# 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. Measurement of fluid speed based on Bernoulli’s principle. (a) A manometer is connected to two tubes that are close together and small enough not to disturb the flow. Tube 1 is open at the end facing the flow. A dead spot having zero speed is created there. Tube 2 has an opening on the side, so the fluid has a speed v across the opening; thus, pressure there drops. The difference in pressure at the manometer is 1 2 ρ v 2 2 , so h is proportional to 1 2 ρ v 2 2 . (b) This type of velocity measuring device is a Prandtl tube , also known as a pitot tube.

## 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? Pressure in the nozzle of this fire hose is less than at ground level for two reasons: The water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

## 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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