Chapter 9 Fluids
States of Matter - Solid, liquid, gas.
 Fluids (liquids and gases) do not hold their shapes.
 In many cases we can think of liquids as being incompressible.
 Liquids do not change their volume (appreciably) when they are heated.
 Gases do not have a definite volume or shape.
Pressure
A fluid particle collides with the surface. The change in momentum is caused by an impulse that
acts to the right on the particle. By Newton’s third law, the particle pushes to the left on the
surface.
Definition of pressure
Pav 
F
A
Surprisingly, pressure is a scalar and not a vector.
Pressure is measured in N/m2 which is also called a pascal
(Pa). There are a zillion other units: atmosphere, lbs/square
inch, torr, bar, etc.
The atmosphere exerts pressure. The pressure of one
atmosphere is
1atm  1.013  105 Pa
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Pascal’s Principle
“A change in pressure at any point in a confined
fluid is transmitted everywhere throughout the
fluid.” (p. 320) Pascal’s principle is a
consequence of the incompressibility of fluids.
Pascal’s principle is the basis of hydraulics.
Hydraulics are the most effective way to transmit
a force.
Notice that the displacement in the narrow tube
is much greater that the displacement in the wide
tube. We will return to this observation later.
The force on the right (F1) changes the pressure in the fluid. By Pascal’s principle, the change in
pressure is transmitted throughout the fluid.
P1  P2
F1 F2

A1 A2
A 
F2   2  F1
 A1 
If A2>A1, then the F2>F1.
Modern Marvels: https://www.youtube.com/watch?v=74VEL22U_0o
Excavator advertisement: https://www.youtube.com/watch?v=VBMUvAUPTGM
9-2
Density is the mass per unit volume. It is defined as

m
V
Density is measured in kg/m3.
As we descend down into a fluid the amount of fluid above us increases. That additional fluid
pushes down and the pressure increases with depth. At depth d the pressure has increased
P2  P1  gd
Measuring Pressure
A manometer consists of a U-shaped tube containing some mercury. When both sides are open
to the atmosphere, the height in both arms are the same. When one side is connected to the
pressure to be measured, the heights are different.
The pressures at B and B' are the same,
PB  PC  gd
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Usually, the manometer is open to the atmosphere. It will measure pressures relative to
atmospheric. The gauge pressure is the pressure relative to an atmosphere.
Pgauge  Pabs  Patm
“Blood pressure is measured with a sphygmomanometer. The oldest kind of
sphygmomanometer consists of a mercury manometer on one side attached to a closed bag—the
cuff. The cuff is wrapped around the upper arm at the level of the heart and is then pumped up
with air. The manometer measures the gauge pressure of the air in the cuff.
“At first, the pressure in the cuff is higher than the systolic pressure—the maximum pressure in
the brachial artery that occurs when the heart contracts. The cuff pressure squeezes the artery
closed and no blood flows into the forearm. A valve on the cuff is then opened to allow air to
escape slowly. When the cuff pressure decreases to just below the systolic pressure, a little
squirt of blood flows past the constriction in the artery with each heartbeat. The sound of
turbulent blood flow past the constriction can be heard through the stethoscope.
“As air continues to escape from the cuff, the sound of blood flowing through the constriction in
the artery continues to be heard. When the pressure in the cuff reaches the diastolic pressure in
the artery—the minimum pressure that occurs when the heart muscle is relaxed—there is no
longer a constriction in the artery, so the pulsing sounds cease. The gauge pressures for a
healthy heart are nominally around 120 mm Hg (systolic) and 80 mm Hg (diastolic).” (p. 327)
Buoyant Force
When an object is submerged in a fluid, the fluid pushes up on the object. The buoyant force is
given by
FB  gdA  gV
Archimedes’ principle
A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight
of the volume of fluid displaced by the object.
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We still need to use free body diagrams! The force F1 is the force of
the fluid above the block pushing down and the force F2 is the force of
the fluid below the block pushing up. The buoyant force is



FB  F2  F1
FBy  F2 y  F1 y
FB  F2  F1
The specific gravity is defined as the ratio of the density of the
material to the density of water.
S.G. 

