Continuum Mechanics Lecture 4 Fluid dynamics

Prof. Romain Teyssier http://www.itp.uzh.ch/~teyssier

Continuum Mechanics 30/04/2013 Romain Teyssier Outline

- Fluid kinematics - Mass and momentum conservation laws - The energy equation - Real fluids - Ideal fluids - Incompressible fluids

Continuum Mechanics 30/04/2013 Romain Teyssier What is a fluid ?

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.

Continuum Mechanics 30/04/2013 Romain Teyssier Fluid kinematics

For each particle in the system, the velocity is defined as ￿x(t +dt) ￿x(t) ￿v(t)= lim − dt 0 dt where the trajectory of the particle→ is ￿x ( t ) . We consider the mapping between the initial positions ￿y = ￿x(t = 0) and the final position, defined as ￿x ( t )= −→φ ( ￿y, t ). The velocity field is the vector field defined by d￿x 1 ￿v ( ￿x, t )= ( ￿y, t ) where ￿y = −→φ − (￿x, t) dt We can now obtain the trajectory directory from the velocity field as t ￿x(t)=￿x(0) + ￿v(￿x(u),u)du ￿0 Streamlines are also used to visualize the flow. dx dy dz They are defined as = = vx(￿x, t) vy(￿x, t) vz(￿x, t) Only for a stationary flow are streamlines and trajectories equal.

Continuum Mechanics 30/04/2013 Romain Teyssier Kinematics of a volume element

Consider a scalar field Ψ ( ￿x, t ) . Along a trajectory, we have Ψ x 0 ( t )= Ψ ( ￿x ( t ) ,t ) . dΨ ∂Ψ DΨ Using the chain rule, we have x 0 ( t )= + ￿v −→ Ψ = dt ∂t · ∇ Dt This is the Lagrangian derivative of the scalar function Ψ. We consider the scalar «color» function Ψ ( ￿x )=1 for ￿x V and Ψ(￿x)=0 ￿x/V ∈ t ∈ t Using this color function, we can show that for any scalar field α we have: d ∂α d3xα(￿x, t)= d3x + −→ (α￿v) dt ∂t ∇ · ￿Vt ￿Vt ￿ ￿ This is Reynold’s transport theorem. d d ∂Ψ ∂α Proof: d3xα(￿x, t)= d3xΨ(￿x, t)α(￿x, t)= d3x α + Ψ dt dt 3 3 ∂t ∂t ￿Vt ￿R ￿R ￿ ￿ DΨ ∂Ψ d ∂α We have = + ￿v −→ Ψ =0 so d3xα(￿x, t)= d3x Ψ α￿v −→ Ψ Dt ∂t · ∇ dt 3 ∂t − · ∇ d ￿Vt ￿R ∂α￿ ￿ Integrating by parts, we have d3xα(￿x, t)= d3x Ψ + Ψ−→ (α￿v) dt 3 ∂t ∇ · ￿Vt ￿R ￿ ￿ Continuum Mechanics 30/04/2013 Romain Teyssier Mass conservation

The total mass in the volume element is obtained using α(￿x, t)=ρ(￿x, t)

Since the total mass is conserved within the Lagrangian volume Vt, we have dM d ∂ρ = d3xρ(￿x, t)= d3x + −→ (ρ￿v) =0 dt dt ∂t ∇ · ￿Vt ￿Vt ￿ ￿ Since this equation is true for any Lagrangian volume, we have proven the continuity equation (or mass conservation in continuous form) ∂ρ + −→ (ρ￿v)=0 ∂t ∇ ·

We now consider the mass variation in a fixed Eulerian volume V0 dM d ∂ρ(￿x, t) 0 = d3xρ(￿x, t)= d3x = d3x−→ (ρ￿v) dt dt ∂t − ∇ · ￿V0 ￿V0 ￿V0 Using the divergence theorem, we have finally the equation for mass conservation in integral form d d3xρ(￿x, t)= ρ￿v ￿ndS dt − · ￿V0 ￿S0

Continuum Mechanics 30/04/2013 Romain Teyssier Lagrangian derivative

We now compute the variation of a scalar field per unit mass β(￿x, t) d ∂β ∂ρ α = ρβ dx3ρβ = dx3 ρ + β + −→ (ρβ￿v) −→ dt ∂t ∂t ∇ · ￿Vt ￿Vt ￿ ￿ Differentiating the product, we have −→ (ρβ￿v)=β−→ (ρ￿v)+ρ￿v −→ β ∇ · ∇ · · ∇ and using the continuity equation, we have d ∂β Dβ dx3ρβ = dx3ρ + ￿v −→ β = dx3ρ dt ∂t · ∇ Dt ￿Vt ￿Vt ￿ ￿ ￿Vt

We have used the Lagrangian derivative of the scalar field defined by Dβ ∂β = + ￿v −→ β Dt ∂t · ∇ 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 e = ρ￿ The fluid velocity is equivalent to a specific momentum.

