## What is Bernoulli’s theorem?

Bernoulli’s theorem states that –

*“Total energy of a liquid particle in a perfect in-compressible liquid flowing in a continuous stream in a pipe remains constant.”*

- Bernoulli’s theorem is associated with the energy of a flowing fluid.
- Bernoulli’s theorem statement is based on the assumption that, there are no energy loss due to
*friction*in the pipe.

Therefore, according to the Bernoulli’s statement –

\quad \left ( Z + \frac {p}{w} + \frac {v^2}{2 g} \right ) = \text {Constant} .

Or, \quad \text {Total energy of liquid} = \text {Potential energy} \ ( Z ) + \text {Pressure energy} \left ( \frac {p}{w} \right ) + \text {Kinetic energy} \left ( \frac {v^2}{2 g} \right ) = \text {Constant} .

## Energy of a fluid

*Energy of a fluid body is defined as its capacity of doing work.*

A fluid body can possess various forms of energies of which the following types of *energies* are most important.

- Potential energy.
- Pressure energy.
- Kinetic energy.

### Potential energy

*Potential energy is the energy possessed by a fluid body by virtue of its position or location in space.*

Consider about ( W ) kg of a fluid kept at a height of ( Z ) meters above a datum line. The potential energy of that fluid will be equal to ( WZ ) \ \text {kg-m}

Thus, the potential energy per kg of the fluid body will be –

\left ( \frac {W Z}{W} \right ) \left ( \frac {\text {kg-m}}{\text {kg}} \right ) = Z \ \text {mtr.} .

- We can say that,
*pressure head*due to potential energy of the fluid is ( Z ) \ \text {mtr.}

### Pressure energy

*Pressure energy is the energy possessed by a fluid by virtue of the pressure at which it is maintained.*

Consider that, ( W ) kg of a fluid body is under a pressure of ( p ) \ \text {kg-m}^{-2} . Then the pressure energy of the fluid will be –

W \left ( \frac {p}{w} \right ) \ \text {kg} \left ( \frac {\text {kg-m}^{-2}}{\text {kg-m}^{-3}} \right ) = W \left ( \frac {p}{w} \right ) \ \text {kg-m}

Where ( w ) is the *specific weight* of the fluid in \text {kg-m}^{-3} .

Therefore, pressure head per kg of the fluid under pressure will be \left ( \frac {p}{w} \right ) \ \text {mtr.}

- We can say that,
*pressure head*due to pressure energy of the fluid is \left ( \frac {p}{w} \right ) \ \text {mtr.}

### Kinetic energy

*Kinetic energy is the energy possessed by a fluid body by virtue of its motion. *

Consider that, ( W ) kg of a fluid is moving at a *velocity* of ( v ) \text {m-s}^{-1} .

Then, the kinetic energy of fluid will be –

W \left ( \frac {v^2}{2 g} \right ) \ \text {kg-m}

Therefore, kinetic energy per kg of the fluid body will be –

\left ( \frac {v^2}{2 g} \right ) \ \text {mtr.}

- We can say that,
*pressure head*due to kinetic energy of the fluid is \left ( \frac {v^2}{2 g} \right ) \ \text {mtr.}

### Total energy

If a fluid body of *weight* ( W ) \ \text {kg} is at a height of ( Z ) \ \text {mtr.} above a datum line, and at a pressure intensity of ( p ) \ \text {kg-m}^{-2} and moving with a velocity of ( v ) \ \text {m-s}^{-1} , then the total energy of the fluid body will be –

W \left ( Z + \frac {p}{w} + \frac {v^2}{2 g} \right ) \ \text {kg-m}

Therefore, total energy per kg of the fluid body will be –

\left ( Z + \frac {p}{w} + \frac {v^2}{2 g} \right ) \ \text {mtr.}

- We can say that, the pressure head due to total energy is –

\left ( Z + \frac {p}{w} + \frac {v^2}{2 g} \right ) \ \text {mtr.}

- It is also known as
**total energy head.**

## Equation of cotinuity

Consider about a perfect in-compressible liquid flowing through a non-uniform pipeline as shown in figure. In this, consider about two sections AA \text {and} BB of the pipe. Assume that, the pipe is running full and there is a continuity of flow between the two sections.

