Transcript SPH3UW - The Burns Home Page

Fluids


At ordinary temperature, matter exists in one of three
states (5 if you include plasma and Bose-Einstein
condensate).
 Solid - has a shape and forms a surface
 Liquid - has no shape but forms a surface
 Gas - has no shape and forms no surface
What do we mean by “fluids”?
 Fluids are “substances that flow”…. “substances
that take the shape of the container”
 Atoms and molecules are free to move.
 No long range correlation between positions.
Fluids

What parameters do we use to describe fluids?
 Density
Mass
m


Volume V
units :
kg/m3 = 10-3 g/cm3
If any substance is
more dense than
water, it will sink
(water) = 1.000 x 103 kg/m3
= 1.000 g/cm3
(ice)
= 0.917 x 103 kg/m3
= 0.917 g/cm3
(air)
= 1.29 kg/m3
= 1.29 x 10-3 g/cm3
(Hg)
= 13.6 x103 kg/m3
= 13.6 g/cm3
Fluids

What parameters do we use to describe fluids?
 Pressure
Force F
P

Area
A
units :
1 N/m2
1 bar
1 mbar
1 torr
= 1 Pa (Pascal)
= 105 Pa
= 102 Pa
= 133.3 Pa
1 atm = 1.013 x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/ in2 (=PSI)
 Any force exerted by a fluid is perpendicular to a surface of
contact, and is proportional to the area of that surface.
 Force (a vector) in a fluid can be expressed in terms of
hydrostatic pressure (a scalar) as:
n
F  pAnˆ
A
Practice
An 80 kg metal cylinder, 2.0 m long and with each end
having an area of 25 cm2. It stands vertically on one of the
ends. What pressure does the cylinder exert on the floor?
25cm2
80kg
25cm2
Solution
An 80 kg metal cylinder, 2.0 m long and with each end
having an area of 25 cm2. It stands vertically on one of the
ends. What pressure does the cylinder exert on the floor?
F
p
A
mg

A

N
80kg   9.8 
kg 


2.5 103 m2
 3.1105 Pa
80kg
25cm2  5cm  5cm
  5 102 m    5 102 m 
 25 104 m2
 2.5 103 m2
Practice
A waterbed is 2m on a side and 30cm deep. Find
a) Its weight.
b) The pressure the waterbed exerts on the floor.
Waterbed
Solution
A waterbed is 2m on a side and 30cm deep. Find
a) Its weight.
b) The pressure the waterbed exerts on the floor.
Water density is 1000 kg/m3
M

V
M  V
Let’s rearrange
for Mass, M
Now, we write weight in
terms of density ,
volume and g
W  Mg
 Vg
kg 
m


 1000 3   2.00m  2.00m  0.300m   9.80 2 
m 
s 


 1.18 104 N
Solution
A waterbed is 2m on a side and 30cm deep. Find
a) Its weight.
b) The pressure the waterbed exerts on the floor.
P
F
A
The force is the weight
of the bed.
1.18  10 4 N

2m  2m
N
 2.95 103 2
m
Surface area of contact
between the floor and the
bed.
Weight is 1.18 x 104 N
Specific Gravity
An old-fashioned but still very common way of expressing
the density of a substance is to relate it to the density of
water.
The ratio of the density of any substance to the density
of water is known as the specific gravity of the
substance.
substance
sp.gr 
 water
If the specific gravity is
less than 1, it will float in
water, if it is greater than
1, it will sink.
Act 3
A cork has a volume of 5 cm3 and weighs 0.03N. What is
the specific gravity of cork?
Act 4 Solution
A cork has a volume of 5 cm3 and weighs 0.03N. What is
the specific gravity of cork?
First let’s find the mass
w 0.03 N
m 
 3.06  103 kg
g 9.8 N
kg
Now for the density
3
m 3.06 103 kg  100cm 
kg
cork  

