This document outlines the goals and topics covered in a fluid mechanics course. The course covers fluid statics, fluid flow concepts, flow of incompressible and compressible fluids, pumping systems, fluid mixing, and fluidization. Key concepts discussed include fluid properties like density, viscosity, compressibility, and pressure. Laminar and turbulent flow regimes are defined. The continuity, Bernoulli, and Reynolds equations are introduced. Example problems are provided to help understand concepts like pressure, viscosity, flow rate calculations, and fluid flow analysis.

1. Fluid Mechanics
CHEM 218
Zin-Eddine Dadach
2006-2007

2. Goals of the course
• 1) Fluid statics and its application
• 2) Fluid Flow concepts and related
experiments
• 3) Flow of incompressible fluids in pipes
and related experiments
• 4) Flow of compressible fluids
• 5) Pumping and metering systems of
fluids
• 6) Fluid mixing and fluidization—principles
and applications

3. Fluid Mechanics?
• 1) Fluid Mechanics is the study of the
behavior of fluids at rest ( Fluid statics) or
in motion ( Fluid dynamics)
• 2) In fluid statics, specific weight ( unit
weight) is the important property
• 3) In fluid dynamics, density and viscosity
are the predominant properties

4. Fluid?
• 1) Capable of flowing and conform to the
shape of the container
• 2) Fluid can be liquid or gas
• 3) Liquids are incompressible whereas
gases are compresible

5. Effects of temperature and
pressure on the density of fluids?
• 1) Density of all fluids depends on the
temperature and pressure
• 2) Incompressible fluids: slight change on
the density –> example 1.4
• 3) Compressible fluids:
• T ↑→ V↑→ d↓
• P↑ →V↓ → d↑

6. What is compressibility?
• 1) Change in volume V of a fluid with
change of pressure
• 2) The bulk modulus E is used:
E= (-ΔP) / (ΔV/V)
Examples for values of E at 20°C:
Ethyl Alcohol : 130.000
Water : 316.000
Mercury : 3.590.000

7. What is density?
• 1) Amount of mass per unit volume
• ρ = m/ V
• ASTM method of measure : pycnometers
• Units :
SI system: ( Kilograms per cubic meter)
US system: (slugs per cubic foot)

8. What is specific weight?
• Amount of weight per unit volume
• γ= ω/ V
• V is the volume of a substance having the
weight ω
• SI system: newtons per cubic meter
• US system: pounds per cubic foot

9. What is specific gravity?
• Definitions:
1) Ratio of the density of a substance to the
density of water at 4°C
2) Ratio of the specific weight of a substance to
the specific weight of water at 4°C
3) Mathematically:
sg= ( γs / γw at 4°C) = (ρs/ρw at 4°C)
4) γw at 4°C= 9.81 kN/ m3 or 62.4 lb/ft3
5) ρw at 4°C = 1000kg/m3 or 1.94 slugs / ft3

10. Density and specific gravity
relationship ?
• γ=ρg with g= acceleration of gravity
• Do examples 1.5 to 1.9

11. Class work
• Do problems : 1.17 -1.20 page 22
• Do problems : 1.48-1.54 page 23
• Do problems : 1.58-1.67 page 23
• Do problems : 1.70-1.76 page 24

12. What is pressure?
Amount of force exerted on a unit area of a
substance
p=F/A Units: SI Pascal or US lb/ft2
Absolute and gage pressure:
Pabs=Pgage + Patm
Do examples 3.1 to 3.4

13. Pressure and elevation relationship?
For homogeneous liquid at rest:
Δp= γ.h where
Δp = Change in pressure
γ = specific weight of liquid
h = change in elevation
Do examples 3.5 to 3.7

14. Example 3.7 on page 56
From figure 3.3 on page 57 -
Calculate the pressure at:
Point A: Point B:
Point C: Point D:
Point E: Point F:

15. What is a manometer?
• Pressure measurement device
• Uses the relationship : Δp=γ.h where γ is the
specific weight and h is the height
• Simplest kind of manometer is the U-tube
(Figures 3.9 and 3.10 on pages 55 and 56)
• The tube contains a liquid called the gage fluid
which should not mix with the fluid which
pressure is to be measured
• Fluid can be water, mercury and colored light
oils

16. Calculation of pressure using
the U-tube
• In calculations , it is convenient to start
with the open end
• Class work:
• Given the γmercury= 132.8 kN/m3 and
γwater = 9.81 kN/m3
What is the gage pressure at point A of
Figure 3.10 page 63?

