on the numerical simulation of a laser-induced cavitation bubble

Introduction Mathematical model Equations of state (EOS) Numerical Results Future work References

On the Numerical Simulation of a Laser-InducedCavitation Bubble with Phase Transition

Ali Zein

Supervisor: Prof. Dr. Gerald Warnecke

Co-supervisor: Dr. Maren Hantke

Institut fur Analysis und Numerik

Otto-von-Guericke-Universitat Magdeburg

Fifth Workshop "Micro-Macro Modelling and Simulation of

Liquid-Vapour Flows"

Strasbourg, France

April, 14-16, 2010


Outline of Topics

1 Introduction

2 Mathematical model

3 Equations of state (EOS)

3 Numerical ResultsTests for vapor bubbleTests for gas-vapor bubble

4 Future work

5 References

2


Outline

1 Introduction




4 Future work

5 References


Introduction

Experiment:

- A single bubble is induced by laser pulses (in clean distilled water).

- Surrounding liquid: Pressure = 1 bar, Temperaure = 20 ◦C

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

t (µs)

Rad

ius

(mm

)

Figure 1: Experimental data from Muller et al. (2009).

4


Modeling issues

Compressibility: A shock wave is emitted in the liquid at the finalstage of the collapse, experiments, Lauterborn et al. (2000).

Phase transition: Strong influences, Computations of Fujikawa andAkamatsu (1980).

Non-condensable gas: Strong influences, Akhatov et al. (2003) andDreyer et al. (2009).

A wide range of temperature: The temperature at the interfaceexceed the critical point, Akhatov et al. (2003)

5


Extra difficulties

The initial state inside the bubble is unknown (?)We use several assumptions

Validation of the results (?)we use

- Experiments- Physical explanations

⇒ Using a validated model

The mathematical model

Vapor bubble (two phases)

Gas-vapor bubble (three phases)

6


Outline

1 Introduction




4 Future work

5 References


Mathematical model

Averaged models (Diffuse interface)

The seven-equation model: full non-equilibrium, Saurel and Abgrall (1999)

The six-equation model: single velocity, Kapila et al. (2001)

The five-equation model: single velocity and single pressure , Kapila et al.

(2001)

An equilibrium is achieved by relaxation procedures.Validated for a wide range of applications

Phase transition

The five-equation model: Saurel et al. (2008)

Severe numerical difficulties - special procedure

The seven-equation + six-equation models: Zein, Hantke & Warnecke

(2010)

Inserting the heat and mass transfer through the relaxation procedures

Applications in metastable liquids


Modeling phase transition for the seven-equation model

The Saurel and Abgrall model (7EQ model) without heat and mass transfer in 1D

∂α1

∂t+ uI

∂α1

∂x= µ(p1 − p2),

∂α1ρ1

∂t+

∂(α1ρ1u1)

∂x= 0,

∂α1ρ1u1

∂t+

∂(α1ρ1u21 + α1p1)

∂x= pI

∂α1

∂x+ λ(u2 − u1),

∂α1ρ1E1

∂t+

∂(α1(ρ1E1 + p1)u1)

∂x= pI uI

∂α1

∂x+ µpI (p2 − p1) + λuI (u2 − u1),

∂α2ρ2

∂t+

∂(α2ρ2u2)

∂x= 0,

∂α2ρ2u2

∂t+

∂(α2ρ2u22 + α2p2)

∂x= −pI

∂α1

∂x− λ(u2 − u1),

∂α2ρ2E2

∂t+

∂(α2(ρ2E2 + p2)u2)

∂x= −pI uI

∂α1

∂x− µpI (p2 − p1) − λuI (u2 − u1).

αk is the volume fraction (α1 + α2 = 1), ρk the density, uk the velocity, pk the pressure, Ek = ek +u2k

2the total

specific energy, where ek is the specific internal energy.

pI and uI are the interfacial pressure and interfacial velocity respectively.

The parameters λ and µ > 0 are the relaxation parameters for the velocity and the pressure.

9


The solution of the model is obtained by the Strang splitting (1968).

Un+1j = L

∆t2

s L∆th L

∆t2

s Unj

where U = (α1, α1ρ1, α1ρ1u1, α1ρ1E1, α2ρ2, α2ρ2u2, α2ρ2E2)T ,

L∆th : the solution operator of the hyperbolic part over ∆t,

A Godunov-type scheme is used

L∆t2

s : the solution operator of the following system of ODE.

dU

dt= S

The vector S represents the relaxation terms and can be decomposed as

S = SV + SP + SQ + Sm,

where SV and SP , represent the velocity and pressure relaxation terms

SV =

2

6

6

6

6

6

6

6

4

00

λ(u2 − u1)λuI (u2 − u1)

0−λ(u2 − u1)−λuI (u2 − u1)

3

7

7

7

7

7

7

7

5

and SP =

2

6

6

6

6

6

6

6

4

µ(p1 − p2)00

µpI (p2 − p1)00

−µpI (p2 − p1)

3

7

7

7

7

7

7

7

5

10


Source and Relaxation Operators

The vectors SQ and Sm represent the relaxation terms of thetemperature and Gibbs free energy ⇒ that have to be modeled.