  612.245 3
3
V
5cm
m
 1m 
Finally the specific gravity
sp.gr 
cork
 water
kg
3
m

 0.61
kg
1000 3
m
612.245
Pressure vs. Depth
Incompressible Fluids (liquids)
The Bulk Modulus of a substance measures the substance’s
resistance to uniform compression. It is defined as the pressure
increase needed to cause a relative decrease in volume.
 When the pressure is much less than the bulk
modulus of the fluid, we treat the density as a
constant independent of pressure:
1.
LIQUID: incompressible (density almost
constant)
2.
GAS: compressible (density depends a lot on
pressure)
 For an incompressible fluid, the density IS the same
everywhere, but the pressure is NOT
Bulk Modulus
B
P
( V / V )
Pressure vs. Depth
For a fluid in an open container:
• pressure same at a given depth
independent of the container
y
p(y)
• fluid level is the same everywhere in a
connected container (assuming no surface
forces)
• Why is this so? Why, in equilibrium,
does the pressure below the surface
depend only on depth?
• Imagine a tube that would connect two regions at the same depth.
•If the pressures were different, fluid would flow in the tube!
• However, if fluid did flow, then the system was NOT in equilibrium,
since no equilibrium system will spontaneously leave equilibrium.
•The horizontal forces acting at the left and right side of a cylinder at
specific depth are PAL and PAR. Since the cylinder is in equilibrium PAL=
PAR Therefore The pressure, P at either side is the same.
ACT 5
1. What happens with two different fluids?? Consider a U
tube containing liquids of density 1 and 2 as shown:
 Compare the densities of the liquids:
A) 1 < 2
B) 1 = 2
C) 1 > 2
2
dI
1
I
ACT 5 Solution
At the depth of the interface, the pressures in each side
must be equal.
d2
2
dI
Since there’s more liquid above this depth on the left side, p
that liquid must be less dense!
C) 1 > 2
1
I
d1
Pressure Measurements: Barometer



Invented by Torricelli
A long closed tube is filled with
mercury and inverted in a dish of
mercury
 The closed end is nearly a
vacuum
Measures atmospheric pressure as
One 1 atm = 0.760 m (of Hg)
F mg Vg
P 

  hg
A
A
A
Pa   gh
h
pa
g
Pressure vs. Depth
Incompressible Fluids (liquids)


Due to gravity, the pressure depends on
depth in a fluid
Consider an imaginary fluid volume (a
cube, each face having area A)
 The sum of all the forces on this volume
must be ZERO as it is in equilibrium.
» There are three vertical forces:
The weight (mg)
 The upward force from the pressure
on the bottom surface (F2)
 The downward force from the
pressure on the top surface (F1)
 Since the sum of these forces is
ZERO, we have: F2-F1=mg
p
0
y1
p1
F1
y2
A
Same fluid)
p
2
mg F2

F  0
 F1  mg  F2  0
mg  F2  F1
F2  F1  mg
Pressure vs. Depth
Incompressible Fluids (liquids)
p
0
F2  F1  mg
P2 A  P1 A  mg
P   gh
y2
p1
Now since
Area
P2 A Force=pressure×
P1 A  Vg
Now since Mass=Density× Volume
P2 A  P1 A    y2  y1  Ag
P2  P1    y2  y1  g
F1
y1
A
Same fluid
p
2
mg F2
Volume is area x height
P  P0   gh
The pressure at the bottom of a vertical column is equal
to the pressure at the top of the column (Po , Usually
atmospheric) , plus the pressure due to all the water in
the column of height , h. (density x gravity x height).
GAUGE Pressure





One must be careful when we are talking about
pressure. Do we want to include atmospheric pressure
in our calculations?
If we want the pressure over and above atmospheric
pressure, we have P0=0. This is called the Gauge
Pressure.
If we need to include atmospheric pressure we use:
P0=1x105N/m2 = 100 kPa
Most pressure gauges register the pressure over and
above atmospheric pressure.
For example a tire gauge registers 220 kPa, the actual
pressure within the tire is 220 kPa +100 kPa = 320 kPa
Gauge Pressure in a Tank Filled with Gasoline and Water
What is the pressure at point A? At point B?
kg
G  680 3
m
kg
W  1000 3
m
At point A:
PA  P0  G ghG
kg 
m

