The Bernoulli’s principle formulated by Daniel Bernoulli states that as the speed of a moving fluid increases (liquid or gas), the pressure within the fluid decreases. Although, Bernoulli deduced the law, it was Leonhard Euler who derived Bernoulli’s equation in its usual form in the year 1752.

The Bernoulli’s principle can be derived from the principle of conservation of energy. This states that the total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure and the kinetic energy of the fluid motion, remains constant.

According to Bernoulli’s principle,

\(p+\frac{1}{2}\rho V^{2}+\rho gh=constant\)

Where,

p = pressure

\(\rho\)

V = velocity

h = elevation

g = gravitational acceleration

p = pressure head

\(\frac{1}{2}\rho V^{2}\)

\(\rho gh\)

Bernoulli’s equation gives a great insight into the balance between pressure, velocity and elevation.

When we are standing on a railway station and a train comes we tend to fall towards the train. This can be explained using Bernoulli’s principle as the train goes past, the velocity of air between the train and us increases.

Hence, from equation we can say that the pressure decreases so the pressure from behind pushes us towards the train. The working of an aeroplane is also based on Bernoulli’s effect.

It is designed such that the speed of air above the wing is faster than the speed beneath the wing and hence the pressure beneath is more which lifts the aeroplane. Another Principle is Principle of continuity.

**Principle of Continuity:**

According to this if the fluid is in streamline flow and is in-compressible then we can say that mass of fluid passing through different cross sections are equal.

From the above situation we can say mass of liquid inside the container remains the same.

Rate of mass entering = Rate of mass leaving

Rate of mass entering = \(\rho A_{1}V_{1}\Delta t\)

Rate of mass entering = \(\rho A_{2}V_{2}\Delta t\)

Using the above equations,

\(\rho A_{1}V_{1} =\rho A_{2}V_{2}\)

This equation is known as Principle of continuity. Suppose we need to calculate the speed of efflux for the following setup.

Using Bernoulli’s Equation at point 1 and point 2,

\(p+\frac{1}{2}\rho v_{1}^{2}+\rho gh=p_{0}+\frac{1}{2}\rho v_{2}^{2}\\ \\ v_{2}^{2}=v_{1}^{2}+2p-p_{0}/\rho +2gh\)

Generally, A_{2} is much smaller than A_{1}; in this case, v_{1}^{2} is very much smaller than v_{2}^{2} and can be neglected. We then find,

\(v_{2}^{2}=2\frac{p-p_{0}}{\rho }+2gh\)

Assuming A_{2}<<A_{1}, We get, \(v_{2}=\sqrt{2gh}\)

Hence the velocity of efflux is \(\sqrt{2gh}\)

“In fluid dynamics, for a non – viscous, incompressible, non – conducting fluid, an increase in the velocity of the fluid during its flow (laminar), results in a simultaneous decrease in pressure (or potential energy) of the fluid. The converse of this is also true.”

**How is it based on the principle of conservation of energy? **

Bernoulli’s equation is based on the Bernoulli’s Principle for fluid flow. According to Bernoulli’s equation for a fluid, we can see that the total energy of the system is always constant. It is expressed as per the relation given below.

\(p+\frac{1}{2}\rho V^{2}+\rho gh=constant\)

where,

P is the Pressure Energy

\(\frac{1}{2}\rho V^{2}\)

ρgh is the Potential Energy per unit volume

Look at the image given below. It shows us the flow in a venturi tube.

During this kind of flow, at the point where the pressure across the tube decreases, the velocity increases. Similarly, when the pressure increases at the tube (because of increase in diameter), the velocity at this point will decrease. That is why if you partially cover over the outlet of the water pipe with your thumb, the water comes out with a greater force. You reduce the area of the outlet, it makes up for it by increasing the velocity.

**Let us try the following application to understand this concept.**

Say water is flowing through a pipe. You have the following information at hand.

**Profile at one of the pipe**: Point A**Pipe pressure**= 150,000 Pa**Speed**= 5 m/s**Height**= 0 m**Profile at other end of pipe**: Point B**Speed**= 10 m/s**Height**= 2 m

The density of water is 1000 kg/m3. Your mission should you chose to accept it, is to figure out the pressure at the other end of the pipe (point B). How are you going to solve this?

**Hint**: Based on conservation of energy and Bernoulli’s principle, if the total energy of the system is constant then \(P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho gh_{1} = P_{2} + \frac{1}{2} \rho v_{2}^{2} + \rho gh_{2}\)

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