Laboratoire de Mécanique des Fluides et d’Acoustique LMFA UMR CNRS 5509

ECL - UE-FLE - September 2016

Fluid dynamics and energy Christophe Bailly, Julian Scott, Mikhael Gorokhovski & Lionel Soulhac

Ecole Centrale de Lyon & LMFA - UMR CNRS 5509

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x Course organization q Outline of the course General introduction Chapter 1 : Kinematic properties and fundamental laws (2h) Chapter 2 : Newtonian viscous fluid flow (2h) Chapter 3 : Dimensional analysis - Reynolds number (2h) Chapter 4 : Regimes and flow structures as a function of the Reynolds number (2h) Chapter 5 : Turbulent flows (2h) Chapter 6 : Vorticity and basis of aerodynamics (3h) Chapter 7 : Energy, thermodynamics and compressible flows (3h) Chapter 8 : Heat transfer (2.5h) Chapter 9 : Mixing of fluids (2.5h) Chapter 10 : Combustion and flame (3h)

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2 - Newtonian viscous fluid flow

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2 - Newtonian viscous fluid flow Viscous flow

Equations of fluid dynamics

Newtonian law Deformation of a fluid particle Hydrostatic pressure and Archimedes boyancy force Plane Couette flow Kinetic energy and viscous dissipation Integral form Viscous dissipation Dynamic viscosity Force derived from potential

General case Navier-Stokes equations (homogeneous incompressible flow) Gravitation Boundary conditions Inviscid model Euler’s equations of motion Bernouilli’s theorem Energetics of continuous-flow system Mechanical energy budget Energy head and energy loss

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x Viscous flow q Newtonian law Equation for the conservation of momentum, see Chapter 1, Eq. (4) ρ

DU = ∇ · σ + ρg Dt

where σ is the stress tensor (internal forces inside the fluid) Fluid at rest (hydrostatics) or inviscid fluid flow, σ = − pI (only pressure) In general, σ = − pI + τ where τ is the viscous stress tensor DU ρ = −∇ p + ∇ · τ + ρg Dt

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  −∇ p pressure force ∇·τ

viscous force exerted on the fluid particle

∂( pδij ) ∂τij ∂σij ∂p ∂τij =− + =− + ∇·σ i = ∂x j ∂x j ∂x j ∂xi ∂x j

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x Viscous flow q Newtonian law Constitutive relation for common fluids such as air or water τ = 2µD + λ(∇ · U ) I µ dynamic (shear) viscosity λ second viscosity

)

fluid properties

D deformation tensor (or velocity strain tensor) Dij =

1 2



∂Ui ∂Uj + ∂x j ∂xi



This relation reflects the local and instantaneous link between the viscous stress and the deformation tensor (velocity gradients) Dij is directly associated with the deformation of fluid particles The fluid will be assumed to be a Newtonian fluid for the remainder of this course.

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x Viscous flow q Deformation of a fluid particle Taylor series for the velocity inside the fluid particle xP (at a given time t), Ui ( x) = Ui ( xP ) + 1 ∂Ui = ∂x j 2 |





1 ∂Ui ∂Uj + + ∂x j ∂xi 2 {z } | Dij



∂Ui ( x j − xP j ) + · · · ∂x j



∂Ui ∂Uj = − ∂x j ∂xi {z } ωij

ωij is associated with rotation of the fluid particle

(

D symmetric part of ∇U ω antisymmetric part of ∇U



0 − Ω3 Ω2   ω =  Ω3 0 − Ω1  0 − Ω2 Ω1

U ( x) ≃ U ( xP ) + D · ( x − xP ) + ω · ( x − xP ) | {z } | {z } | {z } translation

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deformation



rotation at Ω

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Ω=

1 ∇×U 2

Neither translation nor rotation deform the fluid particle : deformations are only induced by the (deformation) tensor D

x Viscous flow q Deformation in shear flow, U1 = Sx2 and U2 = U3 = 0 













0 1 0 0 1 0 0 1 0 ∂Ui S S       = S  0 0 0  =  1 0 0  +  −1 0 0  ∂x j 2 2 0 0 0 0 0 0 0 0 0 {z } | {z } | ωij Dij

