numerical simulation of compressible two -phase...
TRANSCRIPT
1
Numerical simulation of compressible Numerical simulation of compressible
twotwo--phase flowsphase flows
F. PetitpasSMASH Team - MARSEILLE
2
Flows zoology – Position of the topic
Compressibles Incompressibles
Single phase Multi-phase
Single velocity Multi-velocity
• Interface problems resolution
• Two-phase mixtures in mechanical equilibrium
• Non equilibrium two-phase mixtures
• General models
3
Single velocity flows / Interface problems
σ
Single contact interface
Media B
Media A
Steam
Evaporation front
Liquid
Detonation
σ
σ
4
Some words about discretisation scales
ug
ul
“Macro” scale : Multi-velocity flow
An averaged model with two velocities is needed
« Micro » scale : Single velocity flow
A reduced model for interface problems is enough
5
On the choice of the method
� Sharp interface methods� Lagrangian methods with moving meshes, ALE (Arbitrary
Lagrangian Eulerian)
� Front tracking, VOF, Level Set
Good for solid weak deformation
Not adapted for fluids computation with extreme deformations
Very impressive results
Generally non conservative regarding mass and energy
Heavy numerical treatment
6
On the choice of the method
� Diffuse interface methodsThese methods authorize numerical diffusion of interfaces. This presents several
advantages:
� Interfaces are not tracked or reconstructed, they are captured by the numerical scheme as artificial diffusion zone.
� By the way disappearance or apparition of interfaces are naturally obtained
� Conservative
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Volume Fraction of phase 1
Zone de mélange = interface
Liquide pur
Gaz pur
εα −>1liqεα <liq
7
On the choice of the model
� Euler equations with liquid-vapor Equation of state for evaporation problemsSingle phase model with equilibrium EOS (T,p,g,u)- Able to compute liquid-vapor mixtures at Thermodynamical equilibrium- But metastable states are omitted- Unable to treat liquid-gas interfaces
� Multi-phase models� 4-equation : Euler + mass equationThermal and mechanical equilibrium (T,p,u)- Largely use for gas mixtures where thermal equilibrium condition is not so restrictive- But unable to treat simple contact interface (interface condition of equal pressure and velocity are
violated)� 5-equation : 4 equations + volume fraction equation (Kapila et al., 2001)Mechanical equilibrium (p,u)- Able to treat interfaces between non miscible fluids (liquid-gas)- Able to treat mixture evolving in mechanical equilibrium
� 6-equation modelVelocity equilibrium (u)
� 7-equation (Baer & Nunziato, 1986)Total disequilibrium- Able to solve a large scale of problems- Difficult to solve numerically
8
Outline : The 5-equation model
� Topic 1 : Origins and properties of the 5-equation model
� Topic 2 : Numerical resolution� The Euler equations
� The 5-equation model
� Topic 3 : Phase transition with the 5-equation model
� Topic 4 : Other extensions� Capillary effects
� Compaction effects
� Low Mach computing
� Etc.
9
Topic 1 : Origins and properties of the 5-equation model
10
Starting point : origin of the 5-equation model (The 7-equation model)
)uu.()PP(P .P)u)PE(.(t
Ek'k'kkIkIkkkkk
kkk �������−σλ+−µ−α∇σ=+ρα∇+
∂ρ∂α
)uu(P))IPuu(.(t
uk'kkIkkkkk
kkk �������
−λ+α∇=+⊗ρα∇+∂ρ∂α
)PP(.t 'kkkk −µ=α∇σ+
∂∂α ��
0)u.(t kkk
kk =ρα∇+∂
ρ∂α ��
2u�� =σ 1I PP =and Baer & Nunziato (1986)
21
1221I ZZ
PZPZP
++=
21
2211
ZZ
uZuZ
++=σ��
�Saurel & al. (2003)
Chinnayya & al (2004)and
Each phase obeys its own thermodynamics (pressure, density, internal energy) and has its own set of equations :
11
Example of interface problem evolving to a two-velocity mixture
Simulation : Jacques Massoni, SMASH team
12
Asymptotic reduction of the 7-equation model
� Why it is interesting to use the 5-equation model instead of the 7-equation one ?� It is difficult to solve and implies heavy costs regarding CPU and memory.
