Fluids
Buoyancy
Fluid Dynamics
Bernoulli’s Equation
Lana Sheridan
De Anza College
April 12, 2017
Last time
• introduction to static fluids
• pressure and depth
• Pascal’s principle
• measurements of pressure
ure at a given distance above level 1 in terms of the atmospheric pressure p1 at level 1
ssuming that the atmospheric density is uniform over that distance). For example, to
nd the atmospheric pressure at a distance d above level 1 in Fig. 14-3, we substitute
Warm Up Question
y1 ! 0,
p1 ! p0
and
y2 ! d,
Fig. 14-3 The
with depth h belo
according to Eq. 1
p2 ! p.
shows
hen withThe
r ! rfigure
air, we obtain
four containers of olive oil. Rank them according
! p0 "h,
rairgreatest
gd.
to the pressure at pdepth
first.1
CHECKPOINT 1
The figure shows four
containers of olive oil.
Rank them according
to the pressure at depth
h, greatest first.
h
(a)
(b)
(c)
A a, b, c, d
B a, d, c, b
C a, c, d, b
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 363.
(d)
ure at a given distance above level 1 in terms of the atmospheric pressure p1 at level 1
ssuming that the atmospheric density is uniform over that distance). For example, to
nd the atmospheric pressure at a distance d above level 1 in Fig. 14-3, we substitute
Warm Up Question
y1 ! 0,
p1 ! p0
and
y2 ! d,
Fig. 14-3 The
with depth h belo
according to Eq. 1
p2 ! p.
shows
hen withThe
r ! rfigure
air, we obtain
four containers of olive oil. Rank them according
! p0 "h,
rairgreatest
gd.
to the pressure at pdepth
first.1
CHECKPOINT 1
The figure shows four
containers of olive oil.
Rank them according
to the pressure at depth
h, greatest first.
h
(a)
(b)
(c)
A a, b, c, d
B a, d, c, b
C a, c, d, b
D All the same
1
←
Halliday, Resnick, Walker, 9th ed, page 363.
(d)
Overview
• buoyancy and Archimedes’ principle
• fluid dynamics
• the continuity equation
• Bernoulli’s equation
• Torricelli’s law
• other implications of Bernoulli’s equation
Barometers
Barometers are devices for measuring local atmospheric pressure.
Typically, simple barometers are filled with mercury, which is very
dense.
The weight of the mercury in the tube exerts the same pressure as
the surrounding atmosphere. On low pressure days, the level of the
mercury drops. On high pressure days it rises.
Mercury Barometer
The pressure at points A and B
is the same.
P!0
h
The pressure at B is P0 .
Above the mercury in the tube is
a vacuum, so pressure at A is
ρHg gh.
(ρHg = 13.6 kg/m3 )
P0
A B
Therefore, P0 = ρHg gh.
h ∝ P0
Pressure is sometimes quoted in
“inches of mercury”.
a
1
Diagrams from Serway & Jewett, 9th ed.
mmon baromedManometer
at one end
g. 14.6a).
The being measured, P, can be compared to atmospheric
The pressure
p of the
merpressure
P0 by measuring the height of the incompressible fluid in
pointtheA,U-shaped
due tube. a
to the atmoP0
uld move merherefore, P 0 5
f the mercury
h
P
olumn varies,
B
A
us determine
P 0 5 1 atm 5
b
If h ismpositive, P > P0 , if “negative”, P < P0 .
5 0.760
Figure 14.6 Two devices for
P − P0 is called measuring
the gauge pressure:
pressure. (a) a mercury
o be the presbarometer and (b) an open-tube
Buoyancy
Astronauts training in their spacesuits:
The total mass of NASA’s EMU (extravehicular mobility unit) is
178 kg. Why does training underwater make maneuvering in the
suits easier?
1
Picture from Hubblesite.org.
Buoyancy
The apparent weight of submerged objects is less than its full
weight.
For an object that would float, but is held underwater, its apparent
weight is negative!
There is an upward force on an object in a fluid called the buoyant
force.
Buoyancy
Why does this force exist? Where does it come from?
Buoyancy
Why does this force exist? Where does it come from?
We know pressure depends on depth, so an object that’s not
completely flat will have different pressure on different parts of its
surface.
Consider a rectangular object of height h and base area A with its
top edge at a depth d.
Buoyancy
Why does this force exist? Where does it come from?
We know pressure depends on depth, so an object that’s not
completely flat will have different pressure on different parts of its
surface.
