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 dx 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 + −→ (αv) 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 αv −→ Ψ Dt ∂t · ∇ dt 3 ∂t − · ∇ d Vt R ∂α Integrating by parts, we have d3xα(x, t)= d3x Ψ + Ψ−→ (αv) 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 + −→ (ρv) =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 + −→ (ρv)=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−→ (ρv) 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)= ρv 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 + −→ (ρβv) −→ dt ∂t ∂t ∇ · Vt Vt Differentiating the product, we have −→ (ρβv)=β−→ (ρv)+ρv −→ 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 ρvdV = ρ−→F dV + σndS 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 σndS = −→ σdV ∇ · St Vt So finally, we get the Euler equations in Lagrangian form Dv 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 = σn vdS = σv ndS = −→ σv dV · · · ∇ · St St St Vt we deduce the Lagrangian derivative of the specific total energy D v2 ρ + = −→ σv + ρv −→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 + −→ Ev σv = ρv −→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 + −→ (ρv)=0 = −→ v ∂t ∇ · ρ Dt −∇ · ∂ρvi Dv + −→ (ρviv −→σi )=ρFi ρ = ρ−→F + −→ σ ∂t ∇ · − Dt ∇ · ∂E D + −→ Ev σv = ρv −→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 !).