17. CLASS WORK
• Work examples : 3-8 to 3-13
• Work Problems : 3.52, 3.54, 3.56, 3.57,
3.58 and 3.59

18. What is a gravity decanter?
• Used for continuous separation of two
immiscible liquids (Figure 2.5 from note):
A) Feed enters at one end
B) slow flow through the decanter
C) Separation into two layers
D) Discharge trough overflow lines on the
other hand

19. Fluid Statics analysis of
gravity decanter
• Hypothesis:
A) Large overflow lines to neglect flow
resistance
B) the discharge is at the same pressure
as the one in the gas space above the
liquid in the vessel:

20. Calculation of a gravity decanter
At point 1: P1 =0
At point 4: P4 =P1+ ρA. ZA2= ρA. ZA2
At point 5: P5=P4= ρA. ZA2
At point 2: P2= P5 - ρA. ZA1= ρA. (ZA2- ZA1)
At point 3: P3= 0= P2 - ρB. ZB
P3= ρA. (ZA2- ZA1) - ρB. ZB=0
ZT=ZA1 + ZB =} ρA. (ZA2- ZA1) - ρB.( ZT-ZA1)=0
And finally : ZA1= [ZA2 – ZT (ρB/ ρA)]/ [1- (ρB/ ρA)]

21. Separation time of the
two fluids in the decanter
ZA1= [ZA2 – ZT (ρB/ ρA)]/ [1- (ρB/ ρA)]
This equation shows that the position of liquid-
liquid interface depends on:
A) Ratio of densities
B) Elevation of overflow lines
By definition the time of separation is calculated
by: t= (100.μ) / (ρA- ρB)
If ρA≈ ρB → t approaches ∞

22. What is a centrifugal decanter?
• We need it when the difference between
the density of the two liquids is too small
• Centrifugal decanter is based on
centrifugal force created by a vertical
rotating cylinder
• Bowl at rest ( Figure 2.6a)
• Bowl rotating ( Figure 2.6b)

23. Goal 2: Fluid Flow
• Fluid flow concept:
 One dimensional flow
 Laminar Flow
 Turbulent Flow

24. One dimensional flow
A characteristic of this flow is that the velocity becomes invariant in the flow
direction as shown in Figure
It is readily seen that velocity at any location depends just on the radial
distance from the centerline and is independent of distance, x or of the
angular position. This represents a typical one-dimensional flow.

25. Laminar Flow
• Fluid flows uniformly with low velocity and
with little or no mixing from one layer to
another layer ( Figures 8.1 and 8.2)
• Velocity profile is parabolic

26. Velocity Profile in Laminar Flow
• To get the kinetic energy of laminar flow in a tube, an average of the
square of the velocity must be taken to account for the velocity profile.
•
•
• V = Vm ( 1- (r/R)2) where Vm is the velocity at r=0
•
• V= 2Va (1- (r/R)2) where Va is the average velocity
•
• vm = 2va
•

27. Turbulent Flow
• Occurs at higher velocities
• Mixing is chaotic within the stream
• Flat velocity Profile ( Figures 8.3 and 8.5)

28. Flows and velocity gradients
• Velocity gradient is a measure of velocity changes or shear rate and
is defined as:
Δv/ Δy where y is the direction of the flow
• Example: a fluid is placed between two parallel plates that are 1.0
cm apart, the upper plate moving at a velocity of 1.0 cm/sec and
the lower plate fixed.
• The velocity gradient is the rate of change of velocity with distance
from the plates.
• This simple case shows the uniform velocity gradient with shear rate
(v1 - v2)/h = shear rate = (cm/sec)/(cm/1) = 1/sec.
• Hence, shear rate units are reciprocal seconds or sec-1

