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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
INTRODUCTION
Fluid is important in life especially to human because 95% of human body and 2/3 of earth
surface is water. In Fluid Mechanics aspect, it’s involved in many activities done in our lives
like weather and climate, vehicles, environment, physiology and medicine, and many more.
As an engineer, Fluid Mechanics is important to study in order to properly design many
facilities such as:
i. Water Supply System
ii. Wastewater Treatment Facilities
iii. Dam Spillways
iv. Valves, Flow meters, Hydraulic Shock absorbers and Brakes
v. Air Craft ,ships ,submarines, and rockets
vi. Windmills ,turbines and pump
vii. Heating and air conditioning, etc.
1.1 WHAT IS FLUID?
Fluid can be defined as:
Material continuum that is cannot to withstand a static shear stress but instead deforms
continuously under the influence of such forces, irrespective to its magnitude.
Fluid also responds with an irrecoverable flow and unable to retain any unsupported
shape (takes the shape of any solid body which comes into contact).
Fluids can be appearing to solids when it has a very small deformation rates when
flowing under their own weight such as ice.
It can be including gases (or vapour) or liquid phases of the physical forms.
It is also can be categorised into Newtonian Fluid and Non-Newtonian Fluid.
What are the differences between liquids and gases?
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Liquids and gases can be classified as a Fluid. The major differences between liquids and
gases are stated in Table below.
Table 1 The differences between liquids and gases.
Liquids (oil,water,etc) Gases (oxygen,air,etc)
Incompressible – if all pressure removed
(except its own vapour pressure). The
cohesion between molecules holds them
together, so it’s does not expand
indefinitely.
Compressible and easily deformed when the
external pressure is removed, its tend to
expand indefinitely.
Fixed volume Volume is changes respect to
pressure/temperature.
Have free surface if the volume of container
is greater than that liquid
Cannot form a free surface
Irrespective of the shape/size of its
container
Completely fill any vessels in which it is
place.
What is a vapour?
A vapour is a gas where the temperature and pressure it is very near the liquid phase.
Example is steam because its state and condition is not from that of water. Water vapour is
responsible for humidity. A highly superheated vapour can classified as a gas where its state
is far removed from the liquid phase. Air is a one of example of highly superheated vapour
and considered as a gas due to its state very far from the liquid air. Changes in pressure and
temperature or both will affect the volume of gas and vapour and significant changes in
temperature and phase will involve during hot phenomena or known as thermodynamics.
What are the differences between Fluid and Solid ?
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STRESS
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
The differences of fluid and solid is state in following table :
SOLID FLUID
It is ‘hard’ and not easily deformed and
molecules of a solid are close together.
It is ‘soft’ and easily deformed (we can
readily move through air)
Attractive forces between molecules large
and tends to retain its shape
E.g: steel ,concrete, etc.
Attractive forces between molecules are
smaller.
An ideal elastic solid deform under load and
deformation continue as long as load applied
if not it will return to original state. E.g :
Span , spring,etc.
While for plastic solid, deformation will
continue under sufficient load and can
prevent from rupture but after load is
removed it does not return to its original
state.
Intermolecular cohesive forces in fluid are
weak to hold various elements of the fluid
together. Therefore, fluid can flow under
action of the slightest stress and continue
flow as long as the stress is present.
E.g : water, oil etc.
1.2 WHAT IS FLUID MECHANICS?
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Fluid mechanics is the study of liquids behaviour and gases at rest and in motion.Fluid
Mechanics can be categories into three classes as following:
Figure 1 Three categories of Fluid Mechanics.
1.3 UNITS AND DIMENSION
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FLUID MECHANICS
FLUID STATICS
the study of the mechanics of fluids at
rest , ζ =0
KINEMATICS deals with velocities
and streamslines without considering forces and energy
FLUID DYNAMICS
concerns with the relations with
between velocities and accelerations and the forces exerted by
or upon fluids in motion
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
In Fluid Mechanics, student will be able to study and understanding the fluid properties.