Our assumptions

We assume that the mechanical properties relax much faster thanthe temperature and Gibbs free energy.

Also we will assume that the relaxation time for the temperature ismuch smaller than that of the Gibbs free energy.

For physical justifications, see Zein et al. (2010).

During the temperature relaxation the pressure stays in equilibrium.

During the Gibbs free energy relaxation the pressure andtemperature stay in equilibrium.

11


The heat transfer is modeled through the temperature relaxation.

The mass transfer is modeled through the Gibbs free energyrelaxation.

The phase transition occurs at the interface (Figure 2).

0

0.2

0.4

0.6

0.8

1

x

Liquid

Pure LiquidPure Gas

Volume fraction of liquid

Interfacial Zone

Gas

αl=εα

l=1−ε

ε<αl<1−ε

Figure 2: An interface location as diffuse zone. Typically, ε = 10−6 is added for

numerical reasons. ε > ε, ε = 10−4.

12


Heat Transfer and Temperature Relaxation

The heat source vector SQ is modeled as

SQ = (Q

κ, 0, 0, Q, 0, 0,−Q)T .

Where the new variable κ has to be determined.

Then to take into account for heat transfer we have to solve

dU

dt= SQ . (1)

To find the expression for κ we use (pressure stays in equilibrium)

∂p1

∂t=

∂p2

∂t.

13


From this condition with system (1), we obtain

κ =

ρ1c21

α1+

ρ2c22

α2

Γ1

α1+ Γ2

α2

−

Γ1

α1p1 + Γ2

α2p2

Γ1

α1+ Γ2

α2

.

Here Γk denotes the Gruneisen coefficient, Γk = 1ρk

(

∂pk

∂ek

)

ρk

With this modification, the equation of mixture entropy becomes

∂ρs

∂t+

∂ρsu

∂x= θ(1 +

peq

κ)(T2 − T1)

2

T1T2≥ 0

where ρs = α1ρ1s1 + α2ρ2s2, p1 = p2 = peq and u = u1 = u2.The heat transfer Q is modeled as Q = θ(T2 − T1), where θ > 0 isthe temperature relaxation parameter.

Temperature relaxation procedure with θ → ∞ was proposed inZein et al. (2010).

14


Mass Transfer and Gibbs Free Energy Relaxation

For mass transfer we have to solve

dU

dt= Sm

Let us assume that Sm is given in the model as

∂α1

∂t=

m

,

∂α1ρ1

∂t= m,

∂α1ρ1u1

∂t= uI m,

∂α1ρ1E1

∂t= (ei +

u2I

2)m,

∂α2ρ2

∂t= −m,

∂α2ρ2u2

∂t= −uI m,

∂α2ρ2E2

∂t= −(ei +

u2I

2)m.

The new variables and ei have to be determined.

15


To find the new variables we use the following assumptions

∂p1

∂t=

∂p2

∂t,

∂T1

∂t=

∂T2

∂t.

=

φ

ρ1c21

α1

+ρ2c2

2

α2

!

− φ

„

Γ1

α1

p1 +Γ2

α2

p2

«

+ ψ

0

B

B

B

B

@

ρ21

„

∂e1

∂ρ1

«

T1

α1ρ1Cv1

+

ρ22

„

∂e2

∂ρ2

«

T2

α2ρ2Cv2

1

C

C

C

C

A

φ

c21

α1

+c22

α2

!

− φ

„

Γ1

α1

h1 +Γ2

α2

h2

«

+ ψ

0

B

B

B

B

@

e1 + ρ1

„

∂e1

∂ρ1

«

T1

α1ρ1Cv1

+

e2 + ρ2

„

∂e2

∂ρ2

«

T2

α2ρ2Cv2

1

C

C

C

C

A

,

ei =

e1 + ρ1

„

∂e1

∂ρ1

«

T1

α1ρ1Cv1

+

e2 + ρ2

„

∂e2

∂ρ2

«

T2

α2ρ2Cv2

φ−

ρ21

„

∂e1

∂ρ1

«

T1

α1ρ1Cv1

+

ρ22

„

∂e2

∂ρ2

«

T2

α2ρ2Cv2

φ,

where Cvk is the specific heat capacity at constant volume, φ =1

α1ρ1Cv1

+1

α2ρ2Cv2

, ψ =Γ1

α1

+Γ2

α2

and hk = ek +pk

ρk

is the specific enthalpy for phase k.

16


The mixture entropy satisfies the second law of thermodynamics, i.e.

