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# A mixture-energy-consistent numerical method for compressible two-phase ﬂow with interfaces, cavitation, and evaporation waves

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### interfaces, cavitation, and evaporation waves

Keh-Ming Shyue

Institute of Applied Mathematical Sciences National Taiwan University

Joint work with Marica Pelanti at ENSTA, Paris Tech, France

(2)

### Outline

Compressible 2-phase solver for matastable fluids: application to cavitation & flashing flows

1. Motivation

2. Constitutive law for metastable fluid

3. 6-equation single-velocity 2-phase flow model Model with & without heat & mass transfer 4. Stiff relaxation solver

Mixture-energy-consistent method means total energy conservation& pressure consistency

Flashing flowmeans a flow with dramatic evaporationof liquid due to pressure drop

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### Motivation: Dodecane 2-phase Riemann problem

Riemann data for metastable dodecane modelled by SG EOS Liquid phase: Left-hand side (0 ≤ x ≤ 0.75m)

v, ρl, u, p, αv)L = 2kg/m3, 500kg/m3, 0, 108Pa, 108 Vapor phase: Right-hand side (0.75m < x ≤ 1m)

v, ρl, u, p, αv)R = 2kg/m3, 500kg/m3, 0, 105Pa, 1 − 108 ,

Liquid Vapor

← Membrane

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### Dodecane 2-phase problem: Sample solution

0 0.2 0.4 0.6 0.8 1

100 101 102 103

t = 0 t=473µs

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−50 0 50 100 150 200 250 300 350

t = 0 t=473µs

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105 106 107 108

t = 0 t=473µs

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

t = 0 t=473µs

Vapor volume fraction

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

t = 0 t=473µs

Vapor mass fraction

4-wave structure:

Rarefaction, contact, phase, shock

(5)

### Dodecane 2-phase problem: Sample solution

Allphysical quantities arediscon- tinuous across phase boundary

(6)

### Expansion wave problem: Cavitation test

Liquid-vapor mixture(αvapor = 102) for water with pliquid = pvapor = 1bar

Tliquid= Tvapor = 354.7284K< Tsat

ρvapor = 0.63kg/m3> ρsatvapor, ρliquid= 1150kg/m3> ρsatliquid gsat> gvapor > gliquid

Outgoing velocity u = 2m/s

← −~u ~u →

← Membrane

(7)

### Expansion wave problem: Sample solution

Cavitation pocket formation &

mass transfer

(8)

### Expansion wave problem: Sample solution

0 0.2 0.4 0.6 0.8 1

102 103 104

t = 0 t=3.2ms

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−2

−1.5

−1

−0.5 0 0.5 1 1.5 2

t = 0 t=3.2ms

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105

t = 0 t=3.2ms

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1 1.2 1.4x 10−4

t = 0 t=3.2ms

Vapor mass fraction

0 0.2 0.4 0.6 0.8 1

0 2 4 6 8 10 12x 104

t = 0 t=3.2ms

colorblue gv− gl(J/kg)

Equilibrium Gibbs free energy inside cavitation pocket

(9)

(10)

### Constitutive law: Metastable fluid

Stiffened gas equation of state (SG EOS) with Pressure

pk(ek, ρk)= (γk−1)ek−γkpk−(γk−1)ρkηk

Temperature

Tk(pk, ρk)= pk+ pk

k−1)Cvkρk

Entropy

sk(pk, Tk)= Cvklog Tkγk

(pk+ pk)γk1 + ηk Helmholtz free energy ak = ek−Tksk

Gibbs free energy gk = ak+ pkvk, vk = 1/ρk

(11)

