Physical Sciences 2
Nov. 3 – 10, 2015
Physical Sciences 2: Assignments for Nov 3 – 10, 2015
Homework #8: Surface Tension, Gases, and Fluid Flow
Due Tuesday, Nov. 10, at 9:30AM
After completing this homework, you should…
• Understand surface tension.
• Be able to explain how fluid surfaces exert forces.
• Be able to qualitatively and quantitatively describe the pressure inside curved surfaces.
• Be able to explain capillary rise and the wetting of solid surfaces.
• Be able to explain the effect of surfactants on liquids and some important applications.
• Understand why the pressure of a liquid varies with depth.
• Understand why the pressure in a gas varies exponentially with height.
• Be able to use the exponential pressure expression to solve problems involving gases.
• Know what causes fluids to flow.
• Understand why the volume flow rate is constant for incompressible fluids, and use the
expression to solve for flow characteristics in a pipe of varying dimension.
• Be able to qualitatively explain the formation of aneurysms and its connection to fluid
dynamics.
Physical Sciences 2
Nov. 3 – 10, 2015
Here are summaries of this module’s important concepts to help you complete this homework:
Physical Sciences 2
Nov. 3 – 10, 2015
Physical Sciences 2
Nov. 3 – 10, 2015
1. Surface energy (2 pts).
Let’s use the concept of surface tension as surface energy per unit
area to see if we can estimate, at least to the correct order of
magnitude, the surface tension of water.
a) Water has a molar mass of 18 g/mol and a density of
1000 kg/m3 (or 1 g/cm3). Based on this data, estimate the
number of water molecules per unit surface area of water.
Based on the density and the molar mass, the volume occupied per water molecule is
3
−29
3
1 mol
! 0.018 kg $ !
$ ! 1 m $ 3 × 10 m
=
#"
&#
&
mol % " 6 × 10 23 molecules % #" 1000 kg &%
molecule
The molecules don’t have regular structure, but we can estimate distances by assuming that
they are arranged in a cubical lattice. A cube of water can be seen as being formed by stacks
of water sheets, each with a thickness of one molecule diameter. The average distance
between adjacent water sheets is then (from Wikipedia) d = 0.3 nm.
Therefore the number of water molecules per unit area at the surface of the water is:
molecules molecules
1molecule
1×1019 molecules
−10
=
d=
(3×10
m)=
.
area
volume
3×10 −29 m 3
m2
b) The coordination number of water (i.e., the average number of “neighbors” each water
molecule has) in the liquid state is 4. Neighboring water molecules attract each other via
hydrogen bonds, each of which has a binding energy of roughly 10–20 J (although this
number depends relatively strongly on temperature). Use this information to estimate the
surface tension of water. How does your estimate compare to the observed figure (
γ water = 0.072 N/m )?
On average, bringing a molecule from the bulk to the surface requires breaking half of the
bonds between that molecule and its neighbors (since a molecule at the surface is only halfsurrounded by other water molecules). The energy required to break those bonds is
" 10 −20 J %
1
Esurface − Ebulk = ( 4 bonds ) $
= 2 × 10 −20 J.
2
# bond '&
The surface tension is the surface energy required per unit area:
energy $ 2 × 10 −20 J ' $ 1019 molecules '
γ =
=&
= 0.2 J/m 2 = 0.2 N/m .
2
)
&
)
area
m
% molecule ( %
(
Physical Sciences 2
Nov. 3 – 10, 2015
2. Xylem and phloem (1 pts)
Water and nutrients are delivered from the roots to the leaves in trees via tube-like tissues known
as xylem and phloem. If these tissues have radii around 100 μm, calculate the capillary rise for
water in the xylem of a tree. Does this answer make sense? If not, propose another hypothesis
for the transport of sap in trees. Hint: what are the radii of the pores, or stomata, in the leaves?
