Basic fluid mechanics for civil engineers

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Basic fluid mechanics for civil engineers

Maxime Nicolas

To cite this version:

Maxime Nicolas. Basic fluid mechanics for civil engineers. Engineering school. France. 2016. �cel-01440549�

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Basic fluid mechanics for civil engineers

Maxime Nicolas

maxime.nicolas@univ-amu.fr

D´epartement g´enie civil

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Course 1 outline

1 _Preamble Course schedule Online Working advices Course outline

2 _{Introduction and basic concepts}

Description of a fluid Maths for fluid mechanics

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Preamble

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Preamble Course schedule

Course syllabus

Schedule:

10 lectures 10 workshops Assessment and exam:

activity percentage

homework 20%

bonus +1 if written in english

final exam (Dec. 5th) 80%

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Preamble Online

Online

This course is available on ENT/AmeTice :

Sciences & technologies � Polytech � G´enie civil � [16] - S5 - JGC51B - M´ecanique des fluides (Maxime Nicolas)

with slides

workshops texts equation forms

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Preamble Working advices

Working advices

personal work is essential

read your notes before the next class and before the workshop be curious

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Preamble Course outline

Course outline

1 Introduction and basic concepts vector calculus

2 Statics hydrostatic pressure, Archimede’s principle

3 Kinematics Euler and Langrage description, mass conservation

4 Balance equations mass and momentum cons. equation

5 Flows classification and Bernoulli Venturi e↵ect

6 The Navier-Stokes equation Poiseuille and Couette flows

7 The Stokes equation Flow Sedimentation

8 Non newtonian fluids Concrete flows

9 Flow in porous media Darcy

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Introduction and basic concepts

INTRODUCTION AND BASIC CONCEPTS

Description of a fluid

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What is fluid mechanics?

●○

Physics solid mech. continuum mech. fluid mech. aerodynamics viscous flows waves supersonic flows

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What is fluid mechanics?

○●

Fluid mechanics is the mechanical science for gazes or liquids, at rest or flowing.

Large set of applications : blood flow

atmosphere flows, oceanic flows, lava flows pipe flow (water, oil, vapor)

flight (birds, planes) pumping

dams, harbours . . .

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Large atmospheric phenomena

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FM for civil engineering: dams

Hoover dam, 1935

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FM for civil engineering: wind e↵ects on structures

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FM for civil engineering: concrete flows

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Introduction and basic concepts Description of a fluid

What is a fluid?

move with the plate continuously at the velocity of the plate no matter how small the force F is. The fluid velocity decreases with depth because of fric-tion between fluid layers, reaching zero at the lower plate.

You will recall from statics that stressis defined as force per unit area and is determined by dividing the force by the area upon which it acts. The normal component of the force acting on a surface per unit area is called the

normal stress,and the tangential component of a force acting on a surface per unit area is called shear stress(Fig. 1–3). In a fluid at rest, the normal stress is called pressure.The supporting walls of a fluid eliminate shear stress, and thus a fluid at rest is at a state of zero shear stress. When the walls are removed or a liquid container is tilted, a shear develops and the liquid splashes or moves to attain a horizontal free surface.

In a liquid, chunks of molecules can move relative to each other, but the volume remains relatively constant because of the strong cohesive forces between the molecules. As a result, a liquid takes the shape of the container it is in, and it forms a free surface in a larger container in a gravitational field. A gas, on the other hand, expands until it encounters the walls of the container and fills the entire available space. This is because the gas mole-cules are widely spaced, and the cohesive forces between them are very small. Unlike liquids, gases cannot form a free surface (Fig. 1–4).

Although solids and fluids are easily distinguished in most cases, this dis-tinction is not so clear in some borderline cases. For example, asphalt appears and behaves as a solid since it resists shear stress for short periods of time. But it deforms slowly and behaves like a fluid when these forces are exerted for extended periods of time. Some plastics, lead, and slurry mixtures exhibit similar behavior. Such borderline cases are beyond the scope of this text. The fluids we will deal with in this text will be clearly recognizable as fluids.

Intermolecular bonds are strongest in solids and weakest in gases. One reason is that molecules in solids are closely packed together, whereas in gases they are separated by relatively large distances (Fig. 1–5).

The molecules in a solid are arranged in a pattern that is repeated through-out. Because of the small distances between molecules in a solid, the attrac-tive forces of molecules on each other are large and keep the molecules at

3 CHAPTER 1 Fn Ft F Normal to surface Tangent to surface Force acting on area dA dA FIGURE 1–3

The normal stress and shear stress at the surface of a fluid element. For fluids at rest, the shear stress is zero and pressure is the only normal stress.

Free surface

Liquid Gas

FIGURE 1–4

Unlike a liquid, a gas does not form a free surface, and it expands to fill the entire available space.

(a) (b) (c)

FIGURE 1–5

The arrangement of atoms in different phases: (a) molecules are at relatively fixed positions in a solid, (b) groups of molecules move about each other in the liquid phase, and

(c) molecules move about at random in the gas phase.

Shear stress: t !Ft

dA

Normal stress: s !Fn

dA cen72367_ch01.qxd 10/29/04 2:31 PM Page 3

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Main concepts

density

stresses and pressure viscosity

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density

density = weight per unit volume unit : kg⋅m−3 fluid density in kg⋅m−3 air 1.29 water 1 000 concrete 2 500 molten iron ≈ 7 000

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Stress

Elementary force �→F applying on an elementary surface S.

𝛿F 𝛿S n Ratio is �→ = �→F S

thestress vector.

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Stress

The surface element S is oriented by a unit vector �→n . �→_{n is normal (perpendicular) to the tangential plane.}

�→ = �→n+ �→t with �→_n_{= (�→ ⋅ �→}_n_)�→_n �→_t_{= �→ − �→}_n_{= (�→ ⋅ �→t)�→t} 𝛿S n t

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Pressure

The pressure is a normal stress.

