Continuum Mechanics Lecture 4 Fluid dynamics - UZHteyssier/km_2013_lectures/cm_lecture4.pdf · Continuum Mechanics Lecture 4 Fluid dynamics ... In continuum mechanics, ... From the

Romain TeyssierContinuum Mechanics 30/04/2013

Continuum Mechanics

Lecture 4

Fluid dynamics

Prof. Romain Teyssier

http://www.itp.uzh.ch/~teyssier


- Fluid kinematics

- Mass and momentum conservation laws

- The energy equation

- Real fluids

- Ideal fluids

- Incompressible fluids

Outline

In thermodynamics or kinetic theory, a fluid is a collection of atoms or molecules, in liquid or gaseous form.

In continuum mechanics, a fluid is a system that flows.

The central property is the fluid velocity.

In solid mechanics, we have studied various equilibrium solutions, for which the stress was related to the strain (static deformation): the elastic regime. Above a given threshold (the yield strength), the solid enters the plastic regime and eventually breaks.

For a fluid, the deformation can be arbitrarily large. The stress is related to the rate of strain, not the strain.

Note that the boundary between solid and fluid is fuzzy: waves, tooth paste, glaciers... It depends on the exact time scale and length scale we are interested in. Example: on short time scale, a glacier is a solid, but on longer time scale (months), it behaves like a fluid.

A fluid will be described as a continuum if length scales are much larger than the atomic or molecular mean free path.


What is a fluid ?

Streamlines are also used to visualize the flow.

They are defined as

Only for a stationary flow are streamlines and trajectories equal.

For each particle in the system, the velocity is defined as

where the trajectory of the particle is .

We consider the mapping between the initial positions

and the final position, defined as .

The velocity field is the vector field defined by

where

We can now obtain the trajectory directory from the velocity field as

�v(t) = limdt→0

�x(t+ dt)− �x(t)

dt

�y = �x(t = 0)

�x(t) = �x(0) +

� t

0�v(�x(u), u)du


Fluid kinematics

�x(t)

�x(t) =−→φ (�y, t)

�v(�x, t) =d�x

dt(�y, t) �y =

−→φ −1(�x, t)

dx

vx(�x, t)=

dy

vy(�x, t)=

dz

vz(�x, t)

Consider a scalar field . Along a trajectory, we have .

Using the chain rule, we have

This is the Lagrangian derivative of the scalar function Ψ.

We consider the scalar «color» function for and

Using this color function, we can show that for any scalar field α we have:

This is Reynold’s transport theorem.

Ψ(�x, t) Ψx0(t) = Ψ(�x(t), t)

Ψ(�x) = 1 Ψ(�x) = 0�x ∈ Vt �x /∈ Vt

d

dt

�

Vt

d3xα(�x, t) =d

dt

�

R3

d3xΨ(�x, t)α(�x, t) =

�

R3

d3x

�∂Ψ

∂tα+

∂α

∂tΨ

�

DΨ

Dt=

∂Ψ

∂t+ �v ·−→∇Ψ = 0

d

dt

�

Vt

d3xα(�x, t) =

�

R3

d3x

�Ψ∂α

∂t− α�v ·−→∇Ψ

�

d

dt

�

Vt

d3xα(�x, t) =

�

R3

d3x

�Ψ∂α

∂t+Ψ

−→∇ · (α�v)�


Kinematics of a volume element

d

dt

�

Vt

d3xα(�x, t) =

�

Vt

d3x

�∂α

∂t+

−→∇ · (α�v)�

Proof:

dΨx0

dt(t) =

∂Ψ

∂t+ �v ·−→∇Ψ =

DΨ

Dt

We have so

Integrating by parts, we have

The total mass in the volume element is obtained using

Since the total mass is conserved within the Lagrangian volume Vt, we havedM

dt=

d

dt

�

Vt

d3xρ(�x, t) =

�

Vt

d3x

�∂ρ

∂t+

−→∇ · (ρ�v)�

= 0

∂ρ

∂t+−→∇ · (ρ�v) = 0

dM0

dt=

d

dt

�

V0

d3xρ(�x, t) =

�

V0

d3x∂ρ(�x, t)

∂t= −

�

V0

d3x−→∇ · (ρ�v)

d

dt

�

V0

d3xρ(�x, t) = −�

S0

ρ�v · �ndS


Mass conservation

α(�x, t) = ρ(�x, t)

Since this equation is true for any Lagrangian volume, we have proven the continuity equation (or mass conservation in continuous form)

We now consider the mass variation in a fixed Eulerian volume V0

Using the divergence theorem, we have finally the equation for mass conservation in integral form

Scalar field per unit mass are also called specific quantities.

For example, the internal energy per unit volume e and the internal energy per unit mass (or the specific energy) ε are related by

The fluid velocity is equivalent to a specific momentum.