Therefore, it is important for student to know the various systems of units and dimension
used in Fluid Mechanics and engineering. There is two unit standards applied in this
subjects, include S.I (International system of units) and British Engineering (or FPS) systems
of units.
International System of Units (S.I)
The basics measurement used in this systems are length (metres, m), mass (kilogramme,
kg), and time (seconds, s).
British Engineering (or FPS) Systems of Units
Meanwhile in this system, the fundamental mechanical dimensions used are length (foot,
ft), mass (pound, lb), and time (seconds, s).
Basic Dimensions
In order to solving varies problem in Fluid Mechanics, there are three
basic dimension is used which is length (L), time (T) and mass (M).
The common dimensions and systems of units is the basic knowledge which needs to be
remembered by engineering student. Table 2 has listed the important common dimension
and units used in Fluid Mechanics.
Table 2 List of basic dimensions and units applied in Fluid Mechanics.
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Dimension (MLT
system)
SI units To convert
from SI to FPS
multiply by
FPS units
Acceleration LT-2 m.s-2 3.281 ft.s-2
Area L2 m2 1.076E+1 ft2
Density ML-3 kg.m-3 1.940E-3 or
6.243E-2
Slugs.ft-3 or
lbm.ft-3
Energy ML2T-2 N.m or J 7.376E-1 lb.ft
Force MLT-2 N 2.248E-1 lb
Length L m 3.281 ft
Mass M kg 6.852E-2 or
2.205
Slug or lbm
Power ML 2T-3 N.m.s-1 7.376E-1 ft.lb.s-1
Pressure ML-1T-2 N.m-2 2.089E-2 lb.ft-2
Volume flowrate L3T-1 m3.s-1 3.531E+1 ft3.s-1
Velocity LT-1 m.s-1 3.281 ft.s-1
Viscosity
(dynamic)
ML -2T N.s.m-2 2.089E-2 lb.s.ft-2
Viscosity
(kinematic)
L2T-1 m2s-1 1.076E+1 ft2s-1
Specific Weight M.L-2T-2 N.m-3 6.366E-3 lb.ft-3
Temperature θ °C TF=1.8Tc +32° °F
(R.Munson, F.Young, & H.Okiis, 2006)
Example 1
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Determine the dimensions using MLT systems, for:
a) The product of forces times volume
b) The product of pressure times mass divided by area
c) Moment of a force divided by velocity
Solution
a) Forces x Volume = N x m3
= MLT-2 x L3 = ML 4 T -2
b) Pressure x mass /area = N.m-2 x kg /m2
= ML-1T-2x M /L2 = M 2 L -3 T -2
c) Moment of force /velocity = Nm x m3
= ML 2T-2/LT-1 =MLT -1
1.4 FLUID PROPERTIES – DENSITY
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
1.4.1 Mass Density, ρ (rho)
In Fluid Mechanics , Mass density can be describe as mass per unit volume :
ρ (rho)= m /V
= γ /g (kg/m3 )
whereas : ρ = kg/m3 .In international system , a typical values for water is 1000 kg/m3 at
4°c. (See Appendix 1, in Table 1)
1.4.2 Specific Weight ,γ(gamma).
Specific weight is the weight per unit volume, γ(gamma). The equation for specific unit
weight is :
γ = ρg = W/V (N/m3)
whereas ; γ = N/m3. Usually, the values for water is 9.81 x103 N/m3 and other additional
values can be refer in Appendix 1, Table 1.
1.4.3 Specific Gravity (or Relative Density), SG .
Specific gravity or relative gravity, SG is defined as dimesionless ratio of its mass density t
or weight of fluid to the mass density or weight of a substance taken as a standard. In
this matter, solid or liquid are referred to water as standard at the temperature of 4°C
(39.2°F) which the density of water at this temperature is 1000 kg/m3.
SG of substance = density of substance /density of water@4°C
= weight of substance / weight of equal volume of water
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
= specific weight of substance/specific weight of water
Whereas : SG = dimensionless and no unit.The typical values for water,1.0 ; benzene,0.879
and other substances values can be refer to Appendix 2, Table 2.
