Bernoulli's Equation

Bernoulli’s Equation is one of the fundamental equations in fluid dynamics. It applies the conservation of energy to fluid mechanics. It only applies to ideal fluid flow, i.e. nonviscous, laminar, incompressible flow.

Occasionally, it’s known as Bernoulli’s Principle.

Definition §

Formula §

Bernoulli's Equation

$$\frac{\rho v^2}{2}+\rho gy+P=\text{constant along a streamline}$$

$$\frac{\rho (v_1{)}^2}{2}+\rho g{y}_1+P_1=\frac{\rho (v_2{)}^2}{2}+\rho g{y}_2+P_2$$

Terms §

• $\rho$ = Density (in ${\text{kg/m}}^3$)
• $v$ = Velocity of fluid (in $\text{m/s}$)
• $g$ = Acceleration due to Earth’s gravity $\approx 9.8{\text{ m/s}}^2$
• $y$ = Relative height (to a base heigh of $y=0$) (in $\text{m}$)
• $P$ = Pressure (in $\text{Pa}$)
• Definition of a Streamline

Implications §

Pressure Velocity Relation §

If the change in height between two points in a streamline is negligible, then we can ignore those terms in the equation:

$$\frac{\rho (v_1{)}^2}{2}+\cancel{\rho g{y}_1}+P_1=\frac{\rho (v_2{)}^2}{2}+\cancel{\rho g{y}_2}+P_2$$

This shows us that if the pressure increases at a point, the velocity of the fluid decreases:

$$v\uparrow ⟺P\downarrow (\text{if height is same})$$

Alternatively, this can be interpreted as saying areas of high pressure have low velocity and vice versa.

Derivation §

Bernoulli’s equation can be easily derived from the conservation of energy in an isolated system.

Let’s start with a some fluid in a pipe, with pressure $P_1$, flowing at a velocity of $v_1$. Furthermore, let’s assume the Earth is part of this system, just out of view. The pipe as at a height $y_1$ above the Earth.

Bernoulli’s Equation .excalidraw

The total energy in this system is the sum of the potentials energies in the system and the kinetic energy of each body in the system:

$$K_{fluid}=\frac{m_{fluid}\cdot (v_1{)}^2}{2}$$

The potential energy arises from the Gravitational Potential Energy, which we can approximate assuming that $m_{fluid}\ll m_{Earth}$:

$$U_{sys}=\frac{m_{fluid}\cdot m_{Earth}\cdot G}{(y_1{)}^2}\approx m_{fluid}\cdot g\cdot y_1$$

So the net energy of the system, in this current configuration is:

$$E_{sys}=\frac{m_{fluid}\cdot (v_1{)}^2}{2}+m_{fluid}\cdot g\cdot y_1$$

We can express the equation in terms of density:

$$E_{sys}=\frac{V_{fluid}\cdot \rho (v_1{)}^2}{2}+V_{fluid}\cdot \rho \cdot g\cdot y_1$$

Even if we change the configuration of the system, the conservation of energy states that this the net energy must be zero.

If we want to change this system to represent a new configuration, we must do so without applying any external energy to the system. All energy transfers must be internal. Let’s see what would happen if we try to increase $v_1$, and decrease $y_1$:

600

We know that the energies must be the same: #TODO Pressure as a potential energy?!