Fluid Mechanics 2011

Transilvania University of BrasovThermodynamics and Fluid Mechanics Department

Angel HUMINIC

FLUID MECHANICSTheory and Applications

2007

1CONTENTSINTRODUCTION ..... 31. FLUIDS ..... 31.1 Description of fluids .. 31.2 Continuum Hypothesis. Concepts of homogenous and isotropic medium .. 5

1.3 Models for the fluid description ... 7

2. PROPERTIES OF FLUIDS 82.1 Pressure . 8

2.2 Density .... 10

2.3 Specific weight ... 11

2.4 Isothermal compressibility 12

2.5 Velocity of sound ... 13

2.6 Mach number .... 13

2.7 Viscosity ..... 14

2.8 Applications of fluid properties 20

3. FUNDAMENTALS OF FLUID STATICS . 243.1 Forces on fluids ..... 24

3.2 Eulers equation for fluids statics ... 25

3.3 Particular forms for the fundamental equation of statics . 27

4. FLUID FORCES ON SUBMERGED SURFACES . 334.1 Fluid forces on plane surfaces ...... 34

4.2 Fluid forces on curved surfaces .. 36

4.3 Floating bodies .. 39

4.4 Static pressure measurement by manometers 40

4.5 Examples of fluid statics .. 44

5. IDEAL FLUID DYNAMICS ... 525.1 Description of the flow field .. 52

5.2 Acceleration in a flow field ... 55

5.3 Equations of the fluid motion ... 55

6. APLICATIONS OF THE BERNOULLIS EQUATION .. 626.1 Flow through small orifices .. 62

6.2 Time for a tank discharge .... 63

6.3 Flow over notches and weirs ... 64

6.4 Mixing of fluids. The ejectors .. 65

6.5 Fluid metering .... 67

7. APPLICATIONS OF THE MOMENTUM EQUATION ... 727.1 Hydrodynamic forces on a flat plate ... 72

27.2 Hydrodynamic force on a pipe nozzle 73

7.3 ENERGY FROM WIND. The Axial Momentum Theory Betzs Theory . 74

8. REAL (VISCOUS) FLUIDS FLOW ..................................................... 788.1 Laminar and turbulent flows. Experiments of Reynolds .. 78

8.2 Velocity distribution on laminar and turbulent flows .... 79

8.3 Steady flow in pipelines. Head loss and Bernoullis augmented equation .. ... 81

8.4 Friction factor computation .. 83

8.5 Transient flow in pipelines. Water hammer ... 91

9. BOUNDARY LAYER THEORY 929.1 General description of the boundary layer 92

9.2 Boundary layer separation .. 96

9.3 Navier-Stokes equations . 98

9.4 The notion of resistance, drag, and lift .. 100

9.5 Applications of boundary layer theory 102

10. ROTODYNAMIC MACHINERY . 10610.1 Introduction to rotodynamic machinery .. 106

10.2 Centrifugal pumps .. 106

10.3 Axial flow machinery .. 110

10.4 Introduction to hydraulic turbines . 112

11. DIMENSIONAL ANALYSIS 11611.1 Introduction to dimensional analysis 116

11.2 Dimensions and units 116

11. 3 Dimensional homogeneity 117

11.4 Results of dimensional analysis 118

11.5 Buckingham's theorems 11811.6 Common groups . 12111.7 Scale models testing .. 122

12. REFERENCES .. 126

31. INTRODUCTIONAs its name suggests, Fluid Mechanics is the branch of Applied Mechanics concerned with the

fluids. The analysis of the behaviour of fluids is based on the fundamental laws of Mechanics,

which relate continuity of mass and energy with force and momentum, together with their physical

properties.

Fluid Mechanics is involved in nearly all areas of Engineering either directly or indirectly, being one

of the oldest and broadest fields of Engineering. It encompasses aerodynamics, hydrodynamics,

gas dynamics, flows in turbomachineries, computational fluid dynamics (CFD), convection heat

transfer, acoustics, biofluids, physical oceanography, atmospheric dynamics, wind engineering a.o.

Fluid Mechanics has practical importance in almost all human activities, from meteorology (flow in

atmosphere) and, in fact, even astronomy (motion of interstellar gas) up to medicine (flow of fluids

inside human body). Modern design of aircraft, spacecraft, automobiles, ships, land and marine

structures, power and propulsion systems, or heat exchangers is impossible without a clear

understanding of the relevant Fluid Mechanics.

Also, fluids are involved in transport and/or conversion of the energy in almost its forms. The flow

problems are present if we study fuel combustion on one side and water evaporation on the other

side in a boiler, flow of heat-transporting medium in a nuclear reactor, in the blade system of a

turbine, in recycling and coolant pumps and also in the processes taking place in the cylinders of

the internal combustion engines and others.

The aims of this course are to introduce Fluid Mechanics and to establish its relevance in

engineering. It consists of some llectures which are presenting the concepts, theory and practical

applications. Worked examples will also be given to demonstrate how the theory is applied.

1.1. FLUIDS

1.1.1 Description of fluids

We normally recognise three states of matter: solid, liquid and gas. Name fluids is a generalizeddescription meant to include both liquids and gases. There are two aspects of Fluid Mechanics,

which make it different to Solid Mechanics.

1. The nature of a fluid is much different to that of a solid

2. In fluids we usually deal with continuous streams of fluid without a beginning or end. In

solids we only consider individual elements.

4A solid can resist a deformation force while at rest, this force may cause some displacement but

the solid does not continue to move indefinitely.

In contrast to solids the fluids do not have the ability to resist deformation. Because a fluid cannot

resist the deformation force, it moves, it flows under the action of the force. Its shape will change

continuously as long as the force is applied. The deformation is caused by shearing forces, which

act tangentially to a surface. Referring to the figure below, we see the force F acting tangentially on

a rectangular (solid lined) finite element ABCD. This is a shearing force and produces the (dashed

lined) rhombus element A'B'CD, defined by the angular displacement :

yx)( tg =

Fig 1.1 Angular displacement of a fluid element due to shear deformationIn consequence it can say that:

A fluid is a substance that deforms continuously, or flows,

when it is the subject to shearing (tangential) stresses (forces).

As will be discussed later on (see, Fluids Properties, Viscosity), the magnitude of the stress

depends on the rate of the angular deformation, in opposition as solids for what the magnitude of

the stress depends on the magnitude of the angular deformation. This definition implies a very

important point, which state that:

If a fluid is at rest there are no shearing forces acting.

At rest all forces must be perpendicular to the planes which they are acting.

When a fluid is in motion shear stresses are developed if the particles of the fluid move relative to

one another. When this happens adjacent particles have different velocities. If fluid velocity is the

same at every point then there is no shear stress produced: the particles have zero relative

velocity.

5This is very well illustrated by the case of the rivers flow. At the bottom of the river, the velocity of

the water is zero and will increase toward the top of the river. This change in velocity across the

direction of flow is known as velocity profile and is shown graphically in the Figure 1.2:

Fig 1.2 Velocity profile in a river cross section

Because particles of fluid next to each other are moving with different velocities there are shear

forces in the moving fluid i.e. shear forces are normally present in a moving fluid. On the other

hand, if a fluid is a long way from the solid boundary, nearly to the top of the river in the presented

case, all the particles are travelling with the same velocity, the velocity profile would look closely

like in Figure 1.3.

Fig 1.3 Velocity profile in uniform flows

As presented in Figure 3 there will be no shear forces if all particles have zero relative velocity. In

practice we are concerned with flow past solid boundaries; aeroplanes, cars, pipe walls, river

channels etc. and shear forces will be present.

1.1.2 Continuum Hypothesis. Concepts of homogenous and isotropic medium

All fluids are composed of molecules discretely spaced and in continuous motion. In the definitions

used to describe fluids, this discrete molecular structure was ignored and the fluid was considered

as a continuum. This means that all dimensions in a fluid space are taken as large compared to the

molecular spacing.

In Fluid Mechanics, the continuum concept is expressed through that, for any point )z,y,x(P of a

fluid, we can associate a continuum description for each quantity (variable) that characterizes the

fluid, e.g. pressure, temperature, velocity etc., for a certain time t :

- pressure )t,z,y,x(pp = ,- temperature )t,z,y,x(TT = ,- velocity )t,z,y,x(vv = ,

6and these function are continuous ones, consequently

they are derivative. The reason one uses the continuum hypothesis is that it allows us to use

differential calculus to analyze the properties of a fluid and its behaviour.

Continuous changes, or gradients, in physical properties and forces define mechanical problems,

and differential calculus is the mathematical tool that treats such gradients. Such a view is

reasonable from a modelling point of view as long as the mathematical model generates results

which agree with experiment. Practicaly we can assume that, the ratio of the mean free path length

of the molecules of a fluid is small, the continuum hypothesis is more accurate (see details about

Knudsen number in bibliography). Thus, the continuum concept is not always applicable. For

example, in the cases of highly compressible flows or flows in very dilute gases there will be

regions of discontinuity, such as the shock waves on the surfaces of an airplane which is flying

close to the sound speed, 1M = (Mach: ratio between fluid velocity v and sound speed c , seeLecture 2, Fluids Properties), see Figure 1.4.

Fig 1.4 Shock wave on airplane breaking the sound barrier

A fluid is considered homogenous if its density is constant for constant conditions of temperature

and pressure.

ctctT,p == Off course, no real fluid is homogeneous in an absolute sense. Homogeneity must be specified

relative to a characteristic length, e.g. the size of the probe used to measure the properties of the

fluid in experiments.

A fluid medium can be considered as isotropic if it has the same properties in all directions around

a point.

