# 12.3 The most general applications of bernoulli’s equation  (Page 3/3)

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

Each term has a clear physical meaning. For example, $\text{PQ}$ is the power supplied to a fluid, perhaps by a pump, to give it its pressure $P$ . Similarly, $\frac{1}{2}{\mathrm{\rho v}}^{2}Q$ is the power supplied to a fluid to give it its kinetic energy. And $\rho \text{ghQ}$ is the power going to gravitational potential energy.

## Making connections: power

Power is defined as the rate of energy transferred, or $E/t$ . Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.

## Calculating power in a moving fluid

Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of $0\text{.}\text{700}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . What power does the pump supply to the water?

Strategy

Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water's kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by $0\text{.}\text{92}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ (from $0.700×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ to $1.62×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ ).

Solution

As discussed above, the power associated with pressure is

$\begin{array}{lll}\text{power}& =& \text{PQ}\\ & =& \left(\text{0.920}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\right)\left(\text{40}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}\right)\text{.}\\ \text{}& =& 3\text{.}\text{68}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{W}=\text{36}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kW}\end{array}$

Discussion

Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water's pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.

## Summary

• Power in fluid flow is given by the equation $\left({P}_{1}+\frac{1}{2}{\mathrm{\rho v}}^{2}+\rho \text{gh}\right)Q=\text{power}\text{,}$ where the first term is power associated with pressure, the second is power associated with velocity, and the third is power associated with height.

## Conceptual questions

Based on Bernoulli's equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli's equation.)

Water that has emerged from a hose into the atmosphere has a gauge pressure of zero. Why? When you put your hand in front of the emerging stream you feel a force, yet the water's gauge pressure is zero. Explain where the force comes from in terms of energy.

The old rubber boot shown in [link] has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of leak 1? Explain your responses in terms of energy.

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.

## Problems&Exercises

Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300 MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of $\text{650}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}$ . (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility's average of 680 MW?

(a) $\text{9.56}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{W}$

(b) 1.4

A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) (a) At takeoff, an aircraft travels at 60.0 m/s, so that the air speed relative to the bottom of the wing is 60.0 m/s. Given the sea level density of air to be $1\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$ , how fast must it move over the upper surface to create the ideal lift? (b) How fast must air move over the upper surface at a cruising speed of 245 m/s and at an altitude where air density is one-fourth that at sea level? (Note that this is not all of the aircraft's lift—some comes from the body of the plane, some from engine thrust, and so on. Furthermore, Bernoulli's principle gives an approximate answer because flow over the wing creates turbulence.)

The left ventricle of a resting adult's heart pumps blood at a flow rate of $\text{83}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\text{/s}$ , increasing its pressure by 110 mm Hg, its speed from zero to 30.0 cm/s, and its height by 5.00 cm. (All numbers are averaged over the entire heartbeat.) Calculate the total power output of the left ventricle. Note that most of the power is used to increase blood pressure.

1.26 W

A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of $3.00×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem.

## Test prep for ap courses

A horizontally oriented pipe has a diameter of 5.6 cm and is filled with water. The pipe draws water from a reservoir that is initially at rest. A manually operated plunger provides a force of 440 N in the pipe. Assuming that the other end of the pipe is open to the air, with what speed does the water emerge from the pipe?

1. 12 m/s
2. 19 m/s
3. 150 m/s
4. 190 m/s

(a)

A 3.5-cm-diameter pipe contains a pumping mechanism that provides a force of 320 N to push water up into a tall building. Upon entering the piston mechanism, the water is flowing at a rate of 2.5 m/s. The water is then pumped to a level 21 m higher where the other end of the pipe is open to the air. With what speed does water leave the pipe?

A large container of water is open to the air, and it develops a hole of area 10 cm 2 at a point 5 m below the surface of the water. What is the flow rate (m 3 ⁄s) of the water emerging from this hole?

1. 99 m 3 ⁄s
2. 9.9 m 3 ⁄s
3. 0.099 m 3 ⁄s
4. 0.0099 m 3 ⁄s

(d)

A pipe is tapered so that the large end has a diameter twice as large as the small end. What must be the gauge pressure (the difference between pressure at the large end and pressure at the small end) in order for water to emerge from the small end with a speed of 12 m/s if the small end is elevated 8 m above the large end of the pipe?

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