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Fluids:
Example:
Application of Fluid Physics: The Sinking of the Titanic (15 April, 1912)
Fact #1: The Titanic sailed at a mass of about 4.2x107 Kg. It was 268.85 m long, 28 m wide,
31.6 m high (bridge to keel) and it floated. Where was the waterline?
Fact #2: The Titanic struck an iceberg which created damage to the side below the water
surface. The bow became very heavy due to the ship taking on water. The Titanic sank after 2
hours, 40 minutes. How big was the hole in the side of the Titanic?
Fact #3: The sinking was complex: First the boat tilted forward, and then it broke between the
third and forth smoke stacks. The forward section sank bow first to the bottom and remains
largely intact today. The rear of the vessel tilted back almost to level, then, as it filled in with
water, it sank vertically with the stern high. It arrived at the bottom in many pieces strewn
over a large area. On the way to the bottom of the ocean, why did the bow section remain
intact while the stern exploded?
What is a fluid?
A Fluid is a collection of molecules that are randomly arranged and held together by weak
cohesive forces or by forces exerted by the walls of its container. Fluids can be either
liquids or gasses.
A Fluid is a substance that flows
Examples?
Water, air, alcohol, oxygen.
Fluid mechanics is the study of stationary (statics) and moving (dynamics) fluids.
For the first 4 sections of chapter 15, we shall deal only with static fluids
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Fluid Statics
Fluid statics: Read Chapter 15, sections 15-1 to 15-4
Section 15-1: Fluids
Density ρ for an object is defined as the mass per unit volume:
ρ = m/v
SI units (kg/m3)
Density of fresh water = 1000 kg/m3
= 1000× (1000g/kg) × 1kg / [(1×106 cm3/m3) ×1 kg] = 1g/cm3
See Table 15-1 on page 446 of your text for densities of a variety of fluids.
What is the mass of air in this lecture room?
Assume the volume has a triangular shape with height = 6 m, length = 16m, width =25m
Then we get M = ρV = ρ× ½ length×height×width = HW 1.28 kg/m3× 6m×16m×25m = 1229 Kg
Section 15-2: Pressure
How do we apply a force to a fluid? What happens when you push on a fluid? How does it push
back on you?
Fluids cannot sustain shearing (sliding) or tensile (pulling) stresses. The force exerted by a
fluid is always perpendicular to the walls of the vessel.
How do we deal with forces in fluids? Answer: we use the concept of pressure.
Consider a cube immersed in a static fluid. What forces are exerted by the fluid on the cube?
Note that pressure is a scalar, not a vector. A fluid exerts
pressure in all directions. At any surface the pressure is
exerted perpendicular to that surface- this is because a fluid
cannot sustain shearing forces.
For an area A on the top or bottom surface of the cube (preferably the bottom or the top since
the force is constant across), the pressure is defined as
P = F/A
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Note that Pressure is a scalar quantity (i.e. it has a magnitude but no direction) whereas Force is a
vector (with magnitude and direction).
Since the force is often dependent on position, it is often better to use the definition:
P=dF/dA for an infinitesimal area on the surface of the object.
The standard units of pressure are the Pascal.
1 Pascal = 1 Pa = 1 N/m2
Variation of pressure with depth
Consider a stationary incompressible fluid of density ρ in a container open to the
atmosphere. (We want to measure the pressure as a function of depth in the fluid.) Note the
pressure due to air in the atmosphere at sea level is Po = 1.013x105 Pa = 1 atm. This corresponds
to the weight of air (in Newtons) above a 1m2 square.
PoA
HH
h
We marked a cylinder of water in the beaker with dark
lines. These lines could be made from a very thin light
material. What would happen to the system if we
removed the box?
Mg
Why do we expect the pressure to vary with depth?
Water in lower parts of the beaker has to bear the weight
of water above it.
PA
+
We consider an imaginary cylinder of fluid of depth h and cross sectional area A. The fluid
enclosed by the cylinder has mass M = ρV = ρAh.
Choosing upwards as the positive direction, we sum the forces acting on our imaginary cylinder
Σ Fy = PA - Mg - PoA = 0.
So PA - ρAhg - PoA = 0
Note that the pressure of the atmosphere is transmitted from
the surface of the liquid all the way down to the bottom of the
container.