The concepts of homogenous and isotropic medium are useful in study of Fluid Mechanics

because they allow us to consider that the Equations established for an elementary volume of fluid

(fluid particle) are applicable for entire fluid.

71.1.3 Models for the fluid description

It is necessary to emphasize that generally, fluids are very complicated objects of study and as a

consequence, it is almost always necessary to work with certain simplified fluid models. Such

models usually neglect properties not important in the mechanical context, as taste, smell, colour,

pH value and others. Also, in order to obtain mathematical simplifications, many models often

neglect even mechanical properties as long as they are of secondary consequences for a

phenomenon under study.

Common models for fluid are the following ones:

Newtonian fluids: named after Sir Isaac Newton, who studied fluid motion, are fluids having

directly proportionality between stress and rate of angular deformation

starting with zero stress and zero deformation. In these cases, the constant

of proportionality is defined as the absolute or dynamic viscosity (see

Properties of fluids). The more common fluids like air and water are

Newtonian ones.

Non-Newtonian fluids:fluids having a variable proportionality between stress and deformation rate

(basically everything other than Newtonian fluid). A vast number of fluids

which are not commonly encountered but which are extremely important,

nevertheless, are non-Newtonian: plasmas, liquid crystals, foams a. o. Also,

some of the plastics behave like a solid and above a stress magnitude they

have fluidic behavior. Rheology is the subject that treats plastics and non-

Newtonian fluids.

Ideal fluids: fluids having no viscosity (Euler model). The effect of the shear stress

between the adjacent layers of fluid is neglected.

Light fluids: Fluids having small mass density, as gasses and vapours. The effect of their

own weight can be neglected.

Incompressible fluids: fluids which have not significant variation of their volume with the pressure

changes, and as a consequence with constant mass density (Pascal model).

It is used in study of the liquids (heavy fluids).

When a subject is theoretically studied, is desirable to use the simplest possible model of fluid and

to apply a more complicated one only when is absolutely necessary.

82. PROPERTIES OF FLUIDS

A property is a characteristic of a substance which is invariant when the substance is in a particular

state. In each state the condition of the substance is unique and is described by its properties.

The properties outlines below are general properties of fluids, which are of interest in Engineering.

The symbol usually used to represent the property is specified together with some typical values in

SI units for common fluids. Values under specific conditions (temperature, pressure etc.) can be

found in reference books. V

2.1 PRESSURE, p

If a volume of fluid is isolated as a free body, the force system acting on the volume includes

surface forces over every element of area bounding the volume. In general, a surface force will

have components perpendicular and parallel to the surface. At any point, the perpendicular

component per unit area is called the normal stress. If this is a compression stress, it is called

pressure intensity, or simply pressure.

Thus, the pressure is defined as (pressure) force F per unit area A applied on a surface in a

direction perpendicular to that surface. Mathematically, it is expressed as:

dAdF

AFlimp

0A==

, (2.1)

in differential form, or simply:

AFp = (2.2)

Also, in a fluid pressure may be considered to be a measure of energy per unit volume or energy

density by means of the definition of work. For a force exerted on a fluid, this can be seen from the

definition of pressure:

VolumeEnergy

VW

dAdF

AFp ==

== (2.3)

Pressure is a scalar quantity, and its unit in the SI is ][N/m2 , or Pascal [Pa] science 1971, after

the name of French mathematician, physicist and philosopher Blaise Pascal (1623 1662). The SI

multiplies are:

the kilopascal, ,Pa 10kPa 1 3= and the megapascal Pa 10MPa 1 6= .

9A non-SI multiple, still in use in many parts of the world, is the bar:

Pa 10bar 1 5= .

Other units in use include the following:

the technical atmosphere having as symbol: at

== 242 mN 1080665.9

cmkgf1at 1 510

2mN 1= [ ]bar . (2.4)

the standard (physical) atmosphere (atm or At ): is an establishedconstant and it is approximately equal to the air pressure at earth

mean sea level; the first one who measured the atmospheric pressure

was the Italian physicist and mathematician Evangelista Torricelli

(1608 1647). In 1643 he created a tube by 1 meter long, sealed at

the top end, filled it with mercury, and set it vertically into a basin of

mercury (see Figure 2.1); the column of mercury fell to about 70 cm,

leaving a vacuum above, this was the first barometer and the unit of

pressure torr ( mmHg 1torr 1 = ) was named after him; as we nowknow, the height of column fluctuated with changing atmospheric

pressure and the acceleration due to gravity on Earth's surface; the 4th

Resolution of the 10th General Conference of Weights and Measures stated that:

[ ]mmHg 760mN 101325atm 1 2 =

= (2.5)

Because pressure is commonly measured by its ability to displace a column

of liquid in a manometer, pressures are often expressed as a depth of a

particular fluid (see Figure 2.2). The most common choices are mercury

(Hg) and water (or alcohol); water is nontoxic and readily available, while

high density mercury allows for a shorter column (and so a smaller

manometer) to measure a given pressure. The pressure exerted by a

column of liquid of height h and density is given by the hydrostaticpressure equation:

[ ]fluid of column h g p = . (2.6)

23

233

2 mN 81.9m10

sm81.9

mkg10OH mm 1 ==

23

23 mN 875.7m10

sm81.9

mkg803alc mm 1

10

2mN 322.133Hg mm 1 =

Accordingly with two previous shown situations, pressures can be measured:

relative to an absolute zero value (E. Torricelli) (called absolute pressure absp ) or,

relative to the atmospheric pressure at the location of the measurement atmp , called gaugepressure gp or simply p ; an example of this is the air pressure in an automobile tire, which

might be said to be "2.2 bar, but is actually 2.2 bar above the atmospheric pressure; gauge

pressure is a critical measure of pressure wherever one is interested in the stress on

storage vessels and the plumbing components of fluidics systems:

atmabs ppp = (2.7)While pressures are generally positive, from the previous Equation we can see that there are

situations in which a negative pressure may be encountered if atmabs pp < (vacuummetricpressure).

O the basis of reaction to pressure, a subdivision of fluids into two main classes, either

compressible or incompressible, can be made. All gases and vapours are highly compressible.

Liquids by comparison are only slightly compressible. As we shall see, compressibility introduces

thermodynamic considerations into fluid flow problems. If incompressibility can be assumed, it is

much easier to describe the state of the fluid and its behaviour in motion. With some important

exceptions, liquids usually are treated as incompressible for practical purposes. Gases, on the

other hand, can be treated as incompressible only if the change in pressure is small throughout the

flow system.

2.2 DENSITY, The density of a substance is the quantity of matter contained in a unit volume of the substance.

Mathematically, it is defined as the rate of mass per volume and has the SI unit [ ]3m/kg :Vd

dm=

3m

kg . (2.8)

A given amount of matter is said to have a certain mass that is treated as an invariant. Thus it

follows that density is a constant so long as the volume of the given amount of matter is unaffected

i.e., for a gas, so long as pressure and temperature conditions are the same, or generally for

homogeneous fluids. In this case:

Vm=

3mkg . (2.9)

11

The previous formula is applicable also, for the average density of an matter. Generally, the

density of a substance depends on point of measurement, pressure p and temperature T . This is

holding good especially for gasses. Their density can be computed from the Equation of State,

applied for two conditions, one of them being known (as reference):

TT

pp 00

0 = , (2.10)

where: terms with index 0 are the reference parameters.

The Equation (2.9) express the absolute density. In order to facilitate the measurement of the

substance density, sometimes is used the relative density r , defined as the ratio of mass densityof a fluid to some standard (reference) mass density ref :

.ref

fluidrfluid )(

= (2.11)

For liquids, which are treated generally as fluids with constant density, this reference is the

maximum density for water, 3apa kg/m 1000= . For gases, the reference is the air density instandard conditions: 30 kg/m 225.1air = for Hg mm 760p air0 = and K 15.288T air0 = .

The reciprocal of density is the specific volume, v .

1=v (2.12)

Associated with density, there can be defined another parameter, respectively the:

2.3 Specific weight, Specific weight (or less often specific gravity) is defined as the rate of weight per volume and has

the SI unit [ ]3m/N :Vd

dG=

3m

N(2.13)

For homogeneous fluids can be defined as the force exerted by gravity upon a unit volume of the

substance:

Vg m=

3mN

(2.14)

The Relationship between and g can be determined by Newton's 2nd Law, since: g= (2.15)

12

2.4 ISOTHERMAL COMPRESSIBILITY, k AND BULK MODULUS OF ELASTICITY, Compressibility is the measure of change of volume and density when a substance is subjected to

normal pressures. It is defined as relative change in volume (or density) for a given pressure

change:

dp1dk

0VV=

Nm2 . (2.16)

where: 0V is the original volume.

The negative sign indicates a decrease in volume with an increase in pressure, as shown in Figure

below.

Fig. 2.3 Variation of volume in a cylinder due to pressure change

The reciprocal of compressibility is known as the bulk modulus of elasticity:

0

ddp

k1

VV

==

2m

N .(2.17)

If we take the derivative of Equation .ctVm == , we find that:

dddd0dddm ===+=

VVVVVV , (2.18)

and the Equations (2.16), (2.17) can be rewritten as:

dpd1k =

Nm2 . (2.19)

ddp=

2mN . (2.20)

Compressibility data for liquids are usually given in terms of , as determined experimentally.Theoretically, should depend on the manner or process in which the volume or density changeis effected, e.g., isothermally, adiabatically, etc. For the common gases (such as oxygen), these

two processes give (see also the thermodynamics properties of gases):

p= (for isothermal process); (2.21)

13

p= (for adiabatic process); (2.22)where:

vp c/c= ratio of specific heat at constant pressure to that at constant volume;p absolute pressure.

The effect of the process is smaller for liquids than for gases.