 0   680 3  9.8 2  10m 
m 
s 

N
 66640 2
m
At point B:
10m
PB  PA  W ghW
N 
kg 
m

1000
9.8
3m 

2
3 
2 
m 
m 
s 
N
 96040 2
m
 66640
3 m
Pascal’s Principle


So far we have discovered (using Newton’s Laws):
 Pressure depends on depth: P =  g y
Pascal’s Principle addresses how a change in
pressure is transmitted through a fluid.
Any change in the pressure applied to an enclosed fluid is
transmitted to every portion of the fluid and to the walls of
the containing vessel.
 Pascal’s Principle explains the working of hydraulic lifts
 i.e., the application of a small force at one place can result
in the creation of a large force in another.
 Will this “hydraulic lever” violate conservation of energy?
 No
Pascal’s Principle

Consider the system shown:
 A downward force F1 is applied to the piston of area A1.
 This force is transmitted through the liquid to create an upward
force F2.
 Pascal’s Principle says that increased pressure from F1 ,
[P=(F1/A1)] , is transmitted throughout the liquid.
P
F1
F
 2
A1 A 2
F2  F1
Yes, but what
gives?
Free Energy?
Where did this
extra force
come from?
A2
A1
Since A2 is larger
than A1, we have
a force multiplier.
Pascal’s Principle
Let’s check the
if the work done
by F1 equal the
work done by F2
F1
W1  F1d1
F2
 F2
A1
d1
A2
 F2
V1
A2
 F2
V2
A2
d2
d1
A1
A2
F2  F1
 F2 d 2
F1d1  F2 d 2
d2 
d1 A1
A2
A2
A1
Displaced
volumes
are the
same,
so…
 W2
Therefore
energy is
conserved
With a hydraulic lever, a given force applied over a given
distance can be transformed to a greater force applied
over a smaller distance.
ACT
In the hydraulic system shown below in the diagram, the 200kg
cylinder has a cross sectional area of 100 cm2. The cylinder on
the right has a cross sectional area of 10.0 cm2
a) Determine the amount of weight (F) you must apply on the
right side of the system to hold the system in equilibrium (Stop
the left side from moving down).
b) If the left cylinder is pushed down 5.0cm, determine the
distance F will move.
c) Determine the mechanical advantage of the system
ACT Solutions
In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional
area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2
a)
Determine the weight (F) required to hold the system in equilibrium
b)
If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will
move.
c)
Determine the mechanical advantage of the system
Apply Pascal’s Principle for F1
F1 F2

A1 A2
F1 mg

A1 A2
F1

2
10.0cm
 200kg   9.80

100cm 2
m

s2 
m

2
10
cm
200
kg
9.80






s2 

F1 
100cm2
 196 N
ACT 7 Solutions
In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional
area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2
a)
Determine the weight (F) required to hold the system in equilibrium
b)
If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will
move..
c)
Determine the mechanical advantage of the system
The volume of fluid displaced by the 100
cm2 cylinder equals the change in the
volume of fluid in the right hand cylinder
V  Ad
W1  W2
F1d1  F2 d 2
V1  V2
A1  d1  A2  d 2
d1 
A2  d 2
A1
or
100cm   5.0cm 


2
d1
10.0cm2
 50 cm
d1 
F2 d 2
F1

N
200
kg

9.8

  5.0cm 
kg


196 N
 50cm
ACT 7 Solutions
In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional
area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2
a)
Determine the weight (F) required to hold the system in equilibrium
b)
If the left cylinder is pushed down 5.0cm, determine the distance F will move.
c)
Determine the mechanical advantage of the system
Determine the Ideal Mechanical
Advantage (IMA) of the system.
IMA=(output Force/ Input Force)
IMA 