29. What is shear Stress?
 IS the force per unit area required to sustain a constant rate of fluid
movement. Mathematically, shear stress can be defined as:
•
• If a fluid is placed between two parallel plates spaced 1.0 cm apart, and
a force of 1.0 dyne is applied to each square centimeter of the surface of
the upper plate to keep it in motion, the shear stress in the fluid is
1 dyne/cm2 at any point between the two plates

30. For Common Fluids
• Water, oil, gasoline, air ……
 the shearing stress and rate of
shearing strain ( velocity gradient) can be
related by :
τ= μ ( dv /dy)
Where μ is the dynamic viscosity with unit
as [μ] = N.s/m2 or Pa.s and [τ] = N/m2

31. What is viscosity?
• The viscosity of a fluid is a very important
property in the analysis of liquid behavior
and fluid motion near a solid boundary.
• The knowledge of viscosity is often
necessary for proper design of required
temperatures for storage, pumping or
injection of fluids.

32. Dynamic viscosity
The dynamic viscosity of a fluid is its resistance to
shear or flow and is a measure of the fluids
adhesive/cohesive or frictional properties.
τ= μ ( dv /dy)
Where μ is the absolute or dynamic viscosity with
unit as [μ] = N.s/m2 or Pa.s and [τ] = N/m2

33. Kinematic viscosity
• A coefficient which describes the diffusion
of momentum.
• Let the dynamic viscosity be μ, then
ν= μ/ρ
• the unit of kinematic viscosity is the Stoke,
equal to 1 cm2 s-1.
•

34. • A force of 3 Newtons is used to move a
plate of 1cm2 section. The variation of
velocity is 3 cm s-1. The two plates are 1
cm apart.
• A) calculate the dynamic viscosity
• B) If the density is 1000 kg/m3, calculate
the kinematic viscosity

35. For turbulent flow
• The dynamic viscosity is replaced by the
Eddy viscosity ( Ev) and then
τ= Ev ( dv /dy)

36. Newtonian Fluid ?
A fluid that has a constant viscosity at all shear
rates at a constant temperature and pressure,
and can be described by a one-parameter
rheological model.
•
Water, sugar solutions, glycerin, silicone oils,
light-hydrocarbon oils, air and other gases are
Newtonian fluids.

37. Non Newtonian flow?

38. Reynolds Number
 The Reynolds Number is important in analyzing any type of flow when there is
substantial velocity gradient - shear.
 The Reynolds Number indicates the relative significance of the viscous effect
compared to the inertia effect.
 The Reynolds number is proportional to inertial force divided by viscous force.
• Reynolds Number can be express
• The viscosity above is dynamic viscosity also called absolute viscosity. For
a pipe or duct the characteristic length is the pipe or duct diameter.

39. Values of NRe
• For laminar Flow :
NRe < 2000
• For turbulent Flow:
NRe > 4000
 Do examples 8.1 to 8.3 of the book
 Do problems 8.1 to 8.4 page 247

40. Continuity Equation
• When a fluid is in motion, it must move in
such a way that mass is conserved.
• To see how mass conservation places
restrictions on the velocity field, consider the
steady flow of fluid through a duct (that is,
the inlet and outlet flows do not vary with
time).
• The inflow and outflow are one-
dimensional, so that the velocity V and
density are constant over the area A (figure
14).