Associated with these two parameters, there can be defined another one, respectively the:

2.5 VELOCITY OF SOUND, c

It is given by the Equation:

k1

ddpc ===

sm (Newton). (2.23)

The magnitude of c is the velocity with which small-amplitude pressure signals will be transmitted

through a fluid of infinite extent or through a fluid confined by completely rigid walls. The density

change caused by an infinitely small pressure wave occurs almost frictionlessly and adiabatically.

For the liquids the speed of sound is usually determined from experimental values of and .Using the adiabatic bulk modulus, Equation (2.22), the following Equation can be applied quite

accurately for the common gases:

p c =

sm . (2.24)

2.6 Mach number, Ma

Mach number is a dimensionless measure of relative speed. It is defined as the speed of an object

relative to a fluid medium, v , divided by the speed of sound in that medium c ; as it is defined as a

ratio of two speeds, it is a dimensionless number:

cvMa = [ ] . (2.25)

The Mach number is named after Austrian physicist and philosopher Ernst Mach (1838 1916).

The Mach number is commonly used both with objects travelling at high speed in a fluid, and with

high-speed fluid flows inside channels such as nozzles, diffusers or wind tunnels. At a temperature

of 15 degrees Celsius, Mach 1 is 340.3 m/s in the atmosphere. The Mach number is not a

constant, being temperature dependent.

According with Mach value, the following classification of fluid flows can be made:

25.0Ma < Subsonic incompressible flows;8.0Ma25.0

14

5Ma2.1 Hypersonic flows.

2.7 VISCOSITY

Viscosity is the property of a fluid, due to cohesion and interaction between molecules, which offers

resistance to sheer deformation. Different fluids deform at different rates under the same shear

stress. Fluid with a high viscosity such as oil, deforms more slowly than fluid with a low viscosity

such as water. It is one of the most important properties of fluids due to it stand on the base of the

principle of fluids motion. The reciprocal of viscosity is fluidity.

2.7.1 Newton's Law of Viscosity

As mentioned before, fluids do not have the ability to resist deformation. Its shape will change

continuously as long a stress is applied. These deformations are caused by shearing forces, which

act tangentially to a surface. Referring to the figure below, we see the force F acting tangentially on

a rectangular (solid lined) finite element ABCD. This is a shearing force (stress) and produces the

dashed lined rhombus element A'B'CD.

Fig 2.4 Angular displacement of a fluid element due to shear deformation

A particle in point E (Figure 2.4) moves under the shear stress to point 'E , in time t , on distance

x . The deformation that this shear stress causes is measured by the size of the angle , which isknow as shear strain:

yx)( tg = (2.26)

In a solid, shear strain is constant for a fixed shear stress . In a fluid, increases as long as is applied. It has been found experimentally that the rate of shear strain (shear strain per unit time)

is directly proportional to the shear stress. For small deformations we can write the shear strain as:

yx= , (2.27)

and rate of shear strain:

15

dydv

dy1

dtdx

dtd == . (2.28)

where:

dydv is the change in velocity with y, in the direction perpendicular to the layers,

or the velocity gradient.

Using the experimental results the proportionality between shear stress and rate of shear strain

was expressed by the following Equation (Sir Isaac Newton):

dydv = , (2.29)

where:

is the constant of proportionality, known as the dynamic (molecular) viscosity.

Equation (2.29) is known as Newton's law of viscosity.

The relationship between the shear stress and the velocity gradient can be easelly understanding

from the following experiment, ilustrated below.

Fig 2.5 Laminar shear of fluid between two plates

Let consider two plates separated by a homogeneous fluid, spaced at a distance h . The two

plates are considered very large, with the area A , such that the edge effects may be ignored, and

that the lower plate is fixed. If a constant force F is applied to the upper plate, it will be put in

motion with the constant speed v . The first layer of fluid, atached on moving plate has same

velocity as the plate and this velocity will decrease in depth to 0v = , velocity of the layer attachedon fixed plate. This decreasing of the velocity in depth is due to shear stress, , between layers offluid, that ultimately opposes to any applied force: ( )AF= . On the other hand, the applied forceis proportional to the velocity of the plate, vF , (higher velocity means higher force/stress) andinversely proportional to the distance between the plates yF (lower distance means higherforce/stress). Combining these three relations results in the equation (2.29), expressed in terms of

shear stress.

16

2.7.2 Dynamic viscosity, The dynamic viscosity is defined as the shear force, per unit area, (or shear stress), required to

drag one layer of fluid with unit velocity past another layer a unit distance away.

dndv =

2msN or

smkg (2.30)

where:n : normal unitary vector.

Equation (2.30) shows that the velocity distribution in a fluid is a continuous field. Otherwise, the

shear stress becomes infinite for a finite velocity change between two adjoining points.

In Technical System the units of is P (Poise): scm

g 1P 1 = . Typical values are:

skg/m 1014.1 3water = ;skg/m 1078.1 5air = (for the standard atmosphere);

skg/m 9.1lparaffinoi = ;skg/m 552.1mercury = .

2.7.3 Kinematic viscosity, Kinematic viscosity is defined as the ratio of dynamic viscosity to mass density:

=

sm2 . (2.31)

In Technical System the units of is St , Stokes (named after Sir George Gabriel Stokes, (1819 -1903), an Irish mathematician and physicist, who at Cambridge made important contributions to

fluid dynamics, including the Navier-Stokes equations, optics, and mathematical physics):

scm 1St 1

2= .

Typical values are: s/m 1014.1 26water = ;s/m 1046.1 25air

= (for the standard atmosphere);s/m 10375.2 23lparaffinoi

= ;s/m 10145.1 24mercury

= .

17

2.7.5 Viscosity in gases and liquidsViscosity in gases arises principally from the molecular diffusion that transports momentum

between layers of flow. It is independent of pressure and increases as temperature increases.

In liquids, the additional forces between molecules become important. This leads to an additional

contribution to the shear stress. Thus, in liquids viscosity is independent of pressure (except at

very high pressure) and tends to fall as temperature increases (for example, water viscosity goes

from 1.79 cP to 0.28 cP in the temperature range from 0 C to 100 C).

The dynamic viscosities of liquids are typically several orders of magnitude higher than dynamic

viscosities of gases. The Sutherland's formula can be used to compute the dynamic viscosity of

various materials as a function of the temperature:

+

+= 23

0

00 T

TCTCT

smkg , (2.32)

where:

0 dynamic viscosity for standard atmosphere (at 0p and 0T );C Sutherland's constant (in the case of air K 111C = ).

For water, the Poiseuilles Equation can be used to compute, with good accuracy, the kinematic

viscosity:

2

6

t00022.0t0337.011078.1

++=

sm2 , (2.33)

wheret : is the temperature of water in Celsius degrees.

2.7.6 Classification of Newtonian / Non-Newtonian FluidsFluids obeying Newton's law where the value of is constant are known as Newtonian fluids. If is constant the shear stress is linearly dependent on velocity gradient. This is true for most

common fluids. Fluids in which the value of is not constant are known as non-Newtonian fluids.There are several categories of these, briefly outlined below. These categories are based on the

relationship between shear stress and the velocity gradient (rate of shear strain) in the fluid.

1 - Ideal fluid: this is a fluid that is assumed to have no viscosity. This is a useful concept

when theoretical solutions are being considered - it does help achieve some

practically useful solutions;

2 Solids: no rate of shear strain for shear stress;

3 Newtonian fluids: linear dependency between rate of shear strain and shear stress;

18

4 - Dilatant fluids: viscosity increases with rate of shear e.g. quicksand.

5 - Pseudo-plastic: no minimum shear stress necessary and the viscosity decreases with rate of

shear, e.g. colloidal substances like clay, milk and cement,

6 - Bingham plastic: shear stress must reach a certain minimum before flow commences; above

this value of shear stress they have a Newtonian behaviour, not applicable

for real plastics.

Fig 2.6 Shear stress versus rate of shear strain

Each of these curves can be represented by the equation:n

dydvBA

+= (2.34)

where:

A , B and n are constants. For Newtonian fluids: 0A = , =B , 1n = .

2.8 THERMODYNAMIC PROPERTIES

2.8.1 Specific heat, c

Specific heat or specific heat capacity is the ratio of quantity of heat flowing into a substance per

unit mass, to the change in temperature. It is determined experimentally or can be computed using

molecular theory.

The specific heat for gases and vapours depends on how the change in state is effected: at

constant volume (or density), vc , or at constant pressure, pc . For a perfect gas, a gas which

obeys the equation of state (2.41), both are related through the relations:

Rcc vp += [J/kg K]; (2.35)

=v

p

cc

. (2.36)

Thus:

R1

cp = ;

1Rcv = . (2.37)

where:R [J/kg K]: constant of the studied gas;

[-] adiabatic constant for the gases; 4.1= for air.

19

2.8.2 Specific internal energy, u

The specific internal energy, measured as energy per unit mass, is due to the kinetic and potential

energies bound into a substance by its molecular activity. It depends primarily on temperature. For

a perfect gas, it can be computed according to the relation:

dT cdu v= [J/kg]. (2.38)2.8.3 Specific enthalpy, hThe specific enthalpy represents the sum:

puh += [J/kg]. (2.39)

For a perfect gas the enthalpy depends of temperature only and can be computed according to the

equation:

dT cpuddh p=

+= . (2.40)

2.8.4 Equation of state for gasesFor gases the pressure (as absolute value) p , the mass density , and the temperature T arecorrelated. A useful approximation is the theoretical equation for perfect gasses (Clapeyron -

Mendeleev):

T M

pT RpT R m p R=== V (2.41)where:

R [J/kg K]: constant of the studied gas;

R = 8314.3 [J/kmolK]: universal constant of the gasses;

M [kg/kmol]: molar mass (molecular weight).