### Metastable fluid: SG EOS parameters

Ref: Le Metayer et al. , Intl J. Therm. Sci. 2004

Fluid Water

Parameters/Phase Liquid Vapor

γ 2.35 1.43

p (Pa) 109 0

η (J/kg) −11.6 × 103 2030 × 103

η (J/(kg · K)) 0 −23.4 × 103

Cv (J/(kg · K)) 1816 1040

Fluid Dodecane

Parameters/Phase Liquid Vapor

γ 2.35 1.025

p (Pa) 4 × 108 0

η (J/kg) −775.269 × 103 −237.547 × 103

η (J/(kg · K)) 0 −24.4 × 103

Cv (J/(kg · K)) 1077.7 1956.45

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### Metastable fluid: Saturation curves

Assume two phases in chemicalequilibrium with equal Gibbs free energies (g1 = g2), saturation curveforphase transitionsis G(p, T )=A+B

T +ClogT+Dlog(p+p∞1)−log(p+p∞2) = 0 A= Cp1−Cp2+ η2−η1

Cp2−Cv2

, B= η1 −η2

Cp2−Cv2

C = Cp2−Cp1

Cp2−Cv2

, D= Cp1−Cv1

Cp2−Cv2

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### Metastable fluid: Saturation curves

Assume two phases in chemicalequilibrium with equal Gibbs free energies (g1 = g2), saturation curveforphase transitionsis G(p, T )=A+B

T +ClogT+Dlog(p+p∞1)−log(p+p∞2) = 0 A= Cp1−Cp2+ η2−η1

Cp2−Cv2

, B= η1 −η2

Cp2−Cv2

C = Cp2−Cp1

Cp2−Cv2

, D= Cp1−Cv1

Cp2−Cv2

or, from dg1 = dg2, we get Clausius-Clapeyron equation dp(T )

dT = Lh

T (v2−v1) Lh = T (s2−s1): latent heat of vaporization

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### Metastable fluid: Saturation curves (Cont.)

Saturation curves for water & dodecane in T ∈ [298, 500]K

300 350 400 450 500

0 5 10 15 20 25 30

water dodecane Pressure(bar)

300 350 400 450 500

0 500 1000 1500 2000 2500

water dodecane Latent heat of vaporization (kJ/kg)

300 350 400 450 500

10−4 10−3 10−2

water dodecane Liquid volume v1(m3/kg)

300 350 400 450 500

10−2 10−1 100 101 102 103

water dodecane Vapor volume v2(m3/kg)

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### Compressible 2-phase flow: 6-equation model

6-equation single-velocity 2-phase model with stiff mechanical relaxation of Saurel et al. (JCP 2009) reads

t1ρ1) + ∇ · (α1ρ1~u) = 0

t2ρ2) + ∇ · (α2ρ2~u) = 0

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇ (α1p1+ α2p2) = 0

t1ρ1e1) + ∇ · (α1ρ1e1~u) + α1p1∇ ·~u=µpI(p2−p1)

t2ρ2e2) + ∇ · (α2ρ2e2~u) + α2p2∇ ·~u=µpI(p1−p2)

tα1 + ~u · ∇α1 =µ (p1−p2)

µ: relaxation parameter for volume-transfer rate as p1 → p2; assume stiff µ → ∞ limit

pI: interfacial pressure,

pI = p1/Z1+ p2/Z2

1/Z1+ 1/Z2

, Zi = ρici

(16)

### 6 -equation model (Cont.)

Include conservation of mixture total energy also

tE + ∇ · (E~u + p~u) = 0

for purpose of maintaining numerical conservation of total energy

Phasic entropy equations for sk are αkρkTk

dsk

dt = µ (p1−p2)2 Zk

Z1+ Z2

≥0 for k = 1, 2, yielding nonnegative variationof mixture entropy

s = Y1s1+ Y2s2

Model is hyperbolicwith monotone speed of sound cf as c2f = Y1c21+ Y2c22, Yk= αkρk

ρ

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### 6 -equation model: Reduced model

6-equation model approaches to reduced 5-equation model asymptotically as µ → ∞

t1ρ1) + ∇ · (α1ρ1~u) = 0

t2ρ2) + ∇ · (α2ρ2~u) = 0

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇p= 0

tE + ∇ · (E~u +p~u) = 0

tα1+ ~u · ∇α1 =

 ρ2c22−ρ1c21 ρ1c211 + ρ2c222



∇ ·~u Mixture pressure p determined from total internal energy

ρe = α1ρ1e1(p, ρ1) + α2ρ2e2(p, ρ2)

Model is hyperbolicwith non-monotone sound speed cp: 1

ρc2p = α1

ρ1c21 + α2

ρ2c22

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### 6 -equation model: Reduced model (Cont.)