Consider a rectangular object of height h and base area A with its
top edge at a depth d.
The force on each of the four sides will be equal.
The force on the bottom will be (P0 + ρg (h + d))A.
The force on the top will be (P0 + ρgd)A.
Buoyancy
Why does this force exist? Where does it come from?
We know pressure depends on depth, so an object that’s not
completely flat will have different pressure on different parts of its
surface.
Consider a rectangular object of height h and base area A with its
top edge at a depth d.
The force on each of the four sides will be equal.
The force on the bottom will be (P0 + ρg (h + d))A.
The force on the top will be (P0 + ρgd)A.
There will be a net upward force from the pressure
difference!
Buoyancy
How big will the upward force be?
Fbuoy = Fup − Fdown
Buoyancy
How big will the upward force be?
Fbuoy = Fup − Fdown
= (P0 A + ρg (h + d)A) − (P0 A + ρgdA)
= ρghA
= ρgVobj
because the volume of the submerged block is Vobj = hA.
Buoyancy
How big will the upward force be?
Fbuoy = Fup − Fdown
= (P0 A + ρg (h + d)A) − (P0 A + ρgdA)
= ρghA
= ρgVobj
because the volume of the submerged block is Vobj = hA.
Notice that ρVobj = mf , the mass of the displaced fluid.
Buoyancy and Archimedes’ Principle
Archimedes’ Principle
The buoyant force on an object is equal to the weight of the fluid
that the object displaces.
Logically, if a brick falls to the bottom of a pool it must push an
amount water equal to its volume up and out of the way.
Buoyancy and Archimedes’ Principle
For a fully submerged object the buoyant force is:
Fbuoy = ρf Vobj g
where ρf is the mass density of the fluid and Vobj is the volume of
the object.
ρf Vobj is the mass of the water moved aside by the object.
Buoyancy and Archimedes’ Principle
An object that floats will displace less fluid than its entire volume.
For a floating object:
Fbuoy = ρf Vsub g
where Vsub is the volume of the part of the object underneath the
fluid level only.
Sinking and Floating
Will a particular object sink or float in a particular fluid?
• If the object is less dense than the fluid it will float.
• If the object is more dense than the fluid it will sink.
• If the object and the fluid have the same density if will neither
float or sink, but drift at equilibrium.
Sinking and Floating
Since the relative density of the object to the fluid determines
whether it will sink or float, we sometimes use the notion of
specific gravity.
The specific gravity of an object relates its density to the density of
water (or occasionally other liquids):
Specific gravity, SG
of a sample is the ratio of its density to that of water.
SG =
Often referenced in brewing!
ρsample
ρwater
Sinking and Floating
A floating object displaces a mass of fluid equal to its own mass!
(Equivalently, a weight of fluid equal to its own weight.)
This also means that ρf Vsub = mobj .
Questions
Military ships are often compared by their displacements, the
weight (or mass, depending on context) of water they displace.
The USS Enterprise was an aircraft carrier (now decommissioned).
Displacement: 94,781 tonnes (metric tons), fully loaded.
1 tonne = 1000 kg
What is the mass of the fully loaded USS Enterprise in kgs?
2
Hewitt, page 246.
Questions
Military ships are often compared by their displacements, the
weight (or mass, depending on context) of water they displace.
The USS Enterprise was an aircraft carrier (now decommissioned).
Displacement: 94,781 tonnes (metric tons), fully loaded.
1 tonne = 1000 kg
What is the mass of the fully loaded USS Enterprise in kgs?
m = 94,781,000 kg
2
Hewitt, page 246.
Questions
Military ships are often compared by their displacements, the
weight (or mass, depending on context) of water they displace.
The USS Enterprise was an aircraft carrier (now decommissioned).
Displacement: 94,781 tonnes (metric tons), fully loaded.
1 tonne = 1000 kg
What is the mass of the fully loaded USS Enterprise in kgs?
m = 94,781,000 kg
Another Problem2
Your friend of mass 100 kg can just barely float in fresh water.
Calculate her approximate volume.
2
Hewitt, page 246.
Questions
Military ships are often compared by their displacements, the
weight (or mass, depending on context) of water they displace.
The USS Enterprise was an aircraft carrier (now decommissioned).
Displacement: 94,781 tonnes (metric tons), fully loaded.
1 tonne = 1000 kg
What is the mass of the fully loaded USS Enterprise in kgs?
m = 94,781,000 kg
Another Problem2
Your friend of mass 100 kg can just barely float in fresh water.
