Cours(Gervais_2013)

1.2.2 Velocity in one point The velocity of the point P in a reference frame R is given by: Let us consider a fluid volume V, a point P of the surface Σ of this volume, an element of thi

Fluid Mechanics

November 2013

of stresses to obtain a given deformation.

Problems encountered are varied It is often important to know how the fluidflows around solid bodies, and the knowledge of pressures and flow velocities inthe vicinity of the walls is particularly useful One also often seeks to deter-mine the stresses, the heat fluxes, the fluxes of chemical species, or integratedquantities such as hydrodynamic forces or moments

One is interested in the movements of sets (material volumes) made of a veryhigh number of molecules The characteristic dimensions of the problems con-sidered are very large compared to the average distance between the molecules(mean free path)

Let us consider a continuous fluid medium inside a volume δV around a point

P, δm being the mass of this elementary volume The density, or more precisely

the volume mass (mass per unit of volume) is defined by:

The matter is then considered as being formed of an endless number of cles, a particle being the smallest material volume on which it is still possible to

parti-apply the general laws and balance equations of classical mechanics (Mechanics

of continuous media lemma).

1.2.2 Velocity in one point

The velocity of the point P in a reference frame R is given by:

Let us consider a fluid volume V, a point P of the surface Σ of this volume, an

element of this surface (elementary surface) dσ around P of external normal−→n

The fluid external to V exerts forces (strengths, efforts) on dσ, which generatestresses The latter are defined by

One gives here a short qualitative review of some types of flows

Incompressible fluid: One considers a fluid incompressible if its mass per unit ofvolume (density) is constant or varies only a little with pressure or temperature

(⇔ ρ = cst) Water flows are an example as they are generally considered as

incompressible

Compressible fluid: the opposite As a general example, gases (at highvelocities) are considered as compressible fluids

1.3.2 Laminar and turbulent flows

One says that a flow is laminar when the motion of the particles is regular andordered The flow is turbulent when the displacement is irregular and whenfluctuations are superimposed to the mean motion of the fluid A well knownexample is given by the natural convection flow of the smoke of a cigarette (seefigure)

In order to precise the conditions of turbulent flow achievement a few sical experiments can be shortly described

clas-Flow in a duct: Reynolds experiment (1883)

Let us consider a duct connected to a tank containing a fluid under pressure,and a thin tube allowing the injection of coloured fluid into this duct (see figure).When the velocity of the flow is low, the streamlines of the fluid are stable,regular and ordered If one increases the velocity of the fluid, the motion of theparticles becomes irregular, the particles being of course convected by a meanvelocity

The experience shows that the problems depends in fact of 4 parameters:the density ρ, the velocity V, the diameter of the duct D and the (dynamic)viscosity of the fluid µ

From these 4 parameters it is possible to define the following dimensionless

number, the Reynolds number:

Re = ρVD

µ

(1.4)

One can notice that in this situation the transition from the laminar to the

turbulent regime takes place when Re is about 2000 to 2500:

• If Re < 2000, the flow remains laminar,

• From Re = 2000 to 2500, some turbulent bursts appear,

• If Re > 3000, the flow is fully turbulent.

Flow around a cylinder

It is still possible to define the Reynolds number, based here on the diameter

of the cylinder

• From Re about 10 to 40, one can notice two areas of re-circulation in the

wake of the cylinder These two areas are stretched (and lie) when theReynolds number increases,

• From Re about 40, on can notice the development of an instability within

the wake Beyond this value of the Reynolds number, one can observealternated vortices in the wake,

• At Re about 100, the vortices separate from the cylinder at regular

in-tervals (Von Karman street) When the Reynolds number becomes veryhigh the wake is fully turbulent

1.3.3 Steady and unsteady flows

A flow is steady if all the variables that describe the motion are independent oftime On the opposite, if one variable depends on time the flow is unsteady Allthe wave propagation phenomena are unsteady, as well as atmospheric flows.All the turbulent flows are unsteady by definition However, one can considerthat a turbulent flow is steady if the mean variables that describe the flow areindependent of time

One says that a flow is unidimensional if the variables are constant on eachsection of the flow In other words, all the variables are dependent of a spacecoordinate only

In practical applications, many flows can be considered unidimensional in afirst approximation, leading to interesting simplifications especially when con-sidering in duct flows

Let us consider a fluid volume V bounded by a surface Σ of external normal−→n

The momentum equation for this fluid volume at rest is:

on V are normal forces Let us notice here that p is nothing else that pressure.