Volume-fraction equationis differential form ofpressure equilibriumcondition

p11, s1) = p22, s2)

Assume K = (ρ2c22−ρ1c21) / (ρ1c211+ ρ2c222) < 0, i.e., ρ2c22 < ρ1c21 (phase 1 less compressible)

Compaction effect (K ∇ · ~u > 0)

α1 increases when ∇ · ~u < 0 (compression or shock waves) Relaxation effect (K ∇ · ~u < 0)

α1 decreases when ∇ · ~u > 0 (expansion waves) No effect

α1 remains unchanged when ∇ · ~u = 0 (contacts)

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### p relaxation: Subcharacteristic condition

Mechanical equilibrium sound speed cp ≤ cf (frozen speed) 1

ρc2p =

2

X

k=1

αk

ρkc2k & ρc2f =

2

X

k=1

αkρkc2k

0 0.2 0.4 0.6 0.8 1

101 102 103 104

frozen p relax

αwater

cp&cf

Non-monotonic cp

difficulties in numerical solver, e.g., positivity- preserving in volume fraction

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### 6 -equation model: Phasic-total-energy-based

Alternative 6-equation model based on phasic total energy is

t1ρ1) + ∇ · (α1ρ1~u) = 0

t2ρ2) + ∇ · (α2ρ2~u) = 0

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇ (α1p1+ α2p2) = 0

t1E1) + ∇ · (α1E1~u + α1p1~u) + B (q, ∇q) = µpI(p2−p1)

t2E2) + ∇ · (α2E2~u + α2p2~u) − B (q, ∇q) = µpI(p1−p2)

tα1+ ~u · ∇α1 = µ (p1−p2)

B(q, ∇q) is non-conservative product (q: state vector) B= ~u · [Y1∇(α2p2) − Y2∇(α1p1)]

Model is hyperbolicwith monotonic speed of sound cf as well

& is basis for mixture-energy-consistent method

(21)

### 6 -equation model (Cont.)

6-equation model in compact form

tq + ∇ · f (q) + w (q, ∇q) = ψµ(q) where

q= [α1, α1ρ1, α2ρ2, ρ~u, α1E1, α2E2, α1]T f = [α1ρ1~u, α2ρ2~u, ρ~u ⊗ ~u + (α1p1+ α2p2)IN,

α1(E1+ p1) ~u, α2(E2+ p2) ~u, 0]T w= [0, 0, 0, B (q, ∇q) , −B (q, ∇q) , ~u · ∇α1]T

ψµ= [0, 0, 0, µpI(p2−p1) , µpI(p1−p2) , µ (p1−p2)]T

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### Relaxation scheme

Fractional-step method is employed to solve 6-equation model That is,

1. Non-stiff hyperbolic step

Solve hyperbolic system without relaxation sources

tq + ∇ · f (q) + w (q, ∇q) = 0 usingstate-of-the-art solver over time interval ∆t 2. Stiff mechanical relaxation step

Solve system of ordinary differential equations

tq = ψµ(q)

with initial solution from step 1 as µ → ∞

(23)

### Stiff mechanical relaxation step

Look for solution of ODEs in limit µ → ∞

t1ρ1) = 0

t2ρ2) = 0

t(ρ~u) = 0

t1E1) = µpI(p2−p1)

t2E2) = µpI(p1−p2)

tα1 =µ (p1−p2)

with initial condition q0 (solution after non-stiff hyperbolic step) & under mechanical equilibrium condition

p1 = p2 = p

(24)