∂ρs

∂t+

∂ρsu

∂x= ν

(g2 − g1)2

Teq

≥ 0,

where Teq is the equilibrium temperature, T1 = T2 = Teq .

The mass transfer is modeled as m = ν(g2 − g1), where ν > 0 is therelaxation parameter of the Gibbs free energy.

Gibbs free relaxation procedures with ν → ∞ were proposed in Zeinet al. (2010).

17


The modified seven-equation model

∂α1

∂t+ uI

∂α1

∂x= µ(p1 − p2) +

Q

κ+

m

,

∂α1ρ1

∂t+

∂(α1ρ1u1)

∂x= m,

∂α1ρ1u1

∂t+

∂(α1ρ1u21 + α1p1)

∂x= pI

∂α1

∂x+ λ(u2 − u1) + uI m,

∂α1ρ1E1

∂t+

∂(α1(ρ1E1 + p1)u1)

∂x= pI uI

∂α1

∂x+ µpI (p2 − p1)

+ λuI (u2 − u1) + Q + (ei +u2

I

2)m,

∂α2ρ2

∂t+

∂(α2ρ2u2)

∂x= −m,

∂α2ρ2u2

∂t+

∂(α2ρ2u22 + α2p2)

∂x= −pI

∂α1

∂x− λ(u2 − u1) − uI m,

∂α2ρ2E2

∂t+

∂(α2(ρ2E2 + p2)u2)

∂x= −pI uI

∂α1

∂x− µpI (p2 − p1)

− λuI (u2 − u1) − Q − (ei +u2

I

2)m,

where Q = θ(T2 − T1) and m = ν(g2 − g1).

Relaxation steps: mechanical relaxation ⇒ temperature relaxation ⇒ Gibbs freeenergy relaxation.

18


Modeling phase transition for the six-equation model

The six-equation model without heat and mass transfer in 1D

∂α1

∂t+ u

∂α1

∂x= µ(p1 − p2),

∂α1ρ1

∂t+

∂(α1ρ1u)

∂x= 0,

∂α2ρ2

∂t+

∂(α2ρ2u)

∂x= 0,

∂ρu

∂t+

∂(ρu2 + α1p1 + α2p2)

∂x= 0,

∂α1ρ1e1

∂t+

∂α1ρ1e1u

∂x+ α1p1

∂u

∂x= µpI (p2 − p1),

∂α2ρ2e2

∂t+

∂α2ρ2e2u

∂x+ α2p2

∂u

∂x= −µpI (p2 − p1).

The interfacial pressure pI is assumed as in Saurel et al. (2003)

pI =Z2p1 + Z1p2

Z1 + Z2. (2)

where Zk = ρkck is the acoustic impedance. Here ck is the speed of sound

c2k =

pk

ρ2k

−

“

∂ek∂ρk

”

pk“

∂ek∂pk

”

ρk

(3)

.


Modeling procedure

The same procedure proposed for the seven-equation model is used, or

The reduction procedure of Chen at al. (1994) is applied to the fullseven-equation model.

The six-equation model with heat and mass transfer is given as

∂α1

∂t+ u

∂α1

∂x= µ(p1 − p2) +

1

κQ +

1

m,

∂α1ρ1

∂t+

∂(α1ρ1u)

∂x= m,

∂α2ρ2

∂t+

∂(α2ρ2u)

∂x= −m,

∂ρu

∂t+

∂(ρu2 + α1p1 + α2p2)

∂x= 0,

∂α1ρ1e1

∂t+

∂α1ρ1e1u

∂x+ α1p1

∂u

∂x= µpI (p2 − p1) + Q + ei m,

∂α2ρ2e2

∂t+

∂α2ρ2e2u

∂x+ α2p2

∂u

∂x= −µpI (p2 − p1) − Q − ei m,

where Q = θ(T2 − T1) and m = ν(g2 − g1).

20


Both modified models were validated on a number of test problems.

In general, both models give almost the same results.

The six-equation model is less expensive and easier to be extendedto multiphase flows.

Thus the six-equation model is adopted for the collapsing bubble

All details of modeling, physical justifications, numerical method andvalidation are given in

A. Zein, M. Hantke, and G. Warnecke. Modeling phase transition forcompressible two-phase flows applied to metastable liquids. J. Comput.

Phys., 229(8):2964-2998, 2010.