41. FIGURE
•

42.  M1=M2 This is the continuity
equation

43. Volume Flow rate
 volume of fluid passing trough a
section A per unit time
 Q = A.V [Q] = m3/s
Example: If a section is 6.356.10-4m2 and
the velocity equal 3m/s . Calculate Q
Response : Q= 1.907.10-3 m3/s

44. Specific weight flow rate
• W = γ. Q
Example: If the specific weight of water is
9.81 kN / m3 and its volume flow rate is
0.01 m3/s
W= 9.81x 0.01 = 0.0981 kN/s

45. Mass Flow rate
• M =ρ.Q
Example : If the density of water is 1000
kg/m3 and the volume flow rate is 0.01
m3/s
Mass Flow rate  1000 x 0.01 = 10 kg/s

46. Class work #1
• From figure given , The inside diameters
of the pipe in sections 1 and 2 are 100mm
and 50 mm respectively. Water at 200C
flows with an average velocity of 2m/s at
section 1, Calculate the following:
* Velocity at section 2 ( 8m/s)
* Volume flow rate ( 0.0157m3/s)
* Weight flow rate ( 0.154 kN/s)
* Mass Flow rate ( 15.7 kg/s)

47. Class work #2
• A) Determine the maximum velocity of a
gasoline and water flowing at 200C in laminar
flow manner in a 20mm pipe?
• Kinematic viscosity of gasoline and water are
respectively 6.48 10-7m2/s and
• 1.02 10-6 m2/s
• B) Determine the type of flow occuring in a 12”
pipe when water at 600F flows at velocity of 3.50
ft/s ? Same question for gasoline
• ( νwater= 1.217 10-5 ft2/s and νgasoline =221 10-5
ft2/s)

48. Class work #3
• A) Water flows into a pipe of diameter 0.1 m
with a velocity of 4 m/s . Determine the
diameter of the other hand of the pipe if the
velocity becomes 2m/s
• B) If the velocity of liquid is 1.65ft/s in a 12”
diameter pipe ..what would be the velocity at
the other end where the diameter is 3”
• C) 2000L/min of water flows through 300 mm
diameter pipe that reduces to 150 mm diameter
.calculate the velocity at the two ends.

49. Review #2
1) Use Figure 3.57 page 80 and calculate
the pressure PA.
2) Page 247
3) Page 187

50. Review
I) Analyze the figure given and calculate the
differential pressure ( PA-PB)
II) Determine the minimum velocity of a
gasoline and water flowing at 250C in
turbulent flow manner in a 50mm pipe?
Kinematic viscosity of gasoline and water are
respectively 6.48 10-7m2/s and 1.02 10-6
m2/s
III) 1m3/min of water flows through 200 mm
diameter pipe that reduces to 50 mm
diameter .calculate the velocity at the two
ends

51. Bernoulli equation
• Bernoulli's Equation
• The Bernoulli equation states that,
• Restrictions:
• the fluid has constant density,
• the flow is steady, and
• there is no friction.

52. Bernoulli equation

53. Bernoulli equation
• Consider the steady, flow of a constant density
fluid in a converging duct, without losses due to
friction The flow therefore satisfies all the
restrictions governing the use of Bernoulli's
equation.
• Upstream and downstream of the contraction we
make the one-dimensional assumption that the
velocity is constant over the inlet and outlet
areas.

54. Bernoulli equation

55. Example
• Water at 100C is flowing from section 1 to
section 2. At section 1, the gage pressure
is 345 kPa, the diameter is 25 mm and the
velocity of water is 3 m/s. Section 2 is
above section 1 by 2 m and has a
diameter of 50 mm.
• Calculate P2?

56. Bernoulli equation
• P1 + ½ ρ v1
2 + ρgh1= P2 + ½ ρ v2
2 + ρgh2
• P2= P1 + ½ ρ[ v1
2 - v2
2] + ρg[h1-h2]
• Calculate v2?
• Continuity equation  A1V1= A2V2
• V2 = [A1/ A2] V1
• A1= ¼ ΠD1
2= ¼ x3.14x (25)2= 491 mm2
• A2 = ¼ ΠD2
2= ¼ x3.14x (50)2= 1963 mm2
• V2 = [ 491/1963] x 3 = 0.75 m/s
• ρ = 1000 kg/m3

