1. Chapter - 11
Fluids

2. Fluids 11.1 Mass Density 11.2 Pressure
11.3 Pressure and Depth in a Static Fluid
11.4 Pressure Guages
11.5 Pascal’e Principle
11.6 Archimedes’ Principle
11.7 Fluid in Motion
11.8 The Equation of Continuity
11.9 Bernoulli’s Equation
Applications of Bernoulli’s Equation
Viscous Flow

3. Definition of Mass Density
The mass density of a substance is the mass of a
substance per unit volume:
SI Unit of Mass Density: kg / m3

4. 11.1 Mass Density

5. Example-1: Blood as a Fraction of Body Weight
11.1 Mass Density
Example-1: Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3 m3 of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.

6. 11.1 Mass Density
(a)
(b)

7. Definition of Weight Density
11.1 Mass Density
Definition of Weight Density
The Weight density of a substance is the Weight of a
substance per unit volume:
SI Unit of Weight Density: N / m3

8. Specific Gravity Is relative density.
It is a ratio of the densities of two materials.
Dimensionless.
Using water as a reference can be convenient, since density of water is (approximately) 1000 kg/m³ or 1 g/cm³ at 4C.
Specific gravity of Mercury is 13.6, means it is 13.6 times massive or heavier than water.

9. SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal

10. Pressure Pressure is force per unit area. P = F/A
Pressure is not a vector quantity.
To calculate the pressure we only consider the magnitude of the force.
The force generated by the pressure of the static fluid is always perpendicular to the surface of the contact.
Pa = N/m2 = x 10-4 lb/in2 (very small quantity)
bar = 105 Pa = 100 kPa
Atmospheric pressure is kPa = bar = lb/in2 = 760 torr
lb/in2 or psi = x 103 Pa

11. Pressure causes Perpendicular Force
The forces of a liquid pressing against a surface add up to a net force that is perpendicular to the surface.
Components parallel to
the surface cancels out
Components vertical to
the surface adds up

12. Example-2: The Force on a Swimmer
11.2 Pressure
Example-2: The Force on a Swimmer
Suppose the pressure acting on the back
of a swimmer’s hand is 1.2x105 Pa. The
surface area of the back of the hand is
8.4x10-3 m2.
Determine the magnitude of the force
that acts on it.
(b) Discuss the direction of the force.

13. Since the water pushes perpendicularly
11.2 Pressure
Since the water pushes perpendicularly
against the back of the hand, the force
is directed downward in the drawing.

14. Shape of the Dam Dam-B Dam-A Would you prefer Dam-A or Dam-B?
Justify your answer.
Dam-B
Dam-A

15. Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere
Weight of the column of the air of 1m2 = x105 N
Mass of the column of the air of 1m2 = x105 N/g = kg

16. 11.3 Pressure and Depth in a Static Fluid
Relation between Pressure and Depth
Since the column is in equilibrium,
The sum of the vertical forces
equal to zero

17. 11.3 Pressure and Depth in a Static Fluid
The pressure increment = gh = weight density x height
Pressure at a depth, h = P1 + gh = Atmospheric Pressure + Liquid Pressure
So, Liquid Pressure at a depth, h = gh = weight density x depth

18. Pressure
Water pressure acts perpendicular to the sides of a container, and increases with increasing depth.

19. 11.3 Pressure and Depth in a Static Fluid
Conceptual Example-3: The Hoover Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Dam still be needed
to contain the water, or could a much less
massive structure do the job?

20. 11.3 Pressure and Depth in a Static Fluid
Example-4: The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface
of the water. Find the pressure at each of these two locations.

