# 14.6 Bernoulli’s equation  (Page 4/8)

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

becomes

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

Thus pressure ${p}_{2}$ over the second opening is reduced by $\frac{1}{2}\rho {v}_{2}^{2}$ , so the fluid in the manometer rises by h on the side connected to the second opening, where

$h\propto \frac{1}{2}\rho {v}_{2}^{2}.$

(Recall that the symbol $\propto$ means “proportional to.”) Solving for ${v}_{2}$ , we see that

${v}_{2}\propto \sqrt{h}.$

Part (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air-speed indicators in aircraft.

## A fire hose

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change.

## Calculating pressure: a fire hose nozzle

Fire hoses used in major structural fires have an inside diameter of 6.40 cm ( [link] ). Suppose such a hose carries a flow of 40.0 L/s, starting at a gauge pressure of $1.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}{\text{N/m}}^{2}$ . The hose rises up 10.0 m along a ladder to a nozzle having an inside diameter of 3.00 cm. What is the pressure in the nozzle?

## Strategy

We must use Bernoulli’s equation to solve for the pressure, since depth is not constant.

## Solution

Bernoulli’s equation is

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

where subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$ . Since $Q={A}_{1}{v}_{1}$ , we get

${v}_{1}=\frac{Q}{{A}_{1}}=\frac{40.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\text{m}}^{3}\text{/}\text{s}}{\pi {\left(3.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\text{m)}}^{2}}=12.4\text{m/s}.$

Similarly, we find

${v}_{2}=56.6\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

This rather large speed is helpful in reaching the fire. Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${p}_{2}$ :

${p}_{2}={p}_{1}+\frac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right)-\rho g{h}_{2}.$

Substituting known values yields

$\begin{array}{cc}\hfill {p}_{2}& =1.62\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}+\frac{1}{2}\left(1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left[{\text{(12.4 m/s)}}^{2}-{\text{(56.6 m/s)}}^{2}\right]\hfill \\ & \phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}\left(1000\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)\hfill \\ & =0.\hfill \end{array}$

## Significance

This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus, the nozzle pressure equals atmospheric pressure as it must, because the water exits into the atmosphere without changes in its conditions.

## Summary

• Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
${p}_{1}+\frac{1}{2}\rho {v}_{1}^{2}+\rho g{h}_{1}={p}_{2}+\frac{1}{2}\rho {v}_{2}^{2}+\rho g{h}_{2}.$
• Bernoulli’s principle is Bernoulli’s equation applied to situations in which the height of the fluid is constant. The terms involving depth (or height h ) subtract out, yielding
${p}_{1}+\frac{1}{2}\rho {v}_{1}^{2}={p}_{2}+\frac{1}{2}\rho {v}_{2}^{2}.$
• Bernoulli’s principle has many applications, including entrainment and velocity measurement.

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