Download Characteristics Method applied to the shock tube problem

Characteristics Method applied to
the shock tube problem
1
Expansion wave
diaphragm
gas at rest
void
The gas on the left is initially at rest, and at t = 0 a diaphragm at x = 0 is punctured. The gas flows into
the right region, initially void. All fluid particles will retain their initial entropy, which was constant in the
entire ”left” region. Thus the flow will be isentropic and everywhere:
P
P0
= const. = γ
ργ
ρ0
The Characteristics method shows that:
2
a = Const. along C (1) : dx
u − γ−1
dt = u − a
2
u + γ−1
a = Const along C (3) : dx
dt = u + a
(1)
with a =
γP
ρ
(2)
The initial perturbation coming from (x = 0, t = 0), corresponding to the rupture of the diaphragms, travels
(1)
upstream into the gas at rest along the C0 characteristic at the speed
dx
γP0
= u − a = −a0 = −
(3)
dt
ρ0
any particle below this trajectory is still at rest. The C (3) characteristics originating from the left all take the
same value of the Riemann invariant:
R+ = u +
or
u=
2
2
a=
a0
γ−1
γ−1
(4)
2
(a0 − a)
γ−1
As the gas expands, one expects the speed of sound to decrease, hence u > 0. Since the speed of these
characteristics is larger than that of the fluid particles, that is: dx
dt = u + a > u, the information given by (4)
travels faster than gas particles, this means that (4) holds throughout the flow domain (for any (x, t) point) since
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all C (3) characteristics carry the same value of R+ which is taken from the initial state at rest R+ = γ−1
a0 .
(1)
Any Cλ characteristics carries with it:
R(1) = u −
2
a = Cons(λ)
γ−1
(1)
Cons(λ) is used here to express that this Riemann invariant is a constant along Cλ but that a different
(1)
(1)
”β”characteristic Cβ carries a different value of the Riemann invariant R(1) =Cons(β). Because Cλ characteristics travel more slowly than fluid particles, not all of them can originate from the fluid at rest on the left
(1)
of the diaphragm. Along Cλ one can combine the relations:
2
2
a=
a0
γ−1
γ −1
2
u−
a = Cons(λ)
γ−1
u+
(this holds everywhere)
(1)
(this holds along Cλ )
1
(5)
thus
1
2
u=
a0 + Cons(λ)
2 γ−1
1
γ−1
a0 −
Cons(λ)
a=
2
2
(1)
(6)
(1)
Hence u and a are constant along Cλ and it follows that the speed of Cλ is also constant:
dx
(λ) = u − a
dt
(1)
then Cλ is a straight line. This generates an expansion fan. Indeed the initial characteristic originating from
the (x, t) region where the gas at rest has the slope dx
dt = −a0 then as the gas expands, the velocity u increases
while the speed of sound a decreases, eventually the slope becomes zero, then positive. The expansion fan
originates from the discontinuity at the origin. Thus the slope of a characteristic is also given by:
dx
xλ
=
=u−a
dt
tλ
using again:
u+
2
2
a=
a0
γ−1
γ−1
(7)
everywhere
2
dx
xλ
=u−a=
=
(a0 − a) − a
dt
tλ
γ−1
xλ
2
γ+1
=
a0 −
a
tλ
γ−1
γ−1
This means that at any point (x, t) in the expansion fan the speed of sound can be determined from
x
γ−1
2
a=
a0 −
γ+1 γ −1
t
(8)
(9)
2
then pressure and density are determined by using the isentropic relation, and u is determined by u = γ−1
(a0 −a)
At a given time t , a decreases linearly with x from a = a0 at x = −a0 t (location of the initial perturbation),
2
to a = 0 at x = γ−1
a0 t. Indeed, the fastest characteristic of the expansion fan is obtained for a = 0 with the
slope:
xλ
2
=
a0
tλ
γ−1
2
this corresponds to the trajectories of the first gas particles travelling at speed u = γ−1
a0 with vanishing
pressure and density. Similarly, the variations in time for a fixed x position can be obtained. A special case is
obtained for x = 0. Indeed (9) yields:
a=
and (4) yields:
2
u=
γ−1
2
a0
γ+1
2
2
− 1 a0 =
a0
γ+1
γ+1
Thus the conditions at x = 0 are constant in time , and the Mach number is M = u/a = 1. This is called a
sonic point. To the left of x = 0 the flow is subsonic M < 1, while it is supersonic (M > 1) on the right. In the
subsonic region the C (1) are able to travel upstream, while on the right both C (1) and C (3) travel downstream.