Calculate her approximate volume. 0.1 m3
2
Hewitt, page 246.
Question
Quick Quiz 14.43 You are shipwrecked and floating in the middle
of the ocean on a raft. Your cargo on the raft includes a treasure
chest full of gold that you found before your ship sank, and the
raft is just barely afloat. To keep you floating as high as possible in
the water, should you
(i) leave the treasure chest on top of the raft,
(ii) secure the treasure chest to the underside of the raft, or
(iii) hang the treasure chest in the water with a rope attached to
the raft?
(Assume throwing the treasure chest overboard is not an option
you wish to consider.)
A option (ii) is the best
B option (iii) is the best
C options (ii) and (iii) would be the same, better than (i)
D All would be the same
3
Serway & Jewett, page 425.
Question
Quick Quiz 14.43 You are shipwrecked and floating in the middle
of the ocean on a raft. Your cargo on the raft includes a treasure
chest full of gold that you found before your ship sank, and the
raft is just barely afloat. To keep you floating as high as possible in
the water, should you
(i) leave the treasure chest on top of the raft,
(ii) secure the treasure chest to the underside of the raft, or
(iii) hang the treasure chest in the water with a rope attached to
the raft?
(Assume throwing the treasure chest overboard is not an option
you wish to consider.)
A option (ii) is the best
B option (iii) is the best
C options (ii) and (iii) would be the same, better than (i)
D All would be the same
3
Serway & Jewett, page 425.
←
Buoyancy in Air
Buoyancy in air works the same way as in liquids:
Fbuoy = ρf Vobj g
If an object is less dense than air, it will float upwards.
However, in the atmosphere, the density of air varies with height.
Buoyancy in Air
1
Photo by Derek Jensen, Wikipedia.
Buoyancy in Air
By roughly how much is your weight reduced by the effects of the
air you are submerged in?
Buoyancy in Air
By roughly how much is your weight reduced by the effects of the
air you are submerged in?
Suppose you have a mass of 100 kg and volume of 0.1 m3 .
ρair = 1.20 kg/m3 (at room temperature and atmospheric
pressure)
Buoyancy in Air
By roughly how much is your weight reduced by the effects of the
air you are submerged in?
Suppose you have a mass of 100 kg and volume of 0.1 m3 .
ρair = 1.20 kg/m3 (at room temperature and atmospheric
pressure)
About 1.18 N.
Fluid Dynamics
When fluids are in motion, their behavior can be very complex.
We will only consider smooth, laminar flow.
Fluid Dynamics
When fluids are in motion, their behavior can be very complex.
We will only consider smooth, laminar flow.
Laminar flow is composed of streamlines that do not cross or curl
into vortices.
Streamline
The lines traced out by the velocities of individual particles over
time. Streamlines are always tangent to the velocity vectors in the
flow.
Fluid Dynamics
1
Image by Dario Isola, using MatLab.
Fluid Dynamics
A diagram of streamlines can be compared to Faraday’s
representation of the electric field with field lines. In fluids, the
vector field is instead a field of velocity vectors in the fluid at every
point in space and time, and streamlines are the field lines.
1
Image by Dario Isola, using MatLab.
Fluid Dynamics
We will make some simplifying assumptions:
1
the fluid is nonviscous, ie. not sticky, it has no internal
friction between layers
2
the fluid is incompressible, its density is constant
3
the flow is laminar, ie. the streamlines are constant in time
4
the flow is irrotational, there is no curl
Fluid Dynamics
We will make some simplifying assumptions:
1
the fluid is nonviscous, ie. not sticky, it has no internal
friction between layers
2
the fluid is incompressible, its density is constant
3
the flow is laminar, ie. the streamlines are constant in time
4
the flow is irrotational, there is no curl
In real life no fluids actually have the first two properties.
Flows can have the second two properties, in the right conditions.
Question(14-25)
(kg/s).The
Equation
14-25 a pipe and gives the volume flow rate (in cm3 /s)
figure shows
14-15 each
second
must of flow for all but one section. What are the
and the
direction
cond. volume flow rate and the direction of flow for that section?
(Assume that the fluid in the pipe is an ideal fluid.)
2
/s) and the diw rate and the
4
5
8
A 11 cm3 /s, outward
ple Problem
B 13 cm3 /s, outward
3
C 3 cm /s, inward
am narrows
as it falls
om
D cannot be determined
1
KEY IDEA
Halliday, Resnick, Walker, 9th ed, page 373.