We now compute the variation of a scalar field per unit mass d

dt

�

Vt

dx3ρβ =

�

Vt

dx3

�ρ∂β

∂t+ β

∂ρ

∂t+

−→∇ · (ρβ�v)�β(�x, t)

d

dt

�

Vt

dx3ρβ =

�

Vt

dx3ρ

�∂β

∂t+ �v ·−→∇β

�=

�

Vt

dx3ρDβ

Dt

Dβ

Dt=

∂β

∂t+ �v ·−→∇β

α = ρβ −→


Lagrangian derivative

Differentiating the product, we have−→∇ · (ρβ�v) = β

−→∇ · (ρ�v) + ρ�v ·−→∇β

and using the continuity equation, we have

We have used the Lagrangian derivative of the scalar field defined by

e = ρ�

We define the specific volume so that

Using the Lagrangian derivative, we have ρDV

Dt=

1

V

DV

Dt=

−→∇ · �v

1

ρ

Dρ

Dt= −−→∇ · �v

ρV = 1 1

ρ

Dρ

Dt= − 1

V

DV

Dt= −−→∇ · �v


Variation of the volume

We now used the scalar field .α(�x, t) = 1 The total volume is trivially Vt =

�

Vt

dx3

From Reynold’s transport theorem, we have: dVt

dt=

�

Vt

dx3−→∇ · �v

Vt =

�

Vt

dx3ρVV = 1/ρ

An easy check: we have so we obtain

Equivalently, the Lagrangian derivative of the density writes

The velocity divergence is equal to the rate of change of the specific volume.

ρD�v

Dt= ρ

−→F +

−→∇ · σ

ρ

�∂vi∂t

+ �v ·−→∇vi

�= ρFi +

−→∇ ·−→σi

∂ρvi∂t

+−→∇ · (ρvi�v −−→σi) = ρFi

d

dt

�

V0

dx3ρvi +

�

S0

ρvi�v · �ndS −�

S0

−→σi · �ndS =

�

V0

dx3ρFi

d

dt

�

Vt

ρ�vdV =

�

Vt

ρ−→F dV +

�

St

σ�ndS

d

dt

�

Vt

ρvidV =

�

Vt

ρDviDt

dV�

St

σ�ndS =

�

Vt

−→∇ · σdV


Momentum conservation

We use the dynamical equilibrium equation we derived in the previous lectures

From the definition of the Lagrangian derivative, we get

Using the divergence theorem, we get

So finally, we get the Euler equations in Lagrangian form

It writes in Eulerian form:

Using the continuity equation, we derive the conservative form for the momentum

Momentum conservation in integral form in a fixed Eulerian volume:

dE

dt=

�

Vt

dx3ρD�

Dt=

�

Vt

dx3ρDq

Dt+

�

Vt

dx3Tr(σ �̇)

�̇ij =1

2

�∂vi∂xj

+∂vj∂xi

�

ρD�

Dt= ρ

Dq

Dt+Tr(σ �̇)

ρD�

Dt= Tr(σ �̇)


The internal energy equation

From the first principle of thermodynamics, we have derived in the previous lecture

dE = δQ+

�

VTr

�σ δ�

�dV

For the small displacement field , we define the rate of strain tensor δ�u = �vdt

The Lagrangian variation of the total internal energy is

We derive the internal energy equation in continuous form

In case there is no external heat source or sink (adiabatic system), we have

Warning: do not get confused between the specific energy and the strain tensor !

dE

dt+

dK

dt=

δWext

dt+

δQ

dtδWext

dt=

�

Vt

dx3ρ−→F · �v +

�

St

−→T · �vdS

E = ρ

��+

v2

2

�

�

St

−→T · �vdS =

�

St

σ�n · �vdS =

�

St

σ�v · �ndS =

�

Vt

−→∇ ·�σ�v

�dV

ρD

Dt

��+

v2

2

�=

−→∇ ·�σ�v

�+ ρ�v ·−→F

∂E

∂t+

−→∇ ·�E�v − σ�v

�= ρ�v ·−→F


The total energy equation

From the kinetic energy theorem, we have

The work per unit time for external forces is

Expressing the work of the stress field as a volume integral (see previous lectures)

we deduce the Lagrangian derivative of the specific total energy

Define the total energy per unit volume as

We have the total energy conservation law in continuous form

Exercise: derive the total energy conservation in integral form in a fixed Eulerian volume.

with in general

The pressure is given by the Equation-of-State (EoS) and depends on 2 thermodynamical quantities or

σ = fonction (ρ, �, ∂vi/∂xj , ...)

