continuum mechanics lecture 4 fluid dynamics - uzhteyssier/km_2013_lectures/cm_lecture4.pdf ·...
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Romain TeyssierContinuum Mechanics 30/04/2013
Continuum Mechanics
Lecture 4
Fluid dynamics
Prof. Romain Teyssier
http://www.itp.uzh.ch/~teyssier
Romain TeyssierContinuum Mechanics 30/04/2013
- 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.
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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)�
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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 −→
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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 �̇)
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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, �)
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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)
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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
Romain TeyssierContinuum Mechanics 30/04/2013
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