6
4
Question(14-25)
(kg/s).The
Equation
14-25 a pipe and gives the volume flow rate (in cm3 /s)
figure shows
14-15 each
second
must of flow for all but one section. What are the
and the
direction
cond. volume flow rate and the direction of flow for that section?
(Assume that the fluid in the pipe is an ideal fluid.)
2
/s) and the diw rate and the
4
A 11 cm3 /s, outward
ple Problem
B 13 cm3 /s, outward
5
8
←
cm3 /s,
C 3
am narrows
as it inward
falls
om
D cannot be determined
1
KEY IDEA
Halliday, Resnick, Walker, 9th ed, page 373.
6
4
Bernoulli’s Principle
A law discovered by the 18th-century Swiss scientist, Daniel
Bernoulli.
Bernoulli’s Principle
As the speed of a fluid’s flow increases, the pressure in the fluid
decreases.
This leads to a surprising effect: for liquids flowing in pipes, the
pressure drops as the pipes get narrower.
Bernoulli’s Principle
Why should this principle hold? Where does it come from?
1
Something similar can be argued for compressible fluids also.
Bernoulli’s Principle
Why should this principle hold? Where does it come from?
Actually, it just comes from the conservation of energy, and an
assumption that the fluid is incompressible.4
Consider a fixed volume of fluid, V .
In a narrower pipe, this volume flows by a particular point 1 in
time ∆t.
However, it must push the same volume of fluid past a point 2 in
the same time. If the pipe is wider at point 2, it flows more slowly.
1
Something similar can be argued for compressible fluids also.
Bernoulli’s Principle
V = A1 v1 ∆t
also, V = A2 v2 ∆t
Bernoulli’s Principle
V = A1 v1 ∆t
also, V = A2 v2 ∆t
This means
A1 v1 = A2 v2
The “Continuity equation”.
Bernoulli’s Equation
Bernoulli’s equation is just the conservation of energy for this fluid.
The system here is all of the fluid in the pipe shown.
Both light blue cylinders of fluid have the same volume, V , and
same mass m.
We imagine that in a time ∆t, volume V of fluid enters the left
end of the pipe, and another V exits the right.
At t ! 0, fluid in the blue
Bernoulli’s Equation
is moving past
It makes sense that portion
the energy
of the
S fluid might change: the fluid
point 1 at velocity v1.
is moved along, and some is lifted up.
A1
S
v1
"x1
Point 2
a
Point 1
A2
S
v2
"x2
Afteron
a time
"t,
How does it change? Depends
theinterval
work done:
the fluid in the blue
Wportion
= ∆Kis moving
+ ∆U past
S
The path taken by
velocity of the particl
A set of streamlines
particles cannot flow
lines would cross one
Consider ideal flui
ure 14.16. Let’s focus
shows the segment at
8a and 14.18b are equal because the fluid is incompressible.) This work
because the force on the segment of fluid is to the left and the displacepoint of application of the force is to the right. Therefore, the net work
e segment by these forces in the time interval Dt is
Bernoulli’s Equation
The work done is the sum of the
work done on each end of the
fluid by more fluid that is on
either side of it:
W 5 (P 1 2 P 2)V
The pressure at
point 1 is P1.
P1A1 ˆi
W
S
y1
!x1
v1
Point 2
The pressure at
point 2 is P2.
a
= P1 A1 ∆x1 − P2 A2 ∆x2
S
v2
"P2A2 ˆi
Point 1
!x2
y2
b
1
= F1 ∆x1 − F2 ∆x2
Diagram from Serway & Jewett.
(The “environment fluid” just to
the right of the system fluid does
negative work on the system as it
must be pushed aside by the
system fluid.)
Bernoulli’s Equation
Notice that V = A1 ∆x1 = A2 ∆x2
W
= P1 A1 ∆x1 − P2 A2 ∆x2
= (P1 − P2 )V
Conservation of energy:
W = ∆K + ∆U
Bernoulli’s Equation
Notice that V = A1 ∆x1 = A2 ∆x2
W
= P1 A1 ∆x1 − P2 A2 ∆x2
= (P1 − P2 )V
Conservation of energy:
W = ∆K + ∆U
(P1 − P2 )V
=
1
m(v22 − v12 ) + mg (h2 − h1 )
2
Bernoulli’s Equation
Notice that V = A1 ∆x1 = A2 ∆x2
W
= P1 A1 ∆x1 − P2 A2 ∆x2
= (P1 − P2 )V
Conservation of energy:
W = ∆K + ∆U
(P1 − P2 )V
=
1
m(v22 − v12 ) + mg (h2 − h1 )
2
Dividing by V :
1 2
ρv + ρg (h2 − h1 )
2 2
1
= P2 + ρv22 + ρgh2
2
P1 − P2 =
1
P1 + ρv12 + ρgh1
2
Bernoulli’s Equation
1
1
P1 + ρv12 + ρgh1 = P2 + ρv22 + ρgh2
2
2
This expression is true for any two points along a streamline.
