PHYS 1443 – Section 001
Lecture #22
Monday, May 2, 2011
Dr. Jaehoon Yu
•
•
•
•
Absolute and Gauge Pressure
Buoyant Force and Archimedes’ Principle
Equation of Continuity
Bernoulli’s Principle
Announcements
• Planetarium extra credit sheets due today!!
• Quiz #5 this Wednesday, May 4
– Covers from CH. 11 – CH13.14
• Final comprehensive exam
– 11am, Monday, May 9, in SH103
– Covers: Chapter 1.1 – CH13. 14 + appendices
– Review: Wednesday, May 4th, in the class after the quiz
• Attendance will be taken
• Colloquium this Wednesday at 4pm
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
Absolute and Relative Pressure
How can one measure pressure?
P0
P
h
One can measure the pressure using an open-tube manometer,
where one end is connected to the system with unknown
pressure P and the other open to air with pressure P0.
The measured pressure of the system is
P  P0  gh
This is called the absolute pressure, because it is the
actual value of the system’s pressure.
In many cases we measure the pressure difference with respect to the
atmospheric pressure to avoid the effect of the changes in P0 that
G
depends on the environment. This is called gauge or relative pressure.
P  P  P0  gh
The common barometer which consists of a mercury column with one end closed at vacuum
and the other open to the atmosphere was invented by Evangelista Torricelli.
Since the closed end is at vacuum, it does
not exert any force. 1 atm of air pressure
pushes mercury up 76cm. So 1 atm is
P0  gh  (13.595103 kg / m3 )(9.80665m / s 2 )(0.7600m)
 1.013105 Pa  1atm
Monday,
May 2,the
2011tire pressure with a PHYS
Spring
3
If one
measures
gauge1443-001,
at 220kPa
the2011
actual pressure is 101kPa+220kPa=303kPa.
Dr. Jaehoon Yu
Finger Holds Water in Straw
You insert a straw of length L into a tall glass of your favorite
beverage. You place your finger over the top of the straw so that
no air can get in or out, and then lift the straw from the liquid. You
find that the straw strains the liquid such that the distance from the
bottom of your finger to the top of the liquid is h. Does the air in the
space between your finger and the top of the liquid in the straw
have a pressure P that is (a) greater than, (b) equal to, or (c) less
than, the atmospheric pressure PA outside the straw? Less
pinA
What are the forces in this problem?
Gravitational force on the mass of the liquid
Fg  mg   A L  h g
Force exerted on the top surface of the liquid by inside air pressure Fin  pin A
mg
Force exerted on the bottom surface of the liquid by the outside air Fout  pA A
Since it is at equilibrium F
out
p AA
Cancel A and
solve for pin
Monday, May 2, 2011
 Fg  Fin  0
pin  pA   g  L  h
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
 pA A   g  L  h A  pin A  0
So pin is less than PA by ρg(L-h).
4
Buoyant Forces and Archimedes’ Principle
Why is it so hard to put an inflated beach ball under water while a small
piece of steel sinks in the water easily?
The water exerts force on an object immersed in the water.
This force is called the buoyant force.
How large is the The magnitude of the buoyant force always equals the weight of the
buoyant force? fluid in the volume displaced by the submerged object.
This is called the Archimedes’ principle. What does this mean?
Let‘s consider a cube whose height is h and is filled with fluid and in its
equilibrium so that its weight Mg is balanced by the buoyant force B.
pressure at the bottom of the cube
B  Fg  Mg The
is larger than the top by  gh.
h
Mg B
Therefore,  P  B / A   gh
B  PA  ghA  Vg
B  Vg  Mg  Fg
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
Where Mg is the weight
of the fluid in the cube.
5
More Archimedes’ Principle
Let’s consider the buoyant force in two special cases.
Case 1: Totally submerged object Let’s consider an object of mass M, with density  0, is
fully immersed in the fluid with density  f .
The magnitude of the buoyant force is
B   f Vg
The weight of the object is Fg  Mg  0Vg
h
Mg B
Therefore total force in the system is
What does this tell you?
Monday, May 2, 2011
F  B  Fg   f

 0 Vg
The total force applies to different directions depending on the
difference of the density between the object and the fluid.
1. If the density of the object is smaller than the density of
the fluid, the buoyant force will push the object up to the
surface.
2. If the density of the object is larger than the fluid’s, the
object will sink to the bottom of the fluid.
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
6
More Archimedes’ Principle
Case 2: Floating object
h
Mg B
Let’s consider an object of mass M, with density  0, is in
static equilibrium floating on the surface of the fluid with
density  f , and the volume submerged in the fluid is Vf.
The magnitude of the buoyant force is
The weight of the object is
Therefore total force of the system is
Since the system is in static equilibrium
What does this tell you?
Monday, May 2, 2011
B   f Vf g
Fg  Mg  0V0 g
F  B  Fg   f Vf g  0V0 g  0
 f Vf g  0V0 g
Vf
0