Notation : p

S.I. unit : pascal (Pa) 1 Pa = 1 N⋅m−2 = 1 kg⋅m−1⋅s−2

basic interpretation: normal force applied on a surface

The pressure in a fluid is an isotropic stress: its intensity does not depend on the direction.

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Pressure examples

p(H) p(H) p m= 1500 kg with load S = 0.5 m2 H

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viscosity

A macroscopic view on viscosity :

F U h fixed boundary moving plate velocity profile u(y)

Tangential (shear) stress: t= ⌧ = F_A

Shear rate: ˙ =U_h For a newtonian fluid :

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viscosity

Standard unit: [⌘]=Pa⋅s 1 Pa⋅s=1 kg⋅m−1⋅ s−1

fluid ⌘ (Pa⋅s)

air 1.8 10−5

water 10−3

blood 6 10−3

honey 10

fresh concrete 5–25Bnon-newtonian fluid

Also useful : kinematic viscosity

⌫= ⌘

⇢

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superficial tension

The superficial tension applies only at the interface between 2 di↵erent fluids (e.g. water and air).

The molecules of a fluid like to be surrounded by some molecules of the same kind.

A drop of liquid on a solid surface does not flatten completely under gravity:

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superficial tension and wettability

symbol:

unit: [ ]=N⋅m−1

order of magnitude: 0.02 to 0.075 N⋅m−1

most common: water�air = 0.073 N⋅m−1

When the fluid molecules are preferring the contact with a solid surface rather than the surrounding air, it is said that the fluid is wetting the solid.

wetting non-wetting

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drops and bubbles

When the water/air interface is curved, the surface tension is balanced with a pressure di↵erence, according to Laplace’s law:

p= pint− pext = � 1 R1 + 1 R2� p p_int p ext int pdrop =2_R pbubble =4_R

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capillary rise

The capillary rise is a very common phenomena (rise of water in sils, rocks or concrete), and can be illustrated with a single tube:

𝛥h

wetting → curvature → pressure di↵erence → rise

h= 4 cos ✓

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INTRODUCTION AND BASIC CONCEPTS

Maths for fluid mechanics

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Introduction and basic concepts Maths for fluid mechanics

Maths for fluid mechanics

scalar, vector, tensor scalar fields f(x,y,z) vector fields�→A(x,y,z)

di↵erential operators : gradient, divergence, curl, laplacian partial di↵erential equations

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scalars and scalar field

A scalar is a one-value object. mass, volume, density, temperature. . .

A scalar field is a multi-variable scalar function p(x,y,z) = p(�→r)

Without time, stationary scalar field p(�→r)

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Vectors

A vector is a multi-value object. Useful to represent forces, velocities, accelerations. In 3 dimensions, �→_A _{= (A} x,Ay,Az) = � � � Ax Ay Az � � � example of the gravity acceleration:

�→_g ₌�_� � 0 0 −g � � �

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Vector field

A vector field is a set of scalar functions, each function is a component of a vector. �→_A_{(x,y,z) =}�_� � Ax(x,y,z) Ay(x,y,z) Az(x,y,z) � � � and for an unstationary vector field

�→_A_{(x,y,z,t) =}�_� � Ax(x,y,z,t) Ay(x,y,z,t) Az(x,y,z,t) � � �

The value of the vector has to be computed at each space point and for each time.

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Vector field

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Review of vector and di↵erential calculus

derivative definition for a single variable function:

d dtf(t) =

f(t + t) − f (t)

t , as t → 0

but many useful functions in fluid mechanics are multi-variables functions (pressure, velocity).

Partial derivative: @f(x,y,z,t)

@y =

f(x,y + y,z,t) − f (x,y,z,t)

y , as y → 0 Important implication : @2f @x@y = @2f @y @x @2 _(x2 _{+ 1)}

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Integration of a partial derivative

Let’s define

@f(x,y,z)

@y = k(x,y,z)

Integrating along a single coordinate (here y ) gives

f(x,y,z) = � k(x,y,z)dy + C(x,z)

The integration constant C does not depend on the integration coordinate.

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A very useful di↵erential operator

Let’s define for (x,y,z) coordinates �→_{∇ =}�_�� @ @x @ @y @ @z � �� nabla or del

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gradient

The gradient operator applies to a scalar function: ��→ gradf = �→∇f = � �� @f @x @f @y @f @z � �� scalar ��→ grad ��→ vector

Consequence: the gradient of a scalar field is a vector field.

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divergence

The divergence of a vector field is a scalar field: �→_{∇ ⋅ �→}_A _{= div�→}_A ₌�_�� @ @x @ @y @ @z � �� ⋅ � � � Ax Ay Az � � �= @Ax @x + @Ay @y + @Az @z

vector �→ scalardiv

Example 1: compute �→∇ ⋅ �→A with �→A = (x,y,z) Example 2: compute �→∇ ⋅ �→A with �→A = (y,z,x)

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Why �

∇ is not a true vector?

→

Let’s compare �∇ ⋅ �→→ A and�→A⋅ �→∇ �→_{∇ ⋅ �→}_{A (the divergence) of}�→_{A is a scalar} �→_A _{⋅ �→}_{∇ is an scalar di↵erential operator:}

�→_A_{⋅ �→}_{∇ = A} x @ @x + Ay @ @y + Az @ @z Obviously �→∇ ⋅ �→A ≠ �→A⋅ �→∇

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curl

The curl of a vector field is

�→_{∇ × �→}_A _{= �→}_{∇ ∧ �→}_A _{= ��→}_curl�→_A _{= �→}_rot�→_A ₌�_�� @ @x @ @y @ @z � �� × � � � Ax Ay Az � � � �→_{∇ × �→}_A ₌�_�� @Az @y − @Ay @z @Ax @z − @Az @x @Ay @x − @Ax @y � �� vector ��→ curl ��→ vector

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curl

Alternate method: �→_{∇ × �→}_A _{= det}�_�� →_e_x �→_e_y �→_e_z @ @x @y@ @z@ Ax Ay Az � �� Example: calculate �∇ × �→→ A for

�→_A ₌�_� � x2− y2 y2− z2 z2− x2 � � �

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laplacian

The laplacian is the divergence of the gradient: f = �→∇ ⋅ �→∇f = ∇2f

and for a (x,y,z) coordinate,

f = @ 2_f @x2 + @2f @y2 + @2f @z2 scalar�→ scalar

But a laplacian can also apply to a vector: �→_A ₌�_� � Ax Ay Az � � �

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Useful formulae

The curl of a gradient is always zero : �→_{∇ × �→}_{∇f = 0} Prove it!

