# 2.2 Bernoulli’s equation  (Page 5/7)

 Page 5 / 7

Give an example of entrainment not mentioned in the text.

Many entrainment devices have a constriction, called a Venturi, such as shown in [link] . How does this bolster entrainment?

Some chimney pipes have a T-shape, with a crosspiece on top that helps draw up gases whenever there is even a slight breeze. Explain how this works in terms of Bernoulli’s principle.

Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer.

Why is it preferable for airplanes to take off into the wind rather than with the wind?

Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.

Why does a sailboat need a keel?

It is dangerous to stand close to railroad tracks when a rapidly moving commuter train passes. Explain why atmospheric pressure would push you toward the moving train.

Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect. Explain in terms of energy how the water can emerge from the nozzle against the opposing atmospheric pressure.

A perfume bottle or atomizer sprays a fluid that is in the bottle. ( [link] .) How does the fluid rise up in the vertical tube in the bottle?

If you lower the window on a car while moving, an empty plastic bag can sometimes fly out the window. Why does this happen?

## Problems&Exercises

Verify that pressure has units of energy per unit volume.

$\begin{array}{lll}P& =& \frac{\text{Force}}{\text{Area}},\\ \left(P{\right)}_{\text{units}}& =& {\text{N/m}}^{2}=\text{N}\cdot {\text{m/m}}^{3}={\text{J/m}}^{3}\\ & =& \text{energy/volume}\end{array}$

Suppose you have a wind speed gauge like the pitot tube shown in [link] (b). By what factor must wind speed increase to double the value of $h$ in the manometer? Is this independent of the moving fluid and the fluid in the manometer?

If the pressure reading of your pitot tube is 15.0 mm Hg at a speed of 200 km/h, what will it be at 700 km/h at the same altitude?

184 mm Hg

Calculate the maximum height to which water could be squirted with the hose in [link] example if it: (a) Emerges from the nozzle. (b) Emerges with the nozzle removed, assuming the same flow rate.

Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of $\text{220}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ ? Typical air density in Boulder is $1\text{.}\text{14}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$ , and the corresponding atmospheric pressure is $8\text{.}\text{89}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

$2\text{.}\text{54}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$

(a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6.00 m/s parallel to its front surface and 3.50 m/s along its back surface. Take the density of air to be $\text{1.29 kg}{\text{/m}}^{3}$ . (The calculation, based on Bernoulli’s principle, is approximate due to the effects of turbulence.) (b) Discuss whether this force is great enough to be effective for propelling a sailboat.

(a) What is the pressure drop due to the Bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire hose while carrying a flow of 40.0 L/s? (b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)

(a) $1\text{.}\text{58}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$

(b) 163 m

(a) Using Bernoulli’s equation, show that the measured fluid speed $v$ for a pitot tube, like the one in [link] (b), is given by

$v={\left(\frac{2\rho \prime \mathrm{gh}}{\rho }\right)}^{1/2},$

where $h$ is the height of the manometer fluid, $\rho \prime$ is the density of the manometer fluid, $\rho$ is the density of the moving fluid, and $g$ is the acceleration due to gravity. (Note that $v$ is indeed proportional to the square root of $h$ , as stated in the text.) (b) Calculate $v$ for moving air if a mercury manometer’s $h$ is 0.200 m.

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