 # Bernoulli’s Theorem | Principle, Equation , Derivation and Solved examples .

Bernoulli’s principle is a kind of energy conservation applied to fluid flow. It is applied primarily to incompressible fluid in steady flow .

The principle states that the sum of the static head , kinetic energy per unit volume and gravitational energy per unit volume at any point along the flow line is constant .

## Bernoulli’s Equation

According to Bernoulli’s theorem , along a streamline in a steady flow of an ideal fluid the sum of total energy ( kinetic energy, potential energy and pressure energy ) remains constant . That is ,

P + ρv²/2 + ρgh = constant

Where

• p is the pressure exerted by the fluid

• v is the velocity of the fluid

• ρ is the density of the fluid

• h denotes the height of the container .

An ideal fluid is inviscid and incompressible. In other words ,there is no dissipation of energy due to internal friction between adjacent layers of the fluid and density of this fluid remains constant . In other words, the sum of total energy per unit volume ( pressure + kinetic + potential ) is constant for an ideal fluid .

## OTHER FORMS OF BERNOULLI’S EQUATION

If the kinetic, potential and pressure energies are expressed in terms of per unit mass or per unit weight, then the Bernoulli’s equation may be written as

р/ρ+v²/2+gy = constant (per unit mass)

p/ρg + v²/2g + y = constant ( per unit weight )

The Bernoulli’s equation in per unit weight form is the most commonly used expression. Each term in this equation has the dimensions of length. The term v²/2g is known as the velocity head , p/pg is known as the pressure head and y is known as the potential head .

## Assumption and Limitations of Bernoulli’s equation

– The flow is assumed to be steady, i.e., there is no change in pressure, velocity and density of the fluid at any point with respect to time.

– However, in the problems of unsteady flow with gradually changing conditions, Bernoulli’s equation can be applied without appreciable error. For example, a problem of emptying a large tank can be solved by applying Bernoulli’s equation .

– The fluid is assumed to be incompressible. Since liquids are incompressible, Bernoulli’s equation can be applied to all liquids. However, it can be applied to the problems of gas flow when there is little variation in pressure, velocity and temperature so that density of gas can be assumed to be constant .

## Proof of Bernoulli’s Equation

Figure.1 shows flow of an ideal fluid through a tube of varying cross section and height . The fluid at the left end exerts a force P1A1 , which moves the fluid through a distance ∆x1 and therefore does work W1 = F1∆x1= P1A1∆x1 , on the fluid. At the same time an equal volume of fluid at the upper right end is displaced through ∆x2 . So the work done on this fluid volume is

W2 = -F2∆x2 = – P2A2∆x2

At right end of tube the work done on this fluid element is negative because the force on it is in the opposite direction to the displacement. Such a way the total work done on the entire fluid element will be

W = W1+W2 = P1A1∆x1– P2A2∆x2

Using equation of continuity

A1∆x1 = A2∆x2= ∆V

W =(P1-P2)∆V

Between the left and right end , the change in Kinetic energy is

∆K = ∆mv22/2 – ∆mv12/2

And the change in potential energy is

∆U = ∆mgh2 – ∆mgh1

Using energy conservation,

∆W = ∆K +∆U

( P1-P2 )∆V = ( ∆mv22/2 – ∆mv12/2 ) + ( ∆mgh2 – ∆mgh1 )

Let’s divide the equation by ∆V and substituting ρ = ∆m/∆V

we find

P1 + ρv12/2 + ρgh1 = P2 + ρv22/2 + ρgh2

P + ρv2/2 + ρgh = constant ….( i )

Equation (i) is called Bernoulli’s equation . This equation is actually the conservation of energy in fluid .

There are various applications of Bernoulli’s Equation ( Theorem ) –  pitot tube , Atomizer , siphon tube , lifting of aircraft,  Blowing off the roof during wind storm , Magnus effect ,Venturimeter , Velocity of efflux etc .

## Solved Examples

• Problem 1. Calculate the rate of flow of glycerine through the conical section of a pipe that have radii of 0.1 m and 0.04 m at both end and the pressure drop across its length is 10 N/m². Given that density of glycerine is 1.25 x 10³ kg/m³ .

Solution – From continuity equation ,
A1v1 = A2v

v1/v2 = A2/A1  = πR22/ πR12
= ( r2/r1

= ( 0.04/0.1 )²  = 4/25   …( i )

From Bernoulli’s equation ,
p1 + ρv12/2 = p2 + ρv22 or

v22 – v12 = 2 ( p1 – p2 ) / ρ

v22 – v12 = 1.6× 10-2 m²/s²   …( ii )

Solving equation ( i ) and ( ii ) ,

we get v2 = 0.128 m/s
Rate of volume flow through the tube

Q = A2v2 = ( πR22 ) v2

= π( 0.04 )²( 0.128 ) = 6.43 × 10-4 m³/s

• Problem 2 . A pipe of non-uniform cross-section area , placed horizontally , allows water to pass through it with a velocity of 1 ms−1 when pressure is at this point is 50 kPa . What will be pressure at that point where the velocity of flow is 2 ms−1 ?

Solution – Considering both the point as point 1 and point 2 , Let’s apply Bernoulli’s equation at point 1 and point 2 ,

P1​1​V12​/2=P2​+ρ2​V22​/2

50×10³+1×10³(1)²/2 = P2​+1×10³×4

5×10⁴+500−2000=P2

P2​=48.5 kPa