57. Bernoulli equation
• P2 = P1 + ½ x1000 x [ 32- 0.752 ] +
1000X 9.81 x (-2)
• P2=P1 – 15.4 kPa
• P2 = 345 – 15.4 = 329.6 kPa

58. General Energy Equation
• Objective: We will apply Energy equations
to real systems including pumps, fluid
motors, turbines and energy losses from
friction, valves and fittings
 Analyze the energy in fluids flow
systems by adding terms to Bernoulli
equation

59. What is a turbine?
• Turbines like fluid motors act in opposite
way than pumps
• It takes energy from the fluid and deliver
it in form of work to cause rotating of a
shaft or a linear movement of a piston

60. What is fluid friction?
• A Fluid in motion will lose some energy due to
frictional resistance to flow
• The magnitude of this energy loss depends on:
* properties of fluid
* flow velocity
* Pipe size
* Smoothness of pipe walls
Some of this energy is lost as heat through the
pipe walls

61. What is valve and Fittings?
• Mechanical devices that control the
direction of the flow
• they cause some energy loss usually
small called minor losses

62. Energy Nomenclature
• We will call:
 hA= Energy added to the fluid with a
mechanical device such as a pump often called
total head of the pump.
 hR = Energy removed from the fluid with a
mechanical device such as a turbine or fluid
motor
 hL = Energy losses from the system due to
friction in pipes or minor losses due to valves or
fittings

63. General Energy Equation
• E’
1+ hA-hR-hL= E’
2
• The terms E’
1 and E’
2 are the energy
possessed by the fluid per unit weight as seen in
Bernoulli equation
•
g
v
z
P
E
2
'
2



64. Be Carefull!!!!!
• This general energy equation can be
applied only in the direction of the flow
 An element of fluid at section 1 and
having an energy per unit weight E’
1 may
have energy added ( hA) from a pump , an
energy removed (-hR) from a turbine or
energy loss ( -hL) before it reaches section
2 ( Example see Figure 7.6 page 197)

65. Class work
• Example 7.1 page 198
• Example 7.2 page 199

66. POWER REQUIRED BY PUMPS
• PA= hA. W
where W is called the weight flow rate and is
expressed as N/s
• Since W=γ.Q  PA= hA. γ.Q
• Units: SI 1Watt = 1N.m/s = 1 joule/s
US 1 hp= 550 lb.ft/s
1hp= 745.7 W
1lb.ft/s = 1.356 W

67. Mechanical efficiency of pump
• is the ration between the power delivered
to the fluid PA and the power received by
the pump PI
• eM= PA/PI
• Example 7.3 page 203

68. Fluid motors and turbines
• Power delivered  PR= hR.γ.Q
• Mechanical efficiency is equal to the ratio
of the power output from motor or turbine
Po to the power delivered by the fluid PR
• eM= Po/PR
• Example 7.4 page 205

69. Energy Losses Due to Friction
• Darcy’s equation:
• hL is the energy loss from the system
• One component of the energy loss is due to
friction in the flowing fluid
g
v
z
p
hhh
g
v
z
p
LRA
22
2
2
2
2
2
1
1
1



70. Friction in the flowing fluid
• Friction is proportional to:
* velocity head of the flow v2/2g
* Ratio of the length to the diameter of the
flow stream L/D
• Mathematically, we can write:
g
v
D
L
fhL
2
..
2


71. • hL is the energy loss due to friction (
N.m/N or lb.ft/lb or m or ft)
• L is the length of the flow stream ( m or
ft)
• D is the pipe diameter ( m or ft)
• v is the average velocity of flow ( m/s or
ft/s)
• f friction factor ( dimensionless)

72. When can we use Darcy’s
equation?
• The Darcy’s equation can be used to
calculate energy loss due to friction in
long straight sections of round pipes for
both laminar and turbulent flow
• The difference between laminar and
turbulent flow is in the evaluation of the
friction factor f

73. Hagen- Poiseuille equation
• The parameters involved in energy loss in laminar
flow are:
* Fluid properties : viscosity and specific gravity
* Dynamics of the flow
* geometrical features of length and diameter
• The Hagen- Poiseuille equation:
2
32
D
Lv
hL