### Stiff mechanical relaxation step (Cont.)

We find easily

t1ρ1) = 0 =⇒ α1ρ1 = α01ρ01

t2ρ2) = 0 =⇒ α2ρ2 = α02ρ02

t(ρ~u) = 0 =⇒ ρ~u = ρ0~u0

t1E1) = µpI(p2−p1) =⇒ ∂t(αρe)1 = −pItα1

t2E2) = µpI(p1−p2) =⇒ ∂t(αρe)2 = −pItα2

Integrating latter two equations with respect to time Z

t(αρe)k dt = − Z

pItαk dt

=⇒ αkρkek−α0kρ0ke0k = −¯pI αk−α0k

or

=⇒ ek−e0k = −¯pI 1/ρk−1/ρ0k

(use αkρk = α0kρ0k) Take p¯I = (p0I + p)/2or p, for example

(25)

### Stiff mechanical relaxation step (Cont.)

We find condition for ρk in p, k = 1, 2

Combining that with saturation condition for volume fraction α1+ α2 = α1ρ1

ρ1(p) + α2ρ2

ρ2(p) = 1

leads to algebraic equation (quadratic one with SG EOS) for relaxed pressure p

With that, ρk, αk can be determined & state vector q is updated from current time to next

(26)

### Stiff mechanical relaxation step (Cont.)

We find condition for ρk in p, k = 1, 2

Combining that with saturation condition for volume fraction α1+ α2 = α1ρ1

ρ1(p) + α2ρ2

ρ2(p) = 1

leads to algebraic equation (quadratic one with SG EOS) for relaxed pressure p

With that, ρk, αk can be determined & state vector q is updated from current time to next

Relaxed solution depends strongly on initial condition from non-stiff hyperbolic step

(27)

### Expansion wave problem: p relaxation

Mechanical-equilibrium solution at t = 3.2ms

0 0.2 0.4 0.6 0.8 1

103.02 103.03 103.04 103.05

t = 0 t=3.2ms

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−2

−1.5

−1

−0.5 0 0.5 1 1.5 2

t = 0 t=3.2ms

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

103 104 105

t = 0 t=3.2ms

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

t = 0 t=3.2ms

Vapor volume fraction

(28)

### Dodecane 2-phase Riemann problem: p relaxation

Mechanical-equilibrium solution at t = 473µs

0 0.2 0.4 0.6 0.8 1

100 101 102 103

t = 0 t=473µs

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−20 0 20 40 60 80 100 120 140 160

t = 0 t=473µs

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105 106 107 108

t = 0 t=473µs Pressure(bar)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

t = 0 t=473µs

Vapor volume fraction

(29)

### 6 -equation model: With heat & mass transfer

6-equation single-velocity 2-phase model with stiff mechanical, thermal, & chemical relaxationsreads

t1ρ1) + ∇ · (α1ρ1~u) =m˙

t2ρ2) + ∇ · (α2ρ2~u) =−m˙

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇ (α1p1 + α2p2) = 0

t1E1) + ∇ · (α1E1~u + α1p1~u) + B (q, ∇q) = µpI(p2−p1) + Q + gI

t2E2) + ∇ · (α2E2~u + α2p2~u) − B (q, ∇q) = µpI(p1−p2) − Q − gI

tα1+ ~u · ∇α1 =µ (p1−p2) + Q qI

+ m˙ ρI

Assume Q = θ (T2−T1), m = ν (g˙ 2−g1)

(30)

### 6 -equation with heat & mass transfer (Cont.)

µ, θ, ν → ∞: instantaneous exchanges (relaxation effects) 1. Volumetransfer via pressure relaxation: µ (p1−p2)

µexpresses rate toward mechanical equilibrium p1 → p2,

& isnonzero in all flow regimes of interest

2. Heattransfer via temperature relaxation: θ (T2−T1) θexpresses rate towards thermal equilibrium T1→ T2,

3. Mass transfer viathermo-chemical relaxation: ν (g2−g1) ν expresses rate towards diffusive equilibrium g1 → g2, &

isnonzero only at 2-phase mixture& metastable state Tliquid> Tsat

(31)