Then P = Po + ρgh
Titanic Problem: The stern of the Titanic contained large water-tight compartments which
contained air at atmospheric pressure. At the site of the sinking, the ocean floor was 3.7
km below the surface. At the ocean floor, what would be the force on a 100 m2 area (i.e.
10m x 10m) on the outside of one of the compartments?
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Pressure in water at the ocean floor: P = Po + ρgh; Pressure in the compartment = Po
(Note that the force on the inside walls of the compartment is PoA whereas that on the outside is
PA. P is a lot bigger than Po.)
Force F on 100 m2 = (P - Po) A= ρghA =(1030 kg/m3)(9.81 m/s2)(3700m)(100m2) = 3.7x109N
Equivalent to 335 atm pressure.
Po
Weight of Titanic = (4.2x 107kg)(9.81 m/s2) = 4.1x108 N
Then about 10 times the weight of the Titanic was exerted on a single 10mx10m section of the
stern. It imploded, then exploded when the air escaped.
One car = 10,000N,
One Titanic = 40,000 cars
Pascal's Principle: Pressure applied to a confined fluid is transmitted undiminished to
every point of the fluid and to the walls of the containers.
The classic example where Pascal's Principle is applied is the hydraulic lift.
Fin, Ain, Pin
Fout, Aout, Pout
Remember the previous example where the
force due to the atmospheric pressure was
transmitted throughout the compartment.
Here we replace the atmosphere with an
applied force.
Fin is typically applied by a pump and transmitted throughout the fluid.
The pressures Pin and Pout are equal (provided the fluid level is the same on both). Then
Fin/Ain = Fout/Aout and Fout = Fin(Aout/Ain). This acts like a force multiplier.
Note also that the distance traveled by the fluid level on the right will be Ain/Aout that traveled by
the fluid level on the left. (There is no free ride, the work done on both sides is equal.) Note also:
This is how your car brakes work.
Sealed and moveable
Fixed
Po
P
Definition: gauge pressure = P - Po is the absolute pressure less the pressure due to the
atmosphere.
Gauge pressure often comes up because when we measure pressure, it is often the difference
between the absolute pressure and atmospheric pressure. (e.g in a bicycle or car tire gauge)
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Section 15-4: Buoyant Forces and Archimedes Principle
Consider a cube of unspecified solid material, side h, located inside a volume of a fluid of
density ρ.
Note: that here we are not concerned about the
weight of the object, only in the net force exerted
by the fluid.
We can assume that the horizontal forces are
balanced since the pressure changes only in the
vertical direction. Also, if there was a right/left
pressure difference, the box would move
horizontally
P1
P2
What is the net force exerted vertically on the object by the fluid?
Net Force = P2A - P1A, where A = h2.
Recall that P2 = P1 + ρgh,
then net force = ρgh3 = ρgV = weight of water displaced =buoyant force.
Generally, regardless of the object, Archimedes' Principle holds:
The buoyant force exerted by the fluid on a body is equal to the weight of fluid
displaced by the body.
Titanic problem: Where was the waterline on the Titanic?
The Titanic was L= 268.85 m long and W=28 m wide. We assume that it had a rectangular
shape. (not a good assumption but easier for calculation of volume)
L
+
h
Vf
ρVf g
Mg
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Sum forces
Buoyant force: ρVf g = Mg Where Vf is the volume of the Titanic under the water line.
(Recall that the buoyant force is equal to the weight of water displaced)
Then ρLWhg = Mg
And h = M/ρLW = (4.2 x 107 Kg)/[(1030 Kg/m3)(268.85m)(28m)] = 5.4m
So the distance from the bottom of the Titanic to the waterline was 5.4m.
Of course, for a more reasonable shape (which would be more triangular than rectangular), the
width would not be constant with depth and the estimated water line would be higher.
At the maximum designated displacement of the Titanic (66,000 tons), the height would have
been 7.8m. This makes sense since another record indicates that the height from the ship bottom
to the decks was 10.6m.
The Ice Cube Question
What happens to
the water level
when the ice
cube melts?
The Buoyant force is
equal to the weight of water
displaced.