The divergence of a curl is always zero: �→_{∇ ⋅ (�→}_{∇ × �→}_A_{) = 0} The double curl:

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Memo

scalar ��→ grad ��→ vector vector �→ scalardiv vector ��→ curl ��→ vector scalar�→ scalar vector�→ vector

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Other coordinate systems

The cartesian (x,y,z) is not always the best.

Flow in a pipe: �→v(�→r,t) and p(�→r,t)

�→_v_{(r,✓,z,t), p(r,✓,z,t)}

In this course, only the cartesian and cylindrical coordinate systems will be used.

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Di↵erential operators in cylindrical coordinates

�→_{∇f =}�_� � @f @r 1 r @f@✓ @f @z � � � �→_{∇ ⋅ �→}_A ₌ 1 r @ @r(rAr) + 1 r @A✓ @✓ + @Az @z �→_{∇ × �→}_A ₌�_�� 1 r @Az @✓ − @A✓ @z @Az @r − @Ar @z 1 r @r@(rA✓) − 1 r @Ar @✓ � �� f = @ 2_f @r2 + 1 r @f @r + 1 r2 @2f @✓2 + @2f @z2

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Basic fluid mechanics for civil engineers

Lecture 2

Maxime Nicolas

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Lecture 2 outline

1 _{Force balance for a fluid at rest}

2 _{Pressure forces on surfaces}

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Force balance for a fluid at rest

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Cube at equilibrium

Hypothesis: homogeneous fluid at rest under gravity. Imagine a cube of virtual fluid immersed in the same fluid :

x y

z somewhere in the fluid

�→

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Continuous approach

weight of an infinitesimal volume of fluid V of mass m: �→

W = m�→g = �

V⇢�

→_{g dV} pressure forces acting on surface S, boundary of V :

�→_F p= − � Sp(M)dS�→n at equilibrium,�→W + �→Fp= 0, written as � _V ⇢�→g dV − � Sp(M)dS�→n = 0

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Useful theorem

The gradient theorem

�_Sf dS�→n = � V �→_{∇f dV} Thus � _V ⇢�→g dV − � V �→_{∇p(M)dV = 0} and � _V�⇢�→g dV − �→∇p(M)dV � = 0 finally �→_{∇p − ⇢�→}_g _{= 0}

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integration

for �→g = (0,0 − g) and p = p(z), −⇢g −dp_dz = 0 which gives p(z) = p0− ⇢gz

with p0 the reference pressure at z = 0.

if z = 0 is the free water/air surface, then p0= patm, and the relative

pressure is

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The hydrostatics (( paradox ))

Pressure does not depend on the volume.

A B C D E F

M

H

pA= pB = pC = pD = pE = pF

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numerical example

z 0 for z = −10 m, prel = p − patm= ⇢gz = 103× 10 × 10 = 105 Pa ≈ 2 105

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pressure measurements: the manometer

p_atm p A p B z 1 2 h

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Pressure forces on surfaces

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Pressure force on a arbitrary surface

The total pressure force acting on a surface S in contact with a fluid is �→_F

p= − �

Sp(M)�→n dS

Bremember �→n is an outgoing unit vector

M n p

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Pressure force on a vertical wall

n

z

0 p

_atm

p(z)

x

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pressure center

definition: the pressure center C is defined by �→

OC× �→Fp= − � S

��→

OM× (p�→n)dS, M,P ∈ S

applying �→Fp on P does not induce rotation of the surface.

�→

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Pressure center on a vertical wall

n z 0 p_atm p(z) x n z 0 p_atm x Fp C G H_C

H: height of the wetted wall, L= width of the wetted wall

pressure center located at 2/3 of the depth

h=2

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Pressure center on a vertical wall

n z 0 p_atm p(z) x n z 0 p_atm x Fp C G H_C

H: height of the wetted wall, L= width of the wetted wall pressure center located at 2/3 of the depth

h=2

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pressure center and barycenter

the pressure center is always below the gravity center (barycenter). It can be proved that

HC = HG + I

HGS

with

HC: depth of the pressure center

HG: depth of the gravity center

S: wetted surface I : moment of inertia see Workshop #2

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Archimedes

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Archimedes

The buoyancy principle

In Syracuse (now Sicily), in -250 (est.), Archimedes writes:

A body immersed in a fluid experiences a buoyant vertical force upwards. This force is equal to the weight of the displaced fluid.

z

FA G

This force applies at the buoyancy center: barycenter of the immersed volume.

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Archimedes

Modern formulation of the principle

the pressure force acting on the surface S of a fully immersed body is �→_F

p= − �

Sp�

→_{n dS} from the gradient theorem,

�→

Fp= − �

V

�→_{∇p dV}

and combining with the hydrostatics law �→∇p = ⇢�→g , we have �→_F

p= − �

V⇢�

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Archimedes

a density di↵erence

Writing ⇢s the solid density of the body, its weight is

�→

W = �

V ⇢s

�→_{g dV} and the weight + the pressure force is

�→_R _{= �→}_W _{+ �→}_F

A = (⇢s− ⇢)V �→g

z FA

G

W

this�→R force may be positive or negative (the sign of the density di↵erence

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Archimedes

pressure center of an immersed body

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Archimedes

Example: how to avoid buoyancy

Consider a hollow sphere made of steel, outer radius R and wall width w . Find the width w for which the sphere does not sink nor float.

