CE2134 (Ppt) 1. Basic Concepts of Fluids

CE2134 Hydraulics Basic Concepts of Fluids 1 - 1

Basic Concepts of Fluids

1. Symbols and units

A dimension is the measure by which a physical variable is expressed quantitatively.

A unit is a particular way of attaching a number to the quantitative dimension.

The International System of Unites (S.I. Units) is adopted in this module.

Quantity Unit Symbol

length meter (m)

mass kilogram (kg)

time second (s)

electric current ampere (A)

temperature degree Kelvin (0K)

force Newton (N=kg m/s2)

pressure pascal (Pa=N/m2)

stress pascal (Pa=N/m2)

work, energy joule (J=N m)

power watt (J/s)

frequency hertz (/s)

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Prefix Symbol Multiply factor

peta P 1015

tera T 1012

giga G 109

mega M 106

kilo k 103

centi c 10-2

milli m 10-3

micro 10-6

nano n 10-9

pico p 10-12

femto f 10-15

CE2134 Hydraulics Basic Concepts of Fluids

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a constant

specific weight (g) (N/m3)

boundary layer thickness (m)

dimensionless ratio (z/)

angular measure (rad)

ratio of model to prototype variables (dimensionless)

dynamic viscosity coefficient (kg/m)

mass density (kg/m3)

surface tension (N/m)

shear stress (N/m2)

kinematic viscosity coefficient (m2/s)

< > symbol for time averaging

..


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1.1 Dimensional homogeneity

Fundamental laws of physics representing a physical phenomenon must be valid for any system of units and hence should be dimensionally homogeneous.

Newtons second law states

From the requirement of dimensional homogeneity, the units of the force are ML/T2.

The basic system of units would be represented by the mass (M), the length (L) and time (T).

Stokes-Oseen formula for the drag force on a sphere of diameter D moving with velocity V in a fluid of density and viscosity , respectively as given by

F ma=

2 22 23 2 2

2 29316ML M L MLLT M L MLLT T T L

L T T

F DV V D = +


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The volumetric flow rate Q through a hole, distance h from the liquid surface and of diameter D, on the side of a tank as given by

However, for whichever system of units that is being adopted, the equation is only valid for the same liquid.

Example 1.1: The drag force FD exerted on a car travelling at a constant speed V

depends on a dimensionless drag coefficient CD, the density of the air , the car velocity V and the projected frontal area A of the car. Based on unit consideration alone, obtain a relation for the drag force.


3 1322

2

20.68L

L LT L LTT

Q D gh

=

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Solution:


( ) ( ) ( ) ( )

( ) ( ) ( )

1 2 3

1 2 3

depends on , , ,

, , ,D D

D D D

assumed functional form

D D

F C A V

F f C A V C A V

F C A V

= =

=

( )

( ) ( ) ( )

1 3 1 2 312

3

1 2 3

3 22

2 3

1

3 2

1 2 3

2

For equation to be dimensionally homogeneous

12 11 3 2

D D

D

ML M L M LLT L T T

F C A V

C AV

+ + = =

= = == + +

=

=

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2. Characteristics of fluids

A fluid is a substance which changes shape continuously so long as shear forces, however small, are present.

A solid is a substance which can resist a shear force even when at rest. The shear force may result in some displacement but does not continue to move indefinitely.

When a fluid is at rest, no shear forces are present in it.

A fluid owes its shape to the container it is in and constrained by the forces exerted on it.

All substances are composed of huge numbers of molecules separated from each other by empty space.


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2.1 The continuum

In engineering practice, we are generally interested in the average values of a measurable property at a given location. The property may be density, pressure or temperature.

Unlike the mathematical point, the location is assumed to have an extremely small volume sufficient to contain enough molecules to reflect the property to be measured.

A mole of gas has about 6.0241023 molecules and under normal conditions occupies a volume of 22.4 liters or 0.26891020 molecules to every cm3. One micron is 110-6 m or 10-4 cm so that one cubic micron contains 0.26891020(104)3 2.689107 molecules.


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Example 2.1: At the molecular level, the diameter of an oxygen molecule is about 31010

m and its mass is 5.31026 kg. What is the average number of molecules in a volume of 109 mm3 at the given pressure and temperature? What is the average density of oxygen based on this volume?


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A fluid particle may be thought of to be the collection of molecules within the very small volume of the order of one cubic micron.

In fact, for elemental volumes below a volume, say 109 mm3, there is considerable uncertainty in establishing the mass density of the gas

Beyond a certain volume, 10-9 mm3 say, we obtain consistent results for the mass density.


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The velocity of a fluid particle can now be defined to be the average of the momenta of all the molecules constituting the particle divided by the mass of the particle.

a fluid particle may be seen to consist of Ni molecules belonging to the ith substance in the particle. The subscript i is just an index which is used to indicate the ith substance.


