13
FLUIDS
Q13.1. Reason: Density does not depend on the volume. That is, 1 g of mercury would have the same density
as 1000 g of mercury, and 1 g of water would have the same density as 1000 g of water.
Table 13.1 shows the density of mercury to be 13,600 kg/m3 and that of water to be only 1000 kg/m3.
The density of 1 g of mercury is 13.6 times as much as the density of 1000 g of water.
Assess: It is important to get used to the idea that density is a ratio of mass to volume, so different samples of
the same substance would have the same density.
Q13.2. Reason: Equations 13.1 and 13.2 apply.
(a) Using Equation 13.1,
ρ=
m
3 kg
=
= 3 × 103 kg/m3
V 1 × 10−3 m3
(b) The mass becomes m =
(3 × 103 kg/m3 )(2 × 10−3 m3 ) =
6 kg.
ρV =
(c) Density only depends on the type of material. The density will not change when the volume is doubled. Using
the mass calculated in part (b), we can explicitly calculate the density as ρ =
m/V =
6 kg/2 × 10−3 m3 =
3 × 103 kg/m3. This is the same as the original density, as expected.
Assess: Density is an intrinsic property of a material. It does not change depending on the actual volume or
mass of material.
Q13.3. Reason: If the chunk is heavy (dense) enough to sink in water you would put the chunk into a known
volume of water in a graduated cylinder and note the rise; this would give the volume of the chunk. A simple
measurement of the chunk’s mass on a pan balance would then allow you to use ρ = m / V .
Assess: If the chunk floats in water then you would have to find a way to submerse it to find the volume.
Q13.4. Reason: Density is given by Equation 13.1.
(a) The size and shape of the two objects is the same, so the volume of the two objects is the same. The second
object has twice the mass as the first so its density =
is ρ 2 2=
m / V 2 ρ1. The second object has twice the density of
the first.
(b) The third object has a size in all three dimensions twice that of the first object. Its volume is eight times that
of the first object. Since its mass is the same, the density of the third object =
is ρ3 m=
/8V 18 ρ1. The density of the
third object is one eighth that of the first.
Assess: To make the volume increase in part (b) more concrete, consider increasing the length of each side of a
cube by a factor of two. The volume increases by a factor of 2 × 2 × 2 =
8.
Q13.5. Reason: It all has to do with pressure increasing with depth, which is a result of the gravitational force.
If, when receiving a transfusion, the bag were held below your body the blood would not flow up into your body,
and if, when donating blood, the bag were held above your body, no blood would flow up into it. The blood
wants to flow down.
Assess: It is kind of neat to notice things like this and think physics thoughts, even while getting a transfusion or
donating blood!
Q13.6. Reason: The pressure at a depth of 10 m is
p =+
p0 ρ gh =
1.013 × 105 Pa + (1000 kg/m3 )(9.80 m/s 2 )(10 m) =
2 × 105 Pa
This is almost twice atmospheric pressure. It will be difficult for Tom to inhale.
13-1
13-2
Chapter 13
Assess: Note that atmospheric pressure is approximately the gauge pressure at a depth of 10 m of water.
Q13.7. Reason: The pressure only depends on the depth from the opening. Since A, B, and C are all at the same
depth below the opening at E the pressure is the same for each.
A=B=C
Assess: While there is a taller column of water over B, you can think that the top of the container at D and F
pushes down on the fluid too, so the pressure at A, B, and C is the same.
Q13.8. Reason: The pressure only depends on the depth from the opening. Since point D is the deepest and
point E the highest then
pD > pF > pE
Assess: A point halfway between point E and point B would have a pressure about the same as the pressure at
point D.
Q13.9. Reason: The ratio of the absolute pressures is smaller. The absolute pressure is the gauge pressure plus
atmospheric pressure. For pA > pB,
pA ( pg ) A + 1atm ( pg ) A
=
<
pB ( pg ) B + 1 atm ( pg ) B
Assess: This is easy to see if you plug in a couple of numbers with pA > pB, say pA = 4 atm, pB = 3 atm,
( pg ) A = 3 atm, and ( pg ) B = 2 atm.
4
3
< 32 .
Q13.10. Reason: (a) The pressure at the bottom of either tank is given by Equation 13.5. The pressure at the
bottom of each tank will be the same, since the height of water in each tank is the same. The area of the bottom
of tank A is larger than the area of the bottom of tank B. From Equation 13.3, the force on the bottom of tank A
will be larger.
(b) The pressure at each height in both tanks is the same, since the depth of water is the same. Since the area of
the sides indicated in the diagram in Figure Q13.10 is the same for each tank, the force on the side of each tank is
the same also.
Assess: This makes sense. Since there’s more water in tank A, the force of the water on the bottom of tank A is
larger.
Q13.11. Reason: As the balloons rises less pressure is exerted on them by the air outside as the air gets
thinner, so the helium inside makes them expand. They look full when they are at high altitude. If the balloons
were fully inflated before launch they would pop when they rose to high altitudes and regions of low outside
pressure.
Assess: The helium-filled toy rubber balloons kids buy at the parade pop when they rise several thousand feet for
this reason.
Q13.12. Reason: The density of the heated water is less, but water increases in height so the pressures are the
same. The pressure at the bottom of the beaker can be calculated with Equation 13.5. Consider a beaker whose
cross-sectional area is A. The volume of water in the cool beaker is V = Ah1. For the cool water the pressure at the
bottom of the beaker is,
 m 
m
p1 =
p0 + ρ gh1 =
p0 + 
p0 +   g
 gh1 =
Ah
 A
 1
For the hot water,
 m 
m
p2 =
p0 + ρ gh2 =
p0 + 
p0 +   g
 gh2 =
 A
 Ah2 
The pressure at the bottom of the beaker is the same for the hot water and cold water.
Assess: This result makes sense, since the mass of the water is the same. The total weight of water exerting
force on the bottom of the beaker is the same, so the pressure will be the same.
Fluids
13-3
Q13.13. Reason: Whatever increase in pressure there is due to the ship is distributed evenly at equal depths.
Therefore pA = pB.
Assess: While a column above point A contains the ship, it also contains less water than a similar column above
point B.
Q13.14. Reason: Mentally picture the beaker of water on the balance. The system of interest is the beaker of
water. Two external forces are acting on the system of interest. These forces are the pull of the earth Mg and the
normal force the balance exerts on the beaker of water NB. Since the system is in equilibrium, we may write NB =
Mg. The scale reading (let’s call it scale reading #1, SR1) is equal to NB. That is SR1 = NB.
Next picture the sphere tied to a string and suspended in the beaker of water. The system of interest is the beaker
of water with the sphere imersed in the water. The external forces acting on this system are the pull of the earth
(M+m)g, the normal force the balance exerts on the system N B' , and the tension in the string T. Since the system
is equilibrium, we may write
N B' +T = (M+m)g, or N B' =(M+m)g – T = SR1+mg-T
The scale reading (let’s agree to call it scale reading #2, SR2) is equal to N B' . That is SR2 = N B' . Since mg is
greater than T, we see that the scale reading #2 is greater than the scale reading #1. That is SR2>SR1.
Assess: It is of interest to note that scale reading #3 is less than scale reading #2 by the tension in the string. That
is, SR
=
SR 2 − T . Demonstrating that this is true is a great laboratory activity.
3
Q13.15. Reason: The fraction of the object below the surface of the liquid is the object’s density as a fraction
of the liquid’s density (if we can ignore the density of the air). Since A has the greatest fraction below the surface
it is the most dense. The least dense (B) floats with the largest fraction of its volume above the fluid level (the
smallest fraction below the liquid level).
ρ A > ρC > ρ B
Assess: You've heard that only 10% of an iceberg is visible above the surface of the ocean. That means 90% of
the iceberg is below the surface. Therefore the density of ice is 90% the density of seawater. Verify this by
looking in Table 13.1 and finding on the Web the density of ice, ρice = 917 kg/m3.
Q13.16. Reason: Archimedes’ principle states that the buoyant force on an object is equal to the weight of the
fluid displaced by the object. Each object displaces exactly the same amount of fluid since each is the same
volume. So the buoyant force on all three objects is the same.
Assess: Note that the buoyant force does not depend on the mass or location of the object.
Q13.17. Reason: For objects that are completely submerged the buoyant force is proportional to the volume
of the object. If the densities of objects A, B, and C are the same, then the objects with greater mass (A and C)
must also occupy a larger volume. Thus, A and C will experience a larger buoyant force than B.
FA = FC > FB
Assess: A refresher of Example 13.2 would be very valuable here.
Q13.18. Reason: The air in the room produces a buoyant force on you. As the pressure in the room decreases
so does the density of air, so the buoyant force decreases. The scale reading will increase.
Assess: Note that the scale is surrounded by air so the decrease in pressure does not change the force of the air
on the scale.
Q13.19. Reason: The scale will read less than your true weight. The scale reads the normal force needed to
support your weight, but in the swimming pool your weight is partly supported by the buoyant force. The normal
force required from the scale is reduced.
Assess: One way to approach problems like this is to “take limits” and see what the extreme cases tell you. If
you and the scale were on a moveable platform, think what the scale would read if it were raised to the surface.
The scale reads the normal force, but its magnitude would be the same as your true weight when you are
completely out of the water. On the other hand, let the platform slowly lower until you barely have your nose out
of water; the scale would read very little. Indeed, if you did this in the Great Salt Lake or the Dead Sea the
13-4
Chapter 13
buoyant force of the salt water would be enough to support your whole body weight and the scale would read
zero if the platform is lowered sufficiently.
Q13.20. Reason: Adding salt to the water increases its density. When the density of the water matches that of
the egg, the egg becomes neutrally buoyant and floats.
Assess: This makes sense. An egg consists essentially of water mixed with the substance of the embryo, which
makes the density of the egg slightly larger than that of water.
Q13.21. Reason: Since the volume of the rigid submerged submarine isn’t changing, then the buoyant force
isn’t changing either, but by forcing water out of the tanks the submarine becomes light enough that the buoyant
force is greater than the weight. (The submarine with the tanks full of air weighs less than the submarine with the
tanks full of water.) The net upward force makes the submarine accelerate upward.
Assess: Equivalently, think of this question in terms of density. To float, the density of the submarine must
become less than the density of the water. The two ways to do this are to increase the volume or to decrease the
mass. Since the hull of a submarine is rigid the volume doesn’t change, but by expelling the mass of the water the
submarine becomes less dense—enough to float.
Q13.22. Reason: As the fish swims deeper, the density of the water increases because of the increase in