There are the following dependences between density and pressure for a gas:

Constant volume process - ctV = .0ct == (2.42)

Isothermal process - ctT = :

0

0pctp == (2.43)

Adiabatic process - 0Q = 00pctp == (2.44)

Polytrophic process (general process)

n0

0n

pctp == (2.45)where:

n polytrophic exponent for the gases; 3.1n = for air.

20

2.8 APPLICATIONS OF FLUID PROPERTIES

WORKED EXAMPLES

Exercise 1For checking (or calibration) of manometers, the following experimental set-up (see Figure 2.7),

with a pressure pump, can be used:

Fig. 2.7

It consists on the piston 1 moving in the cylinder 2, through the whirling of the screw piston rod 3.

The pump is feed with oil from the tank 4, through the pipe 5, where are attached the manometers,

first one used as reference 6, and the tested one 7. Compute the number of revolutions of the

piston rod necessary that the reference manometer to show a pressure 2cm/kg 75p = . Thefollowing are known:

mm 36d = inner diameter of the cylinder;mm 8.0h = pitch of the screw piston rod;

3cm 2000=V initial volume of oil;N/m 1085.4k 210= oil compressibility.

Solution:

It is recommended to convert all the quantities in the International System of Units (if is necessary):

m 036.0 m1036 mm36d 3- === ;m 0008.0 m108.0 mm8.0h 3- === ;

3363 m 002.0m 102000cm 2000 === V ;

Nm 1085.4k

210= ;

26

242 mN 1080.9

mN

1081.9100

cmf kg 100p === .

21

Through the whirling of the screw piston rod, the piston is moving by a distance hnl = (number ofrevolutions x pitch). The oil will be compressed in the cylinder and pipe, due to an increasing of

pressure p , from 0 (initial gauge pressure shown by manometer) to 2cm/kg 100p = .Using Equation of the isothermal compressibility,

p k =VV , (2.46)

and replacing the volume variation with:

4d h n

4d lV

22 == (2.47)it found that:

2

2

d hp k 4np k

4d h n

VV

== (2.48)

srovolution 7.11036.0108.0

1081.91085,41024n 236103=

=

Exercise 2

A plate having the surface 2m0.8 S = and mass kg2 m = is sliding on an inclined plane with the angle= 30 which is covered with an oil film by thickness mm2 = . Oil mass density is 3kg/dm 0.9 =

and the kinematic viscosity stokes0.4 = . Compute the velocity of the plate in uniform motion.

Fig. 2.8

Solution:2m 8,0S = ;

kg 0,2m = ;= 30 ;

m0.002 m 102mm 2 -3 === ;3333 m/kg 900m/kg 109,0kg/dm 9,0 === ;s/m1040,0s/cm0,40 stokes 40,0 242 === .

22

Due to the action of the tangential component of the weight

sinGGT = , plate is sliding on the inclined plane with the velocityv , which is constant in uniform motion (when viscous friction force is

equal with TG ).

From the Newtons Equation of tangential stress in a fluid: Fig. 2.9

v

SGT == (2.49)

where: = is the dynamic viscosity.

S sin g mvv

S sin g m

== (2.50)

m/s 681.00,8104,0109,030 sin10281,92v 43

3=

=

Exercise 3Determine the effect of the water temperature on magnitude of sound speed knowing the density and the

modulus of elasticity:3

water kg/m 1000= and 29water N/m 10914.1 = for C 4twater = ;3

water kg/m 26.999= and 29water N/m 10020.2 = for C 20twater = .

Solution:

Using the Newtons Equation for the sound speed in fluid:

== ddpc (2.51)

it found that:

m/s 13881000

10914.1c9

water == if C 4twater = ;

m/s 142226.99910020.2c

9

water == if C 20twater = .

In consequence, the sound speed in water is increasing with the temperature rise.

SELF-ASSESSMENT EXERCISES

Exercise 4

A piston is moving with the constant velocity s/cm1.0v = in a pressure cylinder having the innerdiameter mm50D = and length cm10l = , which is filled with a liquid with the modulus of

23

elasticity 24 cm/daN 102 = . Compute the time and the displacement x [mm] of the piston if thepressure in cylinder increase from zero to bar200p = . Make a sketch.

Exercise 5The velocity distribution of a viscous liquid flowing over a fixed plate is given by the Equation:

2yy68.0u =where;

u is velocity in [m/s]

y is the distance from the plate in [m].

What are the shear stresses at the plate surface and at m0.34 y = , if the dynamic viscosity of thefluid is 2ms N 9.0= ? Draw the variation )y( = for m 0.34)0(y = .

Exercise 6In a fluid the velocity measured at a distance of mm0 5 from the boundary is m/s 1 . The fluid has

dynamic viscosity Pas2 and relative density 0.8 , to the water density. What is the velocity

gradient, and shear stress at the boundary assuming a linear velocity distribution? What is the

kinematic viscosity of the fluid in St (Stokes)? Make a sketch.

Exercise 7

Compute the sound speed in air for a temperature C 20tair = , for an adiabatic dependencebetween density and pressure. The molar mass of the air is kg/kmol 96.28Mair = and theadiabatic exponent is 4.1= . The universal constant of the gasses is [J/kmolK]8314.3 =R .

Exercise 8Explain why the viscosity of a liquid decrease while that of a gas increases with a temperature rise.

24

3. FUNDAMENTALS OF FLUID STATICSFUNDAMENTAL EQUATION OF FLUID STATICS

Characteristic for the Fluid Statics is that the fluids are at rest, with no relative motion between fluid

elements, so there are no shear forces (or shear stress) acting on a fluid. In consequence real

fluids at rest can be treated as ideal ones, having no viscosity.

The fundamental problems of Fluid Statics are the following:

determination of the distribution of pressure in a homogeneous fluid; computation of the fluid forces on submerged surfaces; pressure measurement using manometers

3.1 FORCES ON FLUIDS

Two types of forces are acting on a fluid, having a mass m in a

volume V bounded by a surface S (see Figure 3.1):

mass forces and surface forces.

In the following, only the external forces will be considered, due to the

internal ones are balancing each other.

The elementary (differential) mass forces acting on a fluid element is

expressed by the Equation:

Vd fdm fFd mmm rrr == (3.1)

where: mfr

specific mass force (unitary mass force); it has dimension of acceleration; generally,

for fluids on the gravity field of Earth gfmrr = (gravity acceleration).

The resultant mass force is:

=V

Vd fF mm rr

(3.2)

The surface forces represent the result of the interaction of fluids with surrounding environment

(other substances, either solids or fluids), through the surface S :

dS fFd SSrr = (3.3)

where: Sfr

specific surface force (unitary surface force); it depends of surface position vector

Fig. 3.1

25

rr

, normal unitary vector on surface nr

(pointed outwardly from the surface) and

time t : )t,n,r( ff SSrrrr =

Generally, a surface force will be at an angle to the surface, thus having both normal and

tangential components: nFdr

, respectively Fdr

:

dS f cos dS f cos FdFd nSSnrrrr === , (3.4)dS f sin dS f sin FdFd SS

rrrr === , (3.5)where nf

rnormal unitary surface force; it defines the pressure p on the surface dS :

npfnrr =

fr

tangential unitary surface force; it defines the shear stress on the surface dS ;

3.2 THE EULERs EQUATION FOR FLUID STATICS

The equation for fluid statics can be obtained from the equilibrium of the forces on an arbitrary

elementary volume of fluid:

0FF pm =+rr

. (3.6)

If the fluid is at rest the tangential component of the surface force is zero 0Fd =r

, thus the

resultant surface force is:

=S

dS n pFSrr

. (3.7)

In this way Equation (3.6) becomes:

0dS n pd f0FF mSm =+=+ SV

Vrrrr

. (3.8)

Using first Gauss theorem:

=VS

Vd p dS n pr

. (3.9)

The vector differential operator is (in Cartesian coordinates Oxyz ):

zk

yj

xi

++

= rrr (3.10)

where: k ,j ,irrr

are the unitary vectors on Ox , Oy and Oz directions.

From Equations (3.6) and (3.7):

0d p d fm = VV

VVr (3.11)If the volume 0V , then previous Equation reduces to:

0p grad 1f0p 1fp f mmm === rrr

0 (3.12)

26

Equation (3.12) represents the Eulers equation for fluid statics in vectorial form. In Cartesian

coordinates it can be writing as:

=

=

=

0zp 1f

0yp 1f

0xp 1f

mz

my

mx

(3.13)

where mxf , myf , mzf are the scalar components of mfr

.

Multiplying the equation (3.12) by rdr

and taking into account that rdpr is the total differential of

pressure:

dpdzzpdy

ypdx

xprdp =

++

= r , (3.14)

we obtain the Euler s Equation of fluid statics (sum of two differential continuous functions):

0ddp1dp1rdfm =+= rr

, (3.15)

where )z,y,x( is the potential of the mass forces (the potential energy due to the mass forces).If the components of unitary mass forces are known, then:

xUfmx = ,

yUfmy = ,

zUfmz = ,

( ) ++=== dzfdyfdxfrdfd mzmymxm rr , (3.16)By integration, Equation (3.15) leads to:

ct.dp =+ . (3.17)The Equation (3.17) is known also as the fundamental Equation of fluid statics. It describes the

pressure field in fluids, either compressible or incompressible.

The first term, dp , represents the potential of the pressure forces. In order to solve this integral,

the relationship between pressure and density must be know for gasses.

For fluids having the density as function of pressure, the equipotential surfaces ct= representalso surfaces of constant pressure, such of those which are correspond to the interfaces to the

fluids. As consequence, the free surfaces of fluids are surfaces of constant pressure. The shape

and orientation of these surfaces depend on the unitary mass force mfr

. They are normal

(perpendicular) on mfr

.

