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AP Physics B
Fluid Dynamics
1
College Board Objectives
. FLUID MECHANICS AND THERMAL PHYSICS
A.
Fluid Mechanics
B.
Hydrostatic pressure
Students should understand the concept of
pressure as it applies to fluids, so they can:
a) Apply the relationship between pressure,
force, and area.
b) Apply the principle that a fluid exerts
pressure in all directions.
c) Apply the principle that a fluid at rest exerts
pressure perpendicular to any surface that
it contacts.
2
Fluid mechanics, cont.
d) Determine locations of equal pressure in a
fluid.
e) Determine the values of absolute and gauge
pressure for a particular situation.
f) Apply the relationship between pressure and
depth in a liquid, DP = r g Dh
3
Fluid Mechanics, cont.
Buoyancy
Students should understand the concept
of buoyancy, so they can:
a) Determine the forces on an object
immersed partly or completely in a
liquid.
b) Apply Archimedes’ principle to
determine buoyant forces and densities
of solids and liquids.
4
Fluid Mechanics, cont.
Fluid flow continuity
Students should understand the
equation of continuity so that they can
apply it to
fluids in motion.
Bernoulli’s equation
Students should understand Bernoulli’s
equation so that they can apply it to
fluids in motion.
5
Equations :
P = F/ A
m
r
V
Sp.Gr. = ρ substance / ρwater
P = ρhg
_____________________________
Ptotal = Patm + P liquid
p2  p1  rgh
6
Homework :
Chapter 9 Solids and Fluids
Summaries
2 Examples per section
2 End of Chapter 9 problems per section
2 PROBLEMS IN THE COLLEGE BOARD
https://apstudent.collegeboard.org/apcourse/ap-physics-b/exampractice
7
10.1&2
Density & Specific Gravity
1. The mass density r of a substance is the mass of
the substance divided by the volume it occupies:
unit: kg/m3
r for aluminum 2700 kg/m
mass can be written as m =
3
or 2.70 g/cm3
rV
and
rVg
Specific Gravity: r substance / r water
weight as mg =
m
r
V
8
2. A fluid - a substance that flows and conforms to the
boundaries of its container.
3. A fluid could be a gas or a liquid; however on the AP
Physics B exam fluids are typically liquids which are
constant in density.
9
4. An ideal fluid is assumed
• to be incompressible (so that its density does not
change),
• to flow at a steady rate,
• to be nonviscous (no friction between the fluid
and the container through which it is flowing),
and
• flows irrotationally (no swirls or eddies).
10
5. Turpentine has a specific gravity of
0.9 . What is its density ?
11
5. Turpentine has a specific gravity of
0.9 . What is its density ?
Specific Gravity=r substance / r water
Ρsubstance = sp.gr. X ρwater =
0.9 X 1000kg/m3 = 900 kg / m3
12
• 6. A cork has volume of 4 cm3 and weighs 10-2 N . What is the
specific gravity of cork ?
13
• 6. A cork has volume of 4 cm3 and weighs 10-2 N . What is the
specific gravity of cork ?
Fg = Weight = mg = 10-2 N = m (10 m/s2)
m = 10-2 N / 10m/s2 = 10-3 kg
ρcork = m/v= 10-3 kg / 4cm3 ( 100 cm)3 / m3 = 250 kg / m3
sp. Gr. = ρsubstance /ρwater = 250kg /m3 / 1000kg /m3
sp. Gr.
= 0.25
14
10.3
5. Pressure
Any fluid can exert a force perpendicular to its
surface on the walls of its container. The force is
described in terms of the pressure it exerts, or
force per unit area:
Units: N/m2 or Pa (1 Pascal*)
dynes/cm2 or PSI (lb/in2)
1 atm = 1.013 x 105 Pa or 15 lbs/in2
*One atmosphere is the pressure
exerted on us every day by the
earth’s atmosphere.
F
p
A
15
6. The pressure is the same in every
direction in a fluid at a given depth.
7.Pressure varies with depth.
P = F/A = mg/A = ρVg/A
P = F = rAhg so P = rgh
A
A
16
8. A FLUID AT REST EXERTS
PRESSURE PERPENDICULAR TO
ANY SURFACE THAT IT
CONTACTS. THERE IS NO
PARALLEL COMPONENT THAT
WOULD CAUSE A FLUID AT REST
TO FLOW.
17
PROBLEM 10-9
9.
(a) Calculate the total force of the atmosphere
acting on the top of a table that measures
1.6 m  2.9 m.
(b) What is the total force acting upward on the
underside of the table?
,
18
PROBLEM 10-9
9.
(a) Calculate the total force of the atmosphere
acting on the top of a table that measures
1.6 m  2.9 m.
Patmosphere = 1.013 X 105 N/m2
(b) What is the total force acting upward on the
underside of the table?
a. The total force of the atmosphere on the table
will be the air pressure times the area of the
table.