pressure. The fish must decrease the volume of its swim bladder.
Assess: This result makes sense. Decreasing volume decreases buoyant force.
Q13.23. Reason: The sphere is floating in static equilibrium, so the upward buoyant force exactly equals the
sphere’s weight, FB = w. But according to Archimedes’ principle, FB is the weight of the displaced liquid. That is,
the weight of the missing water in B is exactly matched by the weight of the added ball. Thus the total weights of
both containers are equal.
Assess: Because the sphere is half below the water line and half above, its density is half that of water. If the
sphere were twice as dense (that is, the same density as water) with the same mass, it would occupy half the
volume. A hemisphere of water would weigh as much as the original sphere. The hemisphere of water would just
fill in the depression left by removal of the sphere, leaving the water height the same as A.
Q13.24. Reason: The force that the block exerts on the water is equal to the buoyant force of the water on the
block, which is exactly equal to the weight of the water displaced. The weight of the water that has run off is
matched by the force the block exerts on the water. The scale reading remains the same.
Assess: Note that the block must be floating for the reading to remain the same.
Q13.25. Reason: Ship A will ride higher because the bottom of A displaces water more efficiently, having a
greater volume at lower depths. Less of A’s total height needs to be submerged to displace a weight of water
equal to the weight of the ship.
Assess: It is a popular activity in middle school classrooms to give students a block of non-hardening clay that
sinks and ask the students to re-shape the clay into a boat that floats. They often even have contests to see whose
boat can hold the most paper clips without sinking. Generally the flat-bottomed boats win.
If the above is true, one might wonder why real ships are not shaped like ship A. The answer is that some are,
such as sampans, barges, and rafts. But ocean-going vessels generally use a keel for stability.
Q13.26. Reason: The lower the velocity of fluid, the higher the pressure at a given point in the pipe. The
velocities from highest to smallest are 2, 1, 3.
Assess: This can be seen from Bernoulli’s principle, Equation 13.14. The kinetic energy of a fluid element
balances with the work done by the pressure force.
Q13.27. Reason: The Bernoulli effect says that the pressure is higher at the point where the fluid is moving slower,
and lower where the fluid is moving faster. The pressure is clearly lowest at point 3 and highest at point 2, so those are
the positions of the fastest and slowest fluid, respectively.
v3 > v1 > v2
Assess: The equation of continuity would then say that the inner diameter of the pipe is greatest at point 2 and
smallest at point 3.
Fluids
13-5
Q13.28. Reason: (a) From the equation of continuity, 13.12, as the cross-sectional area of the pipe increases
the velocity of the fluid decreases. The area at point 1 is the smallest, so the velocity is the greatest. The area is
larger at point 2, so the velocity is smaller there. At point 3, the area is the same as point 2, so the velocity is the
same as at point 2. The area at point 4 is the largest, so the velocity is the smallest there. The speeds in order of
highest to lowest are 1, 2 and 3, 4.
From Bernoulli’s equation, the pressure will be highest when the velocity is lowest. The pressures in order from
highest to lowest are, 4, 3 and 2, 1.
Assess: The continuity equation holds for incompressible fluid, so the answer to (a) is the same whether the
fluid is viscous or not.
Q13.29. Reason: The pressure is reduced at the chimney due to the movement of the wind above. Thus, the
air will flow in the window and out the chimney.
Assess: Prairie dogs ventilate their burrows this way; a small breeze above their mound lowers the pressure
there and allows the air in the burrow to move between openings of different types or heights.
Q13.30. Reason: The walls of a tube create drag. The circular tube requires less pressure to move fluid through
the tube because the fluid is farther away from the wall.
Assess: This is discussed in the section on the derivation of Poiseuille’s equation.
Q13.31. Reason: Since both blocks are more dense than water they will both submerge and not float (so D is not
the answer). However, while the two blocks have the same mass, they do not have the same volume. The
aluminum (with lower density) will have a larger volume. For submerged blocks, the one with the larger volume
(the aluminum) will experience a greater buoyant force.
Since the string is massless and the pulley is massless and frictionless then the tension T in the string is the same
everywhere. Now draw a free-body diagram for each block. For each block the downward weight force is the
same and the upward force of tension in the string is the same. But because the buoyant force is not the same for
the two blocks then there is a net upward force on the aluminum block and a net downward force on the copper
block.
The correct answer is A.
Assess: To the extent that the buoyant force of air is not negligible the same effect would occur out of water.
Q13.32. Reason: The blocks are at the same height so the pressure on the fluid due to each block is the same.
Since the area of the piston for block B is twice that of A, the force block B exerts on its piston is twice as large
as the force block A exerts on its piston. Block B has a larger weight. The correct choice is B.
Assess: In hydrostatic equilibrium the pressure is the same at all points on the same horizontal line.
Q13.33. Reason: Since the question talks about the “extra pressure” we will ignore the air pressure above the
water; you have an equal amount of air pressure on the inside. The 7 N quoted is the “increased pressure.”
Therefore the equation we need is Equation 13.5 without the po term,
p = ρ gd
where we want to solve for d . We will also use p = F / A and round g to 10 m/s 2 for one significant figure accuracy.
d
=
p
F /A
7 N/(7 × 10−5 m2 )
= =
= 10 m
ρ g ρ g (1000 kg/m3 )(10 m/s 2 )
The correct answer is D.
Assess: This one significant figure calculation can easily be done in the head without a calculator. The answer
seems plausible, and the other choices seem too small. The units cancel appropriately to leave the answer in m.
Q13.34. Reason: We can determine what will happen to the ball by determining its density and comparing it
to the density of water. The density of the ball may be determined by
=
r
M
M
(8 lb)(1 N / 0.2248 lb)
=
=
= 840 kg / m3
V
(4π r 3 / 3) (4π )[(8.5 in)(2.54 cm/in)(m / 100 cm)]3
Since the density of the ball is less than the density of water, it will float in water.
Assess: From the above calculation we not only know that the ball will float but also that it will float with 84%
of its volume below the surface.
13-6
Chapter 13
Q13.35. Reason: We’ll assume the ball is not being forcibly held under the water (because then the net force
would be zero). A free-body diagram shows only two forces on the ball: the downward force of gravity
( w = mb g ), and the upward buoyant force, whose magnitude equals the weight of the displaced fluid (the weight
of 8.0 × 10−3 m3 of water).
2
The downward gravitational force has a magnitude of=
w m=
(0.50 kg)(9.8 m/s
=
) 4.9 N.
bg
The upward buoyant force has a magnitude of =
w m=
( ρwV=
(1000 kg/m3)(8.0 × 10−3 m3)(9.8m/s 2)
wg
w)g
= 78.4 N.
Subtracting gives the magnitude of the net force, 78.4 N − 4.9 N = 73.5 N ≈ 74 N.
The correct answer is B.
Assess: This is enough force to cause the basketball to accelerate upward fairly impressively, as you have
probably seen in the swimming pool (if you haven’t, have your teacher assign you a field trip to the pool to try
this out).
Q13.36. Reason: Since the object is floating, the buoyant force equals the weight of the water displaced. The
volume of water displaced is 75% of the volume of the object, so rr
gV = water g ((0.75)=
V ). So rr
(0.75)
=
water
750 kg/m3. The correct choice is B.
Assess: This makes sense. The density of the object must be less than the density of water in order for the object
to float.
Q13.37. Reason: We’ll use the equation of continuity, Equation 13.12, and make all the assumptions inherent
in that equation (no leakage, etc.). We’ll also use A = π r 2. Call the plunger point 1 and the nozzle point 2, and solve
for v1.
v1
=
v2 A2 v2π r 22 v2 r 22 (10 m/s)(1.0 × 10−3 m) 2
=
= =
= 0.10 m/s
A1
r 21
(1.0 × 10−2 m) 2
π r 21
The correct answer is B.
Assess: The answer seems to be a reasonable slow and steady pace. The ratio of the speeds is the square of the
ratio of the radii. The last step (the computation) can easily be done without a calculator.
Q13.38. Reason: The question says to ignore viscosity, so we do not need Poiseuille’s equation; Bernoulli’s
equation should suffice. And because the pipe is horizontal we can drop the ρ gy terms (because they will be the
same on both sides).
∆p = p2 − p1 =
(
1
ρ v12 − v22
2
)
We are
given r1 0.040
=
=
m, r2 0.02 m, and v1 = 1.3 m/s.
We will also use the equation of continuity to solve for v2.
=
v2
A1
r 12
(0.040 m) 2
=
v1 =
v
= 5.2 m/s
(1.3 m/s)
2 1
A2
r2
(0.020 m) 2
Putting it all together
(
)
1
1
∆p =p2 − p1 = ρ v 12 − v 22 = (1000 kg/m3 )[(1.3 m/s) 2 − (5.2 m/s) 2] =−12,700 Pa
2
2
The magnitude of this is 12,700 Pa, so the correct answer is D.
Assess: Since Pa = N/m2 the units work out.
Q13.39. Reason: Assume an ideal fluid. In this problem the fluid is moving (hydrodynamics), so we don’t
want any hydrostatic formulas. Poiseuille’s equation relates the variables given with the pressure difference.
Fluids
=
Q v=
ave A
13-7
p R 4∆p
8η L
Equation 13.13 defines the volume flow rate Q =∆ V / ∆ t =vA. For an ideal viscous fluid we replace v with vave. The
problem states that the flow rate
=
Q v=
L/s 0.0012 m 2/s.
ave A 1.2=
We are also given R = 2.5 cm/2 = 0.0125 m and L = 15 m. Look up the viscosity of the water in Table 13.3,
η =×
1.0 10−3 Pa ⋅ s.
Solve the equation for ∆p.
=
∆p
Q8η L (0.0012 m3/s)(8)(1.0 × 10−3 Pa ⋅ s)(15 m)
=
= 1877 Pa ≈ 1900 Pa
R4
(0.0125 m) 4
pp
This is the difference in pressure between the ends of the hose. The open end of the hose is open to the
atmosphere so the water is at a pressure of 1 atm there. The gauge pressure at the spigot end is the absolute
pressure minus one atmosphere, but that is precisely ∆p given above. So the gauge pressure is 1900 Pa.
The correct answer is A.
Assess: The m and s cancel where needed, leaving Pa. While there are a number of variables in Poiseuille’s
equation, the strategy for this question is quite straightforward.
Problems
−3
P13.1. Prepare: =
In SI Units 1 L 10
=
m3 and 1 g 10−3 kg.
Solve: The density of the liquid is
m 0.120 kg
0.120 kg
=
=
= 1200 kg/m3
V
100 mL 100 × 10−3 × 10−3 m3
ρ
=
Assess: The liquid’s density is a little more than that of water ( 1000 kg/m3 ) and is a reasonable number.
P13.2. Prepare: The volume of the helium gas in container A is equal to the volume of the liquid in container
B. That is, VA = VB. We will use the definition of mass density ρ = m / V .
Solve: Because VA = VB,
mA
ρA
=
mB
ρB
⇒
mHe
ρ He
=
7600 mHe
ρB
⇒ ρ B = (7600) ρ He = (7600)(0.166 kg/m3 ) = 1260 kg/m3
Referring to Table 13.1, we find that the liquid is glycerin.
Assess: Since the gas and the liquid have the same volume, the ratio of the densities is the same as the ratio of
the masses.
P13.3. Prepare: The volume of the sphere will be reduced by a factor of 8 when its radius is halved. We will use
