Continuum Mechanics Lecture 4 Fluid dynamics

Transcription

1 Continuum Mechanics Lecture 4 Flui ynamics Prof.

2 Outline - Flui kinematics - Mass an momentum conservation laws - The energy equation - Real fluis - Ieal fluis - Incompressible fluis

3 In thermoynamics or kinetic theory, a flui is a collection of atoms or molecules, in liqui or gaseous form. In continuum mechanics, a flui is a system that flows. The central property is the flui velocity. What is a flui? In soli mechanics, we have stuie various equilibrium solutions, for which the stress was relate to the strain (static eformation): the elastic regime. Above a given threshol (the yiel strength), the soli enters the plastic regime an eventually breaks. For a flui, the eformation can be arbitrarily large. The stress is relate to the rate of strain, not the strain. Note that the bounary between soli an flui is fuzzy: waves, tooth paste, glaciers... It epens on the exact time scale an length scale we are intereste in. Example: on short time scale, a glacier is a soli, but on longer time scale (months), it behaves like a flui. A flui will be escribe as a continuum if length scales are much larger than the atomic or molecular mean free path.

4 Flui kinematics For each particle in the system, the velocity is efine as x(t +t) x(t) v(t) = lim t 0 t where the trajectory of the particle is x(t). We consier the mapping between the initial positions an the final position, efine as x(t) = φ (y, t). y = x(t = 0) The velocity fiel is the vector fiel efine by v(x, t) = x where t (y, t) y = φ 1 (x, t) We can now obtain the trajectory irectory from the velocity fiel as x(t) =x(0) + Streamlines are also use to visualize the flow. x They are efine as v x (x, t) = y v y (x, t) = z v z (x, t) Only for a stationary flow are streamlines an trajectories equal. t 0 v(x(u),u)u

5 Kinematics of a volume element Consier a scalar fiel Ψ(x, t). Along a trajectory, we have Ψ x0 (t) =Ψ(x(t),t). Ψ Using the chain rule, we have x0 (t) = Ψ t t + v Ψ = DΨ Dt This is the Lagrangian erivative of the scalar function Ψ. We consier the scalar «color» function Ψ(x) =1 for x V t an Ψ(x) =0 Using this color function, we can show that for any scalar fiel α we have: α 3 xα(x, t) = 3 x t V t V t t + (αv) x/ V t This is Reynol s transport theorem. 3 xα(x, t) = Ψ 3 xψ(x, t)α(x, t) = 3 x t V t t R 3 R t α + α Proof: t Ψ 3 DΨ Dt = Ψ t + v Ψ =0 3 xα(x, t) = 3 x Ψ α t V t R t αv We have so Ψ 3 3 xα(x, t) = 3 x Ψ α Integrating by parts, we have t V t R t + Ψ (αv) 3

6 Mass conservation The total mass in the volume element is obtaine using α(x, t) =ρ(x, t) Since the total mass is conserve within the Lagrangian volume Vt, we have M = ρ 3 xρ(x, t) = 3 x t t V t V t t + (ρv) =0 Since this equation is true for any Lagrangian volume, we have proven the continuity equation (or mass conservation in continuous form) ρ t + (ρv) =0 We now consier the mass variation in a fixe Eulerian volume V0 M 0 = 3 xρ(x, t) = 3 ρ(x, t) x = 3 x (ρv) t t V 0 V 0 t V 0 Using the ivergence theorem, we have finally the equation for mass conservation in integral form 3 xρ(x, t) = ρv ns t V 0 S 0

7 Lagrangian erivative We now compute the variation of a scalar fiel per unit mass β(x, t) x 3 ρβ = x 3 ρ β t V t V t t + β ρ t + α = ρβ (ρβv) Differentiating the prouct, we have (ρβv) =β (ρv)+ρv β an using the continuity equation, we have β x 3 ρβ = x 3 ρ t V t V t t + v β = x 3 ρ Dβ V t Dt We have use the Lagrangian erivative of the scalar fiel efine by Dβ Dt = β t + v β Scalar fiel per unit mass are also calle specific quantities. For example, the internal energy per unit volume e an the internal energy per unit mass (or the specific energy) ε are relate by e = ρ The flui velocity is equivalent to a specific momentum.

8 We efine the specific volume Variation of the volume We now use the scalar fiel α(x, t) =1. The total volume is trivially V t = From Reynol s transport theorem, we have: V t t = x 3 v V t V =1/ρ so that V t = V t x 3 ρv V t x 3 Using the Lagrangian erivative, we have ρ DV Dt = 1 V DV Dt = v The velocity ivergence is equal to the rate of change of the specific volume. Equivalently, the Lagrangian erivative of the ensity writes An easy check: we have ρv =1 1 Dρ ρ Dt = v so we obtain 1 Dρ ρ Dt = 1 V DV Dt = v

9 Momentum conservation We use the ynamical equilibrium equation we erive in the previous lectures t ρvv = ρ F V + σns V t V t S t From the efinition of the Lagrangian erivative, we get ρv i V = t V t Using the ivergence theorem, we get σns = σv S t V t So finally, we get the Euler equations in Lagrangian form It writes in Eulerian form: ρ Dv Dt = ρ F + σ vi ρ t + v v i = ρf i + σ i V t ρ Dv i Dt V Using the continuity equation, we erive the conservative form for the momentum ρv i + (ρv i v σ i )=ρf i t Momentum conservation in integral form in a fixe Eulerian volume: x 3 ρv i + ρv i v ns σi ns = x 3 ρf i t V 0 S 0 S 0 V 0

10 The internal energy equation From the first principle of thermoynamics, we have erive in the previous lecture E = δq + Tr σδ V For the small isplacement fiel δu = vt, we efine the rate of strain tensor ij = 1 vi + v j 2 x j x i The Lagrangian variation of the total E t = x 3 ρ D internal energy V t Dt = x 3 ρ Dq is V t Dt + x 3 Tr(σ ) V t We erive the internal energy equation in continuous form ρ D Dt = ρdq Dt ρ D Dt V +Tr(σ ) In case there is no external heat source or sink (aiabatic system), we have =Tr(σ ) Warning: o not get confuse between the specific energy an the strain tensor!

