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Riemann solutions of balance systems with

phase change for thermal ﬂow in porous media.

Adviser: Dan Marchesin

Co-adviser: Johannes Bruining

Student: Wanderson Jos

e Lambert

May, 2006

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Contents

1 Introduction 9

2 General Formulation 15

2.1 Physical situations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.1 Accumulations and ﬂuxes for a physical situation . . . . . . . . . 21

2.1.2 Riemann problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.1.3 Wave Sequences and Riemann Solution . . . . . . . . . . . . . . . 24

2.2 Characteristic speeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.3.1 Shocks and the Rankine-Hugoniot condition . . . . . . . . . . . . 29

2.3.2 The Rankine-Hugoniot locus and shock waves . . . . . . . . . . . 30

2.3.3 Extension of the Bethe-Wendroff theorem . . . . . . . . . . . . . . 37

2.4 Regular RH locus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.4.1 Small shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.4.2 Other degeneracies of RH locus . . . . . . . . . . . . . . . . . . . . 47

2.4.3 The sign of u

on regular Rankine-Hugoniot loci . . . . . . . . . . 47

3 Mathematical modelling of thermal oil recovery with distillation 50

3.1 Physical and Mathematical Models . . . . . . . . . . . . . . . . . . . . . . 51

3.1.1 Balance Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.1.2 Reduced system of equations . . . . . . . . . . . . . . . . . . . . . 55

3.2 Physical situations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.2.1 Three phases. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.2.2 Two phases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2.3 One phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4 The Riemann solution for the injection of steam and nitrogen 63

4.1 Physical model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

4.2 The model equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

ads:

4.3 Physical situations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.3.1 Single-phase gaseous situation - spg ................. 67

4.3.2 Two-phase situation - tp ........................ 68

4.3.3 Single-phase liquid situation - spl ................... 69

4.3.4 Primary, secondary and trivial variables . . . . . . . . . . . . . . . 69

4.4 General theory of Riemann Solutions . . . . . . . . . . . . . . . . . . . . . 70

4.4.1 Characteristic speeds in each physical situation . . . . . . . . . . 70

4.4.2 Shock waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.4.3 Bifurcation loci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.4.4 Admissibility of shocks . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.5 Elementary waves in the single-phase gaseous situation . . . . . . . . . 81

4.5.1 Characteristic speed analysis . . . . . . . . . . . . . . . . . . . . . 81

4.5.2 Shocks and contact discontinuities . . . . . . . . . . . . . . . . . . 83

4.6 Contact discontinuities in the single-phase liquid situation . . . . . . . . 84

4.7 Elementary waves in the two-phase situation . . . . . . . . . . . . . . . . 85

4.7.1 Characteristic speed analysis . . . . . . . . . . . . . . . . . . . . . 85

4.7.2 Shocks in the two-phase situation . . . . . . . . . . . . . . . . . . 90

4.8 Shocks between regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

4.8.1 Shock between the gaseous and the two-phase situations . . . . . 92

4.8.2 Shock between the two-phase and the liquid situations . . . . . . 94

4.9 The Riemann solution for geothermal energy recovery . . . . . . . . . . . 96

4.9.1 Subdivision of tp ............................. 96

4.9.2 The Riemann solution . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.9.3 V

in III and IV .............................100

5 Riemann solution for steam and water ﬂow 101

5.1 Mathematical and Physical model . . . . . . . . . . . . . . . . . . . . . . 103

5.1.1 The model equations . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.1.2 Physical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.2 Regions under thermodynamical equilibrium . . . . . . . . . . . . . . . . 105

5.3 Equations in conservative form . . . . . . . . . . . . . . . . . . . . . . . . 106

5.4 Elementary waves under thermodynamical equilibrium . . . . . . . . . . 106

5.4.1 Steam region - sr .............................107

5.4.2 Boiling region - br ............................109

5.4.3 Water region - wr ............................110

5.5 Shocks between regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.5.1 Water Evaporation Shock . . . . . . . . . . . . . . . . . . . . . . . 111

5.5.2 Vaporization Shock . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5.5.3 Condensation Shock . . . . . . . . . . . . . . . . . . . . . . . . . . 114

5.5.4 Steam condensation front . . . . . . . . . . . . . . . . . . . . . . . 116

5.6 The Riemann Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.6.1 Riemann Problem A . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5.6.2 Riemann Problem B . . . . . . . . . . . . . . . . . . . . . . . . . . 122

5.6.3 Riemann Problem C . . . . . . . . . . . . . . . . . . . . . . . . . . 126

5.6.4 Riemann Problem D . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5.6.5 Riemann Problem E . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6 Riemann solution for problem of Chapter 4 for V

in III and IV. 145

6.1 V

in L

......................................145

6.2 V

in L

......................................146

7 Summary and Conclusions 148

A Physical quantities; symbols and values for the Nitrogen Problem 154

A.1 Temperature dependent properties of steam and water . . . . . . . . . . 156

A.2 Constitutive relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

B Physical quantities; symbols and values for the steam injection 158

B.1 Temperature dependent properties of steam and water . . . . . . . . . . 158

B.2 Constitutive relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

B.3 Approximation for T

+ M

(T)/C

in the sr ................159

B.3.1 Rarefaction wave behavior . . . . . . . . . . . . . . . . . . . . . . . 161

C Applications 163

C.1 Hyperbolic waves in a physical situation . . . . . . . . . . . . . . . . . . . 163

C.2 Numerical Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

C.2.1 General Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . 167

C.2.2 Numerical Method for the System . . . . . . . . . . . . . . . . . . 167

C.3 Travelling waves and viscous proﬁles. . . . . . . . . . . . . . . . . . . . . 168

As duas mulheres da minha vida na ordem em que

apareceram: Ivete e Milene.

Abstract

This work encompasses two types of results. First we present a new general the-

ory which deals with Riemann solution for a large class of balance equations. This

class of equations has interest because it can be applied to model thermal ﬂow with

mass interchange between phases in porous media, with important applications in oil

recovery.

As applications of this theory, we present three models of steam injection in a

horizontal porous media. The systems of equations are based on mass balance, energy

conservation andDarcy law of force. We neglect compressibility, heat conductivity and

capillarity effects.

In ﬁrst example we consider steam/water/oil ﬂow in porous medium. We only

prepare the formalism for this class of equations.

In second example we consider steam/water/nitrogen ﬂow into a porous medium.

We develop the general theory for a 4

× 4 system of balance equations. We solve the

Riemann problem with application to recovery of geothermal energy. A rarefaction

evaporation wave is observed.

In third example we consider the steam/water injection into a porous medium. We

completely solve the Riemann problem associated with this model. We obtain a rich

class of bifurcations in the solutions. A new type of shock, the evaporation shock, is

identiﬁed in the Riemann solution.

Keywords: Porous medium, steamdrive, Riemann solution, balance equa-

tions, multiphase ﬂow, distillation

Resumo

Este trabalho abrange dois tipos de resultados. Primeiramente n

os apresentamos

uma nova teoria geral que trata da soluc¸

ao de Riemann para uma grande classe de

equac¸

oes de balanc¸o. Esta classe de equac¸

oes tem interesse porque pode ser aplicada

ao modelamento de ﬂuxos t

ermicos em meios porosos com transfer

encia de massa

entre fases, com importantes aplicac¸

oes na recuperac¸

ao de

oleo.

Como aplicac¸

oes desta teoria, n

os apresentamos tr

s modelos de injec¸

ao de vapor

em meios porosos horizontais. Os sistemas de equac¸

oes s

ao baseados no balanc¸o de

massa, na conservac¸

ao de energia e na lei de Darcy da forc¸a. N

os negligenciamos a

compressibilidade, a condutividade do calor e efeitos de capilaridade.

No primeiro modelo, n

os consideramos ﬂuxo de vapor/

agua/

oleo no meio poroso.

os somente preparamos o formalismo para esta classe de equac¸

oes para resolvermos

o problema de Riemann como um trabalho futuro.

No segundo modelo, n

os consideramos ﬂuxo de vapor/

agua/nitrog

enio num meio

poroso. N

os desenvolvemos uma teoria geral para um sistema 4

× 4 de equac¸

oes de

balanc¸o. Resolvemos o problema de Riemann com aplicac¸

a recuperac¸

ao de energia

geot

ermica. Uma onda de rarefac¸

ao de evaporac¸

e observada.

No terceiro exemplo, consideramos a injec¸

ao de vapor/

agua em um meio poroso.

os resolvemos completamente o problema de Riemann associado com este modelo.

Obtemos uma rica classe de bifurcac¸

oes para as soluc¸

oes. Um novo tipo de choque, o

choque da evaporac¸

ao, identiﬁcado na soluc¸

ao de Riemann.

Palavras-chaves: meios porosos, injec¸

ao de vapor, soluc¸

ao de Riemann,

equac¸

oes de balanc¸o, ﬂuxo multif

asico, destilac¸

ao.

Agradecimentos

Agradecimentos prescindem de ordem. Aos que eu agradec¸o, n

ao h

a aqueles que

tenham sido mais importantes. Todos foram imprescind

ıveis nos momentos certos,

pois uma tese n

ao se erige em quatro anos. Uma tese comp

oe-se de gr

aos: dos estudos

anteriores, de cada momento dispensado, de cada impress

ao da vida. Colaboram uns

mais, outros menos: desde os professores da inf

ancia at

e o orientador.

Portanto s

ao muitos a serem agradecidos e se eu, injustamente, esquecer de algu

em,

desculpem-me.

Agradec¸o sobretudo a Deus que em sua eterna sabedoria soube esconder a matem

atica

nos improv

aveis rec

onditos do Universo.

A matem

atica pela poesia e arte que se amalgamam entre seus furtivos segredos.

Ao meu orientador pela quase inﬁnita paci

encia com a qual me orientou na tese,

no ingl

es e na vida.

Ao Professor Johannes Bruining que me deu boas noc¸

oes de termodin

amica e

procurou sempre sanar minhas impertinentes d

uvidas.

Aos meus antigos orientadores da UNICAMP: Petr

onio e Marcelo. O primeiro me

enveredou para a matem

atica e o segundo para as leis de conservac¸

ao.

A banca, por ter me ajudado a melhorar a vers

ao ﬁnal.

Ao IMPA, aos seus funcion

arios e ao incompar

avel suporte de pesquisa que a mim

foi propiciado.

Aos professores que me ﬁzeram enxergar o mundo com outros olhos, a todos eles.

Aos funcion

arios do laborat

orio de ﬂuidos pelo apoio, pelo suporte, pela ajuda ou

simplesmente pela amizade.

Aos novos amigos que adquiri no Rio (de alguma forma ou de outra todos atrav

do IMPA) e que ajudaram a amainar a inexor

avel saudade: da namorada, da fam

ılia

e dos velhos amigos.

A minha fam

ılia pelo apoio. A esse micro-cosmo que

e meu suporte. Em especial

a minha m

ae,

a tia Francinete, aos meus irm

aos e irm

as, cunhadas e cunhados e as

crianc¸as que muitas vezes atrapalharam meus estudos e sempre me faziam participar

de suas travessuras.

A Milene, uma impressionante mulher que me preencheu de amor. Que mesmo

longe, afugentava meus medos e me afagava com sua terna m

ao em minhas lembranc¸as.

Agradec¸o a ela pela paci

encia e pelas brigas...pois sempre era doce o retorno.

Agradec¸o aos meus amigos de Cambu

ı. Que s

ao apenas amigos mas poderiam ter

sido irm

aos. Agradec¸o por cada riso, pelas discuss

oes, pelo compartilhar de id

eias e

da partilha das inquietaes terrenas que por vezes experimentamos.

Por ﬁm agradec¸o ao CNPq que parcialmente ﬁnanciou esta pesquisa e a ANP pelo

suporte ﬁnanceiro e a oportunidade de desenvolver este trabalho.

CHAPTER 1

Introduction

A particular class of balance equations are the conservation laws. The physical

interpretation of these laws is that for any limited spatial domain

D ⊂ R

the change

the total amount of any quantity (for example, mass, momentum, energy) is due ex-

clusively to the ﬂow of this quantity through the boundary of the domain.

In the Mathematics literature, a system of conservation laws in one spatial dimen-

sion is usually deﬁned as:

+ F

= 0, (1.1)

where U

∈ Ω ⊂ R

represents a state vector ﬁeld of dependent variables. The func-

tion F :

= F(U) :=(F

(U), F

(U),···, F

(U))

: Ω −→ R

is the ﬂux of such quantities,

generically, it is a

function. The derivatives U

= ∂U/∂t and F

(U)=∂F/∂x are

interpreted in the distribution sense, so that

(1.1) is regarded to be in the weak form;

this is important because of the presence of shocks.

An important problem associated to Eq.

(1.1) appeared in the work of Riemann

in the XIX century. He was interested in the evolution of two gases with different

pressures separated by a thin membrane within a tube. Suddenly, the membrane

is broken. He analyzed the evolution of the gas ﬂow. In this model he observed

rarefaction waves and discontinuous solutions, which are called shocks nowadays.

This work originated the study of an important class of Cauchy problems, the so

called Riemann problem, governed by equations of type

(1.1) with initial data of the

form:

(x, t = 0) :=



, if x < 0

, if x > 0

. (1.2)

Riemann problems have the important property that solutions are invariant under

the scaling x

→ ax, t → at, for a > 0. For non-linear problems (1.1), they play

the role of fundamental solutions, in a way analogous to sines and cosines for linear

problems.

Conservation laws have been used to model a variety of physical phenomena (such

as ﬂuid dynamics, ﬂow in porous medium, road trafﬁc, electromagnetism, etc...) and

therefore the theory for this class of equations is well developed, see for example

[60,20,59,45] and for numerical methods [46,36,34].

In most applications it is useful to utilize a more general form of conservation

laws:

+ F

= 0, (1.3)

where G :

= G(U) :=(G

(U), G

(U),···, G

(U))

: Ω −→ R

also are C

. G is the

accumulation term; the equations are interpreted in the distribution sense.

In classical models of ﬂow in porous medium, the system of equations can be writ-

ten in the form, see

[12]:

+(uF)

= 0, (1.4)

where u is the total or Darcy speed. However, incompressibility is often assumed in

these models, which implies that u is constant spatially. As a result, in such ﬂows

(1.4) reduces to the form (1.3).

In many problems, such as transport of hot ﬂuids and gases undergoing mass gain,

loss or transfer, ﬂows involving chemical reactions such as combustion balance laws

generalizing

(1.3) are required to describe the ﬂow. The classical form of balance laws

is:

+ F

= Q. (1.5)

Here Q :

= Q(U) :=(Q

(U), Q

(U),···, Q

(U))

: Ω −→ R

are C

functions that

represent sources, sinks or transference of the physical quantities. Although there

are many applications for this class of equations, its theory is not well consolidated.

In our work, we are interested in models for ﬂow in porous media, for such ﬂows

Q represents mass transfer of chemical components between phases with no net gain

or loss of component mass, as well as conservation of total energy. The variable u

representing the Darcy speed also appears, but only in particular way within the ﬂux

term:

(

)

= Q. (1.6)

The variables in

(1.6) are V∈Ω ⊂ R

and u ∈R. In this model, the variable u is not

constant, generically.

System of type

(1.6) model thermal compositional ﬂows in porous media. The

variable u is called speed because this is its interpretation in many applications; the

pair

(V, u) in R

n+1

is called state variable. G and F are the vector-valued functions

G =(G

,···,G

m+1

)

: Ω −→ R

m+1

and F =(F

,···,F

m+1

)

: Ω −→ R

m+1

where u

is the ﬂux for the conserved quantity G

and ∂G

/∂t is the correspond-

ing accumulation term, for i

= 1,2,···,m + 1. On the right hand side in Q =

(

, Q

,···, Q

m+1

)

: Ω −→ R

m+1

, the ﬁrst m terms represent mass transfer and

m+1

represents energy conservation. Generically for a thermally isolated system,

the ﬁrst m equations represent mass balance for different chemical components in

different phases and the

(m + 1)-th equation represents the conservation of total en-

ergy, which usually can be represented by setting Q

m+1

= 0. The functions G, F

and Q are continuous in the whole domain Ω, but later we will see that they are

only piecewise C

. The equations are supplemented by relationships expressing local

thermodynamic equilibrium in each of the

parts.

The solution

V(x, t) and u(x, t) for x ∈ R and t ∈ R

needs to be determined. The

terms Q often generate fast variations in the solution due thermodynamic processes;

this is another reason to regard the solution as distributions and

(1.6) must be taken

in the weak sense. Despite the fact that the variable u does not appear in the accumu-

lation term, but only in the ﬂux term and despite the presence of the transfer term Q

formally breaking scale invariance, we are able to solve the complete Riemann prob-

lem associated to Eq.

(1.6). In fact, a general theory to deal this class of equations is

proposed in this work, extending recent works of Bruining et. al.

[5, 6, 8, 9, 10, 11].

This theory encompasses and generalizes the bifurcation phenomena for systems that

change type discovered over the last two decades, see

[24,25,26].

In Chapter 2, we deﬁne the physical situations, in which one or several phases or

chemical components are missing, determining the expression of the prevailing local

thermodynamic equilibrium. In each of these physical situations, by combining the

original equations the balance system

(1.6) reduces to simpler systems of conserva-

tion laws with fewer equations of the form:

∂

∂t

(V)+

∂

∂x

(V)=0, (1.7)

where the transfer terms Q has cancelled out. In a physical situation governed by

(1.7), the corresponding set of variables V is a subset of the set of variables V, while

F and G are obtained from

F and G; they have n + 1 components, n ≤ m. The physical

domain in each physical situation is denoted by Ω

for j = I, II, III, ··· and it is a sub-

set of Ω. There are three groups of variables: the basic variables V, called “primary

variables”; the variable u, called “secondary variable” because it is obtained from the

primary variables; and the “trivial variables” are constant or can be recovered from

other variables in a simple way. Notice that the numbers of primary, secondary and

trivial variables always add up to m

+ 1.

[15], Colombo et. al studied a problem with phase transitions in 2 × 2 systems

of type

(1.1). They considered that the physical domain is formed by two disjoint

sub-domains

and X

. The domains are physical situations, which Colombo called

phases. The phase transition is the jump in the solution with left and right states

belonging to different phases. Our model is physically more adequate because it in-

cludes also inﬁnitesimally small phase transitions, so that

adjoins X

, as they

often are do in actual physics.

The class of equations studied in this work was motivated by practical problems of

steam injection in porous medium. Mathematical models for steam injection can be

found in the works of Coats

[13], Marx et. al. [50], Lake et. al. [23] and Pope et. al.

[54]. The rigorous analysis we are pursuing started in the works of Bruining et. al.

[5, 6, 8, 9, 10, 11]. Following these initial ideas, we focus our interest in Riemann

problems for variants of steam injection. Techniques based on steam injection have

much interest because they are used in oil industry. For half a century, direct of cyclic

steam injection have been applied as a method for enhancing recovery of oil, mainly

for heavier oil. Heavy oil at low temperatures is highly viscous and so recovery is

very difﬁcult. Steam injection is used to increase the oil temperature, improving

recovery both because the oil viscosity is lowered and because the volatility of light

oil is increased, concentrating it in certain regions and allowing it to act as a liquid

solvent that displaces the heavy oil, see

[4 ] and [11].

Steam ﬂood is part of the so called thermal methods, see

[55]. Thermal methods

are divided into two large groups. One group encompasses the techniques in which

heat is injected into the porous medium, a category encompassing steam injection.

In other group heat is generated locally in the porous medium, usually by means of

combustion. In oil recovery, combustion utilizes a small part of the oil in the reser-

voir. Air is injected and the oxygen participates in the chemical reactions. As steam

injection, combustion phenomena are modelled by balance equations, see

[1 ], and the

works of Coats

[14], Crookston et. al. [16]. Modern theory of conservation laws was

used for the study of combustion by Mota et. al.

[27, 28, 29, 30, 31, 32], focusing on

the study of the internal structures of combustion front which is a classical problem

in engineering, see Souza et. al.

[62].

In Chapter 3, we present the mathematical and physical model for ﬂow of water,

steam, volatile and dead oil. Dead oil is a heavy oil with negligible vapor pressure

while volatile oil is a light oil with non-negligible vapor pressure. A partial Riemann

solution of this problem was found by Bruining and Marchesin

[11].

In the last decades, steam injection has been adapted for clean-up of organic con-

taminants from the subsurface, the so called groundwater remediation. The organic

pollutants are referred to as non-aqueous phase liquids; they are divided into two

large groups: DNAPL and LNAPL. The DNAPL’s are denser than water; examples of

DNAPL are chlorinated solvents. It is very difﬁcult to decontaminate soils containing

these substances because they can enter deeply underground. The LNAPL’s are the

substances less dense than water, such as gasoline and oil.

Traditional remediation of contaminated sites consists of groundwater extraction

and clean up of the drained liquid; this technique is called pump-and-treat, see

[21],

[49] and references therein. The material is treated and discharged in another place.

This method is very expensive and time-consuming, for example, in the Visalia test

site, in California, polluted with creosote (a NAPL ﬂuid), the estimated costs for

pumping and treating are US$26,000 per gallon of creosote and the expected time

for this procedure is 3250 years. It is inefﬁcient to clean up sites contaminated with

DNAPL’s (dense non aqueous phase liquids). So removal of contaminants with steam

is considered as an alternative; for the same site in Visalia, the estimated cost per

gallon of creosote removed with steam is US$130 and the time to clean up is approxi-

mately 3 years, see

[17].

[18,19], Davis described the mechanism of steam injection. For clean up, steam

is injected in the soil by using wells. Initially the steam heats the formation around

the wells. The steam condenses as the latent heat of vaporization of water is trans-

ferred from the steam to the region around the wells. As more steam is injected, the

hot water moves into the formation, pushing the water initially present in the rock,

which is at the ambient temperature. When the porous medium at the injection re-

gion has absorbed enough heat to reach the temperature of the injected steam, steam

itself actually enters the medium, pushing the cold water and the bank of condensed

steam in front of it. When these hot liquids reach the region that contains the volatile

contaminant, the contaminant is displaced. Since the temperature of the volatile con-

taminant increases part of this contaminant evaporates originating a gaseous ﬂow

with organic components.

Notice that in this model there appear several physical situations and that there

is mass transfer between the phases. The total energy is conserved, a fact that is

modelled by using the conservation law for energy. The conservation of mass in the

ﬂow together with Darcy’s law for porous media have motivated the introduction of

equations of type

(1.6). In chapter 2, we propose a general formalism for solving this

class of equations. We consider the different physical situations as simple connected

subsets state of

. Physical situations are adjacent in state space. There exists a

linear map E that takes the system of balance laws

(1.6) into a system of conservation

laws

(1.7) associated to each physical situation. After this elimination, further sim-

pliﬁcations using thermodynamical constraints may occur in each physical situation

is described by a set of variables V that is a subset of the set of variables

V. Of course,

each of the systems of type

(1.7) has fewer equations than the complete system (1.6).

We are interested in the Riemann-Goursat problem where x

= 0 is the injection

point and the data in

(1.2) for x < 0 is replaced by the injection boundary data at

= 0. It is well known that the Eqs. (1.1), (1.3) with initial data of (1.2) exhibits

solutions that are constant along lines x

/t, i.e., self-similar solutions. By physical

considerations, we assume that in our model rarefaction waves occur only within each

physical situation, where Eq.

(1.6) reduces to equations of type (1.7). Since there is

fast transfer of mass in the thin space between the physical situations, we propose

shocks linking them. In Sections

[2.2] and 2.3, we prove that the solution states are

constant along the lines x

/t. We also show that we can obtain the rarefaction and

shock waves ﬁrst in the space of primary variables, and that the secondary variable

u can be recovered in terms of the primary variables.

In Proposition 2.3.1 we summarize the theory showing that the Riemann problem

can be completely obtained in the space of primary variables. Moreover, if a sequence

of waves and states solves the Riemann problem in the primary variables, for a given

> 0 (or u

> 0), then it is also a solution for any other u

> 0 (or u

> 0), i. e., if

the Darcy speed u

in the initial data is modiﬁed while V

and V

are kept ﬁxed, the

Darcy speeds along the Riemann solutions are rescaled in the

(x, t) plane, while the

wave sequences and the Riemann solution in

V(x, t) are kept unaltered. As tools, we

prove three generalization of the Triple Shock Rule

[25]. These rules are important to

compare the speed of shocks of different families and obtain the Riemann solution.

In Section 2.3.3 we generalize the Bethe-Wendroff theorem for shocks between

physical situations for equations of type

(1.6). As in the classical case, extrema of

shock speeds are related to shock/rarefaction resonance. This theorem is very im-

portant for the identiﬁcation of bifurcation loci in the solution. These structures are

important because the Riemann problem changes when the left (or right) data is pre-

scribed in different regions in the physical domain relatively to these structures.

In Section 2.4, we deﬁne the notion of regular Rankine-Hugoniot (RH) locus, which

are curves that parametrize shocks. For the class of equations

(1.6), we deﬁne the

notion of strictly hyperbolic physical situation and show a generalization of the Lax

theorem for shocks within a physical situation, Proposition 2.4.1. In Section 2.4.3,we

prove that the speed u does not change sign in connected branches of the RH locus.

In Appendix C, we show some applications of the general theory. In Appendix

C.1, we prove that the formalism for the class of equations

(1.6) can be applied to

the simpler classical equation

(1.3). In Appendix C.2, we obtain an efﬁcient quasi-

implicity numerical scheme for equations

(1.7) without knowing the variable u.In

Appendix C.3, we show that the viscous proﬁle for equations

(1.6) can be studied only

in the space of primary variables V, without knowing the speed u.

In Chapter 3, we present a physical model for steam and gaseous volatile oil injec-

tion into a horizontal, linear porous medium ﬁlled with oil and water. We neglect com-

pressibility, heat conductivity and capillarity effects and present a physical model for

oil/water/steam injection is based on the mass balance and energy conservation equa-

tions as well as on Darcy’s. We present the main physical deﬁnitions and summarize

the equations of the model. This system of equations is a representative example of

ﬂow in porous media with mass interchange between phases (in the absence of grav-

itational effects) Eq.

(1.6). The Riemann solution was partially solved in [11]. Our

purpose is to show how this model ﬁts within the formalism.

In Chapter 4, using the model proposed by Bruining et. al.

[6 ], we formulate a

system that describes the ﬂow of nitrogen, steam and water in a porous medium.

The physical model for nitrogen/steam/water injection is based on the mass balance

and energy conservation equations. We describe the formalism for solving Riemann

problem associated to the a 4

×4 system of balance equations. As example of Riemann

solution, we show an application for the recovery of geothermal energy. Injecting

water and nitrogen into a hot porous rock, the water in the mixture evaporates and

rock thermal heat can be recovered even if the rock is not very hot. A new fact is the

presence of a rarefaction wave associated to evaporation in the two-phase situation.

In Chapter 5, we obtain the solutions for the basic one-dimensional proﬁles that

appear in the clean up problem or in recovery of geothermal energy. We consider the

injection of a mixture of steam and water in several proportions into a porous rock

ﬁlled with a different mixture of water and steam. We describe completely all possible

solutions of the Riemann problem. We ﬁnd several types of shock between regions

and a rich structure of bifurcations. A new type of shock, the evaporation shock, is

identiﬁed in the Riemann solution. This model generalizes the work of Bruining et.

al.

[6 ], where the condensation shock appeared.

Each chapter can be read separately in this thesis. We put in each chapter a

brief introduction explaining the content of each chapter, so that they can be read

separately in this thesis. More precisely, Chapter 3 requires Chapter 2. Chapters 4

and 5 can be read on their own.

Much remains to be studied for models of type

(1.6) as discussed in the conclusion,

Chapter 7.

CHAPTER 2

General Formulation

Mass interchange between different phases occurs typically in multiphase ﬂows in

porous media. Such ﬂows are not represented by conservation laws or hyperbolic

equations both because they have source terms and because they are not evolutionary

in all the variables. In the absence of gravitational effects, such ﬂows can be modelled

by systems of m

+ 1 equations:

∂

∂t

G(V)+

∂

∂x

F(V)=Q(V) , (2.1)

where

V∈Ω ⊂ R

and u ∈R. The variable u is called speed because this is its

interpretation in many applications; the pair

(V, u) in R

n+1

is called state variable.

G and F represent the vector-valued functions G =(G

,···,G

m+1

)

: Ω −→ R

m+1

and F =(F

,···,F

m+1

)

: Ω −→ R

m+1

, where uF

is the ﬂux for the conserved

quantity

and ∂G

/∂t is the corresponding accumulation term, for i = 1,2,···,m + 1.

On the right hand side Q

=(Q

, Q

,···, Q

m+1

)

: Ω −→ R

m+1

, the ﬁrst m components

represent mass transfer and Q

m+1

represents energy transfer due to phase change.

Generically for a thermally isolated system, the m ﬁrst equations represent mass bal-

ance for different chemical components in different phases and the

(m + 1)-th equa-

tion represents the conservation of total energy, which usually can be represented by

setting Q

m+1

= 0. The functions G, F and Q are continuous in the whole domain Ω,

but later we will see that they are only piecewise smooth.

The solution

V(x, t) and u(x, t) for x ∈ R and t ∈ R

needs to be determined.

The terms Q often generate fast variations in the solution, so that in such cases the

solution may be regarded as distributions and the equation

(2.1) must be taken in the

weak sense. This equation has an important feature: the variable u does not appear

in the accumulation term, but only in the ﬂux term.

Generically, the equations

(2.1) model ﬂows where there are phase changes giving

rise to mass exchange between different phases. Such ﬂows encompass particular

physical situations, where one or several phases or chemical components are missing.

The balance system

(2.1) reduces in each particular physical situation to simpler

systems of conservation laws with fewer equations of the form:

∂

∂t

(V)+

∂

∂x

(V)=0, (2.2)

where the source term Q is absent. In each physical situation, the corresponding set

of variables V is a subset of the set of variables

V; F and G are obtained from F and

G; they have n + 1 components, n ≤ m.

Typically, the laws of thermodynamics play a central role in models

(2.1). Each

physical situation where the system of balance equations

(2.1) reduces to a system of

type

(2.2) is actually a physical situation under local thermodynamical equilibrium

or quasi-equilibrium and the system

(2.2) is a system under local thermodynamical

equilibrium or quasi-equilibrium: this equilibrium is enforced by thermodynamic re-

lationships between the quantities in V. There are three groups of variables in each

physical situation. Generically, when physical changes occur in waves under ther-

modynamical equilibrium, the solution of the system

(2.1) and (2.2) are the same;

however, when there is only local thermodynamical equilibrium, the solution of

(2.1)

is an approximation of the ( 2.2). In a future work, we intend to understand how de-

viations in thermodynamical equilibrium interfere in the approximation of

(2.1). The

basic variables V, called “primary variables”; the variable u, called “secondary vari-

able” because it is obtained from the primary variables; and the “trivial variables” are

constant or they can be recovered from other variables in a simple way. Notice that

the number of primary, secondary and trivial variables always add up to m

+ 1.

2.1 Physical situations

We deﬁne the physical situation Ω

as a subset of Ω that describes the j-th physical

situation. The Roman number index I, II, III,

··· indexes the physical situation; we

represent the set of indices by

PS = {I, II, III, ···}. We denote the points of Ω

, see Examples 2.1.1 and 2.1.2 below.

The sets Ω

satisfy:

Ω

⊂ Ω and Ω =



Ω

. (2.3)

The sets

of primary state variables are obtained from Ω

, one subset for each

physical situation. The primary variables correspond to the unknowns V

of a system

of conservation laws of the form

(2.2). The number of primary variables in P

indicated by n

(j). The primary variables in each P

are denoted by V

following the

notation in

(2.2) to indicate that they depend on the physical situation j. For each

physical situation, there are accumulation and ﬂux functions G and F satisfying an

equation of the form

(2.2); G and F are obtained from G and F. To indicate that

(G, F) depends also on j, we utilize the index j, so the functions are written as G

(

, G

,···, G

(j)+1

)

and F

=(F

, F

,···, F

(j)+1

)

We denote the trivial variables by

and the space of trivial variables by P

Notice that in each physical situation Ω

, the trivial variables are uniquely de-

termined by the primary variables or solely by the physical situation, so there is a

one-to-one correspondence between

and Ω

, which is denoted by i

; its inverse i

−

is deﬁned uniquely, which is the restriction of Ω

to P

Consider a point

∈ Ω with primary coordinates V

∈P

, the restriction i

satisﬁes:

: Ω

−→ P

, i

)=V

,. (2.4)

Notice that this restriction is a projection of

into P

The reverse is a map from

to V

, which is:

−

: P

−→ Ω

, i

−

)=V

. (2.5)

We deﬁne the restriction of

G and F to Ω

by R

: Ω

−→ R

m+1

and R

: Ω

−→

m+1

, respectively. For V

∈ Ω

, they satisfy:

)=G(V

) and R

)=F(V

). (2.6)

The original

G and F are continuous in Ω, but they are not differentiable. However

we assume that the restrictions are smooth, speciﬁcally, they belong to

(Ω

m+1

We will see below that the functions

, F

) for the j-th conservation system can

be obtained as a linear combination of

and R

, so the pairs (G

, F

) belong to

(Ω

m+1

Because we are solving problems in one spatial dimension, certain pairs of physical

situations may occur contiguously. Let Ω

and Ω

with P

and P

represent the

primary variables in one of these pairs, with Ω

on the left of Ω

in (x, t) space; we

use

(− ) for j and (+) for k. Whenever there is a non-empty intersection between a

pair Ω

and Ω

, we denote it by Ω

,i.e.,Ω

= Ω



Ω

. The elements of Ω

are

denoted by

In the space of primary variables, this intersection is denoted by Π

and is given

by:

= i

(Ω

)=i

(Ω

Deﬁnition 2.1.1. Let

V∈



Ω

. We say that V is a private point of some Ω

for

∈PS, if there exists no other l ∈PS, such that V∈Ω

Deﬁnition 2.1.2. Let

V∈



Ω

. We say that V is a frontier point of some Ω

for

∈PS, if there exists at least another l ∈PS, such that V∈Ω

Remark 2.1.1. Notice that if V is a private point, then we can perform the projection

for some j ∈PS, i.e., there exists a V ∈P

, such that V = i

(V); we call such

V too private point. If

V is a frontier point, then in the applications it can belong to

different physical situations depending on the necessity, but there always exists at least

a j

∈PSand a V ∈P

such that V = i

); we call such V too frontier point.

From the continuity of the accumulation and ﬂux functions

F and G in Ω we

obtain:

Lemma 2.1.1. For all

∈ Ω

, R

)=R

) and R

)=R

It is important to deﬁne the notion of open sets in the topology induced by Ω

and

relatively to R

n( j)

. Generically we are interested in solving the system of equations

to obtain the primary variables in each physical situation, so it is useful to deﬁne the

induced topology of Ω

from the induced topology of P

and the inverse function i

−

given by

(2.5). For each P ∈ R

n( j)

and



> 0, the open balls are the sets of x ∈ R

n( j)

that satisfy || x − P||

n( j)



, denoted by B



(P). The norm || · || is the euclidian norm

, with m > n( j) and || · ||

n( j)

is the induced norm in R

n( j)

Deﬁnition 2.1.3. The euclidian norm in the induced topology Ω

or P

is the

restriction of

|| · || in each space, which are denoted by ||·||

Ω

and ||·||

Deﬁnition 2.1.4. Let V

∈P

, the open balls in the induced topology P

are B







(V), and the open balls in the induced topology Ω

are i

−



) where i

and

−

are deﬁned in (2.4) and (2.5).

Deﬁnition 2.1.5. Let D

⊂P

. We say that D is an open subset in the space of

primary variables, if all V

∈ D there exists an



> 0 such that B



(V) ⊂ D. The subset

−

(D) is an open subset of Ω

Deﬁnition 2.1.6. Let V

∈P

. We say that V is an internal point of P

, if there exists



> 0 such that B



(V) ⊂P

. The other points are called boundary points of P

.If

V is an internal (boundary) point for a

, then i

−

(V) is an internal (boundary) point

for Ω

Remark 2.1.2. Notice that the frontier points are boundary points, however, there are

boundary points that are private points. The points that are boundary and frontier

points are called exterior boundary.

Since the primary variable spaces are projections of the Ω

for j ∈PS, it is not

possible to consider the union of different primary variable spaces

. However, we

can consider the union of some Ω

. Since we are interested in physical phenomena

we consider a connected union of the variable spaces Ω

, which is denoted by



Ω

Remark 2.1.3. Since each physical situation contains its frontiers, it is closed in Ω.

Example 2.1.1.

In Chapter 5, we study the problem of steam injection into a porous medium,

proposed originally by Bruining et. al. and partially solved in

[6 ]. In this model

diffusive, capillarity and heat conductivity effects are ignored. The mass balance

equations for liquid water, steam and energy conservation in terms of enthalpies are

written as:

∂

∂t

ϕρ

∂

∂x

= q

g−→ a,w

, (2.7)

∂

∂t

ϕρ

∂

∂x

= −q

g−→ a,w

, (2.8)

∂

∂t



(

)



∂

∂x

(

)=0. (2.9)

Here

is the rock porosity assumed to be constant; s

and s

are, respectively, the

water saturation and steam saturation; u is the Darcy velocity; f

and f

are the

fractional ﬂow functions for water and steam;

and

are the constant rock and

water densities. The steam density

is a function of the temperature T; h

, h

and

are the rock, water and steam enthalpies per unit mass; these properties depend on

temperature (see Appendix B); and q

g−→ a,w

is the water mass transfer term or steam

condensation term. All variables are deﬁned in Appendix B.