F2 d1

F1 d 2
50 cm
5.0 cm
 10
Archimedes’ Principle

Suppose we weigh an object in air (1) and in water (2).
 How do these weights compare?
W 1 < W2
W 1 = W2
 Why?
Since the pressure at the
bottom of the object is greater
than that at the top of the
object, the water exerts a net
upward force, the buoyant
force, on the object.
W1 > W2
W1
W2?
Archimedes’ Principle
W1
W2?
F
y1
1
y2
p
1
A
p
2
F
2

The buoyant force is equal to the difference in the pressures
times the area. FB  F2  F1
 ( p2  p1 )  A
   gy2   gy1   A
  g  y2  y1   A
  gV
FBuoyancy   liquid Vsubmerged  g   M liquid  g  Wliquid
Therefore, the buoyant force is equal to
the weight of the fluid displaced.
Archimedes’ Principle
A 60 kg rock lies at the bottom of a pool. Its volume is
3.0 x 104 cm3. What is its apparent weight?
The buoyant force on the rock due to the water is
equal to the weight of 3.0 x 104 cm3 = 3.0 x 10-2 m3
of water: FB  mwater g  water gV
kg 
m

 1.0 103 3  9.8 2   3.0 102 m3 
m 
s 

 290 N
m

The weight of the rock is:mrock g   60kg   9.8 2 
s 

 590 N
wapparent  w  FB
 590 N  290 N
 300 N
This is as if the rock had
a mass of 31kg
Sink or Float?


The buoyant force is equal to the weight of the liquid that is
displaced.
If the buoyant force of a fully submerged object is larger
than the weight of the object, it will float; otherwise it will sink.
y
FB mg

We can calculate how much of a floating object will
be submerged in the liquid:
 Object
is in equilibrium
FB  mg
Sink of Float?
y
Object is in equilibrium
FB  mg
FB mg
liquid  g  Vdispl.  object  Vobject  g
Vdispl.
Vobject
 object

 liquid
The Tip of The Iceberg:
What fraction of an iceberg is submerged? If the Density of ice is
917 kg/m3 and the density of water is 1024 kg/m3
Vwater displ.
Vice
ice
917 kg/m3


 90%
3
 water 1024 kg/m
Archimedes
Archimedes is said to have discovered his principle
in his bath while thinking how he might determine
whether the king’s new crown was pure gold or
fake.
Gold has a specific gravity of 19.3, somewhat
higher than most metals, but a determination of
specific gravity or density is not easy with a
irregularly shaped object.
Archimedes realised that if the crown is weighed in
air and weighed in water, the density can de
determined.
Archimedes
A 14.7 kg crown has an apparent weight of 13.4 kg when
submerged in water. Is it gold?
The apparent weight of the submerged crown, w`, equals
its actual weight, w, minus the buoyant force, FB.
w '  w  FB
 objectVg  water gV
w '  w  FB
w  w '  FB
  water gV
object objectVg w
w



We can find the specific gravity :
water waterVg FB w  w '
object
 water

N
14.7 kg  9.8 
kg 
w



 11.3
w  w'


N
N
14.7 kg  9.8   13.4kg  9.8 
kg 
kg 


This corresponds to a
density of 11,300 kg/m3
(that of lead)
Alternate solution
Let’s find the
volume of the
displaced water
(this is also the
volume of the
crown)
w  w '  FB
  water gVwater
14.7kg  13.4kg  g   water gVwater

3 kg 
1.3kg  1.000 x 10 3  Vwater
m 

1.3kg
Vwater 

3 kg 
1.000
x
10


m3 

 1.3 103 m3
crown
We can now
determine the
density of the crown
mcrown

Vcrown
14.7 kg

1.3 103 m3
kg
 11307 3
m
Practice

Pb
A lead weight is fastened to a large
styrofoam block and the combination
floats on water with the water level
with the top of the Styrofoam block
as shown.
 If you turn the Styrofoam + Pb
upside down, what happens?
A) It sinks
B)
styrofoam
Pb
C)
styrofoam
Pb
styrofoam
D)
styrofoam
Pb
Solution
Pb
styrofoam
A) It sinks
B)
styrofoam
Pb




