Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit.

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

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


Phase factor x effective amplitude

Group Velocity


Phase factor x effective amplitude

Constructiveinterference in integral when

Summary and example: sound waves in a moving fluid

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



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




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


( ) e + ekz kzP z P


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


( ) 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


( ) 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


( ) 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:


2: Wave phase should be stationary for different wavelengthsin ship’s rest frame:

ships rest frame

( ', , )0

S x y k

k


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


w cos 0kU

k U

=


w w

w w2 2w

( ', , ) cos ' sin

= cos cos

S x y k k x k y

gkg kU k K

U


w cos 0kU

k U

=

Wave phase inship’s frame:

Wavenumber:


w2 2wcos

gk K

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

Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit.

Documents

Transcript of Prof. dr. A. Achterberg, Astronomical Dept., IMAPP, Radboud Universiteit.