F  PA  1.013  10 N m
5
2
 1.6 m  2.9 m   4.7 10 N
5
(b) Since the atmospheric pressure is the same on the underside of the
table (the height difference is minimal), the upward force of air pressure is
19
the same as the downward force of air on the top of the table,
5
4.7 10 N
10. A vertical column made of cement has a base area of 0.5 m2
If its height is 2m , and the specific gravity of cement is 3 , how
much pressure does this column exert on the ground ?
20
10. A vertical column made of cement has a base area of 0.5 m2
If its height is 2m , and the specific gravity of cement is 3 , how
much pressure does this column exert on the ground ?
P = F / A = mg /A = ρVg /A = ρAhg/A = ρhg
Sp. Gr. = ρ cement / ρwater
ρcement = Sp. Gr. X 1000 kg / m3 = 3 X 1000 kg / m3 =
3000 kg/m3
P = ρh g = 3,000 kg/m3 X 2m X 10m/s2 = 60,000 Pa = 60k Pa
21
11. Atmospheric Pressure and Gauge Pressure
p1
p1
p1
h
p2
h
p2
h
p2
• The pressure p1 on the surface of the water is 1
atm, or 1.013 x 105 Pa. If we go down to a depth
h below the surface, the pressure becomes
greater by the product of the density of the water
r, the acceleration due to gravity g, and the
depth h. Thus the pressure p2 at this depth is
p2  p1  rgh
22
12. In this case, p2 is called the absolute(total)
pressure -- the total static pressure at a certain
depth in a fluid, including the pressure at the
surface of the fluid
13. The difference in pressure between the surface
and the depth h is gauge pressure
p2  p1  rgh
Note that the pressure at any depth does not
depend of the shape of the container, only the
pressure at some reference level (like the surface)
and the vertical distance below that level.
p1
p1
p1
23
h
p2
h
p2
h
p2
14. (a) What are the total force and the absolute
pressure on the bottom of a swimming pool 22.0
m by 8.5 m whose uniform depth is 2.0 m? (b)
What will be the pressure against the side of the
pool near the bottom?
(a)The absolute pressure is given by Eq. 10-3c,
and the total force is the absolute pressure times
the area of the bottom of the pool.
24
14. (a) What are the total force and the absolute
pressure on the bottom of a swimming pool 22.0
m by 8.5 m whose uniform depth is 2.0 m? (b)
What will be the pressure against the side of the
pool near the bottom?
(a)The absolute pressure is given by Eq. 10-3c,
and the total force is the absolute pressure times
the area of the bottom of the pool.