19 There are three physical situations: steam situation, labelled by I where there

is only steam above the ﬁxed water boiling temperature; the boiling situation, labelled

by II, where water and steam coexist at boiling temperature; the water situation,

labelled by III, where there is only water below boiling temperature, see Figure 2.1.

In the steam situation the space of primary and trivial variables are

= {T} and

= {s

= 0} with Ω

= {s

= 0,T}; the Darcy speed u is an unknown (secondary

variable). In the boiling situation,

= {s

}, P

= {T

} and Ω

= {S

, T

where T

is the boiling temperature at the prevailing temperature; the Darcy speed

u is constant (secondary variable) and the ﬂow is governed by the Buckley-Leverett

equation. In the water situation,

III

= {T}, P

III

= {s

= 1} and Ω

III

= {s

= 1,T};

the Darcy speed u is constant (secondary variable).

Example 2.1.2.

In Chapter 4, we study ﬂow of nitrogen and steam in a porous medium with water,

proposed originally by Bruining et. al and partially solved in

[10], with same simpli-

ﬁcations as in the Example 2.1.1. The water, steam, nitrogen mass balance equation

and the energy balance equations are:

∂

∂t

ϕρ

∂

∂x

= q

g−→ a,w

, (2.10)

∂

∂t

ϕρ

∂

∂x

= −q

g−→ a,w

, (2.11)

∂

∂t

ϕρ

∂

∂x

= 0, (2.12)

∂

∂t



(

)



∂

∂x



) f



= 0.

(2.13)

The quantities

, s

, f

and f

were deﬁned in the previous example.

In the gaseous phase, the concentration of steam is

(steam mass per unit gas

volume) and the concentration of nitrogen is

; in the presence of liquid water ther-

modynamical considerations specify the dependence of these concentrations solely on

temperature, see Appendix A.

We consider an example illustrating how the accumulation and ﬂux functions G

and F depend on the physical situation. We describe the gaseous physical situation,

where s

= 1 and s

= 0. We assume that nitrogen and steam behave as ideal

gases with densities denoted by

and

, see Appendix A. We also assume that

there are no volume effects due to mixing, namely the volumes of the components are

additive, i.e.,

(T)+

(T)=1. This fact allows to deﬁne the steam and

nitrogen gas composition, respectively, as:

(T) and

(T) , (2.14)

so the compositions are additive, i.e.,

= 1. We use

as the independent

variable. Therefore, there are three unknowns to be determined: temperature T, gas

composition

and speed u. The saturation is trivial, i.e., s

= 1 and s

= 0.

We rewrite equations (2.10)-(2.13) using Eq.

(2.14). Since s

= 0, f

= 0, from Eq.

(2.10) we obtain that q

g−→ a,w

vanishes, so the system is written as:

∂

∂t

−1

∂

∂x

−1

= 0,

∂

∂t

(1 −

−1

∂

∂x

(1 −

−1

= 0,

∂

∂t





∂

∂x





= 0,

where

(T), h

= h

(T), h

= h

(T). The constants are described in

Appendix A. Notice that

/R and M

/R are constants that cancel out of the

ﬁrst two equations

In the single-phase gaseous situation, labelled by I, the spaces of primary and

trivial variables are

= {

, T} and P

= {s

= 0} with Ω

= {s

= 0,

, T};

the Darcy speed u is an unknown (secondary variable). In the two-phase situation,

labelled by II, the spaces of primary and trivial variables are P

= {s

, T} and

= {

(T)} with Ω

= {s

(T), T}; the Darcy speed u is an unknown (sec-

ondary variable). In this case the gas composition is given in terms of temperature by

Clausius-Clapeyron and Raoult’s laws, see

[44,68].

In the single-phase liquid situation, labelled by III, the spaces of primary and

trivial variables are

III

= {T} and P

III

= {s

= 1,

(T)} with Ω

III

= {s

(T), T}; the (secondary variable) Darcy speed is a constant. Even though there

is no gas in this physical situation, it is useful to deﬁne the gas composition by conti-

nuity from Eq

(2.14).

In each physical situation, it is necessary to calculate the solution for the primary

variables. In Figure 2.1, we show the physical situations in the primary state space

variable for I, II and III.InΩ, Eqs.

(2.10)-( 2.13) represent a system with four equa-

tions and four variables: the temperature, the gas saturation, the gas composition

and the speed.

Steam region

Boiling region

Steam regionSteam region

Water region

Figure 2.1: State space, physical situations and their domains in Example 2.1.2. The

bold straight lines represent the nitrogen-free situations, that are: the watersituation

for s

= 1; the boiling situation for s

non-constant; the steam situation for s

= 0.

This Riemann problem is solved in Chapter 5.

2.1.1 Accumulations and ﬂuxes for a physical situation

Generically, there is no unique choice of (G

, F

). As a simple example, consider G

(

, G

,···, G

(j)+1

) and F

=(F

, F

,···, F

(j)+1

) satisfying (2.2) for V

∈ Ω

permutation in the components of (G

, F

) yields the same solution of (2.2). However,

it is necessary to select a choice for the

, F

) for each j ∈PS.

Property 2.1.1. We are interested in solving problems where there exists a linear map

E :

m+1

−→ R

n+1

that satisﬁes:

(Q(V)) = 0, ∀V∈Ω, (2.15)

where Q

(V) is a (m + 1) ×1 vector and 0 is a (n + 1) ×1 vector. So the vector Q(V) ∈

E), where K(E) is the kernel of E.

For each chemical component, this map acts on

(2.1) by taking the equations for

all phases and adding them. The resulting equation

(2.2) represents the conservation

of mass of this component in all phases, so that the source terms add to zero.

Deﬁnition 2.1.7. We say that a linear map E :

m+1

−→ R

n+1

has maximal rank

associated to the system

(2.1),ormaximal rank, if it satisﬁes (2.15) and:

+ 1 = r ank(E)+di m(span{Q(V)}), (2.16)

where span

{Q} is the vector space generated by Q. The rank of E is denoted by n + 1.

We call the linear map E internal transfer matrix.

There exists a class of linear maps E :

m+1

−→ R

n+1

that satisfy (2.15) and (2.16)

and denote by E:

E = {E : R

m+1

−→ R

n+1

that satisfy (2.15) and (2.16)}. (2.17)

From

(2.16), for E ∈E, notice that K(E)=span{Q(V)}.

Deﬁnition of

, F

)

Since this pair of functions is deﬁned only in the physical situation j ∈PSand in

this physical situation we are interested only in the primary variables, it is necessary

to utilize

and R

deﬁned in (2.6) and the relationship between the dependent

variable

and the corresponding primary variable V

. Let E ∈Ebe a linear map,

then for

and the corresponding primary variable V

, the pair (G

, F

) is:

)=ER

) and F

)=ER

). (2.18)

The number n

(j) represents the dimension of (G

, F

), i.e., the number of primary

variables to be determined.

Remark 2.1.4. Notice that the n

(j) is the same for any j ∈PS, so in what follows we

only write n to indicate that it does not depend on the physical situation.

From Lemma 2.1.1 and from the continuity of the linear map E we obtain:

Proposition 2.1.1. Consider two pairs (G

, F

) and (G

, F

) obtained from the same

linear map E from

(2.18), such that Π

= Ø, then:

(V)=G

(V) and F

(V)=F

(V) ∀ V ∈ Π

. (2.19)

Thus we can see that the cumulative and ﬂux terms can be understood only as

functions of V for V belonging to some

Consider the balance system

(2.1) for V

∈ Ω

, in a speciﬁed physical situation.

Applying the linear map E

∈Ein this system and using R

and R

deﬁned in (2.6)

and (2.18), we obtain the following system of conservation laws (2.2) for which the

unknowns are the primary variables V

associated to V

that is:

∂

∂t

∂

∂x

= 0. (2.20)

We would like the solution of the resulting conservation laws not to depend on

∈E, i.e., the solution of (2.2) should be the same for some E ∈E(at this stage, we

are not interested yet in the solution associated to initial and boundary conditions.

We are interested only in the form of the differential equation). To guarantee this

independence, we utilize the following Lemma:

Lemma 2.1.2. Let A and B be two vector spaces; E

: A −→ B and E

: A −→ B two

linear maps. If

K(E

)=K(E

), there exists an isomorphism I : B −→ B such that

= IE

Using Lemma 2.1.2, we obtain the following Proposition for E

∈E:

Proposition 2.1.2. Let E

, E

∈Ebe two different matrices. Then for (V

,u) ∈ Ω

×R

for each j ∈PS, the corresponding (V

,u) ∈P

×R is the solution of

∂

∂t

∂

∂x

= 0. (2.21)

if, only if, is solution of

∂

∂t

∂

∂x

= 0, (2.22)

where V

is the primary variable associated to V

Proof: Since E

, E

∈E, then, by deﬁnition of E, we know that K(E

)=K(E

From Lemma 2.1.2 there exists an isomorphism

I : R

n+1

−→ R

n+1

such that E

= IE

Assume that

,u) is a solution of (2.21). So using Def. 2.18 and E

= I

−1

, Eq. (2.21) can be written as:

−1



∂

∂t

∂

∂x



= 0.

Since I

−1

is also an isomorphism, K(I

−1

)=0, so we obtain that:



∂

∂t

∂

∂x



= 0.

(2.23)

That is

(2.22). Since it is valid for E

and E

in E, the assertion follows. 

Corollary 2.1.1. The solution of (2.2) does not depend on the linear map E ∈E.

Thus we can choose any E

∈Eand since the solution of conservation law does not

depend on E we rewrite

, F

) as (G

, F

Example 2.1.3. For a conservation law in the form

(2.2), the transfer matrix E is any

non-singular matrix.

For the balance system (2.7)-(2.9

) and (2.10)-(2.13) the transfer matrices are :



110

001



and E





1100

0010

0001





2.1.2 Riemann problem

We use W

=(V

,u), for V

∈P

for some j ∈PSto represent the state variables,

where u is the secondary variable and V

are the primary state variables; similarly,

W =(V, u) for V∈Ω represents the state variables, u is the secondary variable

and

V are the primary together with the trivial state variables. We are interested in

the Riemann problem associated to

(2.1), that is the solution of these equations with

initial data



=(V

) if x > 0,

=(V

,·) if x < 0.

(2.24)

Since the system

(2.1) has an inﬁnite characteristic speed associated to u, only one of

the speeds u

or u

is given (in this case we have chosen u

). Later, it will be clear that

the other speed on the right hand side can be determined from the other equations in

the system.

The general solution of the Riemann problem associated to Eq.

(2.2) consists of a

sequence of elementary waves: rarefactions and shocks. These waves may be sepa-

rated by constant states. The solutions are studied in the sections that follow.

2.1.3 Wave Sequences and Riemann Solution

A Riemann solution is a sequence of elementary waves w

(shocks and rarefactions)

for k

= 1,2,···,m and constant states W

for k = 1,2,···,m.

≡W

−→ W

−→···

−→ W

≡W

. (2.25)

We represent any state by W. The wave w

has left and right states W

k−1

and W

and speeds

−

in case of rarefaction waves and v =

−

in case of shock

waves. The left state of the ﬁrst wave w

is (V

) and the right state of w

), where u

needs to be found. In the Riemann solution it is necessary that

≤

−

k+1

; this inequality is called geometrical compatibility. When

−

k+1

there

is a separating constant state W

k+1

between w

and w

k+1

; in this sequence the wave

is indicated by →.If

−

k+1

there is no actual constant state in physical space,

so the wave w

is a composite with w

k+1

; it is indicated by .

Different physical situations are separated by shocks respecting the geometrical

compatibility. When it is useful to emphasize the waves in the sequence

(2.25) rather

than the states, we use the notation:

 w

 ··· w

, (2.26)

where for each w

,  represents → for the case

−

k+1

and  for the case

−

k+1

2.2 Characteristic speeds

Since the cumulative and ﬂux terms are not smooth between physical situations, we

calculate the characteristic speeds only within the physical situation. The system

of conservation reduces to the conservation form

(2.2), thus it is possible to ﬁnd the

characteristic speeds, and then the rarefaction waves. These waves are important to

solve the Riemann problem; the characteristic speeds are also important to propose

adequate numerical methods to solve Eq. (2.1) and condition for numerical stability,

see

[40].

For a smooth wave we differentiate all equations in (2.2) with respect to their

variables, obtaining a system:

∂

∂t





+ A

∂

∂x





= 0; (2.27)

the matrices B and A are the derivatives of G

(V) and uF(V) with respect to W =

(

V,u), where V = V

∈P

are the primary variables for some j ∈PSand u is the

speed. Since G

(V) does not depend on u, the matrix ∂G/∂W has a zero column, ie.

∂G

∂W







∂G

∂V

∂G

∂V

···

∂G

∂V

∂G

∂V

∂G

∂V

···

∂G

∂V

∂G

∂V

∂G

∂V

···

∂G

∂V

∂G

n+1

∂V

∂G

n+1

∂V

···

∂G

n+1

∂V









∂G

∂V





= B, (2.28)

and

∂uF

∂W







∂F

∂V

∂F

∂V

··· u

∂F

∂V

∂F

∂V

∂F

∂V

··· u

∂F

∂V

∂F

∂V

∂F

∂V

··· u

∂F

∂V

∂F

n+1

∂V

∂F

n+1

∂V

··· u

∂F

n+1

∂V

n+1









∂F

∂V





= A, (2.29)

where ∂G

/∂V and u∂F/∂V represent the n ﬁrst columns of B and A, 0 and F represents

the

(n + 1)-th column of B and A.

Remark 2.2.1. Let V

∗

∈P

.IfV

∗

is private point, the derivatives ∂G/∂V and ∂F/∂V

are well deﬁned at V

∗

, because if the point is an internal point the cumulative terms G

and ﬂux terms F are

. On the other hand, if V

∗

is a boundary point the derivatives

are deﬁned laterally.

However, if V

∗

is a frontier point, i.e., there exists another k ∈PS, with k = j,

such that V

∗

∈P

,atV

∗

the derivatives are not well deﬁned, because the derivatives

can be different in neighboring physical situations. Since by Rem. 2.1.3 each physical

situation is closed, and we are interested in the rarefaction waves in a determined

physical situation, we deﬁne the derivatives laterally in each

The eigenvalues

and right eigenvectors



r =(r

,···,r

n+1

)

with n + 1 compo-

nents of the following system are the rarefaction wave speeds and the characteristic

directions, respectively:



r =



r. (2.30)

Hereafter, the word “eigenvectors” means “right eigenvectors”. Similarly the left

eigenvector



 =(

,

,···,

n+1

) satisﬁes:





A =





B or A









To ﬁnd

, we need to solve the characteristic equation

det

(A −

B)=0. (2.31)

Whenever necessary, we write the right and left eigenvectors without the upper arrow,

i.e., r



r and  =



.

Remark 2.2.2. We utilize an index p to indicate a particular characteristic speed fam-

ily. The eigenvalue, right eigenvector and left eigenvector of the p

−family are denoted

, r

=(r

,···,r

) and 

=(

,

,···,

,

We deﬁne the set of indices n as

C = {1,2,···,n + 1}. (2.32)

Deﬁnition 2.2.1. The integral curve of the p-family is the solution of :





= r

, i.e,

= r

for all i ∈C and

= r

. (2.33)

When

increases satisfying:

(V(

),u(

)), (2.34)

the integral curve deﬁnes a rarefaction curve. If in addition to

(2.34), the variable

satisﬁes x =

t the integral curve deﬁnes rarefaction waves in the (x, t) plane.

Lemma 2.2.1. Assume that u

= 0 and that F

(V) = 0 for some ﬁxed k ∈Cfor all V.

The eigenvalue, right and left eigenvectors for the system

(2.27), with B and A given

by Eqs.

(2.28) and (2.29) have the form:

= u

(V), (2.35)

=(g

(V), g

(V) ··· , g

(V), ug

n+1

(V)) and  =(

(V), 

(V), ··· ,

n+1

(V)),

(2.36)

where

, g

and 

(for all i ∈C) are functions that depend only on V. Moreover, there

are at most n eigenvalues and eigenvectors associated to this system of n

+ 1 equations.

Proof: The eigenvalues

of (2.27) are the roots of det(A −

B)=0, for B and A

given by Eq.

(2.28) and (2.29). This characteristic equation is:

det

(A −

B)=det



∂F

∂V

(V) −

∂G

∂V

(V)



(V)



= 0. (2.37)

Since u

= 0, we divide the ﬁrst n columns by u and deﬁne

/u, then after some

calculations Eq.

(2.37) reduces to:

det



∂F

∂V

(V) −

∂G

∂V

(V)



(V)



= 0. (2.38)

Since Eq

(2.38) depends on V only, the eigenvalues

are functions of V, i.e.,

(V),

they do not depend on u. Using that

= u

, the eigenvalues have the form (2.35).

The eigenvectors are solutions of

(A −

B)r = 0. Writing r =(r

,···,r

n+1

)

,we

obtain:



∂F

∂V

(V) −

(V)

∂G

∂V

(V)





(V)



= 0. (2.39)

Since there is an index k such that F

(V) = 0, then we can write r

n+1

as:

n+1

= u

(V)

∑

l=1



∂F

∂V

(V) −

(V)

∂G

∂V

(V)



, (2.40)

Substituting r

n+1

given by Eq. (2.40) into Eq. (2.39), we obtain a linear system in

the unknowns r

for j = 1,2,··· ,n, where we can cancel u, showing that the r

for

= 1,2,···,n depend on V only. So (2.36. a) follows from (2.40).

For  =(

,

,···,

n+1

), we solve:

 ·



∂F

∂V

(V) −

(V)

∂G

∂V

(V)





(V)



= 0,

then

 is solution of the following system:



∂F

∂V

−

∂G

∂V





+ ···+ u



∂F

n+1

∂V

−

∂G

n+1

∂V





n+1

= 0 for l = 1,··· ,n (2.41)



+ F



+ ···+ F

n+1



n+1

= 0. (2.42)

Since u = 0, we divide Eqs. (2.41) by u, then we obtain a system with coefﬁcients that

depend only on the variables V, leading to

(2.36.b). 

Remark 2.2.3. The hypothesis that there exists an index k such that F

(V) = 0 for all

V is sufﬁcient, but it is not necessary. However in the most of problems this hypothesis

is satisﬁed.

Proposition 2.2.1. Assume that locally the eigenvector r associated to a certain family

forms a local vector ﬁeld. Then we calculate the primary variables V on the rarefaction

waves in the

(x, t) plane independently of u, i.e., ﬁrst we obtain the primary variables

in the classical way, i.e., solving

(2.33.b), and then we calculate the secondary variable

u in terms of the primary variables from:

= u

−

ex p(

(

)),

(



−

n+1

(V(

))d

, (2.43)

where

= x/t,

−

) and u

−

is the “ﬁrst” value of u on the rarefaction wave,

i.e., u

= u

−

for

−

Proof: Using

(2.33) and (2.34) and since g

(V)=r

for i = 1,2,···,n we obtain V

independently of u by solving the system of differential equations:



,···,



=(g

, g

,···, g

), with

−

After obtaining V

(

), we use the expression for the last component of r in (2.36.a)

to solve du/d

= ug

n+1

, yielding (2.43). From Lemma 2.2.1, we know that

has the

form u

(V), so using Eqs. (2.34) and (2.43), we obtain

implicitly as:

= u

−

(V(

))ex p(

(

)). (2.44)

Since

depends only on u

−

and V on the integral curves, the results follows. 

Deﬁnition 2.2.2. Since only the ﬁrst n coordinates of eigenvectors are important to

obtain the integral curves in the space V (i.e. independently of u), it is useful to deﬁne

for any p-family the structures, see Rem. 2.2.2:

=(r

,···,r

) and



=(

,

,···,

), (2.45)

and

p,−

(V) :=

(W)/u

−

and

p,+

(V) :=

(W)/u

. (2.46)

Corollary 2.2.1. Assume that u

−

= 0, so we perform the change of variables:

−→

−

, (2.47)

The system

(2.2) can be written in the space of variables (x, T)=(x, u

−

t).

Proof: In V space, one can prove that the eigenvalues and eigenvectors have the

form

−

(V), with r given in Eq. (2.36. a). From Eqs. (2.34) and (2.44), it follows that

−

and

satisfy:

−

) and

(V(

))ex p(

(V(

))). (2.48)

Thus in the

(x, T) plane, the variables V and

do not depend on u, so the rarefaction

curve does not depend on u. The speed keeps the form

(2.43). 

Corollary 2.2.2. If r

n+1

≡ 0 in a region, then from Prop. 2.2.1 u is constant along the

corresponding rarefaction curve.

2.3 Shock waves

Typically, discontinuities or shocks appear in solutions of Riemann (or Cauchy) prob-

lems of non-linear hyperbolic or hyperbolic-elliptic equations (see

[60]). The Rankine-

Hugoniot (RH) condition for shocks needs to be satisﬁed as it expresses conservation

of mass. For a given state W

−

, the set of states W

that satisﬁes the RH condition is

called Rankine-Hugoniot locus (RH locus) of W

−

, which is denoted by RH (W

−

2.3.1 Shocks and the Rankine-Hugoniot condition

The main feature of the class of equations (2.1) is the existence of separate regions

where the system of balance equations (2.1

) reduces to different systems of conserva-

tion equations

(2.2). To obtain the complete Riemann problem solution it is necessary

to link these regions. Because often the term Q gives rise to fast changes between

regions, we consider shocks linking these different regions.

We apply a linear map E given in Def. 2.1.7 on the system

(2.1). Since E[Q]=0,

we obtain a system in conservative form:

∂

∂t

G(V)+

∂

∂x

(

uEF(V)

)

= 0, (2.49)

where

V = V

∈ Ω

for some j ∈PS.

Remark 2.3.1. We should use a single E linking neighboring physical situations. No-

tice that it is possible to use a single E for the complete system yielding the simplest

global formulation. However, we have proved that the weak solution does not depend

on E, so we can use different linear maps E in the interior of each physical situation,

see Proposition 2.1.2.

The RH condition applied to Eq. (2.49) is written as:



G(V

) −EG(V

−

)



= u

EF(V

) − u

−

EF(V

−

), (2.50)

where

−

) is the state on the left of the shock and (V

) is the state on the

right of the shock; v

is the shock speed. We specify the left conditions for the variables

V and u, but as we will see, the right conditions are speciﬁed only for one component

. The speed u

is always obtained from the RH condition (2.50).

The deﬁne the cumulative mass transfer in the shocks can be deﬁned as:

[Q(V)] := u

F(V

) −u

−

F(V

−

) −v



G(V

) −G(V

−

)



. (2.51)

Once the Riemann solution is computed,

(2.51) calculates the rate of transfer of each

chemical component in a shock.

We consider shocks of two different types: shocks within regions in a certain phys-

ical situation and shocks between different physical situations. In both cases V

−

and

are the primary variables associated to V

−

and V

, respectively.

Shock within a physical situation

Since

−

∈ Ω

for some j ∈PS, using the deﬁnition of R

and R

given in Eq.

(2.6) and the relationship (2.18) , the RH condition (2.50) can be written as:



) −G

−

)



= u

) −u

−

). (2.52)

Shock between distinct physical situations

Since

∈ Ω

and V

−

∈ Ω

for some pair j, k ∈PSwith k = j, using the deﬁnition of

, R

and R

given in Eq. (2.6) and the relationship (2.18), the RH condition

(2.50) can be written as:



) −G

−

)



= u

) −u

−

). (2.53)

In the following sections, we will always indicate functions of the shock left state

by the superscript

(− ) and functions of the right state of the shock by (+). This

notation encompasses both

(2.52) and (2.53).

2.3.2 The Rankine-Hugoniot locus and shock waves

Notice that in the RH condition (2.52) or (2.53) for each ﬁxed state (−), there is a

degree of freedom in the variables V, that is a 1-dimensional structure, generically.

This structure is called the RH locus.

Writing Eq. (2.53) as a linear homogeneous system, we obtain:







[

]

−

[

]

−

[

]

−

[

n+1

]

−

n+1

−

n+1











−





= 0, or M





−





= 0, (2.54)

where

] for i ∈Cis given by:

[

]

) − G

−

), F

= F

), F

−

= F

−

), (2.55)

The 3

×3 minors of the matrix M are denoted by M

pqs









−F

−





−F

−

[

]

−





for all distinct p, q and s in

C. (2.56)

The homogeneous linear system

(2.54) has a non-trivial solution, if only if

pqs

= det



pqs



= 0 for all distinct p, q and s in C. (2.57)

Deﬁnition 2.3.1. The RH locus is given by implicit expressions in the primary vari-

ables. For

−

) ﬁxed in some physical situation, the RH locus in the projected

space V consists of the V

that satisfy det(M

pqs

)=0 for all distinct p, q and s in C.

Notice that V

can lie in the same or in a distinct physical situation relatively to V

−

We denote the RH locus of a state V

−

as RH(V

−

Notice that there are

equations satisfying (2.57); however, under certain

hypotheses, the following Lemma guarantees that we only need to solve a set of

(n−1)

equations of type (2.57).

Since we want to characterize the solution of

(2.57), it is useful to deﬁne for each

∈C, the vector

≡D

−

) ≡



[G]

,−F

, F

−



. (2.58)

Using the notation

(2.58) we obtain the simple lemma:

Lemma 2.3.1. Let V

−

be ﬁxed. Assume that there exists two linearly independent

vectors

and D

for l,m ∈Cin a neighborhood B

). If there are (n − 1) dis-

tinct minors

lms

for s ∈C−{l, m} such that det(M

lms

)=0, then it follows that

det

pqs

)=0 for all distinct p, q and s in C.

Proof: There are

(n − 1) indices s ∈C−{l, m} such that det(M

lms

)=0.

Since

and D

are L.I., it follows that for all s ∈Cthe vector D

is a linear

combination of

and D

. Since any minor M

pqs

with distinct p, q and s in C has the

form:

pqs









it follows that det

pqs

)=0 and the lemma is proved. 

Deﬁnition 2.3.2. Consider V

−

∈P

and V

∈P

. Assume that V ∈P

belongs to a

1-dimensional locus where the vectors

and D

are linearly dependent (LD) for some

= q ∈C. We call this locus in Ω a linear degeneracy and denote it by X

−

Example 2.3.1. In Section 4.5, we show that in the single-phase gaseous situation (see

Example 2.1.2), the accumulations and ﬂuxes for steam and nitrogen are:

ϕθ

, F

ϕθ

, G

ϕθ

and F

ϕθ

where

= 1 −

. The constans

are given by by Eq. (4.24). The terms D

and D

are:



ϕθ



−



−

ϕθ

−





ϕθ



−



−

ϕθ

−



Let

−

) be ﬁxed. It is easy to see that in the plane (T,

) the straight line

(T,

−

) is the linear degeneracy X

−

). Of course, this straight line belongs

to the RH locus for

(− ).

Remark 2.3.2. If a linear degeneracy occurs a more detailed analysis is required.

Corollary 2.3.1. Given V

−

∈P

for some j ∈PS, assume that for V

∈RH(V

−

)

and in a neighborhood B

) there exists at least a pair of linearly independent (LI)

vectors

and D

with p,q ∈Cof the form (2.58). Assume also that for all l = m ∈C

and V

∈B

) − X

−

) the vectors D

and D

are linearly independent.

Then the RH locus is a 1-dimensional structure and it is obtained as the solution of

(n − 1) equations:

det









= 0, ∀ r ∈C−{p,q}. (2.59)

Moreover, X

−

) for l = m ∈Cbelongs to the RH locus.

Deﬁnition 2.3.3. The minimum set formed by the triplet of indices that deﬁne implic-

itly the RH locus are called RH indices and they are denoted by

. There are n −1

triplets, numbered from

−→

1 to

−−→

n −1. There is no a unique choice for C

So using Def.

(2.57), the RH locus is:

H = 0 for all l ∈{

−→

2,···,

−−→

n −1}∈C

, (2.60)

where each number l corresponds to a triplet in

Let us assume that RH (V

−

) was found. Now we want to determine u

. Let

p,q

∈C. Given V

−

∈ Ω

for some k ∈PS,ifD

and D

are LI in a neighborhood

), then:

det



] −F



= 0. (2.61)

The shock and speeds are obtained by solving the following system on the RH locus:



] −F







−u

−

−u

−



. (2.62)

In this case v

and u

can be written as function of u

−

and V

−

as:

= u

−

− F

−

] − F

]

and u

= u

−

] − F

−

]

] − F

]

. (2.63)

Of course, we could have chosen to solve v

and u

−

in terms of u

Remark 2.3.3. In the deﬁnitions that follow, all wave structures can be obtained

solely in the space of primary variables V. Using u

:= u

−

) and v

−

), we deﬁne:

−

) :=

−

and

−

) :=

−

)

. (2.64)

Let us index the rarefaction curves in the Riemann solution by l for l

= 1,2,···,

;

the shocks are indexed by m for m

= 1,2,···

. From the Sections 2.2 and 2.3.2,we

have proved the following:

Proposition 2.3.1. Let u

be positive. The primary variables V in the shock and

rarefaction curves do not depend on the left speed u

> 0. If a sequence of waves and

states solve the Riemann problem in the primary variables, for a given u

> 0 (or

> 0), then it is also a solution for any other u

> 0 (or u

> 0):



) if x < 0

,·) if x > 0.

(2.65)

Moreover, assume that for each m there are p, q

∈Csuch that the following inequality

is satisﬁed: F

q,m

p,m

] − F

p,m

q,m

] = 0. Then u

is given by:

= u

∏

l=1

ex p





+,l

−,l

(V(

))d



∏

m=1

−

q,m

p,m

] − F

−

p,m

q,m

]

q,m

p,m

] − F

p,m

q,m

]

. (2.66)

where g

is the (n + 1)-th component of eigenvector r,

−,l

and

+,l

are the ﬁrst and

the last values of

associated to the l-th rarefaction wave.

In the Proposition above, G

j,m

and F

j,m

represent the j-th components ( j ∈C) of G

and F on the m-th shock wave. Similarly ex p





0,l

n+1

(V(

))d



is computed along

the l-rarefaction curve, see

(2.43).

In Proposition 2.3.1, instead of u

we could have prescribed the right speed u

and

obtain u

and the other speeds as function of V and u

Proposition 2.3.2. Assume that u

= 0. If the speed u

in the initial data is modiﬁed

while

and V

are kept ﬁxed, the speed u

as well as the Riemann solutions are

rescaled in the

(x, t) plane, while the wave sequences and the values of V are kept

unaltered.

Proof: The rarefaction case was proved in Corollary 2.2.1. In the shock case,

performing the change of variable

(2.47) with u

−

= u

and using (2.52):

− G

−

. (2.67)

Deﬁning U :

= u/u

and

= v

, we have:

(1) If there exists a rarefaction before u

−

, from Corollary 2.2.1, we see that u

−

T (V), where T (V) := ex p(

(V(

))) depends only on V.

(2) If there is no rarefaction with u

−

= u

, we conclude that (2.67) can be rewritten

as:

− G

−

)=U

(V)F

−U

−

(V)F

−

, (2.68)

where U

−

(V)=T (V) if (1) is satisﬁed or U

−

(V)=1 if (2) is satisﬁed. Since U

(V)

depends only on V, using Corollary 2.2.1 we obtain the result. 

This Proposition leads to an important result:

Corollary 2.3.2. Assume the same hypotheses of Proposition 2.3.2. Then the Riemann

solution can be obtained in each physical situation ﬁrst in the primary variables V.

The solution for the speed can be obtained in terms of

V and u

by Eq. (2.66).The

same is true for rarefaction and shock speeds.

We can now prove a generalization of the Triple Shock Rule

[25]:

Proposition 2.3.3. (Triple Shock Rule.) Consider three states in different physical

situations with two speeds namely





(

)

and

(

−

)

. Assume that

−

∈RH(V

), V

∈RH

(

−

)

and V

∈RH(V

), with shock speeds v

+,−

, v

−,M

and v

M,+

. Then, either:

(1) v

+,−

= v

M,−

= v

+,M

;or

(2) G

) −G(V

−

) and G(V

) −G(V

) are LD.

Proof: The RH conditions can be written as:

+,−

(G(V

−

) −G(V

)) = u

−

F(V

−

) − u

F(V

), (2.69)

−,M

(G(V

) −G(V

−

)) = u

F(V

) −u

−

F(V

−

), (2.70)

M,+

(G(V

) −G(V

)) = u

F(V

) −u

F(V

). (2.71)

Add

(2.71), (2.70) and (2.69). We obtain:

M,+

−v

−,M

)



) −G(V

)



+(v

−,M

−v

+,−

)



) −G(V

−

)



= 0. (2.72)

Thus

(1) or (2) are satisﬁed. 

For future use we deﬁne  := (V

−

∗

) as:

 :=(v

M,∗

−v

−,+

)



) − G(V

−

)



+(v

−,M

−v

M,∗

)



) −G(V

−

)



; (2.73)

The limitation of the triple shock rule is that, in several applications we do not

know the intermediate speed. In such cases we need the following result:

Proposition 2.3.4. (Quadruple Shock Rule I.) Consider two physical situations, with

three primary points and three speeds, determining four states, namely

−

) in Ω

) and (V

∗

) in Ω

and (V

) either in Ω

or Ω

. Assume that there are

shocks between the following pairs of states:

(i)

−

) and (V

) with speed v

−,+

(ii)

−

) and (V

) with speed v

−,M

(iii)

) and (V

∗

) with speed v

M,∗

We conclude that:

R1) If

 = 0 and at least two shock speeds coincide, i.e., one of the following shock

speed equalities is satisﬁed:

either v

−,+

= v

−,M

or v

−,+

= v

M,∗

or v

−,M

= v

M,∗

. (2.74)

Then:

(1) u

∗

= u

;

(2) all three shock speeds are equal:

−,+

= v

−,M

= v

M,∗

. (2.75)

R2) On the other hand, if

 = 0, then u

∗

= u

Proof: Assume ﬁrst G

) − G(V

−

) and G(V

) − G(V

−

) are LI. Disregarding

the indices that indicate the physical situations in the cumulative and ﬂux terms, the

RH condition for

−

)-(V

), (V

−

)-(V

) and (V

)-(V

∗

) can be

written, respectively as:

−,+

(G(V

) −G(V

−

)) = u

F(V

) − u

−

F(V

−

), (2.76)

−,M

(G(V

) −G(V

−

)) = u

F(V

) −u

−

F(V

−

), (2.77)

M,∗

(G(V

) − G(V

)) = u

∗

F(V

) − u

F(V

). (2.78)

Assume now that Eq. (2.74. a) is satisﬁed. Substituting v

−,+

= v

−,M

= v in Eqs.

(2.76

) and (2.77) and subtracting the resulting equations, we obtain:

(G(V

) −G(V

)) = u

F(V

) −u

F(V

), (2.79)

From Eq.

(2.63), notice that the shock speeds depend solely of the primary vari-

ables and speed on the left state and on the primary variable of the right state. From

Eqs.

(2.78) and (2.79), we can see that the left and right states as well as left speed

are the same, so u

∗

= u

. Eq. (2.75) is satisﬁed also because the shock speed depends

just on the initial and ﬁnal states.

To prove R2, we subtract

(2.76) from the sum of (2.77) with (2.78), obtaining:

 =(u

−u

∗

)F(V

 = 0, then u

∗

= u

and the result follows.

The other cases are proved similarly.



Remark 2.3.4. The equality  = 0 is a particular case of the Triple Shock Rule.

Generically, we do not know if V

∈RH(V

−

)



RH(V

). Therefore, we rewrite

Prop. 2.3.4 in the more practical form:

Proposition 2.3.5. (Quadruple Shock Rule II.) Consider two physical situations Ω

and Ω

for k, j ∈PS, with four primary points and four speeds, determining four

states, namely

−

) in Ω

, (V

) in Ω

, (V

) and (V

∗

) either in Ω

Ω

. Assume that there are shocks between the following pairs of states:

(i)

−

) and (V

) with speed v

−,+

(ii)

−

) and (V

) with speed v

−,M

(iii)

) and (V

∗

) with speed v

M,∗

Assume also that the following conditions are satisﬁed:

(I) G

) −G(V

−

) and G(V

∗

) −G(V

) are LI;

(II) two shock speeds coincide, i.e., at least one of the following shock speed equali-

ties is satisﬁed:

either v

−,+

= v

−,M

or v

−,+

= v

M,∗

or v

−,M

= v

M,∗

. (2.80)

(III) V

and V

∗

have a coordinate V

with coinciding values;

(IV) The curve

RH(V

) does not possess two different points with the same coordi-

nate V

having coinciding values.

Then

(1) V

∗

;

(2) u

∗

= u

;

(3) all three shock speeds are equal:

−,+

= v

−,M

= v

M,∗

. (2.81)

Proof: Disregarding the indices that indicate the physical situations in the accu-

mulation and ﬂux terms, the RH conditions for

−

)-(V

), (V

−

)-(V

)

and (V

)-(V

∗

) are written, respectively as:

−,+

(G(V

) −G(V

−

)) = u

F(V

) − u

−

F(V

−

), (2.82)

−,M

(G(V

) −G(V

−

)) = u

F(V

) −u

−

F(V

−

), (2.83)

M,∗

(G(V

∗

) − G(V

)) = u

∗

F(V

∗

) −u

F(V

). (2.84)

Assume that now Eq.

(2.80.a) is satisﬁed. Substituting v

−,+

= v

−,M

= v in Eqs.