Let Vf be the volume of the ice cube below the water level. From Archimedes Law:
ρwaterVf g = Mice g or ρwater Vf = ρice Vice so Vf = ρice/ρwater Vice or about 0.917 Vice
Now, when the ice melts, it changes its density without changing its mass, so the new volume
that it occupies is Vnew = Mice/ρwater = ρiceVice/ρwater = Vf so the melted ice just fills the space that
it previously occupied under water. There will be no change in water level.
Problem from Final Exam in 2001: An iron block containing a number of cavities weighs
6000N in air and 4000N in water. What is the volume of the cavities? The density of iron is 7870
kg/m3 and the density of water is 1000 kg/m3.
We assume that water cannot enter the cavities of the iron block and we neglect the weight of the
air in the cavities. Then the buoyant force is:
FB = 6000N – 4000N = ρwaterVwater g and the volume of the block is Vwater =
2000N/1000kg/m3/9.81m/s2 = 0.2039 m3
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The volume of iron is given by Miron = ρironViron; Viron = Miron/ρiron = 6000N/(9.81m/s2)/7870
kg/m3 = 0.0777 m3 and the volume of the cavities is 0.2039 m3 – 0.0777 m3 = 0.126m3.
Rock in Boat Question: What happens to the water level when the rock is pitched out of the
boat and falls to the bottom of the lake? (Note: the density of rock is about 3x that of H2O)
The water level
goes down
The water level depends upon the water volume displaced by the boat and the rock.
The volume displaced by the rock is as if the
rock had the density of water (true for the boat
also)
The volume displaced by the rock is as if the rock had the density of rock.
Buoyant Force: ρwVf1g = (Mboat+ Mrock)g
Vf2 = Mboat/ρw = water volume displaced by boat alone.
However, the
Change in water displaced by boat = Vf1 - Vf2 = Mrock/ρw
rock now displaces: Vrock = Mrock/ρrock which is ~ 1/3 Mrock/ρw.
Vf1 = (Mboat + Mrock)/ρw = Mboat/ρw + Mrock/ρw
= water volume displaced by boat + rock.
Buoyant force: ρwVf2g = Mboatg
Therefore, the boat plus rock displace less water and the water level falls.
Fluid dynamics
Read Section 15.5
Section 15-4: Flow Rate and the Equation of Continuity
The Ideal Fluid:
1) Non viscous (no internal friction, doesn’t stick to itself or the container walls)
2) Incompressible (Constant density) (not always found in gasses but a good approximation in
most liquids)
3) Steady flow (constant velocity at any point in the fluid) (as opposed to turbulence)
4) No rotational motion
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Equation of continuity
Consider a pipe of variable thickness containing a fluid undergoing steady flow. (The lines
of flow (streamlines) are parallel to the sides of the tube.)
Note that there doesn’t have to be rigid
walls here: this could be inside a volume
of liquid undergoing steady flow.
Since mass is conserved, the amount of fluid
going in, in time ∆t, is equal to the amount
of fluid going out in the same time period.
v2
A2
v1
A1
This tube of flow consists of a set of streamlines. A
streamline is the path followed by a single particle
undergoing steady flow. The particle velocity is
tangent to the streamline at all points.
∆x2
∆x1
Fluid that goes in in
time ∆t.
The rest of the tube is in steady flow. Since nothing is
changing, we need not consider it further.
Fluid that leaves in time
∆t.
In the time interval ∆t, the fluid moves a distance ∆x1 at the entrance and ∆x2 at the exit. v1
= ∆x1/∆t and v2 = ∆x2/∆t.
How much fluid enters the tube in time ∆t?
What is the flow rate?
A1 ∆x1 = A1v1∆t
(a volume)
A1v1
(volume per unit time)
How much fluid leaves the tube in time ∆t?
What is the flow rate?
A2 ∆x2 = A2v2∆t
A2v2
Note that v1 does not equal v2
Since the fluid is incompressible, the amount of fluid entering the tube must be equal to the
amount of fluid leaving the tube.
Then,
A1v1 = A2 v2
the Equation of Continuity
Problem: Blood flows from arterioles to the capillaries. A typical arteriole has diameter 0.030
mm and carries blood at a rate of 5.5x10-6 cm3/s (a) What is the speed of blood in an arteriole?
