CHAPTER 9 FLUIDS: INTRODUCTION TO FLUIDS.ppt

Physics Lecture

1. Chapter 9

Fluids

2. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 2
Chapter 9: Fluids
•Introduction to Fluids
•Pressure
•Measurement of Pressure
•Pascal’s Principle
•Gravity and Fluid Pressure
•Archimedes’ Principle
•Continuity Equation
•Bernoulli’s Equation
•Viscosity and Viscous Drag
•Surface Tension

3. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 3
Pressure
Pressure arises from the collisions between the particles of a fluid
with another object (container walls for example).
There is a momentum
change (impulse) that is
away from the container
walls. There must be a
force exerted on the
particle by the wall.

4. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 4
Pressure is defined as.
A
F
P
The units of pressure are N/m
2
and are called Pascals
(Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
By Newton’s 3
rd
Law, there is a force on the wall due
to the particle.

5. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 5
Example (text problem 9.1): Someone steps on your toe, exerting
a force of 500 N on an area of 1.0 cm
2
. What is the average
pressure on that area in atmospheres?
atm 49
Pa 10013.1
atm 1
N/m 1
Pa 1
N/m 100.5
m 101.0
N 500
52
26
24av


















A
F
P
24
2
2
m 100.1
cm 100
m 1
cm 0.1







A 500N person
weighs about
113 lbs.

6. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 6
Gravity’s Effect on Fluid Pressure
An imaginary
cylinder of
fluid
FBD for the fluid cylinder
P
1
A
P
2A
w
x
y
Imaginary cylinder
can be any size

7. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 7
Apply Newton’s 2
nd
Law to the fluid cylinder. Since the
fluids isn’t moving the net force is zero.

gdPP
gdPP
gdPP
gAdAPAP
wAPAPF









12
12
12
12
12
or
0
0
0
If P
1 (the pressure at the top of the cylinder) is known, then the
above expression can be used to find the variation of pressure
with depth in a fluid.

8. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 8
If the top of the fluid column is placed at the surface of the fluid,
then P
1
= P
atm
if the container is open.
gdPP 
atm
You noticed on the previous slide that the areas canceled out.
Only the height matters since that is the direction of gravity.
Think of the pressure as a force density in N/m
2

9. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 9
Example (text problem 9.19): At the surface of a freshwater lake,
the pressure is 105 kPa. (a) What is the pressure increase in going
35.0 m below the surface?
  
atm 3.4kPa 343
m 35m/s 8.9kg/m 1000

23
atm
atm




gdPPP
gdPP



10. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 10
Example: The surface pressure on the planet Venus is 95 atm.
How far below the surface of the ocean on Earth do you need to
be to experience the same pressure? The density of seawater is
1025 kg/m
3
.
  
m 950
N/m 109.5m/s 8.9kg/m 1025
N/m 109.5atm 94
atm 1atm 95

2623
26
atm





d
d
gd
gd
gdPP




11. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 11

12. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 12
Measuring Pressure
A manometer is a
U-shaped tube that
is partially filled
with liquid,
usually Mercury
(Hg).
Both ends of the
tube are open to the
atmosphere.

13. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 13
A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the atmosphere, a force
will be exerted on the fluid in the U-tube. This changes the equilibrium
position of the fluid in the tube.

14. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 14
Also
atmcPPAt point C
B'BPP
The pressure at point B is the pressure of the gas.
gdP
gdPPPP
gdPPP
BCB
CBB






gauge
atm
'
From the figure:
gauge meas atmP =P -P

15. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 15
A Barometer
The atmosphere pushes on the container of mercury which forces
mercury up the closed, inverted tube. The distance d is called
the barometric pressure.

16. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 16
Atmospheric pressure is equivalent to a column of
mercury 76.0 cm tall.
gdP
A

From the figure atmBA PPP 
and

17. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 17
The Many Units of Pressure
1 ATM equals 1.013x10
5
N/m
2
14.7 lbs/in
2
1.013 bar
76 cm Hg
760 mm Hg
760 Torr
34 ft H
2O
29.9 in Hg

18. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 18
Pascal’s Principle
A change in pressure at any point in a confined fluid is
transmitted everywhere throughout the fluid. (This is
useful in making a hydraulic lift.)

19. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 19
Apply a force F
1
here
to a piston of cross-
sectional area A
1.
The applied force is
transmitted to the piston
of cross-sectional area
A
2
here.
In these problems neglect
pressure due to columns
of fluid.

20. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 20
Mathematically,
1
1
2
2
2
2
1
1
A
A
A
F
A
F
2point at 1point at
FF
PP












21. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 21
Example: Assume that a force of 500 N (about 110 lbs) is applied to
the smaller piston in the previous figure. For each case, compute
the force on the larger piston if the ratio of the piston areas (A
2
/A
1
)
are 1, 10, and 100.
50,000 N100
5000 N10
500 N1
F
2
12
AA
Using Pascal’s Principle:

22. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 22
Archimedes’ Principle
An FBD for an object floating
submerged in a fluid.
The total force on the block due to
the fluid is called the buoyant force.