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Archimedes

Buoyancy of a partially immersed body

z FA G W V V₁ 2 B �→_F A = − � V1 ⇢1�→g dV − � V2 ⇢2�→g dV = −(⇢1V1+ ⇢2V2)�→g

Bthe buoyancy center B is the barycenter of the immersed volume V1 and

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Archimedes

Example: stability of a diaphragm wall

dry sand saturated sand h L l H w: concrete width

Find h for which the structure starts to uplift.

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Basic fluid mechanics for civil engineers

Lecture 3

Maxime Nicolas

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Lecture 3 outline

1 _{Eulerian and Lagrangian descriptions}

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Eulerian and Lagrangian descriptions

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Fluid particle

A fluid particle is a mesoscopic scale containing a very large number of fluid molecules, but much smaller than the macroscopic flow scale.

macro scale meso scale micro scale

O H H O H H fluid particle a ni ce vi ew of St oc khol m

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The travel of a fluid particle

Lagrange’s description of the path of a fluid particle: �→_r _{= �→r(�→r}₀_,t₎ O r(t )0 1 2 3 4 r(t ) r(t ) r(t ) r(t )

BUT TOO MANY FLUID PARTICLES TO FOLLOW except for diluted gas, sprays.

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The travel of a fluid particle

Eulerian description: the motion of the fluid is determined by a velocity field

�→_u _{= �→}_u_(�→r,t) with

�→_u ₌ d�→r dt Integration of �→u gives �→r (if needed)

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Steady flow

A steady flow is such �→u(�→r) only: no time dependence. Bsteady ≠ static !

photo A. Duchesne, MSC lab, Paris

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Flow example of the day

Consider the 2D steady flow

�→_u ₌U0 L � � � x −y 0 � � �

where U0 is a characteristic velocity, and L a characteristic lengths (both

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Flow example of the day

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Flow example of the day

iso-velocity lines (��→u� =constant)

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Acceleration

from t to t+ t, the particle moves from �→r to a new position �→r + �→r and

has a new velocity �→u + �→u

�→_r _{+ �→r = (x + x,y + y,z + z), �→}_u _{+ �→}_u _{= (u}_x_{+ u}_x_,u_y_{+ u}_y_,u_z_{+ u}_z₎ The acceleration (change of velocity) has two origins :

variation of velocity at the same location variation of velocity by a change of location

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Acceleration

since each velocity component is a 4 variables function ux = ux(x,y,z,t)

its total derivative is ux= @ux @x x+ @ux @y y+ @ux @z z+ @ux @t t+ . . . the same for uy and uz:

uy = @uy @x x+ @uy @y y+ @uy @z z+ @uy @t t+ . . . uz= @uz @x x+ @uz @y y+ @uz @z z+ @uz @t t+ . . .

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Acceleration

alternate writing: ux = x@ux @x + y @ux @y + z @ux @z + t @ux @t and the x-acceleration is

ux t = ux @ux @x + uy @ux @y + uz @ux @z + @ux @t or ux t = �ux @ @x + uy @ @y + uz @ @z + @ @t� ux

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particular derivative

the particular derivative is an operator with two terms: D

Dt =

@

@t + �→u ⋅ �→∇ and the particular acceleration is

D�→u Dt = @�→u @t + (�→u ⋅ �→∇)�→u �→_u _{⋅ �→}_{∇ = u}_x @ @x + uy @ @y + uz @ @z

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particular derivative for a steady flow

in the case of a steady flow �→u(�→r), the particular derivative reduces to D�→u

Dt = (�→u ⋅ �→∇)�→u

Plane flow : �→u = (ux(y,z),0,0), then

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Flow example of the day

�→_u ₌U0 L � � � x −y 0 � � � the acceleration is : �→_a ₌

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Flow example of the day

acceleration vector field (in blue)

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Streamlines

Def: In every point of the flow field, the tangent to a streamline is given by the velocity vector �→u .

a streamline is not the path of a single fluid particle

with �→dl a curve element, and �→u the fluid velocity, �→dl and �→u must be

colinear : _�→ dl × �→u = 0 so dx ux = dy uy = dz uz

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Flow example of the day

�→_u ₌U0 L � � � x −y 0 � � � the streamline equation is

dx ux = dy uy so dx x = − dy y

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Flow example of the day

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Stream tubes

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Flow rate

The volume flow rate:

dQ = �→u ⋅ �→n dS

this is the volume of fluid crossing dS during a unit time. dS

u

Integration over a surface give the flow rate

Q= �

SdQ= �S

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Example of a river flow rate

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syringe

piston U1 S U2 S 1 2 flowrate is Q= U1S1= U2S2

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Mass conservation

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Mass conservation

Mass conservation equation

the mass variation in a reference volume is due to the flow (in/out) through the surface of this volume:

@ @t �V ⇢dV = − �S⇢� →_u _{⋅ �→}_{n dS} using Ostrogradski, �_V @⇢_@tdV = − � V �→_{∇ ⋅ (⇢�→}_u_)dV then �_V�@⇢_@t + �→∇ ⋅ (⇢�→u)� dV = 0

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Mass conservation

local mass conservation:

@⇢

@t + �→∇ ⋅ (⇢�→u) = 0

since _�→

∇ ⋅ (⇢�→u) = �→u ⋅ �→∇⇢ + ⇢�→∇ ⋅ �→u , the mass conservation equation is

@⇢

@t + �→u ⋅ �→∇⇢ + ⇢�→∇ ⋅ �→u = 0 or

D⇢

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Mass conservation

Mass conservation for a incompressible flow

Ball fluids are compressible

air = 6.610−5 Pa−1, water = 4.610−10 Pa−1

but the flow may be incompressible = no significant variation of ⇢ during the flow.

