fluid mechanic and turbines i

FLUID MECHANIC AND TURBINES I

F

l

e

nacelle

am

air

A

vBB

h

http://web.univ-ubs.fr/lg2m/PagesPerso/ausias/

Summary

1. Fundamentals 1.1. Introduction 1.2. Definitions2. Fluid statics 2.1. Pressure at a point 2.2. Fundamental principle of fluid statics 2.3. Units and scales 2.4. Archimedes 2.5. Ideal gaz 2.6. Exercices3. Internal flows – Bernoulli laws 3.1. first Bernoulli law 3.2. Pressure head and piezometric head 3.3. second Bernoulli law 3.4. pressure drop 3.5. Euler equation4. Extern flow – aerodynamic and hydrodynamic 4.1. Lift and drag 4.2. Spheres et cylinders 4.3. Bodies with sharp edges

1. Fundamentals 1.1. Introduction

In Engineering* aerodynamic - hydrodynamic* flow in pipe* material processing

We are interested in all movements of a large number of molecules, considering that we are working on a continuous medium.

Everyone naturally distinguished by the sense of touch, three states of matter states-solid, liquid and gaseous-each of which it associates for example rocky shoreline, water waves and sea breeze.

In the deformation of a gas or liquid, the molecules do not keep permanent place against each other: they undergo shifts during which they change their neighbors. Instead, when a steel rod is subjected to bending or twisting the atoms that constitute retain the same neighbors [9].

Fluid mechanics is the extension of rational mechanics to a class of continuous media whose deformations can take values as large as you want [3].

Continuum mechanics

Continuum mechanicsThe study of the physics of continuous materials

Solid mechanicsThe study of the physics of continuous materials with a defined rest shape.

ElasticityDescribes materials that return to their rest shape after an applied stress.

PlasticityDescribes materials that permanently deform after a sufficient applied stress.

RheologyThe study of materials with both solid and fluid characteristics.

Fluid mechanicsThe study of the physics of continuous materials which take the shape of their container.

Non-Newtonian fluids

Newtonian fluids

Different fields of fluid mechanic

Simple fluids Complex fluids

No flow Fluid staticFluid static

Simple flows   Material

processing  

Complex flows Aerodynamichydrodynamic 

1.2. Definitions

Density and specific gravity

V

m

31000 /d

kg m

d : specific gravity ()

: density (kg / m3) m : elementary mass (kg)V

: Compressible and incompressible fluids incompressible: no change in density = 0 = cte

liquid are usually considered as incompressible except for natural convection, in the plastics industry with high pressure …

: elementary volume (m3)

Temp (°C) Density (kg/m³) Temp (°C) Density (kg/m³)

+100     958.4 +20     998.2071

+80     971.8 +15     999.1026

+60     983.2 +10     999.7026

+40     992.2 +4     999.9720

+30     995.6502 0     999.8395

+25     997.0479 −10     998.117

+22     997.7735 −20     993.547

−30     983.854

The density of water changes with temperature

Perfect fluid : zero fluid viscosity (without internal friction)

Newtonian fluid : fluid whose viscosity is constant and equal to .

Laminar and turbulent flowlaminaire characterized by smooth, constant fluid motion turbulent : chaotic eddies, vortices and other flow instabilities

Reynolds number :VD

Re

V : fluid velocity : fluid viscosity : fluid densityD : characteristic dimension perpendicular to the flow

laminar : Re < 2100turbulent : Re > 4000

transitional regime between the two : alternating turbulent and laminar flow.

Steady and un steady flowssteady flow : variables describing the motion are independent of time.

Unidirectional flow Variables depend only on one coordinate space.

Laminar and turbulent flow [10]

Stress and pressure

Consider a fluid volume V bounded by a surface S. The fluid outside [2] of volume V has on it the tensions that are transmitted through the surface S. Suppose that the elementary force exerted on S is proportional to S:

0lim

S

F

St

normal component: normal stress tangential component: shear stress Pressure is called the normal stress

In fluid statics, only involved the pressure forces. The tangential forces appear only in fluid dynamics for viscous fluids.

F t S

n st t n t s

Viscosity

Suppose we shear a fluid between a fixed lower plate and an upper plate which moves at a speed V.