2γ
. If we assume the fluid in the xylem and
ρgr
phloem is mainly water, we can use the value for the air-water surface tension (~7 x 10-2
N/m, from Lecture 17) and a density of 1000 kg/m3. This gives a capillary rise of
N
2(7x10−2 )
€
2γ
m
h=
=
= 0.14m or 14 centimeters
ρgr (1000 kg )(10 m )(10−4 m)
m3
s2
The capillary rise h for a tube of radius r is h =
€
This does not make sense, since trees are obviously taller than 14 cm. Transpirational pull
due to evaporation of water from the leaves and osmosis into the roots due to a concentration
difference of nutrients between the roots and soil are the main causes for water and nutrients
to be transported through the xylem.
Regarding the hypothesis and experiment, there’s no “correct” answer, per se.
Physical Sciences 2
Nov. 3 – 10, 2015
3. Water strider (2 pts)
The end of a water strider’s foot can be treated as roughly
spherical with a radius of r = 20 µm. Let’s look at the forces on a
sphere which is partially submerged in water, as in the diagram
at right. Use γair-water = 0.07 N/m.
a) If the foot is submerged such that the radius to the level of water makes an angle θ with the
vertical, as shown, what is the net force on the foot due to surface tension, in terms of γ, r,
and θ? For what value of θ is this a maximum?
If we look closer at the place where the insect’s foot touches
the water, it would look something like the diagram at left.
The water’s surface curves away from the hydrophobic foot.
If we assume a contact angle of 180°, then at the point
shown, surface tension pulls in the direction indicated:
tangent to both the water and the surface of the foot (sphere),
so at an angle of θ with the horizontal.
Surface tension acts everywhere around the circumference
of the circle where the water touches the sphere, but the
direction is different everywhere: always at an angle of θ
with the horizontal but pointing “outward” at every point.
By symmetry, the horizontal component of the force due
to surface tension cancels out, but the vertical components
add up. To get the vertical component, we need to
multiply by sin θ, and to find the total force due to surface
tension, we take γ times the circumference of the circle,
which is 2πr sin θ. So the net force due to surface tension
is

Fst = 2πγr sin 2 θ, upward
The maximum force comes when sin θ = 1, or θ = 90°
€
Physical Sciences 2
Nov. 3 – 10, 2015
b) Calculating the buoyant force on the sphere requires difficult calculus to determine the
volume of just the underwater portion of the sphere. However, we can at least estimate its
order of magnitude: if exactly half of the sphere is underwater, what is the buoyant force on
the sphere? How does this compare in magnitude to the force due to surface tension in that
position? Use ρwater = 1000 kg/m3.
If half the sphere is underwater, then the buoyant force is:



Fbuoyant = − ρVdisplaced g = − ρ( 2 3 πr 3 ) g
For the numbers given, this has a magnitude of about 170 pN. We can compare this to the
force due to surface tension at this location: when half the sphere is underwater, θ = 90°, so
€
the magnitude of the force due to surface tension is:
Fst = 2πγr = 2π (0.07 N/m)(2 ×10 -5 m) = 8.8 µN
That’s almost five orders of magnitude larger than the buoyant force, so we can ignore
buoyancy.
€
c) A water strider has a mass of about 3 milligrams and has six legs. What will be the angle θ
when the strider is resting on the water’s surface?
We set the force from part a) equal to mg/6 (since each leg supports 1/6 of the weight) and
solve for θ:
mg
= 2πγr sin 2 θ
6
mg
θ = sin−1
= sin−1 0.75
12πγr
or about θ = 49°.
€
Physical Sciences 2
Nov. 3 – 10, 2015
4. Blood flow (2 pts)
When you are resting, your heart typically pumps about 5 L of blood through your body every
minute.
a) Your aorta has a radius of about 0.9 cm. What is the speed of blood flow in your aorta?
We would like to relate the volume flow rate (which
we’ll call Q) of fluid through a pipe to the speed v and
the cross-sectional area A. Consider the diagram at left.