Therefore,
1
P + ρv 2 + ρgh = const
2
is constant along a streamline in the fluid.
This is Bernoulli’s equation.
Bernoulli’s Equation
1
P + ρv 2 + ρgh = const
2
Even though we derived this expression for the case of an
incompressible fluid, this is also true (to first order) for
compressible fluids, like air and other gases.
The constraint is that the densities should not vary too much from
the ambient density ρ.
Bernoulli’s Principle from Bernoulli’s Equation
For two different points in the fluid, we have:
1
1 2
ρv1 + ρgh1 + P1 = ρv22 + ρgh2 + P2
2
2
Bernoulli’s Principle from Bernoulli’s Equation
For two different points in the fluid, we have:
1
1 2
ρv1 + ρgh1 + P1 = ρv22 + ρgh2 + P2
2
2
Suppose the height of the fluid does not change, so h1 = h2 :
1 2
1
ρv1 + P1 = ρv22 + P2
2
2
Bernoulli’s Principle from Bernoulli’s Equation
For two different points in the fluid, we have:
1
1 2
ρv1 + ρgh1 + P1 = ρv22 + ρgh2 + P2
2
2
Suppose the height of the fluid does not change, so h1 = h2 :
1 2
1
ρv1 + P1 = ρv22 + P2
2
2
If v2 > v1 then P2 < P1 .
Bernoulli’s Principle
However, rom the continuity equation A1 v1 = A2 v2 we can see
that if A2 is smaller than A1 , v2 is bigger than v1 .
So the pressure really does fall as the pipe contracts!
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of pipe
according to the volume flow rate through them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
1
Flow
2
3
A 4, 3, 2, 1
B 1, (2 and 3), 4
C 4, (2 and 3), 1
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 375.
4
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of pipe
according to the volume flow rate through them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
1
Flow
2
3
A 4, 3, 2, 1
B 1, (2 and 3), 4
C 4, (2 and 3), 1
D All the same
1
←
Halliday, Resnick, Walker, 9th ed, page 375.
4
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of
pipe according to the flow speed v through them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
1
Flow
2
3
A 4, 3, 2, 1
B 1, (2 and 3), 4
C 4, (2 and 3), 1
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 375.
4
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of
pipe according to the flow speed v through them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
1
2
Flow
3
A 4, 3, 2, 1
B 1, (2 and 3), 4
←
C 4, (2 and 3), 1
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 375.
4
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of pipe
according to the water pressure P within them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
1
Flow
2
3
A 4, 3, 2, 1
B 1, (2 and 3), 4
C 4, (2 and 3), 1
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 375.
4
qs. 14-32, 14-33, and 14-34 yields
"V(Question
y2 # y1) # "V(p2 # p1) ! 12 % "V(v22 # v21).
Water flows
smoothly
through
the pipe
shown
in prove.
the figure,
earrangement,
matches
Eq. 14-28,
which
we set
out to
descending in the process. Rank the four numbered sections of pipe
according to the water pressure P within them, greatest first.
4
hly through the
gure, descending
k the four numpe according to
rate RV through
peed v through
water pressure p
first.
A 4, 3, 2, 1
1
Flow
2
3
←
B 1, (2 and 3), 4
C 4, (2 and 3), 1
D All the same
1
Halliday, Resnick, Walker, 9th ed, page 375.
4
Torricelli’s Law from Bernoulli’s Equation
Bernoulli’s equation can also be used to predict the velocity of
streams of water from holes in a container at different depths.
w blow more strongly and watch the increased pressure dif-
Torricelli’s Law from Bernoulli’s Equation
The liquid at point 2 is at rest, at a height y2 and pressure P.
At point 1 is leaves with a velocity v1 , at a height y1 and pressure
P0 .
in its side at a distance
the atmosphere, and its
e air above the liquid is
uid as it leaves the hole
Point 2 is the surface
of the liquid.
A2
P
h
Point 1 is
the exit
point of
the hole.
A1
y2
her. When the hole is
e pressure P at the top
speed. If the pressure
e extinguisher must be
S
P0
y1
v1
Figure 14.20 (Example 14.9)
A liquid leaves a hole in a tank
0 at
1 2speed v1.
