustrated in the following figure: Due to Torricelli’s theorem we can then affirm that the exit velocity of the liquid through an orifice that is at 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 the city of Faenza, Italy in 1608. Torricelli is credited with the invention of the mercury barometer and in recognition there is a pressure unit called “torr”, equivalent to one millimeter of mercury (mm of Hg).

Article index

• one

Proof of the theorem

• 1.1

Falling object

• 1.2

Liquid coming out of the hole

• two

Solved exercises

• 2.1

Exercise 1

• 3

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

• 3.1

Exercise 2

• 3.2

Exercise 3

• 4

References

## Proof of the theorem

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

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

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

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 frictional losses, it is valid to apply the principle of conservation of mechanical energy. Suppose the falling object has mass m and the height h is measured from the exit level of the liquid.

### 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

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

Ec = ½ mv 2

Since the energy is conserved Ep = Ec from what is obtained:

½ mv 2 = mgh

Solving for the speed v , the Torricelli formula is then obtained: ### Liquid coming out of the hole

Next we will find the exit velocity of the liquid through the hole, in order to show that it coincides with that which was 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 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 cross-sectional area.
• The third represents the gravitational potential energy per unit volume of fluid.

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

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

The figure below shows the Torricelli formula starting from Bernoulli’s principle.

We apply Bernoulli’s formula on the free surface of the liquid denoted by (1) and on the exit hole denoted by (2). The zero head level has been chosen flush with the outlet hole.

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

For this reason V 1 = 0 has been set , the pressure to which the liquid is subjected in (1) is atmospheric pressure and the height measured 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 at the outlet is also atmospheric pressure and the outlet height is zero.

The values ​​corresponding to sections (1) and (2) are substituted in Bernoulli’s formula and set equal. The equality holds because we assume that the fluid is ideal and there are no viscous friction losses. Once all the terms have been simplified, the velocity at the exit hole is obtained. The box above shows that the result obtained is the same as that of a freely falling object, ## I ) The small outlet pipe of a water tank is 3 m below the surface of the water. Calculate the exit velocity of the water.

#### Solution:

The following figure shows how Torricelli’s formula is applied in this case. ### Exercise 2

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

#### Solution:

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

The following figure shows the details of the calculation. ### Exercise 3

III ) Determine how high the free surface of the water is in a container if you know

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 detail of the calculations. ## References

1. Wikipedia. Torricelli’s theorem.
2. Hewitt, P. Conceptual Physical Science . Fifth edition .119.
3. Young, Hugh. 2016. Sears-Zemansky’s University Physics with Modern Physics. 14th Ed. Pearson. 384.