11 The total energy equation E t + K t = δw ext From the kinetic energy theorem, we have + δq t t δw ext = x 3 ρ The work per unit time for external forces is F v + T vs t V t S t Expressing the work of the stress fiel as a volume integral (see previous lectures) T vs = σn vs = σv ns = σv V S t S t S t V t we euce the Lagrangian erivative of the specific total energy ρ D + v2 = σv + ρv F Dt 2 Define the total energy per unit volume as E = ρ + v2 2 We have the total energy conservation law in continuous form E t + Ev σv = ρv F Exercise: erive the total energy conservation in integral form in a fixe Eulerian volume.

12 Flui ynamics equations Equations in Eulerian (conservative) form ρ t + (ρv) =0 ρv i t + (ρv i v σ i )=ρf i E t + Ev σv = ρv F Equations in Lagrangian form 1 Dρ ρ Dt = v ρ Dv Dt = ρ F + σ ρ D Dt =Tr(σ ) To close the previous systems, we nee the material law of the flui σ = fonction (ρ,, v i / x j,...) In general, we efine the pressure an the viscous tensor as σ = p1+τ with in general Tr(τ) =0 The pressure is given by the Equation-of-State (EoS) an epens on 2 thermoynamical quantities p = p(ρ, ) or p = p(ρ,t)

13 Equation-of-State for Aluminum The SESAME library (LANL 1992)

14 Mie-Grüneisen EoS for real gases an fluis A general form for real flui EoS: P P c (ρ) =Γρ ( c (ρ)) The main constituents are the col pressure P c (ρ) an the col energy c (ρ). Both col curves shoul satisfy the thermoynamical consistency conition: The Mie-Grüneisen soun spee is c(ρ) = P c(ρ) ρ 2 c s = P c(ρ)+(γ + 1) P P c(ρ) ρ Simple examples: 1- Ieal gas EoS: 2- Polytropic EoS: P =(γ 1)ρ P = P 0 ρ ρ 0 γ 3- Isothermal EoS: P = ρc 2 0

15 Following the methoology of linear, isotropic an thermoelastic material, we consier a moel where the stress tensor epens linearly on the rate of strain. vi τ ij = λ v δ ij + µ + v j x j x i where λ an µ are the equivalent of the Lamé coefficients (not the same units!). The usual viscosity law is given using the following form (Newtonian fluis): τ ij = ξ( vi v)δ ij + η + v j 2 x j x i 3 ( v)δ ij The coefficient The coefficient Real fluis an gases: linear viscosity ξ = λ µ η = µ is calle the bulk or volumetric viscosity coefficient. is calle the ynamical viscosity coefficient. Usually, the bulk viscosity is zero, or much smaller than the thermal pressure. The ynamical viscosity epens on the flui/gas temperature. For example, for a Coulomb plasma, we have η T 5/2. The exact value of the coefficients can be erive using kinetic theory.

16 Dynamical viscosity coefficients Liquis 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 Vali for T=300 K an P=1 atm

17 Compressible ieal fluis The fluis ynamics equation without viscosity apply for ieal fluis. 1 Dρ ρ Dt = v ρ Dv Dt = ρ F P ρ D Dt = P ( v) where we use the relation (P 1) = P In conservative form, we have ρ t + (ρv) =0 ρv t + (ρv v)+ P = ρ F E t + [(E + P )v] =ρu F

18 Dρ For incompressible fluis, we have or equivalently. Dt =0 v =0 Note that, in general, we on t have ρ=constant (multiple fluis) but in practice, in a single flui we have ρ=ρ0. Dv We are left with only one equation, the Euler equation Dt = F 1 P ρ 0 with a constraint given by v =0 in the volume an the bounary conition on the outer surface v n =0. Incompressible ieal fluis Note that we on t nee to use the Equation of State. The pressure follows from the zero ivergence constraint. v t + v v = F 1 Using the efinition of the Lagrangian erivative, we have P ρ 0 We use the Helmholtz ecomposition of w = F v v = A + φ The scalar fiel Φ satisfies a Poisson equation φ = w with BC φ = w n v Using an the unicity of the Helmholtz ecomposition, we have t =0 v t = v v + F Φ

19 Incompressible real fluis Using the linear viscosity law in the velocity equation, τ ij = ξ( vi v)δ ij + η + v j 2 x j x i 3 ( v)δ ij we obtain the Navier-Stoke equations ρ Dv i Dt = ρf i P +(ξ η) x i ( v)+η v i For incompressible fluis, we have the simpler form ρ Dv i Dt = ρf i P + η v i Together with the constraint v =0 an the BC on the outer surface v =0