### 6 -equation with heat & mass transfer (Cont.)

Modified 6-equation model in compact form

tq + ∇ · f (q) + w (q, ∇q) = ψµ(q) + ψθ(q) + ψν(q) where

q= [α1, α1ρ1, α2ρ2, ρ~u, α1E1, α2E2, α1]T f = [α1ρ1~u, α2ρ2~u, ρ~u ⊗ ~u + (α1p1+ α2p2)IN,

α1(E1+ p1) ~u, α2(E2+ p2) ~u, 0]T w= [0, 0, 0, B (q, ∇q) , −B (q, ∇q) , ~u · ∇α1]T

ψµ= [0, 0, 0, µpI(p2−p1) , µpI(p1−p2) , µ (p1−p2)]T ψθ = [0, 0, 0, Q, −Q, Q/qI]T

ψν = [ ˙m, − ˙m, 0, gIm, −g˙ Im, ˙˙ m/ρI]T

(32)

### 6 -equation with heat & mass transfer (Cont.)

Flow hierarchy in 6-equation model: H. Lund (SIAP 2012)

6-eqns µ → ∞ θν → ∞ µθν → ∞

µθ → ∞

µν → ∞ θ → ∞

ν → ∞

(33)

### 6 -equation with heat & mass transfer (Cont.)

Stiff limits as µ → ∞, µθ → ∞, & µθν → ∞ sequentially

6-eqns µ → ∞ θν → ∞ µθν → ∞

µθ → ∞

µν → ∞ θ → ∞

ν → ∞

(34)

### Relaxation scheme: With heat & mass transfer

Continue from previous algorithm for 6-equation model with stiff mechanical relaxation, 2 sub-steps are included

3. Stiff thermal relaxation step

Solve system of ordinary differential equations

tq = ψµ(q) + ψθ(q)

4. Stiff thermo-chemical relaxation step

Solve system of ordinary differential equations

tq = ψµ(q) + ψθ(q) + ψν(q) Take solution from previous step as initial condition

(35)

### Relaxation scheme: Stiff solvers

1. Algebraic-basedapproach

Saurel et al. (JFM 2008), Zein et al. (JCP 2010), LeMartelot et al. (JFM 2013), Pelanti-Shyue (JCP 2014) Imposeequilibrium conditions directly, without making explicit of interface states pI, gI,. . .

2. Differential-based approach

Saurel et al. (JFM 2008), Zein et al. (JCP 2010) Imposedifferential of equilibrium conditions, require explicit of interface states pI, gI,. . .

3. Optimization-based approach (for mass transfer only) Helluy & Seguin (ESAIM: M2AN 2006), Faccanoni et al.(ESAIM: M2AN 2012)

(36)

### Stiff thermal relaxation step

Assume frozen thermo-chemical relaxation ν = 0, look for solution of ODEs in limits µ & θ → ∞

t1ρ1) = 0

t2ρ2) = 0

t(ρ~u) = 0

t1E1) =µpI(p2−p1) + θ (T2−T1)

t2E2) =µpI(p1−p2) + θ (T1−T2)

tα1 =µ (p1−p2) + θ qI

(T2−T1) under mechanical-thermal equilibrium conditons

p1 = p2 = p T1 = T2 = T

(37)

### Stiff thermal relaxation step (Cont.)

We find easily

t1ρ1) = 0 =⇒ α1ρ1 = α01ρ01

t2ρ2) = 0 =⇒ α2ρ2 = α02ρ02

t(ρ~u) = 0 =⇒ ρ~u = ρ0~u0

tkEk) = θ qI

(T2−T1) =⇒ ∂t(αρe)k= qItαk

Integrating latter two equations with respect to time Z

t(αρe)k dt = Z

qItαk dt

=⇒ αkρkek−α0kρ0ke0k = −q¯I αk−α0k Take q¯I = (qI0+ qI)/2 or qI, for example, & find algebraic equation for α1, by imposing