21


Vapor bubble model

The six-equation model in spherical coordinates and assuming rotational symmetry

∂α1

∂t+ u

∂α1

∂r= µ(p1 − p2) +

1

κQ +

1

m,

∂α1ρ1

∂t+

∂(α1ρ1u)

∂r= m −

2

rα1ρ1u,

∂α2ρ2

∂t+

∂(α2ρ2u)

∂r= −m −

2

rα2ρ2u,

∂ρu

∂t+

∂(ρu2 + α1p1 + α2p2)

∂r= −

2

rρu

2,

∂α1ρ1e1

∂t+

∂α1ρ1e1u

∂r+ α1p1

∂u

∂r= µpI (p2 − p1) + Q + ei m −

2

rα1ρ1e1u −

2

rα1p1u,

∂α2ρ2e2

∂t+

∂α2ρ2e2u

∂r+ α2p2

∂u

∂r= −µpI (p2 − p1) − Q − ei m −

2

rα2ρ2e2u −

2

rα2p2u,

∂(ρe + 12ρu2)

∂t+

∂u(ρe + 12ρu2 + α1p1 + α2p2)

∂r= −

2

r(ρe +

1

2ρu

2 + α1p1 + α2p2)u.

The last equation is the mixture energy equation (redundant)

22


Gas-vapor bubble model

The non-condensable gas is modeled as a third phase.

The full non-equilibrium model for multiphase flows of Saurel-Abgrall typewithout heat and mass transfer is written as8

>

>

>

>

>

>

>

>

>

>

<

>

>

>

>

>

>

>

>

>

>

:

∂αk

∂t+ uI · ∇αk = µ(pk − p),

∂αkρk

∂t+ ∇ · (αkρkuk ) = 0,

∂αkρkuk

∂t+ ∇ · (αkρkukuk) + ∇(αkpk) = pI∇αk + λ(uk − u),

∂αkρkEk

∂t+ ∇ · (αk(ρkEk + pk)uk) = pIuI · ∇αk − µpI (pk − p) + λuI · (uk − u),

where k = 1, 2, ..., N, where N is the number of phases.

u =

NX

k=1

uk

N, p =

NX

k=1

pk

N.

23


Assume stiff velocity relaxation for this model, i.e.

λ =1

ǫwhere ǫ → 0+.

Then following the method of Chen et al. (1994), we get the following reducedmodel

8

>

>

>

>

>

>

>

>

>

>

<

>

>

>

>

>

>

>

>

>

>

:

∂αk

∂t+ u · ∇αk = µ(pk − p),

∂αkρk

∂t+ ∇ · (αkρku) = 0,

∂ρu

∂t+ ∇ · (ρuu) + ∇p = 0,

∂αkρkek

∂t+ ∇ · (αkρkeku) + αkpk∇ · u = −µpI (pk − p),

where ρ =N

X

k=1

αkρk and p =N

X

k=1

αkpk .

24


A model with three phases is enough to investigate the bubble containing vaporand a non-condensable gas8

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

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>

>

>

>

>

>

>

<

>

>

>

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>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

>

:

∂α1

∂t+ u

∂α1

∂r= µ(p1 − p),

∂α2

∂t+ u

∂α2

∂r= µ(p2 − p),

∂αkρk

∂t+

∂(αkρku)

∂r= −

2

rαkρku,

∂ρu

∂t+

∂(ρu2 + p)

∂r= −

2

rρu2,

∂αkρkek

∂t+

∂αkρkeku

∂r+ αkpk

∂u

∂r= −µpI (pk − p) −

2

rαkρkeku −

2

rαkpku,

∂(ρe + 12ρu2)

∂t+

∂u(ρe + 12ρu2 + p)

∂r= −

2

r(ρe + 1

2ρu2 + p)u,

where k = 1, 2, 3, and ρe =3

X

k=1

αkρkek .

25


The three phase model is hyperbolic.

The Riemann solver and the pressure relaxation procedure of thesix-equation model are extended here.

The heat and mass transfer are considered only if the interfaceseparates between the liquid and its vapor (Petitpas et al. (2009)).The interface is located by

ε ≤ α1 ≤ (1 − ε) & ε ≤ α2 ≤ (1 − ε),

where α1 and α2 are the volume fractions of vapor and liquid.

The bubble is identified by the summation of the vapor volumefraction and the non-condensable gas volume fraction.

26


Outline

1 Introduction




4 Future work

5 References


Equations of state (EOS)

A cubic EOS ⇒ loss of hyperbolicity.

To overcome this problem- Each fluid obeys its own EOS as a pure material- These EOS should satisfy certain convexity constraints,

Menikoff and Plohr (1989), Saurel et al. (2008).

The stiffened gas EOS (SG-EOS) contains the main properties of thepure fluids, also satisfies the convexity constraints.

The SG-EOS (Harlo and Amsden (1971))

e(p, ρ) =p + γπ

ρ(γ − 1),

where e is the internal energy, γ and π are characteristic constants.

The parameters of the SG-EOS are determined by using a reference curve(the Hugoniot curve, the saturation curve).

28


Using the Hugoniot curve as a reference

The parameters γ and π are determined by using experimental data.

Experimental relation: us = c0 + aup,us : the shock speed, up : the particle speed, a: constant, c0 is the sound speed.