 water
If S.G. < 1, the object will float. If S.G. > 1, the object sinks.
Archimedes and the golden crown:
http://www.youtube.com/watch?v=hIYdxQuzb60
The story: http://longlongtimeago.com/once-upon-a-time/great-discoveries/eureka-the-story-ofarchimedes-and-the-golden-crown/
We have completed our study of fluids at rest. Now we consider fluids in motion.
Fluid Flow
A fluid moving past a surface can exert a viscous force against the surface. This is similar to the
frictional force of an object sliding over a surface. We will start by assuming the viscous force
to be small.
When flow is steady, the velocity at any point is constant in time. The flow may not be the same
everywhere.
Steady flow is laminar. The streamlines are clearly defined.
As we have done many times this semester, we assume the ideal case first. An ideal fluid is
incompressible, undergoes laminar flow, and has no viscosity.
9-5
The continuity equation
Since the fluid is incompressible, the fluid flows faster in the narrow portions of the pipe.
The mass flow rate is defined as
m
 Av
t
The volume flow rate is
V
 Av
t
The continuity equation for an incompressible fluid equates the volume flow rates past two
different points,
A1v1  A2v2
The continuity equation is a consequence of conservation of mass.
Bernoulli’s Equation
Using energy ideas, the pressure of the fluid in a constriction cannot be the same as the pressure
before or after the constriction. For horizontal flow the speed is higher where the pressure is
lower. This is called the Bernoulli effect.
“The Bernoulli effect can seem counterintuitive at first; isn’t rapidly moving fluid at high
pressure? For instance, if you were hit with the fast-moving water out of a firehose, you would
be knocked over easily. The force that knocks you over is indeed due to fluid pressure; you
would justifiably conclude that the pressure was high. However, the pressure is not high until
9-6
you slow down the water by getting in its way. The rapidly moving water in the jet is, in fact,
approximately at atmospheric pressure (zero gauge pressure), but when you stop the water, its
pressure increases dramatically.” (p. 334)
For a more general situation where the pipe is not
horizontal, we can use energy considerations to derive
Bernoulli’s equation. (The derivation is given on
page 335. I will only quote the result.)
P1  gy1  12 v1  P2  gy2  12 v2
2
2
or
P  gy  12 v 2  constant
Hopefully, this reminds you of
Wnc  mgy1  12 mv1  mgy2  12 mv2
2
2
Bernoulli’s equation relates the pressure, flow speed, and height at two points in an ideal fluid.
Problem 50. Suppose air, with a density
of 1.29 kg/m3 is flowing into a Venturi
meter. The narrow section of the pipe at
point A has a diameter that is 1/3 of the
diameter of the larger section of the pipe
at point B. The U-shaped tube is filled
with water and the difference in height
between the two sections of pipe is 1.75 cm. How fast is the air moving at point B?
Strategy Use the continuity equation to relate the speeds at points A and B. Then, use
Bernoulli’s equation to find the speed of the air at point B.
Solution Relate the air speeds at points A and B.
AB vB  AAv A , so v A 
AB
d B2
vB 
vB  9vB .
AA
(d B 3)2
y A  yB .
Find the speed of the air at point B. Note that
PB  12 v B  gy B  PA  12 v A  gy A
2
2
PB  12 v B  PA  12 v A
2
1
2
2
v B 2  PA  PB  12  (9v B ) 2
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1
2
v B 2  PA  PB  812 v B 2
40 v B  PB  PA
2
The pressure difference causes the height difference in the manometer
PB  PA  W gh
Substituting
40v B  W gh
2
vB 

 W gh
40 
(1000 kg/m 3 )(9.8 m/s 2 )(1.75  10  2 cm )
40(1.29 kg/m 3 )
 1.82 m/s
Problem 46. In a tornado or hurricane, a roof may tear away from the house because of a
difference in pressure between the air inside and the air outside. Suppose that air is blowing
across the top of a 2000 ft2 roof at 150 mph. What is the magnitude of the force on the roof?
P1
P2
Strategy Use Bernoulli’s equation to find the pressure difference at the roof.
Solution Let the region above the roof be labeled 1. The region inside the house near the roof is
labeled 2. Assume the air under the roof is still.
P1  12 v1  gy1  P2  12 v2  gy2
2
2
P1  12 v1  gy1  P2  gy2
2
Now y1 is almost equal to y2 and we can assume that the difference in height has negligible effect
on the pressure.
P1  12 v1  P2
2
P2  P1  12 v1
9-8
2
Which side is at the higher pressure, the inside or outside? The magnitude of the force on the
roof is
1
 v 2A
2 air 1
2
2
2
2
1
 150 mi   1 h   1609 m 
2  1 m   5.0 105 N .
 (1.20 kg m3 ) 
 
 
 (2000 ft ) 

2
 1 h   3600 s   1 mi 
 3.281 ft 
F  PA 
which is equal to 56 tons!
Viscosity
Bernoulli’s equation ignores viscosity. When real
fluids flow, the different layers of fluid drag
against each other. A pressure difference is needed
to maintain the flow. This is similar to needing a
constant force to overcome kinetic friction. Fluid
layers further away from the wall flow faster than
those close to the wall.
Poiseuille’s Law
The volume flow rate of viscous fluid through a
horizontal cylindrical pipe depends on
 Pressure gradient
P
L