� It contains extra unuseful physics to treat interface problems (two velocities and two pressures)
� Extra physical effects are difficult to introduce (as for example phase transition, capillary effects, etc.)
Asymptotic reduction by the Chapman-Enskog method :
∞→ε=µλ 1,
1o fff ε+=
Relaxation parameters tend to infinity
Each flow variable is supporting small variation around an equilibrium state
13
The diffuse interface model (5-equation model)
( )
( )
=∂+ρ∂+
∂ρ∂
=∂
+ρ∂+∂ρ∂
=∂ρα∂+
∂ρα∂
=∂ρα∂+
∂ρα∂
∂∂
ρα+ραρ−ραα=α
0
0
0
0
2222
1111
2221
2112
211
222
211
x
upE
t
Ex
p²u
t
ux
u
t
x
u
t
x
u
cc
cc
dt
d
222111
2211
EEE ρα+ρα=ρρα+ρα=ρ
ppp
uuu
====
21
21
+ Équation d’état de mélange :
This is a mechanical equilibrium but each phase remains in thermal disequilibrium
Equilibre mécanique
Variables de mélange :
( )11
11
2
2
1
1
2
222
1
111
−γα+
−γα
−γγα
+−γ
γα−ρ
=αρ=
∞∞ ,,
k
ppe
,e,pp
4 co
nser
vativ
e eq
uatio
ns
We will come back on thermodynamic closure in the following …
14
Physical meaning of the volume fraction equation
( )
( )
=∂
+ρ∂+∂ρ∂
=∂
+ρ∂+∂ρ∂
=∂
ρα∂+∂
ρα∂
=∂
ρα∂+∂
ρα∂∂∂
ρα+ραρ−ραα=α
0
0
0
0
2222
1111
2221
2112
211
222
211
x
upE
t
Ex
p²u
t
ux
u
t
x
u
t
x
u
cc
cc
dt
d
x
uK
xu
t ∂∂=
∂α∂+
∂α∂ 11
0<K Exemple
1
u0 increases
x
∂ < ⇒ α∂
1
u0 decreases
x
∂ > ⇒ α∂
Liquide : phase 1
Bulles gaz : phase 2
2k kcρ is the Bulk modulus of media k
It traduces the compressibility of a media
-Big when it is weakly compressible
-Small when it is strongly compressible
15
Back to diffuse interface aptitude …
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Volume Fraction of phase 1
Zone de mélange = interface
Liquide pur Gaz pur
εα −>1liqεα <liq
� Diffusion is due to numerical treatment : Artificial diffusion� By the way, an interface looks like a mixture zone : � The interface conditions of pressures and velocities equalities are automatically obtained !
ε−<α<ε 1k
16
5-equation model properties
� Hyperbolic systems : 3 waves speeds.
( )
( )
( )
( )
( )[ ]
=+ρ+∂ρ∂
=∇+⊗ρ+∂ρ∂
=ρα+∂
ρα∂
=ρα+∂
ρα∂
=α
0
0
0
0
2222
1111
1
upEdivt
E
puudivt
u
udivt
udivt
uKdivdt
d
�
���
�
�
�
ww cu,u,cu +−
222
2211
12
1
cccw ρα+
ρα=
ρ
� The speed of sound is those of Wood (1930)
10
100
1000
10000
0 0.2 0.4 0.6 0.8 1
Fraction volumique eau
Vitesse du son de wood dans un melange eau/air (m/s)
17
Topic 2 : Numerical resolution considerations
� Basics for Euler conservative equations
� 5-equation model numerical resolution
18
Basics of numerical resolution for the Euler equations
� Euler equations:
� Mass balance
� Momentum balance
� Total energy balance
� 1D simplification:
0)u(divt
=ρ+∂∂ρ �
0)P(grad)uu(divt
u =+ρ+∂
∂ρ⊗��
�
0)u)PE((divt
E =+ρ+∂
∂ρ �
0x
u
t=
∂∂ρ+
∂∂ρ
0x
P²u
t
u =∂
+∂ρ+∂
∂ρ
0x
)PE(u
t
E =∂
+ρ∂+∂
∂ρ
0x
F
t
U =∂∂+
∂∂
T)E,u,(U ρρρ=T))PE(u,P²u,u(F +ρ+ρρ=
19
Computational mesh
i-1 i i+1 i+2
i-1/2 i+1/2
� The cell ‘i’ is bounded by inlet and outlet sections i-1/2 and i+1/2.