Continuum Mechanics 30/04/2013 Romain Teyssier Variation of the volume

3 We now used the scalar field α ( ￿x, t )=1 . The total volume is trivially Vt = dx V dV ￿ t From Reynold’s transport theorem, we have: t = dx3−→ ￿v dt ∇ · ￿Vt

3 We define the specific volume V =1 / ρ so that Vt = dx ρV ￿Vt DV 1 DV Using the Lagrangian derivative, we have ρ = = −→ ￿v Dt V Dt ∇ · The velocity divergence is equal to the rate of change of the specific volume.

Equivalently, the Lagrangian derivative of the density writes 1 Dρ = −→ ￿v ρ Dt −∇ · 1 Dρ 1 DV An easy check: we have ρ V =1 so we obtain = = −→ ￿v ρ Dt −V Dt −∇ ·

Continuum Mechanics 30/04/2013 Romain Teyssier Momentum conservation

We use the dynamical equilibrium equation we derived in the previous lectures d ρ￿vdV = ρ−→F dV + σ￿ndS dt Vt Vt St ￿ ￿ ￿ d Dv From the definition of the Lagrangian derivative, we get ρv dV = ρ i dV dt i Dt ￿Vt ￿Vt Using the divergence theorem, we get σ￿ndS = −→ σdV ∇ · ￿St ￿Vt So finally, we get the Euler equations in Lagrangian form D￿v ρ = ρ−→F + −→ σ Dt ∇ · ∂v It writes in Eulerian form: ρ i + ￿v −→ v = ρF + −→ σ ∂t · ∇ i i ∇ · −→ i ￿ ￿ Using the continuity equation, we derive the conservative form for the momentum ∂ρv i + −→ (ρv ￿v σ )=ρF ∂t ∇ · i − −→ i i Momentum conservation in integral form in a fixed Eulerian volume: d dx3ρv + ρv ￿v ￿ndS σ ￿ndS = dx3ρF dt i i · − −→ i · i ￿V0 ￿S0 ￿S0 ￿V0 Continuum Mechanics 30/04/2013 Romain Teyssier The internal energy equation

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

dE = δQ + Tr σδ￿ dV ￿V For the small displacement field δ￿ u = ￿v d t , we ￿define￿ the rate of strain tensor 1 ∂v ∂v ￿˙ = i + j ij 2 ∂x ∂x ￿ j i ￿ The Lagrangian variation of the total internal energy is dE D￿ Dq = dx3ρ = dx3ρ + dx3Tr(σ ￿˙) dt Dt Dt ￿Vt ￿Vt ￿Vt We derive the internal energy equation in continuous form D￿ Dq ρ = ρ +Tr(σ ￿˙) Dt Dt In case there is no external heat source or sink (adiabatic system), we have D￿ ρ =Tr(σ ￿˙) Dt Warning: do not get confused between the specific energy and the strain tensor !

Continuum Mechanics 30/04/2013 Romain Teyssier The total energy equation dE dK δW δQ From the kinetic energy theorem, we have + = ext + dt dt dt dt δW The work per unit time for external forces is ext = dx3ρ−→F ￿v + −→T ￿vdS dt · · ￿Vt ￿St Expressing the work of the stress field as a volume integral (see previous lectures)

−→T ￿vdS = σ￿n ￿vdS = σ￿v ￿ndS = −→ σ￿v dV · · · ∇ · ￿St ￿St ￿St ￿Vt we deduce the Lagrangian derivative of the specific total energy￿ ￿ D v2 ρ ￿ + = −→ σ￿v + ρ￿v −→F Dt 2 ∇ · · ￿ ￿ ￿ ￿ v2 Define the total energy per unit volume as E = ρ ￿ + 2 ￿ ￿ We have the total energy conservation law in continuous form ∂E + −→ E￿v σ￿v = ρ￿v −→F ∂t ∇ · − · ￿ ￿ Exercise: derive the total energy conservation in integral form in a fixed Eulerian volume.

Continuum Mechanics 30/04/2013 Romain Teyssier Fluid dynamics equations

Equations in Eulerian (conservative) form Equations in Lagrangian form

∂ρ 1 Dρ + −→ (ρ￿v)=0 = −→ ￿v ∂t ∇ · ρ Dt −∇ · ∂ρvi D￿v + −→ (ρvi￿v −→σi )=ρFi ρ = ρ−→F + −→ σ ∂t ∇ · − Dt ∇ · ∂E D￿ + −→ E￿v σ￿v = ρ￿v −→F ρ =Tr(σ ￿˙) ∂t ∇ · − · Dt ￿ ￿ To close the previous systems, we need the material law of the fluid

σ = fonction (ρ, ￿, ∂vi/∂xj,...)