(2.82

) and (2.83) and subtracting the resulting equations, we obtain:

(G(V

) −G(V

)) = u

F(V

) −u

F(V

). (2.85)

Notice that Eqs.

(2.85) and (2.84) deﬁne implicitly the RH locus from (V

). Since

the RH locus in the variable V depends solely on V

and the accumulation and ﬂux

functions, the RH locus deﬁned by Eqs.

(2.85) and (2.84) coincide. Because (III) and

(IV) are satisﬁed, it follows that V

∗

= V

Now from Eqs.

(2.64), notice that the and shock speeds depend solely on the pri-

mary variables and speed of the left state and on the primary variable of the right

state. From Eqs.

(2.84) and (2.85), we can see that the left and the right states are

the same for each expression and deﬁne the same RH locus, so u

∗

= u

and Eq. (2.81)

is satisﬁed.

The other cases are proved similarly.



2.3.3 Extension of the Bethe-Wendroff theorem

There are loci where the solutions change topology such as: secondary bifurcations,

coincidences, double contacts, inﬂections, hystereses and interior boundary contacts.

So the Riemann problem cannot be solved by the classical construction

[43]. The

Bethe-Wendroff theorem is an important tool to characterize such loci.

We prove an extension of the important Bethe-Wendroff theorem for shocks be-

tween regions in different situations. We recall that the left eigenvector

 does not

depend on u. We recall the notation

[G]=(G

− G

−

), where G

= G(V

) and

−

= G(V

−

Proposition 2.3.6. Let v

−

) be the shock speed between different physical sit-

uations. Assume that



) · [G] = 0. Then v

has a critical point at W

and so

−

) has a critical point at V

, if and only if:

−

p,+

) for the family p, (2.86)

where

p,+

(V) is given by Eq. ( 2.46) and

−

) is given by (2.64.c).

Proof: Assuming that the RH curve can be parametrized by

in a neighborhood

), we can write the RH condition as:

(G(V(

)) −G

−

)=u(

)F(V(

)) −u

−

, (2.87)

where v

:= v

(

). Differentiating (2.87) with respect to

we obtain:

(G(V(

)) −G

−

)+v

∂G(V(

))

∂W

∂

(

)F(V(

))

)

∂W

, (2.88)

Setting

such that (V(

),u(

)) = W(

)=W

=(V

), Eq. (2.88) yields:

[G]

+ v

∂G

∂W

∂

(

)

∂W

, (2.89)

where W

=(V, u). Assume ﬁrst that (2.86) is satisﬁed. Notice that if

−

, then for (V

,u = u

),wehave

= u

and v

−

) (we

dropped the family index p). Substituting them in

(2.89) we obtain at (+) = (V

)

the expression (2.89) with v

:= u

/u. Let  the left eigenvector associated to

;

taking the inner product of

(2.89) at (V

) by , we obtain:

 · [G]

+  ·



∂G

∂W

−

∂

(

)

∂W



= 0. (2.90)

Since

 is an eigenvector associated to

, the second term of (2.90) is zero and it follows

that:

 ·[G]

= 0.

Since by hypothesis

 · [G] = 0, we obtain that dv

= 0 and the shock speed is

critical.

On other hand, assume that v

has a critical point, so dv

= 0 and Eq. (2.89)

reduces to:

∂G

∂W

∂

(

)

∂W

, or



∂

(

)

∂W

−v

∂G

∂W



= 0. (2.91)

Eq.

(2.91) has a solution if, only if,

−

) so

−



Following Oleinik [53] and Liu [47,48], we can obtain a useful result for equations

in the form

(2.2):

Proposition 2.3.7. Let

−

) be the shock speed. Assume that 

·(G

−G

−

) =

0 for , r and

belong to the same family p. Assume that

has a critical point at V

Then the following relations are true:

i) If

has a minimum:

(



·∂

)(

∇

·r

)



·[G]

0. (2.92)

ii) If

has a maximum:

(



·∂

)(

∇

·r

)



·[G]

0, (2.93)

where ∂

(

∂G/∂W

)

(

), 

:= (V

) and r := r(V

). Moreover the in-

equalities

(2.92) and (2.93) do not depend on the speed u.

Proof: Assume that the shock and rarefaction curves can be parametrized by

in a neighborhood of (V

) (if the curves exist, these parametrizations exists in a

small neighborhood of

)).

Differentiating Eq.

(2.88) again we obtain:

(G(V(

)) −G

−

)+2

∂G(V(

))

∂W

+ v

∂

G(V(

))

∂





∂G(V(

))

∂W

∂

(

)F(V(

))

)

∂





∂

(

)F(V(

))

)

∂

, (2.94)

Set

and denote W(

)=W

. Notice that since 

· [G] = 0 and

has a

critical point at V

, by Proposition 2.3.6 the equality dv

= 0 is satisﬁed at V

and dW/d

= r := r

.So(2.94) reduces to:

[G]



∂

(

r,r

)

∂G

∂W



∂

(

)

∂

(

r,r

)

∂

(

)

∂

, (2.95)

We recall that Eq.

(2.30) can be written as:

∂G

∂W

∂

(

)

∂W

r. (2.96)

Differentiating

(2.96) with respect to

along a rarefaction curve starting at (V

we obtain:

∂G

∂W

∂

(

r,r

)

∂G

∂W

∂

(

)

∂

(

r,r

)

∂

(

)

∂W

, (2.97)

where d

= ∇

·r and we set

in (2.97). Manipulations of Eqs. (2.95) and

(2.97) yield:

[G]

−

∂G

∂W



∂G

∂W

∇

·r −

∂

(

)

∂W



−



= 0 (2.98)

We multiply

(2.98) by the left eigenvector . Notice that the last term in (2.98) van-

ishes, i.e.:

 ·



∂G

∂W

−

∂

(

)

∂W



−



= 0, (2.99)

because from Lemma 2.2.1,  is eigenvector associated to

. The system (2.98) re-

duces to:

 ·[G]

−



 ·

∂G

∂W



∇

·r = 0, (2.100)

after some manipulations, we see that the inequalities

(2.92) and (2.93) are satisﬁed.

Notice that the inequalities

(2.92) and (2.93) do not depend on the speed u because:

i)- The matrix ∂

has the form (2.28) and by Lemma 2.2.1, the right eigenvec-

tors r have the form

(2.36.a), so the expression ∂

r is independent of u.

ii)- From Lemma 2.2.1, we know that

 does not depend on the speed u,so ·∂

does not depend on u.

Since G is solely a function of V, the claim follows.



Lemma 2.3.2. Assume that the shock and rarefaction curves can be parametrized by

in a neighborhood B

), with

at (V

). Assume also that G is smooth,

−

is sufﬁciently close to V

and

−

+,p

) for some family p for

, 

and r, then we approximate 

·[G] by:



·[G] A(

−

;

)=

∂

(

−

), (2.101)

where V

−

= V(

−

), V

= V(

) and |

−

| <



Proof: Since G is smooth, we can approximate

[G] by:

[G]=∂

(

−

)+O



(

−

)



Disregarding terms of order 2, since

−

), by Proposition 2.3.6 it fol-

lows that dW

(

)/d

= r(V

), so Eq. (2.101) is satisﬁed and the Lemma follows.



Notice that it is not necessary to know u

−

and u

, because (2.101) does not depend

on the speed u.

Proposition 2.3.8. Assume the hypotheses of Proposition 2.3.7. Then the following

relationships are valid:

i) If

−

) > 0, then

has a minimum.

ii) If

−

) < 0, then

has a maximum.

Proof: First, let V

and V

−

satisfy hypotheses of Lemma 2.3.2. From Eqs. (2.100)

and (2.101), we know that d

, satisﬁes:





(

−

)∇

·r

∇

·r

−

, (2.102)

Observe that if

∇

·r

> 0, then by the geometrical compatibility it is necessary that

−

. On the other hand, if ∇

· r

< 0, then

−

, by the same argument.

Using Eq.

(2.102), we obtain that d

> 0, i.e., v

(and

) is minimum at V

Now let V

−

and V

be such that A(

−

) > 0. Since 

∇

· r

depends

solely on V

, it is necessary to analyze only 

·[G]. Since A(

−

) > 0, 

·[G] has

the same sign as



(

−

), then d

s > 0 and v

(and so

) is minimum

at V

Now let V

−

and V

be such that A(

−

) < 0, then 

·[G] and 

(

−

)

have opposite signs, so d

s < 0 and v

(and

) is maximum at V

. 

Secondary bifurcation manifold

The RH locus is obtained implicitly by

−

)=0 for l ∈C

, where H

: R

−→

, and V

−

and V

are the primary variables. Notice that RH is 1-dimensional lo-

cus, deﬁned by expressions

(2.59). However, sometimes these implicit expressions fail

to deﬁne a curve for V

in the different physical situations, typically the curve self-

intersects. We call these points V

secondary bifurcation and, from the implicit

function theorem, they are the

(+) states for which there exists a (−) state such that

the following equalities are satisﬁed:

= 0 for l ∈{

−→

2,···,

−−→

n −1} (2.103)

and

∂H

∂V

does not have maximal rank for all l ∈{

−→

2,···,

−−→

n −1}, (2.104)

where

is deﬁned implicitly by (2.57) and J

is the Jacobian matrix of the vectors

for l ∈{

−→

2,···,

−−→

n −1} corresponds to a triplet of C

. The Jacobian is an

(

n −1

)

matrix in the following form:







∂

−→

∂V

∂H

−→

∂V

···

∂H

−→

∂V

∂H

−→

∂V

∂H

−→

∂V

···

∂H

−→

∂V

. s

∂

−−→

n−1

∂V

∂H

−−→

n−1

∂V

···

∂H

−−→

n−1

∂V







(2.105)

The Jacobian does not have maximal rank if any two minors of size

(

n −1

)

(

n −1

)

have determinant zero. Following the same ideas to obtain RH locus, we can prove

that:

Corollary 2.3.3. The secondary bifurcation is an

(

n −1

)

dimensional structure in the

space of n variables V

, i.e., it is generically codimension−1 in the space V

2.4 Regular RH locus

Deﬁnition 2.4.1. Given a state V

−

∈P

for j ∈PS, we say that the RH locus for V

−

is regular if for all k ∈PS(including k = j) and for each V

∈P

with V

= V

−

the shock speed v

and the speed u are ﬁnite and there are at least two distinct indices

p,q

∈Cdepending on k and j such that:

] − F

] = 0; (2.106)

] = 0 and the following limit exist:

lim

−→V

−

]

. (2.107)

3. The following inequality is satisﬁed:

−

= F

−

lim

−→V

−

]

. (2.108)

Remark 2.4.1. When V

−

is a frontier point, then the limits in (2.107) and (2.108) may

not be well deﬁned. However, since we are interested in obtaining the solution in the

physical situation

we regard the limits as lateral limits.

2.4.1 Small shocks

A shock between two states W

−

and W

is said to be small if there exists a sufﬁciently

small



> 0 such that |W

−W

−

| <



. For this type of shock we can prove important

results that can be extended to more general shocks.

Speed for regular small shocks

Lemma 2.4.1. Assume that F and G are

for V ∈P

. In a regular system, when the

primary variable V

converges to V

−

the speed u

converges to u

−

Proof:

Let V

−

be ﬁxed. From eq. (2.63.b), we deﬁne for all V

∈P

the expression:

= u

−

] − F

−

]

] − F

]

. (2.109)

We can rewrite

(2.109) as:

−u

−

)



−

] − F

−

]



= u



][G

] − [F

][G

]



. (2.110)

Assume that [G

] = 0. Dividing (2.110) by [G

], we obtain:

−u

−

)



−

]

−



= u



]

[

]

−[

]



. (2.111)

Let V

∈RH(V

−

) tend to V

−

and remember that the limit (2.107) exists and (2.108)

is satisﬁed. From Eqs. (2.106)-(2.108), u

is bounded. Notice that [F

] and [F

] tend

to zero when V

tends to V

−

,sou = u

= u

−

. 

Extension of Lax theory

Often shocks between regions are not weak and the cumulative and ﬂux terms are not

smooth; the Lax theory cannot be extended to this type of shocks. However, within

a physical situation and under certain hypotheses we can obtain a theory for the

Riemann solution similar to Lax’s. We consider a physical situation j

∈PS.

We say that the system

(2.2) represents a strictly hyperbolic physical situation if

there are n distinct eigenvalues

= u

(V) that satisfy:

(V) <

(V) < ···<

(V). (2.112)

We know that if

(2.112) is satisﬁed, there exist n right r

(and n left 

) eigenvectors

that are LI.

From

(2.36) of Lemma 2.2.1, the right and left eigenvectors of the system (2.2) have

the form r

=(g

(V), ··· , g

(V), ug

(V)) and 

=(

(V), ··· ,

(V), 

(V)). No-

tice that neither the right nor the left vectors form bases for the space

(V,u).

Nevertheless, the left and right eigenvectors satisfy the following lemma:

Lemma 2.4.2. Consider

=

, and the associate right and left eigenvectors r

and



, then:



∂

= 0. (2.113)

We can prove now that if the system

(2.2) is satisﬁed in the regular case there

exists a local basis of shock curves in the space of primary variables in each physical

situation.

Proposition 2.4.1. Let W

−

=(V

−

) be any point in phase space with V

−

in the

interior of

. Then there are n C

one-parameter curves; of states V

= V

(

) for

= 1,2,···,n and

sufﬁciently small, where V

(0)=V

−

. The points V

(

) satisfy the

RH condition

(2.52). Moreover the following relationships are satisﬁed:

(

)=V

−

∂

−

)

−

)+O(

) (2.114)

and

−

∂

−

) ·r

−

)+O(

). (2.115)

We can also obtain the speed u

= u

(V(

)) from V

(

) and Eq. (2.64.b),fork =

1,2,··· ,n.

Proof: We follow the same idea of the original theorem of Lax [43]. We can write:

(G(V) −G(V

−

)) =



(G(V

−

(V −V

−

)))d



∂

G(V

−

(V −V

−

))(V −V

−

(2.116)

Similarly, from Eq.

(2.64.a), we can rewrite uF(V) − u

−

F(V

−

). Using u = u

−

, with

−

,V) we deﬁne

W −

−

(

V −V

−

−1

)

, and F as follows:

(V) −u

−

F(V

−

)=u

−





(1 +

(

−1)) F(V

−

(V −V

−

))



= F(

−

), (2.117)

where

F := F(W

−

) is an (n + 1) ×(n + 1) matrix given by:

F := u

−





(1 +

(

−1))∂

F(V

−

(V −V

−

)), F(V

−

(V −V

−

))



and

W −

−

(

V −V

−

−1

)

From Eq.

(2.116) we can see that v

(G(V) − G(V

−

)) does not depend on u. How-

ever, it is necessary rewrite it in the variable

W, so we deﬁne the

(n + 1) × (n + 1 )

matrix G(V) as:

G(V) :=





∂

G(V

−

(V −V

−

),0)



Thus, we can rewrite v

(G(V) −G(V

−

)) as:

(G(V) −G(V

−

)) = v

G(V)(

−W

−

). (2.118)

Using Eqs.

(2.117) and (2.118), the Rankine-Hugoniot condition (2.50) becomes:

(

F(V) −v

G(V)

)

(V −V

−

−1)=0. (2.119)

Using

(2.119), we know that a state V ∈Nbelongs to the RHL locus of V

−

, if, only

if, there exists a k

∈{1,2,··· ,n} such that:

−

;V)=

−

,V). (2.120)

and that

(V −V

−

−1) is a right eigenvector of F(V) −

G(V) associated to

−

,V).

Notice that when V converges to V

−

by the Lemma 2.4.1 the speed u converges to

−

, so the matrices F(V) −→ A and G(V) −→ B, with B and A given by Eq. (2.28) and

(2.29). Since A −

B has n distinct eigenvalues that satisfy (2.112), and the functions

F(V) and G( V) are continuous, there exists a neighborhood of N of V

−

in the topology

and n distinct functions

(V,V

−

) that are the eigenvalues of F(V) −

G(V),

that satisfy

−

We denote the left eigenvectors of the system F( V) −

G(V) by (L

(V))

. Since

(V))

is not a basis in (V,u) space, for (V − V

−

− 1) to be a right eigenvector,

the condition:

(V)G(V −V

−

−1)=0 j = k (2.121)

is only necessary. Notice that the ﬁrst n coordinates of the eigenvectors are uniquely

deﬁned by Eq.

(2.121). Moreover, if V ∈Nbelongs to the RHL of V

−

, the speed u is

given by Eq.

(2.64.a), thus Eq. (2.121) deﬁnes uniquely the eigenvectors of F(V) −

G(V).

Since

G has a zero column, Eq. (2.121) does not depend on

−1, and Eq. (2.121) is

a system with

(n −1) equations in n unknowns V. This system which can be written

as:

(V)=M

(V)(V −V

−

)=0, (2.122)

where

(V)=









(V)





k−1

(V)





k+1

(V)



(

(V)

)







G, (2.123)

with

G =



∂

G(V

−

(V −V

−

))d

Notice that at V

−

we obtain:

−

)=0 and ∂

−

)=M

−

). (2.124)

Since L

−

)=(V

−

) and

G have rank n, the matrix M

(V) also rank n −1, yielding

that the matrix ∂

−

) has rank n − 1. By the implicit function theorem, there

exists a one-parameter family of functions V

(



), for 1 ≤ k ≤ n and |



|≤



small

enough, with

(0)=V

−

, (2.125)

and the n different waves with shock speed satisfying:

lim

V−→V

−

;V)=

−

) for 1 ≤ k ≤ n. (2.126)

Letting



tend to zero in Eqs. (2.121), we obtain that dV(0)/d



is parallel to

−

). After a change in parametrization, we obtain

(0) /d



−

). (2.127)

After obtaining V, Eq.

(2.87) determines u. Differentiating (2.89), using that v

satisﬁes Eq. (2.87), from (2.127) it follows that:







(0)=r

−

) for k = 1,2,··· ,n. (2.128)

We drop the index k in the expressions. Now using Eqs. (2.94) and (2.97) we can prove

the expansion

(2.115). To do so, ﬁrst we subtract (2.94) from (2.97) and set



= 0 we

obtain:







∂G

∂W



−

∂G

∂W

∇

·r +



∂G

∂W

−

∂

(

)

∂W





−





= 0, (2.129)

with

−

) and r = r

−

) for some k = 1,2,··· ,n. Multiplying Eq.

(2.129) by  and using (2.99), we obtain:







∂G

∂W







∂G

∂W



∇

·r, (2.130)

Since

∂

Gr = 0, we get:



∇

·r. (2.131)

Substituting

(2.131) in Eq. (2.129) and noticing that dr/d



= ∂

r ·r, we obtain:



∂G

∂W

−

∂

(

)

∂W





−





. (2.132)

Therefore, there exists a real number

such that













r,u

n+1



n+1



. (2.133)

Changing the parametrization by setting:



−

αθ

. (2.134)

Then using Taylor expansion, Eqs.

(2.126) and (2.131) yield Eq. (2.114):



∇

·r + O(



∇

·r + O(



Eq.

(2.115) is obtained by using Eq. (2.133):

(



)=V

−



(

∂

)



)

−

∂

+ O(



). 

Under certain hypotheses, we can prove the Theorem of Lax to construct the local

Riemann solution within each physical situation.

2.4.2 Other degeneracies of RH locus

In addition to the linear degeneracy 2.3.2, we study another degeneracy that seldom

occurs. To do so, we utilize the notation:

−

) := F

] − F

Proposition 2.3.1 is valid if non zero denominators can be found for Eq.

(2.66). So it is

necessary to study the behavior of the solution when

−

)=0 for all p , q ∈C.

For a ﬁxed pair p, q

∈C, we can follow the same steps of Proposition 4.4.3 and prove:

Proposition 2.4.2. If

−

)=0 for all p,q ∈C, it follows that:

and F

−

, ∀ k ∈C.

where

and

are numbers depending on [G], F

−

and F

. Moreover the shock speed

satisﬁes:

−

) −

−

where

−

) is given by Eq. (2.64.a).

2.4.3 The sign of u

on regular Rankine-Hugoniot loci

In the previous Propositions, it is necessary that u is never zero. Moreover, it is

useful that u and the injection speed u

−

have the same sign. From Eq. (2.43), these

properties are clearly true on the integral curves (and so on rarefaction waves), but

they are not obvious on shock waves. We now prove a sign invariance for the connected

branches of the RHL containing

−

), that we denote by CRH L.

To do so, for a ﬁxed V

−

in some P

, we deﬁne for each V ∈P

(including k = j ) and

for all index pairs p,q

∈C, the following continuous functions Ξ

:= Ξ

(V,V

−

) from

the denominator and numerator of the expression for u given in

(2.64.b):



−

− G

−

) − F

−

−G

−

)



−G

−

) − F

− G

−

)



, (2.135)

where the cumulative and ﬂux terms depend on j and k and G

= G(V), G

−

= G(V

−

= F(V) and F

−

= F(V

−

). The shock curves can occur between different physical

situations, therefore it is useful to represent the states on RHL as belonging to Ω

Using

(2.18), we can write Ξ

:= Ξ

−

Remark 2.4.2. Notice that for

−

and V

∈RHthere exists at least a pair p, q ∈C

such that

−

) = 0 in the regular case. By the hypothesis of regular case,

the speed u is ﬁnite, then using the Eq.

(2.64.b) we can see that the denominator of

expression

(2.64.b) is zero, if only if, its numerator is zero. We conclude that if Ξ

zero for all p, q, then the numerator of

(2.64.b) is zero for any p, q, that is D

= 0 for

all p, q. It is impossible because it violates the hypothesis 1 of Deﬁnition 2.4.1.

It follows from Prop. 2.1.1 that:

Lemma 2.4.3. Ξ

are continuous functions for all p, q ∈C.

Remark 2.4.3. Notice that for u

given by (2.63), the inequality u

−

< 0 is satisﬁed

if, only if, Ξ

−

) < 0.

Remark 2.4.4. For two ﬁxed states

−

and V

∈RH(V

−

) for all pairs p, q ∈Ceither

−

) ≤ 0 or Ξ

−

) ≥ 0. This fact occurs because for points in RH locus

the speed is uniquely deﬁned.

Lemma 2.4.4. For

−

∈ Ω

, the expression Ξ

given by Eq. (2.135) satisﬁes:

1. Ξ

= 0 for V

= V

−

and for all p, q ∈C;

2. There exists an



> 0 and and indices p,q ∈Csuch that for V∈CRHL





−

the inequality Ξ

−

) > 0 is satisﬁed.

Proof: The assertion

(1) is immediate. Because u

= u

−

when V

converges

−

, there exists an



> 0 such that for each V∈CRHL





−

), it follows that

−

,V) ≥ 0. Moreover, since we assume that the system is regular in the sense of

Deﬁnition 2.4.1, for each V

∈ B



−

)



CRH L\V

−

there exists at least a pair p,q ∈C

such that Ξ

−

,V) > 0, and (2) is valid. 

Deﬁnition 2.4.2. For ﬁxed V

−

∈ Ω

, we utilize Ξ

given in (2.135) to deﬁne the two

disjoint sets:

−,0



∈ CRH L



Ω

, for all k = PS, such that for all p, q ∈C

−

) ≤ 0, but there is at least a pair l,m ∈C, such that Ξ

−

) < 0



(2.136)

and

+,0



∈ CRH L



Ω

, for all k ∈PS, such that for all p,q ∈C

−

) ≥ 0, but there is at least a pair l,m ∈C, such that Ξ

−

) > 0



(2.137)

These two sets form a disjoint decomposition of CRHL in Ω as summarized in the

Lemma below:

Lemma 2.4.5.

−,0



+,0

= ∅ and J

−,0



+,0

= CRH L(V

−

)\V

−

Proof: Let a

V∈J

−,0



+,0

.IfV∈J

−,0

then Ξ

−

,V) ≤ 0 for all p, q ∈Cand

there exists at least a index pair l, m such that Ξ

−

,V) < 0; similarly, if V∈J

+,0

then there exists a index pair i, j

∈Csuch that Ξ

−

,V) > 0. From Remark 2.4.4,

we know that it does not occurs, so

−,0



+,0

= ∅ .

Consider now

V∈CRH L(V

−

). It is immediate that V belongs to the J

−,0



+,0

Notice that

−

there is no in J

−,0



+,0

because of Lemma 2.4.4. 

From the continuity of Ξ

−

,V) we obtain the following lemma.

Lemma 2.4.6. The sets J

−,0

and J

+,0

are connected and open in CRHL.

Since CRHL is connected and from Lemma 2.4.4 we know that there is a pair

p,q

∈Csuch that Ξ

−

,V) for V belongs to the CRHL(V

−

) in a neighborhood, the

Lemmas 2.4.5 and 2.4.6 we yield:

Proposition 2.4.3. Let any

−

∈ Ω

. The speed u has the same sign of the speed u

−

for the state V

−

on the CRHL.

We show some applications of the theory developed in Appendix C.

CHAPTER 3

Mathematical modelling of thermal oil recovery with

distillation

We present a physical model for steam and gaseous volatile oil injection into a hori-

zontal, linear porous medium ﬁlled with oil and water. The oil in the porous medium

consists of dead oil and dissolved volatile oil. Dead oil is an oil with negligible va-

por pressure while volatile oil is an oil with substantial vapor pressure. The model

is based on mass balance and energy conservation equations as well as on Darcy’s

law. We present the main physical deﬁnitions and summarize the equations of the

model. This system of equations is a representative example of ﬂow in porous media

with mass interchange between phases (in the absence of gravitational effects). Such

ﬂows are not represented by conservation laws or hyperbolic equations both because

they have source terms and because they are not evolutionary in all the variables.

The model contains a system of 6 equations: 5 equations for mass balance and one

equation for energy conservation, which is generically represented by:

∂

∂t

G(V)+

∂

∂x

F(V)=Q(V) , (3.1)

where

V =(s

, T,v

, y

) ⊂ Ω ⊂ R

represents the 5 dependent variables to be de-

termined for all x and t

≥ 0. The cumulative and ﬂux terms are G = {G

,···,G

}

and F = {F

,···,F

}. Here s

and s

are the water and gas saturations, T is

the temperature, v

and y

are the volatile oil volume fraction in the oleic gaseous

phases, respectively. The real-valued dependent variable u is the Darcy speed. In

(3.1), uF

is the ﬂux for the conserved quantity G

and ∂G

/∂t is the corresponding

accumulation term, in each of the 6 equations. The source term Q

= {Q

, Q

,···, Q

}

represents the water and volatile oil mass transfer between the phases; the compo-

nent Q

vanishes because it represents the conservation of energy total. The functions

and F

are continuous in the whole domain Ω, but later we will see that they are

only piecewise smooth. In the

(x, t) plane the terms Q often generate fast variations

in the variables V(x, t), so that Q(V(x, t)) may be regarded as distribution. Eq. (3.1)

has another important feature: the variable u does not appear in the accumulation

term, but only in the ﬂux term.

Thermodynamics plays a very important role in the physical phenomena modelled

by Eq.

(3.1). The thermodynamical behavior of interest can be divided into systems in

equilibrium or in quasi-equilibrium. For such types of behavior we can use the Gibb’s

phase rule, which determines the number of thermodynamical degrees of freedom:

= c − p + 2, (3.2)

where f is the number of thermodynamical degrees of freedom, c is the number of

components and p is the number of phases. Such ﬂows encompass particular physi-

cal situations, where one or several phases or chemical components are missing. We

study the possible physical component and phase mixture situations in thermody-

namical equilibrium or quasi-equilibrium that appear in the Riemann solution. For

equilibrium situations, Q

= 0. In the situations in quasi-equilibrium Q is not zero

but very small, so that the thermodynamical degrees of freedom still satisfy Eq.

(3.2)

in all physical situations. Finally, Q has fast variation in the regions between the

physical situations. In the model considered in this work, there is a map E that can

be applied to the balance system

(3.1). This map reduces (3.1) to simpler systems of

conservation laws, that is, E eliminates the terms Q; F and G are obtained from

F and

G through E and reduces to equation of form (2.2) After this elimination, further sim-

pliﬁcations may occur in each physical situation, so that each situation is described

by a set of variables V that is a subset of the set of variables

V. Each of the systems

of type

(2.2) has fewer equations than the complete system (3.1). It is useful to deﬁne

=(V, u). Although Eq. (2.2) is a conservation law, it is not a hyperbolic system, be-

cause the variable u appears solely in the ﬂux term. Indeed, Eq.

(2.2) has an inﬁnite

speed mode associated to u. Under certain hypotheses we show that if u

> 0 the Rie-

mann solution in the variables

V does not depend on u. Thus the unknown variables

in the system of equation for each physical situation are called primary variables,to

be determined from the corresponding version of (2.2). Because the variable u can be

obtained from the primary variables, we call it a secondary variable. The variables

that are constant or that are obtained in a simple way from the primary variables are

called trivial variables.

3.1 Physical and Mathematical Models

We consider the injection of steam and volatile oil into a linear horizontal porous rock

cylinder with constant porosity and absolute permeability, containing water and oil

initially. The oil consists of dead oil with or without volatile oil. There are three chem-

ical components: water, volatile oil and dead oil,

(w,v, d) as well as three immiscible

phases: aqueous (liquid water), oleic (liquid oil) and gaseous,

(w,o, g). We use the

following convention in subscripts for physical properties: the ﬁrst subscript

(w,o, g)

refers to the phase and the second subscript (w,v,d) refers to the component.

We use simple mixing rules. We disregard any heat of mixing both in the gaseous

phase between steam and volatile oil and in the oleic phase between volatile oil and

dead oil. Moreover we disregard any volume contraction effects resulting from mixing.

The concentration





of (dead) volatile oil in the oleic phase is denoted as

(

)

. In the same way we deﬁne the concentration of the volatile oil (water vapor) in

the gaseous phase as





. For ideal ﬂuids we obtain

= 1,

= 1, (3.3)

where the capital subscripts indicates that the component is in a pure phase. The

pure liquid densities of volatile and dead oil in the oleic phase, denoted by

[kg/m

], are considered to be independent of temperature; the pure vapor densities of

steam and volatile oil in the gaseous phase, denoted by

and

, are considered

to obey the ideal gas law, i.e.,

, (3.4)

where M

, M

denote the molar weights of water and volatile oil respectively. T is

the variable temperature and the gas constant is R

= 8.31[J/mol /K]. The pressure P

is not a variable in this problem, but the ﬁxed prevailing pressure value.

We deﬁne x

as the mole fraction of volatile oil in the oleic phase and y

the

mole fraction of volatile oil in the vapor phase. In some physical situations, in addi-

tion to temperature and pressure, it is necessary to specify another thermodynamical

variable: either x

or y

, which are explained in the following text.

Using Clausius-Clapeyron law, we can write the the pure water vapor pressure P

as:

(

)

exp



−M







−





, (3.5)

where Λ





[

J/kg

]

is its evaporation heat at its normal boiling temperature T

[

]

at P

, the atmospheric pressure. Eq. (3.5) is valid for any temperature in situations

where there exists liquid water.

We also use Clausius-Clapeyron law for the vapor pressure of pure volatile oil:

(

)

exp



−M







−





, (3.6)

where Λ





is its evaporation heat at its normal boiling temperature T

and atmo-

spheric pressure. Eq.

(3.6) is valid for any temperature below T

. In addition, we use

Raoult’s law, which states that the vapor pressure of volatile oil is equal to its pure

vapor pressure given by Eq. (3.6

) times the mole fraction x

of volatile oil dissolved

in the oil phase. Therefore we obtain

(

)

exp



−M







−





. (3.7)

We assume that the prevailing pressure P is the sum of the two vapor pressures: P

and P

We can use Eq. (3.3.a) and deﬁne the volatile oil volume fraction and the dead oil

volume fraction in the oleic phase respectively, as:

,1−v

(3.8)

Similarly, we can use Eq. (3.3.b) and deﬁne the volatile oil volume fraction and the

steam water in the gaseous phase respectively, as:

,1− y

(3.9)

We can derive an equation that relates the oleic phase density to the mole frac-

tion x

. From the deﬁnition of the mole fraction (moles volatile oil / total moles) we

immediately observe that

. (3.10)

The concentration of steam and volatile oil in the gaseous phase is given by:

. (3.11)

In the physical situations in which there exists volatile oil in the oleic phase, we can

use Eqs.

(3.7), (3.9. a) and (3.11.b), we derive the following expression for the mole

fraction of volatile oil in the vapor phase y

as a function of x

and T:

(

T, x

)

(

)

exp



−M







−





. (3.12)

We assume that the rock is ﬁlled with a mixture of water, oil and gas, i.e.:

+ s

= 1, (3.13)

where s

, s

and s

are the oil, gas and water saturations, i.e., the pore volume fraction

is ﬁlled with the liquid water, gaseous and oleic phases respectively. We usually write

in terms of s

and s

We assume that the ﬂuids are incompressible and that the pressure changes are

so small that they do not affect the physical properties of the ﬂuids. Darcy’s law for

multiphase ﬂow relates pressure gradient with its Darcy speed:

= −

∂p

∂x

, u

= −

∂p

∂x

and u

= −

∂p

∂x

. (3.14)

The relative permeability functions k

, k

and k

are considered to be functions of

the saturations (see Appendix A);

and

are the viscosities of the aqueous,

gaseous and oleic phases. The pressure in the aqueous, gaseous and oleic phases is

taken to be identical, p, because the capillary pressure between the different phase

is neglected. The fractional ﬂow functions for water phase, gaseous phase and oleic

phase are deﬁned by:

, f

and f

, (3.15)

where d

= k

+ k

Using Darcy’s law (3.14) and Eq. (3.15):

= uf

, u

= uf

and u

= uf

, where u = u

+ u

(3.16)

is the total or Darcy speed.

3.1.1 Balance Equations.

We can write the mass balance equations for liquid water, steam in the gaseous phase,

volatile oil in the gaseous phase, volatile oil in the oleic phase and dead oil conserva-

tion respectively as:

∂

∂t

∂

∂x

= q

g−→ a,w

, (3.17)

∂

∂t

∂

∂x

= −q

g−→ a,w

, (3.18)

∂

∂t

∂

∂x

= q

g−→o,v

, (3.19)

∂

∂t

∂

∂x

= −q

g−→o,v

, (3.20)

∂

∂t

∂

∂x

= 0, (3.21)

where q

g−→ a,w

and q

g−→o,v

are the water and the volatile oil condensation rates. The

steam condenses into liquid water and the volatile oil dissolves into the dead oil.

The conservation of energy in terms of enthalpies per unit mass is given as

∂

∂t



(

)





∂

∂x



+ u

(

)





= 0. (3.22)

Using u

, u

and u

given by Eq. (3.16) in Eqs. (3.17)-(3.22), we obtain:

∂

∂t

∂

∂x

= q

g−→ a,w

, (3.23)

∂

∂t

∂

∂x

= −q

g−→ a,w

, (3.24)

∂

∂t

∂

∂x

= q

g−→o,v

, (3.25)

∂

∂t

∂

∂x

= −q

g−→o,v

, (3.26)

∂

∂t

∂

∂x

= 0, (3.27)

∂

∂t



(

)





∂

∂x



+ f

(

)





= 0. (3.28)

3.1.2 Reduced system of equations

The system of balance laws ( 3.23)-(3.28) is written in the conservative form by adding

(3.23) to (3.24) and (3.25) to (3.26):

∂

∂t





∂

∂x



+ f



= 0, (3.29)

∂

∂t





∂

∂x





= 0, (3.30)

∂

∂t

∂

∂x

= 0,

∂

∂t



(

)





∂

∂x



+ f

(

)





= 0.

Eq.

(3.29) is the conservation of component water in the steam and gaseous

phases; Eq.

(3.30) is the conservation of volatile oil in the gaseous and oleic phases.

Notice that it is possible to substitute Eq.

(3.27) by the the conservation of total oil,

which is obtained by adding Eqs.

(3.25)-( 3.27):

∂

∂t





∂

∂x



) f



= 0 (3.31)

The map E is applied to the cumulative and ﬂux vectors generically given by

and F in Eq. (3.1). There is one such map E from (3.23)-(3.28) to ( 3.29), (3.30)(3.27),

(3.28) and another to (3.29), (3.30), (3.31), (3.28). They are:







110000

001100

000010

000001







and E







110000

001100

001110

000001







. (3.32)

The isomorphism (basis change) taking E

to E

and its inverse taking E

to E

are:







100000

010000

001000

000100

001110

000001













10 0 0 00

01 0 0 00

00 1 0 00

00 0 1 00

−1 −110

00 0 0 01







In this work we will use E

3.2 Physical situations

This model exhibits a rich structure of physical situations depending on the injection

and initial conditions. Although certain physical situations seldom occur, for com-

pleteness we mention all of them. In the formalism proposed in Chapter

[2 ], we can

identify 12 physical situations Ω

, 12 space of primary variables P

and the 12 space

of trivial variables

for j = I, II,···, XII.

3.2.1 Three phases.

In this case we have three phases and three components. Using Gibb’s phase rule

(3.2), we see that there are two thermodynamical degrees of freedom. Since the pres-

sure is ﬁxed, the only degree of freedom is the temperature. Thus x

and y

are

uniquely deﬁned by the temperature T, as explained now.

From Eqs. (3.5), (3.7) and P

= P

(

)

(

)

, for the mole fraction in the liquid

phase x

we ﬁnd an expression depending on the temperature:

(

)

P −P

exp



−M

(

)



−





exp



−M

(

)



−





. (3.33)

Since x

is function of temperature, we use Eq. (3.8.a) to obtain v

as a function

of temperature only:

+(1 − x

)

, (3.34)

where x

is obtained from expression (3.33). The system of equations reduces to

(3.29), (3.30)(3.27), (3.28) (alternatively to the system (3.29), (3.30), (3.31), (3.28) ).