P  P0  r gh  1.013  105 N m 2  1.00  103 kg m3

9.80 m s 2
  2.0 m 
 1.21 105 N m 2

F  PA  1.21 105 N m 2

7
22.0
m
8.5
m

2.3

10
N



25
(b) The pressure against the side of the pool, near
the bottom, will be the same as the pressure at the
bottom,
P  1.2110 N m
5
2
26
CW : Hydrostatic Pressure
• 1. What is the hydrostatic gauge pressure at a point 10m
below the surface of the ocean ? The specific gravity of
seawater is 1.025.
27
• 1. What is the hydrostatic gauge pressure at a point 10m
below the surface of the ocean ? The specific gravity of
seawater is 1.025.
Gauge pressure = ρgh = P2 - P1
Sp. Gr. = ρsubstance / ρwater
ρocean = Sp. Gr. X 1000kg/m3 = 1.025 X 1000kg/m3 =
= 1025 kg/ m3
Gauge Pressure = ρgh = 1025 kg/m3 ( 10 m/s2) ( 10m) =
= 102, 500 Pa
28
2. A swimming pool has a depth of 4 m . What is the hydrostatic
gauge pressure at a point 1 m below the surface ?
29
2. A swimming pool has a depth of 4 m . What is the hydrostatic
gauge pressure at a point 1 m below the surface ?
Gauge Pressure = ρhg = 1000kg/m3 ( 1m) (10m/s2) =
= 10,000 Pa
30
15. What happens to the gauge pressure if we double our depth
below the surface of the liquid ? What happens to the total
pressure ?
31
15. What happens to the gauge pressure if we double our depth
below the surface of the liquid ? What happens to the total
pressure ?
Gauge pressure will double.
Gauge Pressure = ρgh
Total Pressure = Patm + ρgh
It will increase based on ρgh and having Patm as constant
32
16. A flat piece of wood , of area 0.5m2, is lying at the bottom of
a lake . If the depth of the lake is 30 m , what is the force on the
wood due to pressure ? ( use Patm= 1 X 10 5 Pa)
33
16. A flat piece of wood , of area 0.5m2, is lying at the bottom of
a lake . If the depth of the lake is 30 m , what is the force on the
wood due to pressure ? ( use Patm= 1 X 10 5 Pa)
F= P/A
P total = P atm + ρgh
= 1 X 105 Pa + (1000 kg /m3) ( 10 m/s2) ( 30m)
= 4 X 10 5 Pa
Force = P A = 4 X 10 5 Pa X 0.5 m2 = 2 X 10 5 N
34
CW : Hydrostatic Pressure
• 3. Consider a closed container , partially filled with a liquid of
density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the
surface of the liquid .
• A. if the space above the surface of the liquid is vacuum , what
is the absolute pressure at point X ?
• B. if the space above the surface of the liquid is occupied by a
gas whose pressure is 2.4 X 104 Pa , What is the absolute
pressure at Point X?
•
35
CW : Hydrostatic Pressure
• Consider a closed container , partially filled with a liquid of
density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the
surface of the liquid .
• A. if the space above the surface of the liquid is vacuum , what
is the absolute pressure at point X ?
• P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m)
•
= 6 X 10 3 Pa
• B. if the space above the surface of the liquid is occupied by a
gas whose pressure is 2.4 X 104 Pa , What is the absolute
pressure at Point X?
36
CW : Hydrostatic Pressure
• Consider a closed container , partially filled with a liquid of
density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the
surface of the liquid .
• A. if the space above the surface of the liquid is vacuum , what
is the absolute pressure at point X ?
• P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m)
•
= 6 X 10 3 Pa
• B. if the space above the surface of the liquid is occupied by a
gas whose pressure is 2.4 X 104 Pa , What is the absolute
pressure at Point X?
• P absolute = P1 +ρgh = 2.4 X 104 Pa + 6 X 10 3 Pa = 3X 104 Pa
37
10.5 Pascal’s Principle
• Pascal’s Principle - if an external pressure is
applied to a confined fluid, the pressure at every
point within the fluid increases by that amount.
Applications: hydraulic lift and brakes
Pout = Pin
And since P = F/a
Fout = Fin
Aout
Ain
Mechanical Advantage:
Fout = Aout
Fin
Ain
38
17. Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
18. The buoyant force is
found to be the upward
force on the same volume
of water:
39
10-7 Buoyancy and Archimedes’ Principle
19. The net force on the object is then the
difference between the buoyant force and the
gravitational force.
40
10-7 Buoyancy and Archimedes’ Principle
20. If the object’s density is less than that of
water, there will be an upward net force on it, and
it will rise until it is partially out of the water.
41
10-7 Buoyancy and Archimedes’ Principle
21. For a floating object, the fraction that is
submerged is given by the ratio of the object’s
density to that of the fluid.
42
10-7 Buoyancy and Archimedes’ Principle
This principle also works in
the air; this is why hot-air and
helium balloons rise.
43
22.A geologist finds that a Moon rock whose
mass is 9.28 kg has an apparent mass of
6.18 kg when submerged in water. What
is the density of the rock?
44
22.A geologist finds that a Moon rock whose
mass is 9.28 kg has an apparent mass of
6.18 kg when submerged in water. What
is the density of the rock?
ρrock = mrock/ Vrock
ρwater = mwater / V water
Vrock = Vdisplaced water = Mdisplaced water /ρ water
= 9.28kg -6.18 kg / 1000 kg/m3
= 0.0031 m3
ρrock = mrock/ Vrock = 9.28 kg / .0031 m3
ρrock =2.99X 103 kg/m3
45
23. A crane lifts the 18,000-kg steel hull of a ship out of the
water. Determine (a) the tension in the crane’s cable when the
hull is submerged in the water, and (b) the tension when the
hull is completely out of the water.
Tension (T)
mg
Fb = Fwater Buoyant Force
46
23. A crane lifts the 18,000-kg steel hull of a ship out of the
water. Determine (a) the tension in the crane’s cable when the
hull is submerged in the water, and (b) the tension when the
hull is completely out of the water.
When the hull is submerged, both the buoyant force and the
tension force act upward on the hull, and so their sum is equal
to the weight of the hull. The buoyant force is the weight of
the water displaced.
T  Fbuoyant  mg 
T  mg  Fbuoyant  mhull g  r waterVsub g  mhull g  r water
mhull
r hull
 r water 
g  mhull g  1 