T2 e2, α02ρ02/(1 − α1) − T1 e1, α01ρ011 = 0

(38)

### Stiff thermal relaxation step: Algebraic approach

Impose mechanical-thermal equilibrium directly to 1. Saturation condition

Y 1

ρ1(p,T)+ Y2

ρ2(p,T) = 1 ρ0 2. Equilibrium of internal energy

Y1 e1(p,T) + Y2 e2(p,T) = e0 Give 2 algebraic equations for2 unknowns p & T

For SG EOS, it reduces to single quadraticequation for p&

explicit computation of T: 1

ρT = Y1

1−1)Cv1

p+ p1

+ Y2

2−1)Cv2

p+ p2

(39)

### Reduced model: Thermal relaxation step

Reduced model after thermal relaxation step is (Saurel et al. 2008, Fl˚atten et al. 2010)

tρ + ∇ · (ρ~u) = 0

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇p = 0

tE + ∇ · (E~u + p~u) = 0

t(ρY1) + ∇ · (ρY1~u) = 0 Mixture entropy ρs satisfy

t(ρs) + ∇ · (ρs~u) = θ



1 + peq

qI

 (T2−T1)2 T1T2

≥0 Mechanical-thermal equilibriumsound speed cpT satisfies

1

ρc2pT = 1 ρc2p+T

 Γ2

ρ2c22

− Γ1

ρ1c21

2  1 α1ρ1cp1

+ 1

α2ρ2cp2



(40)

### Stiff thermo-chemical relaxation step

Look for solution of ODEs in limits µ, θ, & ν → ∞

t1ρ1) =ν (g2 −g1)

t2ρ2) =ν (g1 −g2)

t(ρ~u) = 0

t1E1) =µpI(p2−p1) + θ (T2−T1) + ν (g2−g1)

t2E2) =µpI(p1−p2) + θ (T1−T2) + ν (g1−g2)

tα1 =µ (p1−p2) + θ qI

(T2−T1) + ν ρI

(g2−g1) under mechanical-thermal-chemical equilibrium conditons

p1 = p2 = p T1 = T2 = T

g1 = g2

(41)

### Stiff thermal-chemical relaxation step (Cont.)

In this case, states remain in equilibrium are

ρ = ρ0, ρ~u = ρ0~u0, E = E0, e = e0 but αkρk6= α0kρ0k & Yk6= Yk0, k = 1, 2

Impose mechanical-thermal-chemical equilibrium to 1. Saturation condition for temperature

G(p,T) = 0 2. Saturation condition for volume fraction

Y1

ρ1(p,T)+ Y2

ρ2(p,T) = 1 ρ0 3. Equilibrium of internal energy

Y1 e1(p,T) +Y2 e2(p,T) = e0

(42)

### Stiff thermal-chemical relaxation step (Cont.)

From saturation condition for temperature G(p,T) = 0 we get T in terms of p, while from

Y1

ρ1(p,T) + Y2

ρ2(p,T) = 1 ρ0

&

Y1 e1(p,T) +Y2 e2(p,T) = e0 we obtain algebraic equation for p

Y1 = 1/ρ2(p) − 1/ρ0

1/ρ2(p) − 1/ρ1(p) = e0−e2(p) e1(p) − e2(p) which is solved by iterative method

(43)

Stiff thermal-chemical relaxation step (Cont.)

