Lecture 6

Transcription

Lecture 6

Continuum Mechanics
Lecture 6
Waves in Fluids
Prof. Romain Teyssier
http://www.itp.uzh.ch/~teyssier
Continuum Mechanics 07/05/2013
Romain Teyssier
Outline
- Sound waves
- Jeans instability
- Gravity waves
- Rayleigh-Taylor and Kelvin-Helmholtz instabilities
- Quasi-linear waves and shock formation
- Shock waves and Rankine-Hugoniot relations
Romain Teyssier
Sound waves
We use the fluid equation in one dimension without gravity source term.
In conservative form, they write
∂t ρ + ∂x (ρv) = 0
�
�
∂t (ρv) + ∂x ρv 2 + P = 0
We assume a barotropic EoS P = P (ρ)
Far from any discontinuities, we can also use the quasi-linear form:
∂t ρ + v∂x ρ + ρ∂x v = 0
1
∂t v + v∂x v + ∂x P = 0
ρ
We consider the reference equilibrium state ρ = ρ0 and v = v0 everywhere in space.
Waves are small disturbances of this equilibrium state
ρ(x, t) = ρ0 + δρ(x, t) with δρ � ρ0
v(x, t) = v0 + δv(x, t) with δv � v0 and c0
Romain Teyssier
Sound waves
We linearize the quasi-linear form, dropping high-order terms.
∂t (δρ) + v0 ∂x (δρ) + ρ0 ∂x (δv) = 0
c20
∂t (δv) + v0 ∂x (δv) + ∂x (δρ) = 0
ρ0
where we have used the definition of the sound speed c2 = P � (ρ)
We are looking for monochromatic planar wave solutions:
δρ = ∆ρ expi(kx−ωt)
δv = ∆v expi(kx−ωt)
We obtain the following linear system for the amplitudes:
(−iω + ikv0 )∆ρ + ikρ0 ∆v = 0
c20
ik ∆ρ + (−iω + ikv0 )∆v = 0
ρ0
In order to have a non vanishing solution, the determinant must be zero.
We obtain the dispersion relation for sound waves:
(ω − kv0 )2 − k 2 c20 = 0
ω
The velocities of sound waves are v = = v0 ± c0
k
Romain Teyssier
Riemann invariants and characteristics
The previous linear system of partial differential equations can be written as
∂t W + A∂x W = 0 where the vector of unknowns is W = (δρ, δv)T
�
�
v0 ρ0
The matrix A is given by A =
. The eigenvalues are λ± = v0 ± c0
c20
v0
ρ0
�
�
1
ρ
0
and the eigenvectors components are given by δα+ =
δρ + δv
2
c0
�
�
1
ρ
0
δα− =
δρ − δv
Since the matrix is diagonal in the eigenvector
2
c0
basis, we have :
∂t (δα+ ) + (v0 + c0 )∂x (α+ ) = 0
We define the characteristic curves
(different from the trajectories) as:
∂t (δα− ) + (v0 − c0 )∂x (α− ) = 0
dx ±
= (v0 ± c0 )
dt
δα± are conserved quantities along their corresponding characteristic
curves: they are called Riemann invariants.
Given the initial conditions at t=0, we can reconstruct the final solution by combining
Riemann invariants along each crossing characteristic (in this case straight lines).
Romain Teyssier
Self-gravitating fluids
The fluids equation in conservative form in presence of gravity write:
−
∂ρ →
+ ∇ · (ρ�v ) = 0
∂t
−
→
−
→
−
∂ρ�v →
+ ∇ · (ρ�v ⊗ �v ) + ∇P = ρ F
∂t
In a self-gravitating fluid, the gravitational potential follow the Poisson equation
→
−
→
−
∆Φ = 4πGρ with F = − ∇Φ
→
− →
−
ρ
=
∆Φ
=
−
∇
·
F
We drop the constant 4πG from now on:
�→
�
− →
−
→
− � →
− � �→
− →
−�
For each component, we have ρFx = − ∇ · F Fx = − ∇ · Fx F + F · ∇ Fx
2
We then use the relations ∂x Fx = −∂xx Φ = ∂x Fx
The tidal tensor
∂j Fi = −∂ij Φ
2
∂y Fx = −∂xy
Φ = ∂ x Fy
is symmetric
2
�→
∂z Fx = −∂xz
Φ = ∂ x Fz
− ��2
�
�F �
→
− � →
−� →
− →
−
→
− � →
−�
Finally, we have ρFx = − ∇ · Fx F + F ∂x F = − ∇ · Fx F + ∂x
2
�
�
−
→
− →
−
∂ρ�v →
F2
Momentum conservation:
+ ∇ · ρ�v ⊗ �v + P 1 + F ⊗ F −
1 =0
∂t
2
Romain Teyssier
Jeans instability
