This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 200 s 3.0 m / s bg
b g 120 m 3
= 0.14 m
1200 s 5.0 m / s bg
b g 58. REASONING
a. According to Equation 11.10, the volume flow rate Q is equal to the product of the crosssectional 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 crosssection, its crosssectional area is
A1 = π r12 , where r1 is the radius. Thus, the speed of the blood is v1 = Q1 Q1
3.6 ×10−6 m3 / s
= 2=
= 4.2 ×10−2 m/s
2
A1 π r1 π 5.2 ×10−3 m ( ) (11.10) b. The volume flow rate is the same in the normal and constricted parts of the artery, so
Q2 = Q1. Since Q2 = A2 v2 , the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the
radius of the constricted part of the artery is onethird that of the normal artery, so r2 = 1 r1.
3
Thus, the speed of the blood at the constriction is
v2 = Q1
Q
Q1
= 12 =
A2 π r2 π 1 r
31 () 2 = 3.6 × 10−6 m3 / s ( ) π 1 5.2 ×10−3 m 3 2 = 0.38 m/s 598 FLUIDS 59. REASONING AND SOLUTION Let rh represent the inside radius of the hose, and rp the
radius of the plug, as suggested by the figure below.
hose
plug
rh rp 2
2
Then, from Equation 11.9, A1v1 = A2v2 , we have π rh v1 = (π rh2 − π rp )v2 , or v1
v2 ( rh2 − rp2 ) = 1 − rp 2
= or r h rh2 rp
rh = 1− v1
v2 According to the problem statement, v2 = 3v1, or
rp
rh = 1− v1
3v1 = 2
= 0.816
3 60. REASONING The number N of capillaries can be obtained by dividing the total crosssectional area Acap of all the capillaries by the crosssectional area acap of a single capillary.
2
We know the radius rcap of a single capillary, so acap can be calculated as acap = π rcap . To find Acap, we will use the equation of continuity.
SOLUTION The number N of capillaries is
N= Acap
acap = Acap
2
π rcap (1) 2
where the crosssectional area acap of a single capillary has been replaced by acap = π rcap . To obtain the total crosssectional area Acap of all the capillaries, we use the equation of
continuity for an incompressible fluid (see Equation 11.9). For present purposes, this
equation is
Aaorta vaorta
(2)
Aaorta vaorta = Acap vcap or Acap =
vcap Chapter 11 Problems 599 2
The crosssectional area of the aorta is Aaorta = π raorta , and with this substitution, Equation (2) becomes
Acap = Aaorta vaorta
vcap = 2
π raorta vaorta vcap Substituting this result into Equation (1), we find that N= = Acap
2
π rcap π r2 aorta vaorta vcap r2 v =
= aorta aorta
2
2
rcap vcap
π rcap (1.1 cm )2 ( 40 cm/s ) =
2
( 6 ×10−4 cm ) ( 0.07 cm/s ) 2 ×109 61. SSM REASONING AND SOLUTION
a. Using Equation 11.12, the form of Bernoulli's equation with y 1 = y 2 , we have ( ) 2
2
P − P2 = ρ v2 − v1 =
1
1
2 1.29 kg/m3
2 (15 m/s)2 − ( 0 m/s )2 = 150 Pa b. The pressure inside the roof is greater than the pressure on the outside. Therefore, there is
a net outward force on the roof. If the wind speed is sufficiently high, some roofs are "blown
outward." 62. REASONING We apply Bernoulli’s equation as follows:
2
2
PS + 1 ρ vS + ρ gyS = PO + 1 ρ vO + ρ gyO
24
24
144 2444 144 2444
3
3 At surface of
vaccine in reservoir At opening SOLUTION The vaccine’s surface in the reservoir is stationary during the inoculation, so
that vS = 0 m/s. The vertical height between the vaccine’s surface in reservoir and the
opening can be ignored, so yS = yO. With these simplifications Bernoulli’s equation becomes
2
P = PO + 1 ρ vO
S
2 Solving for the speed at the opening gives 600 FLUIDS vO = 2 ( P − PO )
S ρ = ( 2 4.1×106 Pa
1100 kg/m3 )= 86 m/s 63. SSM REASONING AND SOLUTION Let the speed of the air below and above the
wing be given by v 1 and v 2 , respectively. According to Equation 11.12, the form of
Bernoulli's equation with y 1 = y 2 , we have c P1 − P2 = ρ v − v
1
2 2
2 2
1 1.29 kg / m 3
=
( 251 m / s) 2 − ( 225 m / s) 2 = 7.98 × 10 3 Pa
2 h From Equation 11.3, the lifting force is, therefore, c h F = P1 − P2 A = ( 7 .98 × 10 3 Pa)(24.0 m 2 ) = 1.92 × 10 5 N 64. REASONING The absolute pressure in the pipe must be greater than atmospheric
pressure. Our solution proceeds in two steps. We will begin with Bernoulli’s equation.
Then we will incorporate the equation of continuity.
SOLUTION According to Bernoulli’s equation, as given by Equation 11.11, we have
2
2
P + 1 ρ v1 + ρ gy1 = P2 + 1 ρ v2 + ρ gy2
12
24
1442443 144 2444
3
Pipe Nozzle The pipe and nozzle are horizontal, so that y1 = y2 and Bernoulli’s equation simplifies to
2
2
P + 1 ρ v1 = P2 + 1 ρ v2
12
2 or ( 2
2
P = P2 + 1 ρ v2 − v1
1
2 ) where P1 is the absolute pressure of the water in the pipe. We have values for the pressure
P2 (atmospheric pressure) at the nozzle opening and the speed v1 in the pipe. However, to
solve this expression we also need a value for the speed v2 at the nozzle opening. We
obtain this value by us...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details