# 14.6 Bernoulli’s equation  (Page 2/8)

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We also assume that there are no viscous forces in the fluid, so the energy of any part of the fluid will be conserved. To derive Bernoulli’s equation, we first calculate the work that was done on the fluid:

$dW={F}_{1}d{x}_{1}-{F}_{2}d{x}_{2}$
$dW={p}_{1}{A}_{1}d{x}_{1}-{p}_{2}{A}_{2}d{x}_{2}={p}_{1}dV-{p}_{2}dV=\left({p}_{1}-{p}_{2}\right)dV.$

The work done was due to the conservative force of gravity and the change in the kinetic energy of the fluid. The change in the kinetic energy of the fluid is equal to

$dK=\frac{1}{2}{m}_{2}{v}_{2}^{2}-\frac{1}{2}{m}_{1}{v}_{1}^{2}=\frac{1}{2}\rho dV\left({v}_{2}^{2}-{v}_{1}^{2}\right).$

The change in potential energy is

$dU=mg{y}_{2}-mg{y}_{1}=\rho dVg\left({y}_{2}-{y}_{1}\right).$

The energy equation then becomes

$\begin{array}{ccc}\hfill dW& =\hfill & dK+dU\hfill \\ \hfill \left({p}_{1}-{p}_{2}\right)dV& =\hfill & \frac{1}{2}\rho dV\left({v}_{2}^{2}-{v}_{1}^{2}\right)+\rho dVg\left({y}_{2}-{y}_{1}\right)\hfill \\ \hfill \left({p}_{1}-{p}_{2}\right)& =\hfill & \frac{1}{2}\rho \left({v}_{2}^{2}-{v}_{1}^{2}\right)+\rho g\left({y}_{2}-{y}_{1}\right).\hfill \end{array}$

Rearranging the equation gives Bernoulli’s equation:

${p}_{1}+\frac{1}{2}\rho {v}_{1}^{2}+\rho g{y}_{1}={p}_{2}+\frac{1}{2}\rho {v}_{2}^{2}+\rho g{y}_{2}.$

This relation states that the mechanical energy of any part of the fluid changes as a result of the work done by the fluid external to that part, due to varying pressure along the way. Since the two points were chosen arbitrarily, we can write Bernoulli’s equation more generally as a conservation principle along the flow.

## Bernoulli’s equation

For an incompressible, frictionless fluid, the combination of pressure and the sum of kinetic and potential energy densities is constant not only over time, but also along a streamline:

$p+\frac{1}{2}\rho {v}^{2}+\rho gy=\text{constant}$

A special note must be made here of the fact that in a dynamic situation, the pressures at the same height in different parts of the fluid may be different if they have different speeds of flow.

## Analyzing bernoulli’s equation

According to Bernoulli’s equation, if we follow a small volume of fluid along its path, various quantities in the sum may change, but the total remains constant. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, let us consider some specific situations that simplify and illustrate its use and meaning.

## Bernoulli’s equation for static fluids

First consider the very simple situation where the fluid is static—that is, ${v}_{1}={v}_{2}=0.$ Bernoulli’s equation in that case is

${p}_{1}+\rho g{h}_{1}={p}_{2}+\rho g{h}_{2}.$

We can further simplify the equation by setting ${h}_{2}=0.$ (Any height can be chosen for a reference height of zero, as is often done for other situations involving gravitational force, making all other heights relative.) In this case, we get

${p}_{2}={p}_{1}+\rho g{h}_{1}.$

This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by ${h}_{1}$ , and consequently, ${p}_{2}$ is greater than ${p}_{1}$ by an amount $\rho g{h}_{1}$ . In the very simplest case, ${p}_{1}$ is zero at the top of the fluid, and we get the familiar relationship $p=\rho gh$ . $\left(\text{Recall that}\phantom{\rule{0.2em}{0ex}}p=\rho gh$ and $\text{Δ}{U}_{\text{g}}=\text{−}mgh.\right)$ Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid is $\rho gh$ . Although we introduce Bernoulli’s equation for fluid motion, it includes much of what we studied for static fluids earlier.

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