Last time we introduced the physics of fluids. This is a very complicated subject. Entire
books are dedicated to this topic. The mathematics can be very difficult.
We talked about basic physical definitions and we dealt with fluids at rest. Today
we talk about 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.
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.
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
or
P1  gy1  12 v1  constant
2
Hopefully, this reminds you of
Wnc  mgy1  12 mv1  mgy2  12 mv2
2
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 tofind the speed of the air at point B.
Solution Relate the air speeds at points A and B.
A
d B2
AB vB  AAv A , so v A  B vB 
vB  9vB .
AA
(d B 3) 2
Find the speed of the air at point B. Note that y A  yB .
PB  12 vB  gyB  PA  12 v A  gyA
2
2
PB  12 vB  PA  12 v A
2
1
2
2
vB 2  PA  PB  12  (9vB ) 2
 PA  PB  812 vB
2
40 vB  PA  PB
2
The pressure difference is related to the height difference in the manometer
PA  PB  W gh
Subsituting
40 vB  W gh
2
vB 
W gh
(1000 kg/m 3 )(9.8 m/s 2 )(1.75  10 2 cm)

 1.82 m/s
40 
40(1.29 kg/m 3 )
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. 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
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  1m 
5
 (1.20 kg m3 ) 
(2000
ft
)
 
 


  5.0 10 N .
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 needed 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 , measured in Pa-s. Other units are poise (pwaz) and cP.
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.
Solution Poiseuille’s law is
V  P / L 4

r
t
8 
Solving for P
V  P / L 4

r
t
8 
V 8 L
P 
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 m 3 /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
Unsteady fluid flow. 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.
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
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.
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.
Chapter 10 Elasticity and Oscillations
In this problem
3m
4m
960 N
we assume that the beam is perfectly rigid. The weight hanging from the end will not
effect it. This assumption is not true.
In this chapter we study the physical properties of solids. We studied fluids in the
last chapter.
Some Vocabulary
 A deformation is a change in the size or shape of an object. The deformation
may be too small to see if the object is strong and the deforming force is small.
 When the forces are removed from an elastic object, the object returns to its
original shape and size.
A More General Hooke’s Law
Suppose a force is applied to both ends of a wire.
How does the amount of elongation L depend on the original length of the wire?
If F causes L to increase by L, the same force F should cause 2L to increase by
2L.
Since the amount elongation depends on the original length of the wire, we
defined the strain as
strain 
L
L
Does changing the cross-sectional area of the wire change the effect of F?
A thicker cable could be considered to be made of several thinner cables placed
side by side. The force would be spread over a bigger area and it would be less effective.
We define the stress as
stress 
F
A
Stress is measured in N/m2 or Pa.
We rewrite Hooke’s law in terms of stress and strain. In general
stress  strain
In terms of the definitions
F
L
Y
A
L
The constant of proportionality is called the elastic modulus or Young’s modulus. If
has the same units as stress.
Hooke’s law holds up to a maximum stress called the proportional limit.
Beyond the Proportional Limit
If the stress exceeds the proportional limit, the strain is no longer proportional to the
stress. The solid will return to its original shape when the stress is removed.
Prince Rupert’s drop: http://www.youtube.com/watch?v=xe-f4gokRBs