Gas Dynamics, Lecture 6 (Waves & shocks) see: www.astro.ru.nl/~achterb/ Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP, Radboud Universiteit
Mar 28, 2015
Gas Dynamics, Lecture 6(Waves & shocks)see: www.astro.ru.nl/~achterb/
Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP, Radboud Universiteit
Phase- and group velocity
Central concepts:
Phase velocity: velocity with which surfaces of constant phase move
Group velocity: velocity with which slow modulations of the wave amplitude move
Phase velocity
Definition phase S
Phase velocity
Definition phase S
Definition phase-velocity
Phase velocity
Definition phase S
Definition phase-velocity
Group velocity: the case of a “narrow” wave packet
Fourier phase factoramplitude
Fourier integral
d( , ) ( ) exp( - )
2π
kx t k ikx i t
A
Group velocity: the case of a “narrow” wave packet
0
0
0 0
Narrow packet with :
( )k
k k
k k k kk
Group velocity: the case of a “narrow” wave packet (cntd)
0
0
0 0
0 0 0
Same for wave phase :
( )
= ( )
k
k
S kx t
SS k S k k k
k
k x k t k k x tk
This should vanishfor constructiveinterference!
Group Velocity
Wave-packet, Fourier Integral
Group Velocity
Wave-packet, Fourier Integral
Phase factor x effective amplitude
Group Velocity
Wave-packet, Fourier Integral
Phase factor x effective amplitude
Constructiveinterference in integral when
Summary and example: sound waves in a moving fluid
Summary and example: sound waves
Summary and example: sound waves
Application: Kelvin Ship Waves
Waves in a lake of constant depth
Fundamental equations:
1. Incompressible, constant density fluid (like water!)
2. Constant gravitational acceleration in z-direction;
3. Fluid at rest without waves
constant, =0 Ñ V
dˆ
d
Pg
t
ÑVz
Unperturbed state without waves:
atm( )
ˆ0, constant (!)
P z P g H z
P g
ÑV z =
Small perturbations:
, 0 0t t
x x
Ñ Ñ Ñ xV V V
atm( )
ˆ0, constant (!)
P z P g H z
P g
ÑV z =
Equation of motion small perturbations:
0 Ñ x
t
x
V V
2
2
dˆ
d
P Pg
t t
Ñ x ÑVz
SAME as for SOUND WAVES!
Solve for pressure perturbation first!
2 2 2
2 2 0
P P P
t t
x Ñ x ÑÑ
t
x
V V
0 Ñ x
Solution for pressure perturbation:
t
x
V V
0 Ñ x
2
plane wave part
0 , try: ( , , ) ( ) exp( ) .P P x z t P z ikx i t cc
22
2
d0 ( ) e + e
dkz kzP
k P P z P Pz
Solve equation of motion:
( ) e + ekz kzP z P P
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
2
2 & , , ( ) exp( )
Px z t z ikx i t cc
t
x Ñ
x a
Solve equation of motion:
2
2
e + e
de e
d
kz kzx
kz kzz
a ik P P
Pa k P P
z
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P P
There are boundary conditions: #1
2
2
e + e
e e
kz kzx
kz kzz
a ik P P
a k P P
1. At bottom (z=0) we must have az = 0:
P P
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P P
There are boundary conditions: #2
2. At water’s surface we must have P = Patm :
d0
dz
PP P
z H
2
2
e + e
e e
kz kzx
kz kzz
a ikP
a kP
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P
There are boundary conditions: #2
2
2
e + e
e e
kz kzx
kz kzz
a ikP
a kP
2. At water’s surface we must have P = Patm :
( )P z g H H
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P
Dispersion relation from boundary conditions:
2( ) e e e + ek k k k
z
kgPga P
H H H HH
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P
Dispersion relation from boundary conditions:
2( ) e e e + ek k k k
z
kgPga P
H H H HH
2 e etanh( )
e + e
k k
k kkg kg k
H H
H H H
plane wave part
( , , ) ( ) exp( ) .P x z t P z ikx i t cc
( ) e + ekz kzP z P
( )z H H
Limits of SHALLOW and DEEP lake
2 tanh( )kg k H
1: tanh ,
1: tanh 1 ,
k k k k g
k k kg
H H H H
H H
Shallow lake:
Deep lake:
Universal form using dimensionless variables for frequency and wavenumber:
1/2tanh/g
k
H
H
shallow lake
deep lake
Finally: ship waves
Situation in rest frame ship: quasi-stationary
Case of a deep lake
w wˆ ˆcos sin
kg
k k
k x y
wave frequency:
wave vector:
Ship moves in x-direction with velocity U
1: Wave frequency should vanish in ship’s rest frame:
wcos 0kg kU k U =Doppler:
Case of a deep lake (2)