� The fluxes cross over these cell boundaries.
� The unknowns are computed at the cell center and are piecewise constant functions in the cell.
f
xHow to obtain average cells variables form on time step to an
other one
Example at a given time
20
Numerical approximation
i-1 i i+1 i+2
i-1/2 i+1/2
( )tV
Udiv F dVdt 0
t∆
∂ + = ∂ ∫∫ ( )n 1 n * *
i i i 1/2 i 1/2
tU U F F
x+
+ −∆= − −∆
Integration on cell i over time :
The ‘*’ superscript is for : Solution of the Riemann problem at cell boundary
The Flux F at cell boundaries has been supposed to be constant during integration time step CFL condition
21
The Riemann problem
f
x0
fR
fL
x
At t=0
� The initial data are know at a given time and are constant on the right and left side.
� A discontinuity connects the two state.
� Question: How does the solution evolves at t>0 ?
22
The Riemann problem solution (advection)
02/1>−iu 0
2/1<−iu
x
f
x
f
The two possible solutions are:
x/t = ui-1/2
x
t
fi-1 fi
−>
−<
−=
2/1iut/x if
if
2/1iut/x if
1if
)t/x(f
The solution is:
In the (x,t) diagram
f fu 0
t x
∂ ∂+ =∂ ∂
23
For the linearized Euler equations
0x
W)W(A
t
W =∂∂+
∂∂
T)P,u,(W ρ=
ρρ
ρ=
uc0
1u0
0u
)W(A²
u , cu , cu 0 =λ−=λ+=λ −+
analogue of 0x
fu
t
f =∂∂+
∂∂
Plays the role of a propagation velocity
The eigenvalues of A are the waves speeds:
0x
u
t=
∂∂ρ+
∂∂ρ
0x
P²u
t
u =∂
+∂ρ+∂
∂ρ
0x
)PE(u
t
E =∂
+ρ∂+∂
∂ρ
24
The Riemann problem for linearized Euler equations
x
t
u+cuu-c
Solving the Riemann problem consist in determining the perturbed states WL* and WR* after waves propagation from the known states WL and WR.
WR
WR*WL*
WL
25
The Riemann problem solution for linearized Euler equations
*LLL
*LLLLL ucpucp ρ+=ρ+
*LL
*LLLL ²cp²cp ρ−=ρ−
*RRR
*RRRRR ucpucp ρ−=ρ−
x
t
u+cuu-c
*RR
*RRRR ²cp²cp ρ−=ρ−
*L
*R pp =
*L
*R uu =
From this algebraic set of 6 equations the two intermediate states
WL* and WR* are readily obtained.
26
Riemann problem solution
LR
RLLRRLLR*
ZZ
)uu(ZZpZpZp
+−++=
LR
LLRRRL*
ZZ
uZuZppu
+++−=
²c
pp
R
R*
R*R
−+ρ=ρ
²c
pp
L
L*
L*L
−+ρ=ρ
27
Example: shock tube
P=10 000 bars
u=0 m/s
ρ=10 kg /m3
P=1 bar
u=0 m/s
ρ=1 kg /m3
Exact solution (lines) / Computed results with Godunov (symbols)
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
vite
sse
Espace (m)
1
2
3
4
5
6
7
8
9
10
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
dens
ite
Espace (m)
28
Example – Supersonic flow around plane profil
29
Summary for Euler equations
� The Riemann problem solution is a local solution of the Euler equations between two discontinuous initial states.