A flow is seen as incompressible when

the characteristic velocity of the flow is much lower than the sound velocity: V � csound.

cair = 340 m ⋅ s−1 cwater = 1 500 m ⋅ s−1

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Mass conservation

Mass conservation for a incompressible flow

if ⇢ is a constant during the flow, @⇢

@t = 0 and �→∇⇢ = 0,

and the mass conservation equation reduces to �→_{∇ ⋅ �→}_u _{= 0}

(105)

Mass conservation

Flow example of the day

Mass conservation for

�→_u ₌U0 L � � � x −y 0 � � � �→_{∇ ⋅ �→}_u ₌ U0 L (1 − 1) = 0

(106)

Mass conservation

(107)

Mass conservation

WS2 preparation

Three hydrostatics problems: pressure force on a dam

uplift of an empty swimming pool tunnel A B C 80 m 60 m 100 m 30 m 5 m 2 m

(108)

Basic fluid mechanics for civil engineers

Lecture 4

Maxime Nicolas

(109)

Lecture 4 outline: conservation equations

1 _{A general transport law}

2 _{Mass conservation equation}

3 _{Momentum conservation equation}

(110)

Introduction

AXIOMS CONCERNING LAWS OF

MOTION, in Principia Mathematica (1687)

Mutationem motus proportionalem esse vi motrici impressae, & fieri secundum lineam rectam qua vis illa imprimitur.

(111)

Introduction

Newton’s second law from Principia Mathematica (1687) The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Modern formulation:

F = d

dt(mv)

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Introduction

Newton’s second law for a rigid body: m

W v

d

dt(m�→v) = �→W

How to transpose this law to a fluid particle (infinitesimal volume)?

V S

d

(113)

A general transport law

(114)

control volume

V(t )₀ V(t )₁ V(t )₂

the mass inside the control volume is constant

(115)

control volume

V(t )0 V(t )1 V(t )2

We aim to calculate the variation of f during the transport: d

dt �V(t)f(�→r,t)dV

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A general transport law for a scalar

The variation of f in V has two terms:

the local variation of f (at a fixed location) the flux of f through S, the surface of V

d

dt �V f dV = �V

@f

@t dV + �Sf �

→_u _{⋅ �→}_{n dS} using the divergence theorem,

d

dt �Vf dV = �V �

@f

@t + �→∇ ⋅ [f �→u]� dV known as Reynolds’s theorem.

(117)

Vector transport law

With a vector �→A = (Ax,Ay,Az) transported by the flow-field �→u , each

component (scalar) follows d dt �VAidV = �V� @Ai @t + �→∇ ⋅ [Ai�→u]� dV It follows that d dt �V �→_{A dV} _{= �} V @�→A @t dV + �S �→_A_(�→_u _{⋅ �→}_n_{) dS}

(118)

Mass conservation equation

(119)

Mass conservation as a transport law

Taking f = ⇢, we write

d

dt �V⇢ dV = �V �

@⇢

@t + �→∇ ⋅ [⇢�→u]� dV The mass conservation is

d dt �V ⇢ dV = 0 this implies @⇢ @t + �→∇ ⋅ [⇢�→u] = 0 or D⇢ Dt + ⇢�→∇ ⋅ �→u = 0

(120)

Reminder

Mass conservation equation from lecture 3: D⇢

Dt + ⇢�→∇ ⋅ �→u = 0

For an steady incompressible flow:

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Momentum conservation equation

(122)

Transport of ⇢�

→

u

With the momentum density �→A = ⇢�→u

d dt �V ⇢� →_{u dV} _{= �} V @ @t(⇢�→u) dV + �S⇢� →_u_(�→_u _{⋅ �→}_n_{) dS} We know that @ @t(⇢�→u) = ⇢ @�→u @t + �→u @⇢ @t and the divergence theorem gives

�_S⇢�→u(�→u ⋅ �→n) dS = �

V

�→_{∇ ⋅ (⇢�→}_u _{⊗ �→}_u_)dV

STOP NEW CONCEPT!

�→_u _{⊗ �→}_{u is a rank 2 tensor (= a matrix)} ⊗ means a tensor product

(123)

Maths: tensor product

For two 3-components vectors �→u and �→v : � � � ux uy uz � � �⊗ � � � vx vy vz � � �= � � � uxvx uxvy uxvz uyvx uyvy uyvz uzvx uzvy uzvz � � � and for our need

�→_u _{⊗ �→}_u ₌�_� � u2x uxuy uxuz uyux uy2 uyuz uzux uzuy uz2 � � � Note that �→u ⊗ �→u is symmetric.

(124)

Maths: divergence of a tensor

Let A be a rank 2 tensor. Its divergence is �→_{∇ ⋅ A =}�_�� →_{∇ ⋅ �→}_A x �→_{∇ ⋅ �→}_A y �→_{∇ ⋅ �→}_A z � ��

with �→Ax = (Axx,Axy,Axz) the x-line of A.

(125)

Maths: divergence of �

→

u

⊗ �→

u

Write on the board:

�→_{∇ ⋅ (�→}_u _{⊗ �→}_u_{) = . . .}

�→_{∇ ⋅ (�→}_u _{⊗ �→}_u_{) = �→}_u_(�→_{∇ ⋅ �→}_u_{) + (�→}_u _{⋅ �→}_∇)�→_u and therefore

(126)

back to the ⇢�

→

u transport law

d dt �V ⇢� →_{u dV} _{= �} V @ @t(⇢�→u) dV + �S⇢� →_u_(�→_u _{⋅ �→}_n_{) dS} d dt �V⇢� →_{u dV} _{= �} V��⇢ @�→u @t + �→u @⇢ @t� + �→u(�→∇ ⋅ [⇢�→u]) + ⇢(�→u ⋅ �→∇)�→u� dV = �_V��→u �@⇢ @t + �→∇ ⋅ (⇢�→u)� +⇢ �@�→u @t + (�→u ⋅ �→∇)�→u�� dV

(127)

back to the ⇢�

→

u transport law

Since @⇢ @t + �→∇ ⋅ (⇢�→u) = 0 and @�→u @t + (�→u ⋅ �→∇)�→u = D�→u Dt , then d dt �V⇢� →_{u dV} _{= �} V ⇢ D�→u Dt dV

(128)

Newton’s law for a fluid

(129)

Newton’s law

What does Newton says: The variation of momentum is balanced by the sum of forces applying on the volume V bounded by a surface S.

d dt �V⇢� →_{u dV} _{= �} V ⇢ D�→u Dt dV = �_i �→_F i

Two kinds of forces: volume forces surface forces

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volume force

the weight is the only volume force for a dielectric and non-magnetic fluid.

