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Document: R7.01.11
Models of behavior THHM

Summary:

This note introduces a family of laws of behavior THM for the saturated and unsaturated mediums. One y
described the relations allowing to calculate the hydraulic and thermal quantities, by taking account of
strong couplings between these phenomena and also with the mechanical deformations. Relations presented here
can be coupled with any law of mechanical behavior, subject making the assumption
said effective constraints of Bishop and that the mechanical law of behavior defines constants
rubber bands (useful for the coupled terms). The purely mechanical part of the laws is not presented.
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Count

matters

1 Introduction ............................................................................................................................................ 4
2 Presentation of the problem: Assumptions, Notations .............................................................................. 5
2.1 Description of the porous environment ........................................................................................................... 5
2.2 Notations .......................................................................................................................................... 5
2.2.1 Descriptive variables of the medium ............................................................................................. 5
2.2.1.1 Geometrical variables ............................................................................................. 6
2.2.1.2 Variables of thermodynamic state ............................................................................ 6
2.2.1.3 Descriptive fields of the medium ..................................................................................... 7
2.2.2 Derivative particulate ............................................................................................................. 7
2.2.3 Sizes ............................................................................................................................... 7
2.2.3.1 Sizes characteristic of the heterogeneous medium ..................................................... 8
2.2.3.2 Mechanical magnitudes .............................................................................................. 8
2.2.3.3 Hydraulic sizes ............................................................................................. 9
2.2.3.4 Thermal quantities ............................................................................................. 10
2.2.4 External data ............................................................................................................ 10
3 constitutive Equations ........................................................................................................................ 10
3.1 Conservation equations ............................................................................................................ 10
3.1.1 Balance mechanical ............................................................................................................. 10
3.1.2 Conservation of the fluid masses ........................................................................................ 11
3.1.3 Conservation of energy: thermal equation ................................................................... 11
3.2 Equations of behavior .......................................................................................................... 11
3.2.1 Evolution of porosity ........................................................................................................ 11
3.2.2 Evolution of the contributions of fluid mass ................................................................................ 11
3.2.3 Laws of behavior of the fluids ........................................................................................ 13
3.2.3.1 Liquid ...................................................................................................................... 13
3.2.3.2 Gas ........................................................................................................................... 13
3.2.4 Evolution of the enthali ...................................................................................................... 13
3.2.4.1 Liquid enthalpy ....................................................................................................... 13
3.2.4.2 Enthalpy of the gases ..................................................................................................... 13
3.2.4.3 Contribution of heat except fluids ................................................................................. 14
3.2.5 Laws of diffusion .................................................................................................................... 14
3.2.5.1 Diffusion of heat .............................................................................................. 14
3.2.5.2 Diffusion of the fluids ................................................................................................. 15
3.2.6 Water-steam balance ............................................................................................................ 17
3.2.7 Balance air dissolved dryness-air ................................................................................................. 18
3.2.8 The mechanical behavior ............................................................................................... 18
3.2.9 The isotherm of sorption ........................................................................................................ 19
3.2.10
Summary of the characteristics of material and the user data .............. 19
3.3 The state of reference and the initial state ................................................................................................... 21
3.4 Nodal unknown factors, initial values and values of reference ........................................................ 21
3.5 Effective constraints and total constraints. Boundary conditions of constraint ...................... 22
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3.6 Some numerical values ...................................................................................................... 23
4 Calculation of the generalized constraints ................................................................................................... 23
4.1 Case without dissolved air ..................................................................................................................... 24
4.1.1 Calculation of porosity and it it density of the fluid ..................................................... 24
4.1.2 Calculation of the dilation coefficients ..................................................................................... 25
4.1.3 Calculation of the fluid enthali ............................................................................................... 25
4.1.4 Air and steam pressures ................................................................................................ 25
4.1.5 Calculation of the mass contributions ............................................................................................. 26
4.1.6 Calculation of the heat-storage capacity and heat Q' ............................................................ 26
4.1.7 Calculation of the mechanical constraints ..................................................................................... 27
4.1.8 Calculation of hydrous and thermal flows .............................................................................. 27
4.2 Case with dissolved air ..................................................................................................................... 27
4.2.1 Calculation of porosity ............................................................................................................. 27
4.2.2 Calculation of the dilation coefficients ..................................................................................... 28
4.2.3 Calculation of density and dissolved and dry air, steam pressures .............. 28
4.2.4 Calculation of the fluid enthali ............................................................................................... 30
4.2.5 Calculation of the mass contributions ............................................................................................. 30
4.2.6 Calculation of the heat-storage capacity and heat Q' ............................................................ 31
4.2.7 Calculation of the mechanical constraints ..................................................................................... 31
4.2.8 Calculation of hydrous and thermal flows .............................................................................. 31
5 Calculation of derived from the generalized constraints .............................................................................. 32
5.1 Derived from the constraints ............................................................................................................... 32
5.2 Derived from the mass contributions .................................................................................................. 32
5.2.1 Case without dissolved air ............................................................................................................ 33
5.2.2 Case with dissolved air ............................................................................................................ 34
5.3 Derived from the enthali and heat Q' ................................................................................. 34
5.3.1 Case without dissolved air ............................................................................................................ 34
5.3.2 Case with dissolved air ............................................................................................................ 35
5.4 Derived from the heat flow ........................................................................................................... 35
5.5 Derived from hydrous flows ............................................................................................................ 35
5.5.1 Case without dissolved air ............................................................................................................ 36
5.5.2 Case with dissolved air ............................................................................................................ 37
6 Bibliography ........................................................................................................................................ 41
Appendix 1
Generalized constraints and variables intern ....................................................... 42
Appendix 2
Data material ................................................................................................... 43
Appendix 3
Derived from the pressures according to the generalized deformations ...................... 46
Appendix 4
Derived seconds from air and steam pressures dissolved according to
generalized deformations .............................................................................................................. 47
Appendix 5
Equivalence with formulations ANDRA ............................................................. 49
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1 Introduction

We introduce here a family of laws of behavior THM for the saturated and unsaturated mediums.
We describe the relations allowing to calculate the hydraulic and thermal quantities, in
taking account of strong couplings between these phenomena and also with the mechanical deformations.
The relations presented here can be coupled with any law of behavior
mechanics, subject making the assumption known as of the effective constraints of Bishop and that the law of
mechanical behavior defines constant rubber bands (useful for the coupled terms). For
this reason, the purely mechanical part of the laws is not presented here.
Modelings selected are based on the presentation of the porous environments elaborate in particular by
O. Coussy [bib1]. The relations of behavior are obtained starting from considerations
thermodynamic and with arguments of homogenization that we do not present here, and which
are entirely described in the document of P. Charles [bib2]. In the same way the general writing of
equilibrium equations and conservation is not detailed, and one returns the reader to the documents
[R5.03.01] [bib3] and [R7.01.10] [bib4], which contain definitions useful for the comprehension of
present document.

The mechanics of the porous environments gathers a very exhaustive collection of physical phenomena
concerning with the solids and the fluids. It makes the assumption of a coupling between the mechanical evolutions
solids and fluids, seen like continuous mediums, with the hydraulic evolutions, which
regulate the problems of diffusion of fluids within walls or volumes, and the evolutions
thermics.

Each component of the porous environment thus has a mechanical, hydraulic behavior and
thermics. The theory tries to gather all these physical phenomena. Phenomena
chemical (transformations of the components, reactions producing of components etc…), in the same way
that the radiological phenomena are not taken into account at this stage of the development of
Code_Aster. The mechanical, hydraulic and thermal phenomena are taken into account or not
according to the behavior called upon by the user in command STAT_NON_LINE, according to
following nomenclature:

Modeling
Phenomena taken into account
KIT_HM
Mechanics, hydraulics with an unknown pressure
KIT_HHM
Mechanics, hydraulics with two unknown pressures
KIT_THH
Thermics, hydraulics with two unknown pressures
KIT_THM
Thermics, mechanics, hydraulics with an unknown pressure
KIT_THHM
Thermics, mechanics, hydraulics with two unknown pressures

The document present describes the laws for the most general case said THHM. Simpler cases
are obtained starting from the general case by simply cancelling the quantity absent.
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2
Presentation of the problem: Assumptions, Notations

In this chapter, one mainly endeavors to present the porous environment and its
characteristics.

2.1
Description of the porous environment

The porous environment considered is a volume made up of a more or less homogeneous solid matrix,
more or less coherent (very coherent in the case of concrete, little in the case of sand). Between
solid elements, pores are found. One distinguishes the closed pores which do not exchange anything with theirs
neighbors and the connected pores in which the exchanges are numerous. When one speaks about
porosity, it is well of these connected pores about which one speaks.

Inside these pores are a certain number of fluids (one excludes solidification from these
fluids), present possibly under several phases (liquid or gas exclusively), and
presenting an interface with the other components. To simplify the problem and to take in
count the relative importance of the physical phenomena, the only interface considered is that enters
liquid and the gas, the interfaces solid fluid/being neglected.

2.2 Notations

We suppose that the pores of the solid are occupied by with more the two components, each one
coexisting in two phases to the maximum, one liquidates and the other gas one. Sizes X associated
with the phase J (j=1,2) of fluid I will be noted: X. When there are two components in addition to the solid, it
ij
are a liquid (typically water) and a gas (typically dry air), knowing that the liquid can be
present in gas form (vapor) in the gas mixture and that the air can be present under
form dissolved in water. When there is one component in addition to the solid, that can be one
liquid or a gas.
The porous environment at the current moment is noted, its border. It is noted, at the initial moment.
0
0
The medium is defined by:

· parameters (vector position X, time T),
· variables (displacements, pressures, temperature),
· intrinsic sizes (forced and mass deformations, contributions, heat,
hydraulic enthali, flows, thermics…).

The general assumptions carried out are as follows:

· assumption of small displacements,
· reversible thermodynamic evolutions (not necessarily for mechanics),
· isotropic behavior,
· the gases are perfect gases,
· mix ideal perfect gases (total pressure = nap of the partial pressures),
· balance thermodynamic between the phases of the same component.

The various notations are clarified hereafter.

2.2.1 Descriptive variables of the medium

These are the variables whose knowledge according to time and of the place make it possible to know
completely the state of the medium. These variables break up into two categories:

· geometrical variables,
· variables of thermodynamic state.
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2.2.1.1 Variables
geometrical

In all that follows, one adopts a Lagrangian representation compared to the skeleton (within the meaning of
[bib1]) and co-ordinates X = X (T
S) are those of a material point attached to the skeleton. All them
space operators of derivation are defined compared to these co-ordinates.
ux


Displacements of the skeleton are noted U (X, T) = U Y.


uz

2.2.1.2 Variables of thermodynamic state

In a general way, the following indices are used:

W for liquid water
AD for the dissolved air
have for the dry air
vp for the steam

The thermodynamic variables are:

· pressures of the components: p
,
X
, p
,
X
, p
,
X
, p
,
X
,
have (
T)
vp (
T)
AD (
T)
W (
T)
· the temperature of the medium T (X, T).

These various variables are not completely independent. Indeed, if only one is considered
component, thermodynamic balance between its phases imposes a relation between the pressure of
vapor and pressure of the liquid of this component. Finally, there is only one pressure
independent by component, just as there is only one conservation equation of the mass.
The number of independent pressures is thus equal to the number of independent components.
choice of these pressures is free (combinations of the pressures of the components) provided that them
pressures chosen, associated the temperature, form a system of independent variables.

