Pressure laws and fast Legendre transform - Philippe LeFloch

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Introduction Legendre transform Mixtures thermodynamics Fast Legendre Transform Numerical experiments Conclusion Pressure laws and fast Legendre transform Philippe HELLUY 1 , Hélène MATHIS 1 1 Université de Strasbourg, IRMA Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

Transcript of Pressure laws and fast Legendre transform - Philippe LeFloch

Page 1: Pressure laws and fast Legendre transform - Philippe LeFloch

IntroductionLegendre transform

Mixtures thermodynamicsFast Legendre Transform

Numerical experimentsConclusion

Pressure laws and fast Legendre transform

Philippe HELLUY1, Hélène MATHIS1

1Université de Strasbourg, IRMA

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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IntroductionLegendre transform

Mixtures thermodynamicsFast Legendre Transform

Numerical experimentsConclusion

Motivations

Euler equations for a compressible fluid

∂tρ +∂x(ρu) = 0,

∂t(ρu)+∂x(ρu2+π) = 0,∂t(ρQ)+∂x((ρQ +π)u) = 0.

The pressure is related to e = Q−u2/2 and τ = 1/ρ by a pressurelaw π = p(τ,e). It has to satisfy some convexity properties in orderthat the Euler system is stable (hyperbolic).General framework for constructing pressure laws of fluid mixtureswith phase transition?

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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IntroductionLegendre transform

Mixtures thermodynamicsFast Legendre Transform

Numerical experimentsConclusion

Outlines

1 Legendre transform

2 Mixtures thermodynamics

3 Fast Legendre Transform

4 Numerical experiments

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Mixtures thermodynamicsFast Legendre Transform

Numerical experimentsConclusion

Legendre transform

Let f : Rn→ R∪+∞. In the following, we always suppose that1 domf := x , f (x)<+∞ 6= /0;2 f is minorized by an affine function.

The Legendre transform (or conjugate) f ∗ of f is defined by

f ∗(s) = supx

(s · x− f (x)) .

The conjugate f ∗ also satisfies the conditions 1 and 2.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Mixtures thermodynamicsFast Legendre Transform

Numerical experimentsConclusion

Graphical interpretation

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Example: Conjugate of a regular convex function

If f : Rn→ R is a C 1 strictly convex function with

f (x)‖x‖

→‖x‖→∞

∞,

thenf ∗(s) = sx− f (x), with ∇f (x) = s.

We deduce that∇f ∗(s) = x ⇔ ∇f (x) = s.

It is then easy to check that in this case the Legendre transform isinvolutive

f ∗∗ = f .

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Example: enthalpy

In thermodynamics, the energy E is a function of the volume Vand the entropy S . The temperature is defined by T = ∂SE andthe pressure by p =−∂V E . The partial Legendre transform of Ewith respect to V is formally

E ∗,V = V ′V −E , with V ′ = ∂V E =−p.

ThusE ∗,V =−(E +pV ) =−H

is the opposite of the enthalpy (expressed as a function of (−p,S)).And we have the relation

TdS = dH− vdp.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Main properties [HL01]

f ∗ is convex and lower semi-continuous (lsc).f ∗∗ = co(f ) is the lsc convex enveloppe (or convex hull) of f

∀x ∈ Rn, co(f )(x) = supg(x),g affine function, g ≤ f .

(inf i fi )∗ = supi f ∗i . A useful consequence is thatmax(f ∗,g∗) = (comin(f ,g))∗.For two convex functions f and g , we define theinf-convolution operation by

f g(x) = infy(f (y)+g(x− y)) ,

then(f g)∗ = f ∗+g∗.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Mixtures thermodynamics [Cal85]

We consider two fluids (i), i = 1,2 with volume Vi ≥ 0, massMi ≥ 0 and entropy Si ≥ 0. Each fluid is characterized by its energyfunction

Ei : (Vi ,Mi ,Si ) 7→ Ei (Vi ,Mi ,Si ).

We extend Ei by +∞ outside C = (R+)3. We suppose that Ei is

convex, lsc and “extensive”, i.e.

∀λ ≥ 0,∀W ∈ R3,Ei (λW ) = λEi (W ).

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Example: perfect gas

The perfect gas energy is

E = V(

MV

exp(

SCvM

)where γ > 1 is the polytropic exponent and Cv > 0 is the specificheat at constant volume.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Mixture equilibrium

The mixture energy is E (W1,W2) = E1(W1)+E2(W2).At equilibrium the mixture achieves a minimum of the energy.The mixture volume, mass and entropy are noted respectively V , Mand S . We also note W = (M,V ,S).Because of mass conservation,

M = M1+M2.

The entropy being an additive variable,

S = S1+S2.

For the volume, two cases are possible depending on the miscibilityof the mixture

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Non-miscible case

We setV = V1+V2,

thusW = W1+W2.