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Example 2.2: The velocity of a fluid particle is defined to be the average of the moment of

all the molecules constituting the particle divided by the mass of the particle. Where the fluid particle consists of Ni molecules (i=1, 2, n) for the ith substance in the particle, show that

Solution:


1 - 13 CE2134 Hydraulics Basic Concepts of Fluids

3. Basic fluid properties

Water

Air

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Mass density () 103 (kg/m3)

Viscosity at 20C () 10-3 (Pas)

Kinematic Viscosity at 20C ( = /) 10-6 (m2/s)

Mass density (air) 1.3 (kg/m3)

Viscosity at s.t.p. (air) (1/60 th that of water) 1.710-5 ( Pas)

Kinematic viscosity at s.t.p. (air = air / air) 1.310-5 (m2/s)

Gas constant (R = pair/Tabs) 287 (J/kgK)

Specific heat capacity at constant pressure (Cp) 1000 (J/kgK)

Specific heat capacity at constant volume (C) 715 (J/kgK)

Ratio of specific heat capacities () 1.4

Vapour pressure at 20C 2340 (N/m2)

Bulk modulus of elasticity at 30C 2.25106 kN/m2


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4. Taylor series expansion

Consider f(x, t) to be a continuous function of a single variable x for the time t.

For very small values of x, the higher powers (x)n for n > 1 will be very much smaller and the right hand side of the expansion can be truncated, thus incurring a truncation error which is expected to be very small also.

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

220 0 0 0

0 0 0 0 2

0 00 0 0 0 0 0

, ,, ,

2!,

, , ,

f x t f x t xf x x t f x t x

x xf x t

f x t x f x t f x tx

+ = + + +

+ = +


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Example 4.1: Water flows through a pipe of circular cross section at a steady rate. The

radius of the pipe is R. The flow velocity u(r) at a distance r from the centreline is given by

What is the velocity of the flow at r = 0.5R? What are the velocities of the

flow at r = 0.52R using the formula given above and using Taylors expansion?

Solution:

( ) 221

m

u r ru R

=


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5. Newtons viscosity law

If we plot the velocity u for various points along z, we obtain the velocity profile.

The term u/z is the inverse of the slope of the tangent at z and it is called the velocity gradient. It represents that spatial rate of change of the velocity with z.


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Newton (1642-1727) postulated that when straight and parallel fluid flow takes place, the tangential shear stress between adjacent fluid layers is proportional to the velocity gradient in the direction perpendicular to the layers.

is called the coefficient of viscosity and it has units of kg/ms.

u uz z

=


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Example 5.1: The viscosity of a fluid is to be measured by a viscometer constructed of

two concentric cylinders of length l m. The inner rotating cylinder has a radius of R m and the gap between the two cylinders is h m. The gap is small such that the velocity profile in the gap is linear. The inner cylinder rotates at a constant speed of N revolutions per minute (RPM) in the counter clockwise direction and the torque measured is Nm. Define your coordinate frame and derive the expression for the viscosity of the fluid. Assume that the gap size is small such that the velocity profile across the gap is linear.

Determine the viscosity of the fluid if l = 0.4 m, h = 0.0015 m, R = 0.12 m, N

= 300 RPM and = 1.8 Nm.


Solution:


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Example 5.2: The air hockey game at Rocky Recreation Room has pucks of mass 30 g

with a diameter of 100 mm. If the air hockey table is poorly installed with a surface slope of 2 from the horizontal. A player fires a puck uphill, and the velocity of the puck after the strike is 10 m/s. The air film under the puck is 0.1 mm thick. Determine its speed at the instant it passes the goal 2 m away.

Solution:


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For most fluids, the viscosity is independent of the rate of deformation forming a class of fluids which is called Newtonian.

There also exists a large class of non-Newtonian fluids where the viscosity is not independent of the rate of deformation.


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6. Perfect gas and equation of state


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A temperature reading TC in the Rankine scale can be expressed in Kelvin scale according to the following equation

No real gas is perfect although their properties match quite closely with those of the perfect gas. Most gases obey this equation quite closely.

The equilibrium conditions so specified means that the gas is not undergoing any accelerative motion and free of heat exchanges. For air, the value of R is 287 J/kg K.

0 0 273T K T C= +


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Example 6.1: Calculate the density of air when the absolute pressure and temperature are

respectively 140 kPa and 500C if R = 287 J/kg K.

A hydrogen filled balloon is to expand to a sphere 20 m diameter at a height of 30 km where the absolute pressure is 1100 Pa and the temperature is -400C. If there is to be no stress in the fabric of the balloon, what volume of hydrogen must be added at the ground level where the absolute pressure is 101.3 kPa and the temperature is 150C?

Solution:


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Example 6.2: An artificial atmosphere consists of 20% oxygen and 80% nitrogen by

volume at 1 atmosphere (standard atmosphere defined as 100 kPa) and 150C. What is the density and partial pressure of the oxygen and the density of the nitrogen? R of oxygen is 259.8 J/kg K and R of nitrogen is 296.8 J/kg K.