For the case known as saturated (only one component air or water) we chose the pressure of this single
component.
For the case says unsaturated (presence of air and water), we chose like variables
independent:

· total pressure of the gas p
,
X T =p + p,
gz (
) vp have
· capillary pressure p
,
X T = p - p = p - p - p
C (
) gz
lq
gz
W
AD.
These pressures have a very strong physical interpretation, the total gas pressure for reasons
obvious, and pressure capillary, also called suction, because capillary phenomena
are very important in modeling presented here. It would have been possible also to choose
steam pressure or the percentage of relative moisture (relationship between the steam pressure and the pressure
of vapor saturated at the same temperature) physically accessible. Modeling becomes then
more complex and in any event, capillary pressure, gas pressure and percentage of relative moisture
(relationship between the steam pressure and the saturating steam pressure) are connected by the law of Kelvin.
For the particular case of the behavior “LIQU_GAZ_ATM” one says makes the assumption known as of Richards:
pores are not saturated by the liquid, but the pressure of gas is supposed to be constant and only
variable of pressure is the pressure of liquid.
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2.2.1.3 Descriptive fields of the medium

The principal unknown factors, which are also the nodal unknown factors (noted U (X, T) in this document)
are:

· 2 or 3 (according to the dimension of space) displacements U (,
X T), U (,
X T), U (,
X T) for
X
y
Z
modelings KIT_HM, KIT_HHM, KIT_THM, KIT_THHM,
· the temperature T (X, T) for modelings KIT_THH, KIT_THM, KIT_THHM,
· two pressures p (X, T), p (X, T) (which are p
,
X
, p
,
X
in the case studied) for
gz (
T)
C (
T)
1
2

modelings KIT_HHM, KIT_THH, KIT_THHM,
· a pressure p (X, T) (which is p
,
X
or p
,
X
according to whether the medium is saturated by one
gz (
T)
W (
T)
1
liquid or a gas) for modelings KIT_HM, KIT_THM.

2.2.2 Derived
particulate

This paragraph partly shows the paragraph “derived particulate, densities voluminal and
mass “of the document [R7.01.10]. Description that we make of the medium is Lagrangian by
report/ratio with the skeleton.
Either has an unspecified field on, or X (T
S) the punctual coordinate attached to the skeleton that
we follow in his movement and is X (T
fl
) the punctual coordinate attached to the fluid. One notes

D its
= dt has
& the temporal derivative in the movement of the skeleton:

D its
(xs (T + T has
), T + T
) - has (X (T), T)
has
S
=
= lim
dt
T
0
T

&


da
has
& is called particulate and often noted derivative. We prefer to use a notation which
dt
recall that the configuration used to locate a particle is that of the skeleton by report/ratio
to which a particle of fluid has a relative speed. For a particle of fluid location X (T is
S
)
unspecified, i.e. that the particle of fluid which occupies position X (T at the moment T is not
S
)
even as that which occupies the position xs (you) at another moment you.

2.2.3 Sizes

The equilibrium equations are:

· conservation of the momentum for mechanics,
· conservation of the masses of fluid for hydraulics,
· conservation of energy for thermics.

The writing of these equations is given in the document [R7.01.10] [bib4], which defines also what
we call in a general way a law of behavior THM and gives the definitions of
generalized constraints and deformations. This document uses these definitions. Equations
of balance utilize directly the generalized constraints.
The generalized constraints are connected to the deformations generalized by the laws of behavior.
The generalized deformations are calculated directly starting from the variables of state and theirs
temporal space gradients.
The laws of behaviors can use additional quantities, often arranged in the variables
interns. We gather here under the term of size at the same time the constraints, the deformations and
additional sizes.
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2.2.3.1 Sizes characteristic of the heterogeneous medium

· Porosity eulérienne: .
If one notes the part of the volume occupied by the vacuums in the current configuration, one
a:

=


The definition of porosity is thus that of porosity eulérienne.
· Saturation in liquid: S
lq
If one notes total volume occupied by the liquid, in the current configuration, one has by
lq
definition:

S = lq
lq


This saturation is thus finally a proportion varying between 0 and 1.
· Densities eulériennes of water, the dissolved air, the dry air, of
W
AD
have
vapor, of gas.
vp
gz
If one notes (resp
,) water masses (resp of dissolved air, dry air and of
W
AD
have
vp
vapor) contents in a volume of the skeleton in the current configuration, one has by
definition:

W = S D
AD


W
lq

=
S
D
AD
lq





have =
1

vp
1

have
- S
D
lq

=
vp
- S
D
lq


(
)

(
)

The density of the gas mixture is simply the sum of the densities
dry air and vapor:

= +
gz
have
vp

In the same way for the liquid mixture:
= +
lq
W
AD

0 are noted
0
0
0
, initial values of the densities.
W
AD
vp
have
· Lagrangian homogenized density: r.
At the moment running the mass of volume, M
M =
rd
, is given by:


.
0
0

2.2.3.2 Sizes
mechanics

1
· The tensor of the deformations U
() (X,) = (U
+T
T
U).
2
One will note V = tr ().
· The tensor of the constraints which are exerted on the porous environment: .
This tensor breaks up into a tensor of the effective constraints plus a tensor of
constraints of pressure = +
1. ”
and
are components of the constraints
p
p
generalized. This cutting is finally rather arbitrary, but corresponds all the same to
an assumption rather commonly allowed, at least for the mediums saturated with liquid.
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2.2.3.3 Sizes
hydraulics

· Mass contributions in components m, m m
,
m
,
(unit: kilogram per meter
W
AD
vp
have
cubic).
They represent the mass of fluid brought between the initial and current moments. They form part
generalized constraints.
· Hydraulic flows M, M, M, M (unit: kilogram/second/square meter).
W
AD
vp
have
One could not give very well no more precise definition of the contributions of mass and of
flow, considering that their definition is summarized to check the equilibrium equations
hydraulics:


m
m
Div M
M
W +
vp +
(W + vp) = 0
& &

éq
2.2.3.3-1

m
m
Div M
M
have +
AD +
(have + AD) = 0

& &
We nevertheless will specify the physical direction as of the these sizes, knowing that what
we write now is already a law of behavior.
Speeds of the components are measured in a fixed reference frame in space and time.
One notes v W the speed of water, vad that of the dissolved air, vvp that of the vapor, go that
dry air, and v S =
that of the skeleton.
dt
The mass contributions are defined by:

m =
+ S - S
W
W (1
V)
0
0
0
lq
W
lq
m =
+ S - S
AD
AD (1
V)
0
0
0
lq
AD
lq

éq
2.2.3.3-2
m =
+ - S -
- S
have
have (1
V) (1
lq)
0
0
have
(0
1
lq)
m =
+ - S -
- S
vp
vp (1
V) (1
lq)
0
0
vp
(0
1
lq)

Mass flows are defined by:

M = S v - v
W
W
L (W
S)
M
= S v - v
AD
AD
L (AD
S)
éq
2.2.3.3-3
M = 1 - S v - v
have
have
(
L) (have
S)
M = 1 - S v - v
vp
vp
(
L) (vp
S)
The mass contributions make it possible to define the total density seen compared to
configuration of reference: R = R
, where R indicates the density
0 + m
+ m + m + m
W
AD
vp
have
0
homogenized at the initial moment.
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Other intermediate hydraulic sizes are introduced:

p
· concentration of the vapor in gas
vp
C =
,
vp
pgz
M
M
M
· gas flow:
gz = (1 - C
+ C
. This equation specifies that the speed of
vp)
have
vp
vp



gz
have
vp
gas is obtained by making an average (balanced sum) speeds of different
gas according to their concentration,
· the steam pressure p.
vp

2.2.3.4 Sizes
thermics

· not convectée heat Q (see further) (unit: Joule),
· mass enthali of the components
m
H
(m
m
m
m
H, H, H, H) (unit
:
ij
W
AD
vp
have
Joule/Kelvin/kilogram),
· heat flow: Q (unit: J/S/square meter).

All these sizes belong to the constraints generalized within the meaning of the document [R7.01.10]
[bib4].

2.2.4 Data
external

· the mass force m
F (in practice gravity),
· heat sources,
· boundary conditions relating either to variables imposed, or on imposed flows.

3 Equations
constitutive

3.1
Conservation equations

It is here only about one recall, the way of establishing them is presented in [R7.01.10] [bib4].

3.1.1 Balance
mechanics

By noting the tensor of the total mechanical constraints and R density homogenized of
medium, mechanical balance is written:
(
Div) + m
RF = 0 éq
3.1.1-1
We point out that R is connected to the variations of fluid mass by the relation:

R = R
éq
3.1.1-2
0 + m
+ m + m + m
W
AD
vp
have
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3.1.2 Conservation of the fluid masses
D its
For the fluid the derivative has = dt
& in fact a derivative eulérienne is and the equations which we write
for the fluid comprise terms of transport, even if they can be hidden by the choice of
unknown factors. The conservation equations of the fluid masses are written then:


m
m
Div M
M
W +
vp +
(W + vp) = 0
& &

éq
3.1.2-1

m
m
Div M
M
have +
AD +
(have + AD) = 0

& &
3.1.3 Conservation of energy: thermal equation
m
H m
H m
H m
H m
Q' Div H M
H M
H M
H M
Div Q
W
W +
m
AD
AD +
m
vp
vp +
m
have
have +
+
&
(MW W + mad AD + mvp vp + farmhouse have) + () =
(&
&
&
&

M
M
M
MR. F
W +
AD +
vp +
) m
have
+
éq 3.1.3-1

3.2
Equations of behavior

3.2.1 Evolution of porosity
dp
S dp
D = (B -)
gz -
lq
C

D
3 dT
V - 0
+



Ks
éq
3.2.1-1

In this equation, one sees appearing the coefficients B and K. B is the coefficient of Biot and K is
S
S
the module of compressibility of the solid matter constituents. If K indicates the module of compressibility
0
“drained” of the porous environment, one with the relation:
K
B
0
=1- Ks éq
3.2.1-2

3.2.2 Evolution of the contributions of fluid mass

By using the definition of the contributions of fluid mass and while putting forward arguments purely
geometrical, one finds:

m =
+ S - S
W
W (1
V)
0
0
0
lq
W
lq
m =
+ S - S
AD
AD (1
V)
0
0
0
lq
AD
lq

éq
3.2.2-1
m =
+ - S -
- S
have
have (1
V) (1
lq)
0
0
have
(0
1
lq)
m =
+ - S -
- S
vp
vp (1
V) (1
lq)
0
0
vp
(0
1
lq)
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As example, we show the first relation in the case saturated S
(with =).
lq = 1
lq
W
That is to say an elementary field of porous environment of volume. One notes the volume occupied by
S
solid matter constituents and the volume occupied by the liquid and gas. 0 are noted
, 0
, 0
the same ones
L
S
L
volumes in an initial state. We point out that note the variation of volume of the porous environment and us
V
let us note the voluminal variation of the solid matter constituents.
S
V

One has by definition: =
L


= - = 1 - = 0 1+
S
L
(
)
(Vs)
S


But (1 -) = 0 (1+

V) (1 -)
One deduces some:
0 (1+ 1 - = 0 1+
V) (
)
(Vs)
S

It is enough to write then
0
0
=
to obtain:
S
(0
1 -)