We consider an equilibrium state with ρ = ρ0 , Φ = 0 and v = 0
In this infinite medium, the Poisson equation has to be modified ∆Φ = 4πG (ρ − ρ0 )
The perturbed state satisfies ∆(δΦ) = 4πG(δρ)
→
−
The linearized continuity equation is ∂t (δρ) + ρ0 ∇ · (δ�v ) = 0
−
→
−
c20 →
∂t (δ�v ) +
∇(δρ) + ∇(δΦ) = 0
The Euler equation becomes
ρ0
Taking the partial time derivative of the continuity equation leads to
→
−
∂t2 (δρ) = −ρ0 ∇ · (∂t (δ�v )) = c20 ∆(δρ) + ρ0 ∆(δΦ) = c20 ∆(δρ) + 4πGρ0 (δρ)
We are looking for plane wave solution
δρ = ∆ρ expi(kx−ωt)
2 2
2
=
c
(k
−
k
)
We get the following dispersion relation ω 2 = c20 k 2 − 4πGρ
0
0
J
�
2π
4πGρ0
=
where we have introduce the Jeans length kJ =
λJ
c20
For short wavelength, k > kJ we have propagating waves with v ≤ c0
For long wavelength, k < kJ small perturbations grow exponentially fast.
ω2 < 0
so ω is purely imaginary and δρ ∝ exp±|ω|t → Instability !
Romain Teyssier
Gravity waves
Incompressible fluid dynamics in deep water under constant gravity.
air
z
z = η(x, t)
z=0
sea level
water
z = −H
x
ground
We consider the equilibrium state η(x, t) = 0 and �v (x, z, t) = 0
In air, we have p = p0 and in water, p = p0 − ρ0 gz .
→
−
�
v
(x,
z,
t)
=
∇φ we have in the volume:
Using the second Bernoulli theorem, using
2
v
p(x, z, t)
→
−
∂t φ(x, z, t) +
+ gz +
= C(t) and ∇ · �v = 0 −→ ∆φ = 0
2
ρ0
We add to this the boundary condition at the bottom
and the kinematic boundary condition at the top
vz (z = −H) = 0
∂t η(x, t) + vx ∂x η(x, t) = vz
Romain Teyssier
Kinematic condition on a free surface
For an incompressible inviscid fluid, we have to solve a Poisson equation with a
boundary condition on the outer surface �v · �n = 0 .
When the outer surface is fixed, both the location and the normal vector are
function of space only. This results in a Neumann BC for the potential.
For a free surface that moves, this is more complicated.
�n
z = η(x, t)
A point on the free surface has position and velocity given by
�r(t) = (x(t), η(x(t), t))
�v (t) = (x� , ∂t η + x� ∂x η)
The BC writes (no vacuum between the fluid and the free surface):
(−∂ η, 1)
�v (t) · �n = (vx , vz ) · �n where the fluid velocity is (vx , vz ) and �n = � x
1 + ∂x η 2
The final kinematic boundary condition writes
∂t η + vx (x, η, t)∂x η = vz (x, η, t)
Romain Teyssier
Gravity waves
We linearize the previous set of equations:
p0
∂
φ(x,
0,
t)
+
gη(x,
t)
+
= C(t)
At the upper surface, we have t
ρ0
and
At the bottom, we have
∂t η(x, t) = vz (x, 0, t) = ∂z φ(x, 0, t)
∂z φ(x, −H, t) = 0
We are looking for propagating waves in the x direction
�
η(x, t) = A expi(kx−ωt)
φ(x, z, t) = φ(z) expi(kx−ωt)
The Poisson equation in the volume writes φ�� (z) = k 2 φ(z)
+
+kz
+φ− exp−kz
for which the general solution is φ(z) = φ exp
φ+ exp−kH −φ− exp+kH = 0
� +
�
−
k φ − φ = −iωA
The upper BC gives us the second relation
� +
�
−
Bernoulli relation at the upper BC gives us −iω φ + φ = −gA
The lower BC gives us the first relation
where we absorbed the constants in the velocity potential.
exp+kH − exp−kH
2
= gk tanh (kH)
The dispersion relation writes ω = gk
exp+kH + exp−kH
Romain Teyssier
Gravity waves
Two interesting limiting cases:
1- Deep water:
H � 1/k
2- Shallow water: H � 1/k
kH � 1 and kH � 1
�
dω
1ω
ω = gk
vg =
=
dk
2k
�
�
ω = k gH
v = gH
We use the shallow water equations to derive directly the second result.
�
∂t (δh) + H∂x (δv) = 0