C)
styrofoam
Pb
D)
styrofoam
Pb
If the object floats right-side up, then it also must float upside-down.
It displaces the same amount of water in both cases
However, when it is upside-down, the Pb displaces some water.
Therefore the styrofoam must displace less water than it did when it
was right-side up (when the Pb displaced no water).
Question
Which weighs more:
1. A large bathtub filled to the brim with water.
2. A large bathtub filled to the brim with water with a battleTub of water + ship
ship floating in it.
3. They will weigh the same.
Tub of water
Overflowed water
Weight of ship = Buoyant force =Weight of displaced water
15
Act 8
Suppose you float a large ice-cube in a glass of water, and after you
place the ice in the glass the level of the water is at the very brim.
When the ice melts, the level of the water in the glass will:
1. Go up, causing the water to spill out of the glass.
2. Go down.
3. Stay the same.
Must be same!
FB = Water g Vdisplaced
FB = Weight of ice
= ice g Vice  W g Vmelted_ice
Example Problems
At what depth is the water pressure two atmospheres?
(It is one
atmosphere at the surface. 1.01105 Pa)
What is the pressure at the bottom of the deepest oceanic trench
(about 104 meters)?
Solution:
P2  P0   gd
kg 
m

2.02 105 Pa  1.01105 Pa  1000 3  9.8 2  d
m 
s 

d  10.3m
For d = 104 m:
P2  P0   gd
kg 
m

P2  1.01105 Pa  1000 3  9.8 2  10000m 
m 
s 

 9.81107 Pa
 971Atm
This assumes that water is
incompressible.
If water were compressible,
would the pressure at the
bottom of the ocean be
greater or smaller than the
result of this calculation?
Example Problems (2)
Have you ever tried to submerge a beach ball (r = 50 cm) in a
swimming pool? It’s difficult. How big a downward force must you
exert to get it completely underwater?
Solution:
F   gV
We are displacing
this much water
4
  g  r3
3
kg 
m4
3

 1000 3  9.8 2    0.5m 
m 
s 3

 5131N
  523kg  g
I’m ignoring the weight of the beach ball.
The force is the weight of a 523 kg object.
More Fun With Buoyancy

Two cups are filled to the same
level with water. One of the two
cups has plastic balls floating in it.
Which cup weighs more?
 Archimedes
Cup I
Cup II
principle tells us that the cups weigh the same.
 Each plastic ball displaces an amount of water that is exactly
equal to its own weight.
Still More Fun!
A plastic ball floats in a cup of water with half of its volume
submerged. Oil (oil < ball <water) is slowly added to the
container until it just covers the ball.
Relative to the water level, the ball moves up. Why?

water
 For
oil to cover the ball, the ball must have “displaced” some water.
 Therefore, the buoyant force on the ball increases.
 Therefore, the ball moves up (relative to the water).
 Note that we assume the buoyant force of the air on the ball is
negligible (it is!); the buoyant force of the oil is not.
Archimedes Summary
Archimedes Principle states that the buoyant force on a
submerged object is equal to the weight of the fluid that is
displaced by the object.
If the weight of the water displaced is less than the weight
of the object (buoyant force is less than the weight of
object) , the object will sink. Otherwise the object will float,
with the weight of the water displaced equal to the weight
of the object.
FBuoyancy    fluid Vsubmerged  g
Vsubmerged
Vobject
object

 fluid
Fluids in Motion

Up to now we have described fluids in terms of
their static properties:
 Density 
 Pressure P

To describe fluid motion, we need something
that can describe flow:
 Velocity v

There are different kinds of fluid flow of varying complexity
non-steady
/ steady
compressible / incompressible
rotational
/ irrotational
viscous
/ ideal
Types of Fluid Flow


Laminar flow
 Each particle of the fluid
follows a smooth path
 The paths of the different
particles never cross each
other
 The path taken by the
particles is called a
streamline
Turbulent flow
 An irregular flow
characterized by small
whirlpool like regions
 Turbulent flow occurs when
the particles go above some
critical speed
Types of Fluid Flow