0 (1+ 1 - = 0 1 - 0 1+
V) (
)
(
) (Vs)
From where one deduces:
Vs (
0
1 -) = V (1 -) - (
0
-)

One uses the definition eulérienne density homogenized R (not to be confused with
the Lagrangian definition of the equation [éq 3.1.1-2]):

R = 1
S (-) + lq

and the definition of the mass contribution in liquid:

R
= (R + mlq) 0

0


One obtains:

m
S (1 -)
0
+ lq
= S (
0
1 -) 0
0
0
0
0
+ + lq
lq

that is to say still:

m
S + lq (
1+ V) 0
0
0
0
0
0
0
= S + + lq
S
lq


Using the conservation of the mass of the solid matter constituents:
0
0
= one obtains finally:
S
S
S
S

+ = 0
1
0 + m
lq (
V)
lq
lq
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3.2.3 Laws of behavior of the fluids

3.2.3.1 Liquid
D
dp
W
W
=
-
3
dT éq
3.2.3.1-1

K
W
W
W

One sees appearing the module of compressibility of water K and his module of dilation.
W
W

3.2.3.2 Gas

For the equations of reaction of gases, one takes the law of perfect gases:
pvp
R
=
T

M ol
vp
vp
éq
3.2.3.2-1
p
R
have =
T éq
3.2.3.2-2

M ol
have
have
One sees appearing the molar mass of the vapor,
ol
M, and that of the dry air,
ol
Mr.
vp
have

3.2.4 Evolution of the enthali
3.2.4.1 Enthalpy
liquid
dp
m
p
dh = C dT + 1 -
3 T

éq
3.2.4.1-1
W
W
(
) W
W
W
One sees appearing the specific heat with constant pressure of water:
p
C.
W
By replacing in this expression the pressure of the liquid by its value according to the pressure
thin cable and of the pressure of gas, one a:
-
-
dhm = 1 -
3
+

éq
3.2.4.1-2
W
(
T
W
) dp dp dp
gz
C
AD
C pdT
W
W
While noting
p
C specific heat with constant pressure of the dissolved air, one a:
AD
dhm = C p dT
éq
3.2.4.1-3
AD
AD

3.2.4.2 Enthalpy of gases
dhm = C p dT
vp
vp

éq
3.2.4.2-1
dhm = C p dT éq
3.2.4.2-2
have
have

One sees appearing the specific heat with constant pressure of the dry air
p
C and that of the vapor p
C.
have
vp
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3.2.4.3 Contribution of heat except fluids

It is the quantity Q
who represents the heat received by the system except contribution enthalpic of
fluids.

Q
= 3 K Td
m
+ 3 Tdp
0
- 3 + 3
+
0
0
V
lq
C
(m
m
gz
lq) Tdp
C dT
gz


éq 3.2.4.3-1

One sees appearing several dilation coefficients:
m
m
,
. The coefficient is a data:
0,
lq
gz
0
it corresponds at the same time to the dilation coefficient of the porous environment and to that of the solid matter constituents (which
find being inevitably equal in the theory which we present here).

m
m
, are given by the relations:
lq
gz
1
m
= 1
- +
gz
(Slq) (b)
(Slq)
0
T
3

éq
3.2.4.3-2
m
= S B -

éq
3.2.4.3-3
0 +
S

lq
lq (
)
lq
lq

One also sees appearing in [éq 3.2.4.3-1] the specific heat to constant deformation of
porous environment 0
C, which depends on the specific heat to constant constraint of the porous environment 0
C by
the relation:
0
0
2
C = C - 9TK
0
0
éq
3.2.4.3-4

0
C is given by a law of mixture:
C 0 = (1 -)
S
p
p
C + S

(C + C) + 1 - C M + C éq
3.2.4.3-5
S
lq
W
W
AD
AD
(lq) (
p
p
vp
vp
have
have)

where
S
C

represent the specific heat to constant constraint of the solid matter constituents and
mass
S
voluminal of the solid matter constituents. For the calculation of, one neglects the deformation of the solid matter constituents, one
S
thus confuses with its initial value 0
, which is calculated in fact starting from the specific mass
S
S
initial of the porous environment R by the following formula of the mixtures:
0

(0
1 -) 0
0
0
0
0
= R - (+) S - 1 - S + éq
3.2.4.3-6
0
W
AD
lq
S
(0lq) 0 (0 0
vp
have)

3.2.5 Laws of diffusion
3.2.5.1 Diffusion of heat

One takes the traditional law of Furrier:
T
Q = -
. T
éq
3.2.5.1-1

where one sees appearing the thermal coefficient of conductivity T.
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T
is a function of porosity, saturation and temperature and is given in the form of
product of three functions plus a constant:

T
T
T
T
T
=
(

). (S

).
T
() +
éq
3.2.5.1-2
S
lq
T
cte

3.2.5.2 Diffusion of the fluids

They are the laws of Darcy, to which one adds the law of Fick in the presence of vapor.
The laws of Darcy are written for gas and the liquid:

Mlq
H
= -
+ F
éq
3.2.5.2-1
lq (
m
plq
lq
)
lq
M gz
H
= (
m
- pgz + gzF) éq
3.2.5.2-2
gz
gz
where we see appearing hydraulic conductivities H
and H
for the liquid and gas
lq
gz
respectively.
M
One makes the approximation that
W
H
= -
+ F
W (
m
plq
lq
)
W


The diffusion in the gas mixture is given by the law of Fick thanks to the relation:

M
M
D
p
vp
have
vp
vp
-
= -


éq
3.2.5.2-3


C
1
(
C) p
vp
have
vp
- vp gz

where D is the coefficient of diffusion of Fick of the gas mixture (L2.T-1), one notes thereafter F such
vp
vp
that:
vp
D
vp
F =

éq
3.2.5.2-4
C
1
(- C)
vp
vp

and with
pvp
Cvp =

éq
3.2.5.2-5
pgz

One thus has:
Mvp Farmhouse
-
= - F C éq
3.2.5.2-6
vp
vp


vp
have
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Moreover, one a:
M gz = (
M
M
1 - Cvp) have
vp
+ Cvp
éq
3.2.5.2-7
gz
have
vp
and:
gz = vp + have éq
3.2.5.2-8

For the diffusion of the liquid mixture, the usual writing is as follows:
M - M = - D
éq
3.2.5.2-9
AD
W
AD
AD


where D is the coefficient of diffusion of Fick of the liquid mixture. In order to keep a writing
AD
homogeneous with that of the gas mixture one notes F thereafter such as:
AD
F = D
éq
3.2.5.2-10
AD
AD
And the concentration C corresponds here to the density of the dissolved air:
AD
C =
éq
3.2.5.2-11
AD
AD

M - M = - F C
éq
3.2.5.2-12
AD
W
AD
AD


Concerning the liquid, one admitted that liquid Darcy applies at the speed of liquid water. There is not
thus not to define mean velocity of the liquid.

M W
H
= - p + F
éq
3.2.5.2-13
lq (
m
lq
lq
)
W
and:
= +
éq
3.2.5.2-14
lq
W
AD

By combining these relations, one finds then:

M have
H
= -
+ F +

éq
3.2.5.2-15
gz (
m
p
C F
C
gz
gz
) vp vp vp
have
Mvp
H
= -
+ F - 1
(-
)


éq
3.2.5.2-16
gz (
m
p
C
F
C
gz
gz
)
vp
vp
vp
vp
M W
H
= - p + F
éq
3.2.5.2-17
lq (
m
lq
lq
)
W
H
M
= - p + F - F C éq
3.2.5.2-18
AD
AD
lq (
m
lq
lq
) AD AD
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Hydraulic conductivities H
lq and H
are not directly data and their value is
gz
known starting from the formulas:

K int
(
K rel
).
H
lq (Slq)
=

éq
3.2.5.2-19
lq
µw (T)
K int
(
K rel
).
,
H
gz (S
p
lq
gz)
=
éq
3.2.5.2-20
gz
µgz (T)
int
K is the intrinsic permeability, characteristic of the porous environment and user datum, function
unspecified of porosity;
µ is the dynamic viscosity of water, characteristic of water and user datum, function
W
unspecified of the temperature;
µ is the dynamic viscosity of gas, characteristic of gas and user datum, function
gz
unspecified of the temperature;
rel
K is the permeability relating to the liquid, characteristic of the porous environment and user datum,
lq
unspecified function of saturation in liquid;
rel
K is the permeability relating to gas, characteristic of the porous environment and user datum, function
gz
unspecified of saturation in liquid and the gas pressure.

Note:

Here definite hydraulic conductivities are not inevitably very familiar for
mechanics of grounds, which usually use for the saturated mediums the permeability K,
which is homogeneous at a speed.
H
K
The relation between K and H
is as follows: =
where G is the acceleration of gravity.
lq
lq
G
W

The coefficient of diffusion of Fick of the gas mixture vp
F is a characteristic of the porous environment,
unspecified user datum function of the steam pressure, the gas pressure, of
saturation and of the temperature which one will write like a product of function of each one of these
variables: F (P, P, T, S) = F vp (P). F gz (P). F T (T). F S (S)
vp
vp
gz
vp
vp
vp
gz
vp
vp
one will neglect the derivative by
report/ratio with the steam pressure and saturation. Same manner for the coefficient of diffusion
of Fick of the liquid medium: F (P, P, T, S) = F AD (P). F lq (P). F T (T). F S (S)
AD
AD
lq
AD
AD
AD
lq
AD
AD
, one does not take
in account that the derivative according to the temperature.

3.2.6 Balance
water-steam

This relation is essential and it results in to reduce the number of unknown factors of
pressure.
M is noted
H mass enthalpy of water, m
S its entropy and m
m
m
G = H - Ts its free enthalpy.
W
W
W
W
W
M is noted
H mass enthalpy of the vapor, m
S its entropy and m
m
m
G = H - Ts its enthalpy
vp
vp
vp
vp
vp
free.
Balance water vapor is written:

m
m
G = G éq
3.2.6-1
vp
W
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Who gives:

m
m
H - H = T S - S
éq
3.2.6-2
vp
W
(m m
vp
W)
dp
In addition, the definition of the free enthalpy teaches us that: dg =
- sdT, which, applied to

vapor and with water, compound with the relation
m
m
dg = dg and while using [éq 3.2.6-2] gives:
vp
W

dpvp
dpw
=
+ (hm
m
-

éq
3.2.6-3
vp
W) dT
H


T
vp
W

This relation can be expressed according to the capillary pressure and of the gas pressure:


dp
vp
=
-
-
+
-
éq
3.2.6-4
vp
(dp dp dp
gz
C
AD)
vp (H m
m
vp
W) dT
H

T
W
3.2.7 Balance air dissolved dryness-air

The dissolved air is defined via the constant of Henry K, who connects the molar concentration of dissolved air
H
ol
C (moles/m3) with the pressure of dry air:
AD
p
ol
have
C =

éq
3.2.7-1
AD
K H

with ol
AD
C =
éq
3.2.7-2
AD
ol
M AD

Molar mass of the dissolved air,
ol
M is logically the same one as that of the dry air
ol
Mr. For
AD
have
the dissolved air, one takes the law of perfect gas:

pad
R
=
T
éq
3.2.7-3

M ol
AD
have

The pressure of dissolved air is thus connected to that of dry air by:
p
p
have
=
RT
éq
3.2.7-4
AD
K H

3.2.8 The mechanical behavior

One will write it in differential form:
D = D +
'D 1
p éq
3.2.8-1
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By using a formulation of Bishop [bib10] extended to the unsaturated mediums one writes:

D = - B dp - S dp
éq
3.2.8-2
p
(gz lq c)

In the relation [éq 3.2.8-1] the evolution of the tensor of effective constraint is supposed
to depend that displacement of the skeleton and variables internal. Usual terms related to
thermal deformation are integrated into the calculation of the effective constraint:

of
= F (
D - IdT,
0

D)
éq
3.2.8-3
The reason of this choice is to be able to use any law of traditional thermomechanics for
calculation of the effective constraints, laws which, in more the share of the writings are in conformity with [éq 3.2.8-3].

