
2/25


Shock tube problem

Special Riemann problem with zero initial velocity

tU + xF(U) = 0

U(x, 0) =

UL if x< 0UR if x> 0

t

x

Rarefactionfan contact Shock

Diaphragm

pressurehighpressure

low

1

1

2

2

3

3 4

4

orlesung, 31.5.2006


3/25


General Riemann problem for Euler eq.

Structure of the solution

initial discontinuity at (x, t) = (0, 0) breaks up into 4 regionsof constant state

In the x t-plane, these 4 regions of constant state aredivided by 3 characteristic waves

x

tshock or

rarefaction fandiscontinuity

con ac

rarefactionor

shock

fan

U U

UU1

2 3

4

orlesung, 31.5.2006


4/25


Nonlinear hyperbolic system. Definition

state R3 set of all admissible values for (, u,E) R3.

tU+ xF(U) = 0 .

strictly hyperbolic, i.e., for any state U state, the Jacobian

A(U) =F

U, (A(U))ij =

Fixj

(1 i,j 3)

has m distinct eigenvalues i(U) which are assumed to be ordered

1(U) < 2(U) < 3(U)

With each eigenvalue i(U) we associate a right eigenvector ri(U)

A(U) ri(U) = i(U) ri(U)

and a left eigenvector li(U) (i.e., an eigenvector of (A(U))T)

li(U)TA(U) = i(U) li(U)

T

with (as eigenvalues are distinct)li(

U)

T

rj(U) = 0 i = j

orlesung, 31.5.2006

T bi K
http://find/


5/25


Characteristic wave pattern for Euler eq.

the 2 = u characteristic wave is always a contact

discontinuity ( u and p continuous, but discontinuous) the 1/3 = u a characteristic wave is either a shock

(piecewise constant discontinuity) or a rarefaction wave(continuous, piecewise smooth)

x x

xx

t t

tt

orlesung, 31.5.2006

T bi K
http://find/


6/25


Characterization of elementary waves

characteristics run into the shock i(U

l) > S

i>

i(U

r)

parallel characteristics at contact i(Ul) = Si = i(Ur)

fan-like divergence of characteristics at a rarefactioni(Ul) < i(Ur)

x

trarefaction shockcontact

U

UU

U1

23

4

orlesung, 31.5.2006

Tobias Knopp
http://find/


7/25


Characterization of elementary waves (cont.)

change in wave speed i between states Ul and Ur Rankine-Hugoniot cond.: Ur = Ul + r(Ul) + O(2)

= consider change in i(U) in direction ofri The i-characteristic field is genuinely nonlinear if

i(U) r(i)(U) = 0 U state

= characteristic speed is not constant across the wave The i-characteristic field is linearly degenerate, if

i(U) r(i)(U) = 0 U

= characteristic speed is constant across the wave (as in alinear system with constant coefficients)

i(U) = iu1

,iu2

,iu3

, i(U) r(i) =

3

j=1iuj

r(i)j

orlesung, 31.5.2006

Tobias Knopp
http://find/


8/25


Genuinely non-linear and linear degeneracy

The 2-characteristic field is lin. degenerate: 2 = u= u2/u1

2(U) =

2u1

, 2u2

, 2u3

=

u1

u2

u1,

u2u2

u1,

u3u2

u1

=

u2

u21,

1

u1, 0

=

u

,

1

, 0

= 2(U) r2(U) =u

, 1

, 0

1, u, 12u2T

= 0

The 1, 3-characteristic fields are genuinely non-linear.

nonlinear t

genuinelynonlineardegenerate

genuinelylinearly

Vorlesung, 31.5.2006

Tobias Knopp
http://find/


9/25


Rarefaction wavesWe seek a continuous self-similar weak solution of the form

U(x, t) =

UL , xt k(UL)vxt

, k(U)L

xt k(UR)

UR ,xt k(UR)

If U(x, t) = v(x, t) solves tU+ A(U)xU= 0, then

xt2

vxt

+

1

t

A

vxt

vxt

= 0

Then by setting = x/t, we obtain

( A(v()) I ) v

() = 0

so that either v() = 0, i.e., U is constant (U= Ul or U= Ur), orthere exists an index k 1, . . . ,m such that

v() = () rk(v()) , k(v()) =

orlesung, 31.5.2006
http://find/


10/25

Tobias Knopp


11/25


Riemann invariants

The solution v(x/t) is constant along each ray

x= k(v(x/t))t The wave speed k(v) is constant along each ray

= The characteristic curves are straight lines

x

t

ULU

R

= k

( UL

)x

t

t

x

= k UR( )

orlesung, 31.5.2006

Tobias Knopp
http://find/


12/25

ppDLR Gottingen, AS-NV

k-Riemann invariantsA k-Riemann invariant is a smooth function w : state R,U

w(

U) with

w(U) r(k)(U) = 0 , U state

There exist (locally) (m 1) k-Riemann invariants w1, . . . ,wm1whose gradients are linearly independent, which are independent ofthe set of independent variables chosen.If the k-characteristic field is linearly degenerate, then k is ak-Riemann invariant.The 2-Riemann invariants may be computed from

w( W) r2( W) = 0

w

,

w

u,

w

p

10

0

= 0

i.e., two linear-independent 2-RI are w1(U) = u, and w2(U) = porlesung, 31.5.2006

Tobias KnoppDLR G i AS NV
http://find/


13/25


k-Riemann invariants (cont.)