1 >
2
ρv1 + ρgy1 + P0 = ρv2 + ρgy2
2
2
+P
Torricelli’s Law from Bernoulli’s Equation
in its side at a distance
the atmosphere, and its
e air above the liquid is
uid as it leaves the hole
Point 2 is the surface
of the liquid.
A2
P
h
Point 1 is
the exit
point of
the hole.
A1
y2
her. When the hole is
e pressure P at the top
speed. If the pressure
e extinguisher must be
S
y1
Figure 14.20
P0
v1
(Example 14.9)
1A liquid
leaves a hole in a tank at
ρv12 +
ρgy1 + P0 = ρgy2 + P
speed
v
.
1
2
Rearranging, and using y = h2 − h1 ,
s
2(P − P0 )
v1 =
+ 2gh
ρ
Torricelli’s Law from Bernoulli’s Equation
Notice that if the container is open to the air (P = P0 ), then the
speed of each jet is
p
v = 2gh
where h is the depth of the hole below the surface.
Implications of Bernoulli’s Principle
Bernoulli’s principle also explains why in a tornado, hurricane, or
other extreme weather with high speed winds, windows blow
outward on closed buildings.
Implications of Bernoulli’s Principle
Bernoulli’s principle also explains why in a tornado, hurricane, or
other extreme weather with high speed winds, windows blow
outward on closed buildings.
The high windspeed outside the building corresponds to low
pressure.
The pressure inside remains higher, and the pressure difference can
break the windows.
It can also blow off the roof!
It makes sense to allow air a bit of air to flow in or out of a
building in extreme weather, so that the pressure equalizes.
Implications of Bernoulli’s Principle
Bernoulli’s principle can help explain why airplanes can fly.
Air travels faster over the top of the wing, reducing pressure there.
That means the air beneath the wing pushes upward on the wing
more strongly than the air on the top of the wing pushes down.
This is called lift.
1
Diagram from HyperPhysics.
Air Flow over a Wing
In fact, the air flows over the wing much faster than under it: not
just because it travels a longer distance than over the top.
This is the result of circulation of air around the wing.
1
Diagram by John S. Denker, av8n.com.
Air Flow over the Top of the Wing: Bound Vortex
A starting vortex trails the wing. The bound vortex appears over
the wing.
Those two vortices counter rotate because angular momentum is
conserved.
The bound vortex is important to establish the high velocity of the
air over the top of the wing.
1
Image by Ludwig Prandtl, 1934, using water channel & aluminum particles.
Wingtip Vortecies
Other vortices also form at the ends of the wingtips.
1
Photo by NASA Langley Research Center.
Vortices around an Airplane
1
Diagram by John S. Denker, av8n.com.
Airflow at different Angles of Attack
1
Diagram by John S. Denker, av8n.com.
Implications of Bernoulli’s Principle
A stall occurs when turbulence behind the wing leads to a sudden
loss of lift.
The streamlines over the wing detach from the wing surface.
This happens when the plane climbs too rapidly and can be
dangerous.
1
Photo by user Jaganath, Wikipedia.
Implications of Bernoulli’s Principle
Spoilers on cars reduce lift and promote laminar flow.
1
Photo from http://oppositelock.kinja.com.
Implications of Bernoulli’s Principle
Wings on racing cars are inverted airfoils that produce downforce
at the expense of increased drag.
This downforce increases the maximum possible static friction
force ⇒ turns can be taken at higher speed.
1
Photo from http://oppositelock.kinja.com.
Implications of Bernoulli’s Principle
A curveball pitch in baseball
also makes use of Bernoulli’s
principle.
The ball rotates as it moves
through the air.
Its rotation pulls the air
around the ball, so the air
moving over one side of the
ball moves faster.
This causes the ball to deviate
from a parabolic trajectory.
1
Diagram by user Gang65, Wikipedia.
Summary
• buoyancy and Archimedes’ principle
• fluid dynamics
• the continuity equation
• Bernoulli’s equation
• Torricelli’s law
• other implications of Bernoulli’s equation
Test Wednesday, April 19, in class.
Collected Homework due Wednesday, April 19.
(Uncollected) Homework Serway & Jewett:
• Ch 14, onward from page 435, OQs: 3, 5, 7, 9, 13; CQs: 5, 9,
14; Probs: 25, 27, 29, 35, 36, 43, 49, 53, 60, 65, 71, 77, 85