We then obtain for the whole surface Σ of the volume V:

φ is a mass density of force

We then obtain for the whole volume V:

If this equation is right for a volume V, it is right for any volume V and

especially for an elementary volume dv (lemma of the mechanics of ous media) The previous equation can then be reduced to its following local

continu-expression:

−−−→grad p + ρ−→φ = 0 (2.10)

In this (common) situation, the mass density of force−→

φ reduces to the gravity

k is an ascendant vertical vector of unity norm.

If the fluid is considered as incompressible, its density is constant (ρ = cst)

With the equation (2.10), it comes:

con-b - The quantity p is called the acon-bsolute static pressure Its dimension is: [p]

=ML−1T−2 (with M the mass, L the length and T the time) In the MKS

International System (SI), p is expressed in Pascal, noted Pa.

We also find as unit of pressure:

• The Bar : 1 Bar = 105 Pa

• The Atmosphere : 1 Atm = 1.01325 105Pa

c - One can also define the driving pressure p∗:

p∗= p + ρgz (2.15)The static equation (2.14) then leads to:

for a fluid at rest

d - We call effective pressure the quantity p e = p − p atm involving in that casethe atmospheric pressure to be taken as reference This involves possible nega-

tive values for p e

p eM = pM− p atm= ρe gh (2.18)

The knowledge (or the measurement) of h allows to obtain the effective pressure

at point M

2.2.3 Incompressible fluid in relative equilibrium

For a fluid in relative equilibrium, the relative acceleration is null (−→

γrel = 0)but the driving acceleration is not (−→

γent 6= 0) The Fundamental Principle of

the Dynamics then leads to (per unit of volume, that is for the local form of the

static equation):

−−−→grad p + ρg−→z + ρ−→γent= 0 (2.19)with −→z the vertical ascendant vector.

Example : Water in a tank in linear uniformly accelerated motion in the

direc-tion −→x

The absolute acceleration of the tank−→

γabsreduces to its driving acceleration

As at x = 0 and z = z o the pressure p is equal to the atmospheric pressure

p a, one shows that:

In this situation, the density ρ is no more a constant and we have to take into

account its variations (with z) The static equation is still given by:

−−→

grad p + ρg−→z = 0 (2.23)

with ρ 6= cst Thus:

Case of the perfect gas

The state relation for a perfect gas is given by:

p

ρ

with for air r = 287 s m2K2 (perfect gas constant)

The equation of static then becomes:

dp

dz = −ρg = − rT p g (2.26)which leads to:

rest - Archimède’s theorem

2.4.1 Pressure forces

Consider a body of volume V bounded by a surface Σ of external normal −→n

immersed into a fluid, M being a current point of its surface As the body is at

rest, the stresses and efforts are limited to the pressure only.

The resultant strength of the actions of the pressure is given by:

One can notably notice that if the pressure is constant (and for example equal

to the atmospheric pressure p atmon the (closed) surface Σ), then the resultantand the moment of the actions of pressure are both null (that justifies the fact

that it is often possible to take the atmospheric pressure p atm as reference inmany practical applications)

thrust

Let us consider a plate inclined with a given angle (see figure), with water onone side We want to determine the force that the water exerts on the plate andthe centre of thrust of the actions of the pressure

It is at first necessary to calculate the pressure at a point M of vertical

coordinate zM This is obtained from the application of the static equation (Obeing the origin of the reference frame):

pO+ ρgzO= pM+ ρgzM (2.31)

So, considering pO = p atm taken as reference (and then equal to zero) and

zO= 0, we have pM= p(z) = −ρgz Then, in our situation where−→n is oriented

outwards of the water volume, the resultant−→

F of the actions of pressure of thewater on the plate is then given by:

F of the elementary forces d−→

F It is thendefined by:

2.4.3 The Archimède Theorem

Let us consider a solid body (S) totally immersed in a fluid at rest This solidbody fills the volume V bounded by a surface Σ This solid body is submitted

to actions of pressure from the surrounding fluid As it is in equilibrium, onecan write :

fluid, said removed, is in equilibrium under the actions of its own weight and of

the pressure strength that exerts the surrounding fluid on the surface Σ One

can write for this removed fluid:

−−−−−−−−−−−−−−−−−−−−−−−→

Weight of the solid body (S) =−−−−−−−−−−−−−−−−−−−−−−−→

Weight of the removed f luid

We can then write the Theorem of Archimède: The resultant of the forces of

pressure exerted by a fluid on a solid totally immersed is a vertical ascendantforce equal to the weight and applied at the centre of gravity of the removedfluid

Remark : This result remains available for a solid immersed in different fluids

at rest

One considers a particle P that one follows in its motion P0(x0, y0, z0) are

the coordinates of P at the initial time (t = t0) in the frame of reference(O; −→x , −→y , −→z ) The motion of the fluid is totally defined if one gives a func-tion f of space and time such that −→

OP = f (−−→

OP0, t) The coordinates of P at a

given time are called the Lagrange variables

The Lagrange description is then a material description as one follows the

lo-cation (the movement) of each particle with time Each particle is considered

as an individual entity

Practically, it is not really interesting to try to follow the motion of each particle(movements and displacements can be very large and one is not truly interested

by the history of the particle, where it comes from and what it is going tobecome), and in fact fluid particles have not really individuality as nothingdiffers particles each others It is more interesting to study the time evolution

of the fluid in one or several points of space

value at each time represents the velocity vector of the particle that goes through

M The Euler variables are the components of this vector according to the

principal axis of the reference frame:

The Euler description is then a spatial description as the evolution of the

fluid is studied at different points of space

One calls the trajectory of a particle the line or the set of successive positions

taken by this particle with time If one knows the velocity vector:

dt = v(x, y, z, t) dz

dt = w(x, y, z, t)

(3.1)

then one has just to integrate to determine the equation of the trajectory

Let us consider the domain in which the fluid develops At a given time t, on

can define the velocity vector−→

V (−−→

OM, t) at each point of this domain One calls

streamline every line or curve which at each point and at any time is tangent to

the velocity vector at this point The streamline equations are given from therelation expressing this collinearity:

One calls the emission line of a point A at time t the geometrical locations at this time t of all the particles that gone previously through the point A (see the

smoke of a chimney as an example) To obtain the equation of an emission line

at a time τ it is necessary to integrate the equation of the trajectory providedthat the particle is in M0(x0, y0, z0) at t = τ.

A flow is permanent (or steady) when all the physical quantities which

charac-terise its evolution are independent of time In this situation, streamline areindependent of time and streamline and trajectories are matching

3.2 Deformations in flows

One gives here some information about the deformation and rotation rate of

a fluid particle which are important notions in fluid mechanics The study ofthe general motion of a fluid particle in two or three dimensions allows to findthe elementary notions of translation, rotation, dilatation (expansion) and de-formation

Different motions (see figures)

• translation,

• rotation,

• dilatation (expansion),

• shear

In practice, these elementary motions are combined

One says that a flow is irrotational (−→

rot−→

V = −→

0 ) when the fluid particleshave not rotation on them self

3.3.1 Material derivative - Rate of variation

Let us consider a function 3 of space and time characteristic of a fluid quantity(velocity, density, ) One wants to study the variations of this quantity

One can expresses 3 as a function of Xk and time t (material description

of Lagrange, with Xk the initial position of the particle), so that 3 = 3(Xk , t).

It can also be described as a function of x i and t, with x i = x i(Xk , t) the

spa-tial coordinate of the particle Xk at time t (spatial description of Euler), the

function 3 characteristic of the considered fluid quantity being now written as

as Xk the initial location of the particle is independent of time

If one wants now to express this variation rate as a function of the spatial

coordinates x i(Xk , t) as for the Euler description, one has to write:

∂t is nothing else that the velocity component in the

di-rection i of the considered particle:

dt =

∂−→W

3.3.2 Derivative of a volume integral

Let us consider a volume fluid D (possibly varying with time, D = D(t)) bounded

by a surface Σ of external normal −→n , a function f (−→x , t) of space and timecharacterising a fluid quantity contained into the volume D as for example thedensity, the energy, et cætera We want to determine the variation of the wholequantity contained into this volume, that is:

f (x, t + ∆t)dx −

Z b(t) a(t)

Z b(t+∆t) a(t+∆t)

f (x, t)dx

+

Z b(t+∆t) a(t+∆t)

f (x, t)dx −

Z b(t) a(t)

(f (x, t + ∆t) − f (x, t))dx

+

Z b(t+∆t) b(t)

f (x, t)dx −

Z a(t+∆t) a(t)

f (x, t)dx

#

=

Z b(t) a(t)

The rate of variation of the mass is then given by:

which is the balance equation of mass for the volume D.