The mixture equilibrium energy is then given by an inf-convolutionoperation

E (W ) = minW1∈R3

E1(W1)+E2(W −W1) = E1E2.

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Miscible case

We haveV = V1 = V2,

andZi = (Mi ,Si ), Z = (M,S) = Z1+Z2.

The mixture equilibrium energy is also given by an inf-convolutionoperation

E (V ,Z ) = minZ1∈R2

E1(V ,Z1)+E2(V ,Z −Z1),

E (V , ·) = E1(V , ·)E2(V , ·).

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Intensive variables

The function α : W 7→ α(W ) is “intensive” if

∀λ ≥ 0,α(λW ) = α(W ).

We introduce the following intensive quantitiesspecific volume τi =

1ρi= Vi

Mi, density ρi =

MiVi,

specific entropy si = Si/Mi , volumic entropy σi = Si/Vi = ρi si .

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Intensive energies

We also introduce two intensive energy functions defined byspecific energy

ei (τi ,si ) =1

MiEi (Vi ,Mi ,Si ) = Ei (

Vi

Mi,1,

Si

Mi) = Ei (τi ,1,si ),

volumic energy

εi (ρi ,σi ) =1Vi

Ei (Vi ,Mi ,Si ) = Ei (1,Mi

Vi,Si

Vi) = Ei (1,ρi ,σi ).

The intensive energy functions are convex lsc and

εi (ρi ,σi ) = ρiei (1ρi,σi

ρi).

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Non-miscible case

For a non-miscible mixture, the intensive equilibrium energies aregiven by

e = comin(e1,e2)

andε = comin(ε1,ε2).

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Proof

E ∗i (V′,M ′,S ′) = sup

V ,M,S≥0VV ′+MM ′+SS ′−Ei (V ,M,S)

= supV ,ρ,σ≥0

V(V ′+ρM ′+σS ′− εi (ρ,σ)

)= sup

V≥0V (V ′+ ε

∗i (M

′,S ′))

=

0 if ε∗i (M

′,S ′)≤−V ′,+∞ else.

We introduce Ai = (V ′,M ′,S ′),ε∗i (M ′,S ′)≤−V ′ then E ∗i is theconvex indicator of Ai . Then, E ∗ = E ∗1 +E ∗2 is the convex indicatorof A1∩A2 = (V ′,M ′,S ′),max(ε∗1(M

′,S ′),ε∗2(M′,S ′))≤−V ′ we

deduce that ε∗ =max(ε∗1 ,ε∗2) and then ε = comin(ε1,ε2).

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Miscible case

For a miscible mixture, the equilibrium volumic energy is given by

ε = ε1ε2.

The proof is immediate.

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Thermodynamics relations

The temperature T , pressure p and chemical potential µ aredefined by

TdS = dE +pdV −µdM.

Because E is extensive, we also get from the Euler relation

E (W ) = ∇E (W ) ·W ,

orµ = e+pτ−Ts.

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Conjugate of the volumic energy

We also havedε = µdρ +Tdσ ,

which implies that ∂ρε = µ and ∂σ ε = T . The Legendre transformof ε is thus

ε∗ = µρ +Tσ − ε

considered as a function of (µ,T ). From the expression of µ abovewe deduce that the Legendre transform of the volumic energy is thepressure

ε∗(µ,T ) = p(µ,T ).

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Physical interpretation

non-miscible case:

p(µ,T ) =max(p1(µ,T ),p2(µ,T )).

The coexistence of the two phases corresponds top(µ,T ) = p1(µ,T ) = p2(µ,T ) (saturation curve).miscible case:

p(µ,T ) = p1(µ,T )+p2(µ,T ).

The pressure is the sum of the partial pressures (Dalton’s law).

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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(max,+) (or idempotent) analysis [Mas87]

classical analysis (max,+) analysis

a ·b ab = a+b

a+b a⊕b = max(a,b) (a⊕a = a)∫Ω f (x)dx ⊕

x∈Ωf (x) = max

x∈Ωf (x)

characters: χ(s,x +y) = χ(s,x) ·χ(s,y) χ(s,x +y) = χ(s,x)χ(s,y)

χ(s,x) = exp(−isx) χ(s,x) = s ·x

Fourier: f (s) =∫

f (x)exp(−isx)dx f ∗(s) =⊕xf (x)χ(s,x) = max

xsx + f (x)

Convolution: (f ∗g)(x) =∫y f (x−y)g(y)dy f g(x) = sup

yf (x−y) +g(y)

(f ∗g)∧ = f · g (f g)∗ = f g (f ,g concave usc)

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Mixtures and (max,+) analysis [Mas87]

non-miscible mixture:

ε∗ =max(ε∗1 ,ε

∗2) = ε

∗1 ⊕ ε

∗2 .

miscible mixture:

ε∗ = ε

∗1 + ε

∗2 = ε

∗1 ε

∗2 .

