Introduction to Fluid Mechanics

Contributed by:

Jonathan James Fri, Jan 21, 2022 06:36 PM UTC

• 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

1. Chapter 9
Fluids

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
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 2

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 3

4. By Newton’s 3rd Law, there is a force on the wall due
to the particle.
F
Pressure is defined as P  .
A
The units of pressure are N/m2 and are called Pascals
Note: 1 atmosphere (atm) = 101.3 kPa
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 4

5. Example (text problem 9.1): Someone steps on your toe, exerting
a force of 500 N on an area of 1.0 cm2. What is the average
pressure on that area in atmospheres?
2 A 500N person
2 1m  4 2
weighs about
1.0 cm   1.0 10 m
 100 cm  113 lbs.
F 500 N
Pav  
A 1.0 10 4 m 2
6 2  1 Pa   1 atm 
5.0 10 N/m  2  5 
 1 N/m   1.013 10 Pa 
49 atm
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 5

6. Gravity’s Effect on Fluid Pressure
FBD for the fluid cylinder
An imaginary
cylinder of y
P1A
x
Imaginary cylinder
can be any size
w P2A
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 6

7. Apply Newton’s 2nd Law to the fluid cylinder. Since the
fluids isn’t moving the net force is zero.
 F P A  P A  w 0
2 1
P2 A  P1 A   Ad  g 0
P2  P1  gd 0
 P2  P1 gd
or P2 P1  gd
If P1 (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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 7

8. If the top of the fluid column is placed at the surface of the fluid,
then P1 = Patm if the container is open.
P Patm  gd
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/m2
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 8

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?
P Patm  gd
P P  Patm gd
 3

 1000 kg/m 9.8 m/s  35 m  2

343 kPa 3.4 atm
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 9

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/m3.
P Patm  gd
95 atm 1 atm  gd
gd 94 atm 9.5 106 N/m 2
1025 kg/m 9.8 m/s d 9.5 10
3 2 6
N/m 2
d 950 m
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 10

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

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

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 13

14. From the figure: At point C Pc Patm
Also PB PB'
The pressure at point B is the pressure of the gas.
PB PB ' PC  gd
PB  PC PB  Patm gd
Pgauge gd
Pgauge = Pmeas - Patm
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 14

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 15

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

17. The Many Units of Pressure
1 ATM equals 1.013x105 N/m2
14.7 lbs/in2
1.013 bar
76 cm Hg
760 mm Hg
760 Torr
34 ft H2O
29.9 in Hg
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 17

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.)
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 18

19. The applied force is Apply a force F1 here
transmitted to the piston
to a piston of cross-
of cross-sectional area
sectional area A 1.
A2 here.
In these problems neglect
pressure due to columns
of fluid.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 19

20. P at point 1 P at point 2
F1 F
 2
A1 A 2
 A2 
F2   F1
 A1 
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 20

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 (A2/A1)
are 1, 10, and 100.
Using Pascal’s Principle:
A2 A1 F2
1 500 N
10 5000 N
100 50,000 N
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 21

22. Archimedes’ Principle
y
F1
An FBD for an object floating
submerged in a fluid.
x
w
F2
The total force on the block due to
the fluid is called the buoyant force. FB F2  F1
where F2  F1
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 22

23. Buoyant force = the weight of the fluid displaced
The magnitude of the buoyant force is:
FB F2  F1
P2 A  P1 A
 P2  P1  A
From before: P2  P1 gd
The result is FB gdA gV
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 23

24. 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.
FB gV
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 24

25. Example (text problem 9.28): A flat-bottomed barge loaded with
coal has a mass of 3.0105 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?
Apply Newton’s 2nd Law to the barge:
y
FBD for
FB
 F F B  w 0
the barge FB w
mw g   wVw  g mb g
x
 wVw mb
w  w  Ad  mb
mb 3.0 105 kg
d  1.5 m
 3
 w A 1000 kg/m  20.0 m *10.0 m  
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 25

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.)
Apply Newton’s 2nd Law to
y
the piece of metal:
FBD for FB
the metal F FB  w ma

The magnitude of the buoyant force
x equals the weight of the fluid displaced
by the metal.
w FB  waterVg
FB ρ waterVg  ρ waterV 
Solve for a: a  g  g g   1
m ρ objectVobject ρ V 
 object object 
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 26