The dynamic viscosity is the ratio of shear stress on the shear rate

F

S V

h

Dynamic viscosity of some fluids [ 1]

2. Fluid statics2.1. Pressure at a point

Pressure at a point in a fluid at rest

M is a point in a fluid at rest and an elementary surface S, with n normal outgoing. The surface is small enough to consider that the pressure is uniform on this surface. Whatever the orientation of the surface S, the fluid exerts a pressure on the surface P.

consider two points close to each other, M and M' having pressure equal to :

Pressure gradient

( , , )Mp p x y z

' ( , , )Mp p x dx y dy z dz

x

M y

z

'

x dx

M y dy

z dz

'

dx

MM dy

dz

��������������

'M Mdp p p

Differential of the function p :

, ,,y z x yx z

p p pdp dx dy dz

x y z

. ' dp grad p MM����������������������������

=

Oxyz

p

xp

grad p py

p

z

��������������

Applications – difference of level

Consider a vertical axis z directed upwards

0

0 '

1

dx

g g MM dy dp gdz

dz

��������������

x

y

z

M1

M2

1 2

2 1

0

0

1

dx

g g M M dy dp gdz

dz z z

��������������

2 2 2 2

1 1 1 12 1 2 1

0

0

/

( )p z p z

p z p z

gradp

dp dz

dp gdz dp g dz p p g z z gh

��������������

h : difference of level

2.2. Fundamental principle of fluid statics

The resulting pressure forces

The resulting pressure forces of a fluid on a volume V of material is the same direction as the vector and pressure gradient in opposite direction

Fundamental principle of fluid staticsIf the fluid is in equilibrium, no friction force exists. The volume element dV is subject to:

- Forces of gravity (weight):

- Pressure forces:

fdmg dVg

df grad p dV�������������� z

x

y

dm

dV

fgrad p g��������������

F 0

g

pgrad

(2)

.F dV p

2.2. Fundamental principle of fluid statics

The resulting pressure forces

The resulting pressure forces of a fluid on a volume V of material is the same direction as the vector and pressure gradient in opposite direction

Fundamental principle of fluid staticsIf the fluid is in equilibrium, no friction force exists. The volume element dV is subject to:

- Forces of gravity (weight):

- Pressure forces:

fdmg dVg

df grad p dV�������������� z

x

y

dm

dV

fgrad p g��������������

F 0

g

pgrad

(2)

.F dV p

Exemples

Calculation of vertical pressure gradient in the air at sea level0 = 1,225 kg/m3 : density of air at sea level

grad p = 1,225 * 9,81 = 12,02 Pa/m = 0,0001202 bar/m

Calculation of vertical pressure gradient in water at sea levelgrad p = 1000 * 9,81 = 9810 Pa/m = 0,0981 bar/m

Case of gazconventional atmosphere (aviation) : z = 0P0 = 760 mm of mercury

= 101 325 Pa = 1,01 barT0 = 288 °K = 15 °C0 = 1,225 kg/m3Variation of pressure for a difference of level of 2.5 mdp = -gdz = -1,225 9,81 2,5

= -30 Pawhich is negligible compared to atmospheric pressureIt is therefore considered that the pressure in the air for several meters or tens of meters high is the same.

The mercury barometer

PB = atmospheric pressure

PA = saturated vapor pressure of mercury

= few Pascals

Nearly equal to zero = 0 Pa

h = 759 mm

PB

PA

C

There is still the same fluid between B and C and the difference of level between B and C is zero, so PB = PC = atmospheric pressure = 1013 hPa (conventional atmosphere used in aircraft)

Specific gravity of mercury = 13,6

There are the same fluid between A and BPressure variation between A and Bdp = pB – pA = 101300 Pa = Hggh = 13600 9,81 hh = 0,759 m = 759 mm

What is the height of mercury if the weather forecast announces a pressure of 1030 hPa ?What height would have a water barometer ?

Pascal's principle

Static fluids in an incompressible fluid, the pressures are transmitted in full.Example: hydraulic press, brake hydraulic

2.3. Units and scales

Absolute temperature (K : Kelvin)0 100 200 300 400

273

Temperature (C : Celcius)-273 0 100 200 300

Absolute pression (bar)0 1 2 3

Effective pression (bar)-1 0 1 2

vide ≈ pressionatmosphérique

Absolute pressure and gauge pressureIn principle we use absolute pressures. But in

many areas were measured pressures (pressures) from an original pressure. It usually takes its origin in atmospheric

pressure (pa). Then measured effective pressure (pe) pe = p - pa.