Let –

- ( Z_1 ) is height of section AA above the datum line.
- ( p_1 ) is pressure of fluid at section AA .
- ( v_1 ) is velocity of liquid at section AA .
- ( a_1 ) is area of cross section of the pipe at section AA .

And Z_2, \ p_2, \ v_2, \ a_2 are the corresponding values at section BB .

Let, the liquid between the two sections AA \ \text {and} \ BB move to sections A'A' \ \text {and} \ B'B' .

Let, the length of pipe between the sections AA \ \text {and} \ A'A' is ( dl_1 ) and between sections BB \ \text {and} \ B'B' is ( dl_2 ) as shown in figure.

This movement of liquid between AA \ \text {and} \ BB is equivalent to the movement of the liquid particles between AA \ \text {and} \ A'A' \ \text {to} \ BB \ \text {and} \ B'B' , the remaining liquid between sections A'A' \ \text {and} \ BB remains unaffected.

Let, ( W ) is the weight of the liquid between AA \ \text {and} \ A'A' .

We can apply Bernoulli’s theorem in the flow of liquid between the sections AA \ \text {and} \ BB

- Since the flow is continuous, therefore –

W = w \ a_1 \ dl_1 = w \ a_2 \ dl_2

Or, \quad a_1 \ dl_1 = \frac {W}{w} \quad \text {and} \quad a_2 \ dl_2 = \frac {W}{w}

Therefore, \quad a_1 \ dl_1 = a_2 \ dl_2

### Expression for Bernoulli’s equation

- Work done by the pressure ( p_1 ) at AA , in moving the liquid from AA \ \text {to} \ A'A' is –

\text {Force} \times \text {Distance} = ( p_1 \ a_1 ) \times dl_1

- Similarly, work done by pressure at BB in moving the liquid from BB \ \text {to} \ B'B' is –

\text {Force} \times \text {Distance} = - ( p_2 \ a_2 ) \times dl_2

Minus sign is taken as the direction of ( p_2 ) is opposite to that of ( p_1 ) .

- Total work done by the fluid pressure will be –

( p_1 \ a_1 ) dl_1 - ( p_2 \ a_2 ) dl_2

Or, \quad p_1 \ a_1 \ dl_1 - p_2 \ a_2 \ dl_2

By substituting \quad ( a_2 \ dl_2 ) = ( a_1 \ dl_1 ) , we get –

*Work done*by the fluid pressure is –

p_1 \ a_1 \ dl_1 - p_2 \ a_1 \ dl_1 = a_1 \ dl_1 ( p_1 - p_2 )

= \frac {W}{w} ( p_1 - p_2 )

- Loss of potential energy in moving the fluid from AA \ \text {to} \ BB is –

W ( Z_1 - Z_2 )

- And gain in
*kinetic energy*in moving the fluid from AA \ \text {to} \ BB is –

W \left ( \frac {v_2^2}{2g} - \frac {v_1^2}{2g} \right ) = \frac {W}{2g} \left ( v_2^2 - v_1^2 \right )

By principle of *conservation of energy*, we know that –

\text {Loss of potential energy} + \text {Work done by pressure} = \text {Gain in kinetic energy}

Therefore, \quad W ( Z_1 - Z_2 ) + \frac {W}{w} ( p_1 - p_2 ) = \frac {W}{2g} ( v_2^2 - v_1^2 )

Or, \quad ( Z_1 - Z_2 ) + \frac {p_1}{w} - \frac {p_2}{w} = \frac {v_2^2}{2g} - \frac {v_1^2}{2g}

Or, \quad Z_1 + \frac {p_1}{w} + \frac {v_1^2}{2g} = Z_2 + \frac {p_2}{w} + \frac {v_2^2}{2g}