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Fluid Dynamics and Its Biological and Medical Applications

87 The Most General Applications of Bernoulli’s Equation

Learning Objectives

  • Apply Torricelli’s theorem to calculate the speed of fluid flow from an opening.
  • Use Bernoulli’s equation to analyze changes in pressure, height, and velocity in real-world fluid systems.

Torricelli’s Theorem

Imagine water gushing from a horizontal pipe near the base of a dam, as shown in Figure 87.1 . How fast does the water shoot out of the opening? Surprisingly, the speed of the water depends only on the height it has dropped—not on the size of the opening. This result is a special case of Bernoulli’s principle and is known as Torricelli’s theorem.

To analyze this, we apply Bernoulli’s equation between two points in the system: the surface of the reservoir (point 1) and the opening (point 2):

[latex]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}[/latex]

Since both points are exposed to atmospheric pressure, we can subtract [latex]P_{1} = P_{2}[/latex] from both sides. This simplifies the equation to:

[latex]\frac{1}{2} \rho v_{1}^{2} + \rho g h_{1} = \frac{1}{2} \rho v_{2}^{2} + \rho g h_{2}[/latex]

We can factor out and cancel the density [latex]\rho[/latex] (assuming an incompressible fluid like water), leaving:

[latex]\frac{1}{2} v_{1}^{2} + g h_{1} = \frac{1}{2} v_{2}^{2} + g h_{2}[/latex]

Solving for the speed [latex]v_{2}[/latex] of water exiting the pipe gives:

[latex]v_{2}^{2} = v_{1}^{2} + 2g(h_{1} - h_{2})[/latex]

Letting [latex]h = h_{1} - h_{2}[/latex] be the vertical drop, we obtain the final form of Torricelli’s theorem:

[latex]v_{2}^{2} = v_{1}^{2} + 2gh[/latex]

This is identical in form to the kinematic equation for a falling object, which reinforces the idea that the fluid gains speed from gravitational potential energy. The direction of motion doesn’t matter—just the vertical drop [latex]h[/latex].

Part a of the figure shows a photograph of a dam with water gushing from a large tube at the base of a dam. Part b shows the schematic diagram for the flow of water in a reservoir. The reservoir is shown in the form of a triangular section with a horizontal opening along the base little near to the base. The water is shown to flow through the horizontal opening near the base. The height which it falls is shown as h two. The pressure and velocity of water at this point are P two and v two. The height to which the water can fall if it falls from a height h above the opening is given by h 2. The pressure and velocity of water at this point are P one and v one.
Figure 87.1 (a) Water exiting from a dam. (b) In the absence of resistance, the speed of water exiting at height [latex]h[/latex] is the same as if it had simply fallen that distance. This is Torricelli’s theorem in action.

Example in Practice: Fire Hose Dynamics

Figure 87.2 shows a firefighter using a hose to spray water at a building. To reach the target, the water must travel uphill and through a narrow nozzle. According to Bernoulli’s equation, two factors cause the pressure at the nozzle to be lower than at ground level:

  1. The water is elevated to a higher height, increasing its gravitational potential energy.
  2. The nozzle speeds up the water, increasing its kinetic energy.

Even though the pressure at the nozzle is reduced, the water can still exert a strong force on the fire because of its high speed and momentum. Once the water exits the nozzle and enters the atmosphere, its pressure becomes equal to atmospheric pressure.

Figure shows a fire engine that is stationed next to a tall building. A floor of the building ten meters above the ground has caught fire. The flames are shown coming out. A fire man has reached close to the fire caught area using a ladder and is spraying water on the fire using a hose attached to the fire engine.
Figure 87.2 A firefighter directs water through a hose to reach flames several meters above ground level. Water pressure drops due to height and nozzle narrowing, but its velocity and force increase. Once in the air, the water pressure equals atmospheric pressure.

Up to this point, we’ve seen Bernoulli’s principle applied in simplified scenarios where height or pressure was constant. The fire hose is a more general example where height, speed, and pressure all change—demonstrating the full power of Bernoulli’s equation in real-world applications.