Tr(τ) = 0σ = −p1 + τ

p = p(ρ, �)


Fluid dynamics equations

∂ρ

∂t+−→∇ · (ρ�v) = 0

∂ρvi∂t

+−→∇ · (ρvi�v −−→σi) = ρFi

1

ρ

Dρ

Dt= −−→∇ · �v

ρD�v

Dt= ρ

−→F +

−→∇ · σ

ρD�

Dt= Tr(σ �̇)

Equations in Eulerian (conservative) form Equations in Lagrangian form

To close the previous systems, we need the material law of the fluid

In general, we define the pressure and the viscous tensor as

p = p(ρ, T )

∂E

∂t+

−→∇ ·�E�v − σ�v

�= ρ�v ·−→F


Equation-of-State for Aluminum

The SESAME library (LANL 1992)

P − Pc(ρ) = Γρ (�− �c(ρ))

Pc(ρ) �c(ρ)

��c(ρ) =Pc(ρ)

ρ2

cs =

�

P �c(ρ) + (Γ+ 1)

P − Pc(ρ)

ρ

P = (γ − 1)ρ�

P = P0

�ρ

ρ0

�γ

P = ρc20


Mie-Grüneisen EoS for real gases and fluids

A general form for real fluid EoS:

The main constituents are the cold pressure and the cold energy .

Both cold curves should satisfy the thermodynamical consistency condition:

The Mie-Grüneisen sound speed is

Simple examples:

1- Ideal gas EoS:

2- Polytropic EoS:

3- Isothermal EoS:

Following the methodology of linear, isotropic and thermoelastic material, we consider a model where the stress tensor depends linearly on the rate of strain.

where λ and µ are the equivalent of the Lamé coefficients (not the same units !).

The usual viscosity law is given using the following form (Newtonian fluids):

τij = λ�−→∇ · �v

�δij + µ

�∂vi∂xj

+∂vj∂xi

�

τij = ξ(−→∇ · �v)δij + η

�∂vi∂xj

+∂vj∂xi

− 2

3(−→∇ · �v)δij

�

ξ = λ+2

3µ


Real fluids and gases: linear viscosity

The coefficient is called the bulk or volumetric viscosity coefficient.

The coefficient is called the dynamical viscosity coefficient.

Usually, the bulk viscosity is zero, or much smaller than the thermal pressure.

The dynamical viscosity depends on the fluid/gas temperature.

For example, for a Coulomb plasma, we have .

The exact value of the coefficients can be derived using kinetic theory.

η = µ

η ∝ T 5/2


Dynamical viscosity coefficients

Liquids η (Pa sec) Gases η (Pa sec)

Water 1,00E-03 H20 1,02E-05

Gasoline 3,00E-04 Dry air 1,80E-05

Mercury 1,60E-03 CO2 1,50E-05

Benzen 6,50E-04 Oxygen 2,00E-05

Kerosen 1,90E-03 CH4 1,30E-05

Oil 1,70E-01

Valid for T=300 K and P=1 atm

∂ρ

∂t+−→∇ · (ρ�v) = 0

∂ρ�v

∂t+−→∇ · (ρ�v ⊗ �v) +

−→∇P = ρ−→F

∂E

∂t+−→∇ · [(E + P )�v] = ρ�u ·−→F

1

ρ

Dρ

Dt= −−→∇ · �v

ρD�v

Dt= ρ

−→F −−→∇P

−→∇ · (P1) =−→∇P

ρD�

Dt= −P (

−→∇ · �v)


Compressible ideal fluids

The fluids dynamics equation without viscosity apply for ideal fluids.

where we used the relation

In conservative form, we have

We use the Helmholtz decomposition of

The scalar field Φ satisfies a Poisson equation with BC

Using and the unicity of the Helmholtz decomposition, we have

For incompressible fluids, we have or equivalently .

Note that, in general, we don’t have ρ=constant (multiple fluids) but in practice, in a single fluid we have ρ=ρ0.

We are left with only one equation, the Euler equation

with a constraint given by in the volume and the boundary condition on

the outer surface .

∂�v

∂t+ �v ·−→∇�v =

−→F − 1

ρ0

−→∇P

∆φ =−→∇ · �w

∂�v

∂t= −�v ·−→∇�v +

−→F −−→∇Φ


Incompressible ideal fluidsDρ

Dt= 0

−→∇ · �v = 0

D�v

Dt=

−→F − 1

ρ0

−→∇P

�v · �n = 0

−→∇ · �v = 0

The pressure follows from the zero divergence constraint.

Using the definition of the Lagrangian derivative, we have

−→∇φ = �w · �n

−→w =−→F − �v ·−→∇�v =

−→∇ ×−→A +

−→∇φ

−→∇ · ∂�v∂t

= 0

Note that we don’t need to use the Equation of State.

ρDviDt

= ρFi −−→∇P + (ξ +

1

3η)

∂

∂xi(−→∇ · �v) + η∆vi

ρDviDt

= ρFi −−→∇P + η∆vi

�v = 0


Incompressible real fluids

τij = ξ(−→∇ · �v)δij + η

�∂vi∂xj

+∂vj∂xi

− 2

3(−→∇ · �v)δij

�Using the linear viscosity law in the velocity equation,

we obtain the Navier-Stoke equations

For incompressible fluids, we have the simpler form

Together with the constraint and the BC on the outer surface −→∇ · �v = 0

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