In a short time Δt , the fluid that starts out at the left end
of the pipe travels a distance Δ1 = v1Δt . So during that
time, the volume of fluid entering the pipe is
A1Δ1 = A1v1Δt . The flow rate (volume per unit time) entering the pipe is therefore
A v Δt
Q1 = 1 1 = A1v1 .
Δt
By the equation of continuity, this flow rate is the same anywhere along the pipe, so we can
drop the subscript 1 and just call it Q.
All of the blood pumped by your heart must go through the aorta, so the flow rate in the aorta
is Q = 5 liters per minute. So the average speed of blood in the aorta is
5 × 10 −3 m 3 /60 s
Q
v= =
= 0.33 m/s .
A
π (9 × 10 −3 m)2
(
)
b) Your lungs are designed with tremendous redundancy; they have far more capillaries than
are actually used at any one time. (The remaining capillaries are collapsed and do not
support blood flow.) Each of your 300 million alveoli is surrounded by an estimated 1000
tiny capillaries. Each capillary has a radius of approximately 4 × 10–4 cm, and blood flows in
a capillary at a speed of about 5 × 10–4 m/s. Estimate the fraction of capillaries in your lung
that are actually being used.
The total number of capillaries in your lungs is 300 million (3 × 108) times 1000 (103), or 3 ×
1011. The flow rate in each capillary is
(
Q = Av = π 4 × 10 −6 m
2
) ( 5 × 10
−4
)
m/s = 2.5 × 10 −14 m 3 /s
The total blood flow rate through the heart is 5 liters per minute (from the previous problem),
which is 8.3 × 10–5 m3/s. All of the blood that flows through the aorta must eventually pass
through the capillaries, too, so this must also be the total flow rate through all the capillaries.
So the number of capillaries being used in your lungs is
8.3 × 10 −5 m 3 /s
= 3.3 × 10 9
−14
3
2.5 × 10 m /s
Comparing this to the total number of capillaries, we see that only about 1% of the capillaries
in your lungs are being used.
Physical Sciences 2
Nov. 3 – 10, 2015
5. Breathe easily (2 pts)
Your lungs contain about 300 million tiny alveoli, which are spherical sacs that inflate with air.
You can think of each alveolus as a tiny bubble of air surrounded by fluid. The fluid is ordinary
mucous tissue fluid that has a surfactant added to lower its surface tension; the resulting
surface tension is about γ = 3 mN/m.
a) Newborn babies must create a pressure difference between the inside and outside of the
alveoli of about 30 torr to expand their alveoli for the first time. Estimate the radius of an
alveolus before it is inflated for the first time in a newborn baby.
When the alveolus is uninflated, the walls are not taut so there is no tension. In order to
maintain an inflated alveolus, the pressure inside must exceed the pressure outside by an
amount given by the Laplace formula:
2γ
.
ΔP = Pin − Pout =
R
In order to expand the alveolus while it is already inflated, the pressure difference must be
even larger. Making an inflated alveolus expand is very difficult to do when the radius is
small, so it’s a good thing babies don’t have to breathe for themselves until their lungs are
grown and developed.
We are told that the first time a baby tries to inflate its alveoli, it takes a pressure difference
of 30 torr. Applying the Laplace formula, we can solve for the radius:
R=
2γ 2(0.003 N/m)
=
= 1.5 µm .
ΔP
30 torr
b) Some newborns do not have the surfactant needed in their alveoli, so the surface tension of
the mucous tissue fluid is increased by a factor of about 15. This condition, known as
respiratory distress syndrome, means that tremendous pressure is required in order for the
newborn to take her first breath. What pressure would be required to inflate the alveoli in the
absence of surfactant?
If the surface tension goes up by a factor of 15, the pressure difference required to inflate the
alveolus also goes up by a factor of 15 (for the same radius). So the pressure required would
be
ΔP = 450 torr .
That’s quite a large pressure difference (almost 1 atm), so some surfactant is required to
enable the baby to breathe for the first time.