1.4.4 Specific Volume , v.
It can be defined as the reciprocal of the mass density which is can be described as volume
per unit mass .
ν = 1 /ρ (m3/kg)
whereas : ν = m3/kg .However this property is always used in thermodynamics compare
than fluid mechanics.
Example 2
If 4 m3 of oil weighs 27 kN, calculate its specific weight γ, mass density ρ, and specific
gravity .
Solution
Specific weight , γ = ρg = W/V
=27 kN / 4 m3 = 6.75 kN/m 3
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Mass density , ρ (rho)= m /V = γ/g
= (6.75 x1000 N/m3)/ 9.81 m/s2 = 688.07 kg/m 3
Specific gravity ,SG = specific weight of oil/specific weight of water
= 6.75 kN/m3 / 9.81 kN/m3 = 0.688
1.5 IDEAL GAS LAW
Gases are highly compressible in comparison to fluids, with changes in gas density
directly related to changes in pressure and temperature through the equation
p=ρRT.
The ideal gas equation of state p=ρRT is a model that relates density to pressure and
temperature for many gases under normal conditions.
The pressure in the ideal gas law must be expressed as an absolute pressure which is
measured relative to absolute zero pressure.
The standard sea-level atmospheric pressure is 14.6996 psi (abs) or 101.33kPa (abs).
EXAMPLE 3
A compressed air tank has a volume of 0.84 ft3. When the tank is filled with air at a gage
pressure of 50 psi, determine the density of the air and the weight of air in the tank.
Assume the temperature is 70 °F and the atmospheric pressure is 14.7 psi (abs).
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
SOLUTION 3
PRESSURE VS. WEIGHT
1.6 FLUID PROPERTIES - VISCOSITY OF A FLUID
What is viscosity of a fluid?
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ρ=PRT
=. . ..=0 .0102 slugs/ ft 3
W=ρg (volume )=(0 .0102 slugs/ ft3 ) (32.2 ft /s2) (0 .84 ft 3 )=0 .276 slug⋅ft /s2=0 .276 lb
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
The viscosity of fluid is a measure of the amount of its resistance to a shear force or
extensional stress. There two factors which can affect the resistance of a fluid from flowing,
which is:
a) Cohesive forces between molecules – liquids have stronger cohesive forces than
gases.
b) Molecules interchange between layers of moving fluid – slower moving fluid layers
pulling the faster moving layers and vice versa.
When temperature is increase, viscosities of typical liquids will decrease and viscosities
of all gases will increases. It happens because force of cohesion will decrease with
temperature, predominates with liquids, while with gases the predominating factor is
the interchange of molecules between the layers of different velocities.
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Figure 2 Dynamic (absolute) viscosity of some common fluids as a function of temperature.
What is no-slip condition?
The fluid ‘sticks’ to the solid boundaries so their velocities are zero relative to the solid
boundaries and occurs with all viscous fluids.
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
How to describe the “fluidity (viscosity)” of the fluid?
The bottom plate is rigid fixed, but the upper plate is free to move. If a solid, such as steel,
were placed between the two plates and loaded with the force P, the top plate would be
displaced through some small distance, da. The vertical line AB would be rotated through
the small angle, db, to the new position AB’.
What happens if the solid is replaced with a fluid such as water?
When the force P is applied to the upper plate, it will move continuously with a velocity U.
The fluid “sticks” to the solid boundaries and is referred to as the NON-SLIP conditions. The
fluid between the two plates moves with velocity u=u(y) that would be assumed to vary
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
linearly, u=Uy/b. In such case, the velocity gradient is du / dy = U / b. In a small time
increment, δt, an imaginary vertical line AB would rotate through an angle, δβ , so that
tan δβ≒δβ = δa / b .Sinceδa = U δt it follows that δβ= U δt / b δβ ? →δβ=δβ(P, t).
Defining the rate of shearing strain, g, as .
The shearing stress is increased by P, the rate of shearing strain is
increased in direct proportion,
Figure 3 Linear variation of shearing stress with rate of shearing strain for common fluids.