the definition of mass density ρ = m / V .
Solve: The new density is
ρ=′
m
m
3
= 8 = 8=
ρ 8(1.4 kg/m=
) 11 kg/m3
V/8
V
Assess: If the mass is constant and the volume is reduced by a factor of 8, the density will increases by a factor
of 8.
P13.4. Prepare: The volume of a cylinder is equal to π (radius)2(length). We will use the definition of mass
density ρ = m/V. The mass enclosed by the air remains unchanged.
Solve: (a) Doubling the length doubles the volume. The new density is
ρ=′
m ρ 1.4 kg/m3
= =
= 0.70 kg/m3
2V 2
2
13-8
Chapter 13
(b) If the radius is halved, the volume is decreased by a factor of 4. Thus,
=
ρ′
m
3
= 4=
) 5.6 kg/m3
ρ 4(1.4 kg/m=
V/4
Assess: As expected, increase in volume decreases density and a decrease in volume increases density.
P13.5. Prepare: The densities of gasoline and water are given in Table 13.1.
Solve: (a) The total mass is
mtotal = mgasoline + mwat er = 0.050 kg + 0.050 kg = 0.100 kg
The total volume is
Vtotal = Vgasoline + Vwater =
mgasoline
+
mwater
rr
gasoline
water
⇒ ravg =
=
0.050kg
0.050 kg
+
= 1.235 × 10−4 m3
680kg/m3 1000kg/m3
mtotal
0.100 kg
=
= 810 kg/m3
Vtotal 1.235 × 10−4 m3
(b) The average density is calculated as follows:
mtotal = mgasoline + mwater = rr
waterVwater + gasolineVgasoline
=
⇒ ravg
rr
waterVwater + gasolineVgasoline
=
Vwater + Vgasoline
(50 cm3 )(1000 kg/m3 + 680 kg/m3 )
= 840 kg/m3
100 cm3
Assess: The above average densities are between those of gasoline and water, and are reasonable.
P13.6. Prepare: Equation 13.2 applies.
Solve: The total mass of the mixture and container is
m
rr
waterVwater + alcoholValcohol + mcontainer =
−6
(1000 kg/m )(200 × 10 m ) + (790 kg/m3 )Valchohol + 0.150 kg =
0.512 kg
3
3
Solving for the volume of alcohol,
0.162 kg
V=
2.05 × 10−4 m =
205 mL
=
790 kg/m3
Assess: This is close to the volume of water in the container. This makes sense. The density of ethyl alcohol is
slightly less than that of water and the required mass is slightly less than that of the water in the mixture.
P13.7. Prepare: Vf is the volume of the fish, Va is the volume of air, and Vw is the volume of water displaced
by the fish + air system. The mass of the fish is mf, the mass of the air is ma, and the mass of the water displaced
by the fish is mw.
Solve: Neutral buoyancy implies static equilibrium, which will happen when ρ avg = ρ w. This means
mf + ma ρ f Vf + ρ aVa
=
=ρ w ⇒ ρ f Vf + ρ aVa =ρ wVf + ρ wVa
Vf + Va
Vf + Va
Va ρ w − ρ f 1000 kg/m3 − 1080 kg/m3
⇒ Va ( ρa − ρ w=
=
= 8.0%
) Vf ( ρ w − ρf ) ⇒ =
Vf ρa − ρ w 1.28 kg/m3 − 1000 kg/m3
Assess: Increase in volume of 8.0% seems very reasonable.
Fluids
13-9
P13.8. Prepare: The density of sea water is 1030 kg/m3. Also, 1=
atm 1.013 × 105 Pa.
Solve: The pressure below sea level can be found from Equation 13.5 as follows:
p =p0 + ρ gd =
1.013 × 105 Pa + (1030 kg/m3 )(9.80 m/s 2 )(1.1 × 104 m)
=
1.013 × 105 Pa + 1.1103 × 108 Pa =
1.11 × 108 Pa =
1100 atm
Assess: The pressure deep in the ocean is very large.
P13.9. Prepare: The density of water is 1000 kg/m3 and the density of ethyl alcohol is 790 kg/m3. Because the
tank is cubic, its side is d = (Vwater)1/3.
Solve: (a) The volume of water that has the same mass as 8.0 m3 of ethyl alcohol is
Vwater
=
mwater malcohol ralcoholValcohol  790 kg/m3 
(8.0=
m3 ) 6.32
m3 6.3 m3 to two significant figures
=
=
= 
=
3 
rrr
water
water
water
 1000 kg/m 
(b) The pressure at the bottom of the cubic tank is =
p p0 + r water gd :
p=
1.013 × 105 Pa + (1000 kg/m3 )(9.80 m/s 2 )(6.32)1/3 =
1.194 × 105 Pa =
1.2 × 105 Pa to three significant figures
Assess: Since water is more dense than alcohol, we expect the volume in part (a) to be less. Notice that the
contribution of the water to the pressure at the bottom of the tank is small compared to contribution of the
atmosphere.
P13.10. Prepare: We will use Equation 13.5 for pressure and Equation 13.1 for mass density.
Solve: The pressure at the bottom of the vat is p =
p0 + ρ gd =
1.3 atm. Substituting into this equation gives
1.013 × 105 Pa + ρ (9.8 m/s 2 )(2.0
m) (1.3)(1.013 × 105 ) Pa
=
=
⇒ ρ 1550.5 kg/m3
The mass of the liquid in the vat is
2
=
m ρ=
V ρπ (0.5 m)=
d (1550.5 kg/m3 )π (0.5 m) 2 (2.0 m)
= 2400 kg
Assess: A mass of 2400 kg for the liquid in a rather large vat is reasonable.
P13.11. Prepare: We’ll use Equation 13.5, which gives the pressure of a fluid as a function of depth. We
know that the pressure at a given depth depends on the depth but not the diameter of the container; however, we
2
2
use d = 5.0 cm to compute the area of the bottom because F = pA. =
A π=
19.63 cm 2.
π (2.5 cm)=
Solve:
=
F ( po + ρ gd ) A
pA =
[101.3 kPa + (1000 kg/m3 )(9.8 m/s 2 )(0.35 m)](19.63cm 2 ) =
[101300 Pa + 3430 Pa](19.63 cm 2 )
= [104730 Pa](19.63 ×=
10−4 m 2 ) 210 N
Assess: The weight of the water added a little bit to the weight of the air; you’d have to go down to a
depth of almost 11 m to have the contribution due to the weight of the water equal the contribution due to
the weight of the air.
P13.12. Prepare: We will use Equation 13.5 for pressure. The gauge pressure is p – p0 = 0.40 atm.
Solve: The gauge pressure at the bottom of the cylinder is p − p0 = ρ gd . Because the cross-sectional area of the
second cylinder is greater by a factor of four, the depth will be smaller by the same factor. As the above
13-10 Chapter 13
relationship indicates, the decrease in depth thus reduces the gauge pressure by the same factor. That is, the
gauge pressure at the bottom of the second cylinder is 0.40 atm/4 = 0.10 atm.
Assess: We could also look at this as p = F/A. The force is the weight of the liquid and it does not change.
However since the radius doubles, the area will increase by a factor of four and the pressure will decrease by a
factor of four.
P13.13. Prepare: The density of seawater rseawater = 1030 kg/m 2.
Solve: The pressure outside the submarine’s window is pout
= p0 + rseawater gd , where d is the maximum safe
depth for the window to withstand a force F. This force is F/A = pout – pin, where A is the area of the window.
With pin = p0, we simplify the pressure equation to
pout − p0 =
F
F
1.0 × 106 N
d=
= rs eawat er gd ⇒ d =
= 3153 m = 3.2 km
A
Arp
(0.10 m) 2 (1030 kg/m2 )(9.8 m/s 2 )
s eawater g
Assess: A force of 1.0 × 106 N corresponds to a pressure of
p
=
F 1.0 × 106 N
=
= 310 atm
A p (0.10 m) 2
A depth of 3.2 km is therefore reasonable.
P13.14. Prepare: The pressure required will be the atmospheric pressure exerted by the air on the water in
the container subtracted by the pressure exerted by the column of water.
Solve:
p =patm − ρ gh =1.013 × 105 Pa − (1000kg/m3 )(9.80m/s 2 )(2.0m) =8.2 × 104 Pa
Assess: The principle here is similar to the principle in a barometer.
P13.15. Prepare: Please refer to Figure P13.15. We assume that there is a perfect vacuum inside the cylinders
with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. The flat end of each cylinder has
an area A = πr2 = π (0.30 m)2 = 0.283 m2.
Solve: (a) The force on each end is
Fatm =
p0 A =
(1.013 × 105 Pa)(0.283 m 2 ) =
2.86 × 104 N =
2.9 × 104 N
(b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is
Σ Fy = Fatm − wplayers = 0 N ⇒ wplayers = Fatm = 2.86 × 104 N ⇒ number of players =
2.86 × 104 N
= 29.2
(100 kg)(9.8 m/s 2 )
That is, 30 players are needed to pull the two cylinders apart.
Assess: This problem does an excellent job of helping you understand the ramifications of the fact that we live
at the bottom of a very deep air ocean (i.e. atmospheric pressure).
P13.16. Prepare: Please refer to Figure P13.16. The density of mercury ρ = 13,600 kg/m3.
Solve: From the figure and Equation 13.6 for hydrostatic pressure, we have
pgas + ρ gh = pa tmos ⇒ pgas + (13,600 kg/m3 )(9.8 m/s 2 )(0.10 m) = 1.013 × 105 Pa ⇒ pgas = 88,000 Pa
Assess: Using 1 atm = 1.013 × 105 Pa, the gas pressure is 0.87 atm. This is expected because the mercury level in the
left tube is higher than that in the right tube.
Fluids
13-11
P13.17. Prepare: Please refer to Figure P13.17. Oil is incompressible and has a density of 900 kg/m3.
Solve: (a) The pressure at point A, which is 0.50 m below the open oil surface, is
pA =
p0 + ρ oil g (1.00 m − 0.50 m) =
101,300 Pa + (900 kg/m3 )(9.8 m/s 2 )(0.50 m) =
1.1 × 105 Pa
(b) The pressure difference between A and B is
pB − pA = ( p0 + ρ gd B ) − ( p0 + ρ gd A ) = ρ g (d B − d A ) = (900 kg/m3 )(9.8 m/s 2)(0.50 m) = 4400 Pa
Pressure depends only on depth, and C is the same depth as B. Thus pC – pA = 4400 Pa also, even though C isn’t
directly under A.
Assess: This problem illustrates clearly that the pressure depends only on the depth of the fluid.
P13.18. Prepare: Glycerin and ethyl alcohol are incompressible and do not mix. The alcohol in the left arm
floats on top of the denser glycerin and presses the glycerin down a distance h from its initial level. This causes
the glycerin to rise a distance h in the right arm.
Solve: Points 1 and 2 are level with each other and the fluids are in static equilibrium, so the pressures at these
two points must be equal.
p1 = p2 ⇒ p0 + ρeth gd eth = p0 + ρgly gd gly ⇒ ρeth g (20 cm) = ρgly g (2h)
=
h
1 ρeth
1 790 kg/m3
(20 cm)
(20 cm) 5.81 cm
=
=
2 ρgly
2 1360 kg/m3
You can see from the figure that the difference between the top surfaces of the fluids is
=
∆y 20 cm −=
2h 20 cm − 2(5.81 cm)
=
8.4 cm
Assess: Compared with other heights given in the problem, a difference in height of 8.4 cm is reasonable. Since
glycerin is close to twice as dense as alcohol, we should expect the height of the glycerin (above the point 1 – 2
equilibrium position) to be a little over half the height of the alcohol. If ∆y =
8.4cm, the height of the glycerin
(above the equilibrium position) is 11.8 cm which is close to half the height of the alcohol.
P13.19. Prepare: Water and mercury are incompressible and immiscible liquids. The water in the left arm floats
on top of the mercury and presses the mercury down from its initial level. Because points 1 and 2 are level with
each other and the fluid is in static equilibrium, the pressure at these two points must be equal. If the pressures
were not equal, the pressure difference would cause the fluid to flow, violating the assumption of static
equilibrium.
13-12 Chapter 13
Solve: The pressure at point 1 is due to water of depth dw = 10 cm.
=
p1 patmos + ρ w gd w
Because mercury is incompressible, the mercury in the left arm goes down a distance h while the mercury in the
right arm goes up a distance h. Thus, the pressure at point 2 is due to mercury of depth dHg = 2h.
p2 =
patmos + ρ Hg gd Hg =
patmos + 2 ρ Hg gh
Equating p1 and p2 gives
patmos + ρ w gd=
patmos + 2 ρ Hg gh ⇒=
h
w
1 ρw
1 1000 kg/m3
d=
10 cm
= 3.68 mm
w
2 ρ Hg
2 13,600 kg/m3
Assess: The mercury in the right arm rises 3.68 mm above its initial level. This is a reasonable number due to
the rather large density of mercury compared to water.
P13.20. Prepare: Please refer to Figure 13.12. The density of water is ρ = 1000 kg/m3.
Solve: From the figure and the equation for hydrostatic pressure, we have
p0 + ρ gh =
patmos
Using p0 = 0 atm, and patmos = 1.013 × 105 Pa, we get
0 Pa + (1000kg/m3 )(9.8 m/s 2 )=
h 1.013 × 105 Pa ⇒=
h 10.3 =
m 10 m
Assess: This large value of h is due to water having a much smaller density than mercury.
P13.21. Prepare: Review Example 13.1. Since the system is closed to the air, we will ignore the atmospheric
pressure and simply use ∆p = ρ g ∆d . We will assume that when you are lying down your brain is at the same
level as your heart, so that it is 40 cm above it when you stand.
Solve:
 1 atm  760 mm Hg 
∆p = ρ g ∆d = (1060 kg/m3 )(9.8 m/s 2 )(0.40 m) = 4155 Pa 

