Physics 107 HOMEWORK ASSIGNMENT #9b

Cutnell & Johnson, 7th edition

Chapter 11: Problems 52, 58, 66, 67, 100

52 Concept Simulation 11.1 reviews the concept that plays the central role in this problem. (a) The volume flow rate in an artery supplying the brain is . If the radius of the artery is 5.2 mm, determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 3. Assume that the volume flow rate is the same as that in part (a).

58 See Multiple-Concept Example 15 to review the concepts that are pertinent to this problem. The blood speed in a normal segment of a horizontal artery is 0.11 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery?

*66 The concepts that play roles in this problem are similar to those in Multiple-Concept Example 15, except that the fluid here moves upward rather than remaining horizontal. A liquid is flowing through a horizontal pipe whose radius is 0.0200 m. The pipe bends straight upward through a height of 10.0 m and joins another horizontal pipe whose radius is 0.0400 m. What volume flow rate will keep the pressures in the two horizontal pipes the same?

*67 An airplane has an effective wing surface area of 16 m2 that is generating the lift force. In level flight the air speed over the top of the wings is 62.0 m/s, while the air speed beneath the wings is 54.0 m/s. What is the weight of the plane?

100 Concept Questions Water flows straight down from an open faucet. The effects of air resistance and viscosity can be ignored. (a) After the water has fallen a bit below the faucet, is its speed less than, greater than, or the same as it was on leaving the faucet? (b) Is the volume flow rate in cubic meters per second less than, greater than, or the same as it was when the water left the faucet? (c) Is the cross-sectional area of the water stream less than, greater than, or the same as it was when the water left the faucet? Give your reasoning.

52. REASONING a. According to Equation 11.10, the volume flow rate Q is equal to the product of the cross-sectional area A of the artery and the speed v of the blood, Q = Av. Since Q and A are known, we can determine v.

b. Since the volume flow rate Q2 through the constriction is the same as the volume flow rate Q1 in the normal part of the artery, Q2 = Q1. We can use this relation to find the blood speed in the constricted region.

SOLUTION a. Since the artery is assumed to have a circular cross-section, its cross-sectional area is

A1 =2

1rpi , where r1 is the radius. Thus, the speed of the blood is

( )6 3

21 11 2 231 1

3.6 10 m / s 4.2 10 m/s5.2 10 m

Q Qv

A rpi pi

= = = =

(11.10)

b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q2 = Q1. Since 2 2 2 ,Q A v= the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the radius of the constricted part of the artery is one-third that of the normal artery, so 12 13 .r r= Thus, the speed of the blood at the constriction is

( ) ( )6 3

1 1 12 2 2 21 312 2 13 3

3.6 10 m / s 0.38 m/s5.2 10 m

Q Q Qv

A r rpi pi pi

= = = = =

58. REASONING We assume that region 1 contains the constriction and region 2 is the normal region. The difference in blood pressures between the two points in the horizontal artery is given by Bernoullis equation (Equation 11.12) as 2 21 12 1 1 22 2 = P P v v , where v1 and v2 are the speeds at the two points. Since the volume flow rate is the same at the two points, the speed at 1 is related to the speed at 2 by Equation 11.9, the equation of continuity: A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the artery. By combining these two relations, we will be able to determine the pressure difference.

SOLUTION Solving the equation of continuity for the blood speed in region 1 gives v1 = v2A2/A1. Substituting this result into Bernoullis equation yields

22 2 22 21 1 1 1

2 1 1 2 22 2 2 21

= =

v AP P v v v

A

Since 11 24 = A A , the pressure difference is

( )

( )( ) ( )

22 22 21 1 1

2 1 2 22 2 2124

2312

= = 16 1

1060 kg/m 0.11 m/s 15 96 Pa

v AP P v v

A

= =

We have taken the density of blood from Table 11.1.

66. REASONING AND SOLUTION As seen in the figure, the lower pipe is at the level of zero potential energy.

A 1

v 2

v 1

A 2

h

If the flow rate is uniform in both pipes, we have (1/2)v12 = (1/2)v22 + gh (since P1 = P2) and A1v1 = A2v2. We can solve for v1, i.e., v1 = v2(A2/A1), and plug into the previous expression to find v2, so that

( )( )( )( )

2

2 2 222

21

2 9.80 m/s 10.0 m2 3.61 m/s0.0400 m1 10.0200 m

ghv

AA

pi

pi

= = =

The volume flow rate is then given by

Q = A2v2 = pi r22v2 = pi (0.0400 m)2(3.61 m/s) = 2 31.81 10 m /s

67. REASONING In level flight the lift force must balance the planes weight W, so its magnitude is also W. The lift force arises because the pressure PB beneath the wings is greater than the pressure PT on top of the wings. The lift force, then, is the pressure difference times the effective wing surface area A, so that W = (PB PT)A. The area is given, and we can determine the pressure difference by using Bernoullis equation.

SOLUTION According to Bernoullis equation, we have

2 21 1B B B T T T2 2P v gy P v gy + + = + +

Since the flight is level, the height is constant and yB = yT, where we assume that the wing thickness may be ignored. Then, Bernoullis equation simplifies and may be rearranged as follows:

2 2 2 21 1 1 1B B T T B T T B2 2 2 2 or P v P v P P v v + = + =

Recognizing that W = (PB PT)A, we can substitute for the pressure difference from Bernoullis equation to show that

( ) ( )( ) ( ) ( ) ( )

2 21B T T B2

2 23 212 1.29 kg/m 62.0 m/s 54.0 m/s 16 m 9600 N

W P P A v v A= =

= =

We have used a value of 1.29 kg/m3 from Table 11.1 for the density of air. This is an approximation, since the density of air decreases with increasing altitude above sea level.

100. Concept Questions a. Since the effects of air resistance and viscosity are being ignored, the water can be treated

as a freely-falling object, as Chapter 2 discusses. It accelerates with the acceleration due to gravity. Therefore, it has a greater speed at the lower point than it did upon leaving the faucet.

b. The volume flow rate in cubic meters per second is the same as it was when the water left the faucet. This is because no water is added to or taken out of the stream after the water leaves the faucet.

c. The cross-sectional area of the water stream is less than it was when the water left the faucet. With the volume flow rate unchanging, the equation of continuity applies in the form of Equation 11.9. The volume flow rate is the cross-sectional area of the stream times the speed of the water. When the speed increases, as it does when the water falls, the cross-sectional area decreases.

SOLUTION Using the equation of continuity as stated in Equation 11.9, we have

{ {2 2

1 1 2 2 11Below faucet At faucet

or A v

A v A v Av

= =

To find the cross-sectional area A1, we must find the speed v1. To do this, we use Equation 2.9 from the equations of kinematics:

2 2 21 2 1 22 or 2v v ay v v ay= + = +

In using this result, we choose downward as the positive direction. Substituting into the equation of continuity gives

( )( )( ) ( ) ( )

4 25 22 2

1 2 2 22

1.8 10 m 0.85 m/s9.3 10 m

2 0.85 m/s 2 9.80 m/s 0.10 m

A vA

v ay

= = =

+ +