Theoretical relation: us =

√

c20 + (

γ + 1

4up)2 +

γ + 1

4up

The parameter γ is chosen to give the closest agreement betweenthe experimental data and the theoretical curve.

By using the expression of the sound speed (3), π =ρ0c

20

γ− p0

For details: Cocchi et al. (1996)

29


Using the saturation curves as reference

This method was proposed by Le Metayer et al. (2004)

e(p, ρ) =p + γπ

ρ(γ − 1)+ q, where q is characteristic constant.

Using the relation of the internal energy with Maxwell relations we getthe temperature relation. The SG-EOS for each phase reads

e(p, v) =p + γπ

(γ − 1)v + q, (4a)

h(T ) = CpT + q, (4b)

T (p, v) =p + π

Cv (γ − 1)v , (4c)

s(p, T ) = Cv lnTγ

(p + π)(γ−1)+ q

′, (4d)

g(p, T ) = (γCv − q′)T − CvT ln

Tγ

(p + π)(γ−1)+ q, (4e)

where v =1

ρis the specific volume, h the specific enthalpy, T the temperature, s the specific entropy, g

the Gibbs free energy and q′ is a characteristic constant, Cv the specific heat at constant volume and Cp

the specific heat at constant pressure.

30


Choose two reference states, in a suitable range like 298 − 473 K .

From the linear relation, h(T ) = CpT + q, ⇒ both Cp and q aredetermined.

From v(T ) =(γ − 1)CvT

psat(T ), ⇒ both π and Cv are determined.

Find γ by γ = Cp/Cv .

At thermodynamic equilibrium the Gibbs free energies are equal, gl = gg ,

(γlCvl − q′

l )T − CvlT lnTγl

(p + πl )(γl−1)+ ql =

(γg Cvg − q′

g )T − CvgT lnTγg

(p + πg )(γg−1)+ qg . (5)

Set q′

l = 0 and choose q′

g that provides the best fit to the experimentalcurve.

31


The above method:

Mainly it depends on the linearity of relation h(T ) = CpT + q,which holds in short ranges far from the critical point

The choice q′

g may lead to negative entropy ?

The collapsing bubble:

A wide range of temperature.

We use the same SG-EOS of Le Metayer et al. (2004), with ourestimations for the parameters.

32


Our estimations

Determination of the SG-EOS for the water vapor:

Set πg = 0 ⇒ eg (T ) = CvgT + qg .

We choose Cvg = 1.2 × 103 J/kg/K.

Then we choose qg which gives a good fitting for the experimentalrelation between eg and T (qg = 1995 × 103 J/kg), see Figure 3.

The specific volume is written as

vg (T ) =(γg − 1)CvgT

psat(T ).

We choose γg which provides a good fitting for the experimentalcurve of vg versus T , (γg = 1.327), see Figure 4.

33


300 350 400 450 500 550 600 6500

500

1000

1500

2000

2500

T (K)

eg (

kJ/k

g)

Figure 3: Saturation internal energy for the water vapor. Experimental curves are

shown in lines and the SG-EOS approximation with symbols.

34


300 350 400 450 500 550 600 65010

−4

10−2

100

102

T (K)

vg (

m3/k

g)

Figure 4: Saturation specific volume for the water vapor. Experimental curves are


35


Determination of the SG-EOS for the liquid water

We start from the internal energy e and specific volume v :- The values of e and v for compressed water are approximated bythe saturated ones.

- The internal energy appears explicitly in the models.

Monotonic relation between e and v . We choose the followingtemperatures T1 = 293 K and T2 = 623 K

The experimental data corresponding to T1 and T2 are, seeOldenbourg (1989):

psat (T1) = 2339 Pa el (T1) = 83.94 × 103 J/kg vl (T1) = 0.001002 m3/kg

psat (T2) = 16.514 × 106 Pa el (T2) = 1641.81 × 103 J/kg vl (T2) = 0.00174 m3/kg

36


* By using each group of the experimental data in (4a), we get

83.94 × 103 = (0.001002)2339 + γlπl

γl − 1+ ql ,

1641.81× 103 = (0.00174)16.514× 106 + γlπl

γl − 1+ ql .

* The sound speed in the liquid water is 1482 m/s at T1 = 293 K. Thususing (3), we get

(1482)2 = 0.001002γl(2339 + πl ).

* Solving the the above equations, we get

γl = 2.057, πl = 1.066× 109, ql = −1.994674× 106.

* Substituting the experimental data related to T1 in (4c) we haveCvl = 3.449× 103 J/kg/K.

37


300 400 500 6000

500

1000

1500

2000

2500

3000

T (K)

e l (kJ/

kg)

300 400 500 6000

500

1000

1500

2000

2500

3000

T (K)

e l (kJ/

kg)

Figure 5: Saturation internal energy for the liquid water. Experimental curves are


Left graph: el(T , v) = CvlT + πlv + ql , (the values of v from tables)

Right graph: el(T ) =psat(T ) + γlπl

psat(T ) + πl

CvlT + ql .