�→_F

v = �→W = � V ⇢�

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surface forces

We write the total surface forces as the sum of local forces applying on the

surface S: _�→

Fs= �

S

�→_T_(M,�→_n_)dS

where �→T(M,�→n) is a stress vector for all M ∈ S, for a unit vector �→n on

each element of S. _�→

T = �→n

STOP 2-tensor× vector (see next slide)

and using (again) the divergence theorem �_S�→T(M,�→n)dS = �

S

�→_{n dS}_{= �}

V

(132)

Maths: product of a 2-tensor with a vector

We need to calculate �→_n � � � xx xy xz yx yy yz zx zy zz � � � � � � nx ny nz � � �= � � � xxnx+ xyny+ xznz yxnx+ yyny+ yznz zxnx+ zyny+ zznz � � � which is a vector

(133)

Newton’s law

�_V⇢D�→u Dt dV = �→W + �→Fs �_V ⇢D�→u Dt dV = �V ⇢� → g dV + � S �→_T_(M,�→_n_)dS �_V ⇢D�→u Dt dV = �V ⇢� → g dV + � V �→_{∇ ⋅ dV} and locally, ⇢D�→u Dt = ⇢�→g + �→∇ ⋅

For each component (i = x,y,z),

⇢Dui = ⇢gi+

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Lecture abstract

Mass conservation equation: D⇢

Dt + ⇢�→∇ ⋅ �→u = 0 Momentum conservation equation:

⇢D�→u Dt = ⇢�→g + �→∇ ⋅ remember that D Dt ≡ @ @t + �→u ⋅ �→∇

(135)

The fluid stress tensor

The fluid stress tensor gathers all the information about surface forces: = −pI + D

pressure and shear I is the unity tensor

the tensor D has the information about the rheology of the fluid. F U h fixed boundary moving plate velocity profile u(y) D=F_S, a function of ˙ = U_h

(136)

Basic rheology

shear stress D, as a function of the shear rate ˙ D= f ( ˙ ) D she ar st re ss slope=viscosity

inviscid fluid: idealized fluid of zero viscosity newt onian fl uid 𝛾. 𝜂 Bingha m fluid shear-thinning shear -thicke ning

(137)

Basic fluid mechanics for civil engineers

Lecture 5

Maxime Nicolas

(138)

Lecture 5 outline: Inviscid flows

1 _{Flash test}

2 _{Flows classification}

(139)

Flash test

(140)

Flash test

Flash test rules

5 questions 15 min to answer work for yourself

(141)

Flash test

Flash test: 5 questions

1 what is the Archimede’s force of a 1 m3 sphere of concrete under

water?

2 calculate the relative pressure at a 2 meters depth under fresh

concrete 3 calculate �∇f with f = (y − z)�x→ 2 4 �→_u ₌ U0 L � � � 2x 2y −4z � � � is this an incompressible flow?

5 calculate D�→u�Dt with

�→_u ₌ gt

(142)

Flows classification

Last lecture abstract

⇢D�→u Dt = ⇢�→g + �→∇ ⋅ remember that D Dt ≡ @ @t + �→u ⋅ �→∇

(143)

The fluid stress tensor

The fluid stress tensor gathers all the information about surface forces: = −pI + D

pressure and shear I is the unity tensor

the tensor D has the information about the rheology of the fluid. F U h fixed boundary moving plate velocity profile u(y) D=F_S, a function of ˙ = U_h

(144)

introducing a useful dimensionless number

Momentum conservation equation:

⇢@�→u @t + ⇢(�→u ⋅ �→∇)�→u = ⇢�→g − �→∇p + �→∇ ⋅ D Inertia term: �⇢(�→u ⋅ �→∇)�→u� ∝ ⇢ �U1 L� U = ⇢ U2 L rheology (viscous) term:

��→∇ ⋅ D� ∝ ⌘_LU₂ To compare : �⇢(�→u ⋅ �→∇)�→u� ��→∇ ⋅ D� = ⇢UL ⌘

(145)

The Reynolds number

this dimensionless number is named the Reynolds number(symbol: Re)

[1883 by Osborne Reynolds] Re= ⇢UL ⌘ , 0< Re < ∞ with ⇢ fluid density U characteristic velocity L characteristic length

⌘ fluid dynamic viscosity

(146)

Flows classification: Re

� 1

Flow is dominated by viscous (stress) e↵ects (low velocity or small size flow or large viscosity

(147)

Flows classification: Re

� 1

Flow is dominated by inertia e↵ects (high velocity or large size or low viscosity).

(148)

The fluid stress tensor for an inviscid flow

Assume Re � 1

In this particular case (no D), the stress tensor reduces to = −pI

and

�→_{∇ ⋅ = −�→}_∇p only pressure gradient remains

(149)

Conservation eq. for inviscid flows

under the assumptions of inviscid, steady an incompressible flow, the conservation equations are

�→_{∇ ⋅ �→}_u _{= 0} and

(150)

Bernoulli’s theorem

(151)

rewriting the momentum cons. eq.