r

hull 
3
3


1.00

10
kg
m
5
5
 1.8  10 4 kg  9.80 m s 2   1 

1.538

10
N

1.5

10
N
3
3 
7.8  10 kg m 

47
(b)When the hull is completely out of the
water, the tension in the crane’s cable
must be equal to the weight of the hull.



T  mg  1.8 104 kg 9.80 m s 2  1.764 105 N  1.8 105 N
48
SG  0.50 
24. A 5.25-kg piece of wood
floats on water. What minimum mass of lead, hung from
the wood by a string, will cause it to sink?
For the combination to just barely sink, the total weight of
the wood and lead must be equal to the total buoyant
force on the wood and the lead.
49
SG  0.50 
A 5.25-kg piece of wood
floats on water. What minimum mass of lead, hung from
the wood by a string, will cause it to sink?
34.For the combination to just barely sink, the total weight
of the wood and lead must be equal to the total buoyant
force on the wood and the lead.
Fweight  Fbuoyant  mwood g  mPb g  Vwood r water g  VPb r water g 
mwood  mPb 
mPb  mwood
mwood
r wood
r water 
mPb
r Pb
 r water 
 1
r
 wood   m
wood
 r water 
1  r 

Pb 
r water
 r water 
 r water 
 mPb  1 
 1 
  mwood 
r Pb 

 r wood 
 1

 1  1

1
 SG



0.50
 wood   5.25 kg 
  5.76 kg



1  1 
1 


 1  SG 
11.3



Pb 
50
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
25. If the flow of a fluid is smooth, it is called streamline or
laminar flow (a).
Above a certain speed, the flow becomes turbulent (b).
Turbulent flow has eddies; the viscosity of the fluid is much
greater when eddies are present.
51
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
We will deal with laminar flow.
26.The mass flow rate is the mass that passes a
given point per unit time. The flow rates at
any two points must be equal, as long as no
fluid is being added or taken away.
Flow rate f = A v
A = Cross-sectional area
v = flow speed
(10-4a)
28. This gives us the equation of continuity:
52
10-8 Fluids in Motion; Flow Rate and the
Equation of Continuity
28. If the density doesn’t change – typical for
liquids – this simplifies to
.
Where the pipe is wider, the flow is slower.
Flow speed is inversely proportional to the
cross-sectional area or the square of the
radius of the pipe
53
29. A pipe of non uniform diameter carries
water. At one point in the pipe , the radius is
2 cm and the flow speed is 6 m/s .
a. What is the flow rate ?
b. What is the flow speed at a point where
the pipe constricts to a radius of 1 cm ?
54
A pipe of non uniform diameter carries
water. At one point in the pipe , the radius is
2 cm and the flow speed is 6 m/s .
a.What is the flow rate ?
f = Av = πr2v = π ( 2 X 10 -2m ) 2 (6m/s)
b. What is the flow speed at a point where
the pipe constricts to a radius of 1 cm ?
55
A pipe of non uniform diameter carries water. At one point
in the pipe , the radius is 2 cm and the flow speed is
6 m/s .
a.What is the flow rate ?
f = Av = πr2v = π ( 2 X 10 -2m ) 2 (6m/s)
b. What is the flow speed at a point where the pipe
constricts to a radius of 1 cm ?
π ( 2 X 10 -2m ) 2 (6m/s) = π ( 1 X 10 -2m ) 2 v
v = 24 m/s
56
• 39. (II) A 85 -inch (inside) diameter garden hose is used to
fill a round swimming pool 6.1 m in diameter. How long