On a k-rarefaction wave, all k-Riemann invariants are

constant. The two Riemann invariants of a k-rarefaction are

w1( W) = s= cv ln(p) cp ln()

w2( W) = u+2a

1

= (const) for x= (u a)t

resp. w2( W) = u2a

1= (const) for x= (u+ a)t

Increase in entropy describes the loss in information about aphysical system

For a continuous solution, there is no loss in information For diverging characteristics, s= 0 (for each point (x, t), its

prior history and its future is uniquely determined) For shocks, s> 0: for (x, t) on the shock, its prior history is

not known (did the point run from the left into the shock or

from the right)orlesung, 31.5.2006

Tobias KnoppDLR Gtti AS NV
http://find/


14/25


Isentropic relations for ideas gases

s= cv ln(p) cp ln() = cv ln(p) cv ln() = cv ln

p

s= cv ln(e) Rln()

= cv ln(cvT) (cp cv)ln() = (const) + cv ln

T1

Then for isentropic processes s= 0

p= (const)

T = (const)1

a = (const)(1)/2

orlesung, 31.5.2006

Tobias KnoppDLR Gottingen AS NV
http://find/


15/25


Rarefaction waves for Euler eq.

Here 1 wave: Recalll the 1-Riemann invariant

u+ 2a 1

= (const) fur x= (u a)t

Using x= (u a)t and hence a = u x/t we get

u+

2

1u

x

t

= uL +

2aL 1 = uR +

2aR 1

Solving this equation for u gives the solution for u in the expansion

u(x, t) =2

+ 1xt

+ 1

2

uL + aL = 2+ 1

xt

+ 1

2

uR + aRa(x, t) = u(x, t)

x

t

p= pLa

aL2/(1)

= pRa

aR2/(1)

orlesung, 31.5.2006

Tobias KnoppDLR Gottingen AS NV
http://find/


16/25


Shock tube problem for Euler eq.

Special Riemann problem with zero initial velocity

tU + xF(U) = 0

U(x, 0) =

UL if x< 0UR if x> 0

t

x

Rarefactionfan contact Shock

Diaphragm

pressurehighpressure

low

1

1

2

2

3

3 4

4

orlesung, 31.5.2006

Tobias KnoppDLR Gottingen AS-NV
http://find/


17/25


Shock tube problem for Euler eq. (cont.)

The states to the left and right of the shock U3 and U4 UR are

related bya23a24

=p3

p4

+11 +

p3p4

1 + +11p3p4

u3 = u4 +

a4

p3p4 1

+12

p3p4 1

+ 1

S= u4 + a4

+ 1

2 p3

p4 1 + 1

The states to the left and right of the contact U2 and U3 arerelated by

u3 = u2

p3 = p2orlesung, 31.5.2006

http://find/


18/25

DLR Gottingen, AS NV


The states to the left and right of the expansion U1 = UL and U2

are connected by

u(x, t) =2

+ 1

x

t+

1

2u1 + a1

a(x, t) = u(x, t)x

t=

2

+ 1

x

t+

1

2

u1 + a1 x

t

p= p1

a

a1

2/(1)As u+ 2a/( 1) is constant across the fan

u2 + 2a2 1

= u1 + 2a1 1

The we use the isentropic relations

u2 = u1 +2a1

1

2a2

1

, a2 = a1 p2p1

(1)/2

orlesung, 31.5.2006

http://find/


19/25

g ,


u2 = u1 + 2a1 1

1 (p2/p1)(1)/2

u3 = u1 +

2a1 1

1

p3

p4

p4

p1

(1)/2

We solve this equation for p1/p4 and obtain

p1

p4=

p3

p4

1 +

1

a1(u1 u3)

2/(1)

We finally substitute u3 and get an implicitequation for p3/p4.

We set x p3/p4 and solve for x using Newtons method

p1

p4=

p3

p4

1 +

1

a1

u1 u4

a4

p3p4 1

+12

p3p4 1 + 1

2/(1)

orlesung, 31.5.2006



20/25

g

L = 1kg/m3,

uL = 0m/spL = 100000N/m

2

R = 0.125kg/m3,

uR = 0m/spR = 10000N/m

2

Acoustic wave: u1 a1 = 3.74 102m/s


Contact: u3 = u2 = 2.93 102m/s

Acoustic wave: u2 + a2 = 6.93 102m/s

Shock wave: S = 5.54 102m/s

Acoustic wave: u1 + a1 = 3.35 102

m/sorlesung, 31.5.2006

http://find/


21/25

Test case 1: Shock tube problem

u and p are continuous at the contact largest decrease in p due to rarefaction

orlesung, 31.5.2006



22/25


M< 1: flow is subsonic, no strong (compression) shock jump in (a, and M) at contact discont.

orlesung, 31.5.2006

http://find/


23/25

L = 1kg/m3,

uL = 0m/spL = 100000N/m

2

R = 0.01kg/m3,

uR = 0m/spR = 1000N/m

2



Contact: u3 = u2 = 6.08 102m/s

Acoustic wave: u2 + a2 = 1.14 103m/s

Shock wave: S = 8.87 102m/s

Acoustic wave: u1 + a1 = 3.74 102

m/sorlesung, 31.5.2006

http://find/


24/25


u and p are continuous at the contact largest decrease in p due to rarefaction

orlesung, 31.5.2006

http://find/


25/25


subsonic and supersonic flow regions jump in (a, and M) at contact discont.

orlesung, 31.5.2006
http://find/

Documents

Talk Shock Tube