S Ω=  2



0  0 −1

x2

t + δt

t O

xP

xP translation

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xP

→

deformation

ECL UE FLE - September 2016 - cb1

xP

→

rotation

x1

x Viscous flow q Hydrostatic pressure and Archimedes boyancy force (An illustration of a fluid at rest) Immersed body

n

FA

Fluid at rest U = 0, σ = − pI Conservation of momentum, −∇ p + ρg = 0 ∂p ∂p ∂p = = 0 and = ρg p = p( x3 ) ∂x1 ∂x2 ∂x3

xA

g = ge3

D S body

When the density is constant, p = p0 + ρgx3 For water, ρ ≃ 103 kg.m−3. The pressure increases about one atmosphere for every 10 meters of water depth. Elementary force exerted by the fluid on the body, σ · ndS = − pndS FA =

72

Z

S

− pndS =

Z

D

−∇ pdV = −

Z

D

ρgdV = − g

ECL UE FLE - September 2016 - cb1

Z

D

ρdV = − gM f

x Viscous flow q Hydrostatic pressure and Archimedes boyancy force (cont’d) Archimedes principle : the upward buoyant force FA that is exerted on a body immersed in a fluid is equal to the weight of the fluid M f that the body displaces. Moment applied to the body C=

Z

S

x × (− pn)dS = −

Z

D

x × ∇ pdV = −

Z

D

ρxdV × g

By choosing the center of mass x A of the displaced fluid, as origin of the coordinate system M f xA ≡

Z

D

ρxdV ,

Z

D

ρ( x − x A )dV = 0 and thus, C = 0

The boyancy force FA applies to the (fictif) center of mass x A of the displaced fluid (the so-called center of buoyancy of the immersed body).

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x Viscous flow q Hydrostatic pressure and Archimedes boyancy force (cont’d) Stability of an immersed submarine

Center of mass of the displaced fluid Oscillations (rolling)

Z

Dim

ρ( x − x A )dv = 0

Center of gravity of the body Capsizing ! The center of gravity xG • (F = Ms g) of the body must be below the point x A • (Archimedes force FA ) 74

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Z

D

ρs ( x − xG )dv = 0

x Viscous flow q Archimedes of Syracuse (Greek, 287-212 avant J.-C.) The birth of hydrostatics

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x Viscous flow q Archimedes’ screw (worm screw)

EPR de Flamanville (EDF, mars 2013)

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x Viscous flow q Plane Couette flow (An illustration of a fluid in motion) Ue

Ue constant velocity of the upper plane

n τw U

h

τw force exerted by the moving plane on the fluid

x2 x1





0 Ue /h 0 ∂U   i =0 0 0 ∇U ij = ∂x j 0 0 0

U = (U1 ( x2 ), 0, 0) with U1 = Ue x2 /h







0 Ue /(2h) 0   Dij =  Ue /(2h) 0 0 0 0 0 

− p µUe /h 0   stress tensor σij = − pδij + τij =  µUe /h − p 0  −p 0 0 77

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Ue τ12 = µ = cst h

x Viscous flow q Plane Couette flow (cont’d) Force F exerted by the upper wall on the fluid, with n = (0, 1, 0) the unit outward normal vector 









− p µUe /h 0 0 µUe /h      F = σ · n =  µUe /h − p 0   1  =  −p  0 0 −p 0 0 Rotation of the fluid particle   0 Ue /(2h) 0   ωij =  −Ue /(2h) 0 0 0 0 0

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



0   Ω= 0  −Ue /(2h)

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τw =

µUe h

x Kinetic energy and viscous dissipation q Integral form of the kinetic energy equation    2
D U DU U· ρ =⇒ ρ = ∇ · σ + ρg = U · ∇ · σ + ρg Dt Dt 2 Reynolds theorem (3) with χ = U 2 /2 d dt

Z


U2 ρ dV = U · ∇ · σ + ρg dV + 2 D |D {z } (a) Z

Z

U2 ρ (W − U ) · ndS 2 S

U · (∇ · σ ) = ∇ · (U · σ ) − σ : ∇U = ∇ · (U · σ ) − σ : D

(a) =

79

 Z Z   U · (σ · n) dS + ρU · g dV  Pext ≡ S

   Pint ≡ −

Z

external forces

D

D

σ : D dV

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internal forces

x Kinetic energy and viscous dissipation q Short break to examine notations ! Scalar product between two tensors (inner product) σ:D=

t

∑ ∑σij Dij = tr(σ D ) i

j

Vectorial identity U · (∇ · σ ) = ∇ · (U · σ ) − σ : D ∂σij ∂ ∂ 1 ∂Ui 1 ∂Ui ∂Ui Ui = (Ui σij ) − σij = (Ui σij ) − σij − σji ∂x j ∂x j ∂x j ∂x j 2 ∂x j 2 ∂x j ∂ (Ui σij ) − σij Dij (i ↔ j, σij = σji ) = ∂x j