27

Fig. 3.2

Fig. 3.3

3.3 PARTICULAR FORMS FOR THE FUNDAMENTAL EQUATION OF STATICS

3.3.1 Light fluids

For fluids having small mass density, as gasses and vapours, the effect of their own weight can be

neglected. We can consider that the potential of mass forces is null: 0d = . From the Equation(3.15) we obtain that:

.ctp0dp1 == , (3.18)Thus, the pressure in a finite mass of light fluid is constant.

3.3.2 Fluids on gravitational field of Earth

For a Cartesian reference as in figure below, with the Oz in the sense of the altitude rising (natural

in study of the Earth atmosphere), the scalar components of the specific mass force mfr

acting on a

fluid are the following:

+==

===

ctz gdz g gf 0ff

mz

mymx .

In the case of liquids, fluids with constant density .ct= , the potential of thepressure forces is:

+== .ctpdp1dp (3.19)Thus, the Equation for incompressible fluids at rest can be written as:

ct.z pctz g pctz gp =+=+=+ (3.20)where: is the specific gravity of the fluid;This is fundamental Equation for hydrostatics. It gives the pressure

distribution in a liquid, which is a linear relationship between depth

(measured from the free surface) and pressure for liquids.

The integral constant .ct must be computed knowing the boundary

conditions. In the case of a liquid in an open tank (see Figure 3.3), at

the level of the free surface, hz = , the pressure is the atmosphericone 0p . For this case, the Equation (3.20) gives the integral

constant:

cth pcth g phz 00 =+=+= (3.21)Substituting the constant into Equation (3.20) we found that:

)zh( pp 0 = (3.22)

28

where: 0pp is the gauge pressure; in fluid mechanics we are dealing generally withrelative (gauge) pressure, except for the thermodynamic processes of

gasses. In the following the gauge pressure will be denoted simply as p .

zh is the depth, measured from the free surface.For the shown case, pressure has the maximum magnitude on the bottom of the tank:

h pmax = . (3.23)For the previous case, if the pressure at the surface is mp , as shown in Figure 3.4 (liquid in a

closed tank, under pressure) in a reference with the Oz in the sense of the depth rising (natural in

study of the liquids), then:

h pp mmax += . (3.24)Equation (3.23) expresses the Pascals principle, which

state that any externally applied pressure is transmitted

undiminished in an enclosed static fluid. This is making

possible a large multiplication of pressure force and has

many practical applications, such as hydraulic pressure,

automatic transmissions used in automobile, brakes,

power steering a.o.

The graphical representation of the pressure variation on

the walls of the tank is known as the pressure diagram.

The plane where the (gauge) pressure is zero is called

the manometric plane and its level, from the free surface,

is defined by manometric high:

m

mph = . (3.25)

Thus, Equation (3.24) can be written as:

)h(h p mmax += . (3.26)

3.3.3 Relative equilibrium of liquids

In the previous section, the equilibrium of liquids was treated in absolute sense. Two other cases of

static equilibrium of liquids, having technical applications, are presented in the following:

liquids in tanks on accelerating translation, applicable in transport of liquids in largereservoirs;

liquids in (cylindrical) tanks on rotational motion, applicable in centrifugal casting, designof impellers a.o.

Fig. 3.4

29

Both cases are concerning the relative equilibrium of liquids, in accelerating systems, attached on

carrying tanks.

3.3.3.1 Liquids in tanks on accelerating translationFor this case, lets consider a liquid, having the specific gravity , in an open tank by length l , upto the level h (see Figure 3.5), which start to move with the acceleration cta = .

Fig. 3.5

When the tank accelerates, the specific mass force mf will have two components on considered

reference system:

along Ox , inertial acceleration afmx = ; along Oz , local gravitational acceleration gfmz = .

The free surface of liquids will change its orientation under the action of the mass forces. It comes

to the equilibrium at an angle with the horizontal:

gatg arc= (3.27)

The potential of the mass forces is, from Equation (3.16):

( ) ( ) ctz gx adz gdx adz fdx f mzmx ++=+=+= (3.28)Also, the potential of the pressure forces for liquid ( ct= ) is:

ctpdp1dp +== (3.29)Thus, the Equation for the relative equilibrium of liquids in accelerating translations is:

ctz gx apctUdp =++=+ (3.30)The integral constant .ct can be computed from the following condition:

0pphz2lx =

==

, (atmospheric pressure).

30

Fig. 3.7

In this way:

h g2lact += (3.31)

Replacing (3.31) into (3.30), the Equation for the relative equilibrium of liquids in accelerating

translations becomes (using the gauge pressure):

)zh( gx2lap +

= (3.32)

The maximum magnitude of pressure is obtained for 0zx == :

+= h g2lapmax . (3.33)

Figure 3.6 shows the pressure diagram on the walls of the tank.

Fig. 3.6

3.3.3.2 Liquids in cylindrical tanks on rotational motion

For this case, lets consider a liquid, having the specific gravity , in an open cylindrical tank byradius R , up to the level h (see Figure 3.7), which start to rotate

with the angular speed ct= .In this case the specific mass force mf will have components on

each direction of the reference system:

component of the centrifugal acceleration on Ox and Oy :2

mx xf = and 2my yf = ; along Oz , local gravitational acceleration gfmz = .

From Equation (3.16), the potential of the mass forces is,

( )( )

ctz g2rctz g

2yx

dz gdy ydx x

dz fdx f

222

22mzmx

++=+++=

=++==+=

(3.34)

31

Also, the potential of the pressure forces for liquid ( ct= ) is:

ctpdp += (3.35)Thus, the Equation for the relative equilibrium of liquids in on rotational motion is:

ctz g2rp 22 =+ (3.36)

The integral constant .ct can be computed knowing that the (gauge) pressure is zero on the free

surface, which is a parabolic type. Thus, from the condition = finalinitial VV

minmaxminmax2

max22 hhh 2)hh(R

21hR hR +== (3.37)

Also 0p = for:

cth g2

RhzRr

max2

2

max=+

== (3.38)

cth ghz0r

minmax

=

==

(3.39)

From the system of Equations (3.37), (3.38) and (3.39) we found that:

h g4

Rct 22

+= (3.40)Replacing (3.40) into (3.36), the Equation for the relative equilibrium of liquids on rotational motion

becomes:

)r 2R(4

)zh( g p 222

= (3.41)The maximum magnitude of pressure is obtained for Rr = and 0z = :

4Rh g p

22 += . (3.42)Figure 3.8 shows the pressure diagram on the walls of the tank.

Fig. 3.8

32

3.3.4 Variable-density pressure fields

If the density is not a constant, in order to compute the potential of the pressure forces, the

relationship between pressure and density must be known. In the following it is presented the

solution for an isothermal process, ctT = :pppctp

0

0

0

0 === (3.43)

Thus:

ctp lnpp

dpp

dp0

0

0

0 +== (3.44)For fluids on gravitational field of Earth:

ctz g += (3.45)Finally:

ctz gp lnp0

0 =+ (3.46)

The solutions for adiabatic or polytropic process can be found similarly.

33

4. FLUID FORCES ON SUBMERGED SURFACES

Engineers interested in the storage of fluids must be concerned with the strength of the container

whether it be a dam, a tank, or a pressure vessel. It is not the purpose of this text to consider the

design of such structures or the gates and valves associated with them, but some knowledge

about the resultant force that the fluid exerts is essential to the designer. Its determination is the

subject of this section.

For purposes of stress analysis, a state of zero stress exists when a fluid is subject only to the

uniform pressure of the atmosphere. It is therefore desirable to use gage pressures whenever

possible in this section.

Consider a surface of arbitrary shape subject to fluid pressure (see Figure 4.1). The surface might

be the side of a tank, the face of a dam, the hull of a ship, or even an imaginary surface cut from

within a body of fluid. At any point on this surface we select an element of area dA having the

normal unitary vector nr

(drawn outwardly from the surface, and thus into the fluid). Acting upon

dA is a differential force that is given by:

dAnpFd = rr , (4.1)The negative sign appears because the force exerted by the fluid is directed into the surface, and

opposite in sense to nr

.

Fig. 4.1 Pressure force on elementary area dA

The resultant of all the differential forces on the surface can be represented by a force Fr

passing

through a conveniently chosen reference point, and a couple Mr

.

The force is obtained by integrating the differential forces over the surface area A .

=A

dAnpFrr

, (4.2)

Each differential force has a moment about the reference point, O , given by FdrMd orrr = . The

moment of the couple is obtained by the integration of Mdr

:

34

==A A

o dA n prFdrMvrrrr

, (4.3)

where rr

is the position vector locating dA with respect to the moment center.

In evaluating the above integrals, it must be kept in mind that we are summing vector quantities.

The operation of integration must therefore be carried out on a set of scalar components of the

vectors (the pressure distribution in fluid must be known).

4.1 FLUID FORCES ON PLANE SURFACES

In the cases of the plane surfaces the normal unitary vector is constant, ctn =r , and the Equations(4.2), (4.3) become:

==AA

dApndAnpFrrr

, (4.4)

==AA

o dAprndA n prMrvvrr

, (4.5)

The point where resultant force will act is calling the centre of pressure and has as symbol C (or

CP ). The position vector of the centre of pressure, CPrr

, can be obtained from Varignons

Theorem: the sum of the moments of each elementary force is equals with the moment of the

resultant force:

===A

ACP

ACP

ACP

A dA p

dA p rrdA prndA p rnFrdA p rn

rrrvrvrrrv

(4.6)

4.1.1 Resultant force and center of pressure on a submerged plane surface in a gasFor gasses, having a finite volume, the value of pressure within can be considered constant:

ctp = . Equations (4.4) and (4.6) become:A pFA p ndAp ndS p nF

AA==== rrrr ; (4.7)

CGG

A

A

A

ACP rA

A rdAp

dA rp

dA p

dA p rr

rrrr

r ====

; (4.8)

where: CGrr

is the position vector of the fluid centroid.