w wˆ ˆcos sin
kg
k k
k x y
wave frequency:
wave vector:
Ship moves in x-direction with velocity U
2: Wave phase should be stationary for different wavelengthsin ship’s rest frame:
ships rest frame
( ', , )0
S x y k
k
Case of a deep lake (3)
w w w
vanishes due to"Doppler condition" 0
( , , ) '
cos ' sin cos
x y x yS x y t k x k y t k x Ut k y t
k x k y k U t
Ship moves in x-direction with velocity U
w cos 0kU
k U
=
Case of a deep lake (4)
w w
w w2 2w
( ', , ) cos ' sin
= cos cos
S x y k k x k y
gkg kU k K
U
Ship moves in x-direction with velocity U
w cos 0kU
k U
=
Wave phase inship’s frame:
Wavenumber:
Case of a deep lake (5)
w2 2wcos
gk K
U
Ship moves in x-direction with velocity U
w cos 0kg kU
k U
=
Stationary phase condition for
w w2 2w w
d0 'cos sin 0
d cos
S gx y
k U
w w( ', , ) cos ' sin :S x y k k x k y
Kelvin Ship Waves
Situation in rest frame ship: quasi-stationary
Shocks: non-linear fluid structures
Shocks occur whenever a flow hits an obstacleat a speed larger than the sound speed
Shock properties
1. Shocks are sudden transitions in flow properties such as density, velocity and pressure;
2. In shocks the kinetic energy of the flow is converted
into heat, (pressure);
3. Shocks are inevitable if sound waves propagate over
long distances;
4. Shocks always occur when a flow hits an obstacle
supersonically
5. In shocks, the flow speed along the shock normal changes from supersonic to subsonic
The marble-tube analogy for shocks
coll
shcoll
L Dt
V
D DV V
t L D
Time between two `collisions’
`Shock speed’ = growth velocity of the stack.
Go to frame where the `shock’ is stationary:
Incoming marbles:
Marbles in stack:
1 sh
2 sh
LV V V V
L D
DV V V
L D
1 2
Flux = density x velocity
Incoming flux:
Outgoing flux:
1 1 1
2 2 2
1
1
L Vn V V
L L D L D
D Vn V V
D L D L D
F
F
1 2
Conclusions:1. The density increases across the shock
2. The flux of incoming marbles equals the flux of outgoing marbles in the shock rest frame:
1 2F F
Steepening of Sound Waves:
( 1)/2sC T
Effect of a sudden transition on a general conservation law (1D case)
0t x
Q F
Generic conservation law:
Change of theamount of Q inlayer of width 2e:
flux in - flux out
in out
d d ( ) ( )
x xt x
F
Q F F
F F
t x
Q F
Infinitely thin layer:
What goes in mustcome out : Fin = Fout
Infinitely thin layer:
What goes in mustcome out : Fin = Fout
Formal proof: use alimiting process for 0
in outd xt
Q F F
0lim d ( ) 0x x
Q
in out = F F
Simplest case: normal shock in 1D flow
Starting point: 1D ideal fluid equations in conservative form;x is the coordinate along shock normal, velocity V along x-axis!
Mass conservation
Momentumconservation
Energyconservation
2
2 2
0
( )0
02 ( 1) 2 ( 1)
Vt x
VV P
t x
V P V PV
t x
Flux in = flux out: three jump conditions
Mass flux
Momentum flux
Energy flux
Three equations for three unknowns: post-shockstate (2) is uniquely determined by pre-shock state (1)!
Three conservation laws means three fluxes for flux in = flux out!
1 2
2 2
1 2
2 2
1 22 ( 1) 2 ( 1)
V V
V P V P
V P V PV V
J
F
S
Shock strength and Mach Number
1D case: ss 1
pre-shock flow speedMach Number
pre-shock sound speed
V
C
M
Shocks can only exist if Ms>1 !
Weak shocks: Ms=1+ with << 1;
Strong shocks: Ms>> 1.
Weak shock:
, and are all small!V P
From jump conditions:
1 11 1
21 1 1
21
1
mass conservation: 0
momentum conservation: 2 0
Energy conservation: (
VV V
V
V V V P
P V
PV V
1 1
2 1 11 1 s1
1 1
01)
0 ( 1)
P
P
P PV V C
Weak shock ~ strong sound wave!
2 21 s1P V C
Sound waves:
2s
PP C
P P
Ñ x
Ñ x
Strong shock: P1<< 1V12
Approximate jump conditions: put P1 = 0!
1 1 2 2
2 21 1 2 2 2
2 21 2 2
2
(1)
(2)
(3)2 2 ( 1)
V V
V V P
V V P
J
1 1 2 2
2 21 1 2 2 2
2 21 2 2
2
(1)
(2)
(3)2 2 ( 1)
V V
V V P
V V P
J
21 2
2 2 2 21 2 1 2 1 2
(1) (2)
2(3)
1
PV V
PVV V V V V V
J
J
21 2
2
1
VV V
1 1 2 2
2 21 1 2 2 2
2 21 2 2
2
(1)
(2)
(3)2 2 ( 1)
V V
V V P
V V P
J
21 2
2
1
VV V
Conclusion for a strong shock:
21 12 1 2 1 2 1 1 1 2 1 1
2
1 1 2 , ,
1 1 1
VV V P V V V V
V