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x Kinetic energy and viscous dissipation q Integral form of the kinetic energy equation (cont’d) d dt

Z

U2 ρ dV = Pext + Pint + 2 D

Z

U2 ρ (W − U ) · n d S 2 S

(6)

Let us consider a materiel domain D : the external power ... is split into the increase of kinetic energy in D and the energy required to deform the fluid particles in D Z U2 d ρ dV − Pint Pext = dt D 2 The internal work has two components :

− Pint =

Z

σ : DdV = − D |

Z

p∇ · U dV D {z }

compression

+ |

Z

τ : D dV D {z }

viscous dissipation > 0

The first term is associated with compression of the fluid particle : this term is positive or negative (acts like a ressort) ; null for an incompressible flow. 81

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x Kinetic energy and viscous dissipation q Viscous dissipation The second term, the viscous dissipation, is associated with the degradation of mechanical energy into heat energy by friction. τ = 2µD + λ(∇ · U ) I

(Newtonian fluid)

It is straightforward to show that τ : D = 2µD : D + λ(∇ · U )2

(left it as an exercise)

Therefore, for an incompressible flow, τ : D = 2µD : D =

82

µ 2



∂Ui ∂Uj + ∂x j ∂xi

ECL UE FLE - September 2016 - cb1

2

>0

x Kinetic energy and viscous dissipation q Dynamic viscosity µ

µ(T ) × 105 (SI)

6

Sutherland’s law (1893)  3/2 T T0 + Ts µ ≃ µ0 T0 T + Ts

5 4

Air T0 = 273 K Ts = 111 K µ0 = 1.716 × 10−5 kg/(m.s)

3 2 1 100

White (1988)

T (K)

1000

2000

For common gases, µ = µ( T ) but the kinematic viscosity ν = µ/ρ = ν( p, T ) For air at T = 20o C and p = 1 bar, one has ν = 1.5 × 10−5 m2.s−1 (ν = 10−6 m2.s−1 for water) 83

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x Kinetic energy and viscous dissipation q Force derived from a potential This is the case of the gravity force† g = −∇Ψ( x) gi = − gδ3i Ψ = gx3 (= − g · x) DΨ U · g = −U · ∇ Ψ = − Dt

g x3 O

Kinetic energy balance (see slide 79)   2 D U +Ψ = U·∇·σ ρ Dt 2

x2 x1

(7)

and the corresponding integral form S Pext ≡

d dt

Z  D

2

ρ



Z

S

U · (σ · n) dS

U S + Ψ dV = Pext + Pint + 2 †

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Z  S

2

ρ



U + Ψ (W − U ) · n d S 2

(8)

existence of such a function Ψ requires that ∇ × g = 0

ECL UE FLE - September 2016 - cb1

x Equations of fluid dynamics q General case   ∂ρ   + ∇ · (ρU ) = 0    ∂t   DU ρ = −∇ p + ∇ · τ + ρg   Dt       τ = 2µD + λ(∇ · U ) I

conservation of mass conservation of momentum constitutive relation : Newtonian fluid

System of differential equations for ρ( x, t) and U ( x, t), not closed for the pressure p( x, t) Assumption of incompressible flow : ∇ · U = 0 Conservation of energy (+ Equation of State) in the general case We will assume an incompressible and homogeneous fluid flow up to Chapter 7. 85

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x Equations of fluid dynamics q Incompressible and homogeneous fluid flow Incompressibility condition : ∇ · U = 0 Homogeneous flow : ρ = cst and µ = cst

=⇒ the conservation of mass is thus satisfied. The viscous stress tensor is given by τ = 2µD, and its divergence (that is the viscous force in the momentum equation) may be written as ∇ · τ = µ∇2U. We thus obtained the well-known Navier-Stokes equations (ν = µ/ρ is the kinematic viscosity)  DU 1   = − ∇ p + ν ∇2U + g  Dt ρ    ∇·U = 0 86

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(9)

x Equations of fluid dynamics q Navier & Stokes

Stokes, G. G., 1845, On the theory of the internal friction of fluids in motion, and of the equilibrium and motion of elastic solids, Transactions of the Cambridge Philosophical Society, 8, 287-305.