21. 11.3 Pressure and Depth in a Static Fluid

22. Where to put the pump? Pump-X Pump-Y Pump-X pushed the water up.
Pump-Y removes the air from the pipe and air pressure outside pushes the water up.
Pump-Y has a limit, what is that?
Pump-X
Pump-Y

23. Pressure Gauges
One of the simplest pressure gauges is the mercury barometer used for measuring atmospheric pressure.
760 mm of mercury

24. Point A and B are at the same depth, so, PB = PA Patm = P1 + gh
11.4 Pressure Gauges
Point A and B are at the same depth, so,
PB = PA
Patm = P1 + gh
= 0 + gh

25. Open-tube manometer
The phrase “open-tube” refers to the fact that one side of the U-tube is open to atmospheric pressure.
The tube contains mercury
Its other side is connected to the container whose pressure P2 is to be measured.
PB = PA
P2 = Patm + gh
P2 - Patm = gh
P2 – Patm is called the guage pressure
P2 is called the absolute pressure

26. 11.4 Pressure Gauges
absolute pressure

27. sphygmomanometer
blood-pressure instrument: an instrument used to measure blood pressure in an artery that consists of a pressure gauge, an inflatable cuff placed around the upper arm, and an inflator bulb or pressure pump.
Systolic pressure is 120 mm of mercury
Diastolic pressure is 80 mm of mercury

28. 11.4 Pressure Gauges

29. Any change in the pressure applied
11.5 Pascal’s Principle
Pascal’s Principle
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.

30. 11.5 Pascal’s Principle

31. The input piston has a radius of 0.0120 m
11.5 Pascal’s Principle
Example-7: A Car Lift
The input piston has a radius of m
and the output plunger has a radius of
0.150 m.
The combined weight of the car and the
plunger is N. Suppose that the input
piston has a negligible weight and the bottom
surfaces of the piston and plunger are at
the same level. What is the required input
force?

32. 11.5 Pascal’s Principle

33. 11.6 Archimedes’ Principle

34. 11.6 Archimedes’ Principle
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:

35. 11.6 Archimedes’ Principle
If the object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.

36. 11.6 Archimedes’ Principle
Example-9: A Swimming Raft
The raft is made of solid square
pinewood. Determine whether
the raft floats in water and if
so, how much of the raft is beneath
the surface.

37. 11.6 Archimedes’ Principle

38. 11.6 Archimedes’ Principle
The raft floats!

39. 11.6 Archimedes’ Principle
If the raft is floating:

40. 11.6 Archimedes’ Principle
Conceptual Example-10: How Much Water is Needed
to Float a Ship?
A ship floating in the ocean is a familiar sight. But is all
that water really necessary? Can an ocean vessel float
in the amount of water than a swimming pool contains?

41. 11.6 Archimedes’ Principle

42. Unsteady flow exists whenever the velocity of the fluid particles at a
11.7 Fluids in Motion
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.

43. Fluid flow can be compressible or incompressible. Most liquids are
11.7 Fluids in Motion
Fluid flow can be compressible or incompressible. Most liquids are
nearly incompressible.
Fluid flow can be viscous or nonviscous.
An incompressible, nonviscous fluid is called an ideal fluid.

44. When the flow is steady, streamlines are often used to represent
11.7 Fluids in Motion
When the flow is steady, streamlines are often used to represent
the trajectories of the fluid particles.

45. Making streamlines with dye and smoke.
11.7 Fluids in Motion
Making streamlines with dye
and smoke.

46. 11.8 The Equation of Continuity
The mass of fluid per second that flows through a tube is called
the mass flow rate.

47. 11.8 The Equation of Continuity

48. 11.8 The Equation of Continuity
The mass flow rate has the same value at every position along a
tube that has a single entry and a single exit for fluid flow.
SI Unit of Mass Flow Rate: kg/s

49. 11.8 The Equation of Continuity
Incompressible fluid:
Volume flow rate Q:

50. 11.8 The Equation of Continuity
Example-12: A Garden Hose
A garden hose has an unobstructed opening
with a cross sectional area of 2.85x10-4m2.
It fills a bucket with a volume of 8.00x10-3m3
in 30 seconds.
Find the speed of the water that leaves the hose
through (a) the unobstructed opening and (b) an obstructed
opening with half as much area.

51. 11.8 The Equation of Continuity
(b)

52. Bernoulli’s Equation For steady flow, The speed, Pressure, and
Elevation
of an incompressible and nonviscous fluid are related by Bernoulli’s equation.