These results are also applicable for small volumes of liquids, if the pressure variation is negligible

insight them.

35

4.1.2 Resultant force and center of pressure on a submerged plane surface in a liquid

For this case, it is considered that the plane surface is totally submerged in a liquid of density and inclined at an angle of to the horizontal (see Figure 4.2). Taking pressure as zero at thelevel of free surface and measuring down from the surface, the (gauge) pressure on an element

dA , submerged at distance h , is given by the fundamental Equation of hydrostatics:

h p = (4.9)

Fig. 4.2 Pressure force on a submerged plane surface in a liquid

By replacing of p in Equations (4.4) and (4.6) the followings are obtained:

====AAAA

dA z)sin( ndA )sin( z ndA h ndA pnF rrrrr , (4.10)

where: A

dA z is the 1st moment of inertia for area about the axisoy :

A zdA z CGA

= , (4.11)where: A is the area of the submerged surface;

CGz is the position for centroid of the surface along Oz . Hence:

A h )sin( z A F CGCG == , (4.12)where: CGh is depth to the centroid of the submerged surface, from Figure 4.2.

A z

dA z r

dA z

dA z r

dA h

dA h r

dA p

dA p rr

CG

A

A

A

A

A

A

ACP

====rrrr

r

, (4.13)

Thus, the coordinates of pressure centre in the yoz plane are:

AzI

Az

dA z yy

CG

yz

CG

ACP ==

;

AzI

Az

dA zz

CG

y

CG

A

2

CP ==

; (4.14)

where: yzI is the product of inertia about the axes Oy and Oz ;

yI is the 2nd moment of inertia about the axis Oz .

Note that:

xoy is the manometric plane (where the gauge pressure is null, )0p = ;

36

If yOz is a symmetry plane, than 0Iyz = and 0yCP = . In a reference system having the origin in the centroid CG of the surface, the Equations

(4.14) can be write in the following form (using the Steiners theorem):

A zA z y'I

A zI

yCG

CGCGyz

CG

yzCP

+== ; A z

A z'IA z

Iz

CG

2CGy

CG

yCP

+== ; (4.15)

where: yz'I is the product of inertia about the axes CGy and CGz ;

y'I is the 2nd moment of inertia about the axis CGz .

Note that in the Equation (4.15) the 2nd moment is always positive so that the center of pressure

CP always falls below the centroid CG .

4.2 FLUID FORCES ON CURVED SURFACES

As stated above, if the surface is curved, the forces on each element of the surface will not be

parallel and must be combined using some vectorial method. It is most convenient to calculate the

horizontal and vertical components and combine these to obtain the resultant force and its

direction. Thus, the components of the resultant pressure forces are given by the following

Equations:

=yOzS

yOzx p dS p F (4.16)

=xOzS

xOzy p dS p F (4.17)

=xOyS

xOyz p dS p F (4.18)

where: xOyS , yOzS , xOzS are the algebraic projections of the curved surface on the planes

of the reference system, xOy , yOz and xOz ;

The coordinate of the pressure center on Ox axis is:

=yOz

yOzx

SyOz

SyOz

F C dS p

dS p r

r

rr

(4.19)

The other two coordinates are computed similarly.

4.2.1 Pressure forces on a submerged open curved surface in light fluidsFor gasses, having a finite volume, the value of pressure within can be considered constant:

ctp = . Equations (4.16) (4.19) become:yOzx p dS p F = ; xOzy p dS p F = ; xOyz p dS p F = (4.20)

37

yOzyOz

x p S GyOz

SyOz

F C rS

dS r

rr

rr ==

;

xOzy p S GF C rrrr = ;

xOyz p S GF C rrrr = (4.21)

4.2.2 Pressure forces on a submerged open curved surface in liquids

The pressure insight a liquid is given by zp = . In a reference system having the xOy asmanometric plane and the axis Oz in the sense of depth rising, the Equations (4.16) (4.19)

become:

yOzS GS

yOzx p S z dS z F yOzyOz

== ; xOzS Gy p S z F xOz= ; V dS z FxOyS

xOyz p == (4.22)

yOzS G

SyOz

F C S z

dS z r

ryOz

yOzx p

=

rr

; xOzS G

SxOz

F C S z

dS z r

rxOz

xOzy p

=

rr

; V GF C rr zrr = (4.23)

where:yozGz is the depth of centroid for yozS about the manometric plane xoy (where

the gauge pressure is null, )0p = ;xozGz is the depth of centroid for xozS about the manometric plane;

V is the volume of the liquid between wetted curved surface and its projection

on the manometric plane xoy .

These results can easily understand considering a liquid at rest on top of a curved surface AC

(two-dimensional case), as shown in the diagram below.

Fig. 4.3 Pressure force on a submerged curved surface in a liquid

The element of fluid ABC is at equilibrium (as the fluid is at rest). As a consequence the horizontal

force on AC plane must be equal and in the opposite direction to the resultant force yF on the

curved surface AB . As AC is the projection of the curved surface AB onto the vertical plane

xOz , it can be generalises this to:

The resultant horizontal force of a fluid above a curved surface is equal with the resultantforce on the projection of the curved surface onto a vertical plane.

We know that the force on a vertical plane must act horizontally (as it acts normal to the plane) and

that yF must act through the same point. Generalising:

38

The resultant horizontal force of a fluid above a curved surface acts horizontally through thecentre of pressure of the projection of the curved surface onto a vertical plane

Concerning the vertical force, there is no shear force on the vertical edges, as consequence the

vertical component zF can only be due to the weight of the fluid. Generalising:

The resultant vertical force of a fluid above a curved surface is equal with the weight of fluiddirectly above the curved surface. It will act vertically downward through the centre of gravity

of the mass of fluid.

The overall resultant pressure force is found by combining vectorialy its components:

2z

2y

2x FFFF ++= (4.24)

The angle the resultant force makes to the horizontal is:

=

z

y

FF

arctg (4.25)

4.2.3 Resultant force on a closed curved surface for a constant pressure fluid

For this case, the resultant force on each direction will be zero: 0FFF zyx === (the algebraicprojections of the curved surface on the planes of the reference system are nulls). But the action of

the fluid has as result stresses on the wall of tanks. Knowing the magnitude of these tensions is

useful on computation of the thickness of wall.

For a circular duct, see Figure 4.4, the following Equation is found for computation of the thickness

of wall, , from the balance of the forces of pressure and strength on Oz direction:= L 2L D p (4.26)

a 2D p = (4.27)

where: p is the pressure of fluid;

D is the inner diameter of duct;

a is the allowable stress characteristic for the material of duct.

Fig. 4.4 Pressure force on a closed curved surface for constant pressure fluid

39

4.3 FLOATING BODIES

4.3.1 Buoyancy of floating bodies. Archimedes principleThe vertical component of the fluid static force on a submerged or floating body is called the

buoyant force (see Figure 4.5).

Fig. 4.5 Buoyancy of floating bodies

About 250 B.C. Archimedes discovered that the buoyant force on a solid body is equal in

magnitude to the weight of the fluid displaced by body.

Thus, the equation of the equilibrium of the forces on a body of specific weight b floating on afluid having specific weight f is:

fdfbbAW FF VV == (4.28)where: WF is the weight of the solid;

AF is the buoyant force (Archimedes force);

bV is the volume of the solid;

bV is the volume of the displaced fluid.

The necessary condition of the floating is:

fb < (4.29)

4.3.2 Stability of floating bodiesThe concept of stability relates to the question of whether or not a floating body returns to its initial

state when its equilibrium is disturbed. This is a very important concept in the design of the ships.

The stability of the floating bodies is described by the following equations:

Stable equilibrium:

0eIdf

x >V (4.30)

Neutral equilibrium:

0eIdf

x =V (4.31)

40

Unstable equilibrium:

0eIdf

x

41

Fig. 4.9

4.4.2 The U-tube manometer

Using a U-tube, the pressure of both liquids and gases can be measured with the same instrument.

The U-tube is connected as in the figure 4.8 and filled with a fluid called the manometric fluid (or

manometric liquid). The fluid whose pressure is measured should have a mass density less than

that of the manometric fluid and the two fluids should not be miscible. For the manometer shown in

Figure 4.8 the following Equation can be written:

Fig. 4.8 - The U-tube manometer

pressure in a continuous static fluid is constant at any horizontal level:CB pp = ;

for the left hand arm:1fAB h g pp += ;

for the right hand arm:2lpC h g p =

Thus the (gauge) pressure in point A is:

1f2lpA h g h g p = . (4.45)Note that if the fluid being measured is a gas, the density will probably be very low in comparison

to the density of the manometric fluid, i.e. flp >> . In this case the term 1f h g can beneglected (because 1h1

42

h g)()hh(g pp flp12fBA += . (4.47)Also, if the fluid whose pressure difference is being measured is a gas and flp >> , then theterms involving f can be neglected:

h g pp lpBA = . (4.48)

4.4.4 Advanced U-tube manometer.The U-tube manometer has the disadvantage that the change in height of the liquid in both sides

must be read. This can be avoided by making the diameter of one side very large compared to the

other. In this case the side with the large area moves very little when the small area side move

considerably more.