74. Friction Factor in Laminar Flow
• The Reynolds Number is defined as :
• Therefore  f=64/NRe

..
Re
Dv
N 
g

 

75. Example
• Problem 9.1 page 242

76. Friction Loss in Turbulent Flow
• There is no formula like in laminar flow
• Experimental data have shown that the
friction factor depends on Reynolds
number and on the roughness of the pipe
• Some values of the roughness are given in
table 9.1 page 243

77. MOODY DIAGRAM
FIGURE 9.2 page 244
• the diagram shows that the friction factor
f plotted versus the Reynolds number with
a series of parametric curves related to
the relative roughness D/ε

78. Description of the Moody Diagram
• Both f and NRe are plotted on logaritmic scales
• At the left side of the graph, we have the
laminar flow equation  f=64/NRe
• In the critical region 2000<NRe< 4000 , there
is no plot, the behavior can not be predicted
• Beyond 4000, the family of curves for different
values of D/ε are plotted

79. Moody Diagram: Turbulent section
• For a given NRe : if D/ε increases  f decreases
• For a given D/ε : if NRe increases  f decreases
• In the zone of complete turbulence, the
Reynolds number has no effects on the friction
factor
• As D/ε increases, the value of NRe where the
complete turbulence starts increases also

80. Examples
• Problems 9.2, 9.3, 9.4 in pages 246-248

81. Minor Losses
• What is valve and Fittings?
• Mechanical devices that control the
direction of the flow
• they cause some energy loss usually
small called minor losses

82. • Darcy’s equation:
• hL is the energy loss from the system
• One component of the energy loss is minor
losses due to valves and fittings.
g
v
z
p
hhh
g
v
z
p
LRA
22
2
2
2
2
2
1
1
1



83. Resistance Coefficient K
• Minor energy losses include elbows,
enlargement of pipe, contraction of pipe and
a valve
• These energy losses are proportional to the
velocity head of the fluid:
)2/( 2
gvKhL 

84. Sudden enlargement
• As shown in Figure 10.1 ( page 272), the
minor loss energy is calculated by the
equation:
• If the velocity v1 is close to 1.2 m/s , the
value of K can be estimated by the relation:
)2/( 2
1 gvKhL 
  22
21 /1 DDK 

85. • Use Figure 10.2 ( page 273) for a more
precise value depending on the velocity
• Work example 10.1 page 273

86. EXIT LOSS
• This is the case where a fluid flows from a
pipe to a tank ( Figure 10.3 page 275)
• The energy loss hL can be calculated as:
Work example 10.3 page 275
gvhL 2/2
1

87. GRADUAL ENLARGEMENT
• In this Figure ( Figure 10.4 page 276),
the transition from small pipe to large pipe
is smoother and the energy loss is
therefore smaller.
• The resistance coefficient K depends on
the ratio D1/D2 and on the value of the
cone’s angle θ ( Table 10.2 page 277)
• Work example 10.4 page 277

88. Sudden Contraction
• The energy loss due to sudden contraction
can be calculated by the relation ( Figure
10.6 page 278):
• The value of K are given in Table 10.7 page
279
• Work example 10.5 page 280
)2/( 2
2 gvKhL 

89. Gradual Contraction
• Figures 10.10 and 10.11 ( pages 281-
282) show that K depends on the ratio
D1/D2 and on the angle θ

90. Entrance loss
• The values of K depends on the shape
and geometry of the entrance.
• Figure 10.13 gives the values of K for
different situations
• Work example 10.6 page 284

91. Valves and fittings
• K = fT(Le/D)
• The value of Le/D are given in table 10.4
page 287 and the values of fT are given in
table 10.5 page 288
• Work example 10.7 page 288

92. Class Work
• Work problems 10.1M , 10.4M, 10.5E,
10.6 M, 10.15M, 10.17 E

93. Homework
• Work problems 10.2, 10.3, 10.8 and 10.16