38


Determination of the entropy constants

The entropy of the liquid at T1 = 293 K is sl1 = 0.296 kJ/kg/K. Using(4d) ⇒ q′

l = 35.78 kJ/kg/K.

We choose q′

g to provide a good agreement to the experimentalsaturation curve. (q′

g = 2.41 kJ/kg/K)

300 350 400 450 5000

5

10

15

20

25

30

T (K)

Psa

t (bar

)

300 400 500 600 6500

50

100

150

200

250

T (K)

Psa

t (bar

)

Figure 6: Saturation curve of the water. Experimental curves are shown in lines and

the SG-EOS approximation with symbols.

39


For example, choose q′

l = 35.64 kJ/kg/K and

q′

g =

2.41, T < 573

2.51, 573 ≤ T ≤ 593

2.57, 593 ≤ T ≤ 613

2.61, T > 613

where the unit of T is K and the unit of q′

g is kJ/kg/K.

300 350 400 450 500 550 600 6500

50

100

150

200

250

T (K)

Psa

t (b

ar)

Figure 7: Saturation curve of the water. Experimental curves are shown in lines and

the SG-EOS approximation with symbols.

40


In Summary the parameters of the SG-EOS by this method are given inthe following table

Phase γ π(Pa) Cv (J/kg/K) q(J/kg) q′(J/kg/K)

vapor 1.327 0 1.2 × 103 1995 × 103 2.41 × 103

liquid 2.057 1.066 × 109 3.449 × 103−1994.674 × 103 35.78 × 103

Table 1: EOS parameters for vapor and liquid water by the present method.

41


Outline

1 Introduction




4 Future work

5 References


Numerical Results

Initial conditions:- In liquid pl = 1 bar, Tl = 293 K.- Inside the bubble: Saturation state (Tv = 293 K, pv = 2339 Pa),

change in pressure, change in temperature.

Rmax = 0.75 mm, domain of computations [0,99]mm.

The CFL number is fixed to 0.6 and a uniform grid is used.

NI : The number of cells that cover the maximum radius of thebubble at the initial state.

NI 250 500 750 1000# of cells 33,000 66,000 99,000 132,000

Table 2: The concept of NI .

Note: The mass transfer is activated from the beginning of theevolution until the critical state is exceeded.

43


Tests for vapor bubble

0 30 60 90 120 150 1800

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150 180−3000

−2000

−1000

0

1000

2000

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 150 1800

2000

4000

6000

8000

t (µs)

Tem

pera

ture

(K)

0 30 60 90 120 150 18010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

100 120 140720

760

Figure 8: Without mass transfer (dashed line), with mass transfer (solid line),

experimental radius (dots). Temperature and pressure at the center, velocity at the

interface. NI = 500 cells, Tv = 293 K and pv = 2339 Pa.


With no mass transfer the temperature after reaching 1500 Kincreases rapidly and then the collapse occurs.

With mass transfer the temperature after reaching 515 K increasesrapidly and then the collapse occurs.

0 10 20 30 40 50 60200

300

400

500

600

700

800

900

t (µs)

Tem

pera

ture

(K)

0 10 20 30 40 50 6010

−2

10−1

100

101

t (µs)

Pre

ssur

e (b

ar)

Figure 9: Without mass transfer (dashed line), with mass transfer (solid line). Zoom

of the temperature and pressure graphs in Figure 8.

45


0 0.01 0.02 0.03 0.040

1

2

3

4x 10

−3

r (mm)

Vapo

r mas

s fra

ctio

n

0 0.15 0.3 0.45 0.6 0.750

1

2

3

4x 10

−3

r (mm)

Vapo

r mas

s fra

ctio

n

0 0.01 0.02 0.03 0.04200

300

400

500

600

700

800

r (mm)

Tem

pera

ture

(K)

0 0.15 0.3 0.45 0.6 0.75200

300

400

500

600

700

800

r (mm)

Tem

pera

ture

(K)

Figure 10: The vapor mass fraction and temperature versus radial direction if the

phase transition is included at time t = 180 µs. With NI = 500 cells, Tv = 293 K and

pv = 2339 Pa.


0 30 60 90 120 150 1800

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150 180−1500

−1000

−500

0

500

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 150 1800

2000

4000

6000

8000

t (µs)

Tem

pera

ture

(K)

0 30 60 90 120 150 18010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

69 70 710

1000

2000

3000

69.4 7010

2

104

106

NI=250 cells

NI=500 cells

NI=750 cells

Experiment

NI=250 cells

NI=500 cells

NI=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

Figure 11: Vapor bubble without mass transfer, comparison using several grids.