Fig. 4.10 - Advanced U-tube manometer

Assume that the manometer is arranged as above, to measure the pressure difference of a gas

(with negligible density) and that pressure difference is 21 pp ( 21 pp > ). The datum line indicatesthe level of the manometric liquid when the pressure difference is zero. The volume of liquid

transferred from the left side to the right is:

4d h

4D h

2

2

2

1 ==V .

The fall in level of the left side is:2

21 Ddhh

= .

The height different in the two columns gives the pressure difference:

+=

+=

2

2lp

2

22lp21 Dd1h g

Ddhhg pp . (4.49)

If D is much larger than d ( dD >> ) then 2)D/d( is very small ( )1)D/d( 2

43

Fig. 4.11 - Manometer with a tilted arm

The pressure difference is given by the height change of the manometric fluid but by placing the

scale along the line of the tilted arm and taking this reading large movements will be observed. The

pressure difference is then given by:

sinl g h g pp 21 == . (4.51)The sensitivity to pressure change can be increased by a greater inclination of the manometer arm.

Alternatively, the density of the manometric fluid may be changed.

4.4.5 Choice Of ManometerSome disadvantages of manometers:

Slow response they are useful for very slowly variations of pressure. For the U-tube manometer two measurements must be performed simultaneously to get the

difference in high value. This may be avoided by using a tube with a much larger cross-

sectional area on one side of the manometer than the other one.

It is often difficult to measure small variations in pressure; alternatively an inclinedmanometer may be used.

Some advantages of manometers:

They are very simple. No calibration is required - the pressure can be calculated from first principles.

44

4.5 FLUID STATICS EXAMPLES

WORKED EXAMPLES

Exercise 1

Determine the pressure intensity at A (Pa and OH mm 2 ) if the pressure at B is at 1.0pB = ,m 2.0h1 = , m 1.0h2 = , m 4.0h3 = , 3water0 kg/m 1000== and 3oil2 kg/m 800== .

Fig. 4.12

SOLUTION:

Step 1:

Convert all the quantities in the International System of Units (if is necessary)

)OmmH10( Pa 1081.9Pa 1081.91.0at 1.0p 2334

B ====Pa 81.91081.91000OmmH 1 32 ==

Step 2:=++ 3120B3210A h g h g p)hh(h g p

++= 31310BA h g )hh( g pp =++= Pa 8.255614.081.9800)2.01.0(81.910001081.9p 3A

OmmH 1280p 2A =

45

Exercise 2

A closed tank, having the shape as in Figure below, is holding water under the pressure mp

(manometric pressure).

Fig. 4.13

The following are known: m 5.1H = depth of water in tankm 5.0R = the curvature of the thank

m 0.1L = width of the tank3mkg/ 1000= mass density of the water

at 1.0pm = (technical atmospheres)

Draw the pressure diagram on the wetted walls of the tank. Find the magnitude of the force on the vertical wall )AB( Find the magnitude of the resultant force on the curved wall )BC( .

SOLUTION:

Step 1:Convert all the quantities in the International System of Units (if is necessary)

Pa 1081.91.0at 1.0p 4m ==

Step 2:Find the level mh of the (horizontal) manometric plane above the free surface of the liquid.

At the level of the manometric datum the absolute pressure is the atmospheric one, respectively

the gauge pressure is zero, 0p = , thus:= mm h g p

46

m 181.910001081.91.0

g ph

4m

m ===

Chose the reference system xOyz that:

xoy is the manometric plane;

oz in the sense of the depth increasing.

Fig. 4.14

Step 3:

Compute the (gauge) pressure at the level of each characteristic point:

Pa 1081.9ph g h g p 3mmAA ==== (pressure in gasses is constant)

+=+== )RH( g p)]RH([h g h g p mmBB Pa 1062.190.5)-(1.5 81.910001081.9p 33B =+=

+=+== H g p)H(h g h g p mmCC Pa 1053.241.5 81.910001081.9p 33C =+=

Draw the pressure diagram on the walls of thank, as shown in Figure 4.15.

47

Fig. 4.15 - Pressure diagram on the walls of thank

Step 4:Compute the magnitude of the force on the (plane) wall )AB( , )AB(F . Denote hRH = (seeFigure 4.16).

4.1 Using the integral Equation:

+=+=+== AA

mAA

mA

mA

)AB( dA zg dApdA z g dA pdA z) g p(dA pF

+=

+=+= dx2

hg L h h g dx 2zg L h h g dx dz zg A pF

L

0

2

m

L

0

h

0

2

m

L

0

h

0(AB)m)AB(

L h 2hh L

2hg L h h g F m

2

m)AB(

+=+=

4.2 Using Equation (4.12):

( ) N 1471511 0.519810 L h 2hh A h F m(AB)(AB) CG)AB( =+=

+==

Note that the (pressure) force is simply the volume of the pressure distribution. It acts through the

centroid of the pressure diagram.

48

Step 5:

Compute the magnitude of the force on the (curved) wall )BC( : )BC(F . Spit the resultant force into

its components, horizontal )BC(yF and vertical )BC(zF (see Figure 4.16). According with Equations

(4.16):

11036.25 N10.5 0.25)-1.5(19810 L R 2RHh Sh F mxozGy xoz =+=

+==

N 21546.191 114

0.5 9810 L hh4R V F

2

m

2

z =

++=

++==

where: xozS is the projection of the curved area )BC( on the datum plane xoz ;

LRSxoz =V is the volume of the liquid between wetted curved surface and its projection

on the manometric plane xoy (where the gauge pressure is null).

L R )hh(4R V m

2

++= .Hence:

N24208.20 FFF 2z2y =+= .

Fig. 4.16

49

Fig. 4.17

Exercise 3

The density of liquids can be determined experimentally using a hydrometer as shown below. It

consists on a weighted bulb and an attached tube whit a

calibrated cross section. Compute the density of a fluid fif the stem is submerged to a depth of mm 50h =relative to the equilibrium position in water.

The followings are known:

gf 20G = : weight of hydrometer;mm 8d = : diameter of the calibrated section;

3w kg/m 1000= : density of the water

Solution:

Step 1

Convert all the quantities in the International System of

Units (if is necessary)

N 1962.0kg0.02 sm9.81gf 20G 2 === ;

m 108mm 8d -3== .Step 2

The buoyancy principle is the basis of the hydrometer. The Equations of the equilibrium for the

floating situations are:

in water: dwwA V g GFG w ==

in other liquid:

+=== h

4d V g V g GFG

2

dwldllAl

Thus:

+=

+

=

+

=h d g G4

G4

h4d

g G g

G

h4d V g

G 2

ww2

w

2

dw

l

+= h d g G4 G4

2w

wl

+=

0.05008.01415.381.910000.196240.19624 1000 2l

3l mkg 334.888=

50

SELF-ASSESSMENT EXERCISES

Exercise 4An inclined tube manometer consists of a vertical cylinder of 35 mm diameter. At the bottom of this

is connected a tube 5 mm in diameter inclined upward at an angle of 15 to the horizontal. The top

of the vertical tube is connected to an air duct. The inclined tube is open to the air and the

manometric fluid has relative (to the water) mass density 0.785. Determine the pressure in the air

duct if the manometric fluid moved 50 mm along the inclined tube. What is the error if the

movement of the fluid in the vertical cylinder is ignored? Make a sketch.

Exercise 5A U tube-manometer (see Figure 4.18) is used to measure the acceleration of a motorcar.

Compute the magnitude of the acceleration if the deflection of the fluid is mm 30h = . Draw thecalibration curve of the accelerometer.

Fig. 4.18

Exercise 6A hydraulic tachometer consists on a U tube-manometer (see Figure 4.19) which is used to

measure the speed (revolutions per minute). Compute the speed if the deflection of the fluid is

mm 60h = . Draw the calibration curve of the tachometer.

Fig. 4.19

51

Exercise 7A closed tank (see Figure 4.20), filled with gasoline, is carried by a truck. Compute the pressure

force pF on the back wall of the tank if the truck has acceleration 2m/s 2a = . Draw the pressure

diagram on the walls of the tank. The followings are known: m 8.1h = , m 6.1b = , m 6l = ,3

gasoline gram/cm 78.0= .

Fig. 4.20

Exercise 8Find the magnitude of the resultant force on the vertical wall of a tank that has oil, of relative

density 0.8, floating on water as shown in Figure 4.21. Draw the pressure diagram. The width of

the tank is m 0.1L = .

Fig. 4.21

Exercise 9

A cylinder, mm 100 in diameter, mm 250 long and mass density 3kg/m 800= , floats in waterwith vertical axis. Determine the stability of the cylinder.

52

5. IDEAL FLUID DYNAMICSThe topics of this lecture are concerning with the analysis of the ideal fluids in motion - fluid

dynamics. The majority of the engineering problems involving fluids occur in flow processes:

y the air flow over a road vehicle;y the lift of an airplane on the flow of air over its wings;y the trust of a rocket engine depends on the flow of gasses through a nozzle;y the cooling of the electronic devices depends on the air flow over their components, a.o.

5.1 DESCRIPTIONS OF THE FLOW FIELDS

The motion of fluids can be predicted in the same way as the motion of solids are predicted,

respectively using the fundamental laws of physics together with the physical properties of the

fluid. But, when a fluid flow, the elements of fluid do not maintain their relative position. A complete

description of their motion is therefore inherently more complex than a description of the rigid

bodies motion. It is not difficult to envisage a very complex fluid flow as:

y wake behind a car,y waves on beaches,y hurricanes and tornadoes, or any other atmospheric phenomenon.

