A homogeneous model for compressible and immiscible two...

IntroductionThe thermodynamical behaviour of the mixture

The full multiphase-flow modelSpecificity for steam/water flows

Some remarks

A homogeneous model for compressible and immiscibletwo-phase flows: 1. Closure laws

Olivier Hurisse†

†EDF R&D MFEE, 6 quai Watier, 78400 Chatou, [email protected]

GDR MANU,Roscoff, July 2-6, 2018.

Olivier Hurisse† A homogeneous two-phase flow model. 1



Some remarks

Plan

1 Introduction

2 The thermodynamical behaviour of the mixturePhasic quantitesThe mixture entropyThe Gibbs relation for the mixtureModeling the exchanges between the phases

3 The full multiphase-flow modelThe closed system of equationsThe intensive form of the model

4 Specificity for steam/water flowsSpecificity of the steam/water mixtureThe heating of the phases

5 Some remarksOther possible relaxation source terms ?Energy fraction or entropy fraction ?




Some remarks

Plan

1 Introduction








Some remarks

Outline of this part:

We first focus on the thermodynamical behaviour of a small amount of an-phase mixture (n ≥ 1).

Then the dynamical behaviour of this mixture volume is described (alwaysfor n ≥ 1 phases).

At last, the specificity of two-phase steam-liquid mixture is introduced(n = 2).

��

��

Liq.

Vap.

Motion

Heat

Isolated system

Vap.

Liq.

External forces

First step Second step




Some remarks

Phasic quantitesThe mixture entropyThe Gibbs relation for the mixtureModeling the exchanges between the phases

Plan

1 Introduction








Some remarks


Plan

1 Introduction








Some remarks


For each phase k we have the following extensive quantities [Callen,85]:

a volume Vk (in m3);a mass Mk (in kg);a energy Ek (in J).

Then for the mixture we have:

the mixture volume V = ∑k Vk (since the phases are immiscible !);the mixture mass M = ∑k Mk ;and the mixture internal energy E = ∑k Ek .

Remark: The surface tension and the topological arrangement of the phases arenot taken into account.

Within each phase we assume that an extensive entropy ηk is given (an EOS)on (R+)3:

Wk 7→ ηk(Wk) (in J/K), where Wk = (Vk ,Mk ,Ek).

We assume that the entropy ηk is such that :

Wk 7→ ηk(Wk) is C2;Wk 7→ ηk(Wk) is concave;∀a ∈ R+, ∀Wk ∈ (R+)3, ηk(aWk) = aηk(Wk);

∀Wk ,∂ηk

∂Ek |Vk ,Mk> 0.




Some remarks


If we assume that the “Classical Irreversible Thermodynamic” (CIT) holds, thefollowing Gibbs relation is fulfiled for each phase:

Tkd (ηk) = d (Ek) +Pkd (Vk)−µkd (Mk) ,

where we have:

Tk is the temperature within phase k, with 1Tk

= ∂ηk

∂Ek |Vk ,Mk;

Pk is the pressure of phase k, with PkTk

= ∂ηk

∂Vk |Mk ,Ek;

and µk is the Gibbs enthalpy of phase k, with Mkµk = Ek +PkVk −Tkηk .

Remark: Since we have assumed that ∂ηk

∂Ek |Vk ,Mk> 0, we get that Tk > 0.

The thermodynamical behaviour of each phase is now defined.

Let’s turn to the thermodynamical behaviour of the mixture.




Some remarks


Plan

1 Introduction








Some remarks


We assume that the extensive mixture-entropy η, is the sum of the extensivephasic-entropies (surface tension is neglected):

η(W ) = ∑k

ηk(Wk), where W = (W1,W2, ...).

Thanks to the properties of the phasic entropies ηk , one can prove that

W 7→ η(W ) is C2;

W 7→ η(W ) is concave on (R+)3n;

∀a ∈ R+, ∀W ∈ (R+)3n, η(aW ) = aη(W ).

Moreover, the restriction η̃ of η on H (M )⊂ (R+)3n is strictly concave:

H (M ) =

{W ∈ (R+)3n;∑

k

Mk = M

}.




