11.10

Applications of Bernoulli’s Equation

When a moving fluid is contained in a horizontal pipe, all parts of it have the same elevation (), and Bernoulli’s equation simplifies to

| (11.12)

Thus, the quantity remains constant throughout a horizontal pipe; if v increases, P decreases and vice versa. This is exactly the result that we deduced qualitatively from Newton’s second law at the beginning of Section 11.9, and Conceptual Example 14 illustrates it.

| Conceptual Example 14 | | | Tarpaulins and Bernoulli’s Equation

A tarpaulin is a piece of canvas that is used to cover a cargo, like that pulled by the truck in Figure 11-32. When the truck is stationary the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. Account for this behavior.

Figure 11-32 The tarpaulin that covers the cargo is flat when the truck is stationary but bulges outward when the truck is moving.

Reasoning and Solution The behavior is a direct consequence of the pressure changes that Bernoulli’s equation describes for flowing fluids. When the truck is stationary, the air outside and inside the cargo area is stationary, so the air pressure is the same in both places. This pressure applies the same force to the outer and inner surfaces of the canvas, with the result that the tarpaulin lies flat. When the truck is moving, the outside air rushes over the top surface of the canvas. In accord with Bernoulli’s equation (Equation 11.12), the moving air has a lower pressure than the stationary air within the cargo area. The greater inside pressure generates a greater force on the inner surface of the canvas, and the tarpaulin bulges outward.
Related Homework: Problem 57

Example 15 applies Equation 11.12 to a dangerous physiological condition known as an aneurysm.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 15 | | | An Enlarged Blood Vessel

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of an aneurysm, the cross-sectional area A1 of the aorta increases to a value of . The speed of the blood () through a normal portion of the aorta is . Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure P2 in the enlarged region exceeds the pressure P1 in the normal region.

The physics of an aneurysm.

Reasoning Bernoulli’s equation (Equation 11.12) may be used to find the pressure difference between two points in a fluid moving horizontally. However, in order to use this relation we need to know the speed of the blood in the enlarged region of the artery, as well as the speed in the normal section. The speed in the enlarged region can be obtained by using the equation of continuity (Equation 11.9), which relates it to the speed in the normal region and the cross-sectional areas of the two parts.
Knowns and Unknowns The given data are summarized as follows:

Description | Symbol | Value | Comment
Normal cross-sectional area of aorta | A1 |  | No value given.
Enlarged cross-sectional area of aorta | A2 | 1.7 A1 |
Density of blood |  | 1060 kg/m3 |
Speed of blood in normal portion | v1 | 0.40 m/s |
of aorta |  |  |
Unknown Variable |  |  |
Amount by which pressure P2 |  | ? | P2 refers to enlarged region
exceeds pressure P1 |  |  | and P1 to normal region.

Modeling the Problem

| Step 1 Bernoulli’s Equation
| In the form pertinent to horizontal flow, Bernoulli’s equation is given by Equation 11.12:

Rearranging this expression gives Equation 1 at the right, which we can use to determine . In this result, the density and the speed v1 are given. However, the speed v2 is unknown, and we will obtain a value for it in Step 2.

| (1)

| Step 2 The Equation of Continuity
| For an incompressible fluid like blood, the equation of continuity is given by Equation 11.9 as

This expression for v2 can be substituted into Equation 1, as shown at the right.

Solution Combining the results of each step algebraically, we find that
Since it is given that , this result for reveals that

This positive answer indicates that P2 is greater than P1. The excess pressure puts added stress on the already weakened tissue of the arterial wall at the aneurysm.
Related Homework: Problems 58, 66 and 101

The impact of fluid flow on pressure is widespread. Figure 11-33, for instance, illustrates how household plumbing takes into account the implications of Bernoulli’s equation. The U-shaped section of pipe beneath the sink is called a “trap,” because it traps water, which serves as a barrier to prevent sewer gas from leaking into the house. Part a of the drawing shows poor plumbing. When water from the clothes washer rushes through the sewer pipe, the high-speed flow causes the pressure at point A to drop. The pressure at point B in the sink, however, remains at the higher atmospheric pressure. As a result of this pressure difference, the water is pushed out of the trap and into the sewer line, leaving no protection against sewer gas. A correctly designed system is vented to the outside of the house, as in Figure 11-33b. The vent ensures that the pressure at A remains the same as that at B (atmospheric pressure), even when water from the clothes washer is rushing through the pipe. Thus, the purpose of the vent is to prevent the trap from being emptied, not to provide an escape route for sewer gas.

Figure 11-33 In a household plumbing system, a vent is necessary to equalize the pressures at points A and B, thus preventing the trap from being emptied. An empty trap allows sewer gas to enter the house.

The physics of household plumbing.