� It is the cornerstone of all numerical schemes used in gas dynamics, shallow water and modern multiphase codes
� Recommended literature:
E.F. Toro (1997) Riemann solvers and numerical methods for fluid dynamics. Springer Verlag
30
The diffuse interface model (5-equation model)
( )
( )
=∂+ρ∂+
∂ρ∂
=∂
+ρ∂+∂ρ∂
=∂ρα∂+
∂ρα∂
=∂ρα∂+
∂ρα∂
∂∂
ρα+ραρ−ραα=α
0
0
0
0
2222
1111
2221
2112
211
222
211
x
upE
t
Ex
p²u
t
ux
u
t
x
u
t
x
u
cc
cc
dt
d
222111
2211
EEE ρα+ρα=ρρα+ρα=ρ
ppp
uuu
====
21
21
+ Équation d’état de mélange :
This is a mechanical equilibrium but each phase remains in thermal disequilibrium
Equilibre mécanique
Variables de mélange :
( )11
11
2
2
1
1
2
222
1
111
−γα+
−γα
−γγα
+−γ
γα−ρ
=αρ=
∞∞ ,,
k
ppe
,e,pp
4 co
nser
vativ
e eq
uatio
ns
We will come back on thermodynamic closure in the following …
31
Numerical resolutions: issues
1) Volume fraction positivity: How to treat the non-conservative term in the volume fraction equation when shocks or strong rarefaction waves are present ?
Difficulty to guarantee that
2) The volume fraction varies across acoustic waves: Riemann solver difficult to construct.
u.cc
)cc(.u
t���� ∇
αρ+
αρ
ρ−ρ=α∇+∂α∂
2
222
1
211
211
222
11
10 1 <α<
32
� The pressure equilibrium 5-equation model is obtained from this 6-equation model in the asymptotic limit of stiff pressure relaxation coefficient,
� The speed of sound is monotonic,
� The volume fraction is constant through right- and left-facing waves when relaxation effects are absent
� Previous difficulties are circumvented using a pressure non-equilibrium model
2211 ρα+ρα=ρ
222111 EEE ρα+ρα=ρ)pp(
xu
t 2111 −µ=
∂α∂+
∂α∂
0x
u
t1111 =
∂ρ∂α
+∂
ρα∂
0x
u
t2222 =
∂ρ∂α
+∂
ρα∂
0x
)pp(u
t
u 22112
=∂
α+α+∂ρ+
∂ρ∂
)pp(px
up
t
ue
t
e21I11
111111 −µ−=∂∂α+
∂ρα∂
+∂ρα∂
)pp(px
up
t
ue
t
e21I22
222222 −µ=∂∂α+
∂ρα∂
+∂ρα∂
0x
)pp()u2
1eYeY(u
t
)u2
1eYeY( 2211
22211
22211
=∂
α+α+++ρ∂+
∂
++ρ∂6 equations + 1 redundantequation (coming from the summation of energies):
222
211
2f cYcYc +=
).( 0=µ
6+1-equation model
33
a) The (6+1)-equation model is solved without relaxation effects: Godunov-type scheme,
b) Stiff pressure relaxation procedure,
c) Energies reset (in order to ensure energy conservation)
The 5-equation model is solved
3-step methods
34
1st step: Godunov-type scheme
� Without relaxation terms, the 6+1 equation model becomes:
0x
ut
11 =∂α∂+
∂α∂
0x
u
t1111 =
∂ρ∂α
+∂
ρα∂
0x
u
t2222 =
∂ρ∂α
+∂