 = 31.2 mm Hg
1 atm
 101300 Pa 

Subtract the difference from the original to get the final value, 120 mm Hg – 31 mm Hg = 89 mm Hg. This is in the
range considered low.
Assess: Most people have experienced this a few times when they have stood quickly. If you stand slowly it
gives the blood vessels time to constrict and keep the pressure up.
P13.22. Prepare: The buoyant force on the sphere is given by Archimedes’ principle.
Fluids
13-13
Solve: The sphere is in static equilibrium because it is neutrally buoyant. That is,
∑ Fy = FB − w = 0 N ⇒ ρ lVl g − ms g = 0 N
The sphere displaces a volume of liquid equal to its own volume, Vl = Vs, so
=
rl
ms
=
Vs
4
3
ms
=
π r s3
4
3
0.0893 kg
= 790 kg/m3
π (0.03 m)3
A density of 790 kg/m3 in Table 13.1 identifies the liquid as ethyl alcohol.
Assess: If the density of the fluid and an object are equal, we have neutral buoyancy.
P13.23. Prepare: We know that the barge displaces an amount of seawater that weighs the same as the barge.
This will allow us to compute the weight of the barge. Then, as the barge moves into the fresh water we know
that it will also displace an amount of water that weighs the same as the barge. Note that the weight of the
displaced seawater, the weight of the displaced fresh water, and the weight of the barge are all the same; their
masses are also equal, call that value m. Because fresh water is less dense than seawater (see Table 13.1), the
barge will displace a greater volume of fresh water, and the barge will ride lower in the water.
Solve: The volume of seawater displaced is
Vsea =3.0 m × 20.0 m × 0.80 m =48 m3
The mass of that volume of seawater (and therefore also the mass of the barge) is
=
m ρ=
(1030 kg/m3 )(48
=
m3 ) 49440 kg
seaVsea
Now, the volume of fresh water that the barge displaces is
V
=
fresh
m
=
rfresh
49440 kg
= 49.44 m3
1000 kg/m3
Lastly, since the area of the barge has not changed, we solve for the new depth.
=
d
Vfresh
49.44 m3
=
= 0.824 m ≈ 0.82 m
A
3.0 m × 20.0 m
In the fresh water the barge rides 2 cm lower than in the seawater.
Assess: The answer is precisely what we expected: The barge rides a bit (2 cm) lower because the fresh water is
less dense than the seawater.
In fact, a shortcut would be to see that seawater is 3% more dense than fresh water, so the ship will ride 3%
deeper in the fresh water. This gives exactly the same answer: 0.80 m × 103% =
0.0824 m.
P13.24. Prepare: The buoyant force on the wood block is given by Archimedes’ principle. The density of
water is 1000 kg/m3 and the density of seawater is 1030 kg/m3. A floating object is in static equilibrium.
13-14 Chapter 13
Solve: The volume of displaced fluid is Vf = Ah. In static equilibrium,
FB =
w ⇒ ρf Vf g =
ρ woodVwood g ⇒ ρf ( Ah) =
ρ woodVwood
ρ
  Vwood   ρ wood  (0.10 × 0.10 × 0.10m3 )  ρ wood 
=
⇒ h  wood
=
= 