The mole fraction of volatile oil in the vapor phase y

is obtained using Eq. (3.12) .

The primary variables to be determined are s

, s

, T,soP

= {s

, T}; the speed

u is not constant, it is the secondary variable; and the trivial variables are v

and y

= {v

, y

}. The physical situation is characterized by Ω

= {s

, T,v

, y

Notice for this physical situation i

, T,v

, y

) :=(s

, T); the terms R

and R

are functions with 6 components given by:



ϕρ

(

)





and



+ f

(

)





3.2.2 Two phases

In this section, the number of phases is two, i.e., p = 2 in Gibb’s phase rule (3.2).

Steam phase and liquid water phase.

There is one component. Using Gibb’s phase rule

(3.2), we can see that T = T

constant. There is no oil, so v

is not deﬁned a priori and y

= 0. If it is necessary to

obtain the complete solution, we deﬁne the variables using limits in the variable v

from the contiguous situations.

The primary variable to be determined is s

,soP

= {s

}; the speed u is constant;

the trivial variables are s

= 0, T = T

, y

= 0 and v

,soP

= {s

= 0,T =

, y

= 0,v

}. The physical situation is characterized by Ω

= {sw, s

= 0,T =

, y

= 0}.

The system of equations reduces to a single conservation law, Buckley-Leverett’s

equation:

∂

∂t

+ u

∂

∂x

= 0. (3.35)

Notice for this physical situation i

, T,v

, y

) :=(s

); the terms R

and R

are functions with 6 components given by:



ϕρ

,0,0,0,H



and



,0,0,0,f

+ f



Gaseous and oleic phases.

There are three components, water, volatile and dead oil. Using Gibb’s phase rule

(3.2), in addition to the temperature T, we need to ﬁnd another thermodynamical

variable, v

. However, notice that we could utilize as a primary variable the quantity

= 1 − v

. Generically, there is no unique choice between primary and trivial

variables.

Inverting Eq.

(3.34), we obtain x

as function of v

+(1 −v

)

. (3.36)

In this physical situation, y

is given by Eq. (3.12) as a function of T and x

. The

variable y

can be obtained using Eqs. (3.36), (3.8.a) and (3.8.b); notice that x

can

be written as function of v

. Thus from Eq. (3.12) y

depends on v

and T.

The primary variables to be determined are s

, T, v

.soP

III

= {s

, T,v

};

the speed u is not constant; ﬁnally the trivial variables are s

= 0 and y

,so

III

= {s

= 0, y

(T,v

)}. The physical situation is characterized by Ω

III

= {s

0,s

, T,v

, y

(T,v

)}.

Using Eqs.

(3.9.a) and (3.9.b), we can rewrite the system (3.29), (3.30) , (3.27),

(3.28) in the variables s

, T, v

and u , and obtain the following system of conservation

for steam, volatile oil, dead oil and energy:

∂

∂t

∂

∂x

= 0, (3.37)

∂

∂t





∂

∂x



+ v



= 0, (3.38)

∂

∂t

(1 − v

∂

∂x

(1 − v

)uf

= 0, (3.39)

∂

∂t



(

(1 − v

)





∂

∂x



(

(1 − v

)





= 0,

(3.40)

where

= y

=(1 − y

)

and y

depend on T and v

A particular case occurs in the absence of dead oil. Then v

= 1 trivially and Eq.

(3.39) disappears.

Notice for this physical situation i

, T,v

, y

) :=(s

, T,v

); the terms R

and R

are functions with 6 components given by:



ϕρ

(1 − v

(

(1 − v

)





and



(1 − v

) f

(

(1 − v

)





Liquid oleic and aqueous phases.

There are three components. Using the Gibb’s phase rule

(3.2), in addition to the

temperature T, we have another thermodynamical degree of freedom; we choose v

to be this variable. In this physical situation y

is meaningless, but if it is necessary,

it can be obtained for continuity in phase space as function of T and v

using Eq.

(3.12).

The primary variables to be determined are s

, T, v

,soP

= {s

, T,v

}; the

speed u is not constant; and the trivial variables are s

= 0 and y

,soP

= {s

0, y

}. The physical situation is characterized by Ω

= {s

= 0,T,v

, y

The system

(3.29), (3.30)(3.27), (3.28) reduces to the following system of conser-

vation for water, volatile oil, dead oil and energy:

∂

∂t

∂

∂x

= 0, (3.41)

∂

∂t

∂

∂x

= 0, (3.42)

∂

∂t

(1 − v

∂

∂x

(1 − v

)uf

= 0, (3.43)

∂

∂t



(

(1 − v

)



∂

∂x



+ f

(

(1 − v

)



= 0. (3.44)

Two particular cases arise: one of them corresponds to absence of dead oil, the other

to the absence of volatile oil.

Liquid water phase and gaseous steam / volatile oil phase.

There are two components, in the absence of dead oil. Using Gibb’s phase rule

(3.2),

we see that there are two thermodynamical degrees of freedom. Since the pressure is

ﬁxed, the only degree of freedom is the temperature. Since there is no liquid volatile

oil v

is not deﬁned. On the other hand, since P

(T)=P − P

(T), y

is uniquely

deﬁned by the temperature T, see Sec. 3.2.1.

The primary variables to be determined are s

, s

, T,soP

= {s

, T}; the speed

u is not constant; and the trivial variables are v

= 0 and y

,soP

= {v

, y

The physical situation is characterized by Ω

= {s

, T,v

= 0, y

The system

(3.29), (3.30)(3.27), (3.28) reduces to the following system of conser-

vation laws for water, volatile oil and energy:

∂

∂t

∂

∂x



+ f



= 0, (3.45)

∂

∂t

∂

∂x

= 0, (3.46)

∂

∂t





∂

∂x



+ f





= 0.

(3.47)

The steam fraction in the gaseous phase is always positive. One particular case arises,

corresponding to the absence of volatile oil.

3.2.3 One phase

In the following subsections, the physical situations consist of a single phase, so p = 1

in the Gibb’s phase rule

(3.2).

Liquid oleic phase.

There are two components. Using the Gibb’s phase rule

(3.2) we can see that in

addition to the temperature T there is another thermodynamical degree of freedom.

The suitable thermodynamical variable is v

. As in the previous physical situation,

is meaningless, but it can be obtained from continuity in phase space as a function

of T and v

using Eq. (3.12).

The primary variables to be determined are T, v

,soP

= {T, v

}; the speed u

is not constant; the trivial variables are s

= 0, s

= 0 and y

,soP

= {s

= 0,s

0, y

}. The physical situation is characterized by Ω

= {s

= 0,s

= 0,T,v

, y

The system of equations

(3.29), (3.30)(3.27), (3.28) reduces to the conservation

laws for volatile oil, dead oil and energy:

∂

∂t

∂

∂x

= 0, (3.48)

∂

∂t

(1 − v

)

∂

∂x

(1 − v

)

= 0, (3.49)

∂

∂t



+(1 −v

)



∂

∂x



+(1 −v

)



= 0.

(3.50)

Liquid aqueous phase.

There is one component. Using the Gibb’s phase rule

(3.2), the only thermodynamical

variable is the temperature T. Since there is no oil in this situation, v

and y

are

not deﬁned in principle, but we obtain them as limits of Eqs.

(3.33) and (3.12).

Since

does not depend on T, one can show that u is constant spatially.

The primary variable to be determined by the system of equations is T,soP

VII

{

T}; the speed u is constant; and the trivial variables are s

= 1 , s

= 0 and v

and

so P

VII

= {s

= 1,s

= 0,v

, y

}. The physical situation is characterized by

Ω

VII

= {s

= 1,s

= 0,T,v

, y

The system of equations

(3.29), (3.30), (3.27), (3.28) reduces to the single scalar

equation for conservation laws of energy:

∂

∂t



ϕρ



+ u

∂

∂x

= 0. (3.51)

Gaseous steam/volatile oil.

There are two components. Since there is no liquid oil v

is not deﬁned, but we

obtain it as a limit of Eq.

(3.33). Using the Gibb’s phase rule (3.2), in addition to the

temperature T, we have another thermodynamical degree of freedom; we choose y

as this variable.

The primary variables to be determined are T, y

,soP

VIII

= {T, y

}; the speed

u is not constant; the trivial variables are s

= 0, s

= 1 and v

, P

VIII

= {s

= 0,s

1,v

}. The physical situation is characterized by Ω

VIII

= {s

= 0,s

= 1,T,v

, y

The system of equations governing this ﬂow reduces to the conservation laws of

steam, of volatile oil in the gaseous phase and of energy:

∂

∂t

(1 − y

)

∂

∂x

(1 − y

)

= 0, (3.52)

∂

∂t

∂

∂x

= 0, (3.53)

∂

∂t



(1 − y

)

+ y



∂

∂x



(1 − y

)

+ y



= 0.

(3.54)

Pure steam phase.

In this case there are one phase and one component. Using the Gibb’s phase rule

(3.2), the only degree of freedom is the temperature. Since there is no oil, v

and y

are not deﬁned, but they can be obtained as limits of Eqs. (3.33) and (3.12).

The primary variables to be determined is T,so

= {T}; the speed u is not

constant; and the trivial variables are s

= 0, s

= 1, v

and y

,soP

= {s

0,s

= 1,v

, y

}. The physical situation is characterized by Ω

VIII

= {s

= 0,s

1,T,v

, y

The system reduces to the conservation laws of steam and of energy:

∂

∂t

∂

∂x

= 0, (3.55)

∂

∂t



ϕρ



∂

∂x

= 0. (3.56)

The next cases consists of single phase and single components, so in the Gibb’s

phase rule p

= 1 and c = 1, so there is only two thermodynamical degrees of freedom.

Since the pressure is ﬁxed, the temperature T is chosen. This will be the primary

variable.

Pure liquid volatile oil.

Since there is only volatile oil in the oleic phase, v

= 1. Since there is no gaseous

phase y

is not deﬁned, but they are obtained by continuity as a limit process.

The primary variable is T,so

= {T}. The speed u is constant; the trivial

variables are s

= 0, s

= 0, v

= 1 and y

,soP

= {s

= 0,s

= 0,v

= 1, y

The physical situation is characterized by Ω

= {s

= 0,s

= 0,T,v

= 1, y

The system of equations reduces to the conservation laws of energy:

∂

∂t



ϕρ



+ u

∂

∂x

= 0. (3.57)

Pure gaseous volatile oil.

Since there is only volatile oil in the gaseous phase, thus y

= 1; Since there is no

liquid oil v

is not deﬁned, but can be obtained by continuity as a limit process.

The primary variable is the T,so

= {T}. The speed u is not constant; the

trivial variables are s

= 0, s

= 0, v

and y

= 1,soP

= {s

= 0,s

= 0,v

, y

1}. The physical situation is characterized by Ω

= {s

= 0,s

= 0,T,v

, y

= 1}.

The system of equations reduces to the conservation laws of gaseous volatile oil

and of energy:

∂

∂t

∂

∂x

= 0, (3.58)

∂

∂t



ϕρ



∂

∂x

= 0. (3.59)

Pure liquid dead oil.

Since the the vapor pressure is negligible in the dead oil, it occurs just in the oleic

phase and y

is not deﬁned, but it is obtained by continuity as a limit process.

The primary variable is T,so

XII

= {T}. The speed u is constant; the trivial

variables are s

= 0, s

= 0, v

= 0 and y

,soP

= {s

= 0,s

= 0,v

= 0, y

The physical situation is characterized by Ω

= {s

= 0,s

= 0,T,v

= 0, y

The system of equations reduces to to the conservation laws of energy:

∂

∂t



ϕρ



+ u

∂

∂x

= 0. (3.60)

CHAPTER 4

The Riemann solution for the injection of steam and

nitrogen

The frequent and widespread occurrence of contamination due to spills and leaks

of organic materials, such as petroleum products, that occur during their transport,

storage and disposal constitute a menace to our high-quality groundwater resources.

In spite of increased awareness of the environmental impacts of oil spills, it appears

to be impossible to avoid these accidents, so it is necessary to develop techniques for

groundwater remediation.

Steam injection is widely studied in Petroleum Engineering, see [10]. In Chapter 5,

we study the steam and water injection in several proportions into a porous medium

saturated with a different mixture of steam and water. A disadvantage of pure steam

injection is the ecological impact of high temperatures. This can be alleviated if we co-

inject nitrogen leading to a lower temperatures. Within of such a mixture, the boiling

temperature of water depends on nitrogen concentration and it is lower than water

regular boiling temperature.

Another application is the recovery of geothermal energy. Injecting water and

nitrogen into a hot porous rock, the water in the mixture evaporates and rock thermal

heat can be recovered even if the rock is not very hot.

We present a physical model for steam, nitrogen and water ﬂow based on mass

balance and energy conservation equations. We present the main physical deﬁnitions

and equations of the model. We study the three possible physical phase mixture

situations. For each physical situation, we reduce the original four balance equations

system to be presented in Section 4.2 to systems of conservation laws of type

(2.2):

∂

∂t

(V)+

∂

∂x

(V)=0,

supplemented by appropriate thermodynamic constraints between variables, such as

Raoult’s law and others described in Appendix A. Here V

=(V

) : R ×R

−→ D ⊂

is a subset of the variables: gas saturation s

, steam composition

and temper-

ature T and it represents the unknowns in each physical situation; G

=(G

, G

) :

D −→ R

and F =(F

, F

) : D −→ R

are the accumulation vector and the ﬂux

vector, respectively; u is a total velocity.

The state of the general system is represented by

, T,u). Systems of conser-

vation equations of the type

(2.2) have an important feature: the variable u does not

appear in the accumulation terms, it appears isolated in the ﬂux terms, thus such sys-

tems have an inﬁnite speed mode associated to u; despite this fact and the presence of

source terms in the original balance equations we are able to solve the associated Rie-

mann problem, which is a mathematical novelty. Another surprise in this model is the

presence of a rarefaction wave associated to evaporation in the two-phase situation.

In Section 4.1, we present the physical model that describes the injection of steam

and nitrogen in a one-dimensional horizontal porous rock initially ﬁlled with water

liquid. In Section 4.2, we present the model equations for balance of water, steam,

nitrogen and energy. The full governing equations are not a system of conservation

laws, since there is a water mass transfer term, which is the condensation or evapora-

tion of water between the liquid and gaseous phases. In Section 4.3, we describe three

physical situations in which the balance system is rewritten as conservation systems

of type

(2.2). In Section 4.4, we study the type of waves that appear in the solution

and some general results for 3

×3 systems; a more complete theory is established in

[2 ]. In Secs. 4.5, 4.6 and 4.7, we obtain the waves in each physical situation, it is

remarkable that there is an evaporation rarefaction wave in the two-phase situation.

In Section 4.8, we study the shocks between different regions. In Section 4.9, we show

an example of the Riemann solution for a mixture of steam and nitrogen in the spg

injected into a rock initially ﬁlled with water.

4.1 Physical model

We consider the injection of steam and nitrogen in a one-dimensional horizontal

porous rock core initially ﬁlled with water at temperature T

. The core consists

of the rock with constant porosity

and absolute permeability k (see Appendix A).

We assume that the ﬂuids are incompressible and that the pressure variations along

the core are so small that they do not affect the physical properties of the ﬂuids.

Darcy’s law for multiphase ﬂow relates the pressure gradient in each ﬂuid phase

with its seepage speed:

= −

∂p

∂x

, u

= −

∂p

∂x

. (4.1)

The water and gas relative permeability functions k

and k

are considered to be

functions of their respective saturations (see Appendix A);

and

are the vis-

cosities of liquid and gaseous phases; p

and p

are the pressures in the liquid and

gaseous phases. The capillary pressure P

and the capillary diffusion coefﬁcient Ω

are:

= P

)=p

− p

and Ω = −f

≥ 0. (4.2)

The fractional ﬂow functions for water and steam are deﬁned by:

+ k

, f

+ k

. (4.3)

Using Darcy’s law (3.14) and the deﬁnition of P

in Eq. (4.2.a), Eqs. (4.2) and (3.15)

yield:

= uf

−Ω

∂s

∂x

, u

= uf

−Ω

∂s

∂x

, where u

= u

+ u

(4.4)

is the total or Darcy velocity. We will see that Ω acts as a capillary diffusion coefﬁ-

cient; s

is the water saturation, i.e., the pore fraction ﬁlled with water; similarly, s

is the gas saturation.

4.2 The model equations

Neglecting molecular diffusion effects, we write the equations of mass balance for

water, steam and nitrogen as:

∂

∂t

ϕρ

∂

∂x

=+q

g−→ a,w

, (4.5)

∂

∂t

ϕρ

∂

∂x

= −q

g−→ a,w

, (4.6)

∂

∂t

ϕρ

∂

∂x

= 0, (4.7)

where q

g−→ a,w

is the water mass source term (i.e., the condensation rate between

the steam and water phases); here

is the water density, which is assumed to be

constant,

(

) denote the concentration of steam (nitrogen) in the gaseous phase

(mass per unit gas volume); the concentration dependence on temperature is speciﬁed

by Raoult’s laws and other thermodynamical relationships given in Appendix A. The

saturations s

and s

add to 1. By (4.3), the same is true for f

and f

Since there are 4 unknowns (T, s

and u) and 3 equations, it is necessary

to specify another equation, representing energy conservation, based on an enthalpy

formulation, see

[2,3]. We include longitudinal heat conduction, but neglect heat

losses to the surrounding rock. We ignore adiabatic compression and decompression

effects. Thus the energy conservation is given by:

∂

∂t





∂

∂x



+ u

(

)



∂

∂x



∂T

∂x



, (4.8)

here H

is the rock enthalpy per unit volume and

= H

; h

, h

and h

are the

enthalpies per unit mass of water in the liquid aqueous phase, water and nitrogen

in the gaseous phase; these enthalpies depend on temperature as described by Eqs.

(A.3) and (A.4). The composite conductivity of the rock/ﬂuid system

> 0 is

given by:

+ s

). (4.9)

We deﬁne H

and H

, the effective water and gas enthalpies per unit volume as:

and H

, (4.10)

so from

(4.4),(4.10), we can rewrite (4.5)-(4.8) as:

∂

∂t

ϕρ

∂

∂x

∂

∂x



Ω

∂s

∂x



+ q

g−→ a,w

, (4.11)

∂

∂t

ϕρ

∂

∂x

∂

∂x



Ω

∂s

∂x



−q

g−→ a,w

, (4.12)

∂

∂t

ϕρ

∂

∂x

∂

∂x



Ω

∂s

∂x



, (4.13)

∂

∂t



+ H



∂

∂x



+ H



∂

∂x





− H



Ω

∂s

∂x



∂

∂x



∂T

∂x



. (4.14)

We are interested in scales dictated by ﬁeld reservoirs. The effect of spatial second

derivative terms (capillary pressure, heat conductivity, etc) is to widen the heat con-

densation front as well as other shocks, while the convergence of the characteristics

tries to sharpen them. The balance of these effects yields the width of these fronts.

In the ﬁeld this width is typically a few tenth of centimeters; on the other hand, the

distance between injection and production wells is of the order of 1000 meters. Thus

this width is negligible, so we can set it to zero and simplify our analysis with no error

of practical importance. From now on we disregard the diffusive terms in (4.11)-(4.14),

resulting in the following balance system:

∂

∂t

ϕρ

∂

∂x

=+q

g−→ a,w

, (4.15)

∂

∂t

ϕρ

∂

∂x

= −q

g−→ a,w

, (4.16)

∂

∂t

ϕρ

∂

∂x

= 0, (4.17)

∂

∂t



+ H



∂

∂x



+ H



= 0. (4.18)

4.3 Physical situations

There are three regions in different physical situations. Initially, there is only water

in the core; we call this situation the single-phase liquid situation, spl. We can inject a

superheated gas to form a region containing only gaseous steam and nitrogen; we call

this the single-phase gaseous situation, spg. There is also the two-phase situation,

tp, where there is a mixture of liquid water, gaseous nitrogen and steam at boiling

temperature. The latter depends on the concentration of nitrogen in the gas. We

assume that each situation is in local thermodynamical equilibrium, so we can use

Gibbs’ phase rule

(3.2).

4.3.1 Single-phase gaseous situation - spg

In this situation there are two components and one gaseous phase, i.e., c = 2 and

= 1, so from Gibbs’ phase rule (3.2) there are three thermodynamical degrees of

freedom. Since in our thermodynamical model the pressure is ﬁxed, there are two

unknown thermodynamical variables: temperature and gas composition.

We assume that nitrogen and steam in the gaseous phase behave as ideal gases

with densities depending on T denoted by

and

, see (A.9) in Appendix A. We

use Raoult’s law (see

[68]) and assume that there are no volume effects due to mixing,

so that the volumes of the components are additive:

(T)+

(T)=1. (4.19)

This fact allows to deﬁne the steam and nitrogen gas composition, respectively, as:

(T) and

(T) , so

= 1, (4.20)

thus when the steam or nitrogen composition is known, the other composition is ob-

tained trivially from (4.20.c). We use

as the basic state variable. The composi-

tions

and

are functions of temperature and composition; they are obtained

from

(4.20.a) and (4.20.b).

Therefore, there are three unknowns to be determined: temperature T, gas com-

position

and speed u. The saturations are trivial, s

= 1 and s

= 0.

We rewrite equations (4.15)-(4.18) using Eqs.

(4.20.a) and (4.20.b). Since s

0, f

= 0, so Eq. (4.15) disappears and we obtain that q

g−→ a,w

vanishes; thus the

remaining system becomes:

∂

∂t

ϕθ

−1

∂

∂x

−1

= 0, (4.21)

∂

∂t

ϕθ

−1

∂

∂x

−1

= 0, (4.22)

∂

∂t





∂

∂x





= 0; (4.23)

we have substituted

and

from Eq. (A.9), and deﬁned

, H

(T) and

(T):

, H

(T)=

and H

(T)=

, (4.24)

where M

, M

are the molar masses of nitrogen and water; R is the universal gas

constant; p

is the atmospheric pressure; h

and h

are functions of T given in

Appendix A.

Remark 4.3.1. The spg does not exist for any pair

(T,

), because of thermodynamic

constraints. The tp situation occurs for states where the composition is given in terms

of the temperature T as:

(T)=

(T)/

(T) , (4.25)

where

and

given by Eqs. (A.8. a), (A.9. a). The composition

(T) is the pro-

jection of the tp on the plane

{T,

}. As we assume thermodynamical equilibrium,

the solution of Eq.

(4.25) deﬁnes the boundary of the physical region in the plane

{T,

}, see Fig. (4.1. a).

We deﬁne the spg physical region as:



(T,

) ∈ spg that satisfy

≤

(T)/

(T)



. (4.26)

For each

, the temperature that satisﬁes (4.25) is called saturation temperature.

The curve ∂Γ is formed by the pairs

(T,

(T)) that satisfy (4.25).

260 280 300 320 340 360 380 400 420

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Temperature /K

Steam Composition

Physical Region

Water boiling temperature

∂ Γ

Steam region

Boiling region

Steam regionSteam region

Water region

Figure 4.1: a) Left: The physical region lies to the right of the curve deﬁned by (4.25).

The continuous graph represents the saturation temperature for the mixture of water

and nitrogen. b) Right: Phase space for V

=(s

, T) and physical situations. The

bold lines represent the physical regions for water and steam without nitrogen, see

Chapter 5.

4.3.2 Two-phase situation - tp

In the tp there are two components, c = 2, and two-phases, p = 2 ; using Gibbs’

phase rule

(3.2), there are two thermodynamical degrees of freedom, temperature

and pressure (which is ﬁxed). The compositions in Eqs.

(4.27) and (4.28) depend on

temperature, and vice-versa inverting Eq.

(4.25). As the pressure is given, the boiling

temperature of water is speciﬁed by these compositions. When pure steam is injected

the condensation temperature is constant T

= 373.15, see Appendix A. We have three

variables to be determined: temperature, saturation and total Darcy velocity.

There is a linearly dependent equation in the system (4.15)-(4.18). We add (4.15)

and (4.16) and replace (4.15)-(4.18) by:

∂

∂t





∂

∂x





= 0, (4.27)

∂

∂t





∂

∂x

(

)=0, (4.28)

∂

∂t



+ H



∂

∂x

+ H

)=0, (4.29)

The compositions

and

are functions of temperature only that can be obtained

from Eqs.

(A.8. a) and (A.8.b) in Appendix A.

4.3.3 Single-phase liquid situation - spl

In the spl there is one component and one phase, so from Gibbs’ phase rule (3.2),

there are two thermodynamical degrees of freedom, temperature and pressure (which

is ﬁxed). Since s

= 1 and s

= 0,wehavef

= 1 and f

= 0. From (4.12), q

g−→ a,w

vanishes. One can prove that the total Darcy velocity u is independent of position, so

(4.15)-( 4.18) become:

∂

∂t



+ H



+ u

∂H

∂x

= 0, (4.30)

where we use u

to indicate that the velocity is spatially constant in the spl; V is

just the temperature T. We assume that rock and water enthalpy depend linearly on

temperature, so we can rewrite

(4.30) as:

∂

∂t

∂T

∂x

= 0, where

, (4.31)

and C

is the water rock heat capacity and

is the rock heat capacity C

divided by

. All quantities are given in Appendix A.

4.3.4 Primary, secondary and trivial variables

Fig. 4.1.b shows the 3 physical situations in the variables V =(T,

). As basic

variable, we choose s

instead of s

for convenience.

For each physical situation there are three groups of variables. One group en-

compasses the variables V in the system

(2.2). We call the variables in V “primary

variables” or basic variables. The variable u is called “secondary variable” because we

will see that it is obtained from the primary variables. The trivial variables are the

variables that depend on primary variables in a simple way (for example, in the tp the

steam composition depends on temperature) or are obtained trivially (for example, in

the spg the gas saturation is trivial and its value is 1). In Table 4.3.4 we summarize

these variables.

Physical situation \ Variable types Primary Secondary Trivial

Single-phase gaseous situation T,

u s

= 1

Two-phase situation s

, T u

(T)

Single-phase liquid situation T u s

= 0,

Table 1: Classiﬁcation according to situation.

4.4 General theory of Riemann Solutions

We are interested in the Riemann problem associated to (4.15)-(4.18) , that is the so-

lution of these equations with initial data



, T,u)

if x > 0

, T,·)

if x < 0,

(4.32)

The speed u

> 0 is speciﬁed at the injection point. In the next sections we show that

can be obtained as function of u

and the primary variables.

The general solution of the Riemann problem associated to Eq.

(2.2) consists of

a sequence of elementary waves, rarefactions and shocks; they are studied in the

sections that follow,

[60]. We assume that the cumulative and ﬂux functions have

continuous second derivative (

) within each physical situation and that are only

continuous between the physical situations.

4.4.1 Characteristic speeds in each physical situation

System of conservation laws in different forms must be used to ﬁnd the characteristic

speeds. If we assume that the solution is sufﬁciently smooth, we differentiate all

equations in ( 2.2) with respect to their variables, obtaining a system of the form 2.27

)

∂

∂t





+ A

∂

∂x





= 0, (4.33)

where the matrices B and A (which depend on V) are the derivatives of G

(V) and

(V) with respect to the variables V and u. Since G(V) does not depend on u, the

last column in the matrix B is zero. The characteristic values

(V,u) and vectors



(V,u), (where i is the label of each eigenvector) for the following system are the

rarefaction wave speeds and directions:



where

is obtained by solving det(A −

B)=0. (4.34)

Similarly the left eigenvectors





=(

,

) satisfy:





A =





B, for the same

. (4.35)

Remark 4.4.1. We consider here a general 3 × 3 system because it is the simplest

non-trivial system of type

(2.2). Generically, the right and left eigenvectors have three

components: two primary variables and one secondary variable. However, in the spl

situation we do not utilize this formalism because the system

(2.2) reduces to a single

equation.

Hereafter , the word “eigenvectors” represents “right eigenvectors”. For brevity,

we will write the right and left eigenvectors without the upper arrow, i.e., we replace



r by r and





by .

Remark 4.4.2. In Lemma

[2.2.1], we prove that the left eigenvectors 

do not depend

on the Darcy speed u.

Deﬁnition 4.4.1. For each i, the integral curves in the

(V,u) plane are solutions of





= r, i.e,

= r

and

= r

. (4.36)

When

satisﬁes:

(V(

),u(

)), (4.37)

the integral curves deﬁne rarefaction curves. If in addition to

(4.37), the variable

increases and satisﬁes x =

t the integral curves deﬁne rarefaction waves in the (x,t)

plane.

Remark 4.4.3. In Secs. 4.5 and 4.7, we calculate the eigenvalues and eigenvectors for

the nitrogen-steam-water model and show that they have the following form:

= u(

)

(V(

)) and r =

(

(V(

)),

(V(

)),u(

)

(V(

)

, (4.38)

where

and

for j = 1,2,3 do not depend on u. It is useful to deﬁne:

i,−

(V) :=

−

i,+

(V) :=

and

=(r

) for i = 1,2, (4.39)

where u

−

is the Darcy speed for V

−

and u

is the Darcy speed for V

. The primary

variables V

−

and V

are the ﬁrst and last value of V in the rarefaction wave. Then we

can see from Eq.

(4.36) that (V, u) along the rarefaction wave satisfy:







(V(

)),

(V(

))



and

= u

(V(

)). (4.40)

Since V

=(V

) does not depend on u we can solve ﬁrst (4.40. a) obtaining V(

) and

then substitute it in

(4.40.b) to obtain u(

(

)=u

−

ex p





−

(V(

))d



. (4.41)

Here u

−

is the ﬁrst value for u on this integral curve, and

−

is chosen such that

−

). Notice that

is given implicitly by (4.37); using (4.41) this implicit

relation is:

= u

−

(V(

))ex p





−

(V(

))d



. (4.42)

depends on u

−

and V, but it does not depend on the Darcy speed in the rarefaction

curve. When V

= V

, i.e., the last V in the rarefaction wave, we denote

Remark 4.4.4. Assume that u

−

> 0, so we perform the change of variable:

−→

−

, (4.43)

The system

(4.33) can be written in terms of the independent variables (x, T)=(x,u

−

t).

In this space, one can prove that the eigenvalues and eigenvectors have the form

i,−

(V);

and r

are the ﬁrst two components of the vector r

in Eq. (4.38.b). From Eqs. (4.37)

and (4.42), it follows that

−

and

satisfy:

−

), and

(V(

))ex p





−

(V(

))d



. (4.44)

It follows that in

(x, T) the variables V and

do not depend on u, so the rarefaction

curves too do not depend on u . The Darcy speed keeps the form

(4.41).

4.4.2 Shock waves

Frequently, discontinuities or shocks appear in solution of the Riemann (or Cauchy)

problems of non-linear hyperbolic or hyperbolic-elliptic equations (see

[60]). The Rankine-

Hugoniot condition for shocks (RH) needs to be satisﬁed.

The main feature of this class of equations is the existence of separated regions

where the system of balance equations (4.15)-(4.18) reduces to different systems of

conservation equations

(2.2). In the nitrogen/steam/water ﬂow problem considered in

this paper, there are three different situations: the spg, the tp and the spl. To obtain

the complete Riemann problem solution it is necessary to link these regions. As the

water mass source term acts in the inﬁnitesimal space between physical situations,

we propose the existence of shocks linking these different regions. Thus we divide

the study of the shocks into different groups: shocks within a physical situation and

shocks between different physical situations.

For a ﬁxed W

−

=(V

−

) the Rankine-Hugoniot Locus, RH locus, parametrizes

the discontinuous solutions of Eq.

(4.15)-( 4.18), ie., it consists of the W

=(V

)

that satisfy the following RH condition and it is denoted by RH(W

−

), see [2.3.1]:



) −G

−

)



= u

) − u

−

), (4.45)

where

) is the state at the right of the shock and (V

−

) is the state at the

left of the shock; v

is the shock speed; G

( G

−

) and F

−

) are the accumulation and

ﬂux terms at the right (left) of the shock, which in general have different expressions.

We specify the state

−

) on the left hand side, but at the right u

is not speciﬁed,

it is obtained from the RH condition

(4.45). From the continuity of F and G at the

interface between different physical situations is clear that W

= W

−

is always a

solution of

(4.45), i.e., the (−) state always belongs to a branch of RH locus.

We deﬁne the function

H = H(W

−

) as:

H := v



) −G

−

)



−u

)+u

−

), (4.46)

notice that the RH locus of W

−

are the states W that satisfy H( W

−

) := 0.

Since in this work we are interested in connected branches (i.e., that contain the

(− ) state), we use the following criterion for admissibility of shocks instead of the

viscosity proﬁle criterion:

Deﬁnition 4.4.2. (T.P. Liu.

[47,48]) We call shock curve the parts V of Rankine-

Hugoniot curve where the shock speed decreases when V moves away from V

−

. When

we consider the waves in

(x, t) also, each point of the shock curve represents shock

wave. The shock curve parametrizes the

(+) states of admissible shocks waves (−),

(+).

Rankine-Hugoniot locus

For a ﬁxed W

−

=(V

−

), we obtain the RH locus, or RH(W

−

), from Eq. (4.45),

which can be written as:

]=u

−u

−

, (4.47)

]=u

−u

−

, (4.48)

]=u

−u

−

, (4.49)

where

]=G

−G

−

, G

= G

) and F

= F

) for i = 1,2,3. We rewrite the

system

(4.47)-( 4.49) as:





] −F

−

[

]

−

[

]

−









−





= 0, which requires det





] −F

−

[

]

−

[

]

−





= 0

(4.50)

to have a non-trivial solution. Eq.

(4.50.b) yields the implicit expression H

= 0 with:

:=[G

]( F

−

− F

−

)+[G

]( F

−

− F

−

)+[G

]( F

−

− F

−

). (4.51)

Since we can obtain the RH locus in V, it more useful denote it by

RH(V

−

) instead

RH(W

−

). Generically, the expression H

−

)=0 deﬁnes a 1-dimensional

structure, see Section 4.4.2. Notice that there are only 2 primary variables in V

for

the spg and the tp (see the following Section for deﬁnition). In the spl there is only

one scalar equation with the temperature as the primary variable, so

RH(T

−

) is the

whole temperature physical range.

Solving the system (4.50), we obtain v

and u

as functions of V

−

, V

and u

−

= u

−

− F

−

] − F

]

−

− F

−

] − F

]

−

− F

−

] − F

]

, (4.52)

= u

−

] − F

−

]

] − F

]

−

] − F

−

]

] − F

]

−

] − F

−

]

] − F

]

. (4.53)

Of course, Eqs.

(4.52) and (4.53) are valid if each denominator is non-zero.

Remark 4.4.5. In the deﬁnitions that follow, all wave structures can be obtained in the

space of primary variables V. Using u

= u

−

) and v

:= v

−

we deﬁne:

−

) :=

−

and

−

) :=

−

;V+)

. (4.54)

We deﬁne the set of unordered index pairs for Eqs.

(4.52) and (4.53) as:

P = {{1,2}, {3,1}, {2,3}}. (4.55)

From this Section 4.4.2 and Rems. 4.4.3 and 4.4.4, we have proved the following:

Proposition 4.4.1. Let u

be positive. The primary variables V in the shock and

rarefaction curves do not depend on the left Darcy speed u

> 0. If a sequence of waves

and states solve the Riemann problem in the primary variables, for a given u

> 0 (or

> 0), then it is also a solution for any other u

> 0 (or u

> 0):



,·) if x < 0

,·) if x > 0.

(4.56)

Moreover, assume that for each m there is

{p, q}∈P such that the following inequality

is satisﬁed: F

q,m

p,m

] − F

p,m

q,m

] = 0. Then u

is given by:

= u

∏

l=1

ex p





+,l

−,l

(V(

))d



∏

m=1

−

q,m

p,m

] − F

−

p,m

q,m

]

q,m

p,m

] − F

p,m

q,m

]

, (4.57)

where

is the third component of eigenvector r,

−,l

and

+,l

are the ﬁrst and the last

values of

associated to the l-th rarefaction wave.

In the Proposition above, G

j,m

and F

j,m

represent the j-th components ( j = {1,2,3})

of G and F on the m-th shock wave. Similarly ex p





+,l

−,l

(V(

))d



is computed

along the l-rarefaction curve.

A very important result follows from the Proposition 2.3.2 is:

Corollary 4.4.1. Assume the same hypothesis of Proposition 2.3.2, then the Riemann

solution can be obtained in each physical situation ﬁrst in the primary variables V.

Then the Darcy speed can be obtained at any point of the space

(V,u) in terms of V

and u

by an equation analogous to Eq. (4.57).

Remark 4.4.6. Corollary 4.4.1 states that the secondary variable u can be obtained

from the primary variables, so the former usually will not appear in ﬁgures.