All are example of highly complex fluid flows. Fortunately, we are not generally concerned with the

motion of the individual particles (the Lagrangian approach). It usually suffices to describe the

velocity associated with a point in a specified area of flow field. When the velocity is specified for

each point in a region, a velocity field is defined. Mathematically it can be represent by the

Equation:

k j i vrrrr

wvu ++= (5.1)where: i

r, j

r, k

rare the unit vectors in the x , y and z directions;

u , v , w are the scalar components of the velocity ( 22 wvu ++= 22v ); thelatter can by expressed by the field equation:

t) z, y,x,( vvt) z, y,(x,

t) z, y,(x, t) z, y,(x,

=

===

wwvvuu

(5.2)

The use of velocity fields to describe fluid motion was introduced by Euler (XVIII century), who is

generally credited with the founding of the fluid dynamics. The Eulerian method may seem rather

complex, yet the alternative Lagrangian method of describing the motion of each fluid element as a

53

function of time is much more complex. The simplification achieved by the eulerian method is most

apparent in problems of steady flow.

A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ frompoint to point but do not change with time. (in practice, there is always a slight variations in velocity

and pressure, but if the average values of these are constant, the flow is considered steady).

In contrast with this type of flow is the unsteady flow, when the conditions are changing with time atany point in the fluid.

5.1.1 StreamlinesGraphical representations of physical phenomena are generally helpful in gaming insight into

problems and in communicating with others. This is particularly true when the mathematical

description of a phenomenon is complex. A very useful graphical representation of fluid flow is

based on the concept of a velocity field. Lines are drawn in such a way that the tangent at any

point on a line indicates the direction of the velocity associated with that point. These lines are

called streamlines, and a family of them constitutes a streamline pattern (Fig. 5.1).

Fig. 5.1 Flow pattern around an airfoil

A streamline pattern indicates the direction of the velocity at every point in the field at a given

instant. In steady flow, the streamline pattern is stationary with the frame of reference. It is also

stationary in unsteady flow if only the magnitude of the velocity is changing with time. Furthermore,

the steady flow streamlines represent the path lines for moving particles. Generally, in unsteady

motion, a fluid particle will not remain on the same streamline, and hence the streamlines and path

lines do not coincide.

The differential equations for a streamline are obtained by noting that:

dtdx=u ;

dtdy=v ;

dtdz=w . (5.3)

Hence:

)t ,z ,y ,x(dz

)t ,z ,y ,x(dy

)t ,z ,y ,x(dx

wvu== . (5.4)

Alternatively, since the velocity vector is tangent to a streamline, then )dz,dy,dx(rd||)(vrr

wv,u, , or:

0rdv = rr (5.5)

54

Practically, the injection of dye or smoke into a steady flow can make the streamline pattern visible,

as shown in Figure 5.2.

Fig. 5.2 Flow visualisation around an airfoil

But no matter how carefully the injection is made, a streamline pattern is not always revealed.

When none is found, measurements with sensitive instruments indicate that the flow contains

irregular, high frequency fluctuations and is therefore not truly steady. This type of flow is known as

turbulent flow. With the use of the mean velocity, stationary streamline patterns can be drawn for

turbulent flows.

Some things to know about streamlines:

y Because the fluid is moving in the same direction as the streamlines, fluid can not cross astreamline.

y Streamlines can not cross each other. If they were to cross this would indicate two differentvelocities at the same point. This is not physically possible.

y The above point implies that any particles of fluid starting on one streamline will stay on thatsame streamline throughout the fluid.

5.1.2 Stream-tube control volume

A useful technique in fluid flow analysis is to consider only a part of the total fluid in isolation from

the rest. This can be done by imagining a tubular surface formed by streamlines along which the

fluid flows. This tubular surface is known as a stream-tube (see Figure 5.3).

Fig. 5.3 A 3D stream-tube

The "walls" of a stream tube are made of streamlines. As we have seen above, fluid cannot flow

across a streamline, so fluid cannot cross a stream tube wall. The streamlines are boundaries in

the same sense that real conduit walls are boundaries. Conversely, the boundaries of the real

conduit or any immersed solid are streamlines. Hence, a stream tube can often be viewed as a

solid walled pipe.

55

5.2 ACCELERATION IN A FLOW FIELD

Before we can relate forces and motion within a flow field, we must relate the acceleration of a fluid

element to the velocity field. Referring to Equations (5.1) and (5.2) and taking the x component as

typical, we express the difference by:

zdz

ydy

xdx

dtt

d +

++

= uuuuu (5.6)

The x component of the acceleration can be obtain dividing the velocity by dt :

wuvuuuuuuuuuzyxtdt

zdzdt

ydydt

xdxdt

dttdt

dax +

++

=+

++

== (5.7)

Similarly:

wvvvuvvzyxt

ay +

++

= , wwvwuwwzyxt

az +

++

= . (5.7)

Hence:

wvuzv

yv

xv

tvk aj ai aa zyx

++

+=++=

rrrrrrrr. (5.8)

The first terms, )tv( r , represent the local acceleration in a transient motion; it is zero for steadyflows. The sum of the three remaining terms express the acceleration associated with the motion of

a fluid element in a non uniform velocity field, called also carrying acceleration or convective

acceleration. The Equation (5.8) can be write also (using the differential operators):

( )vvrot

2vgrad

tvvv

2v

tva

v vtvv

zyxtv

dtvda

22 rrrrrrr

rrrrrrr

++=++

=

+=

+

++

== wvu (5.9)

(5.9)

In the Equation (3.6) are emphasized the potential component, ( )2vgrad 2 and the rotational one,vvrot rr , of the convective acceleration. Also, in this form the acceleration is much easy to

integrate.

5.3 EQUATIONS OF THE FLUID MOTION

5.3.1 Continuity and conservation of matterFrom the definition of a streamline, it is apparent that fluid elements in nonturbulent flow cannot

cross a stream-tube surface. If the density of fluid is independent of time, mass does not

accumulate and:

the mass flow rate at every section of a stream tube must be the same.

This is a specialized statement of the continuity principle the law of mass conservation applied

to a flow field.

56

The continuity principle establishes a condition that any description of the flow field must satisfy,

and it thereby provides a relationship between certain properties within the flow field.

Imagine a stream tube having a cross-sectional area dA at an arbitrary position along the tube. As

dA becomes infinitesimal, the stream tube approaches a streamline. With v representing the

velocity of the fluid at this section, the fluid face coincident with dA moves a distance dt v normal

to dA in time dt . The volume of fluid passing the section is thus given by the expression:

dt dA vd =V . (5.10)The mass is given by:

dtdAvdm = , (5.11)and the differential mass flow rate ( )dtdmmdQm == & is:

dAvmd =& . (5.12)The instantaneous mass flow rate m& at any section of a stream tube of finite size is obtained bythe integration of Equation (3.9) over the cross section.

=A

dAvm & . (5.13)The volumetric flow rate Q (more commonly known as discharge) is often used in preference to

the mass flow rate, especially when the density of the fluid is constant:

=A

dAvQ . (5.14)

If the density is uniform over the section (as is usually the case), the volumetric flow rate can be

obtained by dividing the mass flow rate by the density:

mQQ = . (5.15)

The average velocity v , at any section of a stream tube, is defined by the equation:

AQv = . (5.16)

If the properties of a fluid are uniform over a cross section, the flow is said to be one-dimensional.

Many flows may be assumed to be one-dimensional without introducing serious error. When this is

done, the velocity that is used is the average velocity, but the bar notation is generally dropped.

The mass flow rate for one-dimensional flow can thus be expressed as:

A v Qm = . (5.17)When this is combined with the continuity principle, we obtain that:

ttancons)A v (...)A v ()A v (Q n21m ===== . (5.18)The subscripts are referring to two different sections along a stream tube. Whenever the density

can be assumed uniform throughout the stream tube, the volumetric flow rate will be the same at

every section and:

ttanconsA v...A vA v Q nn2211 ===== . (5.19)

57

5.3.2 Eulers equation of motion. Conservation of energy.

The Equation of (ideal) fluid motion can be founded from Newtons second law of motion applied to

an element of fluid with mass m and volume V , bounded by a surface having area A .

pmext FFFa mrrrr +== . (5.20)

where: extFr is the sum of the exterior forces, which are acting on the mass of fluid,respectively the mass forces mF

r and the pressure forces pF

r.

For an elementary fluid element:

===V

dVdtvd

amdVdtvd

dVadmarrrrr

(5.21)

===V

mmmmm dV fFdV fdmfFd rrrrr

(5.22)

===VA

pp p dV dAnp F dAnp Fdrrrr

(5.23)

where: mfr

is the unit mass force (it has dimension of acceleration) and it is expressed as:

kfjfiff mmmmrrrr

z y x ++= .

Generally: xUf x m = ;

yUf ym = ;

zUf z m = U gradfm =

r (5.24)

U is the potential of the mass forces. It represents the mass potential energy of the fluid. When

x mf , ymf and z mf are known:

( )zd fyd fxd f)z y,x,(U z m ymx m ++= (5.25)Substituting the expressions (3.18, 3.19, 3.20) into Equation (3.17) it is obtained that:

=VV

mV

p dVdV f dVdtvd

rr

(5.26)

Note:In the above Equation, there is no any differential operator before integrals, which can

affect the integration. For an arbitrary volume 0V , the Equation becomes: p1f

dtvd p f

dtvd mm ==

rrrr(5.27)

Equation (5.27) is the Eulers Equation of ideal fluid motion, in vectorial form: a fluid in motion is inequilibrium under the actions of inertia forces )dtvd(-

r, mass forces mf

r and pressure forces

/) p( .Taking into consideration the expression (5.9) of the acceleration, the above equation becomes (in

H. Helmholtzs formulation):

grad p1fvvrot 2

vgradtv

m2

=++ rrrr (5.28)

58

In the case of fluids, which for:

y the mass forces are deriving from a potential U gradfm =r

,

y the mass density is constant or known f