3.2.9 The isotherm of sorption

To close the system, there remains still a relation to be written, connecting saturation and the pressures. Us
chose to consider that saturation in liquid was an unspecified function of the pressure
thin cable, that this function was a characteristic of the porous environment and provided in data by
the user.
Since the user can provide a function S very well
p refines per pieces, and being
lq (c)
S

given that the derivative of this function,
lq, plays an essential physical part, we chose
p
C
to ask the user to also provide this curve, remainder with its load to ensure itself of
coherence of the data thus specified.
It is noticed that in the approach present, one speaks about a bi-univocal relation between saturation and
capillary pressure. It is known that for the majority of the porous environments, it is not the same relation which
must be used for the paths of drying and the paths of hydration. It is one of the limits
approach present.

3.2.10 Summary of the characteristics of material and the user data

· The Young modulus E and the drained Poisson's ratio make it possible to calculate it
0
0
modulate compressibility drained of the porous environment by
0
=
E
K
,
0
(31 - 20)
K
· the coefficient of Biot B
0
= 1
allows to calculate the module of compressibility of the grains
Ks
solids K,
S
· the module of compressibility of water K,
W
· the dilation coefficient of water,
W
· the constant of perfect gases R,
· molar mass of the vapor
ol
M,
vp
· molar mass of the dry air
ol
M, (=
ol
M)
have
AD
· specific heat with constant pressure of water p
C,
W
· specific heat with constant pressure of the dissolved air p
C
AD
· specific heat with constant pressure of the dry air p
C,
have
· specific heat with constant pressure of the vapor p
C,
vp
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· the dilation coefficient of the porous environment which is also that of the solid matter constituents,
0
· specific heat with constant constraint of the solid matter constituents S
C,
· the thermal coefficient of conductivity of the solid matter constituents only, T
, unspecified function of
S
the temperature,
· the thermal coefficient of conductivity of the liquid, Tlq, unspecified function of
temperature,
· the thermal coefficient of conductivity of the dry air, Tas, unspecified function of
temperature,
· the coefficient of diffusion of fick for the gas mixture, F, unspecified function of
vp
temperature, of the gas pressure, the steam pressure and saturation
· the coefficient of diffusion of fick for the liquid mixture, F, unspecified function of
AD
temperature and of the pressure of liquid, the pressure of the dissolved air and saturation.
· The constant of Henry K function unspecified of the temperature,
H
· the intrinsic permeability, int
K, unspecified function of porosity,
· the dynamic viscosity of water, µ, unspecified function of the temperature,
W
· the dynamic viscosity of gas, µ, unspecified function of the temperature,
gz
· the permeability relating to the liquid, rel
K, unspecified function of saturation in liquid,
lq
· the permeability relating to gas, rel
K, unspecified function of saturation in liquid and of
gz
gas pressure,
· the relation capillary saturation/pressure, S p, unspecified function of the pressure
lq (c)
thin cable,
· in a general way the initial state is characterized by:
-
the initial temperature,
-
initial pressures from where one deduces initial saturation 0
S p
lq (
0
c),
-
initial specific mass of water 0
,
W
-
initial porosity 0
,
-
initial pressure of vapor 0
p from where one deduces the initial density from
vp
vapor 0
,
vp
-
initial pressure of the dry air 0
p from where one deduces the initial density from L `dry air
have
0
.
have
-
initial density of the porous environment R,
0
- density of the solid matter constituents 0
. This data is useful only for
S
modelings including of thermics, and one will have to take care that it is coherent
with the other data, through the relation
:
(0
1 -) 0
0
0
0
0
= R - (+) S - 1 - S +,
0
W
AD
lq
S
(0lq) 0 (0 0
vp
have)
-
initial enthali of water, the dissolved air, the vapor and the dry air.
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3.3
The state of reference and the initial state

The introduction of the initial conditions is very important, in particular for the enthali.
In practice, one can reason by considering that one has three states for the fluids:

· the state running,
· the state of reference: it is that of the fluids in a free state. Very often one will take for
pressures of water and air atmospheric pressure. In this state of reference, one can
to consider that the enthali are null,
· the initial state: it is important to note that, in an initial state of the porous environment, water is in a state
hygroscopic different from that of interstitial water. For the enthali of water and vapor one
will have to take:

init
init
m
p
p
p
p
W
- ref.
init
L
W
-
hw =
=
atm


W
W
init
m
H
L T

vp =
(init) =
ion
vaporisat
of

latent
heat
init
m
has = 0
init
m
had = 0

Note:

The initial vapor pressure will have to be taken in coherence with these choices. Very often, one
leaves the knowledge of an initial state of hygroscopy. The relative humidity is the report/ratio
between the steam pressure and the steam pressure saturating at the temperature considered.
One then uses the law of Kelvin which gives the pressure of the liquid according to the pressure of
vapor, of the temperature and the steam pressure
0
p
p
R
p

W -
saturating:
W =
T ln
vp

ol
sat


. This relation is valid only for
M
W
vp
p (T)
vp

isothermal evolutions. For evolutions with variation in temperature, the formula
[éq 4.1.4-1] further established gives a more complete expression but in fact incremental.
Knowing a law giving the steam pressure saturating to the temperature T, by
0

0
T -
5
.
273


2
+
7858
.
sat

31 559
.
+ 1354
.
(0T-
)
example:

p (T) 10
, and a degree of hygroscopy HR, one deduces some
vp
0
=
5
.
273
the steam pressure thanks to p (T) = HR psat (T).
vp
0
vp
0

3.4
Nodal unknown factors, initial values and values of reference

We approach here a point which is due more to choices of programming than to true aspects of
formulation. Nevertheless, we expose it because it has important practical consequences.
principal unknown factors which are also the values of the degrees of freedom, are noted:

ux



uy

{U}
U
ddl

Z


=
ddl

1
PRE


ddl
PRE2
ddl
T



Handbook of Référence
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:
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: 22/52

According to modeling, they can have different significances:


LIQU_SATU LIQU_VAPE LIQU_GAZ_ATM
GAZ
LIQU_VAPE_GAZ
PRE1
p
p
p = - p
p
p = p - p
W
W
C
W
gz
C
gz
W
PRE2




p
gz


LIQU_GAZ LIQU_AD_GAZ_VAPE
PRE1
p = p - p
p = p - p - p
C
gz
W
C
gz
W
AD
PRE2
p
p
gz
gz

One will then define the real pressures and the real temperature by:

ddl
init
p = p + p for the pressures
ddl
PRE1 and PRE2 and
init
T = T
+ T for the temperatures, where init
p and
init
T
are defined under key word THM_INIT of command DEFI_MATERIAU.
The values written by IMPR_RESU are the values of {} ddl
U
. The boundary conditions are defined
for {} ddl
U
. Key word DEPL of the key word factor ETAT_INIT of command STAT_NON_LINE defines
initial values of {} ddl
U
. Initial values of the enthali, which belong to the constraints
generalized are definite starting from key word SIGM of the key word factor ETAT_INIT of the command
STAT_NON_LINE. The real pressures and the real temperature are used in the laws of
dp
D dT
behavior, in particular laws of the type S = F p or
=
+
. Initial values of
lq
(c)
p

T
densities of the vapor and the dry air are defined starting from the initial values of the pressures
gas and of vapor (values read under key word THM_INIT of command DEFI_MATERIAU). One
notice that, for displacements, the decomposition
ddl
init
U = U + U is not made: the key word
THM_INIT of command DEFI_MATERIAU thus does not make it possible to define displacements
initial. The only way of initializing displacements is thus to give them an initial value by
key word factor ETAT_INIT of command STAT_NON_LINE.

3.5 Effective constraints and total constraints. Boundary conditions
of constraint

The partition of the constraints in constraints total and effective is written:

= +
'1
p


is the total constraint, a.c. D that which checks:
(
Div) + m
RF = 0
is the effective constraint. For the laws of effective constraints, it checks:
1
D = F (
D -
= +T
0dT,), where
(U U) and represents the internal variables.
2
is calculated according to the water pressures. The adopted writing is incremental and, if one
p
wants that the value of is coherent with the value init
p defined under the key word
p
THM_INIT, it is necessary
to initialize by the key word
p
SIGM of the key word factor ETAT_INIT of command STAT_NON_LINE.

In the files results, one finds the constraints effective under the names of components
SIXX… and under the name
p
SIP. The boundary conditions in constraints are written in constraints
total.
Handbook of Référence
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Models of behavior THHM


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:
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: 23/52

3.6
Some numerical values

We give here some reasonable values for certain coefficients. These values are not
programmed in Code_Aster, they are provided here as an indication:

For perfect gases, one retains the following values:
1
R
3144
.
8
J. -
=
K
3
-
- 1
M ol

vp = 18. 10
kg.mole
3
-
1
M ol

have =
96
.
28
10 kg.
-
mole
For the CO2, the value of the constant of Henry with 20°C is of:
3
1
K
3162
.
-
H =
Pa m mole
For liquid water, one a:
3
= 1000 kg/m
W
K = 2000 MPa
W


The thermal dilation coefficient of water is correctly approached by the formula:
5
= 52
.
9
10 - Ln (T
(
1
-
K)
W
-
)
- 4
273 - 19
.
2
10

The heat-storage capacities have as values:
S
1
-
1
C = 800
-

JKg K
p
1
-
1
Clq = 4180
-
JKg K
p
1
-
1
Cvp =1870
-
JKg K
p
1
-
1
Case =1000
-
JKg K

One also gives a law of evolution of the latent heat of liquid phase shift
vapor:
L (T) = 2500800 -
(
2443 T - 273 15
. ) J/kg.

4
Calculation of the generalized constraints

In this chapter, we specify how are integrated the relations described into chapter 3. More
precisely still, we give the expressions of the constraints generalized within the meaning of
document [R7.01.10] [bib4] when the laws of behaviors THM are called for the option
RAPH_MECA within the meaning of the document [R5.03.01] [bib3]. So that this document follows of readier the command of
programming, we will consider two cases: the case without dissolved air and that with.

The generalized constraints are:

;m, M, hm;m, M, hm;m, M, hm;m, M, hm;Q
p
W
vp
vp
have
AD
AD
AD
'Q,
W
W
vp
have
have

The generalized deformations, from which the generalized constraints are calculated are:

U, (U); p, p; p, p T
; , T
C
C
gz
gz

Handbook of Référence
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:
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: 24/52

The variables intern that we retained are:

In the case without vapor:
, S
W
lq
In the case with vapor and without dissolved air:
, p, S
W
vp
lq
In the case with dissolved vapor and air:
, p, p, S
W
vp
AD
lq

In this chapter, we adopt the usual notations Aster, namely the indices + for the values
quantities at the end of the step of time and indices - for the quantities at the beginning of the step of time.