Some remarks


Plan

1 Introduction








Some remarks


The operator d . of the previous slides can be seen as a derivative along thestreamlines of the flow:

dφk “ = “

(∂φk

∂ t+uk

∂φk

∂x

)dt,

where uk is the velocity of the phase k.

Since we have assumed that uk = U for all k, the operator d . does notdepend on k.

We can thus write:dη = ∑

k

dηk ,

and we can then exhibit a Gibbs relation for this mixture entropy using thephasic Gibbs relations.




Some remarks


We then have:

dη = ∑k

dηk ⇔ dη = ∑k

(1

TkdEk +

Pk

TkdVk −

µk

TkdMk

),

which by using the chain-rule dφk = φ d(φk/φ) + φk/φ dφ yields:

dη = ∑k

(EkE

1Tk

)dE + ∑k

(VkV

PkTk

)dV −∑k

(MkM

µk

Tk

)dM

+∑k

(E 1

Tkd(

EkE

)+V Pk

Tkd(

VkV

)−M µk

Tkd(

MkM

)).

We can then identify:

the mixture temperature T as 1T = ∑k

(EkE

1Tk

);

the mixture pressure P such that PT = ∑k

(VkV

PkTk

);

and the mixture Gibbs enthalpy µ such that µ

T = ∑k

(MkM

µk

Tk

).




Some remarks


The Gibbs relation for the mixture is finally:

Tdη−(dE +PdV −µdM )︸︷︷︸”external exchange”

=T ∑k

(E

Tkd

(EkE

)+

V Pk

Tkd

(Vk

V

)−M µk

Tkd

(Mk

M

))︸︷︷︸

exchange between the phases

.

In order to write a model, we need to define the following derivatives:

dE , dV , dM , d

(EkE

), d

(Vk

V

), d

(Mk

M

).

Let’s start by the terms for the exchange between the phases, which representthe thermodynamical disequilibrium effects




Some remarks


Plan

1 Introduction








Some remarks


We assume in this subsection that the mixture is a closed system, that is :

dE = dV = dM = 0.

We then define D(V ,M ,E )⊂H (M ) as:

D(V ,M ,E ) =

{W ; W ∈H (M ),∑

k

Vk = V ,∑k

Ek = E

}.

We can thus consider the entropy η̃ which is strictly concave on the closedbounded set D(V ,M ,E ).

Hence, we get the two following properties.

(i) There exist a unique point W = (V k ,M k ,E k)k for which η̃ is maximumon D(V ,M ,E ).

(ii) At any point W in D(V ,M ,E ), the tangent plane to η̃ at W is above η̃.

So that we have the inequality:

η̃(W )≤ η̃(W ) + ∇W (η)(W −W

).




Some remarks


If we assume the following models for the exchange term [Barberon-Helluy,CAF, 2005]:

d(

VkV

)= V k−Vk

λ V dt;

d(

MkM

)= M k−Mk

λ M dt;

d(

EkE

)= E k−Ek

λ E dt;

where λ > 0 is a time scale, we have the Gibbs relation on D(V ,M ,E ):

d η̃ = (∇W (η)dW ) = ∇W (η)

(W −W

λ

)dt.

Due to the concavity of η̃ this yields:

d η̃ = ∇W (η)

(W −W

λ

)dt ≥ 1

λ(η̃(W )− η̃(W )) dt,

and since W corresponds to the maximum of the entropy on D(V ,M ,E ):

d η̃ ≥ 1

λ(η̃(W )− η̃(W )) dt ≥ 0,




Some remarks


As a consequence, the models:d(

VkV

)= V k−Vk

λ V dt;

d(

MkM

)= M k−Mk

λ M dt;

d(

EkE

)= E k−Ek

λ E dt;

ensure that the entropy of the mixture η̃ increases in time when the mixture isa closed system (i.e. dE = dV = dM = 0).

They thus fulfill the second law of thermodynamics.

Remark: These models represent one choice among others. See the last sectionfor a discussion on that point.




Some remarks

The closed system of equationsThe intensive form of the model

Plan

1 Introduction








Some remarks


Plan

1 Introduction








Some remarks


From the previous section, it remains to define dE , dV , and dM .

