Bernoulli's Principle: Class 11 Physics Notes, Definition, Equation, & Derivation

Bernoulli’s Principle shows an inverse relationship between velocity and pressure of a moving fluid. This is a fundamental principle of fluid dynamics, stating that when the flow of a moving fluid increases, the pressure within it decreases, or vice versa. 

So, considering air as a fluid here, you can remember like this

Faster air = Lower air pressure

Slower air = Higher air pressure

 

The Bernoulli’s equation follows or derives from the Law of Conservation of Energy.

Think of this law that applies to fluids in this case, where a fluid's total mechanical energy (pressure + kinetic + potential) remains constant. 

So, as L. J. Clancy, in his book, Thermodynamics shows a way to memorise Bernoulli’s Principle through a simple equation

Static pressure + Dynamic pressure = Total pressure

The basic formula for Bernoulli’s Principle is 

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

Conditions for Bernoulli's Principle 

One of the primary conditions for Bernoulli’s Principle to work is that the fluid must be ideal. 

That means the fluid cannot have any friction when flowing. You can think of water that does not have any friction when flowing, and compare that with thick syrup that has friction.  

Secondly, the fluid’s volume should not change under pressure. If the fluid were compressible, the density would change. In Bernoulli’s equation, the pressure is treated as constant. 

Third condition. To understand Bernoulli’s Principle, the only gravity can affect the fluid. 

Master gravitational concepts with NCERT Solutions for Gravitation and score high in your annuals!

 

 

The derivation of Bernoulli’s Principle equation can be based on the Work-Energy Theorem. This states that the net work done = Change in Kinetic Energy + Change in Potential Energy. 

Let’s try to understand the energy changes in a flowing fluid, while you can read about Work in Physics and prepare for your next chapters for your upcoming tests and exams.  

 

You can consider a small mass dm1​ of fluid entering at point S. 

Its mass is

                                                            $d{m}_1=\rho A_{1}d{x}_1=\rho d{v}_1$

where A1​dx1​ is the volume dV1​.

Once this fluid enters S, the surrounding pressure P1​ does work on the fluid. The calculation is 

                                                                  $\begin{array}{rr}& W_{P_1}=F_{1}d{x}_1=P_1{A}_{1}d{x}_1\\ & W_{P_1}=P_{1}d{V}_1\end{array}$

                                                                    $\left\{\because {\mathrm{A}}_1{\mathrm{d}\mathrm{x}}_1={\mathrm{d}\mathrm{V}}_1\right\}$

At the same time the amount of fluid moves out of the tube at point $T$ is

                                                                    ${\mathrm{d}\mathrm{m}}_2=\rho {\mathrm{d}\mathrm{V}}_2$

So, the equation of continuity goes like $\frac{d{m}_1}{dt}=\frac{d{m}_2}{dt}\Rightarrow d{V}_1=d{V}_2=dv$

The work done in the displacement of ${\mathrm{d}\mathrm{m}}_2$ mass at point T

                                                                    $W_{P_2}=P_{2}d{V}_2$

After we apply the Work Energy Theorem, we get

$W_{P_1}+W_{P_2}=\left(K_f+U_f\right)-\left(K_i+U_i\right)$

$\Rightarrow P_{1}dV-P_{2}dV=\left(\frac{1}{2}\rho dV{v}_2^2+\rho dVg{h}_2\right)-\left(\frac{1}{2}\rho dV{v}_1^2+\rho dVg{h}_1\right)$

$P_1-P_2=\frac{1}{2}\rho V_2^2+\rho g{h}_2-\frac{1}{2}\rho V_1^2+\rho g{h}_1$

$P_1+\rho g{h}_1+\frac{1}{2}\rho V_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho V^2\Rightarrow P+\rho gh+\frac{1}{2}\rho V^2=\text{ constant }$

$\frac{P}{\rho g}+h+\frac{1}{2}\frac{v^2}{g}=\text{ constant }$

$\text{where }\frac{P}{\rho g}=\text{ pressure head }$

$h=\text{ Gravitational head }$

$\frac{1}{2}\frac{v^2}{g}=\text{ volume head}$