Formula of Bernoulli’s Theorem

Bernoulli equation derivation

For incompressible fluids, we can find out Bernoulli’s equation in two ways. The first one involves the integration of the Second Law of motion as proposed by Newton. The second method involves the application of energy conservation law through a streamline. The factors like viscosity and compressibility are neglected in this case.

Bernoulli equation derivation through the integration of the Second Law of motion

We neglect the action of gravity in this first method. But one must consider the narrowing and expansion of the pipelines. Let a box of length dx be moved through a x meter long pipe. The cross-section of the pipe is A. Therefore the box’s volume will become A dx. We can assume the box’s mass to be m and its mass density is ρ.

Then by the formula of mass, m = ρ A dx.

The existent pressure’s deviation within the distance dx is dp. 

So, the velocity (v) of flow must be dx/dt. 

Mass x acceleration = force [Newton’s Second Law of motion] 

Here the box’s force = – A d ρ. 

If the pressure goes down, we will get a negative value of dp.

1. dv/dt = F

ρ V dx. dv/dt = – A dp

dv/dt. ρ = – dp/dx

Velocity of flow (v) is not a function of the duration of flow (t).

 When the box covers a distance x and cross section x(t), then

dv/dt = dv/dx. dx/dt = v. dv/dx = d/dx (v2/2)

Keeping the density constant, the equation for motion is given as 

d/dx(ρ. v2/2 + p) = 0

The final Bernoulli equation derivation is done by integration 

v2/2 + p/ρ = C. 

C is Bernoulli’s constant. This value of C varies from one fluid to another. 

Bernoulli theorem derivation by applying energy conservation 

The theorem of work-energy conservation states that – Kinetics energy change of a particular system is equivalent to the network performed on that system. 

Therefore, W = ∆ Ekin

The quantity of the kinetic energy increases in proportion to the amount of work performed. 

The system initially houses the fluid within cross-sectional areas of A1 and A2. Let us consider a period of ∆ t when fluid flows through area A1. Therefore the distance covered is s1. 

We know, speed = distance x time

Therefore, s1 = v1 ∆ t. 

Similarly, if we consider the outflow of fluid through another cross-sectional area A2, then speed s2 = v2 ∆ t.

The volume of the fluid displaced in both the cases is A1s1 and A2s2 respectively. The fluid’s mass density is assumed as ρ. According to mass conservation law, the number of displaced masses (∆ m) within the interval (∆ t) should be identical. 

 ∴ ∆ m = A1s1. ρ = A1v1. ρ  ∆ t

Also,  ∆ m = A2s2. ρ = A2s2. ρ ∆ t

Two forces are performing the work. 

• Pressure acts upon the cross-sections A1 and A2. 

        This work (Wp) equals to F1, p s1 – F2, p s2

• Work is also performed by gravitational force in the opposite direction. It is denoted by ∆ E pot, gravity. Gravity is supposed to act along the z axis. 

Therefore, ∆ E pot, gravity = ∆ m z2g – ∆ m z1g

Work done by the two forces is Wp + Wg (also called ∆ E pot, gravity) 

The kinetic energy increases by ∆ Ekin. 

½ ∆ mv22 – ½ mv12 =  ∆ Ekin

As per work-energy theory, total work W =  ∆ Ekin

∴ ∆ m z2g – ∆ m z1g + ∆m p1/ ρ –  ∆m p2 / ρ = ½ ∆ mv22 – ½ mv12 

Or, ½ ∆m v12 + ∆m z1g + ∆m p1/ ρ = ½ ∆m v22 + ∆m z2g + ∆m p2/ ρ 

Dividing both sides by  ∆m we get, 

v2/2 + gz + ρ/p = C

So this is the step-by-step process to state and prove Bernoulli’s theorem. 

Applications of Bernoulli’s equation derivation 

To study acoustics and to gather information about surface unstable waves in oceanic bodies we require Bernoulli’s theorem. If we know the nature of the flow of a fluid that is present in the surrounding atmosphere of an airfoil, we can use that data to evaluate the lift force of that foil. The carburetor which is installed in many engines comprises a venturi to generate a low-pressure zone. This is done to produce a mixture of air and petrol. Bernoulli’s equation explains the mechanism of this engine. Similarly, the working principle of injectors in steam boilers also follows Bernoulli’s ideology. Also, we can implement a device named as venturi meter to calculate the speed of flow of a liquid within a pipeline. All these are the results of Bernoulli equation derivation. 

Conclusion 

Bernoulli’s theorem explains the rate of flow of unstable fluids. It helps us to draw the relationship between a fluid’s pressure and its potential energy. There are numerous applications of Bernoulli’s theorem in the modern world.