Laminar flow
 Each particle of the fluid
follows a smooth path
 The paths of the different
particles never cross each
other
 The path taken by the
particles is called a
streamline
Turbulent flow
 An irregular flow
characterized by small
whirlpool like regions
 Turbulent flow occurs when
the particles go above some
critical speed
Onset of Turbulent Flow
The SeaWifS satellite image of a
von Karman vortex around
Guadalupe Island, August 20,
1999
Ideal Fluids


Fluid dynamics is very complicated in general (turbulence,
vortices, etc.)
Consider the simplest case first: the Ideal Fluid
 No “viscosity” - no flow resistance (no internal friction)
 Incompressible - density constant in space and time
 Simplest situation: consider
ideal fluid moving with steady
flow - velocity at each point in
the flow is constant in time
 In this case, fluid moves on
streamlines
streamline
A2
A
1
v1
v2
Ideal Fluids





Streamlines do not meet or cross
Velocity vector is tangent to
streamline
Volume of fluid follows a tube of flow
bounded by streamlines
Streamline density is proportional to
velocity
Flow obeys continuity equation
Volume flow rate Q = A·v is constant along flow tube.
A1v1 = A 2v2
Follows from mass conservation if flow is incompressible.
The Equation of Continuity
Q: Have you ever used your thumb to control the water flowing from
the end of a hose?
A: When the end of a hose is partially closed off, thus reducing its
cross-sectional area, the fluid velocity increases.
This kind of fluid behavior is described by the equation of
continuity.
Continuity
A four-lane highway merges down to a two-lane highway. The officer in the
police car observes 8 cars passing every second, at 30 mph.
How many cars does the officer on the motorcycle observe passing every
second? A) 4
B) 8
C) 16
All cars stay on the road…
Must pass motorcycle too
How fast must the cars in the two-lane section be going?
A) 15 mph
B) 30 mph
C) 60 mph
Must go faster, else
pile up!
Exercise

A housing contractor saves
some money by reducing the
size of a pipe from 1cm
diameter to 1/2cm diameter at
some point in your house.
v1
v1/2
 Assuming the water moving in the pipe is an ideal fluid,
relative to its speed in the 1cm diameter pipe, how fast
is the water going in the ½ cm pipe?
(A) 2 v1
(B) 4 v1
(C) 1/2 v1
(D) 1/4 v1
Continuity of Fluid Flow
Fluid In= Fluid Out
• Watch “plug” of fluid moving through the narrow part of the tube (A1)
•Time for “plug” to pass point t = x1 / v1
• Mass of fluid in “plug”
m1 = Vol1 = A1 x1 or m1 = A1v1 t
• Watch “plug” of fluid moving through the wide part of the tube (A2)
•Time for “plug” to pass point t = x2 / v2
• Mass of fluid in “plug”
m2 = Vol2 = A2 x2 or m2 = A2v2t
• Continuity Equation says m1 = m2 fluid isn’t appearing or disappearing
A1 v1 = A2 v2
Faucet Preflight
A stream of water gets narrower as it falls from a faucet (try it & see).
Explain this phenomenon using the equation of continuity
V1
V2
A1
A2
“Since the area inside the faucet is large, the velocity of the water will be low. Due to
the equation of continuity, the area of the water has to be lower due to the
velocity increasing outside of the faucet. ”
“Velocity and area are inversely proportional. Velocity increases, so area decreases.”
Fluid Flow Concepts
•
•
•
•
Continuity: A1 v1 = A2 v2
Mass flow rate: Av (kg/s)
Volume flow rate: Av (m3/s)
i.e., mass flow rate the same everywhere
e.g., flow of river
Since we have an ideal fluid (incompressible), then
we can simply use: A1v1  A2v2
24
Pressure, Flow and Work



Continuity Equation says fluid speeds up going to smaller opening,
slows down going to larger opening
 But, if the fluid is moving faster, then it has more kinetic energy.
 Therefore something did work on the fluid to accelerate it (apply a
force)
W  Fd
Acceleration due to change in pressure. P1 > P2
 P1 A1d
 Smaller tube has faster water and LOWER pressure
Change in pressure does work!
 PV
1 1
 W = F1x1 –F2x2= P1A1x1 - P2A2x2 = (P1 – P2)V
In essence, Bernoulli’s Principle states that where the velocity of a fluid is
high, the pressure is low and where the velocity is low, the pressure is high.
Pressure ACT