Navier, C.L.M.H., 1823, Mémoire sur les lois du mouvement des fluides (lu à l’Académie le 18 mars 1822), Mémoires de l’Académie Royal des Sciences de l’Institut de France, 389-440. Navier (1785-1836) 87

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x Equations of fluid dynamics q Incompressible and homogeneous fluid flow Force derived from potential, such as gravitation g = −∇Ψ( x) 1 DU = − ∇ p⋆ + ν ∇2U Dt ρ

p⋆ = p + ρΨ

The new variable p⋆ contains gravitation effects (volume force no longer appears in NS Eqs). From the hydrostatic balance, one has p⋆ = p0 , which leads to p( x) = p0 − ρΨ( x) If the boundary conditions do not involve stresses (e.g. for instance surface gravity waves), the gravity force does not affect the velocity field U ( x, t), but only the pressure field. The term −ρΨ( x) is then a simple hydrostatic correction of the pressure field.

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x Equations of fluid dynamics q Boundary conditions The flow is determined by solving the Navier-Stokes equations (9), associated with appropriate boundary conditions. Solid wall : no-slip condition Ufluid = Uwall

e.g. plane Couette flow

Interface between two fluids A et B (without mass transfer and surface tension) : no-slip condition, UA = UB continuity of efforts σ A · n = σ B · n impermeability of the interface, W · n = U · n

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A B

n W

x Equations of fluid dynamics q Formulation of the boundary value problem : flow around a body Navier-Stokes equations (with p⋆ = p + ρΨ)  DU 1   = − ∇ p⋆ + ν ∇2U  Dt ρ    ∇·U = 0

U∞

Boundary conditions : U = 0 on the surface of the body, and U → U∞ far from the body. The fluid is characterized by ν and ρ The geometry of the body and the free stream velocity U∞ are involved in the boundary conditions. Gravitation effect appears as a simple correction of the pressure field (p = p0 + ρg · x) After a transient, an established or developed flow is obtained : the initial conditions are forgot. This flow is however not necessary steady (flow fields independent of time) 90

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x Inviscid model q Euler’s equations of motion Every fluid is viscous, but viscous stresses can be neglected in some cases (discussed later in the course). An inviscid model is then often considered with σ = − pI (corresponding to an ideal fluid) The fluid flow is then governed by the Euler equations (1757) (incompressible homogeneous flow)  DU 1   = − ∇p + g  Dt ρ    ∇·U = 0

For a frictionless fluid, the flow has a nonzero tangential velocity : slip boundary condition at the wall, U · n = 0 (impermeability) For an interface between two fluids (see previous Section), U A · n = UB · n 91

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x Inviscid model q Leonhard Euler (1707-1783) Leonhard Euler is, after Isaac Newton (1643-1727), the founder of analytical mechanics (two-body problem for instance), and of fluid dynamics – thanks to the differential calculus also introduced by Gottfried Leibniz (1646-1716).

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x Inviscid model q Bernouilli’s theorem Under the following assumptions 1. inviscid flow (σ = − pI) 2. steady flow (∂t = 0) 3. incompressible flow (∇ · U = 0) 4. forces deriving from potential (g = −∇Ψ here) Conservation of the kinetic energy (Eq. 8)    2  2 D U U =0 + Ψ = U · ∇ · σ =⇒ U · ∇ p + ρΨ + ρ ρ Dt 2 2 The Bernouilli theorem is then derived U2 H ≡ p + ρΨ + ρ = cst along a streamline 2

(10)

but may be different for different streamlines, and also cst for a fluid particle (streamlines are also pathlines). Eq. (10) can be interpreted as an equation for mechanical energy (work done by pressure is balanced by the change in kinetic energy) 93

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x Inviscid model q The Bernoulli family

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x Energetics of continuous-flow system q Mechanical energy budget

D control domain, with S = S1 ∪ S2 ∪ S p

Pm

S p impermeable (but possibly moving) surface, S1 and S2 are fixed surfaces

D

S1

Sp

Pm mechanical power provided to the fluid flow, shaft work (done by a machine)

S2

Sp

Assumptions : incompressible flow, force derived from potential (gravity) and viscous stresses negligible at S1 and S2 (will be justified later) Conservation of mass (incompressible fluid flow) Z

∇ · U dV = D

Z

U · n dS = S

Z

S1 ∪S2

Qv1 = Qv2 = Qv

95

U · n dS = Qv2 − Qv1 = 0

(m3.s−1 )