Solution:


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6.1 Vapour pressure

The ability of the molecules to leave the liquid increases with temperature. When this happens and there is residual liquid left, the space is said to be saturated with the vapour of the liquid and the vapour pressure is called the saturation pressure corresponding to the given temperature.

Thus, at a given temperature, the pressure at which a pure liquid changes phase (to gaseous) is called the saturation pressure.

If there is no residual liquid in the container, that is, all the liquid have evaporated in the container, the partial pressure of the vapour is less than the saturation pressure for the given temperature.

In a mixture of gases, the partial pressure of a gas is the pressure of the gas in the mixture. The atmosphere consists of dry air and water vapour. Atmospheric pressure consists of the partial pressures of various constituents in the dry air and the partial pressure of the water vapour.


Example 6.5: (a) If water vapour in the atmosphere has a partial pressure of 3500 Pa and

temperature of 300C, what is its density? (b) If the barometer reads 102 kPa abs, what is the partial pressure (dry) air

and what is its density? (c) What is the density of the atmosphere (air plus water vapour present)? Solution:


Example 6.6: At approximately what temperature will water boil in Mexico City where the

elevation is 2256 m?

Solution:


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7. Compressibility

All liquids are slightly compressible; however, under many situations, the density changes associated with changes in pressure are very small.

The compressibility effects of liquids are taken into account only when there are abrupt large changes in pressures. The bulk modulus K of a liquid is defined by

is the very small volumetric change (change in volume) associated with p the small increase in the pressure applied to the material.

The density is expressed in terms of mass per unit volume and = m/. Hence, for a fixed mass of the fluid (m is fixed),

/pK =

0/

m

p pK

= + = =

= =


Gases, on the other hand, are generally compressible. However, they are treated as incompressible when the associated density changes are again very small as with some subsonic flows around buildings.

Example 7.1: At normal atmospheric conditions, approximately what pressure must be

applied to water to reduce its volume by 2%? Solution:


Example 7.2: At normal atmospheric conditions, what is the change in the air pressure

when the percentage change in the mass density of the air is of the order of 1% given that the speed of a pressure wave in air is given by and at normal atmospheric conditions it is of the order of 340 m/s?

Solution:

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/c K =


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8. Cohesion and adhesion

Adhesion forces act to hold two separate bodies together whilst cohesion forces act to hold atoms, ions or molecules of the same body together. Capillarity results from the combination of adhesion and cohesion.

For a water drop on a surface, both cohesion and adhesion occur but one is stronger than the other.

Surface tension comes about from the forces between the molecules of a liquid and the forces between the liquid molecules and those of the adjacent substance.


If the water molecules are more strongly attracted to each other than with the surface, they tend to bead up. The quantity of liquid will adjust its shape until its surface area is a minimum.

A drop of liquid in free space will therefore take on a geometrical shape with the minimum surface area (for stable equilibrium) and this shape is a sphere.

If there is stronger attraction to the surface, they spread out as they try to get closer to the surface material.


A liquid surface is always under tension. If one imagines a line on a liquid surface, the liquid on one side of the line pulls on the liquid on the other. The surface tension is the pulling force acting perpendicular to the line.

Water in contact with air has a surface tension of about 0.073 N/m at normal temperature. For mercury, it is about 0.48 N/m. For all liquids, surface tension decreases with temperature. Surface tension of water is reduced when organic solutes such as soap and detergents are present.


is the surface tension between the liquid and glass at room temperature, is the contact angle or liquid angle measured from the contact surface through the liquid to the liquid air interface, is the mass density of the liquid and D is the diameter of the glass tube.

When the bubble is stationary, the balance of forces consists of the downward force due to pressure difference and the resisting forces due to surface tension.

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2

cos4

4 cos 2 cos

D hD g

hgD gr

=

= =

( ) ( )

( )

20

0

2

2

icircumference

projected area

i

p p r r

p pr

=

=


When the bubble is in the air (like the soap bubble), the balance of forces again consists of the downward force due to the pressure difference and the resisting forces due to surface tension only that now the soap bubble has 2 areas of exposure to the air, inside the bubble and outside the bubble.

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( ) ( )

( )

20

0

2 2

4

icircumference

projected area

i

p p r r

p pr

=

=


Example 8.1: An insect of mass M = 3106 kg rests on the water surface. The leg in

contact with the water surface may be approximated by a hemisphere of radius r = 2105 m. If the surface tension of the water with the insect is = 0.072 N/m, find the contact angle. An insect has 6 legs.

Solution:


Basic Concepts of FluidsSlide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 62. Characteristics of fluidsSlide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 133. Basic fluid properties4. Taylor series expansionSlide Number 165. Newtons viscosity lawSlide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 236. Perfect gas and equation of stateSlide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 327. CompressibilitySlide Number 34Slide Number 358. Cohesion and adhesionSlide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41

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CE2134 (Ppt) 1. Basic Concepts of Fluids