Most gauges are graduated in industrial pe. The zero corresponds to atmospheric

pressure.

Units

* SI : Pascal (Pa) = 1Newton / 1mètre²* multiples of Pascal : 1 bar = 100000 Pa, * Meteorology : 1 mbar = 10² Pa = 1 hPa.* kgf/cm² 1 kgf/cm² = 0.98 bar* psi = pound per square inch 1 psi = 6.89 103 Pa* 1 atm = 101325 Pa = 1013 mbar

2.4. Archimedesz

x

y

objectV

g

fgrad p g��������������

o

f fluid

fgrad p g��������������

(2)

(1)pgradVF

F

Theorem of Archimedes: an object with a volume V completely surrounded by fluid density

f

undergoes a vertical thrust F equals the weight of the displaced fluid.

fF V g����������������������������

object weight : oP V g����������������������������

P

If the object is solid (the table) and the fluid gas (air) then: 1000o f

In this case we can neglect the buoyancy to the weight of the object

2V

1V

a air

e water1 2 e aF V g V g

������������������������������������������ solid

negligible1V : immersed volume

An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

nRTpV

2.5. Ideal gaz

At normal conditions such as standard temperature and pressure, most real gases behave qualitatively like an ideal gas. Many gases such as air, nitrogen, oxygen, hydrogen, noble gases, and some heavier gases like carbon dioxide can be treated like ideal gases within reasonable tolerances.

2.6. Exercices

Exercice 2.1. : tube en UUn tube en U de section s = 1 cm2 est ouvert aux deux extrémités. Il contient de l'eau. D'un coté, on verse 10 cm3 d'huile. La différence de niveau des surfaces libres est de 15 mm. Quelle est la densité de l'huile ?

Exercice 2.2. : GradientsSupposons que l’air dans la classe est immobile et que sa température est plus basse près du sol (19,0 °C) et plus élevée près du plafond (21,5 °C) et que les isothermes sont des plans horizontaux. Exprimer et calculer le gradient de température vertical (on prendra l’axe z vertical dirigé vers le haut), puis horizontal. Présenter le résultat sous forme d’un vecteur. Même exercice pour les gradients de pression dans l’air de la classe (les isobares sont des plans horizontaux). Même exercice pour les gradients de pression dans une piscine de profondeur de 2,5 m. Quelle est la pression absolue au fond de la piscine ?

Exercice 2.3. : AquariumCalculer les forces F1 et F2 qui s'exercent sur les parois d'un aquarium rempli d'eau.A.N. : L = 150 cm, l = 75 cm, h = 60 cm.

F1

F2

Réponses

Exercice 2.1. : tube en UQuelle est la densité de l'huile ? d = 0,75

Exercice 2.2. : GradientsExprimer et calculer le gradient de température vertical : 1 °C/mExprimer et calculer le gradient de température horizontal : 0°C/mPrésenter le résultat sous forme d’un vecteur.

Les gradients de pression dans une piscine de profondeur de 2,5 m.

Quelle est la pression absolue au fond de la piscine : 101300 + 9810*2,5 = 125825 Pa = 1,26 bar

Exercice 2.3. : AquariumF1 = 1324 NF2 = 2648 N

0

0 /

1

gradT C m

��������������

0

0 /

9810

gradp Pa m

��������������

h

Axe

Axe

porteporte

l A

nacelle

am

air

Exercice 2.4. : Porte d’écluseUne écluse contient de l’eau sur une hauteur h . Elle est fermée par deux portes tournant autour de deux axes verticaux. Calculer le couple nécessaire à exercer sur l’axe pour maintenir la porte fermée.

Exercice 2.5. flotteurCalculer l’enfoncement d’un cube de bois de coté a et de densité 0,9 dans de l’eau douce, dans du mercure de densité 13,6.