Proposition 4.4.2. (Quadruple Shock Rule.) In the tpand the spgsituations, consider

four primary points and three Darcy speeds, determining four states:

−

) in the

tp,

) in the spg and (V

) and (V

∗

) both in the tp or the spg. Assume

that there are shocks between the following pairs of states:

(i)

−

) and (V

) with speed v

−,+

(ii)

−

) and (V

) with speed v

−,M

(iii)

) and (V

∗

) with speed v

M,∗

such that two speeds coincide, i.e., at least one of the following equalities is satisﬁed:

either v

−,+

= v

−,M

or v

−,+

= v

M,∗

or v

−,M

= v

−,+

. (4.58)

If the following conditions

(I) to (III) are satisﬁed:

(I) G

) −G(V

−

) and G(V

∗

) −G(V

) are LI;

(II) V

and V

∗

have one coordinate V

with coinciding values;

(III) ∂

/∂V

= 0 for j = i for all V ∈RH(V

−

then

(1) V

∗

;

(2) u

∗

= u

;

(3) all the three speeds are equal:

−,+

= v

−,M

= v

M,∗

. (4.59)

Proof: Disregarding the index in accumulation and ﬂux terms, the RH conditions

for

−

)-(V

), (V

−

)-(V

) and (V

)-(V

∗

) are respectively:

−,+

(G(V

) −G(V

−

)) = u

F(V

) − u

−

F(V

−

), (4.60)

−,M

(G(V

) −G(V

−

)) = u

F(V

) −u

−

F(V

−

), (4.61)

M,∗

(G(V

∗

) − G(V

)) = u

∗

F(V

∗

) −u

F(V

). (4.62)

Assume that now Eq.

(4.58.a) is satisﬁed. Substituting v

−,+

= v

−,M

= v in Eqs.

(4.60

) and (4.61) and subtracting Eq. (4.60) from (4.61), we obtain:

(G(V

) −G(V

)) = u

F(V

) −u

F(V

). (4.63)

Notice that Eqs.

(4.63) and (4.62) deﬁne implicitly the RH locus by H

0 in the variables V

and V

. Since the RH locus depends solely on V

and the

accumulation and ﬂux functions, we obtain that both RH locus deﬁned by Eqs.

(4.63)

and (4.62) coincide. Since (I) to (III) are satisﬁed, then V

∗

= V

Now from Eq.

(4.53), we notice for a ﬁxed u

−

that the Darcy and shock speeds

depend solely on V

−

and V

. From Eqs. (4.62) and (4.63), we can see that the (−)

and (+) states are the same for each expression and that they deﬁne the same RH

locus, so u

∗

= u

and Eq. (4.59) is satisﬁed.

The other cases are proved similarly.



Degeneracies of RH Locus

We utilize the notation:

−

)=F

] − F

]. (4.64)

Proposition 4.4.1 is valid if the denominator of Eq.

(4.57) is non zero for some {i, j}∈

P. So it is necessary to study the behavior of the solution when D

−

)=0 for

all

{i, j}∈P. For a ﬁxed pair {i, j}∈P, it easy to prove that:

Lemma 4.4.1. Let

−

} satisfy H

−

)=0, where H

is given by Eq. (4.51).

−

)=0, then one of following conditions is satisﬁed:

(i) F

−

− F

−

= 0, (4.65)

(ii ) D

−

)=D

−

)=D

−

)=0. (4.66)

From this Lemma it follows immediately that:

Corollary 4.4.2. Let

−

} satisfy H

−

)=0.IfD

−

) vanishes for

two index pairs

{i, j}∈P, then it vanishes for all pairs.

Proposition 4.4.3. If

−

)=0 for all {i, j}∈P and (F

, F

) = 0,we

obtain:

and F

−

, for k = 1,2,3. (4.67)

where

and

are constants depending on [G], F

−

and F

. Moreover, for

deﬁned

(4.54.a), the shock speed v

satisﬁes:

= u

−

) −

. (4.68)

Proof: Let u

−

> 0. Since D

−

)=0 ∀{i, j}∈P, it follows that:

] − F

])e

+(F

] − F

])e

+(F

] − F

])e

= 0, (4.69)

where e

for i = 1,2,3 is the canonical basis for R

. Eq. (4.69) can be written as:

, F

) ×([G

],[G

]) = 0, (4.70)

where

× represents the curl. Since Eq. (4.70) is satisﬁed, it follows that [G] is parallel

to F

, so there is a constant

so Eq. (4.67. a) is satisﬁed. Substituting [G]=

into the RH condition (4.45), we obtain:

= u

−u

−

, (4.71)

= u

−u

−

, (4.72)

= u

−u

−

, (4.73)

If F

−

= 0 for some i = 1,2,3, it follows that:

= u

. (4.74)

If F

−

= 0 for all i = 1,2,3, multiplying Eq. (4.71) by F

−

and (4.72) by −F

−

and adding

the results, we obtain:

ϕρ

−

− F

−

)=u

−

− F

−

). (4.75)

Let us assume temporarily that F

−

− F

−

= 0, so Eq. (4.74) is satisﬁed. Sub-

stituting v

given by Eq. (4.74) into Eq. (4.71) we obtain u

−

= 0. Since F

−

= 0,

generically, it follows that u

−

= 0, which is false. So F

−

− F

−

= 0.

Similar calculations for Eqs.

(4.72), (4.73) and (4.71), (4.73) show that:

−

− F

−

= F

−

− F

−

= F

−

− F

−

= 0. (4.76)

Since Eq.

(4.76) is satisﬁed, there exists a constant

such that F

−

. Eq.

(4.68) can be obtained by substituting [G]=

and F

−

in the RH condition

(4.45). 

Remark 4.4.7. If (F

, F

)=0 and ([G

],[G

]) = 0, it is easy to prove that:

−

] and v

= u

−

, (4.77)

where

is a constant that depends on [G] and F

−

Corollary 4.4.3. The states

−

} satisfying the RH condition (4.45), for which

−

)=0 for all {i, j}∈P, satisfy also:

−

− F

−

= 0, ∀{i, j}∈P. (4.78)

We notice that the system

(4.78) has always the trivial solution V

= V

−

4.4.3 Bifurcation loci

Assume that there are no degeneracies any RH locus. For standard conservation laws,

there are loci where the solutions change topology such as: secondary bifurcation,

coincidence, double contact, inﬂection, hysteresis and interior boundary contact. The

proof of this change depends on the Bethe-Wendroff Theorem 2.3.6, that in this model

can be written as:

Proposition 4.4.4. Assume that F and G are

. Let v

−

) be the shock speed

between different physical situations. Assume that



) ·(G

) − G

−

)) = 0.

Then v

has a critical point at W

(and

−

) has a critical point at V

), if and

only if:

−

i,+

) for i = 1 or 2, (4.79)

where

(V) is given by Eq. (4.39) and v

−

) is given by (4.54).

Secondary bifurcation manifold

The RH locus for a ﬁxed V

−

is obtained implicitly by H

−

)=0, where H

−→ R, and V

are the primary variables. At some points, these implicit expression

fail to deﬁne a curve for V

in the space of primary variables. We call the set of

these points the secondary bifurcation locus. From the implicit function theorem,

they are the “

+ ” states for which there exists a “ − ” state such that the following

equalities are satisﬁed:

−

)=0 and

∂

∂V

= 0, for j = 1 and 2, (4.80)

where

is deﬁned by Eq. (4.51) and ∂H

/∂V

for j = 1,2 are:

∂

∂V



∂F

∂V

−

− F

−

∂F

∂V

∂F

∂V

−

− F

−

∂F

∂V

∂F

∂V

−

− F

−

∂F

∂V



[G]



−

− F

−

, F

−

− F

−

, F

−

− F

−





∂G

∂V



, (4.81)

where

[G]=

(

],[G

]

)

and ∂G/∂V =

(

∂G

/∂V,∂G

/∂V

)

Equation

(4.80) yields an equivalent expression for the secondary bifurcation.

Proposition 4.4.5. A state V

−

belongs to the bifurcation manifold for the family i

when there exists a state V

such that:

∈RH(V

−

) with

−

i,+

) and 

) ·[G]=0. (4.82)

Proof: We drop the family index i. Assume that

(4.80) is satisﬁed. Rearranging

the terms, we can rewrite ∂

/∂V

for j = 1,2:

∂

∂V



−

] − F

−

]



∂F

∂V

−



−

− F

−



∂G

∂V



−

] − F

−

]



∂F

∂V

−



−

− F

−



∂G

∂V



−

] − F

−

]



∂F

∂V

−



−

− F

−



∂G

∂V

. (4.83)

Notice that

= u

, so the matrix ∂

(uF) −

∂

G , is written at (V

) as:









∂F

∂V

−

∂G

∂V





∂F

∂V

−

∂G

∂V





∂F

∂V

−

∂G

∂V





∂F

∂V

−

∂G

∂V





∂F

∂V

−

∂G

∂V





∂F

∂V

−

∂G

∂V









(4.84)

Setting

−

) in (4.84) and assuming that D

−

)=F

−

] −

−

] = 0 for all {i, j}∈P, we use Eqs. ( 4.52) and (4.54) to write an equivalent but

convenient expression for

in each matrix element, obtaining:









∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V





∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V





∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V





∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V





∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V





∂F

∂V

−

−F

−

]−F

−

]

∂G

∂V









(4.85)

Since ∂

/∂V

= 0 for j = 1,2, from Eq (4.83) , it follows that for  given by

 =



−

] − F

−

], F

−

] − F

−

], F

−

] − F

−

]



, (4.86)

the inner products of columns 1 and 2 by

 are zero. Since at (V

−

) the expression

−

) vanishes, after some calculations, we obtain:

 · (F

, F

)=0.

First let us consider the case D

−

) = 0 for all {i, j}∈P, so it is clear that 

is a left eigenvector of the system. One can prove that all eigenvectors for

−

) are parallel to . Notice that  · ([G

],[G

]=0, because this equality

satisﬁes the RH locus

= 0, with H

given by (4.51).

−

)=0 for a pair {i, j }∈P, using Lemma 4.4.1, the relationships

−

− F

−

= 0 or D

−

)=0 for all {i, j }∈P are satisﬁed.

Assume that

−

) = 0 for some {i, j}∈P. For concreteness we set i = 3

and j

= 2; the other cases can be proved similarly. Since F

−

] − F

−

] = 0 and

the matrix

(4.85) has the form (a

) for i, j = 1,2,3, and we can write a

, a

and a

respectively as:



∂F

∂V

−

− F

−

] − F

−

]

∂G

∂V



, u



∂F

∂V

−

− F

−

] − F

−

]

∂G

∂V



and F

Substituting

 in (4.86) by the the following vector:

 =



0, F

−

] − F

−

], F

−

] − F

−

]



, (4.87)

it is easy to prove that this vector

 is in the kernel of the transpose of the matrix

(4.85).

Finally assume that

= 0 for all {i, j}∈P. Since  is a left eigenvector of the

matrix

(4.85) it follows that  ·(F

, F

)=0. From Eq. (4.67.a) in the Proposition

4.4.3, we know that

([G

],[G

]) = (

) for any constant

∈ R, so:

 · [G]= ·(

 · F

= 0.

The converse can be proved similarly by reversing the calculations.



Coincidence loci

There are two important types of speed coincidence: coincidence between eigenvalues

and coincidence between eigenvalues and shock speeds. These structures are impor-

tant because the Riemann solution changes when the L (or R) data is prescribed in

different regions relatively to these curves.

Coincidence between eigenvalues in each physical situation. Notice that it is pos-

sible to analyze the system in the variables V without taking into account the Darcy

speed u. At the coincidence locus between eigenvalues the Darcy speeds also coincide,

so the coincidence in the V space consists of the states in a certain physical situation

such that

(V)=

(V).

Double contact curves

It consists of the states V

−

for which there is a state V

such that the shock joining

−

and V

has speed coinciding with the characteristic speed of family i at (−) and

of family j at

(+):

∈RH(V

−

) with

i,−

−

) and

−

j,+

Notice that V

−

and V

can be in the same or in different physical situations.

Inﬂection curves

The rarefaction curves are useful to construct rarefaction waves where the eigenvalue

varies monotonically; the inﬂection curve is the curve where the monotonicity fails,

thus rarefaction curves stop at this curve. Since u varies along the rarefaction wave,

it is impossible to disregard the effect of u. Any state W

=(V,u) on the inﬂection

curve satisﬁes:

∇

(W) ·r

(W)=0. (4.88)

However, the Darcy speed can be isolated in Eq. (4.88

), so it is easy to prove that in

the space of primary variables, the inﬂection curve of family i, for i

= 1,2, consists of

the states V satisfying the equation:

∇

= −

, (4.89)

where

(V) for i = 1,2,3 are the coordinates for the eigenvector r given in Eq.

(4.38.b).

Interior boundary contact (extension of the boundary)

Because of the presence of physical region boundaries it is important to obtain the

states V joined to points V



on the physical boundary by shock waves that are charac-

teristic at V for the family i. These states satisfy:

∈RH(V



) with V



on the boundary and

(V)=

−

(V,V



4.4.4 Admissibility of shocks

Frequently, for a ﬁxed state V

−

there are points in the RH curve of V

−

that separate

potentially admissible shocks from non-admissible shocks. At these points the shock

speed coincides with the characteristic speed and the shock speed has an extremum,

see Proposition 4.4.4, so the Riemann solution usually changes. There are two type of

such shocks: left characteristic and right characteristic shocks. In this structure, the

shock has speed equal to the characteristic speed at the left or at the right state.

Left characteristic shock.

For a ﬁxed state V

−

, these points are the V

such that:

∈RH(V

−

) with

−

j,+

) for family j = 1 or 2.

Notice that V

−

and V

can be in the same or different physical situations.

Right characteristic shock.

For a ﬁxed V

, these points are the V

−

such that:

−

∈RH(V

) with

−

j,+

) for family j = 1 or 2.

Notice that V

−

and V

can be in the same or in different physical situations.

4.5 Elementary waves in the single-phase gaseous

situation

4.5.1 Characteristic speed analysis

We simplify the system (4.21)-(4.23) by substituting (4.21) by ( 4.21) divided by

and (4.21) by a certain linear combination of (4.21) with (4.22):

∂

∂t

ϕψ

−1

∂

∂x

−1

= 0, (4.90)

∂

∂t

−1

∂

∂x

−1

= 0, (4.91)

∂

∂t





∂

∂x

(

)=0, (4.92)

where we have used the equality

= 1 −

We differentiate Eqs. (4.90)-(4.92) and we rewrite it in the form

(4.33) with V =

(

, T) as:

∂

∂t









+ A

∂

∂x









= 0, (4.93)

where (indicating the differentiation relative to temperature by







−1

−

−2

−T

−2

− H

)







, (4.94)





−1

−u

−2

−1

0 −uT

−2

−1

u(H

− H

) u(



)





; (4.95)

(4.94) the constant

is the rock heat capacity divided by

, see Appendix A.

Using

(4.94) and (4.95), we notice that (u −

ϕλ

) is repeated in the ﬁrst column of

−

B. Factoring this term, the determinant of A −

B is (u −

ϕλ

) times the deter-

minant of





−1

−

−2

(u −

ϕλ

)

−1

0 −(u −

ϕλ

−2

−1

− H

)(u −

ϕλ

)(



) −

ϕλ





(4.96)

so an eigenvalue and eigenvector for the system

(4.93) are

= u/

, r

=(1,0,0)

, (4.97)

which correspond to ﬂuid transport and the wave for this eigenvalue is a contact

discontinuity. The composition

changes but the speed and the temperature are

constant. There is a single vector for

, because the matrix (4.96) with

has

rank 3, generically. We denote this eigenvalue by

to indicate that in the associated

wave only the composition changes.

We ﬁnd the other characteristic speed

in (4.96).If

= 0, we add the ﬁrst line

to the second line multiplied by

−

obtaining:





−1

−( u −

ϕλ

−2

−1

− H

)(u −

ϕλ

)(



) −

ϕλ





(4.98)

The determinant of

(4.98) is −T

−2

times



(u −

ϕλ

)(



) −

ϕλ

+(u −

ϕλ

−1

(



. (4.99)

Setting (4.99) to zero we obtain:



−



, (4.100)

where using

(4.24), F := F(T) is given by:

F(T)=



(T)h



(T)+

(T)h



(T)+



T. (4.101)

, h



, h



and

are positive and

and

are non-negative, F(T) is

always positive. The eigenvector associated to

=(0,F, u

). (4.102)

This eigenvector is unique, because substituting

in (4.98) we see that the resulting

matrix has rank 2. The rarefaction wave has constant composition

. Notice that

in the physical range, so we can order these waves: the slowest wave is a

thermal wave, with speed

; the fastest is a compositional wave, with speed

Behavior of the gaseous thermal inﬂection curve

To obtain the rarefaction curves in the spg, we need to study the sign of

∇

·r

. After

a lengthly calculation, we obtain from

(4.100) and (4.102)

∇

·r

Ξ(

, T), where Ξ =



+(1 −

)



. (4.103)

As u,

, T and

are positive, we need to study the sign of Ξ.

The gaseous thermal inﬂection curve is denoted by

; for Γ given by (4.26) it is

deﬁned by:



(T,

) ∈ Γ that satisfy ∇

·r

= 0



. (4.104)

We plot the physical region and

in Figure (4.2. a), showing the signs of Ξ. In Figure

4.2.b, we plot the horizontal rarefaction lines associated to

and the vertical rar-

efaction lines associated to

, see (4.102). The waves associated to

are contact

discontinuities.

4.5.2 Shocks and contact discontinuities

Using (4.45) in Eqs. (4.90)-(4.92), the RH condition is:



−



= u

−u

−

, (4.105)



−1/T

−



= u

−u

−

, (4.106)



+(1 −

−(H

−

+(1 −

−

)



(

+(1 −

) −u

−

(

−

+(1 −

−

(4.107)

where H

)/T

and H

)/T

(4.105)-( 4.107), there is a contact discontinuity with speed (4.97); the Darcy

speed and the temperature are constant on this wave, i.e., u

= u

−

and T = T

−

The other shock occurs with

constant; it is a thermal shock with speed v

and

Darcy speed u

on the right side given by:

−



−

; T

)(T

− T

−

)+T



and u

−

Υ(T

−

; T

Figure 4.2: a)-left: The single-phase gaseous physical region Γ , and inﬂection locus.

b)-right: Rarefaction curves. The horizontal rarefaction curves are associated to

;

we indicate with an arrow the direction of increasing speed. The vertical lines are

contact discontinuity curves associated to

, in which

changes, T and u are con-

stant.

where Υ is deﬁned as follows, with

, H

and H

given by Eq. (4.24):



−h

−



+(1 −

)



−h

−



−



−





−h

−



+(1 −

)



−h

−



. (4.108)

The RH and rarefaction curves are contained in the horizontal lines in Figure

(4.2.b).

4.6 Contact discontinuities in the single-phase liq-

uid situation

Eq. (4.31) is linear in the spl, so a unique wave is associated to

, given by (4.31.b).

This wave is a contact discontinuity and there is no genuine rarefaction wave.

Remark 4.6.1. In the spl there exists a cooling discontinuity between a state with

temperature T

−

and another state with temperature T

. For the Riemann data: T

−

< 0 and T

if x > 0, the solution is T

−

if x/t <

and T

if x/t >

4.7 Elementary waves in the two-phase situation

4.7.1 Characteristic speed analysis

By means of a procedure similar to that in Section 4.4.1, we write the system (4.27)-

(4.29) in the form (4.33) , with V =(s

, T), as:

∂

∂t









+ A

∂

∂x









= 0, (4.109)

with





(

−

)

ϕρ



ϕρ



− H

)

(

+ C

+ s

−C

)) 0





, (4.110)







(

−

)

∂f

∂s



(

−

)

∂f

∂T





+ f

(

−

)

∂f

∂s



∂f

∂T





u(H

− H

)

∂f

∂s



+ f

−C

)+(H

− H

)

∂f

∂T



− H

)+H







(4.111)

where C

= ∂H

/∂T and the constant C

are the gas and water heat capacities given

in Appendix A.

In order to use Eq.

(4.34.b) for system (4.109), we calculate A −

B as:







∂f

∂T



+ f

∂f

∂T



(u −

λϕ

)+(C

−C

)

−

ϕλ

+ u

∂f

∂T

+ H







, (4.112)

where we have deﬁned

(s, T, u)=u

∂f

∂s

−

λϕ

and

(s, T, u)=uf

−

λϕ

, (4.113)

(T)=

−

, and

(T)=H

− H

. (4.114)

Setting

= 0 in (4.112), its determinant vanishes, and we ﬁnd the Buckley-Leverett

characteristic speed and characteristic vector:

∂f

∂s

, r

=(1,0,0)

, (4.115)

On this rarefaction curve T and u are constant, only s

changes, hence the subscript

s in

and r

is of a saturation wave. The eigenvector associated to the eigenvalue

(4.115) is unique: substituting

in (4.112) leads to a matrix with rank 2, generically.

We factor

in ﬁrst column, for

= 0. The other eigenvalue of (4.112) is the root

det







∂f

∂T



+ f

∂f

∂T



(u −

λϕ



−

ϕλ

+ u

∂f

∂T

+ H







= 0. (4.116)

We consider ﬁrst the case

= 0, the case

= 0 is considered later. We divide the

second row by

and we obtain the eigenvalue:



(



−







)

− H

(



−







)



+ C

)+s



(



−







)

− H

(



−







)



(4.117)

with eigenvector

=(−



,¯

,−u



)

, (4.118)

where

∂f

∂s

−



(



−



)

− H

(



−



)



+ C

)+s



(



−



)

− H

(



−



)



, (4.119)

( f

−s

)

)+s



(



−



)

− H

(



−



)



, (4.120)



= ¯











− f

(



−



)





∂f

∂T

, (4.121)







−





, (4.122)

and

is given in (4.114) .

In Figs.

(4.5.a) and (4.5.b) we can see that in the region where

, along the

rarefaction waves the temperature, gaseous water saturation and u increase; and in

the region where

along the rarefaction waves the temperature and u increase

and the gaseous water saturation decreases. We utilize the subscript e in

, r

indicate that this eigenpair is associated to an evaporation wave.

Coincidence curve

The coincidence curve between

and

is:

s,e



(T,s

) ∈ tp such that

(T,s

)



. (4.123)

Differentiating

in (4.117) with respect to s

, equating it to zero and isolating

∂f

/∂s

, we obtain:

Lemma 4.7.1. On the coincidence curve C

s,e

, the derivative ∂

/∂s

vanishes.

Thus coincidence between eigenvalues occurs where

is a stationary. As in Prop.

5.6.4, we can prove the following Proposition:

Proposition 4.7.1. Lines with ﬁxed T intercept the coincidence curve

s,e

twice. At

the lowest value of s

has a minimum and at the higher value

has a maximum.

Thus, in the tp, when the temperature T varies two branches of the coincidence

curve are generated as sketched in Figure

(4.4.a).

Proof: We verify this proposition geometrically. First, we deﬁne

A = −



− H

)



−



− H

)



− H

(



−



(

−

))



(4.124)

One can verify that

A > 0 for all T in the tp physical range. We multiply and divide

from Eq. (4.117) by A obtaining:

− f

−s

, where f

≡

and s

≡





. (4.125)

For the coincidence points with ﬁxed u, it follows from

(4.115):

∂f

∂s

− f

−s

. (4.126)

Eq.

(4.126) deﬁnes the tangency points of the secant from the point (s

, f

) to the

graph of f

. One can prove that s

> 1 and f

> 1. In Figure 4.3, we plot a possible

point

, f

). At the ﬁrst point where

and

coincide the speed is a minimum and

at the second point the speed is maximum.



Behavior of the two-phase evaporation inﬂection curve

To obtain the evaporation rarefaction curves we need to study the sign of

∇

· r

After a lengthly calculation, we get

∇

·r

−s





−A







−C

(



−



)



, (4.127)

where ¯

, ¯

are deﬁned in Eqs. (4.119) and (4.120). The expression for A is given by

(4.124) and A



is its derivative with respect to temperature.

One can verify that



−A



−C

(



−



) > 0, so the points where

∇

·r

vanishes satisfy:

. (4.128)

Figure 4.3: The coincidence between

and

occurs at the tangency point on the

graph of f

from (s

, f

). Since A > 0, s

and f

are positive, the ﬁrst coincidence point

is a maximum while the second point is a minimum for

l tr

andand

and

l tr

and

l tr

and

l tr

andand

Figure 4.4: a)-left: Coincidence curve C

s,e

. Relative sizes of

and

in a piece of

the tp state space. The almost horizontal coincidence curve

is not in scale,

because it is very close to the axis S

= 0. b)-right: A zoom of the region below the

coincidence curve. In both ﬁgures, all curves form the inﬂection curve, subdividing tp

situation in four parts.

We deﬁne the curve I

as:

:= {(T, s

) ∈ tp satisfying Eq. (4.128.b)}.

We denote the inﬂection curve in the tp of

by I

; it is deﬁned by:



(T,s

) satisfying (4.128.a) and (4.128.b)



(4.129)

The following Lemma follows from Eqs.

(4.128) and (4.123).

Lemma 4.7.2. The coincidence curve

s,e

is contained in I

. Moreover I

= C

s,e



The solid curves in Fig. 4.4 are the

, where we also plot the sign of ∇

·r

.In

Figure 4.5, we draw rarefaction curves in the tp.

300 310 320 330 340 350 360 370

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Temperature /K

Gas Saturation

300 310 320 330 340 350 360 370

0.005

0.01

0.015

0.02

0.025

Temperature /K

Gas Saturation

Figure 4.5: a)-left: The rarefaction curves projected in the plane T, s

. The thin curves

without arrows are

s,e

. The bold curve indicates an invariant curve for the rarefac-

tion ﬁeld, which a rarefaction curve that reaches the point P in Fig.

(4.4.b). b)-right:

The rarefaction curves in the regions III and IV shown in the Figure

(4.4.b). The

arrows indicate the direction of increasing speed.

For the case

= 0 but



= 0, the matrix (4.116) reduces to







∂f

∂T



+ f



(u −

λϕ



ϕλ

+ u

∂f

∂T

+ H







. (4.130)

The eigenvalue and eigenvector are



, r





− ¯





+ 1



,1, ¯





. (4.131)

The case

= 0 and



= 0, with no nitrogen, is studied in [6 and Chapter 5.

4.7.2 Shocks in the two-phase situation

As in the spg, we use Eqs. (4.45) and (4.27)-(4.29) to obtain the RH condition:



−s

−

)

+ s

−s

−





− f

−



+ u

−u

−

(4.132)



−s

−



= u

−u

−

, (4.133)



−

+ H

− H

−

+ s

− H

) −s

−

− H

−

)



( f

+ f

) −u

−

( f

−

+ f

−

). (4.134)

The isothermal branch of the RH curve is obtained for constant temperature T,

i.e., T

= T

−

= T. In this case the shock speed is

−

, T) − f

−

, T)

−s

−

On this wave, the Darcy speed is constant. This is the Buckley-Leverett shock; the

RH and rarefaction curves associated to

lie on the same straight line.

The non-isothermal wave is obtained after a lengthly calculation. We ﬁx the left

state and the temperature of the right state, with T

= T

−

. Following [10], we can

obtain the shock speed v

and Darcy speed u

at the right independently of s

.We

rewrite

(4.132)-( 4.134) as:



−u



−



= v

−



−



−u

−



−

+ f

−



+ u

(4.135)



−u



= v

−

−u

−

, (4.136)



−u



− H



= v



−

− H

+ s

−

+ s

−



+ u

−u

−

( f

−

+ f

−

), (4.137)

where we have used s

= 1 −s

and f

= 1 − f

and H

> H

in the temperature physical range and we assume that

= 0 (the term

is zero only at the water boiling temperature), we add (4.135)

divided by

−

to (4.136) divided by

. Similarly, we add (4.136) divided by

to (4.137) divided by H

− H

and we obtain:



−

(

−

)



−u

−



−

+ f

−



+ u

−

−u

−

(4.138)



−

− H

+ s

−

+ s

−



+ u

−u

−

( f

−

+ f

−

)

− H

−

−u

−

(4.139)

From

(4.138) and (4.139), we obtain v

and u

in terms of s

−

, T

−

, u

−

and T

as:

−

(Π

−Π

) − f

−

(

− H

)

(Π

−Π

) −s

−

(

− H

)

(4.140)

and

= u

−



−



−

(

−

) −Π



+ Π

− f

−

(

−

)



, (4.141)

where Π

−

, T

−

), Π

−

, T

−

), Π

−

, T

−

; T

) and Π

−

, T

−

) are given by:

= f

−

+ f

−

, Π

= f

−

+ f

−

, (4.142)

−

− H

+ s

−

+ s

−

and Π

= s

−

(

−

). (4.143)

After ﬁnding v

and u

, we substitute them in (4.133) and divide the resulting

equation by u

, rearrange the terms, obtaining an implicit equation for s

, T

) − f

∗

−s

∗

, (4.144)

where

∗

−



−



−

and s

∗



−



−

with

−

= v

−

and

−

= u

−

As v

and u

do not depend on s

, the solution of Eq. (4.144) is the intersection

of the graph of f

) with a linear function of s

. From the shape of the graph of

, we see that (4.144) can have one, two or three solutions. Each intersection point

belongs to the Rankine-Hugoniot curve

RH(V

−

). When T

varies each intersection

generates a branch of the Rankine-Hugoniot curve.

This is called the non-isothermal Rankine-Hugoniot curve; some examples are

shown in Figure 4.6.

300 310 320 330 340 350 360 370

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Temperature /K

Gas Saturation

(320,0.6)

(370,0.8)

(370,0.5)

(370,0.1)

300 310 320 330 340 350 360 370

0.005

0.01

0.015

0.02

0.025

Temperature /K

Gas Saturation

(320,0.07)

(320,0.04)

(360,0.01)

(336,0.04)

Figure 4.6: a)-left: The RH curves projected in the plane T, s

for (−) points marked.

b)- right: The region shown in Figure

(4.4.b). Each RH locus is formed by a non-

isothermal curve and a vertical line which is the isothermal Buckley-Leverett RH

locus. For the point

(336,0.04) the Rankine-Hugoniot locus reduces to the isothermal

curve only.

4.8 Shocks between regions

We consider now shocks that separate physical situations. In Section 4.4,wegave

the RH condition

(4.45) for shocks between regions. The main feature is that the

accumulation and ﬂux terms have different expressions at each shock side.

4.8.1 Shock between the gaseous and the two-phase situations

For the left state we need to specify steam composition

−

, the temperature T

−

and

the Darcy speed u

−

. For the right state, we specify the temperature T

; the steam

composition is a function of temperature (see

(4.20)) and the Darcy speed is obtained

from the other variables. The RH condition

(4.45) is written as:



−



= u





−u

−

, (4.145)



−



= u

−u

−

, (4.146)



−

+ H

−



= u



+ H



−u

−

, (4.147)

where

−

with H

and H

given by Eq. (4.24).

The system

(4.145)-( 4.147) is similar to (4.132)-(4.134). As in that case, we elimi-

nate the saturation and we obtain two linear equations in two unknowns v

and u



−



−u

−

+ u

−

(

−u

−

)

−

, (4.148)



−

− H

−



+ u

−u

−

− H

(

−u

−

)

−

, (4.149)

We obtain expressions for v

and u

as functions of the left state (

−

, T

−

) and

−



− H



−



−

− H



−Ξ

, (4.150)

−



−

(

−

)



−v

, (4.151)

where Ξ

and Ξ

are given by:



−

− H

−



−

, Ξ



−



−

After ﬁnding v

and u

, we use Eq. (4.146) to obtain an equation similar to (4.144)

for s

, T

) − f

−s

, (4.152)

where

−

and f

−

Geometrically, Eq.

(4.152) represents tangency points of the secant from the point

, f

) to the graph f

for ﬁxed T

with slope v

The Secondary Bifurcation.

The secondary bifurcation consists of the pairs of states for which the Prop. 4.4.4 fails,

see Sec. 4.4.3. The deﬁnition of secondary bifurcation is given by Eqs.

(4.80)-( 4.81).

From Eq.

(4.152), we deﬁne for T

−

in the spg and T

in the tp:

−

; T

)=u

, T

) − v



−



+ u

−

(4.153)

As v

and u

do not depend on s

we obtain a simple expression for ∂F/∂s

; for

brevity, we rewrite T

as T, s

as s

and u

as u:

∂F

∂s



−u

∂f

∂s



so ∂F/∂s

= 0 yields:

∂f

∂s

. (4.154)

Since both v

and u depend on T, the expression obtained for ∂F/∂T is complicated; a

lengthly calculation yields:

∂F

∂T

∂v

∂T



−



+ v



−u

∂f

∂T

−

∂u

∂T

−uf



. (4.155)

Denoting the numerator and denominator of v

−

in Eq. (4.150) by n

and d

respectively, we obtain:

∂v

∂T

−

(

∂n

/∂T

)

−n

(

∂d

/∂T

)

, (4.156)

where ∂d

/∂T =:



−

− H

−





−(C

+ C

)

−



(1 − H

) −



−

−Ξ

∂n

∂T

−





−C



−





−

− H



−



−

−C



Denoting the numerator of u in Eq.

(4.151) by N

, we obtain that:

∂u

∂T

(

∂n

/∂T

)

− N



(

)

, (4.157)

where

= u

−



−



−





−

∂v

∂T

−v

∂Ξ

∂T

4.8.2 Shock between the two-phase and the liquid situations

The spl can be considered a limit case of the tp situation, with s

= 1 and s

= 0.

The left state is

−

, T

−

); in the right state, we specify only the temperature T

;

the Darcy speed is obtained from the other variables and the steam composition is a

function of temperature (see

(4.20)). The RH condition (4.45) is written as:



−



= u

−u

−

−u

−

, (4.158)

−

= −u

−

, (4.159)



−

+ H

− H

−

−s

−



= u

−u

−



−

+ H

−



. (4.160)

From (4.159), we obtain that

−

, (4.161)

Substituting v

given by (4.161) in Eq. (4.158) we obtain :

= u

−

. (4.162)

There is no mass transfer between the tpsituation and the spl situation, so we can

consider the spl as a particular case of the tp.

Lemma 4.8.1. If there is a shock between the tp “

−” and the spl “ + ” with T

= T

−

(i.e., a non-isothermal shock), then

−

+ C

. (4.163)

Proof: We rewrite

(4.160) using (4.161) and (4.162) as:



+ H



−

−u

−



−

+ s

−

+ s

−



−u

−

( f

−

+ f

−

manipulating this equation we obtain



+ H



−

−u

−



−

+(1 −s

−

+ s

−



−u

−

( f

−

+ f

−



− H

−

+ H

− H

−



−



− H

−



−

(

− H

−

)



− H

−

+ H

− H

−



+ C

where we have used that H

= C

(T −T

) and

(T −T

). 

Remark 4.8.1. We call the (−) states that satisfy (4.163) the condensation curve;

on this curve there is non-isothermal mass transfer between the gaseous and liquid

phases. From Lemma 4.8.1 and Eq. (4.128), we obtain the following:

Corollary 4.8.1. In the tp, the inﬂection curve

is the union of coincidence locus C

s,e

with the “condensation curve”.

Lemma 4.8.2. In the set of

(− ) states that satisfy Eq. (4.163), the evaporation char-

acteristic speed in the tp equals the contact discontinuity speed in liquid situation,

i.e.,

Proof: Since f

satisﬁes Eq. (4.163), we substitute f

= s

/(C

) in Eq.

(4.117). After some calculations we obtain the result.

Proposition 4.8.1. The Riemann solution for any

(− ) state, (−)=(s

−

, T

−

,·) that

satisﬁes Eq. (4.163) and

(+) states, (+) = (0, T

,·) in the spl consists of a shock

between

(− ) and (+).

Proof: From Lemma 4.8.1, if the

(− ) belongs to the condensation curve, the shock

speed between regions is

−

+ C

which is the same as the the contact discontinuity speed in the spl.



The following corollary follows immediately:

Corollary 4.8.2. The Riemann solution for left states L

=(s

, T

,·) in regions III

or IV (see Fig. 4.4.b) and right states R

=(0,T

,·) in the spl consists of a Buckley-

Leverett shock from L to

(0,T

,·) followed by a thermal contact discontinuity in the spl

with speed

given by (4.31) and u

= u

4.9 The Riemann solution for geothermal energy re-

covery

We show an example of Riemann solution. We consider the injection of a two-phase

mixture of water, steam and nitrogen into a rock containing superheated steam (

1) at a temperature T

> T





(T), T,u



if x = 0 (the injection point), with u

> 0



= 1,1, T,·



if x > 0

(4.164)

From Prop. 4.4.1, the projection of the solution into the primary variables does not

depend on u

. We can subdivide the tp region in 4 subregions, relatively to the left

state L.ForL in regions I and II of Figure

(4.4.a) there are 3 subregions with the

following property: for any L in a given subregion the Riemann problem with data R

of form

(4.164) has the same sequence of waves, see Figure 4.7.In4.9.3, we utilize

results of Section 5.6.2 to obtain the Riemann solution.

4.9.1 Subdivision of tp

For the data (4.164) with T

ﬁxed, the Riemann solutions have the same wave se-

quence and structure in for L each subregion labelled

to L

of Figure 4.7; subre-

gions III and IV will be divide in Chapter 6. A very important task is to obtain the

curves that bound subregions: the curve E

, see Fig. (4.7. a) and Sec. 4.9.1; the curve

, see Figs. (4.7.a) and (4.7.b) and Sec. 4.9.1; the curve E

, see Fig. (4.7.a) and Sec.

4.9.1.

Figure 4.7: a)-left The 3 subregions in I and II for a Riemann data of form (4.164).

The curves are deﬁned in Secs. 4.9.1, 4.9.1 and 4.9.1. b)-right Zoom of regions II to

IV. Each curve is explained in Section 4.9.1

Remark 4.9.1. The evaporation wave speed

for states in L

and L

in tp is larger

than the thermal wave speed in the spg. From geometrical compatibility, at the right

of the evaporation shock there is no intermediate thermal wave (rarefaction or shock)

in the spg, i.e., the shock between regions should reach a state of the form

(1,

, T

Curve E

This curve is deﬁned as the states V

−

in the tp such that the evaporation rarefaction

speed

−

) equals the speed v

−

) of the shock to V

=(1,T

) in the

spg. Notice that this curve is an extension of the boundary, see Section 4.4.3.