2
t
Initial
perturbation
Fluid at rest
Dry
x
hρ
x
Density and pressure are finally obtained using the isentropic relation (1):
(γ−1)/2
γP
γP0 ργ
γP0 ργ−1
ρ
=
= a0
a=
γ =
γ−1
ρ
ρ ρ0
ρ0 ρ0
ρ0
2/(γ−1)
a
ρ
=
ρ0
a0
2γ/(γ−1)
a
P
=
P0
a0
a
γ−1
2
x
recalling that :
=
−
a0
γ + 1 γ − 1 a0 t
For γ = 1.3 , this gives profiles of the type
1.1
ρ
ρ0
= (A − Bx)6.66
P
P0
= (A − Bx)8.66
Exercise
The equations describing the mean velocity,u, averaged over the depth of a channel, h, are very similar to the
Euler equations for the flow of gas in a pipe. They write:
∂h
∂h
∂u
+u
+h
=0
∂t
∂x
∂x
∂u
∂u
∂h
+u
+g
=0
∂t
∂x
∂x
Apply the characteristics method to this set of equations and show that the Riemann invariants are:
dx
= u + gh
dt
dx
defined by
= u − gh
dt
Along C + defined by
Along C −
R+ = (u + 2 gh) is constant
R− = (u − 2 gh) is constant
A dam is located at x = 0, with on the left h = h0 and u = 0. On the right h = 0. At t = 0 the dam is
broken. Apply the same method as above.
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2
General Shock tube problem
Shocks and contact discontinuity
We now consider the case where the diaphragm initially separates two constant state gases left (pL , ρL ) and
right (pR , ρR ) with pL > pR so that the flow is still from left to right. The flow is no longer isentropic, so the
starting point is the differential form of the Invariants:
dx
=u−a
dt
dx
=u
:
dt
dx
=u+a
:
dt
dp − ρ a du = 0 along C (1) :
dp −
γP
dρ = 0 along C (2)
ρ
dp + ρ a du = 0 along C (3)
2.1
Contact discontinuity
As noted each fluid particle conserves its initial entropy, which is initially constant but different on each side
(2)
of the diaphragm. Thus an entropy discontinuity travels along Csd : dx
dt = u, starting from x = 0.
(2)
Across this characteristic there can be some discontinuity ∆p , ∆u and/or ∆ρ . This Csd characteristic
(3)
(1)
is crossed from left to right by a C and from right to left by a C which tell that ∆p − ρ a ∆u = 0 and
(2)
∆p + ρ a ∆u = 0 . This shows that ∆u = 0 and ∆p = 0. Thus u, and p are constant across Csd and ρ
(2)
(and subsequently a and the internal energy) are discontinuous. In the vicinity of Csd two gas particles flow in
parallel with identical speeds and pressures, but different densities. This is called a contact discontinuity. The
Riemann invariants are obtained as previously but only apply separately to the sub regions left and right of
(2)
Csd , and then matched using continuity of u, and p
2.2
Shock
Potential energy such as pressure is easily transformed into kinetic energy, the reverse is not true. A subsonic flow
is progressively accelerated to a supersonic flow (as seen above), but a supersonic flow can only be decelerated
to subsonic speeds through a shock. The solution established in the case of void on the right of the diaphragm
no longer applies to the present case:
• It was shown that the supersonic flow to the right of x = 0 is entirely determined by the upstream conditions since all three characteristic travel in the downstream direction. There is no way to accommodate
the conditions imposed by the right (pR , ρR ) initial conditions.
(1)
(2)
(3)
• At any point in the supersonic region, three relations are imposed by the three invariants CL , CL , CL
(1)
along three characteristics coming from the initial ”left” gas, and at least one characteristic CR :
dx
dt = u − a coming from the initial ”right” gas. Since there are only three variables, the only solution is
to admit the formation of a shock introducing discontinuities.
The objective of the course being CFD, it is assumed at this point that stationary shock waves are known.
The solution is presented without demonstration (it is the only solution satisfying initial conditions and
thermodynamic principles).
The rupture of the diaphragm causes a shock wave that travels across the stagnant ”right” gas at constant
speed ws , leaving behind a region of constant flow ρ∗ , p∗ , u∗ . Compared to the previous case of a void ”right”
section of the tube, this pressure will limit the expansion wave to a narrower fan as sketched.
The solution can be extended to the case where the ”right” fluid is also in motion. This then provides
a numerical method called Godunov method introduced later on. The diaphragm represents the interface
between to cells contain gas at different pressure, density and velocity. The initial state is replaced by these
known conditions at t = tn and the characteristics method is used to find the resulting state for tn+1 . This is
obtained by using the Riemann invariants as previously, plus relations across the shock, which are only listed
here without demonstration. Let the Mach numbers:
wS
uR
MS =
MR =
aR
aR
4
t
Expansion
fan
Contact discontinuity
s(left) s(right)
Initial
perturbation
u*,p*
constant
Fluid at rest
Shock
u=0
x
p
x
The conditions behind the shock as function of those before the shock are given by the Rankine-Hugoniot
relations (not demonstrated), leading to:
ρ∗
(γ + 1)(MR − MS )2
=
ρR
(γ − 1)(MR − MS )2 + 2
P∗
2γ(MR − MS )2 − (γ − 1)
=
PR
(γ + 1)
The u∗ and shock speed is given by:
(γ − 1)
(γ + 1) P∗
+
2γ PR
2γ
ρR
ρR
u∗ = (1 −
)wS + uR
ρ∗
ρ∗
wS = uR + aR
Next, using the information brought by the Characteristics originating from the ”left” initial region, a
complete explicit solution can be established, and this is the basis of modern (Godunov) numerical methods for
compressible flows briefly introduced in the next lecture ( for further details, see . E.F. Toro ”Riemann Solvers
and Numerical Methods for Fluid Dynamics” Springer 1999).
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