(b) Suppose that an arteriole branches into 340 capillaries, each of diameter 4.0x10-6m. What is
the blood speed in the capillaries?
Arteriole
Equation of continuity
Capillaries
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(a) Aarteriolevarteriole = flow rate = 5.5x10-6 cm3/s = 5.5x10-12m3/s then varteriole = (5.5x1012 3
m /s)/(π(3.0x10-5m/2)2) = 7.8x10-3m/s = 7.8 mm/s
(b) Aarteriolevarteriole = Acapillariesvcapillaries
where Acapillaries = NAcapillary = 340 (π (4.0x10-6m/2)2) = 4.3x10-9m2
Then vcapillaries = Aarteriolevarteriole/ NAcapillary= (5.5x10-12m3/s)/4.3x10-9m2 = 1.3x10-3m/s = 1.3 mm/s
Issues with blood: 1) blood flow is not constant- it follows the heart cycle; 2) Blood is a viscous fluid; it is
also non-Newtonian- its viscosity depends upon its speed. 3) Blood flow is not necessarily laminar- can have
turbulence at certain locations at certain times.
More on this later….
Bernoulli's Equation
v2
Once again, in time ∆t, at the entrance
the fluid moves to the right a distance
∆l1 with a velocity V1 and under a
pressure P1. At the exit, the fluid moves
a distance ∆l2 and against a pressure P2.
P2A2
∆x2
v1
y2
P1A1
y1 ∆x1
+
We aren’t worried about what happens inside; only
external forces do work on the system
Work done = change in kinetic energy + change in potential energy
At 2, the force is opposite the
distance travelled.
Work done = Force x distance = P1A1∆x1 - P2A2∆x2 = (P1 - P2)V
Net change in kinetic energy = 1/2 mv22 - 1/2mv12
Net change in potential energy = mgy2 - mgy1
The tube itself is in steady flow and hence constant
kinetic energy and potential energy. In the time ∆t,
the rest of the world lost a mass m of fluid at the
left end and gained an equal mass m at the right
end. We consider only these two masses and the
change in energy of the external world.
Then (P1 - P2)V = 1/2 mv22 - 1/2 mv12 + mgy2 - mgy1
(Let m = ρV, divide everything by V and collect the 1's on the left and the 2's on the right to get)
P1 + 1/2 ρv12 + ρgy1 = P2 + 1/2ρv22 + ρgy2
Or more generally, P + 1/2ρv2 + ρgy = constant throughout the fluid.
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Titanic Problem (last one). Given that it took 2 Hr 40 min for the Titanic to sink, how large
was the hole in the side of the boat? Previously, we learned that at its load of 4.2x107Kg, the
waterline was 5.4 m from the bottom. The decks were about 10.6 m from the bottom, so if
we add another 4.2x107 Kg, the water will be over the decks and the boat will sink. We wish
to know how large a hole in the side of the Titanic will let in 4.2x107kg of water in 2 Hr 40
min.
Surface
Po
We shall assume that h is 2m. Note that 2 hr
40 min is 2x60+40 =160 min and 60x160
=9600 s.
1
h
v
Po
A
2
Apply Bernoulli's equation at 1 and 2
Po + ρgh + 0 = Po + 0 + 1/2ρv2
v = √2gh Torcelli's law
Flow rate = Av
We choose the centre of the hole as the
reference point for height.
Note that the difference in atmospheric
pressure between 1 and 2 is negligible
since ρair = 1.28 kg/m3 whereas ρwater =
1000kg/m3.
(m/s) For h=2m, v = 6.3 m/s
(m3/s)
Accumulated mass of water inside the Titanic = Av ∆t ρ = 4.2x107 kg at the time of sinking.
Area of damage A = 4.2x107kg/[v ρ(∆t)] =(4.2x107kg)/[(6.3m/s)(1030kg/m3)(9600s)] = 0.7 m2
Note that the above is an approximation since, 1) we don't really know h, 2) once the water level inside reaches the
hole, the flow will slow down, 3) there were likely several holes in the boat, perhaps at different depths. However,
experts agree that the size of the holes added up to around 1 m2.