f
V0
Since the object is floating, its density is smaller than that of the
fluid.
The ratio of the densities between the fluid and the object
determines the submerged volume under the surface.
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
7
Ex.13 – 10 for Archimedes’ Principle
Archimedes was asked to determine the purity of the gold used in the crown.
The legend says that he solved this problem by weighing the crown in air and
in water. Suppose the scale read 7.84N in air and 6.86N in water. What
should he have to tell the king about the purity of the gold in the crown?
In the air the tension exerted by the scale on
Tair  mg  7.84 N
the object is the weight of the crown
In the water the tension exerted

mg
B
T
 6.86 N
water
by the scale on the object is
Therefore the buoyant force B is
B  Tair  Twater  0.98 N
Since the buoyant force B is
B   wVw g   wVc g  0.98 N
The volume of the displaced
water by the crown is
Vc  Vw 
Therefore the density of
the crown is
c

0.98N
0.98

 1.0  10 4 m 3
 w g 1000  9.8
7.84
m c mc g 7.84

 8.0  10 3 kg / m 3


4
1.0  10  9.8
Vc Vc g Vc g
3kg/m
3, this
Monday,
May 2,the
2011
1443-001,
Spring
2011
Since
density of pure gold PHYS
is 19.3x10
crown is not made of pure gold.
Dr. Jaehoon Yu
8
Example for Buoyant Force
What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Fgi  iVi g
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
B  wVw g
Since the whole system is at its
static equilibrium, we obtain
Therefore the fraction of the
volume of the iceberg
submerged under the surface of
the sea water is
iVi g  wVw g
3
917kg
/
m
Vw i

 0.890

3
Vi  w 1030kg / m
About 90% of the entire iceberg is submerged in the water!!!
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
9
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water dynamics?? Hydro-dynamics
•Streamline or Laminar flow: Each particle of the fluid
Two primary follows a smooth path, a streamline
types of flows •Turbulent flow: Erratic, small, whirlpool-like circles called
eddy current or eddies which absorbs a lot of energy
Flow rate: the mass of fluid that passes the given point per unit time m / t
m1
1V1 1 A1l1


 1 A1v1
t
t
t
since the total flow must be conserved
m1 m2
1 A1v1  2 A2v2

t
t
Equation of Continuity
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
10
Ex. 13 – 14 for Equation of Continuity
How large must a heating duct be if air moving at 3.0m/s through it
can replenish the air in a room of 300m3 volume every 15 minutes?
Assume the air’s density remains constant.
Using equation of continuity
1 A1v1  2 A2v2
Since the air density is constant
A1v1  A2v2
Now let’s imagine the room as
the large section of the duct
A2l2 / t
V2
A2 v2
300
2

0.11
m
A1 



3.0  900
v1
v1  t
v1
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
11
Bernoulli’s Principle
Bernoulli’s Principle: Where the velocity of fluid is high, the
pressure is low, and where the velocity is low, the pressure is high.
Amount of the work done by the force,
F1, that exerts pressure, P1, at point 1
W1  F1l1  P1 A1l1
Amount of the work done by the force
in the other section of the fluid is
W2   P2 A2 l2
Work done by the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
12
Bernoulli’s Equation cont’d
The total amount of the work done on the fluid is
W  W1 W2 W3  P1 A1l1  P2 A2 l2 mgy2  mgy1
From the work-energy principle
1
1
2
mv2  mv12  P1 A1l1  P2 A2 l2 mgy2  mgy1
2
2
Since the mass m is contained in the volume that flowed in the motion
and
m   A1l1   A2 l2
A1l1  A2l2
Thus,
1
1
2
2
 A2 l2 v2   A1l1v1
2
2
 P1 A1l1  P2 A2 l2  A2 l2 gy2   A1l1 gy1
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
13
Bernoulli’s Equation cont’d
Since
1
1
 A2 l2 v22   A1l1v12  P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
2
2
We
obtain
1 2 1 2
 v2   v1  P1  P2   gy2   gy1
2
2
Reorganize P1 
1 2
1 2
Bernoulli’s
 v1   gy1  P2   v2   gy2 Equation
2
2
Thus, for any two
points in the flow
1 2
P1   v1   gy1  const.
2
For static fluid P2  P1   g  y1  y2   P1   gh
1
For the same heights P2  P1    v12  v22 
2
Result of Energy
conservation!
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
14
Ex. 13 – 15 for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at the speed of 0.5m/s through a 4.0cm diameter pipe in the basement under
a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter
pipe on the second 5.0m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1v1  r12 v1
 0.020 

v2 

0.5

 1.2m / s
2


A2
 r2
 0.013 
Using Bernoulli’s equation, the pressure in the pipe on the second floor is


1
P2  P1   v12  v22   g  y1  y2 
2
1
5
 3.0 10  1103 0.52  1.22  1103  9.8   5 
2
 2.5 105 N / m2

Monday, May 2, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu

15
Congratulations!!!!
You all are impressive and
have done very well!!!
I certainly had a lot of fun with ya’ll
and am truly proud of you!
Good luck with your exam!!!
Have safe holidays!!