 (0.10m)
 
(0.10 × 0.10) m 2
 ρf   A   ρf 
 ρf 
For fresh water, h = (0.7)(0.10 m) = 0.070 m and for seawater h = 0.068 m. The corresponding values for d are 3.0 cm
and 3.2 cm.
Assess: Objects float better in seawater, so the above result is reasonable.
P13.25. Prepare: The buoyant force on the aluminum block is given by Archimedes’ principle. The density of
aluminum and ethyl alcohol are ρ Al = 2700 kg/m3 and ρ ethyl alcohal = 790 kg/m3. The buoyant force FB and the tension
due to the string act vertically up, and the weight of the aluminum block acts vertically down. The block is
submerged, so the volume of displaced fluid equals VAl, the volume of the block.
Solve: The aluminum block is in static equilibrium, so
Σ Fy= FB + T − w= 0 N ⇒ ρ fVAl g + T − ρ AlVAl g= 0 N ⇒ T= VAl g ( ρ Al − ρ f )
T=
(100 × 10−6 m3 )(9.80 m/s 2 )(2700 kg/m3 − 790 kg/m3 ) =
1.9 N
where we have used the conversion 100 cm3 =
100 × (10−2 m)3 =
10−4 m3.
Assess: The weight of the aluminum block is ρ AlVAl g = 2.7 N. A similar order of magnitude for T is reasonable.
P13.26. Prepare: The buoyant force on the steel block is given by Archimedes’ principle.
Solve: (a) In air the force on the spring is the weight of the block
Fspr ing on blo ck= w= mg= rsteelVblock g= (7900 kg/m3 )(0.10 m)3 (9.8 m/s 2 =
) 77.42 N= 77 N
to two significant figures.
Fluids
13-15
(b) In oil
Fspring on block + FB =
77.42 N
⇒ Fspring on block = 77.42 N − r oilVblo ck g =77.42 N − (900 kg/m3 )(0.10 m)3 (9.8 m/s 2 )
= 68.6
=
N 69 N
Assess: The difference of (77 N – 69 N) = 8 N is due to the oil’s buoyant force.
P13.27. Prepare: The buoyant force on the sphere is given by Archimedes’ principle.
Solve: For the Styrofoam sphere and the mass not to sink, the sphere must be completely submerged and the
buoyant force FB must be equal to the sum of the weight of the Styrofoam sphere and the attached mass. The volume
of displaced water equals the volume of the sphere, so
4π
(0.25 m)3 (9.80
m/s 2 ) 641.4 N
=
3
4
3
2
wStyrofoam r=
(300 kg/m3 )  π (0.25 m)
=
=
StyrofoamVStyrofoam g
 (9.80 m/s ) 192.4 N
3

FB r=
(1000 kg/m3 )
=
waterVwater g
wStyrofoam + mg =
FB,
Because
=
m
FB − wStyrofoam 641.4 N − 192.4 N
=
= 46 kg
g
9.80 m/s 2
Assess: This large mass allows one to appreciate the importance of the buoyant force.
P13.28. Prepare: We can calculate the volume of the man using Equation 13.1. For the buoyant force,
Equation 13.8 applies.
Solve: The volume of the man
is V m=
=
/ ρ man w/ g ρ man. The buoyant force is
FB rr
=
=
airVg
air
rair
1.20 kg/m3
w
(800
N) 0.960 N
=
=
g rr
1000 kg/m3
man
man
w
g
=
Assess: This result makes sense. The weight of the air displaced is very small.
P13.29. Prepare: Treat the water as an ideal fluid. Two separate rivers merge to form one river. We will use
the definition of the flow rate as given by Equation 13.13, 1 L = 10–3 m3.
Solve: The volume flow rate in the Bernoulli River is the sum of the volume flow rate of River Pascal and River
Archimedes.
QB =
5.0 × 105 L/s + 10.0 × 105 L/s =
15.0 × 105 L/s =
1500 m3/s
The flowrate is related to the fluid speed and the cross-sectional area by Q = vA. Thus
=
vB
QB
1500 m3/s
=
= 1.0 m/s
AB 150 m × 10 m 2
Assess: A flow rate of 1.0 m/s is reasonable.
13-16 Chapter 13
P13.30. Prepare: The pipe is a flow tube, so the equation of continuity, Equation 13.12, applies. To work
with SI units, we need the conversion 1 L = 10–3 m3.
Solve: The volume flow rate is
300 L
300 × 10−3 m3
=
= 1.0 × 10−3 m3/s
5.0 min
5.0 × 60 s
=
Q
Using the definition Q = vA, we get
Q 1.0 × 10−3 m3/s
=
= 3.2 m/s
A
π (0.01 m) 2
Assess: This is a reasonable speed for water flowing through a 2.0-cm pipe.
v
=
P13.31. Prepare: We refer to Equation 13.11 and write V = A∆x = A(v∆t ). Solve this for ∆t. We are given
A π=
r2 π
=
V 6000
=
L 6.0 m3, v = 2.1 m/s, and=
(
2.5 cm
2
)
2
2
= 4.9 cm=
4.9 × 10−4 m 2 .
Solve:
t
∆=
V
6.0 m3
=
= 5800 s ≈ 97 min
Av (4.9 × 10−4 m 2 )(2.1 m/s)
Assess: These all seem like real-life numbers. If you want to empty the pool faster you either need to move the
water at a higher speed or get a bigger hose.
P13.32. Prepare: The pipe itself is a flow tube, so the equation of continuity applies. Note that A1, A2, and A3
and v1, v2, and v3 are the cross-sectional areas and the speeds in the first, second, and third segments of the pipe.
Solve: (a) The equation of continuity is
A1v1 = A2v2 = A3v3 ⇒ π r 21v1 = π r 22v2 = π r 32v3 ⇒ r 21v1 = r 22v2 = r 23v3
(0.01 m) 2 v2 =
(0.0025 m) 2 v3
⇒ (0.005 m) 2 (4.0 m/s) =
2
2
 0.005 m 
 0.005 m 
(4.0 m/s ) 1.0 m/s
v3 
(4.0 m/s) 16 m/s
=
⇒ v2 
=
=
=
0.01
m


 0.0025 m 
(b) The volume flow rate through the pipe is
=
Q A1=
v1 π (0.005 m) 2 (4.0 m/s)
= 3.1 × 10−4 m3/s
Assess: Since most of us do not have a good feel for flow rate in m3/s, let’s look at this value in L/s. A flow rate
3.1 × 10−4 m3/s is equal to 0.31L/s . This is a small but reasonable flow rate for a 0.5-cm diameter pipe.
P13.33. Prepare: Please refer to Equation 13.14 (Bernoulli’s equation). Treat the oil as an ideal fluid obeying
Bernoulli’s equation. Consider the path connecting point 1 in the lower pipe with point 2 in the upper pipe a
streamline.
Solve: Bernoulli’s equation is
p2 +
(
)
1 2
1 2
1
p1
p1
ρ v 2 + ρ gy2 =+
ρ v 1 + ρ gy1 ⇒ p2 =+
ρ v 21 − v 22 + ρ g ( y1 − y2 )
2
2
2
Using p1 = 200 kPa = 2.00 × 105 Pa, ρ = 900 kg/m3, y2 – y1 = 10.0 m, v1 = 2.0 m/s, and v2 = 3.0 m/s, we get p2 = 1.1 × 105
Pa = 110 kPa.
Assess: We expect the pressure at point 2 to be less than the pressure at point 1. If this were not the case, the
fluid would not flow from point 1 to point 2.
Fluids
13-17
P13.34. Prepare: We can use Bernoulli’s equation to find the pressure at point B, once the velocity there is
known. We can use the continuity equation to find the velocity of fluid at point B.
Solve: First, the velocity of the fluid at point B is needed. If the pipe did not split, but simply had its diameter
decreased, the velocity of the water would be
vB, no
=
split
AA vA D A2 vA (0.030 m) 2 (2.0 m/s)
=
=
= 18 m/s
AB
D B2
(0.010 m) 2
However, half the water goes into an identical pipe. So the velocity of water at point B is half the previous result,
vB = 9 m/s.
Applying Bernoulli’s equation between points A and B gives
pB =+
pA
1
ρv
2
2
A
−
1
ρv
2
1
1
50 103 Pa + (1000 kg/m3 )(2.0m/s) 2 − (1000 kg/m3 )(9.0 m/s) 2 =
12 kPa
=×
2
2
2
B
Assess: To understand the use of the continuity equation when the tube splits, reconsider the derivation of the
continuity equation in Section 13.5.
P13.35. Prepare: Treat the water as an ideal fluid obeying Bernoulli’s equation.
ρ g ( y2 − y1 ) = ( p1 − p2 ) + ρ ( v12 − v 22 )
1
2
Consider a streamline connecting a point at the surface with a point in the hole. The pressure at the two points is
the same, so p1 – p2 = 0. Further assumptions are that the area of the trough is so large that (1) it doesn’t matter,
and (2) the speed of the water at the top is zero (v1 = 0).
Call y2 – y1 = h = 0.45 m.
Solve: Since the pressures are equal we have
=
ρ gh
(
1
ρ v 12 − v 22
2
)
Now set v1 = 0 and cancel ρ.
gh =
1 2
v2
2
Solve for v2.
=
v2
=
2 gh
=
2(9.8 m/s 2 )(0.45
m) 3.0 m/s
Assess: The result is independent of the area of the trough, as long as it is big enough that we can assume v1 = 0.
The result v2 = 2 gh is known as Torricelli’s theorem.
P13.36. Prepare: We can use Equation 13.15.
Solve: The pressure difference required is given by Equation 13.15.
Lvavg
Lvavg
Lvavg
(2.0 m)(4.0 m/s)
8pη=
8η=
8(0.7 × 10−3 Pa ⋅ s)
=
∆p 8pη=
= 1.8 × 105 Pa
2
2
A
R
(5.0 × 10−4 m) 2
pR
Assess: This seems like a reasonable pressure for such a narrow tube.
P13.37. Prepare: The equation that relates these variables is Poiseuille’s equation. One end of the hose is
open to the air, so the gauge pressure requested is the same as ∆p.
Known
=
Q 0.25 L/s
= 2.5 × 10−4 m3/s
=
R 1.25 =
cm 1.25 × 10−2 m
L = 10 m
1.0 10−3 Pa ⋅ s at 20 °C
η =×
Find
13-18 Chapter 13
∆p
Solve: Solve Poiseuille’s equation for ∆p.
=
∆p
8η LQ 8(1.0 × 10−3 Pa ⋅ s)(10 m)(2.5 × 10−4 m3/s)
=
= 260 Pa
R4
(1.25 × 10−2 m) 4
pp
Assess: This is just a couple of mm of Hg and sounds reasonable. Isn’t it nice how the units cancel to leave Pa?
P13.38. Prepare: Equation 13.15 applies to the water in the syringe.
Solve: Using Equation 13.15,
∆ p = p − patm =
8η Lvavg
R
2
=
8(1.0 × 10−3 Pa ⋅ s)(0.040 m)(10 m/s)
= 3200 Pa
(1 × 10−3 m) 2
Assess: This result seems reasonable.
P13.39. Prepare: For a cubic structure, volume is equal to the cube of the side. Volume is also equal to mass
divided by density.
Solve: The volume of 0.197 kg of gold is
1/3
 (0.197 kg) 
⇒L= 
= 2.17 cm
3 
ρ Au
 19,300 kg/m 
Assess: Large density for gold means that nearly half a pound of gold has a size of ≈ 2.2 cm × 2.2 cm × 2.2 cm.
V = L3 =
M
P13.40. Prepare: n = M / M A. Because the atomic mass number of Cu is 64, one mole of Cu has a mass of
MA = 64 g.
Solve: The volume of the copper cube is 8.0 × 10−6 m3 and its mass is
M = ρV = (8920 kg/m3)(8.0 × 10−6 m3) = 0.07136 kg = 71.36 g
The number of moles in the cube is
 1 mol 
=
n =
 (71.36 g) 1.1 mol
 64 g 
Assess: Since the mass of copper is a little over the mass of one mole of copper, we expect just over one mole of
the element.
P13.41. Prepare: N = ( M / M A ) N A, where N A is Avogadro’s number. Because the atomic mass number of Al
is 27, one mole of Al has a mass of MA = 27 g.
Solve: The volume of the aluminum cube V = 8.0 × 10−6 m3 and its mass is
M = ρV = (2700 kg/m3)(8.0 × 10−6 m3) = 0.0216 kg = 21.6 g
One mole of aluminum (27Al) has a mass of 27 g. The number of atoms is
 6.02 × 1023 atoms  1 mol 
=
= 4.8 × 1023 atoms
N 

