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7/29/2019 Talk Shock Tube
1/25
Tobias KnoppDLR Gottingen, AS-NV
The Riemann problem and the shock tube problem
for the Euler equations
Tobias Knopp
DLR Gottingen, AS-NV
Vorlesung, 31.5.2006
orlesung, 31.5.2006
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias KnoppDLR Gottingen, AS-NV
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
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Tobias Knopp
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Tobias KnoppDLR Gottingen, AS-NV
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
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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
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DLR Gottingen, AS-NV
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
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DLR Gottingen, AS-NV
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
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DLR Gottingen, AS-NV
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/7/29/2019 Talk Shock Tube
16/25
DLR Gottingen, AS-NV
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/7/29/2019 Talk Shock Tube
17/25
DLR Gottingen, AS-NV
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
Tobias KnoppDLR Gottingen, AS-NV
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DLR Gottingen, AS NV
Shock tube problem for Euler eq. (cont.)
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
Tobias KnoppDLR Gottingen, AS-NV
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g ,
Shock tube problem for Euler eq. (cont.)
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
Tobias KnoppDLR Gottingen, AS-NV
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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
Acoustic wave: u2 a2 = 2.22 101m/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
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Test case 1: Shock tube problem
u and p are continuous at the contact largest decrease in p due to rarefaction
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Test case 1: Shock tube problem
M< 1: flow is subsonic, no strong (compression) shock jump in (a, and M) at contact discont.
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L = 1kg/m3,
uL = 0m/spL = 100000N/m
2
R = 0.01kg/m3,
uR = 0m/spR = 1000N/m
2
Acoustic wave: u1 a1 = 3.74 102m/s
Acoustic wave: u2 a2 = 3.55 101m/s
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
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Test case 2: Shock tube problem
u and p are continuous at the contact largest decrease in p due to rarefaction
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Test case 2: Shock tube problem
subsonic and supersonic flow regions jump in (a, and M) at contact discont.
orlesung, 31.5.2006
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