We linearize the quasi-linear form:
ω 2 = k 2 (gH)
∂t (δv) + g∂x (δh) = 0
In shallow waters, the speed increases as the square root of the depth.
Close to the shore, waves tend to
decelerate. Peaks decelerate slower than
the troughs. They tend to catch up. At the
shore, the trough stops, while the next peak
still travels fast: the wave is breaking.
See «formation of a shock wave».
Romain Teyssier
Rayleigh-Taylor instability
2 semi-infinite incompressible fluids separated by an horizontal interface.
z
z = η(x, t)
z=0
ρ1
g
ρ2
x
→
−
In the 2 separate volume, we have �v1 = ∇φ1
→
−
�v2 = ∇φ2
→
−
∇ · �v1 = 0 = ∆φ1
→
−
∇ · �v2 = 0 = ∆φ2
and at z=0 we have ∂t η + vx,1 ∂x η = vz,1
∂t η + vx,2 ∂x η = vz,2 .
The boundary conditions for the velocity field are
and
�v → 0 when z → ±∞
We also impose pressure continuity at the interface p1 = p2 and from Bernoulli:
v12
p1
v22
p2
∂t φ1 +
+ gη +
= C1 ∂ t φ2 +
+ gη +
= C2
2
ρ1
2
ρ2
As usual, we linearized these equations and look for planar wave solutions:
φ1 = φ1 (z) expi(kx−ωt)
φ2 = φ2 (z) expi(kx−ωt)
η = A expi(kx−ωt)
Romain Teyssier
��
2
The Poisson equation in each domain is φ1 = k φ1 and
φ��2 = k 2 φ2
The unique solutions that satisfy the velocity BC at infinity are
φ1 (z) = φ1 exp −kz
φ2 (z) = φ2 exp +kz
Linearizing the Bernoulli equations and imposing equal pressures give:
ρ1 (∂t φ1 + gη) = ρ2 (∂t φ2 + gη)
where we have absorbed the 2 constants C1 and C2 in the velocity potentials.
Linearizing the 2 interface kinematic conditions gives: ∂t η = ∂z φ1 = ∂z φ2
We use the planar wave solutions in the previous equations to get the system:
−iωA = −kφ1 = +kφ2
The dispersion relation follows:
ρ1 (−iωφ1 + gA) = ρ2 (−iωφ2 + gA)
ω2 =
ρ2 − ρ1
gk
ρ2 + ρ1
If ρ2 > ρ1 , we obtain stable gravity waves in deep water (especially if ρ1 � ρ2 )
If ρ2 < ρ1 , the perturbation is unstable.
Romain Teyssier
Romain Teyssier
Kelvin-Helmholtz instability
We consider exactly the same set-up as for the RT instability, except that gravity is
absent and the boundary conditions at infinity are different (shearing flow).
vx,1 → U1 when z → +∞
The planar wave solutions are now
vx,2 → U2 when z → −∞
φ1 (x, z, t) = φ1 exp −kz exp i(kx−ωt) + U1 x
φ2 (x, z, t) = φ2 exp +kz exp i(kx−ωt) + U2 x
η(x, t) = A exp i(kx−ωt)
The boundary conditions at the interface are now more complicated.
The kinematic conditions are linearized as
∂t η + U1 ∂x η = ∂z φ1
and
∂t η + U2 ∂x η = ∂z φ2
Pressure equilibrium and Bernoulli relations (absorbing the constants) give
ρ1 ∂ t φ 1 + ρ1 U 1 ∂ x φ 1 = ρ2 ∂ t φ 2 + ρ2 U 2 ∂ x φ 2
Using the plane wave solutions, we find the system (−iω + ikU1 )A = −kφ1
ρ1 (−iω + ikU1 )φ1 = ρ2 (−iω + ikU2 )φ2
(−iω + ikU2 )A = +kφ2
�
�
√
ρ
ρ
ρ1 U 1 + ρ2 U 2
1 2
±i
|U1 − U2 |
The dispersion relation is: ω = k
ρ1 + ρ2
ρ1 + ρ2
Romain Teyssier
Kelvin-Helmholtz instability
Romain Teyssier
Quasi-linear waves and Riemann invariants
The 1D isothermal fluid equation in quasi-linear form write:
1
∂t ρ + v∂x ρ + ρ∂x v = 0
∂t v + v∂x v + ∂x P = 0
with
ρ
P = ρc20
The 2 quantities α+ (x, t) = v(x, t) + c0 ln ρ(x, t) and α− (x, t) = v(x, t) − c0 ln ρ(x, t)
satisfy ∂t α+ + (v + c0 )∂x α+ = 0 and ∂t α− + (v − c0 )∂x α− = 0 .
They are Riemann invariants along the characteristic curves:
dx+
dx−
+
= v(x (t), t) + c0
= v(x− (t), t) − c0
dt
dt
dx0
Characteristic curves are different than the fluid trajectories:
= v(x0 (t), t)