 (21.6 g)
1
mol
27
g



Assess: A number slightly smaller than Avagadro’s number is expected since we have slighty less than a mole
of aluminum.
P13.42. Prepare: The density of oil ρoil = 900 kg/m3 and the density of water r water = 1000 kg/m3.
Fluids
13-19
Solve: The pressure at the bottom of the oil layer is p=
p0 + ρ oil gd1 , and the pressure at the bottom of the water
1
layer is
p2 =
p1 + rrr
p0 + oil gd1 + water gd 2
water gd 2 =
⇒ p2 = (1.013 × 105 Pa) + (900 kg/m3 )(9.80 m/s 2 )(0.50 m) + (1000 kg/m3 )(9.80 m/s 2 )(1.20 m) = 1.2 × 105 Pa
Assess: A pressure of 1.2 × 105 Pa = 1.2 atm is reasonable.
P13.43. Prepare: The pressure at the bottom of the tank is due to the atmosphere, the oil, and the water.
Solve: The pressure at the bottom of the oil layer is
pbottom =
po + rr
oil gd oil + water gd water
Solving for the height of the oil and inserting values we obtain
d=
( Pbottom − Po − rr
=
g 27cm
oil
water gd water ) / oil
Assess: Compared to the depth of water, this is a reasonable number.
P13.44. Prepare: Equation 13.5 gives the pressure at the hole.
Solve: The pressure at the hole is
p =p0 + ρ gh =
(101.3 kPa) + (1000 kg/m3 )(9.80 m/s 2 )(2.5 m) =
125.8 kPa
where additional significant figures have been kept in the intermediate result. The force on the Dutch boy’s
finger is F =
(125.8 × 103 Pa)(75 × 10−6 m 2 ) =
9.4 N.
Assess: This result seems reasonable. The pressure is not much higher than atmospheric pressure and the hole
has a small area.
P13.45. Prepare: Please refer to Figure P13.45. Assume that the oil is incompressible and its density is
900 kg/m3.
Solve: (a) The pressure at depth d in a fluid is =
p p0 + ρ gd . Here, pressure p0 at the top of the fluid is due both
to the atmosphere and to the weight of the floating piston. That is, p0 = patm + wp/A. At point A,
wP
+ ρ g (1.00 m − 0.30 m)
A
(10 kg)(9.8 m/s 2 )
=1.013 × 105 Pa +
+ (900 kg/m3 )(9.8 m/s 2 )(0.70 m) =185,460 Pa
p (0.02 m) 2
pA = patm +
⇒ FA = pA A = (185,460 Pa)p (0.10 m) 2 = 5800 N
(b) In the same way,
pB = patm +
wP
+ ρ g (1.30 m) = 190,752 Pa ⇒ FB = 6000 N
A
Assess: FB is larger than FA, because pB is larger than pA.
13-20 Chapter 13
P13.46. Prepare: The tire flattens until the pressure force against the ground balances the upward normal
force of the ground on the tire. The area of the tire in contact with
the road is A (0.15
=
=
m)(0.13 m) 0.0195 m 2.
Solve: The normal force on each tire is
=
n
w (1500 kg)(9.8 m/s 2 )
=
= 3675 N
4
4
Thus, the pressure inside each tire is
n
3675 N
14.7 psi
pinside =
=
=
=
188,500 Pa =
1.86 atm ×
27 psi
A 0.0195 m 2
1 atm
Assess: A pressure slightly lower than 30 psi is what is expected.
P13.47. Prepare: Assume that the air bubble is always in thermal equilibrium with the surrounding water, and
the air in the bubble is an ideal gas.
Solve: The pressure inside the bubble matches the pressure of the surrounding water. At 50 m deep, the
pressure is
p1 = p0 + ρwatergd = 1.013 × 105 Pa + (1000 kg/m3)(9.8 m/s2)(50 m) = 5.913 × 105 Pa
At the lake’s surface, p2 = p0 = 1.013 × 105 Pa. Using the before-and-after relationship of an ideal gas,
 5.913 × 105 Pa  293 K 
4π 3 4π
p2V2 p1V1
p T
=
=
r2
(0.005
m)3 
=
⇒ V2= V1 1 2 ⇒

 ⇒ r2 0.0091 m
5
3
3
T2
T1
p2 T1
 1.013 × 10 Pa   283 K 
The diameter of the bubble is 2r2 = 0.0182 m = 1.8 cm.
Assess: The temperature is in degrees kelvin in the ideal gas equation. Because of lower pressure in the bubble
at the surface of the lake compared at a depth of 50 m, we expected the bubble to be bigger.
P13.48. Prepare: The buoyant force on the cylinder is given by Archimedes’ principle. Vcyl is the volume of
the cylinder and Vw is the volume of the water displaced by the cylinder. Note that the volume displaced is only
from the part of the cylinder that is immersed in water.
Solve: The cylinder is in static equilibrium, so FB = w. The buoyant force is the weight ρwVwg of the displaced
water. Thus
FB =ρ wVw g =
w=
mg =ρ cylVcyl g ⇒ ρ wVw =ρcylVcyl ⇒ ρcyl =ρ w
Vw
Vcyl
A(0.04 m)
kg/m3 )
670 kg/m3
=
⇒ ρ cyl (1000=
A(0.06 m)
Assess: ρcyl < ρw for a cylinder floating in water is an expected result.
P13.49. Prepare: The buoyant force on the sphere is given by Archimedes’ principle. The sphere is in static
equilibrium.
Fluids
13-21
Solve: The free-body diagram on the sphere shows that
1
4
Σ Fy = FB − T − w = 0 N ⇒ FB = T + w = w + w = w
3
3
4
3
3
3
3
⇒ rrrr
wVsphere g =
sphereVsphere g ⇒ sphere =
w = (1000 kg/m ) =750 kg/m
3
4
4
Assess: We expected the sphere’s density to be smaller than the water’s because the sphere is tethered to the
bottom.
P13.50. Prepare: The buoyant force on the ceramic statue is given by Archimedes’ principle.
Solve: The statue’s weight is the 28.4 N registered on the scale in air. In water, the weight of the statue is
balanced by the sum of the buoyant force FB and the spring’s force on the statue. That is,
11.4 N msta tue
wst atu e =
FB + Fspring on st atu e ⇒ 28.4 N =
r wVst atu e g + 17.0 N ⇒ Vst atu e =
=
g rr
w
statu e
⇒ r=
st atue
(mst atu e g ) r w (28.4 N)(1000 kg/m3 )
=
= 2500 kg/m3
(11.4 N)
(11.4 N)
Assess: Many solids have a density around 2500 kg/m3, so the value is reasonable.
P13.51. Prepare: The buoyant force on the rock is given by Archimedes’ principle. We will use Newton’s
first law, as the rock is in equilibrium. A pictorial representation of the situation and the forces on the rock are
shown.
13-22 Chapter 13
Solve: Because the rock is in static equilibrium, Newton’s first law is
1