We obtain

numerically. For each V

−

state of the form (S ,

(T), T) in the tp,

we search the state in the RH locus starting at V

−

with right temperature T

. These

restrictions yield a state in the spg with steam composition

as unknown.

Curve E

We are interested in the Riemann solution for left states in I or II. From geometrical

compatibility, to solve the Riemann problem we choose ﬁrst the slowest wave when-

ever possible. In II, the slowest wave is the evaporation wave. Since we are interested

in connecting a left state at lower temperature to a state at higher temperature, we

use an evaporation rarefaction curve instead of a shock, see Section 4.7. We can do

this until the speed of the evaporation rarefaction curve equals the speed of the shock

between regions; this occurs when the rarefaction curve crosses E

The integral curves starting at certain left states do not cross E

, rather they

reach directly the steam region at boiling temperature. We deﬁne the curve E

as the

evaporation rarefaction curve that crosses E

at water boiling temperature. Notice

that for left state in II above E

, the evaporation rarefaction curve always crosses E

Curve E

For states V

−

above the curve E

, the evaporation rarefaction speed is larger than

−

), for V

=(1,T

); in Figures (4.4. a) and (4.8. a) we show the rela-

tives sizes of v

and

. When s

tends to 1 the Buckley-Leverett shock speed is

zero, so there are transition curves, v

= v

, v

= v

and v

= v

. From the

Prop. 4.4.2, we obtain a bifurcation curve E

where v

= v

. For states above

this curve, the Buckley-Leverett shock is slower than v

and v

, so there is no direct

shock between the tp and the spg situations.

300 310 320 330 340 350 360 370

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Temperature /K

Gas Saturation

=λ

300 310 320 330 340 350 360 370

0.005

0.01

0.015

0.02

0.025

0.03

Temperature /K

Gas Saturation

III

=λ

Figure 4.8: a)-left) The relevant curves for the Riemann solution for the left states in

the tp situation. b)-right) The curve B is a bifurcation curve; for left states V out of

this curve, the shock and rarefactions starting in V do not cross B.

4.9.2 The Riemann solution

We use the notation V



(T), T



and V



= 1,1,T



. All the variables

for the intermediate states are written out. The states are labelled by 1, 2, etc. We

use the following nomenclature: evaporation rarefaction is R

, the Buckley-Leverett

shock and rarefaction are S

and R

, the shock between the tp and the spg is S

and the contact compositional discontinuity is S

. We use the notation for wave se-

quences established in Section 2.1.3.

in L

There is an evaporation rarefaction curve from V

up to V

=(s

,1,T

), where T

the boiling temperature. For this state there is no nitrogen and so

= 1. The so-

lution after the intermediate state (s

,1,T

) was found in Chapter 5: an evaporation

shock between the boiling situation to the steam situation appears, denoted by S

The solution consists of the waves R

→ R

 S

−→ V

−−→ V

. (4.165)

in L

There is a rarefaction from V

up to V

=(s

∗

), T

∗

) in tp; V

is a state where

)=v

), with V

=(1,

, T

) in spg. We have described how to ob-

tain

in Section 4.9.1. Finally, there is a compositional contact discontinuity at

temperature T

with speed v

from V

to V

The solution consists of the waves R

 S

→ S

with sequence:

−→ V

−−→ V

−→ V

. (4.166)

Figure 4.9: Riemann solutions in phase space, omitting the surface shown in Fig.

4.1. a) Left: Solution

(4.165) for V

∈L

, Sec. 4.9.2. b) Right: Solution (4.166) for

∈L

, Sec. 4.9.2. The numbers 1 and 2 indicate the intermediate states V in wave

sequence.

in L

Let V

=(1,

, T

) be a state in the spgsituation. Since

) > v

), there

is a shock between V

and V

with speed v

. From the state V

there is compositional

contact discontinuity with speed v

to V

The solution consists of the wave S

→ S

with sequence:

−−→ V

−→ V

. (4.167)

in L

Since v

and v

are smaller than v

, there is a Buckley-Leverett saturation shock

between V

and V

=(1,

), T

). From V

there is a thermal shock with speed

to V

=(1,

), T

) and ﬁnally there is a compositional contact discontinuity

with speed v

to V

The solution consists of the waves S

 S

→ S

with sequence:

−−→ V

−→ V

. (4.168)

Figure 4.10: Riemann solutions in phase space, removing the surface shown in Fig.

4.1. a) Left: Solution

(4.167) for V

∈L

, Sec. 4.9.2. b) Right: Solution (4.168) for

∈L

, Sec. 4.9.2. The numbers 1 and 2 indicate the intermediate states V in wave

sequence.

4.9.3 V

in III and IV

The integral curves or RH locus starting at any point in III or IV do not reach the wa-

ter boiling temperature. Moreover, the RH locus between the tpand the spgsituations

for points in III or IV do not reach temperatures above the temperature T

. So the

only possibility to get to the right state is by crossing the curve B. However, the evap-

oration wave does not cross the curve B; to cross it we need a Buckley-Leverett shock.

The Buckley-Leverett shock for an initial state below B and a ﬁnal state above B is

faster than the evaporation wave

. Moreover, it is faster than the shock between

regions. So with a Buckley-Leverett shock of this type, it is only possible to reach

states of form

(1,

(T), T) with T < T

. Since the thermal wave is slower than the

Buckley-Leverett shock, so it is impossible to construct the Riemann solution satisfy-

ing the geometrical compatibility principle for waves only in the tp situation.

To obtain the Riemann solution, we utilize the results of Section 5.6.2,soitis

postponed to Chapter 6

100

CHAPTER 5

Riemann solution for steam and water ﬂow

In this section, we use part of the theory developed for conservation laws to solve

a system of balance equations for steam and water ﬂow in a porous medium, see

[20], [45], [59] and [60] for hyperbolic systems. The solution exhibits an intriguing

yet systematic structure. It is desirable to obtain a theory for balance equations as

complete as that for conservation laws; combustion phenomena are also modelled by

balance equations, see

[1 ] and references theirein.

This class of balance equations has appeared in mathematical models for clean-up,

see [6]. Soil and groundwater contamination due to spills of non-aqueous phase liq-

uids (NAPL’s) have received a great deal of attention from society, because, in general,

these components can cause damage to the ecosystem and environmental impact to

a large area around the spills. Removal of contaminants with steam is considered

to be an attractive groundwater remediation technique. We consider here a model

for steam injection presented in

[6 ]. Steam injection is widely studied in Petroleum

Engineering, see [6] and references theirein.

In the recovery of geothermal energy, a steam producing well crosses porous rock

containing water at high pressure and temperature. The water usually boils in the

rock, forming a steam vaporization front.Because of the high pressures and pressure

gradients, the nature of the ﬂow equations changes: for single phase steam or water

ﬂow the asymptotic solution depends on x

√

t, see [37]. In many situations, a two-

phase region develops, precisely as in the case analyzed in our work for small pressure

gradients. The precise asymptotic behavior for for ﬂow of water and steam containing

a two-phase region seems to be an open problem ﬂow under high pressures. Tsypkin

et. al.,

[66]. Studies in steam and water injection is a related application in several

works: Bruining

[6 ], K ondrashov [ 35], Kulikoskii [37], [38] and Tsypkin et. al. [63]-

[67].

We consider the constant rate injection of a mixture of steam/water in a speciﬁed

proportion into a porous medium ﬁlled with another homogeneous steam/water mix-

101

ture. We study all possible proportions of steam and water as boundary and initial

conditions for the problem. We present a physical model for steam injection based

on mass balance and energy conservation equations. We present the main physical

deﬁnitions and equations; we refer to

[6 ] for more details. We study the three possible

physical phase mixture situations: single-phase gaseous situation, which represents a

region with superheated steam, called steam region, sr; a two-phase situation, which

represents a region where the water and steam coexist called boiling region, br; and a

single-phase liquid situation which represents a region with water only, called water

region, wr. We reduce the three balance equations system presented in Sec. 5.1 to a

system of conservation laws of form

(2.2):

∂

∂t

(V)+

∂

∂x

(V)=0,

where V

=(V

) : R × R

−→ Ω ⊂ R represents the variables to be determined;

=(G

, G

) : Ω −→ R

and F =(F

, F

) : Ω −→ R

are the accumulation vector

and the ﬂux vector, respectively; u :

R ×R

−→ R , u = u(x, t) is the total velocity.

It is useful to deﬁne U

=(V,u). The vector V represents the water saturation s

and

the temperature T. The state of the system is represented by

, T,u). Eq. (2.2)

has an important feature, the variable u does not appear in the accumulation term, it

appears isolated in the ﬂux therm, therefore this equation has an inﬁnite speed mode

associated to u. Nevertheless we are able to solve the complete Riemann problem

associated to Eq.

(2.2). Moreover, under certain hypotheses it is possible to solve

numerically the problem, in

[40] Lambert et. al. consider a model in the balance form

for nitrogen and steam injection.

[6 ], Bruining et. al. considered as initial condition for the Riemann problem,

a porous rock ﬁlled with water at a temperature T

, in which a mixture of water

and steam at saturation temperature (boiling temperature) in given proportions is

injected. The main feature was the existence of a Steam Condensation Front (SCF),

which is a shock between the br and the wr. The analysis of the shock between each

pair of regions is important because bifurcations occur and frequently non-classical

structures appear in the solution.

In this work, we completely solve the Riemann problem. We study the three possi-

ble physical phase mixture situations: single-phase superheated steam gaseous situa-

tion, in the sr; a two-phase situation where the water and steam coexist in equilibrium

in the br; and a single-phase wr.

The Riemann problem A is the injection of a mixture of water and steam at boiling

temperature in a porous rock ﬁlled with steam at temperature above the boiling tem-

perature (superheated steam). In this case, a new wave, a vaporization shock (VS),

appears between the br and the sr. In the Riemann problem B, we inject liquid water

in a porous rock containing water and steam at boiling temperature. These initial

and boundary conditions are the reverse of those considered in

[6 ]. There is a water

evaporation shock (WES) between the sr and the br. In the Riemann problem C,we

inject superheated steam in a porous rock containing water and steam at boiling tem-

perature. There appears a condensation shock (CS) between the sr and the br. This

102

Riemann solution is the most interesting solution; it has a rich bifurcation structure.

We obtain two bifurcation curves. The ﬁrst bifurcation is the TCS locus, where the

left thermal characteristic speed in the sr coincides with the condensation shock speed

. The second bifurcation is the CSS locus, where the condensation shock speed v

coincides with the right saturation characteristic speed

in the br. These two bi-

furcation curves intersect at a point, the double bifurcation SHB (see Fig. 5.8). This

state is very important because it is an organizing point between several different

phase mixtures. In the Riemann problem D, we inject water at a temperature below

its boiling temperature in a porous rock containing superheated steam. There is a br

between the wr and the sr. So the Riemann solution consists of a combination of the

waves in the Riemann problem B and C. In the Riemann problem E, we inject super-

heated steam in a porous rock containing water below its boiling temperature. As in

the Riemann problem D, there is a br between the wr and the sr. For this Riemann

problem, the solution is obtained combining the Riemann solution B and the solution

F , see

[6 ] and [39].

In Sec. 5.1, we present the mathematical and physical formulations of the injection

problem in terms of balance equations. In Sec. 5.2, we consider separately each region

in different physical situations under thermodynamical equilibrium and we rewrite

the corresponding balance equations in conservative form. In Sec. 5.4, we study the

shock and rarefaction waves that occur in each physical situation separately. In Sec.

5.5, we study the shocks in the transitions between regions. In Sec. 5.6, we present

the solution of the Riemann problem for the ﬁve types of injection. In Appendix A, we

describe notation and physical quantities appearing in the physical model.

5.1 Mathematical and Physical model

5.1.1 The model equations

Ignoring diffusive effects, the mass balance equation for liquid water and steam read:

∂

∂t

ϕρ

∂

∂x

=+q, (5.1)

∂

∂t

ϕρ

∂

∂x

= −q, (5.2)

where

is the rock porosity assumed to be constant; s

and s

are the water and

steam saturations;

is the water density, which is assumed to be constant for sim-

plicity; the steam density

is a function of the temperature T (i.e, we neglect the

effects of gas compressibility) and decreases with temperature; the term q is the mass

transfer between the gaseous and liquid water; u

and u

are the water and steam

phase velocity.

Disregarding heat conductivity, the energy balance equation can be written as:

∂

∂t

(

∂

∂x

+ u

)=0, (5.3)

103

where H

is the rock enthalpy per unit volume and h

and h

are the water and gas

enthalpies per unit mass, respectively, and

= H

. The enthalpies and

are

functions only of temperature and their expressions are found in Appendix A. From

these expressions, one can see that the enthalpies are increasing functions and that

is a convex function.

Remark 5.1.1. In

[6], Bruining et. al have deﬁned the water and gas enthalpies per

unit volume as:

and H

. (5.4)

From the expressions for

, h

and h

in Appendix A, we can prove that there is a

temperature



< T

where H

and H

satisfy:

(i) H

(



)=H

(



);

(ii ) H

(T) < H

(T) if T <



(iii ) H

(T) > H

(T) if T >



The gas and water enthalpies per unit volume increase when the temperature in-

creases. Moreover, h

> h

for all T in the physical range.

5.1.2 Physical Model

To determine the ﬂuid ﬂow rate, we use Darcy’s law for multiphase ﬂow without

gravity and capillary pressure effects:

= −

∂p

∂x

, u

= −

∂p

∂x

, (5.5)

where k is the absolute permeability for rock (see Appendix A); the relative perme-

ability functions k

and k

are considered to be power functions of their respective

effective saturations (see Appendix A);

and

are the viscosity of liquid water and

the viscosity of steam and they are functions of temperature; p is the common pres-

sure of the liquid and gaseous phase. We deﬁne the fractional ﬂow functions for water

and steam depending on the saturation and temperature as follows, see Figure 5.1:

, f

where d

. (5.6)

Using

(5.6) in Darcy’s law (5.5)

= uf

, u

= uf

, where u = u

+ u

, (5.7)

and u is the total or Darcy velocity. The saturations s

and s

add to 1.

104

Increasing T

-4

T/K500300

/(Pa.s)

w F

2x10

-5

T/K500300

/(Pa.s)

Figure 5.1: a) Left: The shape of water fractional ﬂux f

, originating from typical

) and k

), given in (B.6) for different values of the temperature T. The

separation between the curves is tiny in reality. b) Center: Water viscosity

(T). c)

Right: Gas viscosity

(T).

5.2 Regions under thermodynamical equilibrium

As we will see later, the ﬁve zones can be organized in three regions in different

physical situations where the ﬂuids are in thermodynamic equilibrium, which is of-

ten speciﬁed by an equation of state (EOS). Each physical situation determines the

structure of the governing system of equations. One region consists of steam only,

with temperature at least T

(the condensation temperature of pure water, which is

around 373.15K at atmospheric temperature), where we must determine two vari-

ables: temperature and Darcy velocity u. The steam saturation is s

= 1 (s

= 0).

There is a second region consisting of steam and water, with liquid water saturation

and gas saturation s

both less than 1. We must determine two variables: the

velocity and saturation (either s

or s

, because they add to 1); the temperature here

is known and its value is T

= T

. Finally there is a region of liquid water, where

we must also determine two variables: temperature and velocity. The saturation is

known: s

= 0 and s

= 1.

We summarize these regions as follows:

\T T > T

T = T

T < T

= 0 superheated steam zone hot steam zone 3

< 1 1 hot steam-water zone 4

= 1 2 hot water zone cold water zone

Table 1: Classiﬁcation according to saturation and temperature.

We call “steam region” (represented by “sr”) the superheated steam zone. We call

“boiling region” (represented by br) the hot steam zone together with the hot steam-

water zone and the hot water zone. We call “water region” (represented by “wr”) the

hot water zone together with the cold water zone. In

[6 ], there is no region with steam

105

above the boiling temperature; the sr and wr are called “hot region” and “liquid water

region” respectively.

Notice that the hot steam zone at boiling temperature belongs to both sr and br;

also, the hot water zone belongs to both br and wr.

Remark 5.2.1. Because of thermodynamical equilibrium, steam cannot exist at tem-

peratures lower than T

; similarly, there is no liquid water at a temperature above T

Thus the regions with numbers 1 -4 in Table 1 do not exist because of our requirement

of thermodynamic equilibrium. Regions 1- 4 would represent the following unstable

mixtures: (1) superheated steam with water, (2) superheated water, (3) steam below T

and (4) steam-water below T

5.3 Equations in conservative form

From the previous discussion, we notice that in each region under thermodynamic

equilibrium there are two variables to be determined in the system

(5.1)-( 5.3); the

other variables are trivial. For example, in the br the temperature and Darcy speed

are determined by the system of equations, but the saturation is trivial, its value is

= 0. Thus we can rewrite the system (5.1)-(5.3) as a system of two conservation

laws and two variables as follows. We add Eq.

(5.1) to (5.2) and use (5.7):

∂

∂t





∂

∂x





= 0. (5.8)

Using

(5.7) in the energy conservation equation (5.3), it becomes:

∂

∂t

(

∂

∂x

(

)=0. (5.9)

We will use

(5.8)-( 5.9) from now on. Not only this system models the ﬂow in each

region under thermodynamic equilibrium, but it also determines the shocks between

regions (see Sec. 5), when supplemented by appropriate thermodynamic equations of

state.

As initial conditions, we assume that the porous rock is full of a mixture of water

and steam (saturation s

(x, t = 0)=s

) with constant temperature T(x, t = 0)=T

As boundary conditions at the injection point at the left of the porous rock, the total

injection rate u

is speciﬁed as a constant. The constant water-steam injection ratio

needs to be given too, which is

, T

). It is speciﬁed in terms of the water fractional

ﬂow f

, T

) at the injection point.

5.4 Elementary waves under thermodynamical equi-

librium

We consider rarefaction and shocks waves for (5.8)-(5.9) in each region.

106

5.4.1 Steam region -sr

The temperature is high, T > T

, so there is only steam, s

= 0, (notice that from

Eq.

(5.1), q ≡ 0) and the state (s

, T,u) can be represented by (0,T,u ). The system

(5.8)-( 5.9) reduces to

∂

∂t

ϕρ

∂

∂x

= 0, (5.10)

∂

∂t

(

∂

∂x

= 0. (5.11)

Rarefaction wave

Assuming that all dependent variables are smooth, we can differentiate

(5.10) and

(5.11) with respect to their variables:

ϕρ



∂T

∂t

+ u



∂T

∂x

∂

∂x

= 0, (5.12)





+ C



∂T

∂t

+ uC

∂T

∂x

∂

∂x

= 0, (5.13)

where prime denotes derivative relative to T.

We use the notation



= d

/dT for the effective rock heat capacity di-

vided by

and C

= ∂(

)/∂T for the steam heat capacity per unit volume; we as-

sume that the effective rock heat capacity

is constant (see Appendix A). We rewrite

(5.12)-( 5.13) as:

∂

∂t





+ A

∂

∂x





= 0, (5.14)

where



ϕρ



(

+ C

) 0



and A







. (5.15)

The characteristic speed

and the eigenvector



r =(r

)

=(dT, du)

in the fol-

lowing system are the speed of rarefaction waves and the characteristic direction,

respectively:

det

(A −

B)=0, A



r =



r. (5.16)

We ﬁnd only one characteristic speed and vector:

(T,u)=

−



(

+ C

) −



, and





)



(5.17)

where

(T), h

= h

(T), C

= C

(T), and the derivatives relative to temperature

are



(T), c

= h



(T) and

is constant; we used the equality



= −

/T which

follows from Eq.

(B.4). The notation for this wave has subscript T because it is a

thermal wave; the saturation (s

= 0) stays constant, but the temperature T and the

107

speed u change. We obtain the thermal rarefaction curve in (T,u) space from



(5.17):

= u

(T)c

(T))



1 +

(T)c

(T)/



. (5.18)

Remark 5.4.1. Normally



, so we can approximate

by:



. (5.19)

In Appendix A, we obtain a better approximation for the characteristic speed

The rarefaction wave in the

{x, t} plane is the solution of the following equations:







, with

(u(

), T(

)). (5.20)

From Eqs.

(5.17.c) and (5.20.a), dT/d

satisﬁes:

= 1,

so we write T as:

−

+ T

−

, (5.21)

where, from Eqs

(5.20.b),

−

satisﬁes:

−

, T

−

Remark 5.4.2. If we make the same approximation used in Eq. (5.19) from Remark

5.4.1 we obtain for Eq.

(5.18):

−

, (5.22)

where u

−

and T

−

are any points in the br, so we obtain that u/T is constant on rar-

efaction wave, i.e., Eq.

(5.22) deﬁnes a Riemann invariant. So from (5.21), the speed

u in the plane

{x, t} is given by:

u =

−



−

+ T

−



In Appendix A, we obtain a better approximation for the rarefaction wave

(5.18).

Remark 5.4.3. In the sr, the temperature decreases from left to right along the thermal

rarefaction wave. In the Section 5.4.1 we consider a thermal steam shocks; analogously,

on the right of such a shock the temperature is higher than on the left.

108

Thermal steam shock

We assume now that T

> T

−

≥ T

. Let us consider the thermal discontinuity with

speed v

between the (−) state (0,T

−

) and the (+) state (0,T

). For such a

thermal steam shock, Eqs. (5.10)-(5.11) yield the following Rankine-Hugoniot (RH)

condition:

−u

−

(

−

)

−u

−



(

) −(

−

)



, (5.23)

where h

= h

) and

). From the second equality in Eq.

(5.23), we obtain u

as a function of u

−

= u

−

(

−

+ h

−h

−

(

−

+ h

−h

−

; (5.24)

it is easy to see that the denominator of

(5.24) is positive. Moreover u

> u

−

We substitute

(5.24) in Eq. (5.23); since u

is function of u

−

, we obtain v

−

; T

) or v

= v

−

; T

−

−h

−

(

−

+ h

−h

−

−h

−

(

−

+ h

−h

−

. (5.25)

Remark 5.4.4. Notice that we can rewrite

(5.25.a) as:

= v

−

; T

−



−h

−



(

− T

−

)

−



−h

−



(

− T

−

)

Deﬁning

), from convexity of h

(T) it follows that v

−

for T

> T

−

Using

(5.25.b) we see that

< v

−

if T

> T

−

, so the steam shock satisﬁes the

Lax condition. The equality v

−

holds if, only if, T

= T

−

5.4.2 Boiling region - br

Because the temperature is constant and equal to the boiling temperature T

it can be

shown (see [6]), that there is no mass exchange between phases and that the system

(5.8)-( 5.9) reduces to a single scalar equation with ﬁxed u = u

∂

∂t

+ u

∂

∂x

= 0. (5.26)

Eq. ( 5.26

) supports classical Buckley-Leverett rarefaction and shock waves.

109

Saturation shocks

Consider a

(− ) state (s

−

, T

) and a (+) state (s

, T

); we obtain the following

RH condition, where u

is the common Darcy velocity, T = T

and we use the nomen-

clature f

)=f

, T

−

)=v

−

, T

) − f

−

)

−s

−

) − f

−

)

−s

−

. (5.27)

A particular shock for (5.26) separates a mixture of steam and water on the left from

pure water on the right, both at boiling temperature. Following

[6 ] we call it the Hot

Isothermal Steam-Water shock (or HISW) between the

(− ) state (s

−

, T

) and the

(+) state (s

= 1,T

). It has speed v

g,w

given by:

g,w

−

)=v

g,w

−

, T

1 − f

−

)

1 −s

−

)

−

. (5.28)

Another particular shock for (5.26) separates pure water on the left from a mixture

of steam and water on the right, both at boiling temperature. We call it the Hot

Isothermal Water-Steam shock (or HIWS) between the

(− ) state (s

−

, T

) and the

(+) state (s

= 0,T

). It has speed v

g,s

given by

g,s

−

)=v

g,s

−

, T

−

)

−

. (5.29)

Notice that v

g,s

= 0,T

)=v

g,w

= 1,T

)=0.

Saturation rarefaction waves

We will denote by

the speed of propagation of saturation waves in the br.Itis

obtained from Eq. (5.26) as:

, T

∂f

∂s

). (5.30)

5.4.3 Water region - wr

The system (5.8)-(5.9) reduces to a scalar equation, with constant u

and s

= 1:

∂

∂t



(T)+

(T)



+ u

∂

∂x

(T)=0. (5.31)

Between a

(− ) state (1,T

) and a (+) state (1,T, u

), the following RH condi-

tion for the thermal discontinuity is valid:



−h



−(

)

+ C

, (5.32)

110

where u

is Darcy speed in the wr and C

∂h

/∂T; the second equality is ob-

tained taking into account

(B.5).IfT

= T

or T = T

, then u

= u

. From (5.32),

the discontinuity is a contact wave and there is no other characteristic speed in this

region.

5.5 Shocks between regions

Within shocks separating regions there is no thermodynamic equilibrium, so q is not

zero; however we can still use the system

(5.8)-( 5.9), because in each region the num-

ber of variable to be determined in the system

(5.1)-( 5.3) is at most 2. This system

contains another variable, namely the mass transfer term q. However this variable

is not essential to obtain the Riemann solution. It is useful to deﬁne the cumulative

mass transfer function:

Q(x, t)=



,t)d

, (5.33)

where this integral should be understood in the distribution sense. From Eq.

(5.33),

we can write q

= ∂Q(x, t)/∂x.

We also deﬁne

−

(t)=Q(x

−

,t) and Q

(t)=Q(x

,t), where x

−

and x

are the

points immediately on the left and right of the transition between regions. We deﬁne

the accumulative balance as the difference between

(t) and Q

−

(t) and denote it

[Q].

We can rewrite the system

(5.1)-( 5.3) (in distribution sense) as:

∂

∂t

, T)+

∂

∂x

(

uF(s

, T) − Q(s

, T)

)

= 0, (5.34)

where Q

(

Q(V), −Q(V),0

)

; G =(G

, G

)

and F =(F

, F

)

. The compo-

nents of F and G are readily obtained from Eqs.

(5.1)-( 5.3) using (5.7).

The shock waves are discontinuous solutions of Eq.

(5.34) and satisfy the RH

condition:

(G(s

, T

) −G(s

−

, T

−

)) = u

F(s

, T

) −u

−

F(s

−

, T

−

) −[Q], (5.35)

This term should be understood in the sense of distributions.

5.5.1 Water Evaporation Shock

WES - This is the discontinuity between a ( −) state (1,T

−

) in the wr and a (+)

state (s

, T

) in the br. It satisﬁes the following RH conditions for the speed v

WES

obtained from Eqs.

(5.8)-( 5.9):

(

) −

WES

(

)=u

−

WES

, (5.36)





−u

−

= v

WES



−



(5.37)

111

where f

= 1,·)=1, f

= f

, T

) and f

= 1 − f

The Darcy speed u

is found from u

−

using (5.36) and (5.37) as:

= u

−

(

−

)+s



−h

−



+ s



−h

−



(

−

)





+ f





− f





−h

−



+ s



−h

−



(5.38)

(5.38) is always valid because T

−

< T

, so each term in the denominator is pos-

itive. The terms

−

and h

− h

−

are positive because the enthalpies increase

with temperature. The term h

− h

−

is positive because h

> h

and since h

> h

−

the positivity follows.

Since we can write u

as function of u

, we obtain v

WES

= v

WES

−

WES



−h

−



+ f



−h

−



−

+ s



−h

−



+ s



−h

−



. (5.39)

The denominator of v

WES

in (5.39) is never zero because each term in the sum is

positive.

Lemma 5.5.1. Deﬁne

L (T) as:

L (T)=

−h

) −



−h



, (5.40)

where h

= h

(T). There is a unique temperature T

†

< T

such that L(T

†

)=0.

Moreover,

L (T) < 0 for T < T

†

and L(T) > 0 for T > T

†

Proof: We deﬁne:

−

= H

− H

where H

and H

are deﬁned in 5.4. From (i)-(iii) of Remark 5.4, we can see that

> 0. Since L

does not depend on T,ifT is small, we obtain that

−

= L

> h

(

−

)=⇒L(T)=h

(

−

) −L

< 0,

On the other hand, if T

= T

we obtain from Eq. (5.40) that

L (T

)=L

> 0,

so there is almost a temperature T

†

such that L(T

†

)=0. Moreover, the temperature

†

is unique, because if we differentiate L (T):

L (T)

= C

(

−

> 0,

i.e,

L (T) is a monotonic function, so there is only a temperature that satisﬁes L(T

†

0. 

112

Substituting s

= 1 −s

, f

= 1 − f

, using Lemma 5.5.1 for T = T

†

, we rewrite

Eq.

(5.39)

WES

− f

WES

−s

WES

, f

WES

≡

−h

−

)

L (T

−

)

, s

WES

≡

−

−h

−

)

L (T

−

)

(5.41)

Eqs.

(5.41) are the basis for a graphical construction of the WES, see Fig. 5.2.b.

For T

−

< T

†

, f

WES

and s

WES

are negative, while for T

−

> T

†

, f

WES

> 1 and

WES

> 1. When T

−

= T

†

, we obtain that

− h

−



−h

−



, so Eq.

(5.39)

reduces to:

WES

(1,T

−

= T

†



−h

†



+ f



− H

†

−h

†

)



+ s



−

+ C

(5.42)

Notice that if u

−

= u

, v

WES

= v

in the wr, see Eq (5.32).

5.5.2 Vaporization Shock

VS - It is a discontinuity between a (−) state (s

−

, T

−

) in the br and a (+) state

(0,T

> T

) in the sr. The Vaporization Shock satisﬁes the following RH condi-

tions with speed v

obtained from Eqs. (5.8)-(5.9):

(

−

)=u

−u

−

(

−

), (5.43)

− H

−s

−

−s

−

)=u

−u

−

(

−

(5.44)

where h

= h

), h

= h

), H

= H

) and f

−

= f

−

, T

Since T

> T

, we obtain v

as the following fraction, which has positive denomi-

nator:

= v

−

; T

−



−h



+ f

−



−h



−

+ s

−



−h



+ s

−



−h



. (5.45)

We rewrite Eq.

(5.45) in a shorter form. Substituting s

= 1 − s

and f

= 1 − f

we deﬁne

A(T



−h



−



−h



Lemma 5.5.2. If T

> T

, the water and rock enthalpies satisfy the following in-

equality:



−h





−h



Proof: From Remark 5.1.1, we know that h

< h

. Using the expression for A we

obtain:



−h



−



−h





−h



−



−h



−

)



−h

−



113

but (

−

) > 0 because

and (h

− h

) > 0 since h

increases with

temperature.



From above Lemma 5.5.2, we obtain that A < 0. We multiply and divide (5.45) by

A and we obtain:

−

− f

−

−s

, where f

≡



−h



A(T

)

, s

≡

−



−h



A(T

)

;

(5.46)

notice that f

and s

are negative. Eqs. (5.46) are the basis for a graphical con-

struction of VS, see Fig. 5.2. a.

The Darcy speed u

in the sr is the fraction with positive denominator:

= u

−

(

−

− f

−

)+B

(

−

− f

−

)+f

−

)+C



−

+ f

−





−

+ B

−

+ C



(5.47)

where:



−h



, B



−h



, C

−

. (5.48)

5.5.3 Condensation Shock

CS - This is the discontinuity between a (−) state (0,T

−

> T

−

) in the sr and a

(+) state (s

, T

) in the br. It is the reverse of the shock VS. From (5.47):

= u

−



−

+ B

−

+ C

−



−

(

− f

−

)+B

−

(

− f

)+f

−

)+C

−



+ f



(5.49)

−

, B

−

and C

−

are obtained from A

, B

and C

in (5.48) by substituting T

by T

−

Lemma 5.5.3. The denominator of

(5.49) is always positive.

Proof: We calculate:

−

+ B

−



−

−h





−

−h



= h

−

−h

+ h

−

−h

= h

−

( f

+ f

) −(h

+ h

= h

−

−(h

+ h

), (5.50)

where A

−

and B

−

are given in (5.48) and f

= f

and f

= f

Similarly, we calculate:

−

+ B

−

= h

−

−(h

+ h

), (5.51)

114

where s

= s

and s

= s

−

. So, from (5.51) and (5.50):

−

− f

)+B

−

− f

)=h

( f

−s

)+h

( f

−s



( f

−s

)+h

( f

−s

)





−h



− f

), (5.52)

where



−h



> 0.

We rewrite the denominator of

(5.49) as:

−



+ f





−h



− f

)+A

−

+ B

−

; (5.53)

notice that

−



−h



−



−



, (5.54)

where H

and H

are deﬁned in 5.1.1. From the same Remark, we know that H

−

> H

because T

−

> T

, (see (5.54)) and

−

because

is decreasing, we have:

−



−h





−



−



−



= 0, (5.55)

so from

(5.53) and (5.55) we obtain:

−



+ f





−h



− f

)+A

−

+ B

−

= C

−



+ f





−h



+ A

−

+ f

−



−h



> C

−



+ f





−h



+ A

−

> 0. (5.56)

The last equality is because h

> h

and each term is positive. So the denominator of

(5.49) is always positive. 

Since the CS is reverse of the VS and u

is a function of u

−

, we obtain v

−



−

−h



+ f



−

−h



−

+ s



−

−h



+ s



−

−h



or v

− f

−s

(5.57)

Since the CS shock is reverse of the VS, f

and s

are obtained from f

and s

(5.46) by substituting T

by T

−

; notice that the denominator in (5.57.a) is never zero.

115

5.5.4 Steam condensation front

SCF - This is the discontinuity between a (−) state (s

−

< 1,T

−

) in the br and a

(+) state (1,T

) in the wr. It is the reverse of the WES, so v

SC F

= v

SC F

−

; T

)

is given by:

SC F

−



−h



+ f

−



−h



−

+ s

−



−h



+ s

−



−h



−

− f

SC F

−

−s

SC F

. (5.58)

Here

(5.58) is derived as Eq. (5.41), with f

SC F

and s

SC F

obtained from f

WES

and s

WES

in (5.41) by substituting T

by T

−

From Eqs.

(5.36)-( 5.37), we can ﬁnd u

as the following fraction with positive

denominator:

= u

−

(

−

)



−





+ f

−





−

− f

−





−h



+ s

−



−h



−

+ s

−



−h



+ s

−



−h



(5.59)

5.6 The Riemann Solution

The Riemann problem is the solution of (5.8)-(5.9) with initial data



=(s

, T

) if x > 0

=(s

, T

,·) if x < 0,

(5.60)

where s :

= s

is the water saturation. We will see that the speed u cannot be pre-

scribed on both sides. Given a speed on one side the other one is obtained by solving

the system

(5.1)-( 5.3); in this case we have chosen to prescribe u

We consider in this paper the Riemann problem for all initial data; we divide the

data as:

Riemann Problem L state R state

Data A Steam and Water, T

= T

Steam, T

> T

Data B Water, T

< T

Steam and Water, T

= T

Data C Steam, T

> T

Steam and Water, T

= T

Data D Water, T

< T

Steam, T

> T

Data E Steam, T

> T

Water, T

< T

Data F Steam and water, T

= T

Water, T

< T

The Riemann problem with Data F for T

= T

was solved in [6]; in that paper,

is an arbitrary temperature where the water and rock enthalpies were made to

vanish; we do not repeat this Riemann problem here.

The rarefaction waves are denoted by R

for thermal rarefactions in sr and R

for

(Buckley-Leverett) saturation rarefactions in br. The shocks in regions under ther-

modynamic equilibrium are denoted with a single subscript, S

for thermal shocks in

116

sr, S

for (Buckley-Leverett) saturation shocks , S

for HIWS in br and S

for thermal

discontinuities in wr. We recall that the shocks between regions are WES, VS, CS and

SC F.

5.6.1 Riemann Problem A

Water injection.

First, we inject water with temperature T

, i.e., L =(1,T

) in a porous rock ﬁlled

with superheated steam, i.e., R

=(0,T

> T

), which is a sr. In the br, generated

by the L state, the ﬂow is governed by a Buckley-Leverett equation. It is well known

that the Buckley-Leverett rarefaction has speed

given by (5.30) from s

= 1 to

= s

∗

, which is deﬁned by:

∂f

∂s

∗

, T

∗

, T

)

∗

. (5.61)

There is a saturation shock and the solution is continued by a shock in the br.

Since the temperature increases in the rarefaction joining the

(− ) state (0,T

−

)

to the (+) state (0,T

> T

), see Rem. 5.4.3, there is a shock with speed v

given

(5.25).

Lemma 5.6.1. The speed v

g,s

of the HIWS shock between (s

∗

, T

−

) and (0,T

,u =

−

) given by (5.29) is larger than the speed v

of the shock between (0, T

−

) and

(0,T > T

Proof: From Eq.

(5.25) we have u

−

= u

and we can see easily that v

< u

From

(5.29) and (5.61), we have that v

g,s

for s

∗

satisﬁes v

g,s

> u

. So we conclude

that v

< v

g,s

. 

Thus we conclude that there is a shock with speed v

between the br and the sr.

Let f

and s

be given by (5.46); from the Sec. 2.1.3, we ﬁnd a saturation

s > s

∗

deﬁned by the following equality, (Fig. 5.2.a):

(

s, T

) − f

−s

∂f

∂s

(

s, T

). (5.62)

Lemma 5.6.2. The saturations

s and s

∗

satisfy s

∗

Proof: The result follows from shape of f

, i.e., a function convex between s

= 0

and s

inf

; and concave between s

inf

and s

= 1 with f

(1,·)=1.

