Introduction of fluid properties

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?

Prepared by :Siti Fatimah Hj Che Osmi


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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 ?


STRESS


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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?


http://en.wikipedia.org/wiki/File:Stress-strain1.svg


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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


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


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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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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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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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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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= 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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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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SOLUTION 3

PRESSURE VS. WEIGHT

1.6 FLUID PROPERTIES - VISCOSITY OF A FLUID

What is viscosity of a fluid?


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


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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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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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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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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 ;


γ̇= limδt→0

δβδt

=Ub

=dudy

τ∝du /dy


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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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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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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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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


u=3V2 [1−( yh )

2]


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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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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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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


Ev=− dpd V /V

= dpdρ/ ρ


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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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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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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.


Introduction of fluid properties

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