Fn et = T + FB − wr ock = 0 N ⇒ T = rr
r ockVr ock g − water  Vr ock  g =
2


1
1
r water 



  mr ock g  
=
1 −
 mr ock g
r ock −
water Vr ock g =
 rrrr
 r ock − 2 water   rr
2



  r ock   2 r ock 
Using rr ock = 4800 kg/m3 and mr ock = 5.0 kg, we get T = 44 N.
Assess: A buoyant force of (5.0 × 9.8 N – 44 N) ≈ 5 N is reasonable for this problem.
P13.52. Prepare: We have a straightforward combination of a statics and Archimedes’ Principle problem.
The minimum volume of the Styrofoam must be such that the upward buoyant force on the Styrofoam is equal to
the weight of the Styrofoam and the boy.
Solve: Knowing that we have a situation of static equilibrium and that the upward buoyant force on the
Styrofoam is equal to the weight of the Styrofoam and the boy, we may write
Fbouy
= Ws + Wb
Next using Archimedes’ Principle we may write (the subscript fd stands for fluid displaced)
Fboy
M fd=
g ρf Vfd=
g ρf Vs g
= W=
fd
Then expressing the weight of the Styrofoam and boy in terms of known or desired quantities, obtain
=
Ws M
=
ρsVs g
sg
Wb = M b g
Combining the above and solving for the volume of the Styrofoam we have
=
Vs M b ( ρf − ρ=
0.041 m3
s)
Assess: This is a small volume. However, knowing that the boy has a small mass (40 kg) and that a m3 of
Styrofoam can support a mass of nearly 1000 kg, this appears to be a reasonable number.
P13.53. Prepare: The pictorial representation that follows gives the relevant diameters of the syringe.
Solve: (a) Because the patient’s blood pressure is 140/100, the minimum fluid pressure needs to be 100 mm of
Hg above atmospheric pressure. Since 760 mm of Hg is equivalent to 1 atm and 1 atm is equivalent to 1.013 ×
105 Pa, the minimum pressure is 100 mm = 1.333 × 104 Pa. The excess pressure in the fluid is due to force F
pushing on the internal 6.0-mm-diameter piston that presses against the liquid. Thus, the minimum force the
nurse needs to apply to the syringe is
F=
fluid pressure × area of plunger =
(1.333 × 104 Pa)[p (0.003 m) 2 ] =
0.38 N
(b) The flow rate is Q = vA, where v is the flow speed of the medicine and A is the cross-sectional area of the
needle. Thus,
v
=
Q 2.0 × 10−6 m3 /2.0 s
=
= 20 m/s
A π (0.125 × 10−3 m) 2
Assess: Note that the pressure in the fluid is due to F that is not dependent on the size of the plunger pad. Also
note that the syringe is not drawn to scale.
P13.54. Prepare: The kinematic equations can be used to find the velocity of the water as it leaves the pistol,
and Equation 13.13 can be used to find the velocity flow rate.
Fluids
13-23
Solve: The water has no vertical component of initial velocity as it leaves the pistol. It takes a
time t=
−2 y / g to hit the floor. The horizontal distance it travels in this time is x = (vx )i t. Solving for the initial
velocity of the water,
g
(9.80 m/s 2 )
=
= 3.2 m/s
(vx )i x= (1.2 m)
−2 y
−2( − 0.70 m)
The volume flow rate as it leaves the pistol is
2
Q
= vA
= vπ r=
2.5 × 10−6 m3/s
Assess: This result seems reasonable.
P13.55. Prepare: Knowing the volume flow rate, the number of vessels and the diameter of each vessel, we
can use Equation 13.13 to solve the problem. We can either say that each vessel handles 1/2000 of the volume
flow rate or we can say the total area for this flow rate is 2000 times the area of each vessel. Either way we will
get the same result.
Solve: Starting with the definition of volume flow rate and the knowledge that each vessel handles 1/2000 of
this volume, we may write
V 2000
= Av
t
Solving for v
v=
(V t ) / 2000
3 × 10−8 m3 / s
=
= 2 × 10−3 m / s
2
(2000)π (5 × 10−5 m) 2
πr
Assess: At this rate, water could travel to the top of an 18 m tree in a day. That seems reasonable.
P13.56. Prepare: Treat the air as an ideal fluid obeying Bernoulli’s equation.
Solve: (a) The pressure above the roof is lower due to the higher velocity of the air.
(b) Bernoulli’s equation, with yinside ≈ youtside, is
2
p=
poutside +
inside
 130 × 1000 m 
1
1
1
2
2
=
∆p
=
(1.28 kg/m3 ) 
=
rr
 835 Pa
air v ⇒
air v
2
2
2
3600 s


(c) The force on the roof is (∆p ) A =(835Pa)(6.0m × 15.0m) =7.5 × 104 N. The roof will blow outward (up),
because pressure inside the house is greater than pressure on the top of the roof.
Assess: This problem helps one understand why high winds cause extensive roof damage.
P13.57. Prepare: Please see Figure P13.57. We will treat the water as an ideal fluid obeying Bernoulli’s
equation. A streamline begins in the bigger size pipe and ends at the exit of the narrower pipe. Let point 1 be
beneath the standing column and point 2 be where the water exits the pipe.
Solve: (a) The pressure of the water as it exits into the air is p2 = patmos.
(b) Bernoulli’s equation, Equation 13.14, relates the pressure, water speed, and heights at points 1 and 2.
1
1
1
p1 + ρ v 12 + ρ gy1 =
p2 + ρ v 22 + ρ gy2 ⇒ p1 − p2 =ρ v 22 − v 12 + ρ g ( y2 − y1 )
2
2
2
From the continuity equation,
(
)
v1 A1 = v2 A2 = (4m/s)(5 × 10− 4 m 2 ) = v1 (10 × 10− 4 m 2 ) = 20 × 10− 4 m3/s ⇒ v1 = 2m/s
Substituting into Bernoulli’s equation,
1
p1 − p2 = p1 − patmos = (1000 kg/m3 )[(4 m/s) 2 − (2 m/s) 2 ] + (1000 kg/m3 )(9.8 m/s)(4.0 m)
2
=
6000 Pa + 39,200 Pa =
45,200 Pa
But p1 − p2 =
ρ gh, where h is the height of the standing water column. Thus
=
h
45,200Pa
= 4.6m
(1000 kg/m3 )(9.8m/s 2 )
13-24 Chapter 13
Assess: In order to sustain fluid flow, the pressure at point 1 must be greater than the pressure at point 2. As a
result we should expect the height h to be greater than 4.0 m.
P13.58. Prepare: Please refer to Figure P13.58. The ideal fluid (that is, air) obeys Bernoulli’s equation.
There is a streamline connecting points 1 and 2. The air speeds at points 1 and 2 are v1 and v2, and the crosssectional areas of the pipes at these points are A1 and A2. Points 1 and 2 are at the same height, so y1 = y2.
Solve: (a) The height of the mercury is 10 cm. So, the pressure at point 2 is larger than at point 1 by
ρ Hg g (0.10m) = (13,600 kg/m3 )(9.8m/s 2 )(0.10 m) =13,328 Pa ⇒ p2 = p1 + 13,328 Pa
Using Bernoulli’s equation,
p1 +
(
)
1
1
1
2
2
2
2
p2 +
rrrrr
air v 1 + air gy1 =
air v 2 + air gy2 ⇒ p2 − p1 =
air v 1 − v 2
2
2
2
2( p2 − p1 ) 2(13,328 Pa)
2
2
⇒ v=
=
= 20,825 m 2/s 2
1 − v2
(1.28 kg/m3 )
rair
From the continuity equation, we can obtain another equation connecting v1 and v2.
A1v1= A2v2 ⇒ v1=
A2
π (0.005m) 2
v2=
v2= 25 v2
A1
π (0.001m) 2
Substituting v1 = 25v2 in the Bernoulli equation, we get
(25 v2 ) 2=
− v22 20,825 m 2/s 2 =
⇒ v2 5.78=
m/s 5.8 m/s
Thus v1 = 25v2 = 145 m/s = 150 m/s.
(b) The volume flow rate A2=
v2 π (0.005 m) 2 (5.78 m/s)
= 4.5 × 10−4 m3/s.
Assess: As a check on our work we can also determine the flow rate by
A1=
v1 π (0.001 m) 2 (150 m/s)
= 4.5 × 10−4 m3/s.
Since the two values for the volume flow rate agree, we can conclude that our work is correct.
P13.59. Prepare: Please refer to Figure P13.59. The ideal fluid obeys Bernoulli’s equation. There is a streamline
connecting point 1 in the wider pipe on the left with point 2 in the narrower pipe on the right. The air speeds at points 1
and 2 are v1 and v2 and the cross-sectional area of the pipes at these points are A1 and A2. Points 1 and 2 are at the same
height, so y1 = y2.
Solve: The volume flow rate is=
Q A=
A=
1200 × 10−6 m3/s. Thus
1v1
2 v2
=
v2
1200 × 10−6 m3 /s
1200 × 10−6 m3/s
95.49
m/s
3.82 m/s
=
=
v1 =
2
π (0.0020 m)
π (0.010 m) 2
Now we can use Bernoulli’s equation to connect points 1 and 2.
(
)
1 2
1
1
ρ v1 + ρ gy1 =
p2 + ρ v22 + ρ gy2 ⇒ p1 − p2 =ρ v 22 − v12 + ρ g ( y2 − y1 )
2
2
2
1
(1.28 kg/m3 )[95.49 m/s) 2 =
=
− (3.82 m/s) 2 ] + 0 Pa 5826 Pa
2
p1 +
Because the pressure above the mercury surface in the right tube is p2 and in the left tube is p1, the difference in
the pressures p1 and p2 is ρ Hg gh. That is,
p1 −=
p2 5826 =
Pa ρ Hg gh ⇒=
h
5826 Pa
= 4.4 cm
(13600kg/m3 )(9.8 m/s 2 )
Assess: Note that the pressure difference (5826 Pa) is small compared to atmospheric pressure (1.013 × 105 Pa). As a
result we expect the height of the mercury colume to be small compared to 760 mm for atmospheric pressure.
P13.60. Prepare: Treat the water as an ideal fluid obeying Bernoulli’s equation. There is a streamline
connecting point 1 in the wider pipe with point 2 in the narrower pipe.
Fluids
13-25
Solve: Bernoulli’s equation, Equation 13.14, relates the pressure, water speed, and heights at points 1 and 2.
p1 +
1 2
1
ρ v 1 + ρ gy1 =
p2 + ρ v 22 + ρ gy2
2
2
For no height change y1 = y2. The flow rate is given as 5.0 L/s, which means
Q = v1 A1 = v2 A2 = 5.0 × 10−3 m3/s ⇒ v1 =
A
v1 1=
⇒ v=
2
A2
5.0 × 10−3 m3/s
= 0.6366 m/s
π (0.05 m) 2
2
 0.050 m 
= 2.546 m/s
 0.025 m 
( 0.6366 m/s ) 
Bernoulli’s equation now simplifies to
1 2
1 2
ρ v 1 =+
p2
ρv2
2
2
1
1
⇒ p1 = p2 + ρ v 22 − v 12 = 50 × 103 Pa + (1000 kg/m3 )[(2.546 m/s) 2 − (0.6366 m/s) 2 ] = 53 kPa
2
2
Assess: Reducing the pipe size reduces the pressure because it makes v2 > v1.
p1 +
(
)
P13.61. Prepare: Please refer to Figure P13.61. Assume the gas in the manometer is an ideal gas. Assume that
a drop in length of 90 mm produces a very small change in gas volume compared with the total volume of the
gas cell. This means the volume of the chamber can be considered constant.
Solve: In the ice-water mixture the pressure is
p1 = patmos + ρHgg (0.120 m) = 1.013 × 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.120 m) = 1.173 × 105 Pa
In the freezer the pressure is
p2 = patm + ρHgg (0.030 m) = 1.013 × 105 Pa + (13,600 kg/m3)(9.8 m/s2)(0.030 m) = 1.053 × 105 Pa
For constant volume conditions, the gas equation is
p 
p1 p2
1.053 × 105 Pa
= ⇒ T2 =
=
− 28 °C
T1  2  =
(273 K)
245 K =
T1 T2
1.173 × 105 Pa
 p1 
Assess: This is a reasonable temperature for an industrial freezer.
P13.62. Prepare: During the constant-diameter sections, viscosity contributes to a pressure difference
between the beginning and end of the section. During the short transition length, the viscosity won’t contribute as
much as the difference in pressure due to the narrowing of the tube.
Solve: See the following figure.
The velocity of the water as it leaves the rightmost section of the tube is
v=
4
Q 0.020 × 10−3 m3/s
=
= 25.5 m/s
A π (5.0 × 10−4 m) 2
13-26 Chapter 13
An additional significant figure has been kept in this intermediate calculation.
Applying Poiseuille’s equation between points 3 and 4 gives a pressure difference of
p3 − p=
4
8η Lvavg 8η LQ 8η LQ 8(1.0 × 10−3 Pa ⋅ s)(1.0 m)(0.020 × 10−3 m3 /s)
=
=
=
= 8.1 × 105 Pa
R2
R 2A
R4
(5.0 × 10−4 m) 4
pp
Using Bernoulli’s equation during the transition period gives
1 2 1 2
ρ v 4 − ρ v2
2
2
p2 − p=
3
The velocity of the water at point 2 can be found with the continuity equation.
=
v2
A4v4 D42
= =
v4 1.6 m/s
A2
D22
Substituting into Bernoulli’s equation gives
p2 − p3 =
4
  1.0 mm  4 
1 2   D4   1
5
ρ v 4 1 −   = (1000 kg/m3 )(25.5 m/s) 2 1 − 
  = 3.2 × 10 Pa
  4.0 mm  
  D2   2
2