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

σ = p1+τ with in general Tr(τ)=0 − The pressure is given by the Equation-of-State (EoS) and depends on 2 thermodynamical quantities p = p ( ρ , ￿ ) or p = p(ρ,T)

Continuum Mechanics 30/04/2013 Romain Teyssier Equation-of-State for Aluminum

The SESAME library (LANL 1992)

Continuum Mechanics 30/04/2013 Romain Teyssier Mie-Grüneisen EoS for real gases and fluids

A general form for real fluid EoS: P P (ρ)=Γρ (￿ ￿ (ρ)) − c − c

The main constituents are the cold pressure P c ( ρ ) and the cold energy ￿ c ( ρ ) .

Both cold curves should satisfy the thermodynamical consistency condition:

Pc(ρ) ￿￿ (ρ)= c ρ2

P Pc(ρ) The Mie-Grüneisen sound speed is cs = Pc￿(ρ)+(Γ + 1) − ￿ ρ

Simple examples: 1- Ideal gas EoS: P =(γ 1)ρ￿ −

ρ γ 2- Polytropic EoS: P = P 0 ρ ￿ 0 ￿

2 3- Isothermal EoS: P = ρc0

Continuum Mechanics 30/04/2013 Romain Teyssier Real fluids and gases: linear viscosity

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

∂vi ∂vj τij = λ −→ ￿v δij + µ + ∇ · ∂xj ∂xi ￿ ￿ ￿ ￿ 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): ∂v ∂v 2 τ = ξ(−→ ￿v)δ + η i + j (−→ ￿v)δ ij ∇ · ij ∂x ∂x − 3 ∇ · ij ￿ j i ￿ 2 The coefficient ξ = λ + µ is called the bulk or volumetric viscosity coefficient. 3 The coefficient η = µ 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 η T 5 / 2 . ∝ The exact value of the coefficients can be derived using kinetic theory.

Continuum Mechanics 30/04/2013 Romain Teyssier 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

Continuum Mechanics 30/04/2013 Romain Teyssier Compressible ideal fluids

The fluids dynamics equation without viscosity apply for ideal fluids.

1 Dρ = −→ ￿v ρ Dt −∇ · D￿v ρ = ρ−→F −→ P Dt − ∇ D￿ ρ = P (−→ ￿v) Dt − ∇ ·

where we used the relation −→ (P 1) = −→ P ∇ · ∇

∂ρ In conservative form, we have + −→ (ρ￿v)=0 ∂t ∇ · ∂ρ￿v + −→ (ρ￿v ￿v)+−→ P = ρ−→F ∂t ∇ · ⊗ ∇ ∂E + −→ [(E + P )￿v]=ρ￿u −→F ∂t ∇ · ·

Continuum Mechanics 30/04/2013 Romain Teyssier Incompressible ideal fluids

Dρ For incompressible fluids, we have =0 or equivalently −→ ￿v =0 . Dt ∇ · Note that, in general, we don’t have ρ=constant (multiple fluids) but in practice, in a single fluid we have ρ=ρ0. D￿v 1 We are left with only one equation, the Euler equation = −→F −→ P Dt − ρ0 ∇ with a constraint given by −→ ￿v =0 in the volume and the boundary condition on ∇ · the outer surface ￿ v ￿n =0 . ·

Note that we don’t need to use the Equation of State. The pressure follows from the zero divergence constraint. ∂￿v 1 Using the definition of the Lagrangian derivative, we have + ￿v −→ ￿v = −→F −→ P ∂t · ∇ − ρ0 ∇ We use the Helmholtz decomposition of w = −→F ￿v −→ ￿v = −→ −→A + −→ φ −→ − · ∇ ∇× ∇ The scalar field Φ satisfies a Poisson equation ∆ φ = −→ w ￿ with BC −→ φ = w￿ ￿n ∂￿v ∇ · ∇ · Using −→ =0 and the unicity of the Helmholtz decomposition, we have ∇ · ∂t ∂￿v = ￿v −→ ￿v + −→F −→ Φ ∂t − · ∇ − ∇

Continuum Mechanics 30/04/2013 Romain Teyssier Incompressible real fluids

Using the linear viscosity law in the velocity equation, ∂v ∂v 2 τ = ξ(−→ ￿v)δ + η i + j (−→ ￿v)δ ij ∇ · ij ∂x ∂x − 3 ∇ · ij ￿ j i ￿ we obtain the Navier-Stoke equations

Dvi 1 ∂ ρ = ρFi −→ P +(ξ + η) (−→ ￿v)+η∆vi Dt − ∇ 3 ∂xi ∇ · For incompressible fluids, we have the simpler form

Dv ρ i = ρF −→ P + η∆v Dt i − ∇ i

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

Continuum Mechanics 30/04/2013 Romain Teyssier