Bernoulli's equation

See Bernoulli differential equation for an unrelated topic in ordinary differential equations.

In fluid dynamics, Bernoulli's equation, derived by Daniel Bernoulli, describes the behavior of a fluid moving along a streamline.

${v^{2} \over 2}+gh+{p \over \rho }=\mathrm {constant}$

v = fluid velocity along the streamline

g = acceleration due to gravity on Earth

h = height from an arbitrary point in the direction of gravity

p = pressure along the streamline

$\rho$ = fluid density

These assumptions must be met for the equation to apply:

• Inviscid flow − viscosity (internal friction) = 0
• Steady flow
• Incompressible flow − $\rho$ = constant. (There exists a second form of Bernoulli's equation that is applicable for compressible flow, which makes use of the thermodynamic enthalpy.)
• Generally, the equation applies along a streamline. For irrotational flow, it applies throughout the entire flow field.

The decrease in pressure simultaneous with an increase in velocity, as predicted by the equation, is often called Bernoulli's principle.

The equation is named for Daniel Bernoulli although it was first presented in the above form by Leonhard Euler.

Derivation

A streamtube of fluid moving to the right. Indicated are pressure, height, velocity, and cross-sectional area.

The equation can be derived by integrating the Euler equations, or applying the law of conservation of energy in two sections along a streamline, ignoring viscosity, compressibility, and thermal effects. One has that

the work done by the forces in the fluid + decrease in potential energy = increase in kinetic energy.

The work done by the forces is

$F_{1}s_{1}-F_{2}s_{2}=p_{1}A_{1}v_{1}\Delta t-p_{2}A_{2}v_{2}\Delta t.$

The decrease of potential energy is

$mgh_{1}-mgh_{2}=\rho gA_{1}v_{1}\Delta th_{1}-\rho gA_{2}v_{2}\Delta th_{2}$

The increase in kinetic energy is

${\frac {1}{2}}mv_{2}^{2}-{\frac {1}{2}}mv_{1}^{2}={\frac {1}{2}}\rho A_{2}v_{2}\Delta tv_{2}^{2}-{\frac {1}{2}}\rho A_{1}v_{1}\Delta tv_{1}^{2}.$

Putting these together, one gets

$p_{1}A_{1}v_{1}\Delta t-p_{2}A_{2}v_{2}\Delta t+\rho gA_{1}v_{1}\Delta th_{1}-\rho gA_{2}v_{2}\Delta th_{2}={\frac {1}{2}}\rho A_{2}v_{2}\Delta tv_{2}^{2}-{\frac {1}{2}}\rho A_{1}v_{1}\Delta tv_{1}^{2}$

or

${\frac {\rho A_{1}v_{1}\Delta tv_{1}^{2}}{2}}+\rho gA_{1}v_{1}\Delta th_{1}+p_{1}A_{1}v_{1}\Delta t={\frac {\rho A_{2}v_{2}\Delta tv_{2}^{2}}{2}}+\rho gA_{2}v_{2}\Delta th_{2}+p_{2}A_{2}v_{2}\Delta t.$

After dividing by $\Delta t$, $\rho$ and $A_{1}v_{1}$ (= rate of fluid flow = $A_{2}v_{2}$ as the fluid is incompressible) one finds:

${\frac {v_{1}^{2}}{2}}+gh_{1}+{\frac {p_{1}}{\rho }}={\frac {v_{2}^{2}}{2}}+gh_{2}+{\frac {p_{2}}{\rho }}$

or ${\frac {v^{2}}{2}}+gh+{\frac {p}{\rho }}=C$ (as stated in the first paragraph).

Further division by g implies

${\frac {v^{2}}{2g}}+h+{\frac {p}{\rho g}}=C.$

A free falling mass from a height h (in vacuum), will reach a velocity

$v={\sqrt {{2g}{h}}},$ or $h={\frac {v^{2}}{2g}}$.

The term ${\frac {v^{2}}{2g}}$ is called the velocity head.

The hydrostatic pressure or static head is defined as

$p=\rho gh$, or $h={\frac {p}{\rho g}}$.

The term ${\frac {p}{\rho g}}$ is also called the pressure head.

A way to see how this relates to conservation of energy directly is to multiply by density and by unit volume (which is allowed since both are constant) yielding:

$v^{2}\rho +P=constant$ and

$mV^{2}+P*volume=constant$