# Bernoullis 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. The original form, for incompressible flow, is:

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

These assumptions must be met for the equation to apply:

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.

A second, more general form of Bernoulli's equation may be written for compressible fluids, in which case, following a streamline, we have:

$\displaystyle {v^2 \over 2}+ \phi + w =\mathrm{constant}$

Here, $\displaystyle \phi$ is the gravitational potential energy per unit mass, which is just $\displaystyle \phi = gh$ in the case of a uniform gravitational field, and $\displaystyle w$ is the fluid enthalpy per unit mass, which is also often written as $\displaystyle h$ (which conflicts with our use of $\displaystyle h$ in these notes for "height"). Note that $\displaystyle w = \epsilon + \frac{p}{\rho}$ where $\displaystyle \epsilon$ is the fluid thermodynamic energy per unit mass.

The constant on the right hand side is often called the Bernoulli constant and denoted $\displaystyle b$ . For steady inviscid adiabatic flow with no additional sources or sinks of energy, $\displaystyle b$ is constant along any given streamline. Even more generally when $\displaystyle b$ may vary along streamlines, it still proves a useful parameter, related to the "head" of the fluid (see below).

## Derivation

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

Let us begin with the Bernoulli equation for incompressible fluids.

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

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

The decrease of potential energy is

$\displaystyle m g h_{1}-m g h_{2}=\rho g A _{1} v_{1}\Delta t h_{1}-\rho g A_{2} v_{2} \Delta t h_{2} \;$

The increase in kinetic energy is

$\displaystyle \frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}=\frac{1}{2}\rho A_{2} v_{2}\Delta t v_{2} ^{2}-\frac{1}{2}\rho A_{1} v_{1}\Delta t v_{1}^{2}.$

Putting these together, one gets

$\displaystyle p_{1} A_{1} v_{1}\Delta t-p_{2} A_{2} v_{2}\Delta t+\rho g A_{1} v_{1}\Delta t h_{1}-\rho g A_{2} v_{2}\Delta t h_{2}=\frac{1}{2}\rho A_{2} v_{2}\Delta t v_{2}^{2}-\frac{1}{2}\rho A_{1} v_{1}\Delta t v_{1}^{2}$

or

$\displaystyle \frac{\rho A_{1} v_{1}\Delta t v_{1}^{ 2}}{2}+\rho g A_{1} v_{1}\Delta t h_{1}+p_{1} A_{1 } v_{1}\Delta t=\frac{\rho A_{2} v_{2}\Delta t v_{ 2}^{2}}{2}+\rho g A_{2} v_{2}\Delta t h_{2}+p_{2} A_{2} v_{2}\Delta t.$

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

$\displaystyle \frac{v_{1}^{2}}{2}+g h_{1}+\frac{p_{1}}{\rho}=\frac{v_{2}^{2}}{2}+g h_{2}+\frac{p_{2}}{\rho}$

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

Further division by g implies

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

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

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

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

The hydrostatic pressure or static head is defined as

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

The term $\displaystyle \frac{p}{\rho g}$ 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:

$\displaystyle v^2 \rho + P = constant$ and

$\displaystyle mV^2 + P*volume = constant$

The derivation for compressible fluids is similar. Again, the derivation depends upon (1) conservation of mass, and (2) conservation of energy. Conservation of mass implies that in the above figure, in the interval of time $\displaystyle \Delta t$ , the amount of mass passing through the boundary defined by the area $\displaystyle A_1$ is equal to the amount of mass passing outwards through the boundary defined by the area $\displaystyle A_2$ :

$\displaystyle 0 = \Delta M_1 - \Delta M_2 = \rho_1 A_1 v_1 \, \Delta t - \rho_2 A_2 v_2 \, \Delta t$ .

We apply conservation of energy in a similar manner: It is assumed that the change in energy of the volume of the streamtube bounded by $\displaystyle A_1$ and $\displaystyle A_2$ is due entirely to energy entering or leaving through one or the other of these two boundaries. Clearly, in a more complicated situation such as a fluid flow coupled with radiation, such conditions are not met. Nevertheless, assuming this to be the case and assuming the flow is steady so that the net change in the energy is zero, we have

$\displaystyle 0 = \Delta E_1 - \Delta E_2$

where $\displaystyle \Delta E_1$ and $\displaystyle \Delta E_2$ are the energy entering through $\displaystyle A_1$ and leaving through $\displaystyle A_2$ , respectively.

The energy entering through $\displaystyle A_1$ is the sum of the kinetic energy entering, the energy entering in the form of potential graviational energy of the fluid, the fluid thermodynamic energy entering, and the energy entering in the form of mechanical $\displaystyle p\,dV$ work:

$\displaystyle \Delta E_1 = \left[ \frac{1}{2} \rho_1 v_1^2 + \phi_1 \rho_1 + \epsilon_1 \rho_1 + p_1 \right] A_1 v_1 \, \Delta t$

A similar expression for $\displaystyle \Delta E_2$ may easily be constructed. So now setting $\displaystyle 0 = \Delta E_1 - \Delta E_2$ we obtain

$\displaystyle 0 = \left[ \frac{1}{2} \rho_1 v_1^2+ \phi_1 \rho_1 + \epsilon_1 \rho_1 + p_1 \right] A_1 v_1 \, \Delta t - \left[ \frac{1}{2} \rho_2 v_2^2 + \phi_2\rho_2 + \epsilon_2 \rho_2 + p_2 \right] A_2 v_2 \, \Delta t$

Let us rewrite this as:

$\displaystyle 0 = \left[ \frac{1}{2} v_1^2 + \phi_1 + \epsilon_1 + \frac{p_1}{\rho_1} \right] \rho_1 A_1 v_1 \, \Delta t - \left[ \frac{1}{2} v_2^2 + \phi_2 + \epsilon_2 + \frac{p_2}{\rho_2} \right] \rho_2 A_2 v_2 \, \Delta t$

Now, using our previously-obtained result from conservation of mass, this may be simplified to obtain

$\displaystyle \frac{1}{2}v^2 + \phi + \epsilon + \frac{p}{\rho} = {\rm constant} \equiv b$

which is the sought solution.