All fluids are viscous, "Newtonian Fluids" obey the linear relationship given by Newton's law
of viscosity. , which we saw earlier. Where is the shear stress,
Units ;
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γ̇= limδt→0
δβδt
=Ub
=dudy
τ∝du /dy
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Dimensions .
is the velocity gradient or rate of shear strain, and has
Units: ,
Dimensions
is the "coefficient of dynamic viscosity" as below.
i. Dynamic Viscosity (or absolute viscosity), µ (mu)- the shear stress,ζ ,required to drag one fluid layer with unit velocity past another layer located at unit distanced away.
Units: Newton seconds per square metre, or Kilograms per meter per
second, .
(Although note that is often expressed in Poise, P, where 10 P = 1 .)
Typical values:
Water =1.14 , Air =1.78 , Mercury =1.552 ,
Paraffin Oil =1.9 .
ii. Kinematic Viscosity ,ν (nu)
Kinematic Viscosity, , is defined as the ratio of dynamic viscosity to mass density.
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Units: square metres per second,
(Although note that n is often expressed in Stokes, St, where St = 1 .)
Dimensions: .
Typical values:
Water =1.14 , Air =1.46 , Mercury =1.145 ,
Paraffin Oil =2.375 .
Example 3.
The density of oil is 850 kg/m3. Find its relative density and Kinematic viscosity if the
dynamic viscosity is 5 x 10-3 kg/ms.
Solution
ρ oil = 850 kg/m3 , ρwater = 1000 kg/m3
Therefore, SG oil = 850 / 1000 = 0.85
Dynamic viscosity = µ = 5 x10-3 kg/ms
Kinematic viscosity = ν = µ / ρ
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
What is Newton’s equation of viscosity ?
Fluids for which the shearing stress is linearly related to the rate of shearing strain are designated as Newtonian fluids after I. Newton (1642-1727).
Most common fluids such as water, air, and gasoline are Newtonian fluid under normal conditions.
Fluids for which the shearing stress is not linearly related to the rate of shearing strain are designated as non-Newtonian fluids.
It can be divided into the following categories :
Shear thinning fluids- The viscosity decreases with increasing shear rate – the harder the fluid is sheared, the less viscous it becomes. Many colloidal suspensions and polymer solutions are shear thinning. Latex paint is example.
Shear thickening fluids - The viscosity increases with increasing shear rate – the harder the fluid is sheared, the more viscous it becomes. Water-corn starch mixture water-sand mixtures are examples.
Bingham plastic -neither a fluid nor a solid. Such material can withstand a finite shear stress without motion, but once the yield stress is exceeded it flows like a fluid. Toothpaste and mayonnaise are common examples.
Figure 4 Variation of shearing stress with rate of shearing strain for several types of fluids , including common non-Newtonian fluids.
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
EXAMPLE 4
The velocity distribution for the flow of a Newtonian fluid between two wide , parallel
plates is given by the equation
where V is the mean velocity. The fluid has a velocity f 0.04 lb·s/ft2. When V=2 ft/s and
h=0.2 in. determine: (a) the shearing stress acting on the bottom wall, and (b) the shearing
stress acting on a plane parallel to the walls and passing through the centerline (midplane) .
SOLUTION 4
For this types of parallel flow the shearing stress is obtained from
0h
Vy3
dy
du
ft/lb4.14h
Vy3
dy
du
h
Vy3
dy
du
0y2midplane
2
hy2wallbottom
2
------------ eqn (1)
Thus, if the velocity distribution u = u(y) is known, the shearing stress can be determined at
all points by evaluating the velocity gradient, du/dy. For the distribution given
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u=3V2 [1−( yh )
2]
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
0h
Vy3
dy
du
ft/lb4.14h
Vy3
dy
du
h
Vy3
dy
du
0y2midplane
2
hy2wallbottom
2
-------eqn (2)
a) Along the bottom wall y = -h so that (from eqn 2)
dudy
= 3Vh
And therefore the shearing stress is
τ bottomwall=μ( 3Vh )= (2 N .