Thus, the known quantities are:

· generalized constraints, deformations and internal variables at the beginning of the step of time:
-
-
-
-
-
-
-
-
-
m
-
-
m
-
-
m
-
-
m
-
-
;m, M, H;m, M, H;m, M, H;m, M, H;Q

p
W
W
W
vp
have
AD
AD
AD
'Q,
vp
vp
have
have
-
-
U, (-
U) -
-
-
-
-
-
; p, p; p, p T
; , T
C
C
gz
gz
-
-
-
-
-
, p, p
AD
W
vp
· deformations generalized at the end of the step of time:
-
+
U, (+
U) +
+
+
+
+
+
; p, p; p, p T
;
, T
C
C
gz
gz
· The unknown quantities are the constraints, and variables intern at the end of the step of time:
+
+
+
+
-
+
+
+
+
m
+
+
m
+
+
m
+
+
m
+
+
;m, M, H;m, M, H;m, M, H;m, M, H;Q

p
W
W
W
vp
have
AD
AD
AD
'Q,
vp
vp
have
have
-
+
+
+
+
, p, p
AD
W
vp

4.1
Case without dissolved air

4.1.1 Calculation of porosity and it it density of the fluid

The first thing to be made is of course to calculate saturation at the end of the step of time
+
S
S p. porosity is by integrating on the step of time the equation [éq 3.2.1-1].
lq =
lq (+
c)

One obtains then:
+

B -
+
-
+
+
-

p
p
S p
p
ln
= -
+
-
+
-
gz
gz
lq
C
C


3
T
T
éq
4.1.1-1
-
(V - V) + 0 (-) (-) - (-)
B -

-
K



S


The density of the liquid is by integrating on the step of time the equation [éq 3.2.3.1-1].

What gives:

+
+
-
+
-
p - p - p + p
ln W = gz
gz
C
C -
3
T

éq
4.1.1-2
W
- T
-
(+ -)


W
Kw

Handbook of Référence
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:
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: 25/52

4.1.2 Calculation of the dilation coefficients

It is about a simple application of the formulas [éq 3.2.4.3-2] and [éq 3.2.4.3-3], which are evaluated
at the end of the step of time:
+
+
+
1 S
m
1

éq
4.1.2-1
vp
= farmhouse = mgz = (
-
+
- Slq) (B -)
+
+
(+lq)
0 +
+
T
3
+
m
+

S B

éq
4.1.2-2
0
S
W
= lq (
+
-
)
+
+
+ lq
lq

4.1.3 Calculation of the fluid enthali

The fluid enthali are calculated by integration of the equations [éq 3.2.4.1-1], [éq 3.2.4.2-1],
[éq 3.2.4.2-2].
+
-
1

3 T
m
H
H
C T
T
p
p
p
p éq
4.1.3-1
W
= MW + pw (+ -) (
+
-
-
W
)
+
gz -
gz -
+
+
(+ - + - C
C
)
W
+
-
hm
éq
4.1.3-2
vp
= hmvp + C pvp (+
-
T - T)
+
-
hm
éq
4.1.3-3
have
= hmas + C not (+
-
T - T)

4.1.4 Air and steam pressures

On the basis of the relation [éq 3.2.6-4] in which one carries the law of reaction of perfect gases
dp
M ol
vp
vp 1
1
m
m dT
[éq 3.2.3.2-1], one finds
=


dp
that one integrates by one
gz -
dpc + (hvp - hw)


p
RT


vp

T
W
W

path initially at constant temperature (one then considers the density of constant water),
then of -
T with +
T with constant pressures.

+
p
+
ol
ol
T
ln
M
1
M
vp =
vp
vp
m
m dT
p
p
p
p
H
H

-
+
+ (
+
-
+
-


[gz - gz) - (C - c)] + (vp - W)
p
RT
R
2

T
vp
-
T
W

The first term corresponds to the path at constant temperature, the second with the path with pressures
constants. By using the definitions [éq 3.2.4.1-1] and [éq 3.2.4.2-1] of the enthali, one sees that for
an evolution with constant pressures:

hm - hm
hm - - hm-
-
-
-
vp
W
vp
W
(C p CP
vp
W) (T
T)
=
+

2
2
2
T
T
T

One thus has for such a path:

+
T
+
1
1
1
1
m
m dT
-
-
m
m
p
p
T

(H
ln

vp - hw)
=
2
(hvp - hw)


-
+ C
C
T
-
+
(vp - W) -

-


+

-


-
+
-
T
T
T
T
T
T
T
Handbook of Référence
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:
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: 26/52

That is to say finally:

+
p
ol
vp
M
=
vp
1
ln
p
p
p
p
-
+
+ (
+
-
+
-


[gz - gz) - (C - c)] +
p

vp
RT
W
éq
4.1.4-1
M olvp (
ol
+
-
-
M
hmvp -
-
hmw) 1
1
vp
p
p
T
T

-
+
C
C
ln
1
-
+
(vp - W)



+
-

-
+


R
T
T
R
T T

One can then calculate the densities of the vapor and the air by the relations [éq 3.2.3.2-1] and
[éq 3.2.3.2-2]:

ol
+
M
p
+


éq
4.1.4-2
vp =
vp
vp
+
R T
M ol p
have
gz - p
+
(+ +vp)
have =
+
R
T
éq
4.1.4-3

4.1.5 Calculation of the mass contributions

The equations [éq 3.2.2-1] give null mass contributions to moment 0. Way is written
incremental the equations [éq 3.2.2-1]:

+
-
+
m
m

S

S
W =
W +
W (1
+
+ V) + +
-
lq -
W (1
-
+ V) - - lq
+
-
+
m
m


S


S
have =
have +
have (1
+
+ V) + 1
(
+
-
)
-
lq
- have (1 -
+ V) - 1
(
-
-
)
lq

éq 4.1.5-1
+
-
+
m
m


S


S
vp =
vp +
vp (1
+
+ V) + 1
(
+
-
)
-
lq
- vp (1 -
+ V) - 1
(
-
-
)
lq

4.1.6 Calculation of the heat-storage capacity and Q' heat

There are now all the elements to apply at the end of the step of time the formula [éq 3.2.4.3-5]:

+
0
+
S
+
+
+
p
C = 1
(-) C + C M + 1
(- S +
+
+

)
C
+
+ C éq
4.1.6-1
S
lq
W
W
lq
(
p
p
vp
vp
have
have)

One uses of course [éq 3.2.4.3-4] who gives:
0+
0+
+
2
C = C - 9T K
0
0
éq
4.1.6-2
Although variation of heat Q
is not a total differential, it is nevertheless licit of
to integrate on the step of time and one obtains while integrating [éq 3.2.4.3-1].

+
-
1
1
1
Q' = Q' +
3 K T 2
3
3
3
0
0
(+
0
2
2
V - V)
+
- + m T
lq
(+ -
p - p
C
c) (
+
+
mgz + mlq) T (+
-
p - p
gz
gz)
+
+ C T -
(+
-
T)
éq 4.1.6-3
1
+
-
T +
where we noted:
2
T
=
T. We chose here a formula of “point medium” for
2
variable temperature.
Handbook of Référence
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4.1.7 Calculation of the mechanical constraints

The calculation of the effective constraints is done by calling upon the incremental laws of mechanics selected
by the user. One integrates on the step of time [éq 3.2.8-2] and one a:

+
-

B p
p
bS p
p éq
4.1.7-1
p = p -
(+ -
gz -
gz)
+
+ lq (+
-
C -
c)

4.1.8 Calculation of hydrous and thermal flows

It is necessary well could calculate all the coefficients of diffusion:

The coefficient of Fick
+
F = F (+ + +
T, p, p

C
gz)
Thermal diffusivity T +
T
+
T
+
T
+
T
=
(

). (S

).
T
(
) +
S
lq
T
cte
Hydraulic permeabilities and conductivities:
int
+
rel
int
rel

(
).

+
K
K
(
).
,
H
W (+
Slq)
+
+
K
kgz
H
(+ +
S
p
lq
gz)



lq
=
µ
µ
W (
gz
=
+
T)
gz (+
T)
+
p
Vapor concentrations:
+
C = vp
vp
+
pvp

It does not remain any more whereas to apply the formulas [éq 3.2.5.1-1], [éq 3.2.5.2-15], [éq 3.2.5.2-16] and
[éq 3.2.5.2-17] to find:

+
+
T
+
Q = - T éq 4.1.8-1
+
M
+
have = H

p

F
C F
C éq
4.1.8-2
gz
-
+
+
m + vp vp
+
[+gz (+ +
have
vp)
] + + +vp
have
+
Mvp
+
= H

p

F
1
(
C) F
C éq
4.1.8-3
gz
-
+
+
m - - vp vp
+
[+gz (+ +
have
vp)
]
+
+
+
vp
vp
M+w
H +
=
- p+
+
+ F
éq
4.1.8-4
+
lq [
m
lq
W
]
W

4.2
Case with dissolved air

4.2.1 Calculation of porosity

Same manner, the first thing to be made is to calculate saturation at the end of the step of time
+
S
S p. porosity is by integrating on the step of time the equation [éq 3.2.1-1]. One
lq =
lq (+
c)
thus recall that:

+

B -
+
-
+
+
-

p
p
S p
p
ln
= -
+
-
+
-
gz
gz
lq
C
C


3
T
T
-
(V - V) + 0 (-) (-) - (-)
B -

-
K



S

Handbook of Référence
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Code_Aster ®
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Models of behavior THHM


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:
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: 28/52

4.2.2 Calculation of the dilation coefficients

It is about a simple application of the formulas [éq 3.2.4.3-2] and [éq 3.2.4.3-3], which are evaluated
at the end of the step of time:
+
1 - S
m
m+
m+

1


éq
4.2.2-1
vp
= aas = agz = (
+
- Slq) (B -)
+
+
(+lq)
0 +
+
T
3
+
m
+
W = Slq (
+
B -)
+ +
0 + lq Slq
éq
4.2.2-2
+

m
+
S
AD = S (
+
B -)
+ +
0 +
lq
lq

éq
4.2.2-3
+
T
3

4.2.3 Calculation of density and dissolved and dry air, steam pressures

On the basis of the relation [éq 3.2.6-4] in which one carries the law of reaction of perfect gases
[éq 3.2.3.2-1], one finds:

dp
M ol
vp
vp
=
1 dp

éq
4.2.3-1
W + (H m
vp - H m
W) dT




p
RT

vp

T
W

Contrary to the case without dissolved air p is not now known any more.
W
RT
RT
p = p - p = p - p -
p = p - p -
(p - p)
W
lq
AD
gz
C
have
gz
C
gz
vp
K
K
H
H
thus
RT
R
dp = dp - dp -
(dp - dp) -
(p - p) dT éq
4.2.3-2
W
gz
C
K
gz
vp
K
gz
vp
H
H


One integrates [éq 4.2.3.1] while including there [éq 4.2.3.2] by a path initially into constant temperature (one
then consider the density of constant water), then of -
T with +
T with constant pressures. With
final one obtains:

+
p
ol
ol
ol
ln
M
1
1
M
M
vp = vp (
-
vp
vp
) p
p
p
p
p
p
-
-
+
(+gz - - gz) + - (+vp - - vp) - - + (+c - - c)+


p

RT
K
K
RT
vp
W
H
W
H
W
éq 4.2.3-3
ol
ol
T
MR. R
T
M
vp
vp
dT
m
m
p
p ln
H
H
-
(
+
+
+
vp
- +gz)


+
-

(vp - W)


K
T
R
T 2
W
H


-
T
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Code_Aster ®
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Titrate:
Models of behavior THHM


Date:
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Author (S):
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:
R7.01.11-B Page
: 29/52

Contrary to the preceding case, there is here a nonlinear equation to solve. One will make for that
a method of the corrector-predictor type. P is posed
~ such as:
vp

p~
ol
ol
ln
M
1
1
M
vp = vp (
-
vp
) p
p
p
p
-
-
+
(+gz - - gz) - - + (+c - - c)


p

RT
K
RT
vp
H
W
W
éq
4.2.3-4
+
ol
T
M
+ vp
dT

(m
H
H
vp -
m
W)
R
2
-
T
T

and thus

+
ol
ol
T
M
1
1
M
~
-
p
p.

dT
exp
(
) p
p
p
p
H
H

vp =
vp
vp
-
vp
m
m
-
+
(+gz - - gz) - - + (+c - - c)+ (vp - W)
RT
K
RT

2
-


T
T
H
W
W

éq 4.2.3-5
Moreover, as in the section [§4.1.4], one recalls that
+
T
+
1
1
1
1
m
m dT
-
-
m
m
p
p
T

(H
ln
vp - hw)
=
2
(hvp - hw)


-
+ C
C
T
-
+
(vp - W) -

-


+

-


-
+
-
T
T
T
T
T
T
T
+
p
p~
+
p
+
+
~
p

p
p
p
vp
vp -

+
Like ln vp
ln vp ln vp
=
+



vp
vp

and that by D.L ln
= ln 1+

- 1
-
-
~
~

~

~
p
p
p
p
p
p
vp
vp
vp
vp

vp

vp

+
pvp will thus be given by the following linear expression:
+
ol
ol
+
p
M
MR. R
vp = 1+
vp
vp
T
p
p
p
p
ln
~
éq 4.2.3-6
-
(+vp - - vp) + - (+vp - - gz)

-
p
K
K
vp
H
H
W
W
T
from where


+
ol
T
-
-
-
K
M
p
p R ln


H -
vp
W

vp + gz

-
T
+




p

éq
4.2.3-7
vp =
-
KH

+
W
ol
T



- M 1
ln
vp
~

+ R -
pvp

T

From there the other pressures are calculated easily:

+
+
+
p
p
p
have =
gz -
vp
+
+
not
+
p

AD =
RT
K H
+
+
+
+
p
p
p
p
W =
gz -
C -
AD
Handbook of Référence
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:
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: 30/52

One can then calculate the densities of the vapor and the air by the relations [éq 3.2.3.2-1],
[éq 3.2.3.2-2] and [éq 3.2.7-3]:

ol
+
M
p
+


éq
4.2.3-8
vp =
vp
vp
+
R T
M ol p
have
gz - p
+
(+ +vp)


éq
4.2.3-9
have =
+
R
T
+
ol
+


éq
4.2.3-10
AD = p
M
AD
have
+
RT
The density of water is by integrating on the step of time the equation [éq 3.2.3.1-1].

What gives:

+
+
-
+
-
+
-
p - p - p + p - pad + p
ln W =
AD
gz
gz
C
C
-
3
T

éq 4.2.3-11
W
- T
-
(+ -)


W
Kw

4.2.4 Calculation of the fluid enthali

The fluid enthali are calculated by integration of the equations [éq 3.2.4.1-1], [éq 3.2.4.1-3],
[éq 3.2.4.2-1], [éq 3.2.4.2-2].
+
-
1

3 T
m
H
H
C T
T
p
p
p
p
p
p

éq 4.2.4-1
W
= MW + pw (+ -) (
+
-
-
W
)
+
gz -
gz -
+ C - AD +
+
(+ - + - + - AD
C
)
W
+
-
hm
éq
4.2.4-2
AD
= hmad + C pad (+
-
T - T)
+
-
hm
éq
4.2.4-3
vp
= hmvp + C pvp (+
-
T - T)
+
-
hm
éq
4.2.4-4
have
= hmas + C not (+
-
T - T)

4.2.5 Calculation of the mass contributions

The equations [éq 3.2.2-1] give null mass contributions to moment 0. Way is written
incremental the equations [éq 3.2.2-1]:

+
-
+
m
m

S

S
W =
W +
W (1
+
+ V) + +
-
lq -
W (1
-
+ V) - - lq
+
-
+
m
m

S

S
AD =
AD +
AD (1
+
+ V) + +
-
lq -
AD (1
-
+ V) - - lq

éq
4.2.5-1
+
-
+
m
m


S


S
have =
have +
have (1
+
+ V) + 1
(
+
-
)
-
lq
- have (1 -
+ V) - 1
(
-
-
)
lq
+
-
+
m
m


S


S
vp =
vp +
vp (1
+
+ V) + 1
(
+
-
)
-
lq
- vp (1 -
+ V) - 1
(
-
-
)
lq
Handbook of Référence
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:
R7.01.11-B Page
: 31/52

4.2.6 Calculation of the heat-storage capacity and Q' heat

There are now all the elements to apply at the end of the step of time the formula [éq 3.2.4.3-5]:

+
0
+
S
+
+
+
p
+
p
C = 1
(-) C + S (C + C) + 1
(- S +
+
+

)
C
+
+ C éq 4.2.6-1
S
lq
W
W
AD
AD
lq
(
p
p
vp
vp
have
have)

One uses of course [éq 3.2.4.3-4] who gives:

0+
0+
+
2
C = C - 9T K
0
0
éq
4.2.6-2

Although variation of heat Q
is not a total differential, it is nevertheless licit of
to integrate on the step of time and one obtains while integrating [éq 3.2.4.3-1].

+
-
1
1
1
Q' = Q' +
3 K T 2
3
3
3
0
0
(+
0
2
2
V - V)
+
- + m T
lq
(+ -
p - p
C
c) (
+
+
mgz + mlq) T (+
-
p - p
gz
gz)
+
+ C T -
(+
-
T)
éq 4.2.6-3
1
+
-
T +
where we noted:
2
T
=
T. We chose here a formula of “point medium” for
2
variable temperature.

4.2.7 Calculation of the mechanical constraints

The calculation of the effective constraints is done by calling upon the incremental laws of mechanics selected
by the user. One integrates on the step of time [éq 3.2.8-2] and one a:

+
-

B p
p
bS p
p éq
4.2.7-1
p = p -
(+ -
gz -
gz)
+
+ lq (+
-
C -
c)

4.2.8 Calculation of hydrous and thermal flows

It is of course necessary to calculate all the coefficients of diffusion:

Coefficients of Fick
+
F (+
P, +
P, +
T, +
S) and +
F (+
P, +
P, +
T, +
S)
vp
vp
gz
AD
AD
lq

Thermal diffusivity T +
T
+
T
+
T
+
T
=
(

). (S

).
T
(
) +
S
lq
T
cte

Hydraulic permeabilities and conductivities:

int
+
rel
int
rel

(
).

+
K
K
(
).
,
H
W (+
Slq)
+
+
K
kgz
H
(+ +
S
p
lq
gz)



lq
=
µ
µ
W (
gz
=
+
T)
gz (+
T)

+
p
Concentrations out of vapor and air dissolved:
+
C = vp and +
+
C

AD =
vp
+
p
AD
gz
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:
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: 32/52

It does not remain any more whereas to apply the formulas [éq 3.2.5.1-1], [éq 3.2.5.2-15], [éq 3.2.5.2-16],
[éq 3.2.5.2-17] and [éq 3.2.5.2-18] to find:
+
+
T
+
Q = - T éq 4.2.8-1
+
M
+
have = H

p

F
C F
C éq
4.2.8-2
gz
-
+
+
m + vp vp
+
[+gz (+ +
have
vp)
] + + +vp
have
+
Mvp
+
= H

p

F
1
(
C) F
C éq
4.2.8-3
gz
-
+
+
m - - vp vp
+
[+gz (+ +
have
vp)
]
+
+
+
vp
vp
M+w
H +
=
- p+ + (+
+
+) F
éq
4.2.8-4
+
lq [
m
lq
W
AD
]
W
+
+
M

p
(
) F
F
C
éq 4.2.8-5
AD =
H
AD
lq [
+
+
- lq + W +
m
AD
] + +
- AD AD

5
Calculation of derived from the generalized constraints

In this chapter, we give the expressions of derived from the constraints generalized by
report/ratio with the deformations generalized within the meaning of the document [R7.01.10] [bib4], C `be-with-statement terms
who are calculated when the laws of behaviors THM are called for option RIGI_MECA_TANG
within the meaning of the document [R5.03.01] [bib3].
In order not to weigh down the talk, we give the expression of the differentials of the constraints
generalized, knowing that the derivative partial result some directly.

5.1
Derived from the constraints

The calculation of the differential of the effective constraints is left with the load of the law of behavior
purely mechanical, that we do not describe in this document. Differential of the constraint
is given directly by the expression [éq 3.2.8-2].
p

5.2
Derived from the mass contributions

While differentiating [éq 3.2.2-1], one a:
DM
D
W
W
=
(1+ S + D S + 1+ dS + 1+ dS
V)
lq
V
lq
(V) lq (V) lq


W
W
DM
D
AD
AD
=
(1+ S + D S + 1+ dS + 1+ dS
V)
lq
V
lq
(V) lq (V) lq


AD
AD
éq
5.2-1
DM
D
have
have
=
(1+ 1 - S + D 1 - S + 1+ D 1 - S - 1+ dS
V) (
lq)
V
(lq) (V) (lq) (V) lq


have
have
DM
D
vp
vp
=
(1+ 1 - S + D 1 - S + 1+ D 1 - S - 1+ dS
V) (
lq)
V
(lq) (V) (lq) (V) lq


vp
vp
Handbook of Référence
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:
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: 33/52

5.2.1 Case without dissolved air

by taking account of [éq 3.2.1-1] and of [éq 3.2.3.1-1], [éq 3.2.3.2-1], [éq 3.2.3.2-2] and while supposing
1+
one finds:
V 1

DM
S
S
S 2 B
(
)

(
)
W
lq
lq
lq
-



B -



= bS D


3
lq
V +
-
-
dpc + Slq
+
dpgz -
m dT



W
P
K
K





W

C
W
S

K
K
W
S



DM
1
(
)
(
)
(
1
) (
)
vp

Slq
- S S B
lq
lq
-

B -
- Slq
dpvp
m

= B 1
(- S) D

1
(
)

3
lq
V +
-
-
dpc +
dpgz +
- Slq
-
dT
vp









vp

P
K
C
S


K S

pvp


DM

S
1
(
)
(
)
(
1
) (
)
have
lq
- S S B
lq
lq
-

B -
- Slq
dpas
m

= B 1
(- S) D

1
(
)

3
lq
V +
-
-
dpc +
dpgz +
- Slq
-








dT
have
P
K
K
p
have


C
S


S

have


éq 5.2.1-1

One sees appearing the derivative of saturation in liquid compared to the capillary pressure, quantity which
play an essential part.