Using Eq.

(5.62) we can prove the following Lemma:

Lemma 5.6.3. The solutions of

(5.62) are the points where the derivative of v

with

respect to S

vanishes.

117

Proof: The points that satisfy (5.62) are tangency points of the secant from the

point

, f

) to the graph f

, so the derivative vanishes. (An analytic proof for this

Lemma is similar to proof of the Lemma 5.6.6).



Moreover, we can verify the following Proposition:

Proposition 5.6.1. There are two saturation values that satisfy

(5.62) for each T

. The largest value satisﬁes (5.62) and maximizes v

while the other minimizes

; because of geometrical compatibility, we choose the largest value and we denote it

s. It is called “Hot-Bifurcation saturation I” or HBI.

Proof: We verify geometrically this proposition; equation

(5.62) represents tan-

gency points of the secant from the point

, f

) to the graph f

, for each ﬁxed T

−

In ﬁgure 5.2, we plot a possible point

, f

The largest water saturation value satisfying

(5.62) maximizes v

. This value is

denoted

S, and the other saturation minimizes the shock speed v

The analytical proof for this proposition can be obtained following the steps of

Proposition 5.6.4, however this proof is longer than the geometrical veriﬁcation. No-

tice that in Proposition 5.6.4, the chosen water saturation minimizes v

. 

Figure 5.2: The coincidence between v

and

.AsA < 0,soS

< 0 and f

< 0

are negative, the largest water saturation point that satisﬁes

(5.62) maximizes and

the other point minimizes v

It is necessary that v

(

) > v

; otherwise, the geometrical compatibility in Sec.

2.1.3 says that the vaporization shock VS does not exists. This is summarized as

follows:

Lemma 5.6.4. The vaporization shock between

(− ) state (s

−

, T

−

) and (+) state

(1,T

) with speed v

given by (5.45) is larger than v

for s

∗

≤ s

−

≤ 1 and

> T

. (We recall that s

∗

is deﬁned in Eq. (5.61)).

118

Proof: From (5.45), v

is written as:

−



−h



+ f

−



−h



−

+ s



−h



+ s



−h



, (5.63)

where s

= s

−

and s

= s

−

.Astos

∗

≤ s

≤ 1,wehavef

) > s

, so from Lemma

5.5.2 and

(5.63) we obtain:

) >

−



−h



+ f

−



−h



−

+ s



−h



+ s



−h



; (5.64)

using f

+ f

= 1 and S

+ S

= 1 in (5.64) we write ﬁnally:

) >



−h



−



−h



= v

.  (5.65)

From

(5.29), we conclude that v

g,s

and s

decrease together. From (5.25), v

constant. So we expect that there is a saturation s

∗∗

where v

= v

g,s

∗∗

= 0),

which is called “Hot-Bifurcation II saturation”, or HBII.

Remark 5.6.1. Let a,b, c, d

∈ R

∗

,if

−→

c −

d −

a −

b −

, for

= k.

One can verify the following:

Proposition 5.6.2. For each ﬁxed T

, there is a unique saturation s

∗∗

for which

−

; T

)=v

g,s

−

∗∗

)=v

∗∗

−

; T

). (5.66)

Furthermore s

∗∗

is deﬁned in terms of T

by any of the following equivalent equalities:

1. v

−

; T

)=v

g,s

−

∗∗

);

2. v

g,s

−

∗∗

)=v

∗∗

−

; T

);

3. v

−

; T

)=v

∗∗

−

; T

Proof:

(1) First we assume that v

= v

g,s

We set T

= T

−

and T = T

, so we can write

−

−h

−

)

−

−h

−

)

−

∗∗

= v

g,s

, (5.67)

119

where f

= f

∗∗

, T

We multiply and divide the third term of

(5.67) by



−h

−



, which is positive:

−

−h

−

)

−

−h

−

)

−



−h

−



∗∗



−h

−



= v

g,s

. (5.68)

From the Remark 5.6.1, we obtain:

−

−h

−

)+f



−h

−



−

−h

−

)+s

∗∗



−h

−



−

∗∗

= v

g,s

. (5.69)

We multiply and divide the third equation of

(5.69) by



−h

−



,sov

equals:

−

−h

−

)+f



−h

−



−

−h

−

)+s

∗∗



−h

−



−

−h

−

)

∗∗

−

−h

−

)

g,s

. (5.70)

From the Remark 5.6.1, and using that 1

− f

∗∗

)= f

and 1 −s

∗∗

= s

∗∗

we obtain:

−

−h

−

)+f



−h

−



−

+ s

∗∗

−

−h

−

)+s

∗∗



−h

−



= v

g,s

. (5.71)

so v

= v

g,s

implies that v

= v

g,s

= v

(2) We assume that v

g,s

= v

so:

−

−h

−

)+f



−h

−



−

+ s

∗∗

−

−h

−

)+s

∗∗



−h

−



−

∗∗

= v

g,s

(5.72)

Performing the above calculation in reverse order, we see that v

= v

g,s

implies that

= v

g,s

= v

(3) We assume that v

= v

, so:

−

−h

−

)

−

−h

−

)

−

−h

−

)+f



−h

−



−

+ s

∗∗

−

−h

−

)+s

∗∗



−h

−



. (5.73)

From the Remark 5.6.1, we obtain:

−

−h

−

)

−

−h

−

)



−h

−



−

−h

−

)



∗∗



−h

−



−

−h

−

)



. (5.74)

120

Since



−h

−



−

−h

−

) = 0 (see Lemma 5.5.2), then:

−

−h

−

)

−

−h

−

)

∗∗

. (5.75)

so v

= v

implies that v

= v

g,s

= v

. 

Lemma 5.6.5. For any T

> T

in the physical range, the saturation s

∗∗

given by Eq.

(5.66) satisﬁes s

∗∗

Proof: The saturation s

∗∗

satisﬁes Eq. (1) of Proposition 5.6.2. Geometrically,

this Equation represent a line from a point

, f

) passing on the point (0,0)

and intersecting the graph f

for each ﬁxed T

, where s

and f

are given by Eq.

(5.46.c) and (5.46.b) respectively. Notice that this intersection occurs always because

< f

< 0. Moreover, from Figure 5.3 we can notice that s

∗∗

≤

s and s

∗∗

s if,

only if, s

∗

= s

∗∗

s, and from Lemma 5.6.2 follows that s

∗

s,sos

∗∗

s. Since T

arbitrary the Lemma is proved.



Figure 5.3: The saturation s

∗∗

obtained from a line from a point (s

, f

) passing on

the point

(0,0) and intersecting the graph f

Solution. Now we can describe the possible solutions for Riemann Data A:

For

≤ s

≤ 1. The waves R

 VS, with

s given by Eq. (5.62) and the sequence:

=(s

, T

)

−→ (

s, T

)

−→ (0,T

)=R. (5.76)

For s

∗∗

≤ s

s. The wave VS with s

∗∗

given by Eq. (5.66) and the sequence:

=(s

, T

)

−→ (0,T

)=R. (5.77)

121

For s

< s

∗∗

. The waves S

→ S

, with sequence:

, T

)

−→ (0,T

)

−→ (0,T

). (5.78)

Wave behavior on bifurcations

Notice that the Buckley-Leverett rarefaction wave R

disappears on the vertical dot-

ted line from

S that separates region I from II; we call it the “R

− S

boundary

bifurcation”. On the vertical dotted line from S

∗∗

separating region II from III, the

condensation shock S

between regions disappears and a saturation S

and a ther-

mal shock S

appear in the br; we call this line the “S

− S

bifurcation”.

Figure 5.4: The dotted line is the boundary of physical range. The dashed lines are

bifurcation loci. The temperature is higher than the water boiling temperature (su-

perheated steam) or is equal to boiling temperature. The saturations S

∗∗

and

S are

given in Eqs.

(5.66) and (5.62); the vertical lines from

S and S

∗∗

are the “R

− S

bifurcation” and the “S

− S

bifurcation”.

5.6.2 Riemann Problem B

Since the ﬂow in the wr is governed by the linear Eq. (5.31) with constant character-

istic speed v

given by (5.32), this wave is a contact discontinuity.

For the Riemann problem L

=(1,T

≤ T

) and R =(s

, T

), we have the

following:

Proposition 5.6.3. For each

(− ) state (1,T

−

< T

−

) and (+) state (s

, T

there is a water saturation s



= s



−

), such that for s

satisfying s



≤ s

≤ 1, the

shock speeds v

WES

and v

satisfy:

≥ v

WES

, s



≤ s

≤ 1; (5.79)

122

the equality occurs only if s

= 1 or s

= s



. We call s



a “Cold Bifurcation saturation

I” or CBI. Moreover, there is a water saturation s

††

= s

††

−

) satisfying s



< s

††

such

that:

††

)=v

WES

−

††

). (5.80)

Proof: We deﬁne the functions Θ

−

) and Ψ(T

−

) as:

−

) := v

−v

WES

−

), (5.81)

−

) :=

−

) −v

WES

−

). (5.82)

The proposition is true, because from Figure 5.5 the curve Ψ is above the curve Θ.



300 310 320 330 340 350 360 370

0.98

0.982

0.984

0.986

0.988

0.99

0.992

0.994

0.996

0.998

Temperature/K in the Water Region / T

−

ater Saturation in the Boiling Region / s

Figure 5.5: The functions Θ(T

−

) and Ψ(T

−

) in the {T

−

In the nomenclature of

[26], the state s

= 1 is the left-extension of s

††

with speed

. Also, s

††

here coincides with s

††

obtained in [6]. This saturation maximizes v

WES

(and consequently v

SC F

); we call s

††

the “Cold Bifurcation saturation II” or CBII.

Remark 5.6.2. Notice that from Sec. 5.5.1, f

WES

and s

WES

are negative if T

−

< T

†

and

positive if T

−

> T

†

(see Fig. 5.2.b). The solution behavior is the same in both cases.

Remark 5.6.3. Notice that from Section 5.5.1, f

WES

and S

WES

are negative if T < T

†

and positive if T > T

†

(see Figure 5.6). Nevertheless the solution behavior is the same,

i.e., the point T

†

is not a bifurcation.

Prop. 5.6.3 yields the following Corollaries used to obtain the solution in the br:

123

Figure 5.6: We represent two possible points s

WES

, f

WES

).IfT < T

†

, both s

WES

and

WES

are negative; otherwise both are larger than 1.

Corollary 5.6.1. If s

infl

) < s

< s

††

, the solution continues in the br as a rarefac-

tion to s

.Ifs

< s

infl

the rarefaction continues to s

, where s

is deﬁned by the second

equality in:

§,+

∂f

∂s

, T

) − f

, T

)

−s

, (5.83)

where s

infl

= s

infl

) is the inﬂection saturation deﬁned by:

∂

∂s

, T

)



infl

= 0. (5.84)

Proof: From Section 5.4.3 and

[6 ], we know that v

satisﬁes:

+ C

< 1,

the last inequality holds because

and C

are positive, and from Eq. (5.79) in Propo-

sition 5.6.3 we obtain that v

WES

< 1.

If s

> s

∗

where s

∗

satisﬁes (5.61), the rarefaction wave is always a solution but at

∗

we know that:

∂f

(

∗

)

∂s

> 1.

As v

WES

satisfy (5.80) we obtain that:

∂f

††

, T

)

∂s

∂f

∗

, T

)

∂s

124

so if s

infl

< s

there is a rarefaction from s

††

to s

and if s

< s

infl

there is a rarefac-

tion from s

††

to s

that satisﬁes (5.83). 

Corollary 5.6.2. As the left state temperature T

−

tends to the water boiling tempera-

ture, the water saturation s

††

tends to 1, i.e., the limit of s

††

lies in the wr.

Proof: We take the limit T

−

−→ T

in Eq. (5.38), and obtain:

= u

−h

)

−h

)

because h

> h

. Substituting u

= u

in Eq. (5.39) and taking the limit T

−

−→ T

WES

From Eq.

(5.80), after simpliﬁcations, we obtain that at (s

††

, T):

∂f

∂s

1 − f

1 −s

††

, (5.85)

Eq.

(5.85) has two solutions, one of which is s

††

= 1. The other solution s

††,2

satisﬁes

††,2

< s

infl

, so only s

††

= 1 is important for the solution. 

Solution. Now we can describe the possible solutions for Riemann Data B:

For s

> s

††

. As s

††

satisﬁes (5.80), (u

)∂f

, T

)/∂s

< v

WES

, i.e, the shock

WES

is faster than the characteristic speed in the br, the wave sequence is:

=(1,T

)

WES

−−→ (s

, T

)=R. (5.86)

For s

infl

) < s

< s

††

. The waves WES  R

, with sequence:

=(1,T

)

WES

−−→ (s

††

, T

)

−→ (s

, T

)=R. (5.87)

For s

< s

infl

). The waves WES  R

 S

with sequence:

=(1,T

)

WES

−−→ (s

††

, T

)

−→ (s

, T

)

−→ (s

, T

)=R, (5.88)

where s

is given by Eq. (5.83).

Wave behavior on bifurcations

The Buckley-Leverett rarefaction wave R

appears on the horizontal dotted line from

††

that separates region I from II; we call this line the “S

WES

− R

bifurcation”. On

the horizontal dotted line from S

infl

separating the region II from III, a Buckley-

Leverett shock S

appears; we call this line the “R

− S

bifurcation”.

125

Figure 5.7: The dotted line is the boundary of physical range. The dashed lines divides

are the bifurcation loci. The temperature is lower than the water boiling temperature

(liquid water) or is equal to water boiling temperature. The saturations s

††

and s

infl

are given in Eqs. (5.66) and (5.62); the horizontal lines from s

††

and s

infl

are the

“S

WES

− R

bifurcation” and the “R

− S

bifurcation”.

5.6.3 Riemann Problem C

Let L =(0,T

> T

) and R =(s

, T

). In this Riemann Problem, there are

two relevant bifurcation curves. The ﬁrst bifurcation occurs at the points where the

thermal rarefaction speed

(Eq. (5.17.a)) equals the shock speed v

(Eq. (5.57)).

The other bifurcation appears at the points where the speed v

coincides with the

Buckley-Leverett rarefaction speed

(Eq. (5.30)) in the br.

Moreover, there is a point where all previous speeds coincide. It is a double bi-

furcation point in the

−

} plane of left state temperatures T

−

in sr and right

saturation s

in br (see Sec. 5.6.3). This point is denoted by (

T;s

†

) and it should be

understood as the projection of

= 0,

T;s

†

, T

) onto the {T

−

} plane.

Deﬁnition of

T and s

†

Let Υ = Υ(T

−

) be deﬁned as:

−



−



− f

−s

. (5.89)

The fraction u

−

is obtained from Eq. (5.47). This fraction does not depend on u

−

or u

, showing that also Υ does not depend on u

−

(or u

). Moreover, at the points

−

) where Υ = 0, the equality

= v

holds.

126

Now we can deﬁne the thermal coincidence as the curve where the left thermal

wave speed

coincides with the condensation shock v

; it is denoted by TCS locus:

TCS

= {(T, s) | Υ(T, s)=0, for T ∈ sr and s ∈ br }. (5.90)

Analogously, we deﬁne Λ

= Λ(T

−

) as:

− f

−s

−

∂f

∂s

, (5.91)

where f

and s

depend on T

−

, see (5.46) . Now we can deﬁne the CSS locus as the

curve where v

and for each T

−

ﬁxed, v

−

) is understood as a function

of s

that is minimized (see Prop. 5.6.4):

CSS

= {(T

−

) | Λ = 0 | v

is minimum; T

−

∈ sr, s

∈ br}. (5.92)

In Fig. 5.8, the TCS and CSS loci are shown as curves in the plane

{T,s}. The

horizontal axis represents the states in the sr and the vertical axis represents the

states in the br. The two loci intersect transversally at

(

T;s

†

), the double bifurcation

point “SHB”. It can be obtained numerically using root ﬁnders. The temperature

satisﬁes T

T, and the saturation s

†

satisﬁes 0 < s

†

< 1.

For the Riemann solution, we need to study the relationships between T

and

T at

= 0,T

−

), and between s

and s

†

at (s

, T

) .

Let us deﬁne

−

) :=

− f

−s



−

−h



+ f



−

−h



−

+ s



−

−h



+ s



−

−h



(5.93)

the second equality originates from the deﬁnition f

and s

given in Eq. (5.41).

Lemma 5.6.6. The solutions of

(5.92) are the points where ∂Ξ/∂s

is zero:

Proof: We differentiate

(5.93) with respect to s

, equate it to zero and isolate

∂f

/∂s

to obtain the result. We notice that ∂f

/∂s

= −∂f

/∂s

and ∂s

/∂s

= −1.



Lemma 5.6.7. For all 0 ≤ s

≤ 1 the following inequality is valid

≤

∂f

infl

,·)

∂s

(5.94)

Proof: The extremum for f

is obtained for ∂/∂s

( f

)=0; performing this

differentiation, we obtain that f

= ∂f

/∂s

, which yields

≤ max

∂f

∂s

∂f

infl

,·)

∂s

. 

The CSS locus is obtained using the following proposition:

127

Figure 5.8: Schematic phase space. The intersection of the TCS and CSS loci is the

SHB at

(

T;s

†

). The horizontal axis represents the sr and the vertical axis represents

the br,soSHB represents two points:

T is the projection of

= 0;

T) on the sr;

†

is the projection of (s

= s

†

; T

) on the br; between the TCS locus and the water

saturation axis Υ

> 0,so

> v

; in the complementary region

< v

Proposition 5.6.4. There are always two water saturation values satisfying Λ

(T,s)=

0 for each ﬁxed T > T

. The smallest value minimizes Ξ (and consequently v

), while

the other value maximizes Ξ (and v

). We deﬁne s

•

as the smaller water saturation

value.

The point

(T,s

•

) belongs to CSS locus.

Proof: The function ∂f

/∂s

is continuous and satisﬁes the following relationship,

where ∂f

/∂s

is evaluated at T = T

∂f

= 0)

∂s

∂f

= 1)

∂s

= 0. (5.95)

From equation

(5.93) with s

= 0 it follows that:



−

−h



−



−

−h



< 1. (5.96)

From equation

(5.93) with s

= 1 we have:



−

−h



−



−

−h



< 1. (5.97)

128

From Lemma 5.5.2 we know that

−h

), from the fact that f

> s

we obtain that f

< s

and from Lemma 5.6.7 it follows that:



−

−h



+ f



−

−h



−

+ s



−

−h



+ s



−

−h



∂f

infl

)

∂s

−h

)

−

−h

)

∂f

infl

)

∂s

, (5.98)

so from the mean value theorem, for each ﬁxed T, there is at least one point between

= 0 and s

= s

infl

and at least another point between s

= s

infl

and s

= 1 where

(T;s) vanishes.

We need to show that there exist exactly 2 saturations that satisfy Ξ

(T;s)=0.To

do so we note that:

∂

)

∂s

= 0 ⇐⇒ s

= s

infl

. (5.99)

Moreover ∂

/∂s

satisﬁes:







∂

)

∂s

> 0, if s

< s

infl

∂

)

∂s

< 0, if s

> s

infl

(5.100)

but from Lemma 5.6.6 we know that if s

satisﬁes Ξ(T, s)=0, then s

is a extremum

for

(5.93), i.e,

∂f

∂s

⇐⇒

∂Ξ

∂s

= 0, (5.101)

where Ξ is the function deﬁned in

(5.93).

Claim 1: The smaller water saturation for which Ξ

(T,s) vanishes is a minimum

for Ξ

(T,s).

Proof: We differentiate Ξ with respect to s

and obtain after manipulations that:

∂Ξ

∂s

B − A



∂f

∂s

−Ξ



, (5.102)

where A and B are given in

(5.48) and D is the denominator of Ξ. The term D satisﬁes

= 0, and from Lemma 5.5.2 we know that B − A > 0, so the sign of (5.102) depends

only on the sign of:

∆

∂f

∂s

−Ξ, (5.103)

where ∆ is a continuous function.

From

(5.95) and (5.96), we know that ∂f

/∂s

−Ξ < 0 for s

= 0 or 1.IfΞ(T,s)=0

is satisﬁed we obtain that ∂Ξ

/∂s

= 0 and then ∆ = 0; on the other hand if ∆ = 0 this

point satisﬁes Ξ

(T,s)=0 and ∂Ξ/∂s

= 0.

129

We know that there are at least 2 water saturations where (5.92) is satisﬁed. We

denote the smaller one by s

•

, so we obtain that ∆ < 0 for 0 ≤ s

< s

•

At the s

•

where (5.92) is satisﬁed we obtain that:

∂Ξ

∂s

= 0 and

∂

∂s

> 0, (5.104)

so there is a neighborhood V

•

that satisﬁes:



∂f

∂s

, if s

∈ V

•

/ s

< s

•

Ξ <

∂f

∂s

, if s

∈ V

•

/ s

> s

•

(5.105)

So for s

∈ V

•

, s > s

•

the graph of v

lies below the graph of ∂f

/∂s

. The point s

•

minimizes Ξ.

Claim 2: There is another water saturation point s

••

that satisﬁes (5.92); s

••

maximizes Ξ and satisﬁes s

••

> s

infl

Proof: From Eq.

(5.102) and (5.105) we know that Ξ increases for s

> s

•

. From

Eq.

(5.105), Ξ < ∂fw/∂s

for s

∈ V

•

and s

> s

•

,asΞ is continuous we obtain that

••

satisﬁes:

••

> s

infl

. (5.106)

(5.106) is not satisﬁed, s

••

is a extremum point for Ξ and satisﬁes:

∂Ξ

(

••

)

∂s

= 0 and

∂

(

••

)

∂s

> 0. (5.107)

As Ξ

< ∂f

/∂s

for s

> s

•

, and Ξ(s

••

)=∂f

/∂s

, so there is a neighborhood

••

that satisﬁes:

∂Ξ

∂s

∂

∂s

≥ 0 for s

∈

••

/ s

< s

••

. (5.108)

Taking the limit s

−→ s

••

in Eq. (5.108) we obtain that:

∂Ξ

∂s

∂

∂s

≥ 0. (5.109)

For s

••

to satisfy Eq. (5.109), it is necessary s

••

= s

infl

, but from Eq. (5.98) we obtain

that Ξ

infl

) < ∂f

infl

)/∂s

,sos

••

> s

infl

The point s

••

maximizes Ξ, because Ξ increases for s

< s

••

and as ∂Ξ( s

••

)/∂s

= 0

and ∂

••

)/∂s

< 0, so there is a neighborhood V

••

where Ξ satisﬁes:



∂f

∂s

, if s

∈ V

••

/ s

< s

••

Ξ >

∂f

∂s

, if s

∈ V

••

/ s

> s

••

(5.110)

130

For s ∈ V

••

, s > s

••

the graph of Ξ lies above the graph of ∂f

/∂s

, so from (5.102) we

obtain that:



∂Ξ

∂s

> 0, if s

∈ V

••

/ s

< s

••

∂Ξ

∂s

> 0 < 0, if s

∈ V

••

/ s

> s

••

(5.111)

The point s

••

maximizes Ξ.

Claim 3: There is no other point that satisﬁes

(5.92).

Proof: If this point exists, we denote it by s

•••

,; it should minimize locally Ξ because

> ∂f

/∂s

for s

> s

••

, then from (5.102) we obtain that Ξ decreases. Moreover, Ξ

would satisfy:

∂f

∂s

, for s

••

< s

•••

, (5.112)

and as ∂

/∂s

< 0 for all point s

> s

infl

(including s

•••

) and as ∂Ξ(s

•••

)/∂s

= 0

there should exist a neighborhood V

•••

where

∂f

∂s

for s

∈ V

•••

. (5.113)

So from

(5.113), we obtain that Ξ is tangent in s

•••

, moreover from (5.102)

∂Ξ

∂s

< 0, for s

∈ V

•••

. (5.114)

Since Ξ

> ∂f

for s

< s

•••

and as Ξ(s

•••

)=∂f

•••

)/∂s

is is necessary that:

∂Ξ

∂s

∂

∂s

, for

V  V

•••

(5.115)

where

V is a neighborhood where

(5.115) is satisﬁed.

Taking the limit s

−→ s

•••

in (5.115) we obtain

∂Ξ( s

•••

)

∂s

≤

∂

∂s

< 0. (5.116)

so we conclude that there is no other point that satisﬁes

(5.92).

There are only two points that satisfy

(5.92) for each ﬁxed T

−

. 

From Prop. 5.6.4, one can prove:

Corollary 5.6.3. For each ﬁxed T in the br, the solution after s

•

continues as a rar-

efaction in the br.

Proof: From Proposition 5.6.4, s

•

minimizes Ξ. From eq. (5.96) , we know that

= 0) < 1, (5.117)

So for s

= s

•

we obtain:

∂f

∂s

= Ξ < 1 <

1 − f



)

1 −s



∂f



)

∂s

, (5.118)

131



is the smallest value where the rarefaction can be sketched in the br. 

Prop. 5.6.4 implies that for each ﬁxed temperature T

−

there are two saturations

that satisfy Λ

= 0, i.e.

= v

, but we choose the one that belongs to the CSS and

denote it by s



In Fig. 5.9. a, we obtain the water saturation s



, for each T

−

. We represent this

map of T

−

into s



, which is used in Item (2) of the following proposition, as:



:= s



−

). (5.119)

Proposition 5.6.5. If T

−

T, then

> v

. Thus there is a shock between (s

0,T

−

) to (s



, T

). Furthermore, the solution continues in the br as a rarefaction.

Proof: The proof consist of two steps:

Figure 5.9: a) Left: The graph of Λ = 0. For each T

−

, the vertical line crosses the

graph at a saturation s



= s



−

). b)-Right: The value T

) is obtained from the

TCS locus. We plot a horizontal line from s

, the point where this line intersects the

graph Υ is the point

),s

(1) The characteristic speed

is larger than the shock speed v

at (T



). From

Figs. 5.9. a and 5.8, we can see that Υ

> 0 at (T

−



),so

> v

. Thus there is a

shock between

(0,T

−

) and (s



, T

(2) The solution continues as a rarefaction in the br. As s



satisﬁes Λ = 0, this fact

follows from Cor. 5.6.3.



Prop. 5.6.5 yields:

Corollary 5.6.4. When T

−

tends to T

, the saturation s



tends to s

, the connate

water saturation (see Appendix A); thus the solution is continuous when T tends to T

between the sr and the br.

132

Proof: Using Λ = 0, where Λ is given by Eq. (5.91), taking the limit of T

−→ T

we obtain:

, T

)

∂f

, T

)

∂s

. (5.120)

The smallest value where

(5.120) is satisﬁed is s

= s

,sos



= s

and the solution

is continuous between the sr and the br.



Remark 5.6.4. When T

−

= T

, the left state lies in the br. This solution was obtained

in [6]. In that case, there was a region with connate water saturation in the br. From

Cor. 5.6.4, our solution agrees with that in [ 6]. Notice that the connate water in this

region is immobile, but this water evaporates when the vaporization shock advances.

Fix s

< s

†

. Using the TCS locus, we obtain T

) as a function of s

. For each

we draw a horizontal line. We project the intersection of this horizontal line and

the TCS onto the horizontal axis to obtain T

; we denote this mapping by:

:= T

). (5.121)

It is important that T

) is monotone for s

< s

†

Proposition 5.6.6. For s

< s

†

and T

−

T, there is a rarefaction from

(0,T

−

) to

(0,T

).At(0,T

) the following speeds coincide:

)=v

), (5.122)

so there is a left characteristic shock between

(0,T

) and (s

, T

) with speed

Proof: In Fig. 5.9.b, we plot an example of s

and its respective T

). Since the

temperature decreases from left to right along the thermal rarefaction wave, from

Rem.

(5.4.3), this wave is a rarefaction. 

Corollary 5.6.5. When s

tends to 0 in br, the temperature T

converges to T

; thus

the solution is continuous between the br and the sr.

Proof: Taking the limit of s

−→ 0 in Eq. (5.89) we obtain:

= 0)=

−



−

−h





−

−h



. (5.123)

We multiply the second term on the right hand side by

−

− T

)/(T

−

− T

) and we

obtain:

= 0)=

−



−

−h





−

− T





−

−h





−

− T



. (5.124)

133

Taking the limit T

−

−→ T

in (5.124), we obtain:

lim

−

−→T

Υ(S

= 0)=

− lim

−

−→T



−

−h





−

− T





−

−h





−

− T



−

= 0. (5.125)

Thus T

−

= T

belongs to the TCS locus. Moreover, Eq. (5.123) deﬁnes a curve with

= 0 and T

−

in the sr; we can prove that this curve is monotonic (see Figure 5.10),

therefore this is the only solution.



Figure 5.10: Left: plot of Eq. (5.123) for T ∈ [373.15K,380K]. Right: the same plot for

∈ [373.15K,490K] .

In Eq. (5.121), we ﬁnd s



= s



−

); also T

= T

) from Eq. (5.119), see Fig.

5.9.b. Using s



and T

, four possible solution candidates arise:

(i) T

−

< T

, s



< s

−

. A shock from (0,T

−

) to (s



, T

), continuing to

, T

) through a Buckley-Leverett rarefaction.

(ii ) T

−

< T

, s



> s

. A shock from (0,T

−

) to (s

, T

) with speed v

(iii ) T

−

> T

, s



< s

. A rarefaction from (0,T

−

) to (0,T

) followed by a

shock to



, T

) continuing to (s

, T

) through a Buckley-Leverett rarefaction.

(iv) T

−

> T

, s



> s

. A rarefaction from (0, T

−

) to (0,T

), followed by a

shock to



, T

) with speed v

Proposition 5.6.7. For left temperature T

−

and ﬁxed right saturation s

satisfying

−

T and s

< s

†

, (i) and (iii ) do not occur.

134

Proof: From Fig. 5.11. a, we separate the water saturation in the br (the right side

states of the Riemann problem) in two intervals, s

RegI

and s

RegII

. The ﬁrst one is s

∈

†

,]; the second one is s

∈ [s

= 0,s

]. Recall Eq. (5.121), which deﬁnes T

function of s

. So we deﬁne T

RegI

= T

RegI

); similarly, we deﬁne T

RegII

= T

RegII

One can verify from Fig. 5.11.a that T

) is monotone increasing.

Figure 5.11: a)-Left: The map T

deﬁnes T

RegI

and T

RegII

. The saturation lies in the

br, which is subdivided in s

RegI

=[s

†

] and in s

RegII

=[0,s

]. The corresponding

intervals in the sr are T

RegI

and T

RegII

. b)-Right: Mapping from T

RegI

to br. Each

−

∈ T

RegI

deﬁnes a s



in the br (see Prop. 5.6.6); notice that s



satisﬁes s



≥ s

†

, which

is the saturation at SHB.

Fig. 5.11.b represents the mapping from T

RegI

to the br. Each value of T

−

∈ T

RegI

deﬁnes a value for s



in the br through the function s



−

) in Eq. (5.119). The function



−

) is not monotone and all values of s



= s



−

) for T

−

∈ T

RegI

are larger or

equal to s

†

, the saturation at the SHB point. Notice also that the mapping s



−

) for

−

∈ T

RegI



RegII

is onto s

RegI

. So we obtain that s

< s



−

) for T

−

∈ T

RegI

,so(i)

and (iii ) do not occur. 

Solution: (summarized in Fig 5.13).

(I) For T

T and s

> s

infl

> s

†

. The waves R

 CS  R

 S

with sequence:

=(0,T

)

−→ (0,

−→ (s

†

, T

)

−→ (s

, T

)

−→ (s

, T

)=R, (5.126)

where

(

T,s

†

) is the SHB point and s

is given by Eq. (5.83).

(II) For T

T and s

infl

> s

†

. The waves R

 CS  R

with sequence:

=(0,T

)

−→ (0,

−→ (s

†

, T

)

−→ (s

, T

)=R. (5.127)

(III) For T

< T

and s

< s



). The solution is sketched in Fig. 5.12.a, based

on Props. 5.6.7 and 5.6.6; the waves are R

 CS with sequence:

=(0,T

)

−→ (0,T

T,u

)

−→ (s

, T

)=R.  (5.128)

135

Figure 5.12: a)-Left: Riemann Solution. the point T

in the horizontal axis represents

= 0,T

). The rarefaction from (s

= 0,T

) to (s

= 0, T

) is represented by a line

and an arrow to indicate the direction of increasing speed; the rarefaction is followed

by a shock from

= 0,T

) to (s

, T

) with speed v

with construction shown by

dotted lines. b)-Right: the dotted line represents the shock from

= 0,T

) to

, T

) with speed v

. We draw the solution for a left state (s

= 0,T

), which

we represent by T

136

(IV) For T

> T

and s

< s



), the solution is sketched in Fig. 5.12.b. It is the

wave CS with sequence:

=(0,T

)

−→ (s

, T

)=R. (5.129)

(V) For T

T and s

infl

> s



), where the mapping s



= s



) is deﬁned

in Eq.

(5.119). We obtain the waves CS  R

with sequence:

=(0,T

)

−→ (s



, T

)

−→ (s

, T

)=R. (5.130)

(VI) For T

T and s

> s

infl

> s

†

. See (5.130). The waves CS  R

 S

with

sequence:

=(0,T

)

−→ (s



, T

)

−→ (s

, T

)

−→ (s

, T

)=R, (5.131)

where s

is given by Eq. (5.83) with s

= s

Remark 5.6.5. We remark that the CS is a double sonic transitional wave, see

[58].

Wave behavior on bifurcations

Notice that the Buckley-Leverett shock S

disappears on the line that separates I

from II; we call this line “R

− S

bifurcation”. On the line that separates I from VI,

the thermal rarefaction R

in sr disappears; we call this line the “R

− S

bifurca-

tion”. On the line that separates II and V, the thermal rarefaction R

in the br also

disappears, so this line is the “R

− S

bifurcation”. On the line that separates V

from VI the Buckley-Leverett shock S

disappears, so this line is the “R

− S

bifur-

cation”. The “R

− S

bifurcation” is the vertical line from SHB separating I, II from

V, VI and the “R

− S

bifurcation” is the horizontal line from S

infl

separating I, VI

from II, V.

On the line that separates II from III the Buckley-Leverett rarefaction R

disap-

pears; we call this line the “R

− S

bifurcation”. On the curve that separates III

from IV, the thermal rarefaction R

disappears; we call it the“R

− S

bifurcation”.

Finally, on the curve that separates IV from V the Buckley-Leverett rarefaction R

disappears; we call it the “S

SC S

− R

bifurcation”.

5.6.4 Riemann Problem D

We inject pure water at temperature T

< T

, i.e, the left state is (0,T

); on the

right we have pure steam at temperature T

> T

Before describing our proposed Riemann solution, it is necessary to prove that

there is no possible Riemann solution with a direct shock between sr and the wr.

Using the RH condition

(5.36)-( 5.37) with s

= 0 and T

replaced by T

> T

,weob-

tain this hypothetical “complete water evaporation shock”, labelled CWES, with speed

CWE S

. The superscript + (−) in the following equations represents the temperature

137

Figure 5.13: Phase diagram for Riemann Problem C. The dotted line delimits the

physical range, the dashed lines are bifurcation loci. The continuous curves are parts

of the TCS and the CSS bifurcations loci. The horizontal axis represents the left

states

(0,T

) in sr; the vertical axis represents the right states (s

, T

) in br.

The solutions are given in I-VI, Sec. 5.6.3.

−

). The speed of such shock between (s

= 1,T

−

) to (s

= 0,T

> T

,u)

would be:

CWE S

−

= 0,T

−h

−

)

− H

−

−h

−

)

. (5.132)

Proposition 5.6.8. Complete Evaporation. For any T

−

< T

, if there exists

a complete water evaporation shock from

(1,T

−

) to (0,T

> T

) with speed

CWE S

−

= 0,T

) given by (5.132), then this shock satisﬁes:

CWE S

> v

, (5.133)

where v

is the speed of thermal discontinuity given by Eq. (5.32).

Proof: The speed in the liquid water region v

is written as:

+ C

(

− T

−

)

(

− T

−

)

−h

−

)

− H

−

−h

−

)

Using v

CWE S

given by (5.132) and the relationship

− h

−

) >

− h

−

),it

follows that v

CWE S

> v

. 

If this shock exists, it would not satisfy the Oleinik condition for entropy [53].

Therefore we conclude that instead of a shock there is a br between the

(− ) and (+)

state. The solution is constructed using results from Sections 5.6.1 and 5.6.2. Since

= 0, from Sec. 5.6.2 we obtain:

138

Proposition 5.6.9. The saturation s

††

in (5.80) is larger than

s (deﬁned in (5.62)),

thus there is a rarefaction from

††

, T

) to (

s, T

Proof: For s

= 1, we obtain from Eq. (5.39) and (5.58) that v

WES

and v

satisfy:

WES

−h

−

)

−

−h

−

)

+ C

−h

)

−

−h

)

;

we can show as in Proposition 5.6.8 for s

= 1 that

> v

WES

. (5.134)

The inequality Eq.

(5.134) is satisﬁed for s

, since that f

≥ s

We know that s

††

is given by (5.80) and for this saturation f

††

) > s

††

, so Eq.

(5.134) is satisﬁed from s

= 1 to at least s

= s

††

.As∂f

/∂s

increases for s

decreasing from s

= 1 to s

infl

, the water saturation s

††

that satisﬁes (5.80) is larger

than

s, where

s satisﬁes

(5.62), because f

≥ s

and (5.134) is satisﬁed. 