According one more time to the theorem of the divergence in order to form the surface integral of the previous equation into a volume integral, equa-tion (3.9) allows to write:

trans-∂ρ

∂t + divρ−→

which is the local equation of conservation of mass.

It is possible to develop the right hand side of the previous equation toobtain:

dρ

dt + ρ div−→V = 0 (3.11)which leads directly to the following relation for an incompressible fluid (such

n dσ = 0 indicating in fact the mass flow conservation.

In this section, we consider the ideal irrotational flow of an incompressible andinviscid fluid in permanent regime

3.5.1 Definitions

1 - The flow is consider bi-dimensional (contained into a plan), such that:

−

→V



u(x, y) v(x, y)

2 - The flow is irrotational: −→

3 - Streamlines are given by the curves such that ψ = cte, and equipotential

lines by those such that 3 = cst.

Furthermore, as −−→

grad 3 = −−→grad ψ ∧−→z , streamlines and equipotential lines

constitute a perpendicular network

4 - As 3 and ψ respect the Cauchy-Riemman conditions, one can introduce

an holomorphic function with 3 and ψ its real and imaginary parts respectively,

such that f = 3 + iψ defining the complex potential.

One calls W(z), where z = x + iy, the complex velocity This one is then

Let us give a constant velocity vector −→

U∞ collinear to the −→x direction, suchthat: −→

U∞k −→x In this situation one has:

Source or sink at the origin

We consider a purely radial flow from a source or a sink (well) located at theorigin O of the frame of reference (see figure) In this situation, the flow is purelyradial, symmetrical, streamlines are straight lines that converge and cross atthe origin and equipotential lines are concentric circles of centre O Then thevelocity vector reduces to−→

V = v r−→r and the volume flow is given by:

One can express the potential function of this flow In the Cartesian frame

of reference, one can write, with u and v the components of the velocity vector

in the x and y directions respectively:

u = v rcos(θ) = q cos θ

2πr =

q x

2π(x2+ y2) (3.22)

v = v rsin(θ) = q sin θ

2πr =

q y

2π(x2+ y2) (3.23)According to the Cauchy-Riemann conditions, one obtains:

(3.24)

Making use of the cylindrical coordinates and as z = re ıθ, it comes:

3.5.4 Singular vortex in a point

In this situation (see figure), the equipotential lines are straight lines converging

in z0and streamlines are concentric circles of centre z0 As previously, one canshow starting from the relation:

Γ =Zcircle

Γ is called the circulation It is positive for a counter-clockwise rotating vortex

(trigonometrical direction of rotation), negative otherwise

This situation correspond to the combination of a source and a sink, of lent but opposite intensity and separated by a small distance We have in thiscase:

equiva-f (z) = −µ

2π

1

where µ is the dipole intensity

Let us give A and B two points located on the same equipotential line 3 of

curvilinear abscissa s and M a current point of 3 between A and B of coordinates

M(x(s), y(s)) The volume flow between the two streamlines ψ(A) and ψ(B) is

given by:

q =

ZAB

The Fundamental Principle of the Dynamics (FPD) expresses that the rate

of variation of momentum of a given material (fluid for instance) domain is

counterbalanced by (equal to) the external forces For a fluid domain of volume

Vm (t) momentum is given by:

Surface forces: We retain here a very important basic hypothesis as we consider

the fluid as perfect (inviscid) In this situation, the surface forces reduce to

pressure strength Thus:

by making use of the Green-Ostrogradsky theorem

The Fundamental Principle of the Dynamics (4.2) then leads to:

ρ ∂

−

→V

∂t +grad−→

V−→V

!

= −−−→

grad p − ρ−−→grad(gz) (4.9)

Remark: From the mass conservation equation, it is possible to show that:

dt dv (4.10)

4.1.2 Particular cases of the Euler equations - The Bernoulli

theorem

Permanent and irrotational flow, incompressible fluid

When the flow is permanent and the fluid incompressible, the Euler equationsreduce to:

Permanent and rotational flow, incompressible fluid

As previously, as the flow is permanent and the fluid incompressible, the Eulerequations reduce to:

The components of the tangent vector−→

t are (dx, dy, dz), so that:

t , that is along a streamline (only) for a rotational flow.