The expression [éq 3.2.6-4] of the differential of the steam pressure also makes it possible to calculate
pressure of dry air:

M ol p
M ol
M ol
dp
vp
vp
= 1
(-
) dp
vp
-
dp
vp
-
-
éq
5.2.1-2
have
gz
C
(hm m
vp
W) dT
H
RT
RT
RT
T
W
W

One defers [éq 5.2.1-2] and [éq 3.2.6-4] in [éq 5.2.1-1] and one finds:

DM

S

S
S 2 B
(-)

(-)
W
lq
lq
lq
B
= bS D

+

-
-
dp + S
+
dp
m
-
3
dT
lq
V


P

K
K
C
lq


K
K
gz
W


W
C
W
S


W
S

éq 5.2.1-3
DM


1
-
1

vp = B (1 - S + -
-
-

lq)
Slq (Slq) Slq (B
) (Slq)
D
vp
dp
V


p

K
p
C


vp

C
S
vp
lq
éq 5.2.1-4
(B -) (1 - S
1


1
-

lq)
(Slq) vp
m
vp (
Slq) (hm hm
vp
lq)
+
+
dp + -
3
+
dT

K
p
gz
vp



p T


S
vp
lq

vp

DM


1
-
1

have = B (1 - S
+ -
-
+

lq)
Slq (Slq) Slq (B
) (Slq)
D
vp
dp
V


p

K
p
C


have

C
S
have
lq
(B -) (1 - S
1
-

1
-

lq)
(Slq) lq vp
m
vp (
Slq) (hm hm
vp
lq)
+
+
dp + -
3
-
dT

K
p
gz
have



p T


S
have
lq


have

éq 5.2.1-5
Handbook of Référence
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5.2.2 Case with dissolved air

As previously, by taking account of [éq 3.2.1-1] and of [éq 3.2.3.1-1], [éq 3.2.3.2-1],
[éq 3.2.3.2-2], [éq 3.2.7.3] and by supposing 1+
one finds:
V 1

DM
S
2 (-)



(
)
-



W
lq
p
S
B
S

W
lq
lq
p

W
B
Slq p
= bS D +
-
+
dp + S
+
dp
W
m
+
-
3
dT
lq
V

K
P

K
P
C
lq

K P

K
gz


K T
W

W
W
C
S
C


W
gz
S

W

éq 5.2.2-1

DM
S M ol
2 (-)




(
)
-
AD
lq
have
p
S
B
S

have
lq
lq
M ol
p

have
have
B
= bS D +
-
+
dp + S
+
dp
lq
V

K
P

K
P
C
lq

K P

K
gz

AD
AD
H
C
S
C


AD H
gz
S

éq 5.2.2-2
M ol p


+ S
have
have - 3 B
(

0
-) dT
lq K
T




AD
H


DM

S

1
(- S) S B
(-) 1
(- S)
B
(- 1
) (- S) 1
(- S)
vp = B 1
(- S) D
lq
lq
lq
lq
vp
+ -
-
-
dp
lq
lq
vp
+
+
dp +
lq
V


P

K
p
C



K
p
gz


vp

C
S
vp
lq

S
vp
lq

1
(- S) (hm - hm)
m
vp
lq
vp
lq
3

+ - +
dT
vp

p T

vp


éq 5.2.2-3

DM

S

1
(- S) S B
(-) 1
(- S)
have = B 1
(- S) D
lq
lq
lq
lq
vp
+ -
-
+
dp +
lq
V


P

K
p
C


have

C
S
have
lq
éq
5.2.2-4
B
(- 1
) (- S) 1
(- S) -

1
(- S) (hm - hm)
lq
lq
lq
vp

+
dp
m
vp
lq
vp
lq
3

+ - -
dT

K
p
gz
vp



p T


S
have
lq

have



The derivative partial are given in [§Annexe 3].

5.3
Derived from the enthali and Q' heat

There still, we do nothing but point out expressions already provided to chapter 2:

5.3.1 Case without dissolved air

-
dhm = 1 -
3
+

W
(
T
W
) dp dp
gz
C
C pdT
W
W
dhm = C p dT
vp
vp
dhm = C p dT
have
have
Q
'= 3 K Td
m
+ 3 Tdp
0
- 3 + 3
+
0
0
V
lq
C
(m
m
gz
lq) Tdp
C dT
gz


Handbook of Référence
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: 35/52

5.3.2 Case with dissolved air

-
-
dhm = 1 -
3
+
W
(
T
W
) dp dp dp
gz
C
AD
C pdT
W
W
1 -
3 T
p



p



1 -
3


W
AD
AD
p
T p
=
1
dp - 1+
dp + C
W
AD
-
dT








p
gz




p
C
W









T

W
gz
C

W

dhm = C p dT
AD
AD
dhm = C p dT
vp
vp
dhm = C p dT
have
have
Q
'= 3 K Td
m
+ 3 Tdp
0
- 3 + 3
+
0
0
V
lq
C
(m
m
gz
lq) Tdp
C dT
gz



5.4
Derived from the heat flow

One leaves [éq 3.2.5.1-1] and [éq 3.2.5.1-2].

While differentiating [éq 3.2.5.1-2] and while using [éq 3.2.1-1], one finds:

D T
= B
T
T
(-) '(
). (S
T
). T
() D
S
lq
T
v
B
(-) T
T
+
'(
). (S
T
). T
() dp
K
S
lq
T
gz
S



S

(-)

T
T
T
lq
B
+ (

). '(S). T
().
- S
T
T
'(
). (S
T
). T
() dp
S
lq
T
p
lq

K
S
lq
T
C



C
S

+ (T
T
(
). (S
T
). 'T
() - B
T
T
(- 3
). '(
). (S
T
). T
()
S
lq
T
0
S
lq
T
) dT

That is to say finally:

dq = - B
T
T
(-) '(
). (S
T
). T
(
) Td
S
lq
T
v
B
(-) T
T
-
'(
). (S
T
). T
(
) Tdp
K
S
lq
T
gz
S
éq
5.4-1

S

(-)

T
T
T
lq
B
- (
). '(S).
T
().
- S
T
T
'(
). (S
T
). T
() Tdp
S
lq
T


p
lq

K
S
lq
T
C



C
S

- (T
T
(
). (S
T
). 'T
() - B
T
T
(- 3
). '(
). (S
T
). T
()
0

S
lq
T
S
lq
T
) TdT

5.5
Derived from hydrous flows

It is of course necessary to set out again of the equations [éq 3.2.5.2-15], [éq 3.2.5.2-16], [éq 3.2.5.2-17] and [éq 3.2.5.2-18]
that one differentiates.
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Models of behavior THHM


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:
R7.01.11-B Page
: 36/52

5.5.1 Case without dissolved air

M

M - C F C
have
H
m
have
have
vp
vp
vp
H
DM =
+ F D +
D
have
have
gz
have




H
gz


have

lq
H
+ - dp + D F
éq
5.5.1-1
have
gz (
m
gz
vp
)
F
F

vp
vp
+ C
dT +
p

C + cd. F C + C F cd.
have
vp
gz
vp
have
vp
vp
vp
have
vp
vp
vp
T
p


gz



M

M + 1 - C F C
vp
H
m
vp
vp (
vp) vp
vp
H
DM =
+ F D +
D
vp
vp
gz
vp
H
gz




vp

gz
H
+ - dp + D F

vp
gz (
m
gz
have
)
F
F

vp
vp
- 1
(- C)
dT +
p

C + cd. F C - 1
(- C) F cd.
vp
vp
gz
vp
vp
vp
vp
vp
vp
vp
vp
vp
T
p


gz

éq 5.5.1-2
M

M
W
H
m
W
H
DM =
+ F D +
D
W


W lq

W
H
lq


W

lq
éq
5.5.1-3
H
- dp - dp
W lq (
gz
c)

In order to clarify these differentials completely, it is necessary to know the differentials of the masses
p
voluminal of the fluids, as well as the differentials of
vp
C =
and of its gradient. Knowing
vp
pgz
[éq 3.2.6-4], one can then calculate the differential of the pressure of dry air:

-

dp = dp - dp
W
vp
=
dp
vp
+
dp -
-

éq
5.5.1-4
have
gz
vp
gz
C
vp (H m
m
vp
W) dT
H


T
W
W

D
dp
dT
D
dp
vp
vp
dT
By deriving the relation from perfect gases one a:
have
have
=
-
and
=
-
, which, in

p
T

p
T
have
have
vp
vp
using [éq 3.2.6-4] and [éq 5.5.1-4] gives:

2
2
2
-

vp
vp
vp (H m
hm
vp
W)
D =
dp -
dp
vp


+
-
dT
éq
5.5.1-5
vp
p
gz
p
C

Tp
T
W
vp
W
vp
vp


-


-

have
W
vp
have
vp
have
vp (H m
hm
vp
W)
D =
dp +
dp
have


+ -
-
dT éq 5.5.1-6
have
p
gz

p
C


Tp
T
have
W
have
W
have


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:
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: 37/52

[éq 3.2.6-4] allows to express the gradient of the steam pressure:



p
vp
=

-
+
-
éq
5.5.1-7
vp
(p
p
gz
c)
vp (hm
m
vp
lq)
T
H

T
lq

While differentiating [éq 5.5.1-7] one finds:

m
m





-

dp
vp
=

-
+
-




-
+

vp
(D p D
gz
c)
D
p
vp
vp D
2
W (p
p
gz
c)
H
H
vp
W
D T
vp






T

W
W
W



éq
5.5.1-8

hm - hm
+d
vp
W





T
vp


T




The last term of [éq 5.5.1-8] is written:


hm - hm
hm - hm

D
vp
W





T
vp
W

=

T D
vp
+
T
-
-
-


vp
vp
(dhm dhm
vp
W)
vp (H m
hm
vp
W)
dT
T 2


T


T

T
T





éq 5.5.1-9

Knowing the differentials of the its gradient and steam pressure, the expressions of
differentials of C and its gradient are easy to calculate:
vp

dp
p
p
p
vp
vp
cd. =
-
dp which gives:
vp
vp
C =
-
p and which one differentiates in:
vp
2
gz
p
p
vp
2
gz
p
gz
p
gz
gz
gz
p
p
dp
p
p
p
p
vp
vp
vp
gz
vp
vp
vp
cd. = D
-
p =
-
dp

+ 2
p

-
dp -
dp
vp
2
gz
2
vp
3
gz
2
gz
2
gz
p

p
p


p
p
p
p
gz
gz
gz

gz
gz

gz
gz


dp is given by [éq 3.2.6-4] and dp by [éq 5.5.1-8].
vp
vp

5.5.2 Case with dissolved air
M

M - C F C
have
H
m
have
have
vp
vp
vp
H
DM =
+ F D +
D
have
have
gz
have




H
gz


have

gz
H
+ - dp + D F
éq 5.5.2-1
have
gz (
m
gz
vp
)
F
F

vp
vp
+ C
dT +
dp

C + cd. F C + C F cd.
have
vp
gz
vp
have
vp
vp
vp
have
vp
vp
vp
T
p


gz


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Models of behavior THHM


Date:
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Author (S):
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:
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: 38/52

M

M + 1 - C F C
vp
H
m
vp
vp (
vp) vp
vp
H
DM =
+ F D +
D
vp
vp
gz
vp
H
gz




vp

gz
H
+ - dp + D F

vp
gz (
m
gz
have
)
F
F

vp
vp
- 1
(- C)
dT +
dp

C + cd. F C - 1
(- C) F cd.
vp
vp
gz
vp
vp
vp
vp
vp
vp
vp
vp
vp
T
p


gz

éq 5.5.2-2

M

M
W
H
m
W
H
H
DM =
+ F D +
D + - dp + D F éq 5.5.2-3
W


W lq

W
H
lq
W lq (
m
lq
AD
)



W
lq

DM
= - p + F D + -
(p + F) D
AD
(H
H
m
lq
lq
lq
lq
) AD AD (H
m
lq
lq
lq
) Hlq
H
+ - dp + D F



AD
lq (
m
lq
W
)
F
F

AD
AD
-
dT +
dp C - F cd.
C
AD
AD
AD

T
p



C

éq 5.5.2-4

It is necessary to know the differentials of the densities of the fluids, as well as the differentials of
pvp
C =
, C
= and of their gradient. One first of all will calculate the differentials of the masses
vp
p
AD
AD
gz

voluminal by using the derivative partial of pressures given in [§Annexe 3].