Temperature and pressure at the center, velocity at the interface. Initial values

Tv = 293 K and pv = 2339 Pa.


0 0.2 0.4 0.60

0.2

0.4

0.6

0.8

1

r (mm)

Vapo

r vol

ume

fract

ion

0 0.1 0.2 0.30

0.2

0.4

0.6

0.8

1

r (mm)

Vapo

r vol

ume

fract

ion

0 0.1 0.2 0.30

0.2

0.4

0.6

0.8

1

r (mm)

Vapo

r vol

ume

fract

ion

0 0.2 0.4 0.60

0.2

0.4

0.6

0.8

1

r (mm)

Vapo

r vol

ume

fract

ion

t=50 µ s t=69 µ s

t=71 µ s t=120 µ s

Figure 12: Vapor volume fraction profiles of the results in Figure 11.

48


0 30 60 90 120 150 1800

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150 180−800

−600

−400

−200

0

200

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 150 1800

2000

4000

6000

8000

t (µs)

Tem

pera

ture

(K)

0 30 60 90 120 150 18010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

0100200

2000

4000

6000

103

104

105

NI=500 cells

NI=750 cells

NI=1000 cells

Experiment

NI=500 cells

NI=750 cells

NI=1000 cells

NI=500 cells

NI=750 cells

NI=1000 cells

NI=500 cells

NI=750 cells

NI=1000 cells

Figure 13: Vapor bubble without mass transfer, comparison at several grids.


Tv = 293 K and pv = 2339 Pa.


0 20 40 60 80 100 120 1400

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 20 40 60 80 100 120 140

−4000

−3000

−2000

−1000

0

1000

t (µs)

Velo

city

(m/s

)

0 20 40 60 80 100 120 1400

5000

10000

15000

t (µs)

Tem

pera

ture

(K)

0 20 40 60 80 100 120 14010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

0

0.05

0.1

500

1000

1500

0

1000

2000

3000

104

106

NI=250 cells

NI=500 cells

NI=750 cells

Experiment

NI=250 cells

NI=500 cells

Ni=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

Figure 14: Vapor bubble with mass transfer, comparison at several grids.


Tv = 293 K and pv = 2339 Pa.


... Several tests with different pressures and different temperatures ...

Whatever the initial state inside the bubble we conclude that:

There is no rebound if the mass transfer is included.

In all cases at the collapse time the pressure and temperature at thebubble center jump to very high values.

The pressure and temperature inside the bubble before the collapsetime when the mass transfer is included are less than those valueswith no mass transfer.

In the cases of no mass transfer it is noted that the first collapse ismuch more violent than the second one.

51


Tests for gas-vapor bubble

Besides the vapor inside the bubble we assume a percentage ofnon-condensable gas.

Hydrogen and Oxygen are most probably present.

We assume that the non-condensable gas obeys the SG-EOS, i.e. itobeys equations (4).

For both Hydrogen and Oxygen we assume γ = 1.4, π = 0 andq = 0.Hydrogen: Cv = 10.1 kJ/kg/K.Oxygen: Cv = 0.662 kJ/kg/K.

52


0 30 60 90 120 1500

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150−3000

−2000

−1000

0

1000

2000

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 1500

2000

4000

6000

8000

t (µs)

Tem

pera

ture

(K)

0 30 60 90 120 15010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)30 60

300

600

90 120 150440

460

90 120 150700

800

105

Figure 15: Bubble results with mass transfer, vapor bubble (solid line), gas-vapor

bubble (dashed line). Temperature and pressure at the center, velocity at the

interface. NI = 500 cells, initial state inside the bubble: T = 293 K and p = 2339 Pa.

The non-condensable gas is Hydrogen with a mass fraction of 1%.


0 30 60 90 120 1500

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150

−1500

−1000

−500

0

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 15010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

0 30 60 90 120 1500

1000

2000

3000

4000

5000

t (µs)

Tem

pera

ture

(K)

1000

2000

3000

4000

104

106

−1500

−1000

−500

NI=250 cells

NI=500 cells

NI=750 cells

Experiment

NI=250 cells

NI=500 cells

NI=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

NI=250 cells

NI=500 cells

NI=750 cells

Figure 16: Gas-vapor bubble with mass transfer, comparison at different grids.

Temperature and pressure at the center, velocity at the interface. Initial state inside

the bubble: T = 293 K and p = 2339 Pa. The non-condensable gas is Hydrogen with

a mass fraction of 1%.


0 30 60 90 120 1500

0.2

0.4

0.6

0.8

t (µs)

Rad

ius

(mm

)

0 30 60 90 120 150

−1200

−1000

−800

−600

−400

−200

0

t (µs)

Velo

city

(m/s

)

0 30 60 90 120 15010

−2

100

102

104

106

t (µs)

Pres

sure

(bar

)

0 30 60 90 120 1500

1000

2000

3000

4000

t (µs)

Tem

pera

ture

(K)

90 120 150530

535

540

Figure 17: Gas-vapor bubble with mass transfer, the non-condensable gas is Oxygen

with a mass fraction of 2.2%. Temperature and pressure at the center, velocity at the

interface. NI = 500 cells, initial state inside the bubble: T = 293 K and p = 2339 Pa.