will it take to fill the pool to a depth of 1.2 m if water
issues from the hose at a speed of 0.40 m s?
• 39. The volume flow rate of water from the hose,
multiplied times the time of filling, must equal the volume
of the pool.
 Av hose 
Vpool
t
 t
Vpool
Ahose vhose
  3.05 m  1.2 m 
2

 1 5 "  1m  
 2 8 
" 
39.37



2
 4.429  105 s
 0.40 m s 

1day

4.429  10 s 
 5.1 days

 60  60  24 s 
5
57
40. (II) What gauge pressure in the water mains is
necessary if a firehose is to spray water to a height of
15 m?
40. Apply Bernoulli’s equation with point 1 being the water
main, and point 2 being the top of the spray. The
velocity of the water will be zero at both points. The
pressure at point 2 will be atmospheric pressure.
Measure heights from the level of point 1.
P1  12 r v12  r gy1  P2  12 r v22  r gy2 

P1  Patm  r gy2  1.00  103 kg m3


9.8 m s 2 15 m   1.5  105 N m2
58
30. Bernoulli’s Equation
A fluid can also change its
height. By looking at the
work done as it moves, we
find:
This is Bernoulli’s
equation. One thing it
tells us is that as the
speed goes up, the
pressure goes down.
59
Bernoulli’s Equation
• P1 + ρgy1 + ½ ρv12 = P2 + ρgy2 + ½ ρv22
60
31. In the Figure below , a pump forces
Water at a constant flow rate through a pipe whose crosssectional area , A gradually decreases. At the exit point A
has decreased by 1/3 its value at the beginning of the pipe.
If y =60cm and the flow speed of the water just after it
leaves the pump is 1m/s , what is the gauge pressure at
point 1?
P1 + ρgy1 + ½ ρv12 = PATM + ρgy2 + ½ ρv22
61
30. Bernoulli’s Equation
A fluid can also change its
height. By looking at the
work done as it moves, we
find:
This is Bernoulli’s
equation. One thing it
tells us is that as the
speed goes up, the
pressure goes down.
62
31. In the Figure below , a pump forces
Water at a constant flow rate through a pipe whose crosssectional area , A gradually decreases. At the exit point A
has decreased by 1/3 its value at the beginning of the pipe.
If y =60cm and the flow speed of the water just after it
leaves the pump is 1m/s , what is the gauge pressure at
point 1?
P1 + ρgy1 + ½ ρv12 = PATM + ρgy2 + ½ ρv22
P1 – PATM = ( ρgy2-ρgy1) + ½ ρv22 - ½ ρv12
Pgauge = (1000kg/m3)((10m/s2)(0.6m) + ½ (3v1)2 -v12) =
(1000kg/m3)((6 m2 / s2 +4 m2/s2 ) = 10,000 Pa
63
CW: Bernoulli’s Equation
• What does Bernoulli’s Equation tell us
about a fluid at rest in a container open to
the atmosphere?
64
CW: Bernoulli’s Equation
• What does Bernoulli’s Equation tell us
about a fluid at rest in a regular container
open to the atmosphere?
Because the fluid is at rest , both v1 and v2
are zero and Bernoulli’s equation becomes
P1 + ρgy1 = P2 + ρgy2
P1 = Patm = 1X 105 Pa
P2 = Patm + ρg ( y1 – y2) Hydrostatic
Pressure
65
36.
(I) A 15-cm-radius air duct is used to replenish the
air of a room 9.2 m  5.0 m  4.5 m every 16 min. How
fast does air flow in the duct?
We apply the equation of continuity at constant density, Eq.
10-4b. Flow rate out of duct = Flow rate into room
Aduct vduct   r 2 vduct 
Vroom
t to fill
room
 vduct 
Vroom
 r t to fill
2
room

