Unformatted text preview: ngth L of the vessel, and the pressures P1 and P2 at the
ends of the vessel do not change. Thus, applying this result to the dilated and normal vessel,
we find that 4 Rdilated
Rnormal where we have used = Qdilated
Qnormal 4 8η LQdilated
π ( P2 − P )
1
8η LQnormal
π ( P2 − P )
1 =4 Qdilated
Qnormal = 4 2 = 1.19 = 2 , since the effect of the dilation is to double the volume flow rate. 80. REASONING Because the level of the blood in the transfusion bottle is a height h above
the input end of the needle, the pressure P2 at the input end of the needle is greater than one
atmosphere, as we see from P2 = PAtm + ρ gh (Equation 11.4), where PAtm is the pressure at the level of the blood in the bottle, ρ is the density of blood, and g is the magnitude of the
acceleration due to gravity. The pressure at the output end of the needle is atmospheric
(P1 = PAtm), so that there is the pressure difference P2 − P1 necessary to push blood, a
viscous fluid, through the needle. The volume flow rate Q of blood through the needle is
π R 4 ( P2 − P )
1
given by Poiseuille’s law as Q =
(Equation 11.14), where R and L are,
8η L
respectively, the radius and length of the needle, and η is the viscosity of blood.
SOLUTION Solving P2 = PAtm + ρ gh (Equation 11.4) for h, we find that
h= P2 − PAtm ρg (1) 612 FLUIDS Because the pressure P1 at the output end of the needle is equal to PAtm, Equation 11.4
becomes Q = π R 4 ( P2 − PAtm )
8η L . Solving for the pressure difference, we obtain P2 − PAtm = 8η LQ
π R4 (2) Substituting Equation (2) into Equation (1) yields 8η LQ 4
8η LQ
8 ( 4.0 × 10 −3 Pa ⋅ s ) ( 0.030 m ) ( 4.5 × 10 −8 m3 /s )
h = πR =
=
= 0.34 m
ρg
π R4 ρ g
( 2.5 × 10−4 m )4 (1060 kg/m3 ) ( 9.80 m/s 2 )
π 81. SSM REASONING The volume flow rate Q of a viscous fluid flowing through a pipe of π R 4 ( P2 − P )
1
radius R is given by Equation 11.14 as Q =
, where P2 − P1 is the pressure
8η L
difference between the ends of the pipe, L is the length of the pipe, and η is the viscosity of
the fluid. Since all the variables are known except L, we can use this relation to find it. SOLUTION Solving Equation 11.14 for the pipe length, we have ( )(
)( ) 4 π 5.1×10−3 m 1.8 ×103 Pa
π R 4 ( P2 − P )
1
L=
=
= 1.7 m
8η Q
8 1.0 × 10−3 Pa ⋅ s 2.8 × 10−4 m3 /s ( ) 82. REASONING The radius R of the pipe is related to the volume flow rate Q, the length L of
the pipe, the viscosity η of the glycerol, and the pressure difference P2 − P between the
1 π R 4 ( P2 − P )
1
input and output ends by Poiseuille’s law: Q =
(Equation 11.14). Solving
8η L
Equation 11.14 for the radius R, we obtain R4 = 8η LQ
π ( P2 − P )
1 or R=4 8η LQ
π ( P2 − P )
1 (1) The volume flow rate Q is the ratio of the volume V of glycerol that flows through the pipe
to the total elapsed time t:
V
Q=
(2)
t Chapter 11 Problems 613 Substituting Equation (2) into Equation (1) yields R=4 8η LV
π ( P2 − P ) t
1 (3) SOLUTION
Because the output end of the pipe is at atmospheric pressure,
5
P = 1.013 × 10 Pa. The elapsed time t must be converted to SI units (seconds):
1 60 s
t = 55 min 1 min ( ) R=4 83. ( = 3300 s . Therefore, the radius of the pipe is ( 8 ( 0.934 Pa ⋅ s )(15 m ) 7.2 m3 )( ) ) π 8.6 ×105 Pa − 1.013 ×105 Pa ( 3300 s ) = 1.8 × 10−2 m REASONING The speed v of the water in the hose is related to the volume flow rate Q by
v = Q/A (Equation 11.10), where A is the crosssectional area of the hose. Since the hose is
cylindrical (radius = R), A = πR2. Thus, we have that v= Q
Q
=
A π R2 (1) We turn to Poiseuille’s law to express Q in terms of the upstream pressure P2 and the
downstream pressure P1 π R 4 ( P2 − P )
1
Q=
8η L (11.14) In Poiseuille’s law, the upstream pressure P2 is the same for both hoses, because it is the
pressure at the outlet. The downstream pressure P1 is also the same for both hoses, since
each is open to the atmosphere at the exit end. Therefore, the term P2 – P1 is the same for
both hoses.
SOLUTION Substituting Equation 11.14 into Equation (1) gives π R 4 ( P2 − P )
1
R 2 ( P2 − P )
Q
8η L
1
v=
=
=
2
2
8η L
πR
πR 614 FLUIDS To find the ratio vB/vA of the speeds, we apply this result to each hose, recognizing that the pressure difference P2 – P1, the length L, and the viscosity η of the water are the same for
both hoses:
2
RB ( P2 − P )
1
2
vB
RB
8η L
2
=2
= 2 = (1.50 ) = 2.25
vA RA ( P2 − P ) RA
1 8η L 84. REASONING AND SOLUTION
a. Using Stoke's law, the viscous force is
F = 6πη Rv = 6π (1.00 × 10 –3 Pa ⋅ s)( 5.0 × 10 –4 m)( 3.0 m / s) = 2.8 × 10 –5 N b. When the sphere reaches its terminal speed, the net force on the sphere is zero, so that the
magnitude of F must be equal to the magnitude of mg, or F = mg . Therefore,
6 π η Rv T = mg , where v T is the terminal speed of the sphere. Solving for v T , we have vT = mg
6πη R = (1.0 × 10 –5 kg) (9.80 m / s 2 )
= 1.0 × 10 1 m / s
6π (1.00 × 10 –3 Pa ⋅ s) (5.0 ×...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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