This is done in a classical manner with the following assumptions:

the mass is conserved along the streamlines (so that we are in H (M )):

dM = 0;

the variation of volume V is due to the divergence of the velocity field U:

dV = V ∇x · (U)dt;

the variation of the velocity U follows the Newton’s law (we only considerthe force due to the mixture pressure here):

d(MU) =−V ∇x (P)dt;

the first law of thermodynamics applies to the energy E :

dE =−PdV +Qdt.

The term Q will be discussed in an other section, and it is set to zero until thatsection.




Some remarks


Plan

1 Introduction








Some remarks


We first introduce a notation to deal with the dimension of the variables: forinstance for a mass M which is in kg , we set M = M ′Ikg , where M ′ has nodimension and Ikg is equal to 1 kg .

We then apply this to the phasic entropy ηk (in J/K) and we use the property:

∀a ∈ R+,ηk(aVk ,aMk ,aEk) = aηk(Vk ,Mk ,Ek).

We have:

1

Mkηk(Vk ,Mk ,Ek) =

1

Ikgηk

(Vk

M ′k

,Mk

M ′k

,EkM ′

k

)=

1

Ikgηk

(Vk

MkIkg ,Ikg ,

EkMk

Ikg)




Some remarks


We can thus define a specific entropy sk (in J/K/kg), a specific volume τk (inm3/kg) and a specific internal energy ek (in J/kg) as:

sk (τk ,ek) =1

Ikgηk

(τk Ikg ,Ikg ,ek Ikg

); τk =

Vk

Mk; ek =

EkMk

.

This specific entropy sk corresponds to the entropy for one unit of mass, sothat sk is strictly concave with respect to (τk ,ek).

Moreover, we have for the temperature:

1

Tk=

∂ηk

∂Ek |Vk ,Mk

=∂ (ηk/Mk)

∂ (Ek/Mk) |Vk ,Mk

=∂ sk∂ek |τk

,

and for the pressure:Pk

Tk=

∂ sk∂τk |ek

.

By dividing η by M , we can also define a specific mixture entropy s as:

s

(M1

M,τ1,e1,

M2

M,τ2,e2, ...

)= ∑

k

Mk

Msk (τk ,ek) .




Some remarks


We define the volume fractions αk , the mass fractions yk and the energyfractions zk as:

αk =Vk

V, yk =

Mk

M, zk =

EkE

.

We obviously have ∑k αk = ∑k yk = ∑k zk = 1.

Remark: ∑k αk = 1 because of the phases are immiscible !

We define the equilibrium fractions associated with the equilibrium state W :

αk =V k

V, yk =

M k

M, zk =

E k

E.

The specific volume, mixture density and specific internal energy are:

τ =V

M= ∑

k

ykτk , ρ =1

τ= ∑

k

αkρk , e =E

M= ∑

k

ykek .

Then after some calculations ...




Some remarks


dM = 0; ←→ none

d(

VkV

)= V k−Vk

λ V dt; ←→ ∂

∂ t(ραk) + ∇x · (ρUαk) = ρ

αk −α

λ;

d(

MkM

)= M k−Mk

λ M dt; ←→ ∂

∂ t(ρyk) + ∇x · (ρUyk) = ρ

yk −y

λ;

d(

EkE

)= E k−Ek

λ E dt; ←→ ∂

∂ t(ρzk) + ∇x · (ρUzk) = ρ

zk − z

λ;

dV = V ∇x · (U)dt; ←→ ∂

∂ t(ρ) + ∇x · (ρU) = 0;

d(MU) =−V ∇x (P)dt; ←→ ∂

∂ t(ρU) + ∇x · (ρU⊗U +P I3) = 0;

dE =−PdV ; ←→ ∂

∂ t(ρE) + ∇x · (U(ρE +P)) = 0,

with the total energy E = e +U2/2.

We get a multiphase flow model in conservative form, that allows to deal withimmiscible components and accounts for the thermodynamical disequilibriumbetween the phases.




Some remarks

Specificity of the steam/water mixtureThe heating of the phases

Plan

1 Introduction








Some remarks


Plan

1 Introduction








Some remarks


In the following, we set:

phase 1 for the steam

phase 2 for the liquid water

Since we still have:

V = V1 +V2⇒ 0 = d(

V1V

)+d

(V2V

),

M = M1 +M2⇒ 0 = d(

M1M

)+d

(M2M

),

E = E1 +E2⇒ 0 = d(

E1E

)+d

(E2E

),

and thanks to the closure laws for dV , dM and dE , the Gibbs relation for themixture:

Td η̃−(dE +PdV −µdM )︸︷︷︸”external exchange”=0 !