Corollary 5.6.6. The solution in the br consists of a rarefaction from (s

††

, T

) to

(

s, T

Proof: As s

††

s and from Section 5.6.1, we know that the solution consists of a

rarefaction from s

= 1 to s

s in the br, we conclude that exists a rarefaction from

††

s in this region. 

Solution. The solution consists of the waves WES  R

 CS with sequence:

=(1,T

)

WES

−−→ (s

††

, T

)

−→ (

s, T

)

−→ (0,T

)=R. (5.135)

5.6.5 Riemann Problem E

We inject pure steam at temperature T

> T

, i.e, L =(0,T

> T

); on right we

have pure water at T

< T

, i.e., the right state is R =(1,T

< T

). We use results

from Section 5.6.4 to obtain the solution.

We remark that the vaporization shock, VS, is the reverse of the condensation

shock, CS, so there could exist a hypothetical “complete condensation shock”, labelled

CC S, with speed v

CC S

. This shock would be obtained using the RH condition (5.43)-

(5.44) with the (−) state replaced by (sw = 1,T

< T

), and the (+) state replaced

−

= 0,T

−

> T

−

). The speed of such shock would be:

CC S

−

= 1,T

−

−h

−

)

− H

−

−h

−

)

. (5.136)

The following fact indicates that instead of this shock, there is always a br between

the sr and the wr.

139

Proposition 5.6.10. Complete Condensation. For any T

−

≥ T

the complete conden-

sation shock from

= 0, T

−

) to (s

= 1, T

) with speed v

CC S

(T,u

−

= 1)

satisﬁes:

CC S

−

). (5.137)

Proof: We divide the numerator and denominator of

(5.136) by (T

− T

−

CC S

= 1)=



−h

−

)



(

− T

−

)



−h

−

)



(

− T

−

)



−h

−

)



(

− T

−

)



−h

−

)



(

− T

−

)

, (5.138)

the inequality above are valid because h

> h

and c



= ∂

/∂T

< 0 for T

−

> T



Corollary 5.6.7. There is a br between the states (s

= 0,T

−

) and (s

= 1,T

Proof: From Proposition 5.6.10, there exists a br between

= 0,T

−

) and

= 1,T

), because if this region did not exist, as Eq. (5.137) is satisﬁed it would

be possible to sketch a rarefaction from

= 0,T

−

) to (s

= 0,T

) in the br.



In [6], Bruining et al. obtained the solution for injection of water and steam at

boiling temperature in a porous rock with water at temperature T

. In that work, the

water and rock enthalpies were made to vanish at temperature T

, so the formulae

for shock speeds are slightly different from formulae in present paper; however both

are equivalent since the enthalpy is deﬁned in up to a constant. In that paper, two

saturations denoted by s

†

and s

††

were found both satisfying Eq. (5.80); the choice for

††

was also made. In [6], cases I and II are correct, but there are some mistakes that

inﬂuence the solution in case III. The ﬁrst relevant mistake is the statement that the

saturation S

†

maximizes v

SC F

(Remark 11). The correct statement is:

Proposition 5.6.11. There are two saturation values that satisfy Eq. (3.7) in

[6 ] for

each ﬁxed T

< T

. The smallest S

†

minimizes v

SC F

. The other S

††

maximizes v

SC F

, but

it is irrelevant.

Moreover the solution is stable if we vary the left state, it follows that:

Corollary 5.6.8. In the limit as S

tends to zero, the wave speed v

g,w

in the br given

by Eq. (2.8) in

[1 ] converges to zero, so the solution for the Riemann problem reduces a

cooling discontinuity in the liquid water region.

From Proposition 5.6.11 and Corollary 5.6.8, we see that the Figures 3.1 and 3.4 in

[6 ] contain an error; we correct them in Figure 5.14.

140

Figure 5.14: a)-Left: Corrected schematic bifurcation near S

∗

(for ﬁxed u

, T

versus

. The shock speed SCF is minimum at S

†

; the shock speed v

g,w

tends to zero when

tends to 1. b)-Right: The corrected structure of steam-water zone below solid

curve marked by v

, v

SC F

, v

SC F

given in Eqs (2.7), (2.24) with S

= S

†

, Eq. (2.24) with

†

< S

inj

< S

∗

, and Eq. (2.8) respectively. The ﬁgures are not drawn to scale.

The wave speed diagram in Figure 4.5 of

[6 ] contains an error also. The correct

diagrams are given in Figure 5.15. There are two diagrams because the saturation

shock speed and the thermal shock are different for injected saturations larger than

∗

. These diagrams are drawn out of scale for illustrative purposes, because the

characteristic speed v

can be much larger than the other wave speeds. Figs. 3.1 and

3.4 in

[6 ] contain an error also; those ﬁgures were summarized in Figure 4.5 in [ 6],

which is corrected here in Figure 5.15. The Riemann solution for the cases I and II

are correct, however the Riemann solution for case III contains an error. The mistake

is the statement that the saturation shock speed

is faster than v

SC F

in cases I and

II. The correct statement is that v

g,w

speed converges to zero if the water saturation

at the left state tends to zero, as summarized in Corollary 5.6.8 above. The thermal

wave speed is drawn in Figure 5.15.a

) and in 5.15.b) we draw the saturation wave

speed.

141

Figure 5.15: a)-Left: The diagram represents the thermal wave speed; b)-Right: The

diagram represents the saturation wave speed. They correct the diagram presented

in Figure 4.5 of

[6 ]. In both, the solid curve represents the wave speeds used in the

Riemann solution. The characteristic speed v

, Eq. (2.7), is the Buckley-Leverett

characteristic speed in the br; the shock speed v

g,w

is the HISW speed, Eq. (2.8),

and represents the Buckley-Leverett shock for a water saturation s

to s

= 0; v

SC F

is the condensation shock speed between the br and the wr, Eq. (2.24); v

b,0

is the

cooling contact discontinuity speed in the liquid water region, Eq.

(2.15). Notice that

the temperature shock speed tends to v

when water saturation tends to S

= 1

(liquid water region) and the saturation shock speed tends to zero, thus the solution

converges to the solution in the liquid water region.

142

The solution diagram for case III given in Figure 4.4 of [6] must be modiﬁed. The

strength of saturation shock tends to zero when the injection saturation tends to 1,

while the speed of cooling discontinuity does not change. In Figure 5.16 we show the

schematic solution for case III for three different injection saturations.

Figure 5.16: Steam injection for three high water saturation values. We get a constant

state upstream, the Buckley-Leverett saturation shock and the cooling discontinuity.

These waves have distinct speeds; notice that v

g,w

tends to zero when the injection

saturation tends to 1, while the speed of cooling discontinuity does not change.

To complete the Riemann solution, we obtain a relationship between s



and s

†

Proposition 5.6.12. The water saturation s



, deﬁned in (5.119), obtained in the br by

a shock from

(0,T

−

) to (s



, T

) satisﬁes the following inequality:



< s

†

so from



, T

) the solution continues as a rarefaction to (s

†

, T

Proof: We analyze the relationship between the shock speeds v

and v

SC F

.It

is possible show that v

SC F

is faster than v

. Since (T

−



) belongs to CSS so v

satisﬁes:

∂f



, T

)

∂s

143

and v

SC F

satisﬁes Eq. (5.80), we obtain that:

∂f



, T

)

∂s

∂f

†

, T

)

∂s

as ∂f

/∂s

is monotone increasing for s

< s

infl

, we obtain the result. 

Solution. We obtain the Riemann solution using the results in [6], [39] and

(5.131), (5.126).

For T

T. As in

(5.131), there is a shock from L =(0,T

−

) to (s



, T

)

where s



= s



) is given by (5.119) and s

†

satisﬁes Eq. (5.80). The waves are

 R

 SCF with sequence:

=(0,T

)

−→ (s



, T

)

−→ (s

†

, T

)

SC F

−−→ (s

, T

)=R. (5.139)

For T

T. The waves R

 CS  R

 SCF with sequence:

=(0,T

,u)

−→ (0,

T,u)

−→ (s

†

, T

)

−→ (s

†

, T

)

SC F

−−→ (s, T < T

)=R,

(5.140)

where

(

T,s

†

) is the SHB point.

144

CHAPTER 6

Riemann solution for problem of Chapter 4 for V

in III

and IV.

Here we utilize results of Chapter 4 and Chapter 5. First we can identify subre-

gions

and L

in III



IV. The only way to reach spl region from III and IV is by a

Buckley-Leverret shock, see Section 4.9.3. Notice that the boundary between III and

IV is the coincidence curve

, in which

and

are given by Eqs. (4.115) and

(4.117). Moreover, there exists a coincidence curve between the evaporation wave and

the hot isothermal steam water, HISW, with speed v

g,w

and given by Eq. (5.28);we

denote this coincidence curve by

BLE

, see Fig. 6.1.

In next Riemann solutions, s represents the water saturation.

6.1 V

in L

In this region, HISW is slower than the evaporation wave, so we can reach the spl

directly by a HISW. It is easy to prove that:

Lemma 6.1.1. The following inequalities are valid for

, T) in L

g,w

, T) < v

WES

= 1,T; s

††

) < v

, (6.1)

where s

††

is obtained from Eq. (5.80).

The Riemann solution consists of the waves S

→ WES  R

 CS with se-

quence:

=(s

, T

)

−→ (1,T

)

WES

−−→ (s

††

, T

)

−→ (

s, T

)

−→ (0,T

)=R.

(6.2)

145

6.2 V

in L

In this region, HISW is faster than the evaporation wave. From a state in

, we reach

BLE

using a condensation rarefaction. From this state in the coincidence curve, we

can reach the spl directly by a HISW.

The Riemann solution consists of the waves R

 S

→ WES  R

 CS with

sequence:

=(s

, T

)

−→ (s, T,u)

−→ (1,T,u)

WES

−−→ (s

††

, T

)

−→ (

s, T

)

−→ (0,T

)=R,

(6.3)

where

(s, T) are the states in the coincidence curve C

BLE

obtained from the evapora-

tion rarefaction wave from

, T

) and u is the respective speed.

BLE

Figure 6.1: Subregions L

and L

. C

BLE

is the coincidence curve v

g,w

. For states

, we have that v

g,w

; for states in L

, we have that v

g,w

146

Figure 6.2: Riemann solutions in phase space, omitting the surface shown in Fig. 4.1.

a) Left: Solution

(6.2) for V

∈L

, Sec. 6.1. b) Right: Solution (6.3) for V

∈L

Sec. 6.2. The numbers 1 and 2 indicate intermediate states V in wave sequence. The

waves after spl situation are not represented in this ﬁgure, but can be obtained in the

previous Chapter.

147

CHAPTER 7

Summary and Conclusions

We have described a global formalism for balance laws of form (2.1). We show a

systematic theory for the Riemann solution for general Riemann problems for a wide

class of balance equations with phase changes.

We have also obtained the solution of the Riemann problem for the injection of a

mixture nitrogen/steam/water into a porous rock ﬁlled with steam above boiling tem-

perature. We show a systematic theory for the Riemann solution for a 3

× 3 balance

system. The set of solutions depends L

continuously on the Riemann data.

We have also described completely all possible solutions of the Riemann problem

for the injection of a mixture of steam and water in several proportions and tempera-

ture into a core ﬁlled with a different mixture of steam and water in all proportions,

(of course, the temperature must be lower than the thermodynamical critical tem-

perature of water). The set of solutions depends L

continuously on the Riemann

data. We found several types of shock between regions and systematized a scheme to

ﬁnd the solution from these shocks. A new type of shock, the evaporation shock, was

identiﬁed.

As a future work we would like obtain the Riemann solution for the problem de-

scribed in Chapter 3. We are interested also in considering the gravitational and

capillarity pressure effects and the effects of pressure changes in terms of physical

properties.

148

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EWMARK, R. L., AND AINES,R.D.,Dumping pump and treat: rapid cleanups

using thermal technology, AICHE 1997 Spring Meeting, Houston, TX, March 10-

12, 1997.

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LEINIK,O.,Discontinuous Solutions of Nonlinear Differential Equations, Usp.

Mat. Nauk. (N. S.), Vol. 12, pp 3-73, 1957. English transl. in Amer. Math. Soc.

Transl. Ser. 2, 26, 95-172.

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OPE, G.A., CAREY,G.F.,AND SEPEHRNOORI,K.,Isothermal, Multiphase,

Multicomponent Fluid-Flow in Permeable Media, Part II: Numerical Techniques

and Solution, In Situ J., 8 (1), 41-97, 1984.

[55] P

RATS,M.,Thermal Recovery, Society of Petroleum Engineers, Henry L. Do-

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ICHARDSON,C.N.AND TSYPKIN,G.G.,Nonlinear effects in water-vapor ex-

traction from a geothermal reservoir,Journal of Engineering Physics and Thermo-

physics, Vol. 77, No. 2, 2004.

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CHECTER,S.,PLOHR,B.AND MARCHESIN,D., Classiﬁcation of Codimension-

One Riemann Solutions , Journal of Dynamics and Differential Equations, Vol. 13,

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ERRE,D.,Systems of Conservation Laws, 2 volums, Cambridge University

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MOLLER,J.,Shock Waves and Reaction-Diffusion Equations, Springer-Verlag,

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OUZA,A.J.,Stability of Singular Fundamental Solutions Under Perturbations,

Matem. Aplic. Comput., 1992, 11, 73-115.

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OUZA,A.J.,AKKUTLU,I.Y.,AND MARCHESIN,D,Wave Sequences for Solid

Fuel In-Situ Combustion in Porous Media, Computational and Applied Mathe-

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SYPKIN, G.G., AND BREVDO, L., , A phenomenological model of the increase

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SYPKIN,G.G.,Existence of the frontal regime of water-steam phase transition

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SYPKIN,G.G.AND CALORE,C.,Role of capillary forces in vapour extraction

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SYPKIN,G.G.AND WOODS,A.W.,Vapour extraction from a water-saturated

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SYPKIN,G.G.AND WOODS,A.W.,Precipitate formation in a porous rock

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YLEN,G.J.,Thermodynamics, John Wiley and Sons, 1959.

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ENDROFF,B.,The Riemann problems for materials with non-convex equations

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153

154

APPENDIX A

Physical quantities; symbols and values for the Nitrogen

Problem

Table 2, Summary of physical input parameters and variables

Physical quantity Symbol Value Unit

Water, steam fractional functions f

, f

Eq. (3.15) . [m

]

Porous rock permeability k 1.0 × 10

−12

. [m

]

Water, steam relative permeabilities k

, k

Eq. (A.10). [m

]

Water, steam end point permeabilities k



, k



Eq. (A.10. a), (A.10.b). [m

]

Pressure p

1.0135×10

. [Pa]

Water Saturation Pressure p

sat

Eq. (A.7). [Pa]

Water, steam phase velocity u

, u

Eq. (3.14) . [m

/(m

s)]

Total Darcy velocity u u

+ u

, Eq. (3.16) . [m

/(m

s)]

Effective rock heat capacity C

2.029×10

. [J/(m

K)]

Steam and nitrogen enthalpies h

, h

Eqs. (A.3), (A.4). [J/m

]

Water enthalpy h

Eq. (A.2). [J/m

]

Rock enthalpy H

Eq. (A.1). [J/m

]

Water, steam saturations s

, s

Dependent variables. [m

]

Connate water saturation s

0.15. [m

]

Temperature T Dependent variable. [K]

Water, steam thermal conductivity

0.652 , 0.0208. [W/(mK)]

Rock, composite thermal conductivity

1.83 , Eq. (4.9) . [W/(mK)]

Water, steam viscosity

Eq. (A.5) , Eq. (A.6) . [Pa s]

Steam and nitrogen densities

Eq. (A.8. a), (A.8.b). [kg/m

]

Constant water density

998.2. [kg/m

]

Steam and nitrogen gas composition

Dependent variables. [−]

Universal gas constant R 8.31 [J/mol /K ]

Nitrogen and water molar masses M

, M

0.28, 0.18 [kg/mol ]

Rock porosity

0.38. [m

]

155

A.1 Temperature dependent properties of steam and

water

The rock enthalpy H

and

are given by:

=(1 −

(T −

) and

= H

. (A.1)

The water enthalpy per mass unit h

is:

= C

. (A.2)

The steam enthalpy h

[J/kg] as a function of temperature is approximated by

(

)

= −

2.20269×10

+ 3.65317 × 10

T − 2.25837×10

+ 7.3742 T

−1.33437 × 10

−2

+ 1.26913 × 10

−5

−4.9688 × 10

−9

−h

. (A.3)

The nitrogen enthalpy h

[J/kg] as a function of temperature is approximated by

(

)

975.0T + 0.0935 T

−0.476 × 10

−7

−h

. (A.4)

These constants enthalpies

and h

are chosen such that h

(

)

, h

(

)

vanish at

a reference temperature

T = 293K.

The temperature dependent liquid water viscosity

[Pas] is approximated by

= −0.0123274+

27.1038

−

23527.5

1.01425×10

−

2.17342×10

1.86935×10

(A.5)

We assume that that the viscosity of the gas is independent of the composition.

= 1.8264×10

−5





0.6

. (A.6)

The water saturation pressure as a function of temperature is given as

sat

= 10

(− 175.776 + 2.29272T −0.0113953 T

+ 0.000026278 T

−0.0000000273726T

+ 1.13816 × 10

−11

)

(A.7)

The corresponding concentrations

are calculated with the ideal gas law:

sat



− p

sat



, (A.8)

The pure phase densities are:

(T)=

, (A.9)

where the gas constant R

= 8.31[J/mol /K], M

and M

are the the nitrogen and

water molar mass.

For simplicity the liquid water density is assumed to be constant at 998.2.kg

156

A.2 Constitutive relations

The relative permeability functions k

and k

are considered to be power functions

of their respective saturations, i.e.







−s

1−s

−s









−s

1−s

−s



for s

≤ s

≤ 1,

for s

< s

(A.10)

For the computations we take n

= 2 and n

= 2. The end point permeabilities k



are 0.5 and 0.95 respectively. The connate water saturation s

is given in the table.

157

APPENDIX B

Physical quantities; symbols and values for the steam

injection

B.1 Temperature dependent properties of steam and

water

We use reference [1] to obtain all the temperature dependent properties below. The

water and steam densities used to obtain the enthalpies are deﬁned at the bottom.

The steam enthalpy h

[J/kg] as a function of temperature is approximated by

= −2.20269 ×10

+ 3.65317 × 10

T − 2.25837×10

+ 7.3742 T

−1.33437×10

−2

+ 1.26913 × 10

−5

−4.9688 × 10

−9

. (B.1)

We also use the temperature dependent steam viscosity

= −5.46807 ×10

−4

+ 6.89490 × 10

−6

T − 3.39999×10

−8

+ 8.29842 × 10

−11

−9.97060×10

−14

+ 4.71914 × 10

−17

. (B.2)

The temperature dependent water viscosity

is approximated by

= −0.0123274+

27.1038

−

23527.5

1.01425×10

−

2.17342×10

1.86935×10

(B.3)

We assume that the steam is a ideal gas, so the steam density is a function of temper-

ature:

(T)=p

, (B.4)

where M

is the water molecular mass [Kg/m

], p is the pressure atmospheric [Pa]

and R=8.31 [J/mol K]. The quantity pM

/R is a constant.

158

The liquid water density is constant, and the value is 998.2Kg/m

We deﬁne

and the water enthalpy per mass unit h

respectively as:

(T)=(1 −

T and h

= C

. (B.5)

B.2 Constitutive relations

The relative permeability functions k

and k

are considered to be power functions

of their respective saturations i.e.



1 −s



and k













−s

1−s



for s

≥ s

0 for 0

≤ s

. (B.6)

For the computations we take n

= 4, = n

= 2. The connate water saturation s

given in Table 2 below.

Table 2, Summary of physical input parameters and variables

Physical quantity Symbol Value Unit

Water, steam fractional functions f

, f

Eq. (5.6) . [m

]

Porous rock permeability k 1.0 × 10

−12

. [m

]

Water, steam relative permeabilities k

Eq. (B.6) . [m

]

Pressure p 1.0135×10

. [Pa]

Mass condensation rate q Eqs (5.1)-(5.2). [kg /(m

s)]

Water, steam phase velocity u

Eq. (5.5) . [m

/(m

s)]

Total Darcy velocity u u

+ u

,Eq(5.7). [m

/(m

s)]

Water and rock heat capacity C

, C

4.22×10

, 2.029 × 10

. [J/(m

K)]

Steam and water enthalpies h

, h

Eqs. (B.1), (B.5.b). [J/m

]

Rock enthalpy H

(1 −

T. [J/m

]

Water, steam saturations s

Dependent variables. [m

]

Connate water saturation s

0.15. [m

]

Temperature T Dependent variable. [K]

Boiling point of water–steam T

 373.15 . [K]

Water, steam thermal conductivity

0.652, 0.0208. [W/(mK)]

Rock, composite thermal conductivity

1.83 . [W/(mK)]

Water, steam viscosity

Eqs. (B.3) , (B.2). [Pa s]

Water, steam densities

998.2, Eq. (B.4) . [kg/m

]

Rock porosity (constant)

0.38. [m

]

B.3 Approximation for T + M

(T)/C

in the sr

As the temperature in the sr changes only from T

= 373,15K to around T = 400K,

from the expression for c

we can approximate T +(M

by a linear expression

159

aT + b (or, more generally, the by f(T)), where a e b are given in (B.8) .

(T)=aT + b. (B.7)

where

400

−C

400− T

and b =

400C

− T

400

400− T

, (B.8)

and

400

= 400 +

(400)

= T

)

. (B.9)

370 375 380 385 390 395 40

0.5

1.5

2.5

x 10

−6

Normalizade error between the function f(T) and the function 1+(M

′

)/C

Temperature

Normalizade error

Solving (B.9) and (B.8) to ﬁnd a and b we have

∼

0.99952824228873 and b

∼

0.28502504243858. (B.10)

Solving

(5.18) with linear denominator aT + b (see Appendix B) given by (B.7),we

have

= u

√

aT + b, (B.11)

where u

is the speed at the beginning of the rarefaction wave.

We use here

(B.7) in

and



. After manipulations, we have

∼

(T)

(aT + b)



∼

(1,

aT + b

) and c

(T)=

((

a − 1)T + b)

, (B.12)

160

where a e b are given in Eq. (B.7).

Finally, from

and c

(T) given in (B.12) we have

∼

(a −1)T + b

(aT + b)

. (B.13)

So from

(B.12) and (B.13), (B.16) can be written as

370 375 380 385 390 395 40

x 10

−4

approximated graphic

Temperature

370 375 380 385 390 395 40

−2

x 10

−3

Relative

approximated and real difference graphic

Temperature

Relative error

∇



(a −1)T

(aT + b)

, (B.14)

Thus

∇



is negative for temperatures from 373.15K to 400K, so along rarefac-

tion curves deﬁned by

(5.17.a) there are only decreasing temperatures; physically

this means that we are injecting steam at low temperature into the steam at higher

temperatures.

Thus the inﬂection locus in the plane

(T,u) consists of two parts

= 0 and T = 0, (B.15)

which are physically irrelevant.

B.3.1 Rarefaction wave behavior

Lemma B.3.1. If T decreases, so u decreases on rarefaction curve in direction where

the rarefaction exists.

Proof: As du

/dT satisﬁes (5.18. a),so

> 0,

161

because all terms on the right hand side in (5.18. a) are positive.

So, if T increases then u increases; similarly, if T decreases then u decreases. From

Appendix B, equation

(5.4.3) T decreases along the rarefaction curve, so u decreases

on the rarefaction curve.



We study the rarefaction wave behavior. In this study, we ﬁnd the region when

increases and when it decreases along integral curves, and the inﬂection locus where

is stationary. To do so, we need to calculate

∇



. (B.16)

We use

and



given by (5.17.a) and (5.17.b), respectively, in Eq. (B.16):

∇





(T)

(

T + M

(T))

, (B.17)

so in this case we also have that

decreases when the temperature increases.

162

APPENDIX C

Applications

In Section C.1 we show that the formalism developed to class of equations (2.2) can

be applied to a hyperbolic system of form:

∂

(V)

∂t

+ u

∂

(V)

∂x

= 0, u = const. (C.1)

as a particular case. In Section C.2, we show an application to numerical method;

in Section C.3 we ﬁnd the travelling waves independently on the variable u. We will

drop hats.

C.1 Hyperbolic waves in a physical situation

A very nice property of the formalism developed for shocks and rarefaction structures

of Eq.

(2.2) is that it can be applied to the hyperbolic system in the form (C.1),

illustrating that it is a particular system of the more general class

(2.2).

To do so, we consider the system

(C.1) with n state variables, where the cumulative

terms are G

(

, G

,···, G

)

and the ﬂux terms are F =

(

, F

,···, F

)

Assume that there is no degeneracies. The RH condition for Eq.

(C.1) is:

]=u[F

] for i = 1,2,··· ,n, (C.2)

where

]=G

) −G

−

) and [G

]=F

) − F

−

). The RH locus satisﬁes:

][ F

]=[G

][ F

] for all i, j = 1,2,··· ,n. (C.3)

] = 0 and [F

] = 0, then the system of equations in (C.3) reduces to:

][ F

]=[G

][ F

] for j = 2,3,···,n. (C.4)

163

The eigenvalues

and right eigenvectors r of Eq. (C.1) are obtained by solving ( 2.30)

with A and B given by:







∂G

∂V

∂G

∂V

···

∂G

∂V

∂G

∂V

∂G

∂V

···

∂G

∂V

∂G

∂V

∂G

∂V

···

∂G

∂V







and A







∂F

∂V

∂F

∂V

··· u

∂F

∂V

∂F

∂V

∂F

∂V

··· u

∂F

∂V

∂F

∂V

∂F

∂V

··· u

∂F

∂V







We can rewrite

(C.1) in the form (2.2). To do so, we assume that u is the secondary

variable and V are the n primary variables. The system can be written as:

∂G

(V)

∂t

∂uF(V)

∂x

= 0, (C.5)

∂u

∂x

= 0, (C.6)

where Eq.

(C.1) represents the n ﬁrst equations and (C.6) yields that u is constant in

the space. (Notice that in the classical hyperbolic equations u

≡ 1).

The matrix

M deﬁned in (2.54) for the system (C.5)-(C.6) is:







[

]

−

[

]

−

[

]

−

0 −11







. (C.7)

Assuming the same hypotheses of Corollary 2.3.1 and without loss of generality as-

suming that p

= 1 and q = 2, i.e., D

and D

are L.I., the RH locus are the solutions

det





[

]

−

[

]

−

[

]

−





= 0, (C.8)

for i

= 3,4,···,n + 1, where G

n+1

= 0 and F

n+1

= 1.

Notice that these are n

−2 restrictions. For i = n + 1 in Eq. (C.8), we need to solve:

det





[

]

−

[

]

−

0 − 11





= 0,

that yields:

][ F

]=[G

][ F

]. (C.9)

For i

= 3,4,···,n, we can write (C.8) as:

([G

−

−[G

−

)+F

([G

−

−[G

−

)+F

([G

−

−[G

−

)=0 (C.10)

164

After some manipulations we obtain:

([G

][ F

] − [G

][ F

]) + F

([G

][ F

] − [G

][ F

]) + F

([G

][ F

] − [G

][ F

]) = 0. (C.11)

Applying

(C.9) in Eq. (C.11), we obtain:

([G

][ F

] − [G

][ F

]) + F

([G

][ F

] − [G

][ F

]) = 0. (C.12)

(i) First we assume that [F

]=0. From Eq. ( C.12), it follows that

([G

][ F

] − [G

][ F

]) − F

][ F

]=0. (C.13)

] = 0 then [G

]=0. Since D

and D

are assumed to be L.I., then F

= 0 and for

all i

= 3,4,···,n:

][ F

] − [G

][ F

]=0. (C.14)

(C.9) together (C.14) is the RH locus for (C.5), (C.6) and coincides with the RH

locus of

(C.1).If[F

]=0, then it follows that for i = 3,4,··· ,n:



] − F

]



]=0. (C.15)

(ii ) Let us assume that [F

]=0 for all i = 3,4,··· ,n, then (C.4) is satisﬁed. If

] − F

]=0, (C.9) we know that F

−

] − F

−

]=0, implying that D

and

are linearly dependent and contradicting the hypotheses.

Finally assume that

] = 0; substituting [G

] in (C.12) by [G

][ F

]/[F

] it follows

that:



]

[

]

[

] − [G

][ F

]



+ F

(

[

][ F

] − [G

][ F

]

)

= 0,

]

(

[

][ F

] − [G

][ F

]

)

+ F

(

[

][ F

] − [G

][ F

]

)

= 0,

(

[

][ F

] − [G

][ F

]

)



][ F

] − F

]



= 0. (C.16)

][ F

] − F

] = 0, then for all i = 3,4,··· ,n we obtain:

][G

]=[G

][ F

that is the RH locus. On the other hand, if

][ F

] − F

]=0, then:

−

− F

−

= 0 (C.17)

Since

and D

are assumed to be LI, then:

det



[

]

−

[

]

−



= 0 or det



[

]

−

[

]

−



= 0 or det



−

−F

−

−F



= 0.

(C.18)

From

(C.17) and (C.9), we see that (C.18) cannot be satisﬁed. Thus we conclude that

the RH locus of

(C.1) coincides with the RH locus of the system (C.5), (C.6), deﬁned

by Eq.

(C.8).

We have just proved the following Proposition:

165

Proposition C.1.1. Under the same hypotheses of Corollary 2.3.1, the RH locus of

(C.1) deﬁned by Eq. ( C.4) coincides with the RH locus of the system ( C.5), (C.6).

Proposition C.1.2. The eigenvalues, the right and the left eigenvectors of the system

(C.5), (C.6) are

, r =(

r,0

) and  =(, 

n+1

), where

is the eigenvalue and

and

 are the right and left eigenvectors of the system (C.1). The coordinate 

n+1

determined by the system

(C.5), (C.6).

Proof: The eigenvalues of the system

(C.5), (C.6) are the solutions of:

det







∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V

00··· 01







= 0, (C.19)

which reduces to:

det







∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

∂F

∂V

−

∂G

∂V

··· u

∂F

∂V

−

∂G

∂V







= 0, (C.20)

which yields the eigenvalues of the system

(C.1). The right and left eigenvectors

satisfy

(C.19), with r =(

r,0

) and  =(

,

n+1

), where

is the eigenvalue and

r and



are the right and left eigenvectors of the system (C.1) and



n+1



+ ···+



where



for i = 1,2,··· ,n are the coordinates of left eigenvector

. 

From Propositions C.1.1 and C.1.2 we obtain:

Corollary C.1.1. The wave structures for

(C.1) coincide with the wave structures for

(C.5), (C.6) .

C.2 Numerical Method

We propose a numerical method to solve Eq. (2.2) for all contiguous physical situa-

tions. Because the speed u does not appear in G

(V), there is an inﬁnite speed mode

associated to u. Therefore the usual explicit ﬁnite difference numerical methods for

solving hyperbolic equations cannot be applied to Eq. (2.2

). Bruining et. al. have

applied In

[40], we have applied a similar method to the Riemann problem of nitrogen

and steam injection.

166

C.2.1 General Formulation

We represent the spatial position of any quantity by k and the time by symbol n.We

denote the time interval by ∆t and the spatial interval by ∆x.

For the numerical approximation, we use

k,n

= V(k∆x, n∆t) and u

k,n

= u(k∆x , n∆t), k ∈ Z, n ∈ N,

here k∆x and n∆t are, respectively, the position and the time where V and u are

evaluated.

We denote the variables to be found at time n

+ 1 and position k by

= V(k∆x, (n + 1)∆t) and u = u(k∆x,(n + 1)∆t). (C.21)

For simplicity, we abbreviate the known quantities

k,n

)=G

k,n

, F(V

k,n

)=F

k,n

, G = G(V) and F = F(V). (C.22)

C.2.2 Numerical Method for the System

Using the notation of Section C.2.1, we propose an implicit method to approximate

Eq. ( 2.2) given by the following difference scheme:

−G

k,n

∆t

= −

uF + u

k−1,n+1

∆x

. (C.23)

where u and V are the unknowns deﬁned in

(C.21). These equations are utilized to

ﬁnd the unknown V and u at time n

+ 1 and position k.

Proposition C.2.1. If there exists a j such that F

(V) = 0 for each V in the domain,

the variables V can be obtained from the implicit scheme

(C.23) without calculating u.

Proof: We introduce the new variable

u given by:

u∆t

∆x

. (C.24)

So the system

(C.23) can be rewritten as:

(V) − G

k,n

= −

k−1,n+1

. (C.25)

Since by hypothesis there exists a j such that F

(V) = 0, we can obtain u as function

of V as:

k,n

−G

(V)+

k−1,n+1

. (C.26)

Notice that the equation C.26 is valid for some time n. Substituting

(C.26) in the

remaining equations of system

(C.25) we obtain for all i = j:



(V) − G

k,n





k,n

−G

(V)



k−1,n+1

. (C.27)

167

Notice that we know u

0,n

for some time n, because we know the speed of injection

point. From Eq.

(C.26) we can obtain u

k−1,n+1

as function of u

0,n+1

and V. After some

calculations we obtain:

k−1,n+1

∑

k−1

m,n

−G

m,n+ 1

(V)+

0,n+1

k−1,n+1

. (C.28)

Substituting

(C.28) in (C.27), we obtain for each i = j:



(V) − G

k,n



k−1,n+1

−



k,n

− G

(V)



k−1,n+1

−



k−1

∑

m=1

m,n

−G

m,n+ 1

(V)



k−1,n+1

−

0,n+1

k−1,n+1

= 0. (C.29)



Remark C.2.1. Since the matrix ∂G/∂W is singular, it is not possible to use Newton

method to obtain

(V,u) in Eq. ( C.23). An advantage in utilizing the scheme (C.29) is

that u does not appear in this scheme, so we can linearize the reduced system or solve

it by a non-linear method, such as the Newton method.

C.3 Travelling waves and viscous proﬁles.

The system of equations (2.1) generically is obtained by simpliﬁcations from more

complex systems with diffusive terms of form:

∂

∂t

G(V)+

∂

∂x

F(V)=Q(V)+

∂

∂x

∂V

∂x

, (C.30)

The m

×m + 1 matrix B := B(V) is called viscosity matrix; it can represent molecular

diffusion, capillarity effects, heat conduction and other physical phenomena.

The viscosity proﬁle criterion is widely used for admissibility of physical shocks.

As a future work we will utilize this method to solve more complicated models where

elliptic regions appear, see

[24, 25, 26] and [61].

The travelling waves are solutions of

(C.30) that depend only on the parameter

= x − v

t, where v

is the shock speed to be found. We rewrite this similarity

equation as:

W = W(x −v

t), (C.31)

where we recall that

W =(V, u).

For the analysis of travelling waves and viscosity proﬁle, we deﬁne the cumulative

condensation distribution

Q(x, t) and cumulative condensation Q

(t) by

Q(x, t)=



−∞

Q(x



,t)dx



, Q

(t)=

+∞



−∞

Q(x



,t)dx



. (C.32)

168

From Eq. (C.32), we can write Q = ∂Q(x, t)/∂x. We also deﬁne Q

−

(t)=Q (x

−

,t) and

(t)=Q (x

,t), where x

−

and x

are the points immediately on the left and right

of the transition between regions.

We can rewrite Eq.

(C.30) using the parameter

and we obtain:

−v

d(u F)

(C.33)

Integrating

(C.33) em

and setting that the limits satisfy:

lim

−→−∞

(V(

),u(

),Q(V(

))) = (V

), lim

−→−∞

= 0, (C.34)

lim

−→+∞

(V(

),u(

),Q(V(

))) = (V

), lim

−→+∞

= 0. (C.35)

We obtain the shock speed with

[Q]=Q

−Q

−v

(G(V

) −G(V

)) + u

F(V

) −u

F(V

)=[Q] , (C.36)

and the ordinary system of equations that determines the orbit of travelling wave

with

= G( V

) and F

= F(V

d(V(

))

= v

−G(V(

))) + u(

)F(V(

)) −u

−(Q(V(

)) −Q

), (C.37)

= Q. (C.38)

Lemma C.3.1. Applying E

∈Edeﬁned in (2.17), (C.36) reduces to the RH condition

(2.50).

Lemma C.3.2. If there exists an i such that

(V) = 0 for each V (see Remark 2.2.3)

along the orbit of the system

(C.37), then in the variables V this orbit does not depend

explicitly on u.

Proof: The i-th line of the system

(C.37) is:

d(V(

))

= v

−G

(V(

))) + u(

(V(

)) −u

−(Q

(V(

)) −Q

), (C.39)

where

represents the i-th line of the matrix B. Assume that F

= 0 for some i,we

can obtain the Darcy speed as:

(

−B

d(V(

))/d

−v

−G

(V(

))) −(Q

(V(

)) −Q

)

(V(

))

, (C.40)

169

After same calculations, for ﬁxed i and k = 1,2,···,(i −1),(i + 1), ···,m, we can write

(C.37) as:

(V(

)) −

(V(

)) = u



(V(

)) −F

(V(

))





−G

(V(

))F

(V(

))



−



−G

(V(

))F

(V(

))



+(Q

(V(

)) −Q

(V(

)) −(Q

(V(

)) −Q

(V(

)).

(C.41)

Since the number of equations and variables is the same, the proposition is proved.



Corollary C.3.1. Within each physical situation it is possible to apply E ∈E. In this

case, the orbits of the system

(C.41) can be calculated only in the space of primary

variables.

170

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