Exercice 2.6. : MontgolfièreUne montgolfière a la forme d'une sphère, a un diamètre D = 20 m et contient un air chaud de masse volumique am = 0,7 kg / m3. Sous la

montgolfière est suspendu une nacelle dans laquelle prend place l'équipage. L'ensemble est arrimé au sol et l'air qui entoure l'ensemble est dans les conditions standards de l'air au niveau de la mer. La masse de la nacelle plus la masse de l'enveloppe de la montgolfière est égale à Pne = 117 kg. Pour calculer le

poids total de l'ensemble on ajoute le poids de l'air dans la montgolfière, le poids de l'enveloppe, le poids de la nacelle et le poids de l'équipage.a)Calculer la poussée d'Archimède subit par la montgolfière.b)Calculer le poids maximum de l'équipage que peut emporter cette montgolfière dans ces conditions.c)Si l'équipage a une masse de 140 kg, calculer la force ascensionnelle appliquée à l'ensemble.d)Calculer alors le temps nécessaire à la montgolfière pour atteindre l'altitude de 100 m.Figure : Montgolfière

h

Axe

Axe

porteporte

l A

Réponses

Exercice 2.4. : Porte d’écluse

Exercice 2.5. flotteurCalculer l’enfoncement d’un cube de bois de densité 0,9 dans de l’eau douce,

H = 0,9 a

dans du mercure de densité 13,6.

H = 0,074 a

2 2

4

gh lC

3. Internal flows – Bernoulli laws3.1. first Bernoulli law

dS1 dS1’

M1

M2

P1

dm

1

2

t

t + dt

cote z1

cote z2

ligne de courant

pression p2

dS2 dS2’z

Assumptions:- A stream tube belonging to a steady flow,- Ideal fluid (no viscosity, internal friction, no viscous dissipation),- incompressible fluid- Z axis directed vertically upwards

dm is the mass of fluid between dS1 and dS2 to time t, At t + dt, it is between dS1' and dS2'. As the flow is steady, the fluid between dS1' and dS2 and remains in the same state and it is as if the mass of fluid between dS1 and DS1', was passed between dS2 and dS2' during the time interval dt.

The variation of the kinetic energy of the mass dm between t and t + dt is equal to the sum of the work of the forces exerted on it

1

2 22

12dm v v- Kinetic energy variation :

- Work of force of gravity : - dm g (z2 - z1 )

- Work of pressure forces : p1 v1 dS1 dt = p2 v2 dS2 dt

p v dm

v

p v dm

v

pdm

pdm1 1

1

2 2

2

1 2

In term of difference of level, Unit : mv

g

p

gz cte

²

2

²

2

v pz cte

g g

A

VBB

h

v ghB 2

2

2A B B

A B A B a

P v pz z p p p

g g

Exemple : drain tank

Between A and B

3.2. Pressure head and piezometric head

The pressure head at a point is equal to : H

p

gz

V

g

²

2

The piezometric head at a point is equal to : p

gz

Bernouilli :

H = cte

g

Vz

g

p

g

Vz

g

p BB

BAA

A

22

22

Graphic representation :

zA

z'

H

0 x'

Bernoulli : H = cte

1

2

A v

B v

ZB

g

vA

2

2

g

pA

g

pB

g

vB

2

2

Ventury effect

Pitot tube

3.3. Second Bernoulli law

Bernoulli's equation is often used for fluid flows in ducts, pipes, waterworks ...In a real system, the fluid mass dm provides, receives or dissipates energy.

provide turbinereceive pumpdissipation singular or regular pressure drop

2 22 2 1 1

2 12 2 m

v p v pgz dm gz dm W

Wm : exchanged energy

Pump

H

H

Turbine

1 2 1 2

3.4. pressure drop

Real fluid viscosityViscous dissipation = conversion of mechanical energy into heat energy

Regular pressure drop

²

2r

l vJ f

d g

l

v d

p2 p1

1 2

H

The energy dissipated by the losses (in terms of energy per unit mass) can be written:

f : oefficient of friction (coefficient of linear pressure loss)

They are due to singularities on driving: elbow obstacle enlargement.in terms of height of fluid:

g

vKJ s 2

²

Singular pressure drop

Moody's diagram [ 1]

laminar

Blasius Equation

1/ 4

0,3164

Ref

Only for smooth pipes and when Reynolds number is between 2300 and 105.

Regular pressure drop

64(Re)

Ref f

turbulent

(Re, )e

f fD

e

DRelative rugosity

²

2r

l vJ f

d g

Singular pressure dropg

vKJ s 2

²

3.5 Euler equation

A Dynamic Force due to the change in direction of the fluid flow.