It turns out that the evaluated area is not very dependent on h. h=1 A=0.95m2, h=5, A =0.42m2
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Additional problem using Bernoulli's Eqn.
1
Note: The fluid in the h2 tube is
stationary. We also assume that the
fluid in the main reservoir is stationary.
Also note that the fluid is water.
H= 5.0m
h2=?
2
3
A2=4.0 cm2
v2 = ?
Note that this drawing is
not to scale!
A3=2.0cm2
v3 =?
Apply Bernoulli's equation at 1 and 3 to get v3
Po + ρgH + 0 = Po + 0 + 1/2ρv32; v3 = √2gH = √2(9.81 m/s2)(5m) = 9.9 m/s
Apply the Equation of Continuity to get v2:
A2v2 = A3v3; Then v2 = v3A3/A2 = (9.9m/s)(2.0 cm2)/(4.0cm2) = 4.9 m/s
Apply Bernoulli's at 2 and 3 to get P2
P2 + 0 + 1/2ρv22 = Po + 0 + 1/2 ρv32 ; P2 = Po + 1/2ρ(v32- v22) = Po + 1/2(1000
kg/m3){(9.9m/s)2- (4.9m/s)2} = Po + 3.7x104 Pa
Height h2. Apply Bernoulli's at bottom and top of pipe:
Po + 3.7x104 Pa +0+0 = Po + ρgh2
Then h2 = (3.7x104 Pa)/{(1000kg/m3)(9.81m/s2)} = 3.7m
Note: In order to be able to apply
Bernoulli’s equation, the entire region of
fluid must be participating in the flow. The
fluid in h2 is not participating in the flow,
therefore, we cannot simply apply
Bernoulli’s equation at the top of h2 and
the end of the pipe (3). This is a subtle but
important point. The h2 pipe is acting as a
sort of pressure probe.
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Problem 15.65: Air flows through the tube shown in the figure. Assume air is an ideal fluid.
a. What are the air speeds v1 and v2 at points 1 and 2?
b. What is the volume flow rate?
2 mm
1
2
1.0 cm
The pressure here is
ρHggh less than the
pressure on the other
side, neglecting the
pressure gradient in air
10 cm
The pressures are
equal across at the
tip of the arrow.
Hg
We don’t know the velocities at 1 or 2 but we do know that they follow the equation of
continuity: A1v1 = A2v2. We know the difference in pressure between 1 and 2 since it is able to
support a 10 cm column of mercury.
We neglect changes in air pressure with height. Why?
ρair(height) = ρairgh = 1.28 kg/m3 gh
ρHg(height) = ρairgh = 13,600kg/m3 gh
We could of course calculate
the pressure gradient in air
and a problem might arise
where this is important!
There is a ratio of 10,625 (greater than 10,000 times!)
Then P2 – P1 = ρHggh – ρairgh = ρHg gh to the 4th decimal place.
Apply Bernoulli’s at 1 and 2 to get:
P1 + ½ ρairv12 + 0 = P2 + ½ ρairv22
Then: v12 – v22 = v12(1 – A12/A22) = 2/ρair (P2 – P1)
Or: v12 = (2ρHg/ρair)gh/(1- A12/A22)
= 2×13,600 kg/m3/1.28 kg/m3× 9.80× .1m/(1- ((1mm)2/((5mm)2))2) = 20858m2/s2
and v1 = 144 m/s
Then v2 = A1/A2v1 = 1/25v1 = 5.78m/s
Flow rate = A1v1 = π(1×10-3m)2×144 m/s = 4.54×10-4 m3/s
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Giancoli Sections 13-10,11: Real Fluids: Viscosity
Viscosity is caused by internal friction (units N.s/m2)
What causes internal friction?
intermolecular electrical attractive forces,
entanglements between neighbours, ..
Consider two flat plates of area A separated by a fluid of thickness L. The upper plate moves to
the right with velocity v
Note that the viscous force is determined
F
by the velocity gradient.
L
Viscous Force: F = ηAv/L
Coefficient of viscosity, η;
for water, η = 0.001 N.s/m2 at 20oC
Velocity Gradient (=v/L)
For an ideal fluid flowing through a cylindrical tube, the flow speed is constant throughout.