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Discrete conjugate: 1D case

Let f : [a,b]→ R. We extend f by +∞ outside [a,b]. We considera subdivision of [a,b] with N +1 points a = x0 < x1 < · · ·< xN = b.We suppose that f is piecewise linear

f (xi ) = yi

f (x) = yi + si (x− xi ), x ∈ [xi ,xi+1], i = 0 · · ·N−1.

The naive algorithm to compute f ∗(s) for N ′ values of s wouldhave a O(NN ′) complexity.

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Fast algorithm [Luc97]

We know that f ∗ = (cof )∗, thus we first compute the convex hullof f , which is still piecewise linear on [a,b]. Without loss ofgenerality, we can thus suppose that the sequence of slopes(si )i=0···N−1 is strictly increasing. Then

f ∗(s) =

as− f (a) if s < s0xi s− f (xi ) if si−1 < s < si , i = 1 · · ·N−1bs− f (b) if s > sN−1

The global algorithm to compute N ′ values of f ∗ has a O(N +N ′)complexity.We observe that f ∗ is linear outside [s0,sN−1].The algorithm to compute the Legendre transform of such afunction is very similar.

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2D version

Let f : [a,b]× [c ,d ]→ R. We extend f by +∞ (or by linear maps)outside [a,b]× [c ,d ]. We “apply Fubini”

f ∗(s, t) = supx ,y

(sx + ty − f (x ,y)) = supy(ty + sup

x(sx− f (x ,y)))

= supy(ty + f ∗,x(s,y))

= (−f ∗,x)∗,y .

In practice, we apply the 1D algorithm on all the rows and then allthe columns of the samples. For this, we need to sample f ∗(s,y) atthe same points in s for each y .

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Maxwell equal area rule

We consider a van der Waals gas with dimensionless specific energy

e(τ,s) =es

(3τ−1)8/3− 3

τif τ > 1/3,s > 0,

=+∞ else.

The critical point corresponds to Tc = 1, pc = 1 and τc = 1. Thisenergy is not convex. Therefore we replace it by e∗∗ = coe. This isequivalent to the well known Maxwell equal area rule construction.

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Numerical illustration

It is classical to plot the isotherms in the (p,τ) plane.This is very easy practically with the FLT.First we compute f (τ,T ) := e∗,s(τ,T ) = Ts− e(τ,s), withT = ∂se.Then we fix T and plot p = ∂τ f for a varying τ .

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Isotherms of e

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Isotherms of e∗∗

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Maxwell equal area rule

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A simple non-miscible mixture

We consider a mixture of two perfect gases (γ1 > γ2)

ei (τi ,si ) = exp(si )τ1−γii

It is possible to compute analytically e(τ,s) = comin(e1,e2) andp(τ,s).The saturation curve is simply

p = κT

(κ = κ(γ1,γ2) has a complicated expression).

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Saturation curves

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Numerical saturation zone

We simply set e =min(e1,e2) and plot e− e∗∗ in the plane (s,τ).

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Isotherms

7.5

8

8.5

9

9.5

10

10.5

11

11.5

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

Pre

ssur

e

Specific volume

Isotherms

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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A simple miscible mixture

We take the same perfect gases and now construct the misciblemixture

ε = ε1ε2

The isotherms are very similar to those of a perfect gas

2

4

6

8

10

12

0.2 0.4 0.6 0.8 1 1.2 1.4

Pre

ssur

e

Specific volume

Isotherms

T = 6.47 T = 6.95 T = 7.43

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Mass fraction

But the mass fraction of the gas (1) depends on the pressure andthe temperature.

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

2 4 6 8 10 12

Phi

Pressure

Mass fraction

T = 6.47 T = 6.95 T = 7.43

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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Conclusion

we have found a natural mathematical framework for studyingEOS of mixtures;the FLT is a useful numerical tool for practically computetabulated EOS of mixtures.

Next steps:real EOS for more than two fluids;coupling with CFD computations;use the ideas of idempotent analysis for more general mixtures.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform

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H. B. Callen. Thermodynamics and an introduction tothermostatistics, second edition. Wiley and Sons, 1985.

P. Helluy, H. Mathis. Pressure laws and fast Legendretransform. http://hal.archives-ouvertes.fr/hal-00424061/fr/

J.-B. Hiriart-Urruty and C. Lemaréchal. Fundamentals ofconvex analysis. Grundlehren Text Editions. Springer-Verlag,Berlin, 2001.

Y. Lucet. Faster than the fast Legendre transform, thelinear-time Legendre transform. Numer. Algorithms 16 (1997),no. 2, 171–185 (1998).

V. Maslov. Méthodes opératorielles. Éditions Mir, Moscou,1987.

Philippe HELLUY, Hélène MATHIS Pressure laws and fast Legendre transform