12
12
where FF
FFF


B
w
F
2
F
1
x
y

23. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 23
The magnitude of the buoyant force is:
 APP
APAP
FFF
B
12
12
12



gdPP 
12
From before:
gVgdAF
B The result is
Buoyant force = the weight of the fluid displaced

24. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 24
gVF
B
Archimedes’ Principle: A fluid exerts an upward buoyant force
on a submerged object equal in magnitude to the weight of the
volume of fluid displaced by the object.

25. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 25
Example (text problem 9.28): A flat-bottomed barge loaded with
coal has a mass of 3.010
5
kg. The barge is 20.0 m long and 10.0
m wide. It floats in fresh water. What is the depth of the barge
below the waterline?
x
y
w
F
B
FBD for
the barge
Apply Newton’s 2
nd
Law to the barge:


bw
bww
bwww
B
B
mAd
mV
gmgVgm
wF
wFF








0
  
m 5.1
m 10.0*m 0.20kg/m 1000
kg 100.3
3
5



A
m
d
w
b


26. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 26
Example (text problem 9.40): A piece of metal is released under
water. The volume of the metal is 50.0 cm
3
and its specific gravity is
5.0. What is its initial acceleration? (Note: when v = 0, there is no
drag force.)
FBD for
the metal
mawFF
B
VgF
B water

The magnitude of the buoyant force
equals the weight of the fluid displaced
by the metal.
Solve for a: 







 1
ρ
ρ
ρ
ρ
objectobject
water
objectobject
water
V
V
gg
V
Vg
g
m
F
a
B
Apply Newton’s 2
nd
Law to
the piece of metal:
x
y
w
F
B

27. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 27
Since the object is completely submerged V=V
object.
water
gravity specific



where 
water = 1000 kg/m
3
is the
density of water at 4 °C.
Given 0.5gravity specific
water
object



2
objectobject
water
m/s 8.71
0.5
1
1
..
1
1
ρ
ρ





















 g
GS
g
V
V
ga
Example continued:
The sign is minus because gravity acts down. BF causes a < g.

28. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 28
Fluid Flow
A moving fluid will exert forces parallel to the surface over which
it moves, unlike a static fluid. This gives rise to a viscous force
that impedes the forward motion of the fluid.
A steady flow is one where the velocity at a given point in a
fluid is constant.
V
1
=
constant
V
2
=
constant
v
1
v
2

29. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 29
Steady flow is laminar; the fluid flows in layers. The
path that the fluid in these layers takes is called a
streamline.
An ideal fluid is incompressible, undergoes laminar
flow, and has no viscosity.
Streamlines do not cross.
Crossing streamlines would indicate a volume of fluid with
two different velocities at the same time.

30. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 30
The Continuity Equation—Conservation of Mass
The amount of mass that flows though the cross-sectional area A
1

is the same as the mass that flows through cross-sectional area A
2.
Faster Slower

31. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 31
is the mass flow rate (units kg/s)Av
t
m



Av
t
V


 is the volume flow rate (units m
3
/s)
222111 vAvA 
The continuity equation
is
If the fluid is incompressible, then 
1
= 
2.

32. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 32
Example (text problem 9.41): A garden hose of inner radius 1.0
cm carries water at 2.0 m/s. The nozzle at the end has radius 0.20
cm. How fast does the water move through the constriction?
 m/s 50m/s 0.2
cm 0.20
cm 0.1
2
12
2
2
1
1
2
1
2
2211


























v
r
r
v
A
A
v
vAvA


Simple ratios

33. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 33
Bernoulli’s Equation
Bernoulli’s equation is a statement of energy
conservation.

34. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 34
2
222
2
111
2
1
2
1
vgyPvgyP  
Potential
energy
per unit
volume
Kinetic
energy
per unit
volume
Work per
unit volume
done by the
fluid
Points 1 and 2
must be on the
same streamline
This is the most general equation

35. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 35
Example (text problem 9.49): A nozzle is connected to a horizontal
hose. The nozzle shoots out water moving at 25 m/s. What is the
gauge pressure of the water in the hose? Neglect viscosity and
assume that the diameter of the nozzle is much smaller than the
inner diameter of the hose.
2
222
2
111
2
1
2
1
vgyPvgyP  
Let point 1 be inside the hose and point 2 be outside the nozzle.
The hose is horizontal so y
1 = y
2. Also P
2 = P
atm.

36. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 36
2
1
2
2atm1
2
2atm
2
11
2
1
2
1
2
1
2
1
vvPP
vPvP



Substituting:
v
2
= 25 m/s and v
1
is unknown. Use the continuity equation.
2
2
1
2
22
1
2
2
2
1
2
1
2
2
v
d
d
v
d
d
v
A
A
v















































Since d
2
<<d
1
it is true that v
1
<<v
2
.
Example continued:

37. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 37
 
  
Pa 101.3
m/s 25kg/m 1000
2
1
2
1
2
1
2
1
2
1
5
23
2
2
2
1
2
2
2
1
2
2atm1




vvv
vvPP


Example continued:

1
Since v 0

38. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 38
Viscosity
A real fluid has viscosity (fluid friction). This implies a
pressure difference needs to be maintained across the
ends of a pipe for fluid to flow.

39. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 39
Viscosity also causes the existence of a velocity gradient across a
pipe. A fluid flows more rapidly in the center of the pipe and
more slowly closer to the walls of the pipe.
The volume flow rate for laminar flow of a viscous fluid is given
by Poiseuille’s Law.
4
8
r
LP
t
V





where  is the viscosity
4th power

40. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 40
Example (text problem 9.55): A hypodermic syringe attached to a
needle has an internal radius of 0.300 mm and a length of 3.00 cm.
The needle is filled with a solution of viscosity 2.0010
-3
Pa sec; it is
injected into a vein at a gauge pressure of 16.0 mm Hg. Neglect the
extra pressure required to accelerate the fluid from the syringe into the
entrance needle.
(a) What must the pressure of the fluid in the syringe be in
order to inject the solution at a rate of 0.250 mL/sec?
Solve Poiseuille’s Law for the pressure difference:
  
 
Pa 4716
seccm 250.0
cm 103.0
cm 00.3sec Pa 1000.288
3
4
1
3
4












t
V
r
L
P

41. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 41
Example continued:
This pressure difference is between the fluid in the syringe
and the fluid in the vein. The pressure in the syringe is
Pa 6860Pa 4720Pa 2140 


PPP
PPP
vs
vs
5
1.013x10 Pa
16 (mm Hg)× =2132 Pa
760 mm Hg
Conversion:

42. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 42
(b) What force must be applied to the plunger,
which has an area of 1.00 cm
2
?
Example continued:
The result of (a) gives the force per unit area on
the plunger so the force is just F = PA = 0.686 N.

43. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 43
Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
rvF
D6
Where  is the viscosity of the fluid that the sphere is falling
through, r is the radius of the sphere, and v is the velocity of
the sphere.

44. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 44
Example (text problem 9.62): A sphere of radius 1.0 cm is dropped
into a glass cylinder filled with a viscous liquid. The mass of the
sphere is 12.0 g and the density of the liquid is 1200 kg/m
3
. The
sphere reaches a terminal speed of 0.15 m/s. What is the viscosity
of the liquid?
FBD for
sphere mawFFF
BD 
F
D
w
x
y
F
B
Apply Newton’s Second Law to
the sphere
Drag Buoyant

45. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 45
Example continued:
When v = v
terminal
, a = 0 and
sec Pa 4.2
6
06
06
06
0







t
sls
sslt
sllt
slt
BD
rv
gVgm
gmgVrv
gmgVrv
gmgmrv
wFF






Solving for 

46. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 46
Surface Tension
The surface of a fluid acts like a stretched membrane
(imagine standing on a trampoline). There is a force
along the surface of the fluid.
The surface tension is a force per unit length.

47. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 47
Example (text problem 9.70): Assume a water strider has a
roughly circular foot of radius 0.02 mm. The water strider has 6
legs.
(a) What is the maximum possible upward force on the foot
due to the surface tension of the water?
The water strider will be able to walk on water if the
net upward force exerted by the water equals the
weight of the insect. The upward force is supplied by
the water’s surface tension.
N 109
2
62 






 r
r
PAF 


48. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 48
(b) What is the maximum mass of this water strider so that it
can keep from breaking through the water surface?
Example continued:
To be in equilibrium, each leg must support one-
sixth the weight of the insect.
kg 105
6
or
6
1
6

g
F
mwF

49. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 49
Summary
•Pressure and its Variation with Depth
•Pascal’s Principle
•Archimedes Principle
•Continuity Equation (conservation of mass)
•Bernoulli’s Equation (conservation of energy)
•Viscosity and Viscous Drag
•Surface Tension

50. Quick Questions

51. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 51
Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?
1. Rise
2. Fall
3. Remain unchanged

52. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 52
1. Rise
2. Fall
3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?

53. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 53

54. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 54
In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball
1. rises.
2. falls.
3. remains in place.

55. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 55

56. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 56
The weight of the stand and suspended solid iron ball is equal to the
weight of the container of water as shown above. When the ball is lowered
into the water the balance is upset. The amount of weight that must be
added to the left side to restore balance, compared to the weight of water
displaced by the ball, would be
1. half. 2. the same.
3. twice. 4. more than twice.

57. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 57

58. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 58
A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.

59. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 59

60. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 60
Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will
1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.

61. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 61

62. Extras

63. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 63
The density of the
 block of
wood floating in water is
1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.

64. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 64
1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.
The density of the
 block of
wood floating in water is

65. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 65
1. rises.
2. falls.
3. remains in place.
In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball

66. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 66
1. half. 2. the same.
3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the
weight of the container of water as shown above. When the ball is lowered
into the water the balance is upset. The amount of weight that must be
added to the left side to restore balance, compared to the weight of water
displaced by the ball, would be

67. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 67
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is

68. MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 68
1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.
Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will