With a little chunk of maths, we write (�→u ⋅ �→∇)�→u = �→∇ �u

2

2 � + (�→∇ × �→u) × �→u so that the momentum conservation equation is now

�→_{∇ �}u2 2 � + (�→∇ × �→u) × �→u = �→g − 1 ⇢ �→_∇p = −�→∇(gz) − 1 ⇢ �→_∇p

(152)

along a streamline

Following a streamline, with �→dl a small oriented line element of the streamline �→_{∇ �}u2 2 � ⋅ �→dl + [(�→∇ × �→u) × �→u] ⋅ �→dl = �→g ⋅ �→dl − 1 ⇢��→∇p� ⋅ �→dl u u dl ∇ u ∇ u ∇ u ∇ ( ) u Since �→u��→dl , [(�→∇ × �→u) × �→u] ⋅ �→dl = 0 �→_{∇ �}u2 2 � ⋅ �→dl = �→g ⋅ �→dl − 1 ⇢��→∇p� ⋅ �→dl

(153)

along a streamline

�→_{∇ �}u2 2 + gz + p ⇢� ⋅ �→dl = 0 meaning that u2 2 + gz + p ⇢ = C ′ or ⇢u 2 2 + ⇢gz + p = C

(154)

Bernoulli’s theorem

Under the assumptions of inviscid flow steady flow incompressible flow the quantity p+ ⇢u 2 2 + ⇢gz = C

is constant along a streamline.

(155)

Understanding Bernoulli

Along a streamline, the energy density (energy per unit volume) is conserved.

Multiplying by a volume V transported along the streamline: 1 2⇢Vu 2_{+ pV + ⇢Vgz = CV} or 1 2mu 2_{+ pV + mgz = CV}

(156)

Example: emptying a water tank

A B h g st re am li ne

(157)

Example: emptying a water tank

A B h g st re am li ne output velocity: uB = � 2g h known as the Toricelli’s formula (1608–1647)

(158)

Basic fluid mechanics for civil engineers

Lecture 6

Maxime Nicolas

(159)

Lecture 6 outline: the Navier-Stokes equation

1 _{The Navier-Stokes equation}

2 _{Known solutions of steady NS}

(160)

Flash test stats

Bonus stats: Bonus number 0.0 1 0.2 8 0.4 11 0.6 19 0.8 13 1.0 2 Mean bonus is 0.55

(161)

Last lecture abstract

⇢D�→u

Dt = ⇢�→g + �→∇ ⋅ with

(162)

The Navier-Stokes equation

(163)

Newtonian fluids

The Navier-Stokes equation is the momentum conservation equation for 3D newtonian fluids:

linearity between shear stress and shear rate D she ar st re ss slope=viscosity

inviscid fluid: idealized fluid of zero viscosity

newt onian fl uid 𝛾. 𝜂 Bingha m fluid shear-thinning shear -thicke ning

(164)

tensor D

for an incompressible flow

D= 2⌘E with Eij =1 2� @ui @xj + @uj @xi� example : Exy = 1 2� @ux @y + @uy @x �

Note: the E tensor will be presented extensively in Elasticity class (6th semester).

(165)

the divergence of D

We write D D= ⌘ � �� @ux @x + @ux @x @ux @y + @uy @x @ux @z + @uz @x @ux @y + @uy @x @uy @y + @uy @y @uy @z + @uz @y @ux @z + @uz @x @uy @z + @uz @y @uz @z + @uz @z � �� and (�→∇ ⋅ D)x= ⌘(. . .) which is (�→∇ ⋅ D)x = ⌘ ux

and finally the divergence of D is

(166)

The Navier-Stokes equation

Now we write the Navier-Stokes equation for the incompressible flow of a newtonian fluid: ⇢D�→u Dt = ⇢�→g − �→∇p + ⌘ �→u or ⇢@�→u @t + (�→u ⋅ �→∇)�→u = ⇢�→g − �→∇p + ⌘ �→u

(167)

What is needed to solve NS?

NS is a set of 3 partial di↵erential equations (PDEs) coupled with the mass conservation equation.

As any di↵erential equation, the complete solving needs: boundary conditions (BC) for velocity and/or stress boundary conditions for pressure

(168)

Velocity and stress continuity

The velocity must be continuous at an interface: u�_y₌₀₊= u�_y₌₀₋

The tangential stress must be continuous at an interface: ⌘1 @u @y�_y₌₀₊= ⌘2 @u @y�_y₌₀₋ fluid 1 fluid 2 y velocity and stress continuity

(169)

Example of BC

Example: �→u = �→Usolid at a solid non-deformable surface.

fluid solid y no-slip condition no-slip condition U_solid

(170)

Known solutions of steady NS

(171)

Steady NS analytical solutions

Main classification:

plane cylindrical

Boundary-driven Example 1 Example 4

Pressure-driven Example 2 Example 3

(172)

Example 1: plane boundary-driven flow

Flow: �→u = ux(y), no pressure gradient

(173)

Plane boundary-driven flow

Solution:

ux(y) = Uy

h, p= constant

fixed plate moving plate _U

(174)

Example 2: plane pressure-driven flow

Flow: �→u = ux(y), constant pressure gradient along x: @p�@x = K

(175)

Plane pressure-driven flow

Solution: ux(y) = 1 2⌘ dp dx(y − h)y fixed plate fixed plate p p-+ pressure gradient

(176)

Example3: cylindrical pressure-driven flow

Flow: �→u = uz(r), constant pressure gradient dp�dx = K

Boundary conditions: �→u = 0 at r = R p p-+ pressure gradient r z u=0 on the boundary

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Cylindrical pressure-driven flow