53. The fluid accelerates toward the lower pressure regions.
11.9 Bernoulli’s Equation
The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.

54. Bernoulli’s Equation Bernoulli’s Equation from Work-energy theorem:
Work done on an object by external non-conservative forces is equal to the change in total mechanical energy.
The pressure within a fluid is caused by collisional forces, which are non-conservative.
Therefore, when a fluid is accelerated because of a difference in pressures, work is being done by non-conservative forces.

55. 11.9 Bernoulli’s Equation

56. Bernoulli’s Equation
Whenever a fluid is flowing in a horizontal pipe and encounters a region of reduced cross-sectional area, the pressure of the fluid drops.
Can be explained by the Newton’s 2nd law.
When moving from the wider region 2 to the narrower region 1, the fluid speeds up according to the equation of continuity.
According to the 2nd law, th eaccelerating fluid must be subjected to an unbalanced force.
There can be an unbalanced force only if the pressure in region 2 is higher than the pressure in region 1.

57. Bernoulli’s Equation Explanation:
In figure-b on the top surface of the blue fluid element, the surrounding fluid exerts a pressure P.
This pressure give rise to a force, F = P/A
On the botom of the blue fluid element, the pressure is slightly higher, P+P.
As a result, the force on the bottom surface has a magnitude of F+ F = (P+P)/A.
The net force pushing the fluid element up the pipe is F=P/A

58. Bernoulli’s Equation
When the fluid element moves through its own length s, the work done is the
W = F s = P A s = P V
The total work done on the fluid element in moving it from region 2 to region 1is the sum of the small increments of work P V done as the element moves along the pipe.
This sum amounts to Wnc = (P2 – P1)V
Wnc = (P2 – P1)V = E1 – E2

59. fluid speed, and the elevation at two points are related by:
11.9 Bernoulli’s Equation
Bernoulli’s Equation
In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:

60. Bernoulli’s Equation The term
has a constant value at all positions in the flow.

61. Bernoulli’s Equation Bernoulli’s Equation
reduces to the result for static fluids (v1=v2), as it is when the cross-sectional area remains constant.

62. 11.10 Applications of Bernoulli’s Equation
Conceptual Example-14: Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.

63. Bernoulli’s Equation Bernoulli’s Equation reduces to
When all parts have the same elevation (y1 = y2).
The term P+ ½  remains constant throughout a horizontal pipe.
If  increases, P decreases and vice versa.

64. 11.10 Applications of Bernoulli’s Equation

65. 11.10 Applications of Bernoulli’s Equation

66. 11.10 Applications of Bernoulli’s Equation

67. 11.10 Applications of Bernoulli’s Equation
Example-16: Efflux Speed
The tank is open to the atmosphere at
the top. Find and expression for the speed
of the liquid leaving the pipe at
the bottom.

68. 11.10 Applications of Bernoulli’s Equation

69. 11.11 Viscous Flow
Flow of an ideal fluid.
Flow of a viscous fluid.

70. Force Needed to Move a Layer of Viscous Fluid with Constant Velocity
11.11 Viscous Flow
Force Needed to Move a Layer of Viscous Fluid with
Constant Velocity
The magnitude of the tangential force required to move a fluid
layer at a constant speed is given by:
coefficient
of viscosity
SI Unit of Viscosity: Pa·s
Common Unit of Viscosity: poise (P)
1 poise (P) = 0.1 Pa·s

71. The volume flow rate is given by:
11.11 Viscous Flow
Poiseuille’s Law
The volume flow rate is given by:

72. Example-17: Giving and Injection
11.11 Viscous Flow
Example-17: Giving and Injection
A syringe is filled with a solution whose
viscosity is 1.5x10-3 Pa·s. The internal
radius of the needle is 4.0x10-4m.
The gauge pressure in the vein is 1900 Pa.
What force must be applied to the plunger,
so that 1.0x10-6m3 of fluid can be injected
in 3.0 s?

73. 11.11 Viscous Flow

74. 11.11 Viscous Flow

75. Force A

76. Force A

77. Force A

78. Force A