 9.2 m  5.0 m  4.5 m 
2
 60 s 
  0.15 m  16 min  

1
min


 3.1m s
66
• 39. (II) A 85 -inch (inside) diameter garden hose is used to
fill a round swimming pool 6.1 m in diameter. How long
will it take to fill the pool to a depth of 1.2 m if water
issues from the hose at a speed of 0.40 m s?
• 39. The volume flow rate of water from the hose,
multiplied times the time of filling, must equal the volume
of the pool.
 Av hose 
Vpool
t
 t
Vpool
Ahose vhose
  3.05 m  1.2 m 
2

 1 5 "  1m  
 2 8 
" 
39.37



2
 4.429  105 s
 0.40 m s 

1day

4.429  10 s 
 5.1 days

 60  60  24 s 
5
67
40. (II) What gauge pressure in the water mains is
necessary if a firehose is to spray water to a height of
15 m?
40. Apply Bernoulli’s equation with point 1 being the water
main, and point 2 being the top of the spray. The
velocity of the water will be zero at both points. The
pressure at point 2 will be atmospheric pressure.
Measure heights from the level of point 1.
P1  12 r v12  r gy1  P2  12 r v22  r gy2 

P1  Patm  r gy2  1.00  103 kg m3


9.8 m s 2 15 m   1.5  105 N m2
68
Visit the follow website from
Boston University
• http://physics.bu.edu/~duffy/py105.html
• For more information about (choose from left panel)
• Pressure; Fluid Statics
• Fluid Dynamics
• Viscosity
69
At the website complete the following:
1. Read and record important
equations and facts.
2. For each equation write the quantity
for each symbol
3. Write the unit for each quantity
(symbol ok)
70
Demonstrations to View
• http://www.csupomona.edu/~physics/oldsite/de
mo/fluidmech.html
71
Experiment :
Density VS Buoyancy
I.
Purpose :
To determine the density of water .
To determine if 10 objects will float or sink.
To determine the density of the 10 objects
To calculate the weight of the 10 objects .
To calculate the buoyant force exerted on the 10 objects when
submerged in water.
II. Materials :
10 objects
Water
Cylinder
triple beam balance
digital scale
III.Data Table 1: Density of Fluid
Water
Mass in g
V, cc
Trial 1
10 mL
Trial 2
20 mL
Trial 3
30 mL
D=M/V
In g/cc
III. Data Table 2: Float or Sink
Name of the Object
1.
2.
3.
4.
5.
6.
7.
8.
9.
Water
III. Data Table 3 : Mass and Weight
Objects
1.
2.
3.
4.
5.
6.
7.
8.
9.
Mass, g
22.2
Mass,Kg
.0222
Weight ,N
.222
Data Table 4: Densities of Objects
Name
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mass in g
V, cc
D=M/V
In g/cc
Data Table 4: Buoyant Force VS Weight
Objects
Volume of
object , cc
Density of
Fluid
Buoyant
Force , N
Weight
,N
Which is greater Fw or
FB
IV. Calculations: Show all calculations of Densities of Fluids,
Densities of Objects , Weights and Buoyant force.
V. Analysis and Calculations :
Explain how the density of the fluid was verified.
Explain how the density of objects were calculated.
Explain how weight is calculated .
Explain how the Buoyant force is calculated for each object .
How does the density of the object relate to buoyancy ? Explain
How will you prove that density and buoyancy are related .