One of the most spectacular examples of how fluid flow affects pressure is the dynamic lift on airplane wings. Figure 11-34a shows a wing moving to the right, with the air flowing leftward past the wing. Hence, according to Bernoulli’s equation, the pressures on the top of and beneath the wing are both lower than atmospheric pressure. However, the pressure above the wing is reduced relative to that under the wing. This is due to the wing’s shape, which causes the air to travel faster (more reduction in pressure) over the curved top surface and more slowly (less reduction in pressure) over the flatter bottom. Thus, the wing is lifted upward. Part b of the figure shows the wing of an airplane.

Figure 11-34 (a) Air flowing around an airplane wing. The wing is moving to the right. (b) The end of this wing has roughly the shape indicated in part a. (b, Joe McBride/Getty Images)

The physics of airplane wings.

The physics of a curveball. The curveball, one of a baseball pitcher’s main weapons, is another illustration of the effects of fluid flow. Figure 11-35a shows a baseball moving to the right with no spin. The view is from above, looking down toward the ground. Here, air flows with the same speed around both sides of the ball, and the pressure is reduced on both sides by the same amount relative to atmospheric pressure. No net force exists to make the ball curve to either side. However, when the ball is given a spin, the air close to its surface is dragged around with it, and the situation changes. The speed of the air on one-half of the ball is increased, and the pressure there is even more reduced relative to atmospheric pressure. On the other half of the ball, the speed of the air is decreased, leading to less of a reduction in pressure than without spin. Part b of the picture shows the effects of a counterclockwise spin. The baseball experiences a net deflection force and curves on its way from the pitcher’s mound to the plate, as part c shows.*

Figure 11-35 These views of a baseball are from above, looking down toward the ground, with the ball moving to the right. (a) Without spin, the ball does not curve to either side. (b) A spinning ball curves in the direction of the deflection force. (c) The spin in part b causes the ball to curve as shown here.

Need more practice?

Interactive LearningWare11.2
A fountain sends a stream of water 5.00 m straight upward. The pipe from which the water exits has an effective cross-sectional area of . How many gallons per minute does the fountain use? () Related Homework: Problem 61

As a final application of Bernoulli’s equation, Figure 11-36a shows a large tank from which water is emerging through a small pipe near the bottom. Bernoulli’s equation can be used to determine the speed (called the efflux speed) at which the water leaves the pipe, as the next example shows.

Figure 11-36 (a) Bernoulli’s equation can be used to determine the speed of the liquid leaving the small pipe. (b) An ideal fluid (no viscosity) will rise to the fluid level in the tank after leaving a vertical outlet nozzle.

| Example 16 | | | Efflux Speed

The tank in Figure 11-36a is open to the atmosphere at the top. Find an expression for the speed of the liquid leaving the pipe at the bottom. Reasoning We assume that the liquid behaves as an ideal fluid. Therefore, we can apply Bernoulli’s equation, and in preparation for doing so, we locate two points in the liquid in Figure 11-36a. Point 1 is just outside the efflux pipe, and point 2 is at the top surface of the liquid. The pressure at each of these points is equal to the atmospheric pressure, a fact that will be used to simplify Bernoulli’s equation.
Solution Since the pressures at points 1 and 2 are the same, we have , and Bernoulli’s equation becomes . The density can be eliminated algebraically from this result, which can then be solved for the square of the efflux speed v1:

We have substituted for the height of the liquid above the efflux tube. If the tank is very large, the liquid level changes only slowly, and the speed at point 2 can be set equal to zero, so that .

In Example 16 the liquid is assumed to be an ideal fluid, and the speed with which it leaves the pipe is the same as if the liquid had freely fallen through a height h (see Equation 2.9 with and ). This result is known as Torricelli’s theorem. If the outlet pipe were pointed directly upward, as in Figure 11-36b, the liquid would rise to a height h equal to the fluid level above the pipe. However, if the liquid is not an ideal fluid, its viscosity cannot be neglected. Then, the efflux speed would be less than that given by Bernoulli’s equation, and the liquid would rise to a height less than h.

Check Your Understanding 4

Fluid is flowing from left to right through a pipe (see the drawing). Points A and B are at the same elevation, but the cross-sectional areas of the pipe are different. Points B and C are at different elevations, but the cross-sectional areas are the same. Rank the pressures at the three points, highest to lowest. (The answer is given at the end of the book.)

a. | A and B (a tie), C
b. | C, A and B (a tie)
c. | B, C, A
d. | C, B, A
e. | A, B, C

Background: Bernoulli’s equation plays the principal role. The equation of continuity is also pertinent.
For similar questions (including calculational counterparts), consult Self-Assessment Test 11.2. This test is described at the end of Section 11.11.

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