ρα∂
0x
)pp(u
t
u 22112
=∂
α+α+∂ρ+
∂ρ∂
0x
up
t
ue
t
e11
111111 =∂∂α+
∂ρα∂+
∂ρα∂
0x
up
t
ue
t
e22
222222 =∂∂α+
∂ρα∂+
∂ρα∂
0x
F
t
U =∂∂+
∂∂
( )TE,u,)(,)(U ρραραρ= 21
( )Tu)pE(,p²u,u)(,u)(F +ρ+ραραρ= 11
with
0x
up
t
ue
t
e11
111111 =∂∂α+
∂ρα∂+
∂ρα∂
0x
up
t
ue
t
e22
222222 =∂∂α+
∂ρα∂+
∂ρα∂
A conservative part
A non conservative one
0x
ut
11 =∂α∂+
∂α∂
An advection equation
35
1st step: Godunov-type scheme
( ))U,U(F)U,U(Fx
tUU n
ini
*ni
ni
*ni
ni 11
1−+
+ −∆∆−=
( ) ( )
( ) ( ) ( ) ( )( )*2/1i
*2/1i
n
ik*
2/1ik*
2/1ik
n
ik1n
ik
uupeueux
t
ee
−+−+
+
−α+αρ−αρ∆∆−
αρ=αρ
0x
F
t
U =∂∂+
∂∂
( )TE,u,)(,)(U ρραραρ= 21
( )Tu)pE(,p²u,u)(,u)(F +ρ+ραραρ= 11
with
0x
up
t
ue
t
e11
111111 =∂∂α+
∂ρα∂+
∂ρα∂
0x
up
t
ue
t
e22
222222 =∂∂α+
∂ρα∂+
∂ρα∂
Godunov scheme for conservative equations
A non conventional scheme for non conservative internal energies equations
0x
ut
11 =∂α∂+
∂α∂
Godunov scheme for advection equation
( ))uu()u()u(x
t */i
*/i
ni
*/i
*/i
ni
ni 212112112111
11 −+−+
+ −α−α−α∆∆−α=α
36
Riemann solver
ρρ
ρα
ρα
ρ−=
u0c000
0uc000
u00pp
000u00
0000u0
00000u
)W(A
222
211
2121
u , cu , cu 0ff =λ−=λ+=λ −+
0x
ut
11 =∂α∂+
∂α∂
0x
u
t1111 =
∂ρ∂α
+∂
ρα∂
0x
u
t2222 =
∂ρ∂α
+∂
ρα∂
0x
)pp(u
t
u 22112
=∂
α+α+∂ρ+
∂ρ∂
0x
up
t
ue
t
e11
111111 =∂∂α+
∂ρα∂+
∂ρα∂
0x
up
t
ue
t
e22
222222 =∂∂α+
∂ρα∂+
∂ρα∂
0x
W)W(A
t
W =∂∂+
∂∂
T21211 )p,p,u,s,s,(W α=
3 waves speeds
with222
211
2f cYcYc +=
37
Riemann problem solution
LR
RLLRRLLR*
ZZ
)uu(ZZpZpZp
+−++=
LR
LLRRRL*
ZZ
uZuZppu
+++−=
Rk*
Rk ss =
kL*kL α=α
kR*kR α=α
with
222
211
2f
2211
2211
cYcYc
cZ
ppp
+=ρ=
ρα+ρα=ρ
α+α=
Similar to Euler Riemann solverLk
*
Lk ss =
38
2nd step: Pressure relaxation
)pp( x
ut 21
11 −µ=∂α∂+
∂α∂
0x
u
t1111 =
∂ρ∂α
+∂
ρα∂
0x
u
t2222 =
∂ρ∂α
+∂
ρα∂
0x
)pp(u
t
u 22112
=∂
α+α+∂ρ+
∂ρ∂
)pp(p x
up
t
ue
t
e21I11
111111 −µ−=∂∂α+
∂ρα∂+
∂ρα∂
)pp(p x
up
t
ue
t
e21I22
222222 −µ=∂∂α+
∂ρα∂+
∂ρα∂
Already solved by the 1st step )pp(t 211 −µ=
∂α∂
)pp(pt
eI 21
111 −µ−=∂ρα∂
)pp(pt
eI 21
222 −µ=∂ρα∂
Relaxation system
011 =∂
ρα∂t
022 =∂
ρα∂t
0=∂ρ∂t
u
0=∂ρ∂t
E
39
2nd step: Pressure relaxation
0dtdt
dvp
vv
1p̂
t
0
kI0
kkIk =
−= ∫
∆
( ) 0vvp̂ee 0kkIk
0kk =−+−
( ) ( ) 0vYvYp̂vYvYp̂eYeYeYeY 022222I
011111I
02222
01111
k
=−+−+−+−⇒∑
)pp(t 211 −µ=
∂α∂
)pp(pt
eI 21
111 −µ−=∂ρα∂
)pp(pt
eI 21
222 −µ=∂ρα∂ 022 =+
dt
dvp
dt
deI
011 =+dt
dvp
dt
deI
time integration
Using mass equations
( )( ) 0vYvYp̂p̂ee 011112I1I
0 =−−+−I2I1I p̂p̂p̂ ==
pp̂I =Possible choice