Viscosity. The higher the viscosity, the
lower the flow rate.
Radius of the pipe.
The French physician Poiseuille (pwahzoy) formulated his law after studying blood flow
V  P / L 4

r
t 8 
Viscosity is  (Greek letter eta), measured in Pas. Other units are poise (pwaz) (P) and cP.
Conversions are 1 P = 0.1 Pas and 1 cP = 0.001 Pas.
Problem. Water flows through a pipe of radius 8.50 mm. The viscosity of water is 1.005 cP. If
the flow speed at the center is 0.200 m/s and the flow is laminar, find the pressure drop due to
viscosity along a 3.00 m section of pipe.
Strategy Use Poiseuille’s Law to find the pressure drop.
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Solution Poiseuille’s law is
V  P / L 4

r
t 8 
Solving for P
P 
V 8 L
t  r 4
The volume flow rate is related to the area of the tube and the speed of the flow (see the
continuity equation)
V
 Av  r 2v   (8.50 103 m) 2 (0.200 m/s )  4.54 105 m3 /s
t
Viscosity is not in the correct units.  = 1.005×10-3 Pas. The pressure difference is
P 
V 8 L
8 (1.005 103 Pa  s)(3 m)
5
3

(
4
.
54

10
m
/s)
 67 Pa
t  r 4

(8.50 103 m) 4
Turbulence
Turbulence is unsteady fluid flow, not laminar flow. In turbulent flow, swirling vortices appear.
The vortices are not stationary and they move with the fluid. The velocity of the fluid flow can
change direction and magnitude in an uncontrolled way.
Why do golf balls have dimples?
From HowStuffWorks.com: The reason why golf balls have dimples is a story of natural selection. Originally,
golf balls were smooth; but golfers noticed that older balls that were beat up with nicks, bumps and slices in the
cover seemed to fly farther. Golfers, being golfers, naturally gravitate toward anything that gives them an
advantage on the golf course, so old, beat-up balls became standard issue.
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At some point, an aerodynamicist must have looked at this problem and realized that the nicks and cuts were
acting as “turbulators” – they induce turbulence in the layer of air next to the ball (the “boundary layer”). In
some situations, a turbulent boundary layer reduces drag.
If you want to get deeper into the aerodynamics, there are two types of flow around an object: laminar and
turbulent. Laminar flow has less drag, but it is also prone to a phenomenon called “separation.” Once
separation of a laminar boundary layer occurs, drag rises dramatically because of eddies that form in the gap.
Turbulent flow has more drag initially but also better adhesion, and therefore is less prone to separation.
Therefore, if the shape of an object is such that separation occurs easily, it is better to turbulate the boundary
layer (at the slight cost of increased drag) in order to increase adhesion and reduce eddies (which means a
significant reduction in drag). Dimples on golf balls turbulate the boundary layer.
The dimples on a golf ball are simply a formal, symmetrical way of creating the same turbulence in the
boundary layer that nicks and cuts do.
Baseball: http://www.youtube.com/watch?v=oph9BP4lKjs
Magnus effect. Tennis: top spin, slice, side spin; Baseball: curveballs; Golf: slice, hook
http://www.youtube.com/watch?v=Fk2xU8pEIlI&index=2&list=PLB0A49B27192A26D5
Viscous Drag
An object moving through a fluid experiences drag. Clearly the drag depends on the viscosity of
the fluid, the speed of the object, and its size.
Viscous drag is very complicated, but there is a well understood example. For a sphere of radius
r traveling at appropriate speed v (so there is no turbulence), Stoke’s Law holds
FD  6rv
When the viscous drag is equal to a falling object’s weight, the object reaches terminal velocity.
This is how parachutes work.
Surface Tension
The surface of a liquid has special properties not
associated with the interior of the liquid. The
surface acts like a stretched membrane under
tension.
The surface tension () of a liquid is the force
per unit length with which the surface pulls on
its edge.
Soaps break up the surface tension so that water
can reach into small places and clean them.
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“The high surface tension of water is a hindrance in the
lungs. The exchange of oxygen and carbon dioxide
between inspired air and the blood takes place in tiny sacs
called alveoli, 0.05 to 0.15 mm in radius, at the end of the
bronchial tubes. If the mucus coating the alveoli had the
same surface tension as other body fluids, the pressure
difference between the inside and outside of the alveoli
would not be great enough for them to expand and fill
with air. The alveoli secrete a surfactant that decreases
the surface tension in their mucous coating so they can
inflate during inhalation.” (p. 344)
Bubbles
A gas bubble inside a fluid is in equilibrium between the surface tension trying to shrink the
bubble and the pressure inside trying to expand it. The pressure inside the bubble must be
greater than the fluid pressure outside. It can be shown that the pressure difference, P, is
P  Pin  Pout 
2
r
As the bubbles rise to the surface from the bottom, they expand. Pin reduces with the expansion
and r increases as well. The difference, Pin−Pout, becomes smaller.
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