The unsteady Bernoulli theorem

We consider here a perfect fluid, incompressible, in unsteady and irrotationalflow (in this situation, we have ∂(.)

∂t +−−→

grad(p + ρgz +12ρV2) = 0 (4.17)

or again:

∂−→V

Theorem of Bernoulli for a compressible fluid

We still consider a perfect (inviscid) fluid in the gravity field, the flow being

permanent and irrotational We assume the fluid to be barotropic, so that the pressure is only a function of the density: p = p(ρ) In this situation, one can define a function h such that:

dh = dp

ρ

The function h is called the mass enthalpy: h =R dp

ρ

Here the Euler equations can be written as:

Some important remarks

• Generally, for a gas the quantity ρgz is negligible (as ρ is very small),

so that if the gas is submitted to relatively low pressure variations, one

expresses the Theorem of Bernoulli as : p +12ρV2= cst,

• In duct flows; duct approximation : Generally, one can consider a duct as

a streamline, so that the Theorem of Bernoulli may be applied all alongthe duct According to the perfect fluid hypothesis the velocity is constant

in a cross section of the duct If the duct is rectilinear, streamlines arestraight and the pressure is hydrostatic in a cross section of the duct :

p + ρgz = p∗= cst in a cross section of the duct.

If one divides the expression of the Bernoulli Theorem by ρg, leading to:

p

ρg + z +ρV

2

2g = cst (4.25)

each term of the previous equation has the dimension of a length (a height in

fact), and is expressed in meter of the corresponding fluid:

• p

ρg + z is the piezometric height,

• p

ρg + z + V2g2 = H is the hydraulic head

The Bernoulli Theorem then indicates that the head or the total pressure

are constant all along a streamline (for a perfect fluid) Head and piezometric

heights can be represented by the head and piezometric lines (see figure)

Let us consider a cylindrical rectilinear pipe A piezometric tube is a smallduct connected at one end to the main pipe through a very small opening andopened to the atmosphere (for example) at its other end This tube can be ofunspecified form and slope After being connected, the fluid goes up into thepiezometric tube up to obtain the static equilibrium

Inside the piezometric tube, one can apply the static equation between points

The Pitot Tube is a cylindrical tube with a lateral and a frontal pressure inletsconnected to a pressure gauge (U tube for example) The Pitot Tube is placedinto the flow parallel to the streamlines, such that the perturbation it involves

is minimum The flow is permanent and the fluid is incompressible such thatthe hypothesis of the Bernoulli Theorem are respected The flow upstream tothe Pitot tube is assumed to be uniform (see figure)

In this situation, the Bernoulli Theorem can be applied between the points

M and A, M being a current point of the flow on the same streamline that A,the point at the head of the Pitot tube One can then write:

p∗M+1

2ρV

2

M= p∗A

as A is a stagnation point where the velocity is null (VA = 0), pA being the

stagnation pressure.

One can consider that the point A is also on the same streamline that B,and that the Pitot Tube does not perturb the mean flow, so that VB = VM.Thus:

p∗A= p∗B+1

2ρV

2 B

of the static equation inside the U-tube:

Let us give a plan-parallel tank of height h, a point A of its surface S (at the atmospheric pressure p a) and M a point located at the exit of the tank of surface

s (see figure) The dimensions of the tank are supposed to be sufficiently large,and then the displacements of the free surface S sufficiently slow to neglect itsinstantaneous variations, so that the flow can be considered as steady (this is

in fact what is called the quasi-permanent regime) We want to determine the

velocity at the exit of the tank, and the time necessary for its emptying.Streamlines are converging from the free surface (point A) to the exit ofthe tank (point M), so that the Bernoulli theorem can be applied between thepoints A and M:

The static pressures pA and pM in A and M are equal to the atmospheric

pressure p a : pA = pM = p a Furthermore, the velocity of the free surface S ofthe tank (in A) is very small in comparison with the velocity at the exit (in M),

so that the dynamic pressure in A is negligible in comparison to the one in M :

exit of the tank (the point M): z = zA− zM In this situation, the instantaneousvelocity at the exit of the tank is given by:

VM=p2gz

Furthermore, the conservation of the volume flow (as the fluid is assumed to beincompressible) involves that:

sVM= SVAleading to the following differential equation:

the sign - coming from the fact that the free surface is going down

This differential equation may be integrated as:

Real time of emptying (see slide) : In fact there is a contraction of the flow

at the exit of the tank, so that the effective surface of the exit is σ such that