Entropy inequality is verified
40
2nd step: Pressure relaxation
( ) ( ) ( ) 0vvpv,pev,pe 011
01
01
0111 =−+−
( ) ( ) ( ) 0vvpv,pev,pe 022
02
02
0222 =−+−
Closure relation: ( ) ( ) ( ) ( ) 1pvpv 1 2021
0121 =αρ+αρ⇔=α+α
( )pvv 11 =
Zero function to solve ( ) ( ) ( ) ( ) ( ) 1pvpv pf 2021
01 −αρ+αρ=
( )pvv 22 =
Using EOS :
( ) ( ) kkkk vpvp αρ=α→→Then, we determine:
41
� We have in the 2nd step determined:
� We forget the relaxed pressure but keep volume fractions:
� It is then possible to determine mixture pressure by the mixture EOS. By this way, energy conservation is ensured:
� Phasic EOS permits to reset internal energies:
3th step: Internal energy reset
kp α→
11
1p
1p
e
),,e,(p
2
2
1
1
2
222
1
111
21new
−γα+
−γα
−γγα+
−γγα−ρ
=ααρ
∞∞
Conservative and good treatment of wave dynamics from both sides of the interface
),,p(ee kkknewkk αρα=
kk ,v,p α
42
1D example:Water-Air shock tube
EAU AIR
Peau = 10 000 barρeau = 1000 kg/m3
Peau = 1 barρeau = 1 kg/m3
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
0 0.2 0.4 0.6 0.8 1
x (m)
Pression (Bar)
0
50
100
150
200
250
300
350
400
450
500
0 0.2 0.4 0.6 0.8 1
x (m)
Vitesse (m/s)
0
100
200
300
400
500
600
700
800
900
1000
0 0.2 0.4 0.6 0.8 1
x (m)
Densite de melange (kg/m3)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.2 0.4 0.6 0.8 1
x (m)
Fraction volumique air
Solution exacte Solution numérique
43
Résultats 2D - Expériences IUSTI (Layes, Jourdan, Houas)
Shock wave propagation
t = 134µs
t = 274µs
t = 344µs
t = 554µs
t = 137µs
t = 291µs
t = 350µs
t = 577µs
44
3D example : Missile impact on a metal plate
45
Topic 3 : Phase transition with the 5-equation model
46Hyperbolicity is lost into the phase diagram
022 <
∂∂−=
=ctesv
pvc
Hyperbolicity is preserved in the entire domain
Liquid and Vapor isentropes are linked by a kinetic process
Liquid Vapor
p
v
s=cte
s=cte
c2<0
Thermodynamic path using the Van de
Waals representation
p
v
Liquid
Vapor
Kinetic path with the 5-equation
model and two separate EOS
sl=cte
sv=cte
P0
Why is the 5-equation model a good candidate for phase transition ?
47
1) Mass transfer modifies mass equations:
2) Volume fractions change during phase transition:
( )
( ) 112222
111111
Ymudivt
Ymudivt
ɺɺ�
ɺɺ�
ρ−=−=ρα+∂
ρα∂
ρ==ρα+∂
ρα∂
( )I
mAQuKdiv
dt
d
ρ++=α 1
11 ɺ�
ρρα== 111
1 dt
d
dt
dYY Avec ɺ
This « interfacial » density has to be determined in order to
close the model.