F md mV

dt 2 1F q v v

SV

q/2

q/2

q

0 ²F SV i

F

Exercice 3.1. : Canalisation d'arrosageUn réseau d'arrosage est constitué d'un réservoir et d'une canalisation de 100 m de long et d'un diamètre de 75 mm. La hauteur dans le réservoir est de 4,45 m. Le coefficient de perte de charge à l'entrée de la canalisation est de k = 0,5. On suppose que la canalisation est parfaitement lisse.Quel est le débit dans la canalisation ?

4. Extern flow – aerodynamic and hydrodynamic

Flow around a body immersed in the fluid flow with a relative velocity V0 .

Po

Vo

Po : Breakpoint

Assume there is contact adhesive and therefore the fluid velocity is zero at the body wall.

At very low Reynolds number flow is laminar.

laminar

transitoire

turbulentS

zone de séparation

V0

S : separation point

Different types of flow around the object

4.1. lift and drag

The flow of viscous fluid around the body causes shear forces and pressure forces.

For a symmetrical profile with axis of symmetry is parallel to the fluid velocity, the resultant force F is parallel to the fluid velocity.

F

l

e

V0

V0

L

D

1²

21

²2

L

D

L C SV

D C SV

1²

2 PP C SV

1²

2 TT C SV

Usually we need a maximum of lift and a minimum of drag

LC

DC

4.2. spheres et cylinders

²2

1SVCT T

coefficient of friction for a sphere[ 1]

coefficient of friction for a cylinder [ 2]

4.3. Bodies with sharp edges

Some values of coefficient of friction for simple geometries

Exercices 4.1. : CheminéeCalculer la force qu'exerce un vent de 50 km/h sur une cheminée de 25 m de

haut et de 1 m de diamètre. Calculer le moment d'encastrement de la cheminée.

Exercices 4.2. : ParachuteUn parachutiste de 120 kg ne doit pas descendre à plus de 6 m/s. Calculer le

diamètre minimum du parachute.

Exercice 4.3. : Goutte d’eauCalculer la vitesse de chute d'une goutte d'eau de diamètre D = 3 mm en

supposant que celle-ci a la forme d'une sphère et que l'air est dans les conditions standard de l'air au niveau de la mer. On suppose qu’elle a une vitesse

constante.

Exercice 4.5. : AnémomètreUn anémomètre à aubes est constitué d'un axe en rotation sur

lequel sont fixées 4 aubes ayant la forme de demi-sphère creuse

axe

brasvent

D

Seulement deux aubes sont représentées

A

BFB

FA

R

Ve

Ve = 40

km/hR = 5 cmVue de dessus

L'aube A se déplace dans le sens du vent et ressent un vent (vent relatif) plus faible. L'aube B se déplace contre le vent et reçoit un vent (vent relatif) plus important.1) Sachant que les aubes sont à une distance R de l'axe de rotation, calculer VA le vent relatif pour l'aube A et VB le vent relatif pour l'aube B.

2) Si l'anémomètre tourne à vitesse constante , c'est que les forces FA

et FB sont égales. Ecrire FA et FB en fonction de , Ve, R, , S, CTA , CTB .

CTA : coefficient de traînée d'un hémisphère face ouverte face au vent : 1,42

CTB : coefficient de traînée d'un hémisphère face ouverte contre le vent : 0,38

3) Calculer . Quelle est la racine physiquement acceptable de l'équation du second degré à résoudre ?

REFERENCES

[ 1] R.W. FOX et A.T. Mc DONALD, "Introduction to Fluid Mechanics", John Wiley & Son, New York, 1994

[ 2] P. GUTELLE, "Architecture du Voilier", Edition & d'Outre Mer, 2 tomes, 1979

[ 3] Encyclopaedia Universalis, 1997

[ 4] S. CANDEL, "Mécanique des fluides", Dunod Université, Paris, 1990

[ 5] R. COMOLET, "Mécanique expérimentale des fluides", Masson, 3 tomes, Paris, 1969

[ 6] R. OUZIAUX et J. PERRIER, "Mécanique des fluides appliquées", Dunod Université, Paris, 1978

[ 7] M. HANAUER, "Mécanique des fluides", Bréal, Montreuil, 1991

[ 8] D.N. ROY, "Applied Fluid Mechanics", John Wiley & Sons, New York, 1988

[ 9] R. JOULIE, “Mécanique des fluides appliquée”, Ellipse, Paris, 1998.

[ 10] A.H. Techet, cours de mécanique des fluides, MIT, MA.