# 12.3 The most general applications of bernoulli’s equation

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• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.

## Torricelli’s theorem

[link] shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is

${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$

Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure ( ${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving

$\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}=\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$

Solving this equation for ${v}_{2}^{2}$ , noting that the density $\rho$ cancels (because the fluid is incompressible), yields

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

We let $h={h}_{1}-{h}_{2}$ ; the equation then becomes

${v}_{2}^{2}={v}_{1}^{2}+2\text{gh}$

where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem . Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.

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. (See [link] .)

## Calculating pressure: a fire hose nozzle

Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?

Strategy

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

Solution

Bernoulli’s equation states

${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{,}$

where the 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

${v}_{1}=\frac{Q}{{A}_{1}}=\frac{\text{40}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}}{\pi \left(3\text{.}\text{20}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}=\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$

Similarly, we find

${v}_{2}=\text{56.6 m/s}\text{.}$

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

${P}_{2}={P}_{1}+\frac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right)-\rho {\text{gh}}_{2}\text{.}$

Substituting known values yields

${P}_{2}=1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}+\frac{1}{2}\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left[\left(\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}-\left(\text{56}\text{.}6\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}\right]-\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left(9\text{.}80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(\text{10}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m}\right)=0\text{.}$

Discussion

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.

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