P
η=o
P
Flow rate = Q = Av
Viscous flow through a cylindrical tube
η
P1,
Radius, r
P2
Parabolic flow profile.
We shall not derive the
exact expression for
the viscous velocity
profile through a tube
but rather we shall
present an expression
for the flow rate of a
viscous fluid
In an ideal fluid, P1 =
P2 and the flow
profile is flat.
L
The actual flow is maximum in the tube centre and stopped at the walls.
An important equation for understanding viscous flow is Poiseuilles Equation. (We shall
not derive it here.)
Flow rate (m3/s) = Q = (P1 - P2)πr4
8ηL
Note that for a real fluid with viscosity, when the tube is
narrowed but the pressure gradient is maintained constant
the flow rate decreases. In fact it decreases as the fourth
power of the radius! This is most important for small radii.
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Bernoulli's Equation
Equation of Continuity
Poiseuille’s Equation
Ideal fluids, low viscosity
Ideal Fluids, low viscosity
Viscous fluids
Blood Flow
Head and Arms
Lungs
Heart
Abdomen, legs
Average Pressure at the Left Ventricle is 100 torr. (pressure units used in medicine.)
(~120 torr at systole and 80 torr at diastole)
One torr = 1mm of mercury; the pressure required to support a 1mm column of mercury.
One atmosphere (1.013x105Pa) will support
P=0
5
3
2
Po = ρHggh where h = 1.013x10 Pa/(13,600kg/m )(9.81m/s ) = .760m
Then 100 torr corresponds to a pressure of
(100/760)(1.013x105Pa) = 1.33x104 Pa
Note that this is a 'gauge pressure' since atmospheric pressure
acts inward on the arteries
P=Po
There is a gravitational gradient in this pressure.
The head is about 50 cm above the heart.
Phead =Pheart - ρgh = 1.33x104Pa - (1000kg/m3)(9.81m/s2)(0.5m) = 8.42x103Pa = 63 torr
The feet are about 1.30 m below the heart.
Pfeet = Pheart + ρgh = 1.33x104Pa + (1000kg/m3)(9.81 m/s2)(1.3m) = 2.60x104Pa = 195 torr
Blood has a viscosity of about 4x10-3 N.s/m2 (water has η= 1x10-3 N.s/m2)
(The actual situation is more complicated since the viscosity of blood depends upon the
hematocrit (i.e. the number of red blood cells) and also on the blood speed)
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For non-viscous fluid, the pressure is constant across a cylinder of flow tube of constant
diameter. For a real fluid, like blood, pressure drops along a flow tube. This is because
work must be done to overcome viscous forces in the fluid. Hence the 100 torr pressure
provided by the heart has the job of pushing blood all the way around the body.
The flow velocity for blood in the aorta (leaving the heart) can approach 1 m/s.
It depends upon the phase in the heart cycle and the heart output. v
r
P1
P2
L
Consider an artery of length 30 cm and radius 5mm which carries 8 l/m of blood. What is
the pressure drop along this artery?
Use Poiseille's Equation Q = (P1 - P2) πr4
8ηL
Then P1 - P2 = 8QηL/{πr4) = 8(8x10-3/60m3/s)(0.004N.s/m2)(0.3m)/{π(0.005m)4}
P1 - P2 = 652 Pa. Note that this is about 1/20 the total pressure available for blood flow
around the circulation (100 torr or 1.33×104 Pa
Vessel diameter range
Approximate blood pressure drop(%)
Arteries
< 1cm
10
Arterioles
< 20 µm
50
Capillaries veins
< 1cm
4 to 9 µm
30
10
Atherosclerotic Plaques:
In atherosclerotic plaques, buildup of calcium causes severe restriction in artery diameter. For
the same pressure gradient, a halving of the vessel radius results in the flow being restricted by a
factor 1/16! Many older people these days undergo calcium scoring tests whereby the heart is
imaged by x-rays in synchronization with the cardiac cycle. If there is a lot of calcium in the
coronary arteries, the person is advised to change lifestyle and perhaps undergo a cardiac
catheterisation procedure.
Section 15.6 (Knight) Read this chapter. we shall not cover it in detail in the lectures but we
shall make use of the definition of bulk modulus later when we cover the propagation velocities
of longitudinal waves.