Applying Poiseuille’s equation between the beginning and end of the fluid motion in the leftmost section of the
tube gives a pressure difference of
p1 − p2 =
8η Lvavg
R
2
=
8(1.0 × 10−3 Pa ⋅ s)(1.0 m)(1.6 m/s)
= 3.2 × 103 Pa
(2.0 × 10−3 m) 2
The total pressure at point 1 is the sum of these pressure differences. The pressure at point 4 is atmospheric
pressure. So
p=
1.013 × 105 Pa + 8.1 × 105 Pa + 3.2 × 105 Pa + 3.2 × 103 Pa =
1.2 × 106 Pa
Assess: Note that this is over 10 times the atmospheric pressure.
P13.63. Prepare: The equation that relates these variables is Poiseuille’s equation. Solve it for ∆p.
∆p =
8η LQ
p R4
L will be the same in both situations, and we want to keep Q the same as well; this makes us think to use ratios so
these will cancel. Call the first situation 1 and the second (restricted arteries) situation 2. R2 = 0.90 R1. We want to
know ∆p2.
Solve:
∆ p2
=
∆p1
8η2 LQ
η2
p R24
8η1 LQ
R24
4
4
η  R  2.7 × 10−3 Pa ⋅ s  R1 
= η= 2  1=
=


 1.6
η1  R2  2.5 × 10−3 Pa ⋅ s  0.90 R1 
pR
R
4
1
1
4
1
So ∆p2 = 1.6∆p1 = (1.6)(8.0mm Hg) = 13mm Hg.
Assess: For the smoker, the required pressure difference is 60% greater than for the non-smoker. This is a
significant health issue.
P13.64. Prepare: Viscosity must be taken into account while the fluid is in the narrow tube.
Solve: The pressure difference between the end of the tube connected to the beaker and the open end of the tube is
p =patm + ρ gh − patm =ρ gh =(1000 kg/m3 )(9.80 m/s 2 )(0.45 m) =4.4 × 103 Pa
Poiseuille’s equation gives that the volume flow rate in the tube as
Q=
R4
pp
(1.5 × 10−3 m) 4
∆p=
(4.4 × 103 Pa)= 2.0 × 10−8 m3/s
8η L
8(1.0 × 10−3 Pa ⋅ s)(0.10m)
Fluids
13-27
Assess: This result is reasonable for such a narrow tube and small pressure difference.
P13.65. Prepare: Follow the line of reasoning in Conceptual Example 13.5 and the box about giraffes.
∆p =
ρ blood gd . We want to solve for d. The lowest the pressure can be is zero, so

  101.3 × 103 Pa 
1 atm
=
∆p 100=
mm Hg 
 13,300 Pa

1 atm
 760 mm Hg 

Solve:
d
=
∆p
=
ρb loo d g
13,300 Pa
= 1.3 m
(1060 kg/m3 )(9.8 m/s 2 )
The correct answer is B.
Assess: While no humans have their head 1.3 m above their heart, giraffes do, so their blood pressure is higher than
ours. Since Pa expressed in fundamental SI units is equivalent to kg/(m ⋅ s 2 ) the units work out.
P13.66. Prepare: The continuity equation expresses fluid flow in terms of the cross-sectional area and speed of
the fluid. It furthermore establishes that the fluid flow is the same at all points. The continuity equation is A1v1 =
A2v2.
Solve: The fluid flow at the aorta is Aa va = (π d a2 /4)va .
The fluid flow in a single vessel equivalent to all the capillaries is As vs = (π d s2 /4)vs .
Since the fluid flow is constant, we may equate these two expressions to obtain the diameter of the single vessel
equivalent to all the capillaries.
(π=
d a2 /4)va (π d s2 /4)vs =
or d s d=
(2.5 cm ) (60 cm/s)/(0.07
=
cm/s) 73 cm
a va / vs
This is approximately 75 cm. The correct choice is C.
Assess: This calculation points out that the capillary system is impressive.
P13.67. Prepare: Poiseuille’s equation gives the flow rate for viscous flow (such as that in a small blood
vessel).
Q=
p R 4∆p
8η L
Because “there is no change in pressure across the blood vessel” even ∆p cancels, along with all the other variables
except R. Rf = 0.90 Ri.
Solve:
Qf
=
Qi
p Rf4 ∆p
8η L
4
 Rf 
4
= =
=
0.66
 (0.90)
p Ri4 ∆p
R
 i
8η L
The correct answer is A.
Assess: The fourth power dependence is strong. Just a narrowing of 10% restricts the blood flow by almost 35%.
P13.68. Prepare: The continuity equation expresses fluid flow in terms of the cross-sectional area and speed of
the fluid. It furthermore establishes that the fluid flow is the same at all points. The continuity equation is A1v1 =
A2v2.
Solve: In order to make the situation easier to visualize, let’s assume that all the capillaries are the same size and
that we have N capillaries. In this case the continuity equation becomes Aa vo = NAc vc . Looking at this expression,
we see that the flow rate can be increased by expanding the smaller blood vessels to larger diameters. The correct
choice is B.
Assess: The problem is a good companion to Problem P13.66. If you have not worked Problem P13.66, you will
find it enlightening.