Having known Yk & p, T can be solved from, e.g., Y1 e1(p,T) +Y2 e2(p,T) = e0 yielding updateρk & αk

Feasibility of solutions, i.e., positivity of physical quantities ρk, αk, p, & T , for example

Employ hybridmethod i.e., combination of above method with differential-based approach (not discuss here), when it becomes necessary

(44)

### Reduced model: Thermo-chemical relaxation step

Reduced model after thermo-chemical relaxation is

homogeneous equilibrium model (HEM) that follows standard mixture Euler equation

tρ + ∇ · (ρ~u) = 0

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇p = 0

tE + ∇ · (E~u + p~u) = 0 This gives local resolution at interface only

Mixture entropy ρs satisfies

t(ρs) + ∇ · (ρs~u) = ν(g2 −g1)2 Teq

≥0 Speed of sound cpT g satisfies

1

ρc2pT g = 1 ρc2p +T

"

α1ρ1

Cp1

 ds1

dp

2

+ α2ρ2

Cp2

 ds2

dp

2#

(45)

### Equilibrium speed of sound: Comparison

Sound speeds follow subcharacteristic condition cpT g ≤cpT ≤cp ≤cf

Limit of sound speed

αlimk→1cf = lim

αk→1cp = lim

αk→1cpT = ck, lim

αk→1cpT g 6= ck

0 0.2 0.4 0.6 0.8 1

10−1 100 101 102 103 104

frozen p relax pT relax pTg relax

αwater cpTg,cpT,cp&cf

(46)

### 5 -equation model: liquid-vapor phase transition

Modelling phase transitionin metastable liquids Saurel et al. (JFM 2008) proposed

t1ρ1) + ∇ · (α1ρ1~u) =m˙

t2ρ2) + ∇ · (α2ρ2~u) =−m˙

t(ρ~u) + ∇ · (ρ~u ⊗ ~u) + ∇p = 0

tE + ∇ · (E~u + p~u) = 0

tα1+ ∇ · (α1~u) =α1

s

Ks1∇ ·~u+ 1 qI

Q+ 1 ρI

˙ m K¯s= α1

Ks1 + α2

Ks2

1

, Ksι = ριc2ι qI = Ks1

α1

+ Ks2 α2

   Γ1

α1

2

α2



, Q = θ(T2−T1) ρI = Ks1

α1

+Ks2 α2

   c21 α1

+ c22 α2



, m = ν(g˙ 2−g1)

(47)

### Expansion wave problem: p-pT relaxation

Mechanical-thermal-equilibrium solution at t = 3.2ms

0 0.2 0.4 0.6 0.8 1

103.03 103.04 103.05

t = 0 t=3.2ms

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−2

−1.5

−1

−0.5 0 0.5 1 1.5 2

t = 0 t=3.2ms

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105

t = 0 t=3.2ms

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

−1

−0.5 0 0.5 1

t = 0 t=3.2ms

Tv− Tl(K)

0 0.2 0.4 0.6 0.8 1

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

t = 0 t=3.2µs

Vapor volume fraction

0 0.2 0.4 0.6 0.8 1

−1

−0.5 0 0.5 1 1.5

t = 0 t=3.2ms

Vapor mass fraction

(48)

### Expansion wave problem (Cont.)

Comparison p-, pT - & p-pT g-relaxation solution at t = 3.2ms

0 0.2 0.4 0.6 0.8 1

750 800 850 900 950 1000 1050 1100 1150

p relax pT relax pTg relax

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−2

−1.5

−1

−0.5 0 0.5 1 1.5 2

p relax pT relax pT relax

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

103 104 105

p relax pT relax pTg relax

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

−5

−4

−3

−2

−1 0 1x 106

p relax pT relax pTg relax

gv− gl(103K)

0 0.2 0.4 0.6 0.8 1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

p relax pT relax pT relax

Vapor volume fraction

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1 1.2 1.4x 10−4

p relax pT relax pTg relax

Vapor mass fraction

(49)

### Expansion wave problem: ~u = 500m/s

0 0.2 0.4 0.6 0.8 1

0 2 4 6 8 10x 104

p relaxation p−pT p−pT−pTg p−pTg

p[Pa]