=T ∑k

(E

Tkd

(EkE

)+

V Pk

Tkd

(Vk

V

)−M µk

Tkd

(Mk

M

))︸︷︷︸

exchange between the phases

.

becomes:

d η̃ = V(P1T1− P2

T2

)d(

V1V

)+M

(µ2

T2− µ1

T1

)d(

M1M

)+E

(1T1− 1

T2

)d(

E1E

).




Some remarks


Hence, when the equilibrium between the three phases is reached, i.e. whend η̃ = 0, i.e. when W = W , we have:

P1(V1,M1,E1) = P2(V2,M2,E2),T1(V1,M1,E1) = T2(V2,M2,E2),µ1(V1,M1,E1) = µ2(V2,M2,E2),

with the constraints thatV1 +V2 = V ,

M1 +M2 = M ,

E1 +E2 = E ,

where V , M and E are fixed and known because we are in D(V ,M ,E ).

If there is no solution for this system of equations, it means that theequilibrium state for (V ,M ,E ) corresponds to a single phase state:

pure liquid if η2(V ,M ,E ) > η1(V ,M ,E )

pure vapor if η1(V ,M ,E ) > η2(V ,M ,E )




Some remarks


This pressure/temperature/Gibbs enthalpy equality defines the saturationcurve. Indeed, by assuming a change of variable for the Gibbs enthalpy, we getat equilibrium:

P1 = P2,T1 = T2,µ1(P1,T1) = µ2(P2,T2).

The saturation curve is then defined by the relation

µ1(P,T ) = µ2(P,T ),

which implicitely defines Psat(T ) and Tsat(P) as:

T 7→ Psat(T ), such that µ1(Psat(T ),T ) = µ2(Psat(T ),T );P 7→ Tsat(P), such that µ1(P,Tsat(P)) = µ2(P,Tsat(P)).

On this saturation curve, the two phases co-exist.




Some remarks


Plan

1 Introduction








Some remarks


We may have to account for some external heating of the phases (Q was set tozero previously).

We assume that for each phase k:

(i) the specific volume of each phase is constant, d(τk) = 0;(ii) the partial mass of each phase is constant, d(αkρk) = 0;(iii) the internal energy of each phase is such that, d(αkρkek) = αkρkqkdt;

where qk is the specific power (in J/kg/s) received by phase k.

Then we get that the mixture receives the heat Q̃:

d(ρe) = ∑k

αkρkqkdt = ρQ̃dt, and Q̃ = ∑k

ykqk ,

and that the energy fraction of phase k, zk = (αkρkek)/(ρe), is:

ρe d(zk) = αkρkqkdt− zkρQ̃dt = ρ(ykqk − zk Q̃)dt.




Some remarks


Hence, when the phases are heated by an external source qk , system ofequations becomes:

∂

∂ t(ραk) + ∇x · (ρUαk) = ρ

αk −α

λ, k = 1..3,

∂

∂ t(ρyk) + ∇x · (ρUyk) = ρ

yk −y

λ, k = 1..3,

∂

∂ t(ρzk) + ∇x · (ρUzk) = ρ

zk − z

λ+ ρ

ykqk − zk Q̃

e, k = 1..3,

∂

∂ t(ρ) + ∇x · (ρU) = 0,

∂

∂ t(ρU) + ∇x ·

(ρU2 +PI3

)= 0,

∂

∂ t(ρE) + ∇x · (U(ρE +P)) = ρQ̃,

(1)

Remark: If we omit the additionnal source term on zk , we have:

qk =zkyk

Q̃ =⇒ d(αkρkek) = αkρkqkdt = ρzk Q̃dt,

which means that the heat is dispatched between the phases with respect tothe energy fractions !




Some remarks

Other possible relaxation source terms ?Energy fraction or entropy fraction ?

Plan

1 Introduction








Some remarks


Plan

1 Introduction








Some remarks


At fixed τ and e, they must increase the specific mixture-entropy s:

ds = τ

(P1T1− P2

T2

)dα +

(µ2

T2− µ1

T1

)dy + e

(1T1− 1

T2

)dz .