What will happen when I “blow” air between the two plates?
A) Move Apart
B) Come Together
C) Nothing
Bernoulli’s Equation
For steady flow, the speed, pressure, and elevation of an
incompressible and nonviscous fluid are related by an equation
discovered by Daniel Bernoulli (1700–1782).
It is the application of the conservation of energy. P is the pressure of
the fluid at the specified spot. y is the vertical height at the specified
spot, v is the velocity at the specified spot.
Pressure, Flow and Work
Work done = Gain in KE + Gain in PE
1 2 1 2
mv2  mv1  mgh2  mgh1
Work is done by
2
2
the difference in
1
1
P1 A1x1  P2 A2 x2  mv22  mv12  mgh2  mgh1
pressure
2
2
1 2 1 2
PV

PV

mv2  mv1  mgh2  mgh1
Fluid to
1
2
2
2
left gives
This is the work
m
m 1 2 1 2
pressure.
P1  P2  mv2  mv1  mgh2  mgh1
done to move the

 2
2
coloured fluid from
1
1
left to right
P1  P2   v22   v12   gh2   gh1
2
2
1 2
1 2
1
P1   v1   gh1  P2   v2   gh2
P   v 2   gh  constant
or
F1x1  F2 x2 
2
2
2
Bernoulli’s Equation
The Fluid is pushed
from Point 1 to Point 2
1 2
1 2
P1   v1   gh1  P2   v2   gh2
2
2
Kinetic
External + Energy per + Potential
energy per
Pressure Volume
unit Volume
This sum has the same value at all points along a streamline.
Question
The container below is filled to a height h with liquid of density ρ. The atmospheric
pressure is Patm, determine the pressure of the fluid at each labelled point .
Compare the pressure at point 3 with that at point 5 with A3=1/2A5
1
1
P1   v12   gh1  P2   v22   gh2
2
2
1
 v12   gh1 
2
1
  gh   v12   gh1 
2
1
 gh   v12 
2
P1 
patm
h
1
2
5
patm
1
 v52   gh5
2
1
  v52   gh5
2
1
 v52
2
2gh  v52
v5 
p5  patm
p2  p4  p5
3
4
P5 
2 gh
(Cross sectional area are the same, v1=v2=v3, height are the same, h1=h2=h3)
p1  patm   gh
p3  patm  3 gh
Let’s see how
v5 
Question
2 gh
The container below is filled to a height h with liquid of density ρ. The atmospheric
pressure is Patm, determine the pressure of the fluid at each labelled point .
Compare the pressure at point 3 with that at point 5 with A3=1/2A5
1
1
P1   v12   gh1  P2   v22   gh2
A3v3  A5v5
2
2
1

1
1
A
2
5  v3  A5 v5

P


v


gh

P


v52   gh5
3
3
3
5
2


2
2
v3  2v5
P3 
h
1
p1  patm   gh
p2  p4  p5
3
4
P3  Patm 
1
  v52  v32 
2
P3  Patm 
1
  2 gh  8 gh 
2
5
p3  patm  3 gh
(Cross sectional area are the same, v1=v2=v3, height are the same, h1=h2=h3)
p3  patm  3 gh
p5  patm
2
1
1
 v32  P5   v52
2
2
Bernoulli ACT

Through which hole will the water
come out fastest?
P1+gy1 + ½ v12 = P2+gy2 + ½v22
A
Note: Outside all three holes the external pressure is the same: P=1
Atm (the ρgy term accounts for the pressure of water in column)
1 2
1 2
 gy1   v1   gy2   v2
2
2
1
1
gy1  v12  gy2  v22
2
2
y2 > y1  v1 > v2
Smaller y gives larger v. Hole C is fastest
B
y2
y1
C
Exercise