Example 87.1 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 [latex]1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]. 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

[latex]{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{,}[/latex]

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 [latex]{v}_{1}[/latex] and [latex]{v}_{2}[/latex]. Since [latex]Q={A}_{1}{v}_{1}[/latex]
, we get

[latex]{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 (3\text{.}\text{20}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}}=\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}[/latex]

Similarly, we find

[latex]{v}_{2}=\text{56.6 m/s}\text{.}[/latex]

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

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

Substituting known values yields

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

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.

Power in Fluid Flow

In physics and physiology alike, refers to the rate at which work is done or energy is transferred over time. This is important in fluid systems like blood flow, where energy is continuously used to push fluid through vessels. To understand how power relates to fluid motion, we return to Bernoulli’s equation:

[latex]P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}[/latex]

Each term in this equation represents an energy per unit volume:

  • [latex]P[/latex] is pressure energy per unit volume
  • [latex]\frac{1}{2} \rho v^2[/latex] is kinetic energy per unit volume
  • [latex]\rho gh[/latex] is gravitational potential energy per unit volume

If we multiply each of these energy densities by the [latex]Q[/latex] (which has units of volume per unit time), we get units of power (energy per time):

[latex]\left( \frac{E}{V} \right) \left( \frac{V}{t} \right) = \frac{E}{t}[/latex]

Thus, multiplying Bernoulli’s equation by the flow rate [latex]Q[/latex] gives the total power associated with the fluid:

[latex]\left( P + \frac{1}{2} \rho v^2 + \rho gh \right) Q = \text{Power}[/latex]

Each term in this equation corresponds to a specific type of power:

  • [latex]PQ[/latex] is the pressure power—the energy per second supplied to the fluid by a pump or the heart to maintain pressure.
  • [latex]\frac{1}{2} \rho v^2 Q[/latex] is the kinetic power—the energy per second used to move the fluid with a given velocity.
  • [latex]\rho gh Q[/latex] is the gravitational power—the energy per second used to lift the fluid against gravity.

This formulation is particularly useful in cardiovascular physiology. For example, the heart supplies power to the blood through both pressure and velocity, while blood vessels may convert part of this energy into potential energy if the flow is upward (as in blood being pumped to the brain).

Making Connections: Power in Physiology

In any biological system involving fluid flow—such as the circulatory, lymphatic, or respiratory systems—power represents how much energy is being delivered per second. It accounts for multiple forms of energy transformation, including pressure maintenance, fluid motion, and elevation changes in tissues.

Example 87.2 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 [latex]0\text{.}\text{700}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]. 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 [latex]0\text{.}\text{92}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex] (from [latex]0.700×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex] to [latex]1.62×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]).

Solution

As discussed above, the power associated with pressure is

[latex]\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}[/latex]

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

  • In fluid dynamics, the total power transmitted by a flowing fluid can be calculated using the equation:
    [latex]\left(P + \frac{1}{2} \rho v^2 + \rho gh \right)Q = \text{power}[/latex]

    where:

    • [latex]P Q[/latex] is the rate of work associated with fluid pressure,
    • [latex]\frac{1}{2} \rho v^2 Q[/latex] represents the kinetic energy contribution to power, and
    • [latex]\rho gh Q[/latex] accounts for gravitational potential energy per unit time.

    This equation shows how power in fluid systems involves the combined contribution of pressure, fluid velocity, and elevation. It has important implications in cardiovascular physiology, respiratory flow, intravenous fluid delivery, and biomedical engineering applications involving pumps and flow regulation.

Conceptual Questions

  1. 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.)
  2. 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.
  3. The old rubber boot shown in Figure 87.3 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.
    The picture shows a boot filled with water. The water is shown emerging from two leaks in the old boot, one in front and another at the back. The leaks are at the same height. The leaks are labeled as Leak 1 and Leak 2 respectively.
    Figure 87.3 Water emerges from two leaks in an old boot.
  4. 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

  1. 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 [latex]\text{650}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}[/latex]. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility’s average of 680 MW?
  2. 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 [latex]1\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}[/latex], 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.)
  3. The left ventricle of a resting adult’s heart pumps blood at a flow rate of [latex]\text{83}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{3}\text{/s}[/latex], 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.
  4. 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 [latex]3.00×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}[/latex]. (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.

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