ECL UE FLE - September 2016 - cb1

x Energetics of continuous-flow system q Mechanical energy budget From Eq. (8),   Z  Z  2 2 d U U S ρ + Ψ dV = Pext + Ψ (W − U ) · n d S + Pint + ρ dt D 2 2 S By examining each term,  Z Z  S  P = U · (σ · n) dS = | P pU · n dS  m } − ext {z   S S1 ∪S2  through S p  Z

Z

  τ : D dV = −Dν < 0 p∇ · U dV − Pint =    | D {z | D {z } }   incompressibility

Z  S

96

2

ρ



viscous dissipation

U + Ψ (W − U ) · n d S = − 2

Z

S1 ∪S2



2

ρ



U + Ψ U · n dS 2

ECL UE FLE - September 2016 - cb1

x Energetics of continuous-flow system q Mechanical energy budget d dt

Z  D



2

U ρ + Ψ dV = Pm − Dν − 2

Z

HU · n dS } | S1 ∪S2 {z = Φ2 − Φ1

where H = p + ρU 2 /2 + ρΨ is the energy per unit volume. The term Φ2 − Φ1 is the energy flux between S2 (outlet) and S1 (inlet). By taking the time average of the previous equation, ¯ | P {zm }

mean power provided to the fluid flow

¯2−Φ ¯ = | Φ {z 1 } + increase of energy flux between inlet/outlet

¯ν D | {z }

viscous dissipation

¯1−Φ ¯ 2 = D¯ ν ≥ 0 With no shaft work, that is P¯m = 0, one has Φ loss of energy flux induced by viscous dissipation inside the fluid domain D .

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x Energetics of continuous-flow system q Energy head and energy (head) loss One-dimensional model of the system (commonly use) by introducing the energy head H defined as, Z HU · n dS ¯ Φ H = ¯ = ZS Qv U · n dS S

H is an average of the local quantity H in time, and over the cross section dS weighted by the flow rate. The conservation of mechanical energy may be recast in the following compact form Pm + Q¯ v ( H1 − H2 ) = D¯ ν

shaft work

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  pump, fan blade, ...     Q¯ v ( H2 − H1 ) = ( Pm − D¯ ν ) = ηPm

  turbine, ...    Pturbine = − Pm = Q¯ v ( H1 − H2 ) − D¯ ν = η Q¯ v ( H1 − H2 ) ECL UE FLE - September 2016 - cb1

x Appendix q Poussée d’Archimède (problème !) Ω = Ωez eau

?

balle ping-pong

Note that the free surface is parabolic −ρΩ2rer = −∇ p + ρg  ∂p  2  −ρΩ r = −    ∂r    1 ∂p 0=−  r ∂θ      ∂p   0 = − − ρg ∂z

ρΩ2 2 p(r, z) = r − ρgz + cst 2

Dauxois & Raynal (2005)

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x Appendix q Poussée d’Archimède (version 1) Repère (er , eθ , ez )

Ω = Ωez

Position balle, x0 = r0 er + h0 ez Accélération, Ω × (Ω × x0 ) = −Ω2r0 er

eau, ρe r0 h0

fA T

balle ping-pong (m, V )

Équilibre de la balle, −mΩ2r0 er = mg + T + f A Force d’Archimède f A , 0 = f A + ρe V g

Au final, T = (ρe V − m) g − mΩ2r0 er On a donc T · er = −mΩ2r0 < 0, la balle s’est donc écartée de l’axe central ... ... mais ce n’est pas ce que l’on observe !

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x Appendix q Poussée d’Archimède (version 2) Repère (er , eθ , ez )

Ω = Ωez

Position balle, x0 = r0 er + h0 ez Accélération, Ω × (Ω × x0 ) = −Ω2r0 er

eau, ρe r0 h0

T

balle ping-pong (m, V )

Équilibre de la balle, −mΩ2r0 er = mg + T + f A Force d’Archimède f A , − ρ e V Ω2 r 0 er = f A + ρ e V g

Au final, T = (ρe V − m)( g + Ω2r0 er ) Pour m < ρe V (masse de la balle plus légère que le volume d’eau déplacé, ok avec balle de pong-pong !), on a T · er = (ρe V − m)Ω2r0 er > 0 La balle s’est donc rapprochée de l’axe central ... et c’est ce que l’on observe expérimentallement. 101

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