A housing contractor saves some
v1
money by reducing the size of a
pipe from 1cm diameter to 1/2cm
diameter at some point in your
house.
(A) 2 v1
(B) 4 v1
(C) 1/2 v1
v1/2
(D) 1/4 v1
 For equal volumes in equal times then ½ the diameter
implies ¼ the area so the water has to flow four times as fast.
 But if the water is moving four times as fast the it has 16
times as much kinetic energy. Something must be doing work
on the water (the pressure drops at the neck and we recast the
work as W=P V = (F/A) (Ax) = F x )
Practice
A large bucket full of water has two drains. One is a hole in the side of the
bucket at the bottom, and the other is a pipe coming out of the bucket near
the top, which bent is downward such that the bottom of this pipe is even
with the other hole, like in the picture below:
Though which drain is the water spraying out with the highest speed?
1. The hole
2. The pipe
3. Same
Speed is determined by pressure difference where
water meets the atmosphere. Relevant height is where
the hole is. Both are exiting at the same height!
Example
A garden hose w/ inner diameter 2 cm, carries water at 2.0 m/s. To
spray your friend, you place your thumb over the nozzle giving an
effective opening diameter of 0.5 cm. What is the speed of the water
exiting the hose? What is the pressure difference between inside the
hose and outside?
Continuity Equation
A1 v1 = A2 v2
v2 = v1 ( A1/A2)
= v1 ( πr12 / πr22)
= 2 m/s  16 = 32 m/s
Bernoulli Equation
P1+ gy1 + ½
v12 = P2+ gy2 + ½
v22
P1 – P2 = ½ (v22 – v12)
= ½  (1000 kg/m3) (1024 m2/s2 – 4 m2/s2 ) = 5.1  105 Pa
Bernoulli’s Principle

A housing contractor saves
some money by reducing the
size of a pipe from 1cm
diameter to 1/2cm diameter at
some point in your house.
v1
v1/2
1) What is the pressure in the 1/2cm pipe relative to the 1cm
pipe?
(A) smaller
(B) same
(C) larger
Applications of Bernoulli's
Equation
The tarpaulin that covers the cargo is flat when the truck
is stationary but bulges outward when the truck is
moving.
Household Plumbing
In a household plumbing system, a vent is necessary to equalize the
pressures at points A and B, thus preventing the trap from being emptied. An
empty trap allows sewer gas to enter the house.
Curveball Pitch
Applications of Fluid Dynamics




Streamline flow around
a moving airplane wing
Lift is the upward force
on the wing from the air
Drag is the resistance
The lift depends on the
speed of the airplane,
the area of the wing, its
curvature, and the
angle between the wing
and the horizontal
higher velocity
lower pressure
lower velocity
higher pressure
Note: density of flow lines reflects
velocity, not density. We are assuming
an incompressible fluid.
Applications of Fluid Dynamics
The constriction in the
Subclavian artery
causes the blood in the region
to speed up and
thus produces low pressure.
The blood moving
UP the LVA is then pushed
DOWN instead of
down causing a lack of blood
flow to the brain.
This condition is called TIA
(transient ischemic
attack) or “Subclavian Steal
Syndrome.
Flash Files
Pressure
Buoyancy
Fluid Flow
Bernoulli’s Equation
Summary
These are the formulas that you need to know:
Pressure =
Force
Area
Density =
Mass
Volume
P   g y
m  V
F  weight   Vg
W  Fd
 P1 A1d
 PV
1 1
P = P0   gh
Archimedes’ Principle: FBuoyancy    fluid Vsubmerged  g  m fluid g  W fluid
object

 fluid
Vsubmerged
Vobject


Pascal’s Principle:
Continuity Equation:
Bernoulli’s Equation:
Wobject
Wobject  WAppearant Weight Object
Wobject
FBuoyancy
Work :
F1d1  F2 d 2
Volume :
A1d1  A2 d 2
Pr esssure :
F1
F
 2
A1
A2
A1v1  A2v2
P1   gy1 
Av is flow rate
1 2
1
 v1  P2   gy2   v22
2
2