# Bernoulli’s Theorem: Equation, Applications And Solved Exercise

The Bernoulli theorem , which describes the behavior of a fluid in motion, was enunciated by the mathematical and physical Daniel Bernoulli in his Hydrodynamics . According to the principle, an ideal fluid (without friction or viscosity) that is circulating through a closed conduit, will have a constant energy in its path.

The theorem can be deduced from the principle of conservation of energy and even from Newton’s second law of motion. In addition, Bernoulli’s principle also establishes that an increase in the speed of a fluid implies a decrease in the pressure to which it is subjected, a decrease in its potential energy, or both at the same time. The theorem has many different applications, both in the world of science and in people’s daily lives.

Its consequences are present in the lift force of aircraft, in the chimneys of homes and industries, in water pipes, among other areas.

## Bernoulli’s equation Although Bernoulli was the one who deduced that the pressure decreases when the flow rate increases, the truth is that it was Leonhard Euler who actually developed the Bernoulli equation in the form in which it is known today.

In any case, Bernoulli’s equation, which is nothing more than the mathematical expression of his theorem, is the following:

v 2 ∙ ƿ / 2 + P + ƿ ∙ g ∙ z = constant

In this expression, v is the velocity of the fluid through the section considered, ƿ is the density of the fluid, P is the pressure of the fluid, g is the value of the acceleration of gravity, and z is the height measured in the direction of gravity.

It is implicit in Bernoulli’s equation that the energy of a fluid consists of three components:

– A kinetic component, which is the one that results from the speed at which the fluid moves.

– A potential or gravitational component, which is due to the height at which the fluid is.

– A pressure energy, which is that which the fluid possesses as a consequence of the pressure to which it is subjected.

On the other hand, Bernoulli’s equation can also be expressed like this:

v 1 2 ∙ ƿ / 2 + P 1 + ƿ ∙ g ∙ z 1 = v 2 2 ∙ ƿ / 2 + P 2 + ƿ ∙ g ∙ z 2

This last expression is very practical to analyze the changes that a fluid experiences when any of the elements that make up the equation change.

### Simplified form

On certain occasions, the change in the term ρgz of Bernoulli’s equation is minimal compared to that experienced by the other terms, so it can be neglected. For example, this happens in currents experienced by an airplane in flight. On these occasions, the Bernoulli equation is expressed as follows:

P + q = P 0

In this expression q is dynamic pressure and is equivalent to v 2 ∙ ƿ / 2, and P 0 is what is called total pressure and is the sum of the static pressure P and the dynamic pressure q.

## Applications Bernoulli’s theorem has many and diverse applications in fields as diverse as science, engineering, sports, etc.

An interesting application is found in the design of fireplaces. The chimneys are built high in order to achieve a greater pressure difference between the base and the chimney outlet, thanks to which it is easier to extract the combustion gases.

Of course, the Bernoulli equation also applies to the study of the movement of liquid flows in pipes. It follows from the equation that a reduction in the cross-sectional area of ​​the pipe, in order to increase the velocity of the fluid passing through it, also implies a decrease in pressure.

The Bernoulli equation is also used in aviation and in Formula 1 vehicles. In the case of aviation, the Bernoulli effect is the origin of the lift of airplanes.

Aircraft wings are designed with the goal of achieving greater airflow at the top of the wing. Thus, in the upper part of the wing, the air speed is high and, therefore, the pressure is lower. This pressure difference produces a vertically upwardly directed force (lift force) that allows the aircraft to hover in the air. A similar effect is obtained on the ailerons of Formula 1 cars.

## Exercise resolved

A stream of water flows at 5.18 m / s through a pipe with a cross section of 4.2 cm 2 . The water descends from a height of 9.66 m to a lower level with a height of zero elevation, while the cross-sectional area of ​​the tube increases to 7.6 cm 2 .

a) Calculate the speed of the water current at the lower level.

b) Determine the pressure at the lower level knowing that the pressure at the upper level is 152000 Pa.

#### Solution

a) Given that the flow must be conserved, it is true that:

Q upper level = Q lower level

v 1 . S 1 = v 2 . S 2

5.18 m / s. 4.2 cm 2 = v 2 . 7.6 cm ^ 2

Solving for, it is obtained that:

v 2 = 2.86 m / s

b) Applying Bernoulli’s theorem between the two levels, and taking into account that the density of water is 1000 kg / m 3  , it is obtained that:

v 1 2 ∙ ƿ / 2 + P 1 + ƿ ∙ g ∙ z 1 = v 2 2 ∙ ƿ / 2 + P 2 + ƿ ∙ g ∙ z 2

(1/2). 1000 kg / m 3 . (5.18 m / s) 2 + 152000 + 1000 kg / m 3 . 10 m / s 2 . 9.66 m =

= (1/2). 1000 kg / m 3 . (2.86 m / s) 2 + P 2 + 1000 kg / m 3 . 10 m / s 2 . 0 m

Solving for P 2 we get:

P 2 = 257926.4 Pa

## References

1. Bernoulli’s principle. (nd). On Wikipedia. Retrieved on May 12, 2018, from es.wikipedia.org.
2. Bernoulli’s Principle. (nd). In Wikipedia. Retrieved on May 12, 2018, from en.wikipedia.org.
3. Batchelor, GK (1967). An Introduction to Fluid Dynamics . Cambridge University Press.
4. Lamb, H. (1993). Hydrodynamics (6th ed.). Cambridge University Press.
5. Mott, Robert (1996). Applied Fluid Mechanics  (4th ed.). Mexico: Pearson Education.