dt
At any point in space-time (x,t), we can compute the fluid velocity as:
α+ (x1 ) + α− (x2 )
v(x, t) =
2
where x1 and x2 are the starting points in the initial conditions of the 2 characteristics.
What happens if two «right-going» characteristics cross at the same point ?
→ formation of a shock.
Romain Teyssier
A simple non-linear example: Burger’s equation
Burger’s equation writes in quasi-linear form as
∂t v + v∂x v = 0
2
v
and in conservative form as
∂t v + ∂x
=0
2
In this case, characteristic curves are equal to the trajectories and the velocity is
the Riemann invariant. It follows that characteristic curves are straight lines.
We consider the general initial condition v0 (x) .
The solution is given by the implicit equation
v(x, t) = v0 (x − v(x, t)t)
v0�
− vt) and ∂t v = −v
1 + tv0�
→ formation of the shock.
Taking the time derivative leads to ∂t v = (−v +
t∂t v) v0� (x
The solution blows out at the finite time t = −
1
v0�
Romain Teyssier
Shock waves and the Rankine-Hugoniot relations
Shock waves are discontinuities propagating in the flow that arise naturally from
characteristic crossings and non-linear waves steepening.
We consider here the 1D case (perpendicular to the shock surface).
t
t2
t1
x1
x2
The fluid equations write in conservative form:
x
∂t U + ∂x F (U ) = 0
We use a small enough control volume around the moving discontinuity (speed S) so
that the flow quantities can be considered as homogeneous (x1=St1 and x2=St2).
� t2 � x 2
(∂t U + ∂x F ) dxdt = U2 (x2 − x1 ) − U1 (x2 − x1 ) + F2 (t2 − t1 ) − F1 (t2 − t1 ) = 0
t1
x1
F2 − F1 = S(U2 − U1 )
These are conservation laws connecting the upstream and downstream regions.
Romain Teyssier
Rankine-Hugoniot relations
v1 + v2
v2
S
=
Burger’s equation: U = v and F =
the shock speed is
2
2
Isothermal shocks: U = (ρ, ρv) and F = (ρv, ρv 2 + ρc20 )
Mass conservation:
ρ2 v2 − ρ1 v1 = S(ρ2 − ρ1 )
2
2
2
2
Momentum conservation: ρ2 v2 − ρ1 v1 + ρ2 c0 − ρ1 c0 = S(ρ2 v2 − ρ1 v1 )
Change of variables :
w1 = v1 − S
We have now in the frame of the shock:
w2 = v2 − S
ρ2 w 2 − ρ1 w 1 = 0
ρ2 w22 − ρ1 w12 + ρ2 c20 − ρ1 c20 = 0
w1
M
=
and the Mach number
c0
ρ2
ρ1
The shock relations lead to r2 − r(M2 + 1) + M2 = 0 or r = 1 and r = M2
We define the compression ratio r =
ρ2 = M 2 ρ1
1
w2 =
w1
M2
Rankine-Hugoniot relations are a one-parameter (S) family of solutions.
The shock speed is usually determined using boundary conditions downstream.
Romain Teyssier
Examples of shock waves solutions
Shock wave on a wall for an isothermal ideal fluid.
We assume that we have a wall boundary condition on the left v2 = 0 and v1 < 0.
1
w1
2
w
M
=
We don’t know the shock speed yet. We have ρ2 = M ρ1 w2 =
with
1
M2
c0
2
2
Since v2 = 0 , we have S = −w2 which leads to S − v1 S − c0 = 0.
Among the 2 solutions, only one is physically admissible : compressive wave.
�
�
�
1
S=
v1 + v12 + 4c20
2
c20
|v1 |2
(|v
|
�
c
)
1
0 , we have S �
For a strong shock
ρ2 = 2 ρ1
|v1 |
c0
Hydraulic jump: shallow water. We have at the wall of the sink v2 = 0 .
RH relations are: −h1 v1 = S(h2 − h1 )
1
1
−h1 v12 + gh22 − gh21 = S(−h1 v1 )
2
2
h2
r
=
We define the height ratio
.
h1
We have r3 − r2 − r(1 + 2F 2 ) + 1 = 0
√
For F � 1, r � 2F
Froude number
v1
F =√
gh1
�
gh1
S�
2
Romain Teyssier

Lecture 6

Transcription

Similar documents

Verzahnungswerkzeuge Precision solid carbide gear cutting tools