27. Example continued:
Since the object is completely submerged V=Vobject.

specific gravity 
 water
where water = 1000 kg/m3 is the
density of water at 4 °C.
 object
Given specific gravity  5.0
 water
 ρ waterV   1   1 
a g   1 g 
  1 g   1  7.8 m/s2
ρ V   S .G.   5.0 
 object object 
The sign is minus because gravity acts down. BF causes a < g.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 27

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.
V1 = V2 =
constant constant
v1v2
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 28

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

30. The Continuity Equation—Conservation of Mass
Faster Slower
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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 30

31. m
Av is the mass flow rate (units kg/s)
t
V is the volume flow rate (units m3/s)
 Av
t
The continuity equation
is
1 A1v1  2 A2 v2
If the fluid is incompressible, then 1= 2.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 31

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?
Simple ratios
A1v1  A2 v2
 A1   r12 
v2   v1  2  v1
 A2   r2 
2
 1.0 cm 
   2.0 m/s  50 m/s
 0.20 cm 
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 32

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

34. This is the most general equation
1 2 1 2
P1  gy1  v1 P2  gy2  v2
2 2
Work per Potential
unit volume Points 1 and 2
energy Kinetic
done by the must be on the
per unit energy
fluid same streamline
volume per unit
volume
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 34

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.
Let point 1 be inside the hose and point 2 be outside the nozzle.
1 2 1 2
P1  gy1  v1 P2  gy2  v2
2 2
The hose is horizontal so y1 = y2. Also P2 = Patm.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 35

36. Example continued:
1 2 1 2
Substituting: P1  v1 Patm  v2
2 2
1 2 1 2
P1  Patm  v2  v1
2 2
v2 = 25 m/s and v1 is unknown. Use the continuity equation.
  d2  2 
   2
 A2   2   d 
v1   v2    2  v2  2  v2
 A1     d1    d1 
 2 
   
Since d2< MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 36

37. Example continued:
1 2 1 2
P1  Patm  v2  v1
2 2
1 1 2
2
 2
  v2  v1  v2
2 2
 Since v1 0
1

 1000 kg/m 3  25 m/s 
2
2

3.1105 Pa
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 37

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 38

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.
V  P L 4 4th power
 r
t 8 
where  is the viscosity
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 39

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:
P  4 

8L V 8 2.00 10 3 Pa sec  3.00 cm
0

. 250 cm 3
sec
4
r t 
 0.3 10 cm
 1

4716 Pa
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 40

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
P Ps  Pv
Ps Pv  P
2140 Pa  4720 Pa 6860 Pa
Conversion:
1.013x10 5 Pa
16 (mm Hg)× = 2132 Pa
760 mm Hg
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 41

42. Example continued:
(b) What force must be applied to the plunger,
which has an area of 1.00 cm2?
The result of (a) gives the force per unit area on
the plunger so the force is just F = PA = 0.686 N.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 42

43. Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
FD 6rv
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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 43

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/m3. The
sphere reaches a terminal speed of 0.15 m/s. What is the viscosity
of the liquid?
Apply Newton’s Second Law to
y the sphere
FBD for FB F
D
sphere  F F D  FB  w ma
x
w
Drag Buoyant
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 44

45. Example continued:
When v = vterminal , a = 0 and
FD  FB  w 0
6rvt  ml g  ms g 0
6rvt   lVl g  ms g 0
6rvt   lVs g  ms g 0
ms g   lVs g
Solving for   2.4 Pa sec
6rvt
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 45

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 46

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
(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.
 2  2
F PA  r 9 10  6 N
 r 
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 47

48. Example continued:
(b) What is the maximum mass of this water strider so that it
can keep from breaking through the water surface?
To be in equilibrium, each leg must support one-
sixth the weight of the insect.
1 6F
F  w or m  5 10  6 kg
6 g
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 48

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
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 49

50. Quick Questions

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
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 51

52. 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
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 52

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

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 54

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

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 56

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

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 58

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

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 60

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

62.

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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 63

64. 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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 64

65. 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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 65

66. 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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 66

67. 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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 67

68. 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.
MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 68

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