σ = CCs where CC is the coefficient of contraction There is also friction ofthe flow, especially at the exit of the tank where the fluid velocity is important,such that the real velocity is given by V∗M = CfVM where Cf is the frictioncoefficient The real volume flow is then given by:

q r= CCCf q th = φq th

where q th is the theoretical volume flow given by: q th = sVM = s√

2gz The coefficient φ is called the coefficient of the outlet The result is of course the

same for the real time of emptying

Let us consider a pipe with a convergent followed by a divergent section (see

figure) This convergent-divergent configuration forms a Venturi We want to

obtain the volume flow through this pipe The flow is assumed to be nent and the fluid incompressible, so that the Bernoulli theorem can be appliedbetween the sections 1 and 2, respectively just upstream the contraction and atthe neck of the Venturi

The volume flow conservation (ρ = cst) involves that:

S1V1= S2V2= q v

where q v is the volume flow

One can at first notice that when the surface S of the section increases, thevelocity V decreases (by conservation of the volume flow) and the Bernoulli the-

orem indicates that the pressure p∗ increases The previous Bernoulli equationgives:

u 2(p∗1− p∗

2)ρ

Chapter 5

Momentum Theorem and Applications

Let us consider a fluid volume Vm (t) bounded by a surface Σ m (t) of external

normal −→n As already seen, the Fundamental Principle of the Dynamics applied

to this volume is given by:

Remarks :

• For a real fluid it is necessary to take into account the friction that exertsnaturally on the surface Σ of the domain (surface forces does not reduceany more to the pressure strength and it is necessary to introduce theviscosity),

• If the regime of the flow is permanent, so that the material control volume

is fixed (Vm= V and Σm = Σ) and that there is not time variations ofthe physical quantities (∂

5.2.1 Efforts on a duct or a pipe element

One wants to determine the resultant of the forces that the flow exerts on a pipeelement (see figure) Here the fluid is incompressible and inviscid, and the flow

is permanent Furthermore, one neglects the effects of the gravity, so that theterm −R

Vρg−→z dv is assumed to be negligible (the fluid is a gas or the element

of pipe is horizontal for instance) The material control volume V is bounded

by the surface Σ, of external normal−→n , such that:

Σ = Σ1∪ Σ2∪ ΣD

In this situation, the flow momentum theorem is expressed as:

One can now calculate the two terms of this equation:

• Calculation of the term:

−Z

Σ

p−→

n dσ

One can at first notice that it is possible to introduce the effective pressure

p e , defined with the atmospheric pressure p a as reference: p e = p − p a, so

As the whole system is immersed in a surrounding medium at the

atmo-spheric pressure p a, and as Σ is a closed surface that bounds the materialvolume of interest, one has:

−

→

V·−→V −→

n dσ

Finally, the application of the momentum theorem leads to:

• One can write another relation between p e1 and p e2 from the Bernoulli

principle: p e1+12ρV21= p e2+12ρV22, allowing to define the force that thefluid exerts on the element of pipe knowing the pressure at the inlet (or at

the outlet) of the domain and the volume flow q v (with q v= S1V1= S2V2)only

5.2.2 Force of an impinging jet on a flat plate

One wants to calculate the force that an impinging jet exerts on a flat plate (seefigure) We then consider an horizontal jet impinging a vertical flat plate suchthat the jet is deviated at 90◦ The effects of the gravity are neglected so thatone can consider a symmetry of revolution around the −→

x axis The surface Σ1

of the jet at the inlet is of section S1, the impinging surface of the jet on the

plate Σ2has a section S2, the lateral surface is Σ3 and the exit surface Σ4 has asection S4.

The theorem of momentum yields:

Σ 1

ρ

−

→

V2 −→

V2· −→n2dσ+

Z

Σ 3

ρ

Σ 4

ρ

The integral on the surface Σ2and Σ3are null as−→

V2·−→n2= 0 and−→

V3·−→n3= 0.The integral on the surface Σ4is also null because of symmetry reasons.Then the previous equation reduces to:

Σ 2

p2−→n

2dσ −Z

Σ 3

p3−→n

3dσ −Z

5.2.3 Force of a jet on a bucket of a Pelton turbine

The geometry considered here is very close to the previous one, except that inthe present situation the jet is impinging on a bucket and then deviated of 180◦.According to the definition of the different surfaces and the normal vectors of