Phase transition modeling
48
( ) ( )
( ) ( )122222
121111
ggudivt
ggudivt
−ρν−=ρα+∂
ρα∂
−ρν=ρα+∂
ρα∂
�
�
Mechanical relaxation
( )
( )( ) 0
0
=+ρ+∂ρ∂
=∇+⊗ρ+∂ρ∂
upEdivt
E
)p(uudivt
u
�
����
Mass transfer Heat transfer
2
222
1
211
2
2
1
1
12
2
222
1
211
2
22
1
21
12
2
222
1
211
211
222
11
αρ+
αρ
αΓ+
αΓ
−+
αρ+
αρ
α+
α−ρν+
αρ+
αρ
ρ−ρ=α∇+∂
∂αcc
)TT(Hcc
cc
)gg()u(divcc
)cc()(.u
t���
Entropy analysis for each phase and for the mixture
Lois d’état :
at interfaces (thermodynamical equilibrium),H
0 elsewhere (metastable state)
+∞ν =
Kinetic parameters :
( ) 1011
1111 1 ,
, epp
)p,(e +ρ−γ
γ+=ρ ∞
( ) 2022
22 1 ,ep
)p,(e +ρ−γ
=ρ
11
11
2
2
1
1
2
222
1
111
21
−γα+
−γα
−γγα+
−γγα−ρ
=ααρ
∞∞ ppe
),,e,(p
For the liquid
For vapor
222111 eee ρα+ρα=ρ
Two-phase model for interface problems with phase transition
49
Thermodynamic closure
∞γ−−ρ−γ=ρ p)qe()1()e,(p
Repulsive effects (gas, liquids and solids)
Attractive effects (liquids and solids)
Assumption : Each fluid is governed by the stiffened gas EOS:
Other useful forms for the SG EOS :
qTC)T(h V +γ=
q)pp(
TlnTCT)'qC()T,p(g
1VV ++
−−γ= −γ∞
γ'q
)pp(
TlnC)T,p(s
1V ++
= −γ∞
γ
For each fluid EOS : 5 parameters to determine : 'q,q,C,P, V∞γ
50
0
5
10
15
20
25
30
300 350 400 450 500T (K)
Psat (Bar)
0
500
1000
1500
2000
2500
300 350 400 450 500T (K)
Lv (kJ/kg)
0
200
400
600
800
1000
1200
300 350 400 450 500T (K)
hl (kJ/kg)
2000
2100
2200
2300
2400
2500
2600
2700
2800
300 350 400 450 500T (K)
hg (kJ/kg)
Saturation curves for liquid water and vapor water
Le Métayer et al., Int. Journal of Thermal Sciences, 2004
51
1
10
100
1000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Pression (bar)
0
50
100
150
200
250
300
350
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Vitesse (m/s)
1
10
100
1000
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Densite du melange (kg/m3)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x (m)
Fraction massique de vapeur
Rarefactions Phase transfitionfront
Choc
Phase transition frontContact discontinuity
Dodécane liquide Dodécane vapeur
Pliquide = 1 000 barρliquide = 500 kg/m3
Pvapeur = 1 barρvapeur = 2 kg/m3
Saurel, Petitpas & Abgrall, JFM, 2008
52
High pressure diesel injector
53
High speed motion under water
54
Topic 4 : Some possible extensions
� Capillary effects (Perigaud & Saurel, 2005)
� Detonation (Petitpas et al. 2009)
� Gravity, heat conduction, viscosity, turbulence, etc.
55
Detonation problems
Russian experiences done for the DGA
- 7 fluids
- EOS SG, JWL, IG
- Density ratio : 9000
High velocity impact for ignition
(1500 m/s)
Air
Explosive+ Al Particles
Explosive
PMMA
LifIron
56
Ebullition crisis simulation
==
21
21
TT
gg
q
Relaxation
gTqcond ∇λ−=��
σ
Many physical ingredients are required
57
Bubble growth
No gravity Under gravity
58