sm 2 ) (3 )(0.6
ms)
(5mm)(1m /1000mm)
= 720 N/m 2 (in direction of flow)
This stress creates a drag on wall . Since the velocity distribution is symmetrical , the
shearing stress along the upper wall would have the same magnitude and direction .
b) Along the midplane where y = 0 it follows from eqn 2 that
dudy
= 0
And thus the shearing stress is
τmidplane = 0
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
PROBLEM 1
The velocity distribution of a viscous liquid (dynamic viscosity = 0.9 Ns/m2) flowing over a
fixed plate is given by u = 0.68y - y2 (u is velocity in m/s and y is the distance from the plate
in m).
What are the shear stresses at the plate surface and at y=0.34m?
PROBLEM 2
Explain why the viscosity of a liquid decreases while that of a gas increases with a
temperature rise.
The following is a table of measurement for a fluid at constant temperature.
Determine the dynamic viscosity of the fluid.
du/dy (s-1) 0.0 0.2 0.4 0.6 0.8
t (N m-2) 0.0 1.0 1.9 3.1 4.0
Using Newton's law of viscocity
where m is the viscosity. So viscosity is the gradient of a graph of shear stress against
vellocity gradient of the above data, or
1.7 COMPRESSIBILITY OF FLUIDS
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
1.7.1 Bulk Modulus
Liquids are usually considered to be incompressible, whereas gases are generally considered compressible.
Compressible of the fluid?
A property, bulk modulus Ev, is used to characterize compressibility of fluid.
The bulk modulus has dimensions of pressure. FL-2.
1.7.2 Compression and Expansion of Gases
When gases are compressed or expanded, the relationship between pressure and density depends on the nature of the process.
For isothermal process, pρ=constant
>> Ev=p
For isentropic process, pρk
=constant
>> Ev=kp
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Ev=− dpd V /V
= dpdρ/ ρ
CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
Where k is the ratio of the specific heat at constant pressure, cp, to the specific heat at
constant volume, cv.
cp – cv=R=gas constant .
Example 5
A 0.03 m3 of air at an absolute pressure of 101.3 kPa is compressed isentropically to
0.015m3 by the tire pump .
What is the final pressure?
Solution
For an isentropic compression
pi
ρik=
p f
ρfk
Where the subscript i and f refer to initial and final states, respectively. Since we are
interested in the final pressure, pf , it follows that
pf = ( ρf
ρi )k
pi
as the volume , ∀ , is reduced by one half, the density must double , since the mass , m = ρ∀
, of the gas remains constant. Thus , with k = 1.40 for air
pf = (2)1.40 ( 101.3 kPa) = 267 kPa. (abs)
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
1.8 VAPOUR PRESSURE
If liquids are simply placed in a container open to the
atmosphere, some liquid molecules will overcome
the intermolecular cohesive forces and escape into
the atmosphere.
If the container is closed with small air space left
above the surface, and this space evacuated to form
a vacuum, a pressure will develop in the space as a
result of the vapor that is formed by the escaping
molecules.
When an equilibrium condition is reached, the vapor
is said to be saturated and the pressure that the
vapor exerts on the liquid surface is termed the
VAPOR PRESSURE, PV.
Vapor pressure is closely associated with molecular activity; the value of vapor
pressure for a particular liquid depends on temperature.
Boiling, which is the formation of vapor bubbles within a fluid mass, is initiated when
the absolute pressure in the fluid reaches the vapor pressure.
The formation and subsequent collapse of vapor bubbles in a flowing fluid, called
cavitation, is an important fluid flow phenomenon
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CHAPTER 1 FLUID PROPERTIES
DATE OF LECTURE: _________________________
After completing this chapter, student should be able to:
Determine the dimensions and units of physical quantities. Identify the key fluid properties used in the analysis of fluid behaviour. Calculate common fluid properties given appropriate information. Explain effects of fluid compressibility Use the concepts of viscosity, vapour pressure, and surface tension.
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