We need to write: (�→u ⋅ �→∇)�→u = � �� ur@u_@rr +u_r✓@u_@✓r + uz@u_@zr −u 2 ✓ r ur@u_@r✓ +u_r✓@u_@✓✓ + uz@u_@z✓ +ur_ru✓ ur@u_@rz +u_r✓@u_@✓z + uz@u_@zz � �� = � � � 0 0 0 � � � �→_{∇p =}�_�� @p @r 1 r @p @✓ @p @z � �� = � � � 0 0 @p @z � � � �→_u ₌�_�� @2_u r @r2 +_r12@ 2_u r @✓2 +@ 2_u r @z2 +1_r @u_@rr −_r22@u_@✓✓ −u_r2r @2u✓ @r2 +_r12@ 2_u ✓ @✓2 +@ 2_u ✓ @z2 +1_r @u_@r✓ +_r22@u_@✓r −u_r✓2 @2uz @r2 +_r12@ 2_u z @✓2 +@ 2_u z @z2 +1_r@u_@rz � �� = � � � 0 0 1 r @ @r �r @uz @r � � � �

(178)

Cylindrical pressure-driven flow

Solution: uz(r) = − 1 4⌘ dp dz(R 2_{− r}2₎ p p-+ pressure gradient r z u=0 on the boundary

(179)

Cylindrical pressure-driven flow

uz(r) = − 1

4⌘ dp

dz(R

2_{− r}2₎

flow-rate through the pipe:

q = � sudS= 2⇡ � R 0 uz(r)r dr = − ⇡ 8⌘ dp dzR 4 mean velocity: ¯ u= q ⇡R2 = − 1 8⌘ dp dzR 2

(180)

stresses on the wall

Because of the non-slip condition on the wall, the fluid exerts a stress on

the wall. The local shear stress at r = R is

= �⌘@uz

@r �r=R

Then the total viscous force on a pipe of length L is Fv = � L 0 dz2⇡R dz= 2⇡LR × R 2 dp dz = ⇡R 2_Ldp dz

(181)

Example 4: Cylindrical BC flow

Axisymmetric Couette flow between 2 coaxial cylinders:

R₂ R₁ fluid Ω

(182)

Example 4: Cylindrical BC flow

velocity field:

�→_u _{= (u}_r_,u_✓_,u_z₎ Mass conservation eq.:

�→_{∇ ⋅ �→}_u _{= 0 =} @ur @r + 1 r @u✓ @✓ + @uz @z �→_u _{= (0,u}_✓_(r),0)

(183)

Example 4: Cylindrical BC flow

NS: 0 = −@p @r 0 = −1 r @p @✓ + ⌘ @ @r � 1 r @(ru✓) @r � ⇢g = −@p @z with BC: u✓(R1) = ⌦R1 and u✓(R2) = 0

(184)

Example 4: Cylindrical BC flow

solution: u✓(r) = ⌦R 2 1 R2 2− R12 �R22− r2 r �

(185)

CFD

(186)

CFD

beyond the analytical solutions

when no analytical solution is available, Computational Fluid Dynamics (CFD) helps a lot!

cell grid

BC

fluid domain is discretized on grid NS is solved on each grid cell

continuity of velocity, stress and pressure must be checked BC and IC

(187)

CFD

CFD basic principle

Equations are discretized on a grid. Example for a 1D-domain (hydrostatics)

0= −⇢g −dp dz dp dz ≈ p(z + dz) − p(z) dz = pi+1− pi dz pi+1= pi− ⇢gdz BC : p= patm at z = 0 p(−dz) = patm+ ⇢gdz

(188)

CFD

CFD softwares

home-made codes open-source codes commercial softwares Autodesk CFD ComSol Fluent StarCCM+ . . .

(189)

CFD

Example of CFD result

Fire simulation

(190)

Basic fluid mechanics for civil engineers

Lecture 7

Maxime Nicolas

(191)

Lecture 7 outline: the Stokes equation

1 _{The Stokes equation}

2 _{Properties of the Stokes equation}

3 _{Drag force on a sphere}

(192)

The Stokes equation

(193)

The Stokes equation

From Navier-Stokes to the Stokes equation

Navier-Stokes: ⇢@�→u @t + (�→u ⋅ �→∇)�→u = ⇢�→g − �→∇p + ⌘ �→u Reynolds number: Re= �⇢(�→u ⋅ �→∇)�→u� ��→∇ ⋅ D� = � ⇢(�→u ⋅ �→∇)�→u� �⌘ �→u� = ⇢UL ⌘ Hypothesis:

very low Reynolds numbers Re→ 0

steady flow: @�→u�@t = 0

Within this frame , the NS equation reduces to

(194)

The Stokes equation

From NS to the Stokes equation

Writing the pressure as

p = p′− ⇢gz gives

�→_∇p′_{= ⌘ �→}_u

named the Stokes equation.

George Gabriel Stokes (1819–1903), English physicist

(195)

Properties of the Stokes equation

(196)

Properties of the Stokes equation

The stokes equation �∇p→ ′= ⌘ �→u has 4 interesting properties:

1 Unicity of the solution

2 Linearity

3 Reversibility

(197)

Unicity of the solution

Assume that the BC are known (either at infinite or at finite distance from an interface).

If(�→u1,p′1) is a solution and (�→u2,p′2) is another solution, it can be proved

that

(�→u1,p1′) = (�→u2,p′2)

(198)

Linearity

Suppose two solutions of the Stokes equation: (�→u1,p1′) for BC1

(�→u2,p2′) for BC2

The flow

( 1�→u1+ 2�→u2 , 1p1′ + 2p′2)

(199)

Reversibility

No time variable in the Stokes eq.

the flow is instantaneous: no delay between the driving BC or driving force and the flow

the flow is reversible

Experimental evidence of the reversibility:

A movie featuring G.I. Taylor illustrating low-Reynolds number flows www.youtube.com/watch?v=QcBpDVzBPMk

(200)

Reynolds numbers for swimming

Let’s calculate a few Re numbers:

animal speed (m/s) size (m) ⌘ (Pa.s) Re

mako shark 14 4 10−3 107

human 2.5 2 10−3 106

goldfish 1.5 0.05 10−3 104

E. Coli 4× 10−5 3× 10−6 10−3 10−4