0 0.2 0.4 0.6 0.8 1

−500 0 500

p relaxation p−pT p−pT−pTg p−pTg

u[m/s

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

p relaxation p−pT p−pT−pTg p−pTg

αv

0 0.2 0.4 0.6 0.8 1

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

p relaxation p−pT p−pT−pTg p−pTg

Yv

(50)

### Dodecane 2-phase problem: p-pT relaxation

Mechanical-thermal-equilibrium solution at t = 473µs

0 0.2 0.4 0.6 0.8 1

100 101 102 103

t = 0 t=473µs

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−20 0 20 40 60 80 100 120 140 160

t = 0 t=473µs

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105 106 107 108

t = 0 t=473µs

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

−2 0 2 4 6 8 10 12x 105

t = 0 t=473µs

Tv− Tl(K)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

t = 0 t=473µs

Vapor volume fraction

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

t = 0 t=473µs

Vapor mass fraction

(51)

### Dodecane 2-phase Riemann problem (Cont.)

Comparison p-,pT -& p-pT g-relaxation solution at t = 473µs

0 0.2 0.4 0.6 0.8 1

0 100 200 300 400 500

p relax pT relax pTg relax

Density (kg/m3)

0 0.2 0.4 0.6 0.8 1

−50 0 50 100 150 200 250 300 350

p relax pT relax pT relax

Velocity (m/s)

0 0.2 0.4 0.6 0.8 1

104 105 106 107 108

p relax pT relax pTg relax

Pressure(bar)

0 0.2 0.4 0.6 0.8 1

−6

−4

−2 0 2 4 6x 108

p relax pT relax pTg relax

gv− gl(103K)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

p relax pT relax pT relax

Vapor volume fraction

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

p relax pT relax pTg relax

Vapor mass fraction

(52)

### High-pressure fuel injector

Inject fluid: Liquid dodecane containing small amountαvapor

Pressure & temperature are in equilibrium with p = 108 Pa & T = 640K

Ambient fluid: Vapor dodecanecontaining small amount αliquid

Pressure & temperature are in equilibrium with p = 105 Pa & T = 1022K

liquid →

vapor

(53)

v,l

### = 10

−4

Mixture density Mixture pressure

(54)

v,l

### = 10

−2

Mixture density Mixture pressure

(55)

### High-pressure fuel injector (Cont.)

Vapor volume fraction: αv,l = 10−4 (left) vs. 10−2 (right)

(56)

### High-pressure fuel injector (Cont.)

Vapor mass fraction: αv,l = 10−4 (left) vs. 10−2 (right)

(57)

### High-pressure fuel injector: Remark

Numerical solver (to be discussed) uses 400 × 200 uniform Cartesian grid

6-equation single-velocity two-phase model with stiff mechanical relaxation

Consitutive law is stiffened gasequation of state

Model is solved instiff limit towards equilibrium pressure pliquid = pvapor, while admitting different temperatures Tliquid6= Tvapor & entropies sliquid6= svapor

Observation:

Higher αv,l in fluid mixture, higer volume transfer expansion (Is this correct statement ? If so, to what extent)

(58)

### Fuel injector: p-pT relaxation

Vapor mass fraction: αv,l = 10−4 (left) vs. 10−2 (right)

(59)

### Fuel injector: p-pT -pT g relaxation

Vapor volume fraction: αv,l = 10−4 (left) vs. 10−2 (right)

(60)

### References

M. Pelanti and K.-M. Shyue. A mixture-energy-consistent 6-equation two-phase numerical model for fluids with interfaces, cavitation and evaporation waves. J. Comput.

Phys., 259:331–357, 2014.

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.

R. Saurel, F. Petitpas, and R. A. Berry. Simple and efficient relaxation methods for interfaces separating compressible fluids, cavitating flows and shocks in multiphase mixtures. k J. Comput. Phys.,

228:1678–1712, 2009.

A. Zein, M. Hantke, and G. Warnecke. Modeling phase transition for compressible two-phase flows applied to metastable liquids. J. Comput. Phys., 229:2964–2998, 2010.

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### Thank you

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