55


... Several tests with different percentages, different values for Cv , ...

We conclude the following results for the gas-vapor bubble:

The existence of sufficient amount of non-condensable gas isessential for the rebound after the collapse if the mass transfer isincluded.

The behavior of the rebound depends on the percentage of thenon-condensable gas and on the nature of the gas.

56


0.8 0.9 1 1.1 1.2 1.3 1.40

0.1

0.2

0.3

0.4

0.5

Rmax

(mm)

Ma

xim

um

ra

diu

s a

fte

r th

e first

co

llap

se (

mm

)

ExperimentModel

Figure 18: The maximum radius after the first collapse versus the initial bubble

radius. Computations are made with a uniform grid, ∆r = 1.6 × 10−6.

57


Outline

1 Introduction




4 Future work

5 References


Future work

Diffuse interface- Some adaptive discretization- Coupling with the ideas of sharp interface modeling just around thecollapse point.

Modeling the non-condensable gas with vapor as a mixture.

59


Outline

1 Introduction




4 Future work

5 References


References

I. Akhatov, O. Lindau, A. Topolnikov, R. Mettin, N. Vakhitova, and W. Lauterborn. Collapse and rebound

of a laser-induced cavitation bubble. Phys. Fluids, 13:2805-2819, 2003.

G. Chen, C. Levermore, and T.P.Liu. Hyperbolic conservation laws with stiff relaxation terms and entropy.

Comm. Pure. Appl. Math., 47:787-830, 1994.

J. P. Cocchi, R. Saurel, and J. C. Loraud. Treatment of interface problems with Godunov-type schemes.

Shock Waves, 5:347-357, 1996.

W. Dreyer, F. Duderstadt, M. Hantke, and G. Warnecke. On phase change of a vapor bubble in liquid

water. Technical report, WIAS Preprint No. 1424, 2009.

S. Fujikawa and T. Akamatsu. Effects of the non-equilibrium condensation of vapor on the pressure wave

produced by the collapse of a bubble in a liquid. J. Fluid. Mech., 97:481-512, 1980.

F. Harlow and A. Amsden. Fluid Dynamics. Los Alamos Scientific Laboratory Monograph, LA-4700, second

edition, 1971.

R. Hickling and M. Plesset. Collapse and rebound of a spherical bubble in water. Phys. Fluids, 7:7-14, 1964.

A. Kapila, R. Meniko, J. Bdzil, S. Son, and D. Stewart. Two-phase modelling of DDT in granular materials:

reduced equations. Phys. Fluid, 13:3002-3024, 2001.

W. Lauterborn, T. Kurz, C. Schencke, O. Lindau, and B. Wolfrum. Laser-induced bubbles in cavitation

research. In Proceedings of the IUTAM Symposium on Free Surface Flows. Birmingham United Kingdom,2000.

61


O. Le Metayer, J. Massoni, and R. Saurel. Elaborating equations of state of a liquid and its vapor for

two-phase flow models (in french). Int. J. Thermal Sci., 43(3):265-276, 2004.

R. Menikoff and B.J. Plohr, The Riemann problem for fluid flow of real materials. Rev. Mod. Phys.,

61(1):75-130, 1989.

S. Muller, M. Bachmann, D. Kroninger, Th. Kurz and Ph. Helluy, Comparison and validation of

compressible flow simulations of laser-induced cavitation bubbles, Computers & Fluids. 38:1850-1862, 2009.

R. Oldendourg. Properties of water and steam in SI-units. Springer, 1989.

F. Petitpas, J. Massoni, R. Saurel, E. Lapebie, and L. Munier. Diffuse interface models for high speed

cavitating underwater systems. Int. J. Multiphase Flows, 35(8):747-759, 2009.

R. Saurel and R. Abgrall. A multiphase Godunov method for compressbile multifluid and multiphase flows.

J. Comput. Phys., 150(2):425-467, 1999.

R. Saurel, F. Petitpas, and R. Abgrall. Modelling phase transition in metastable liquids: application to

cavitating and flashing flows. J. Fluid. Mech., 607:313-350, 2008.

G. Strang. On the construction and comparison of difference schemes. SIAM J. Num. Anal., 5:506-517,

1968.

A. Zein, M. Hantke, and G. Warnecke. Modeling phase transition for compressible two-phase flows applied

to metastable liquids. J. Comput. Phys., 229(8):2964-2998, 2010.

62


Thank you for your attention

63

on the numerical simulation of a laser-induced cavitation bubble

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