It can be shown that (α,y ,z) 7→ s(α,y ,z ,τ,e) is strictly concave. Hencestarting from a point (α0,y0,z0), the set

C(τ,e)(α0,y0,z0) ={

(α,y ,z) ∈ [0,1]3;s(α,y ,z ,τ,e) >= s(α0,y0,z0,τ,e)}

is convex.Moreover, (α,y ,z) only depends on (τ,e) and (α,y ,z) ∈ C(τ,e)(α0,y0,z0).

The model for the source terms, i.e. for dα, dy , dz must be such that thepath followed by (α,y ,z):

begins at (α0,y0,z0) at t = 0;

ends at (α,y ,z) when t→ ∞;

lies in C(τ,e)(α0,y0,z0);

and is such that ds ≥ 0 along the path.




Some remarks


(α , y, z)

(α0 , y , z )0 0−equilibrium

µ

(τC

,e)(α0 , y , z )0 0

T−equilibrium

P−equi

libriu

m

and e are fixed)(τ

y

z

α

Model chosen here:

Since the equilibrium state does only depend on τ and e:

dα =α−α

λdt ←→ α(t) = βα0 + (1−β )α, with β = e−

∫ t0 1/λds .

This solution is very simple BUT there is only one parameter for the return toequilibrium.




Some remarks


(α , y, z)


µ

(τC

,e)(α0 , y , z )0 0

T−equilibrium

P−equi

libriu

m

and e are fixed)(τ

y

z

α

”Two-fluid” model:

dα = Kα

(P1T1− P2

T2

)dt, Kα > 0;

dy = Ky

(µ2

T2− µ1

T1

)dt, Ky > 0;

dz = Kz

(1T1− 1

T2

)dt, Kz > 0.

It is then possible to specify different time-scales for the pressure equilibrium,temperature equilibrium and Gibbs potential equilibrium.

BUT, in practice, numerical difficulties may arise and in particularvanishing-phase cases are difficult to handle.




Some remarks


(α , y, z)


µ

(τC

,e)(α0 , y , z )0 0

T−equilibrium

P−equi

libriu

m

and e are fixed)(τ

y

z

α

Instantaneous pressure-equilibrium (incomplete proposal):

We first go from (α0,y0,z0) to a point (α ′0,y′0,z′0) such that:

s(α ′0,y′0,z′0,τ,e) = s(α0,y0,z0,τ,e),

P1(α ′0,y′0,z′0,τ,e) = P2(α ′0,y

′0,z′0,τ,e),

y ′0 = y0.

Then, go from (α ′0,y′0,z′0) to (α,y ,z) while keeping the P−equilibrium.

Similar procedures could be proposed for T , µ or for P/T , T/µ, ...

BUT, does this proposal fulfil the criterion ds > 0 ? does (α ′0,y′0,z′0) exist ? Is

it unique ?Olivier Hurisse† A homogeneous two-phase flow model. 40



Some remarks


(α , y, z)


µ

(τC

,e)(α0 , y , z )0 0

T−equilibrium

P−equi

libriu

m

and e are fixed)(τ

y

z

α

Just for fun, a stochastic procedure:

initialization : (α ′,y ′,z ′) := (α0,y0,z0) and C := C(τ,e)(α0,y0,z0);

then for t→ t + δ t : choose randomly a point (α ′,y ′,z ′) in the interior ofC , and compute C := C(τ,e)(α ′,y ′,z ′)

continue until vol(C) > 0.

Convergence towards (α,y ,z) is ensured by the strict concavity of s and thechoice of the new points inside C (no need to compute (α,y ,z) !).




Some remarks


Plan

1 Introduction








Some remarks


energy fraction entropy fraction

(τk ,ek) 7→ sk(τk ,ek) (τk ,sk) 7→ ek(τk ,sk)

αk , yk , zk = ykek/e αk , yk , βk = yk sk/s

dsk = dek/Tk +Pk/Tkdτk dek = Tkdsk −Pkdτk

P/T = ∑αkPk/Tk P = ∑αkPk

1/T = ∑zk/Tk T = ∑βkTk

c2 =“complex formula” c2 = ∑ykc2k


A homogeneous model for compressible and immiscible two...

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