D
dp
dT
D
dp
vp
vp
dT
By deriving the relation from perfect gases one a:
have
have
=
-
and
=
-
, that one

p
T

p
T
have
have
vp
vp
can express in the form:


p

p



p


D
have
have

=
dp
have
+
dp
have
have
have
+
-
dT
have
éq 5.5.2-5
p p
C

p
gz

p
T

T
have
C
gz
have


p

p


p


D
vp
vp

=
dp
vp
+
dp
vp
vp
vp


+
-
dT
vp
éq 5.5.2-6
p p
C

p
gz

p
T

T
vp
C
gz
vp



By using the relation [éq 3.2.3.1-1], one obtains:
p

p

p


D
W
W
=

dp
W
+
dp
W
W
+
- 3 dT éq
5.5.2-7
W
K p
C

p
gz

K
T
W
W

W
C
gz
W

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:
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: 39/52

ol
M
And like
have
dad =
dpas,
K H
M ol
have
not
not
p

D
éq
5.5.2-8
AD =

dpc +
dpgz +
have dT


K H PC
pgz
T


As previously, the expressions are used

dp
p
vp
vp
cd. =
-
dp which gives
vp
2
gz
p
p
gz
gz
1 p

p

1 p

p
vp
vp
vp
vp
cd. =
dp +
dT


+
-
dp éq
5.5.2-9
vp
C
2
gz
p p

T



p
p




gz
C
gz
gz
p

gz


p
p
and
vp
vp
C =
-
p and which one differentiates in:
vp
2
gz
p
p
gz
gz
p
p
dp
p
p
p
p
vp
vp
vp
gz
vp
vp
vp
cd. = D
-
p =
-
dp

+ 2
p

-
dp -
dp
vp
2
gz
2
vp
3
gz
2
gz
2
gz
p

p
p


p
p
p
p
gz
gz
gz

gz
gz

gz
gz


éq 5.5.2-10

with
p

p

p

p
vp
=
p
vp
+
p
vp
+
T
éq
5.5.2-11
vp
p
gz

p
C

T

gz
C
and dp that one differentiates in the following way:
vp
p

p

p

p

p

p

dp = D
vp p + D vp p + D vp T
vp
+
dp
vp
+
dp
vp
+
dT
vp
p
gz

p
C

T

p
gz

p
C

T
gz
C
gz
C


p

p

p


vp

=
p
vp
+
p
vp
+
T dp
p

p
gz

p

p
C

p

T
C

C
gz
C
C
C


p

p

p


vp

+
p
vp
+
p
vp
+
T dp


p

p
gz

p

p
C

p

T
gz


gz
gz
gz
C
gz


p

p

p


vp

+
p
vp
+
p
vp
+
T dT
T

p
gz

T

p
C

T

T


gz
C


p

p

p
vp

+
dp
vp
+
dp
vp
+
dT
p
gz

p
C

T
gz
C

éq 5.5.2-12
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:
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: 40/52

The derivative partial of the second command are developed in [§Annexe 4].

For the dissolved air, one proceeds with the same stages:
p
C = = M ol
AD
.
AD
AD
AD
RT
thus
ol dp
p
AD
AD

cd.
.
who gives
AD = M AD
-
dT
RT
RT 2


ol
1 p
1
1
AD
p

AD
p
P
AD
AD
cd.
.

éq
5.5.2-13
AD = M AD
dpc +
dpgz +
-

dT





2
RT PC
RT pgz
T
RT plq



ol p
p
AD
AD

and C
.
and that one differentiates in:
AD = M AD
-
T
RT
RT 2



ol
1

1 T
pad
pad
pad

cd.
2

AD = M AD
dpad -
dp
2
AD +
T -
dT -
dT
3
2
2

RT
RT
RT
RT
RT

éq 5.5.2-14
with
p

p

p

p
AD
=
p
AD
+
p
AD
+
T éq
5.5.2-15
AD
p
gz

p
C

T

gz
C
and dp that one differentiates in the following way:
AD

p

p

p


AD
=
p
AD
+
p
AD
+
T dp
p

p
gz

p

p
C

p

T
C



C
gz
C
C
C

p


p


p


AD
+
p
AD
+
p
AD
+
T dp
p

p
gz

p

p
C

p

T
gz



gz
gz
gz
C
gz


éq
5.5.2-16
p

p

p


AD
+
p
AD
+
p
AD
+
T dT
T

p
gz

T

p
C

T

T



gz
C

p

p

p

AD
+
dp
AD
+
dp
AD
+
dT
p
gz

p
C

T

gz
C

The derivative partial of the second command are developed in [§Annexe 4].

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Models of behavior THHM


Date:
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:
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: 41/52

6 Bibliography

[1]
O. COUSSY: “Mechanical of the porous environments”. Editions TECHNIP.
[2]
C. CHAVANT, P. CHARLES, Th. DUFORESTEL, F. VOLDOIRE
: “
Thermo hydro
mechanics of the porous environments unsaturated in Code_Aster “. Note HI-74/99/011/A.
[3]
I. VAUTIER, P. MIALON, E. LORENTZ: “Quasi static nonlinear Algorithm (operator
STAT_NON_LINE) “. Document Aster [R5.03.01].
[4]
C. CHAVANT: “Modelings THHM general information and algorithms”. Document Aster
[R7.01.10].
[5]
A. GIRAUD: “Adaptation to the nonlinear model poroelastic of Lassabatère-Coussy to
modeling porous environment unsaturated “. (ENSG).
[6]
J. WABINSKI, F. VOLDOIRE: Thermohydromécanique in saturated medium. Note EDF/DER
HI/74/96/010, of September 1996.
[7]
T. LASSABATERE: “Hydraulic Couplings in porous environment unsaturated with
phase shift: application to the withdrawal of desiccation of the concrete “. Thesis ENPC.
[8]
PH. MESTAT, Mr. PRAT: “Works in interaction”. Hermes.
[9]
J.F. THILUS et al.: “Poro-mechanics”, Biot Conference 1998.
[10]
A.W. BISHOP & G.E. BLIGHT: “Nap Aspects off Effective Stress in Saturated and Partly
Saturated Soils “, Géotechnique n° 3 flight. XIII, pp. 177-197. 1963.
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:
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: 42/52

Internal appendix 1 Contraintes generalized and variables

Constraints:

Number
Name of Aster component
Contents
1
SIXX

xx
2
SIYY

yy
3
SIZZ

zz
4
SIXY

xy
5
SIXZ

xz
6
SIYZ

yz
7
SIP

p
8
M11
m
W
9
FH11X
M
W X
10
FH11Y
M

W y
11
FH11Z
M
W Z
12
m
ENT11
H
W
13
M12
m
vp
14
FH12X
M

vp X
15
FH12Y
M

vp y
16
FH12Z
M

vp Z
17
m
ENT12
H
vp
18
M21
m
have
19
FH21X
M

have X
20
FH21Y
M

have y
21
FH21Z
M

have Z
22
m
ENT21
H
have
18
M22
m
AD
19
FH22X
M

AD X
20
FH22Y
M

AD y
21
FH22Z
M

AD Z
22
m
ENT22
H
AD
23
QPRIM
Q'
24
FHTX
Q
X
25
FHTY
Q
y
26
FHTZ
Q
Z
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Models of behavior THHM


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:
R7.01.11-B Page
: 43/52

The variables intern if one takes account of the saturation (of type “..HH.” or THV):

Number
Name component Aster
Contents
1 V1
0
-
2 V2
0
-
lq
lq
3 V3
0
p - p
vp
vp
4 V4
S
lq

And in the other cases:

Number
Name component Aster
Contents
1 V1
0
-
2 V2
0
-
lq
lq
3 V3
0
p - p
vp
vp

Appendix 2 Données material

One gives here the correspondence between the vocabulary of the Aster commands and the notations used in
present note for the various sizes characteristic of materials.

A2.1 Mot key factor THM_LIQU

RHO
0

lq

UN_SUR_K
1
Klq

ALPHA

lq

CP
p
C
lq

VISC
µ (T
lq
)

D_VISC_TEMP
µ
lq (T)
T


A2.2 Mot key factor THM_GAZ

MASS_MOL
ol
M
have

CP
p
C
have

VISC
µ (T
have
)

D_VISC_TEMP
µ
have (T)
T

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Titrate:
Models of behavior THHM


Date:
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Author (S):
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:
R7.01.11-B Page
: 44/52

A2.3 Mot key factor THM_VAPE_GAZ

MASS_MOL
ol
M
VP

CP
p
C
vp

VISC
µ (T
vp
)

D_VISC_TEMP
µ
vp (T)
T


A2.4 Mot key factor THM_AIR_DISS

CP
p
C
AD

COEF_HENRY
K
H

A2.5 Mot key factor THM_INIT
TEMP
initT

PRE1
init P
1

PRE2
init P
2

PORO
0


PRES_VAPE
0
p
vp

One recalls that, according to modeling, the two pressures P and P
1
2 represent:


LIQU_SATU LIQU_VAPE LIQU_GAZ_ATM GAS
LIQU_VAPE_GAZ
P
p
p
p = - p
p
p = p - p
1
W
W
C
W
gz
C
gz
W
P
p
2




gz


LIQU_GAZ LIQU_AD_GAZ_VAPE
P
p = p - p
p = p - p - p
1
C
gz
W
C
gz
W
AD
P
p
p
2
gz
gz

A2.6 Mot key factor THM_DIFFU

R_GAZ
R

RHO
R
0

CP
S
C

BIOT_COEF
B

SATU_PRES
S p
lq (c)

D_SATU_PRES
S
lq (PC)
p
C

PESA_X
m
F
X

PESA_Y
m
F
y
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Date:
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:
R7.01.11-B Page
: 45/52

PESA_Z
m
F
Z

PERM_IN
int
K ()


PERM_LIQU
rel
K
S
lq (lq)

D_PERM_LIQU_SATU
rel
K
lq (S
lq)
S
lq

PERM_GAZ
rel
K
S, p
gz (lq
gz)

D_PERM_SATU_GAZ
rel
K
gz (S, p
lq
gz)
S
lq

D_PERM_PRES_GAZ
rel
K
gz (S, p
lq
gz)
p
gz

FICKV_T
F T (T)
vp

FICKV_S
F S (S)
vp

FICKV_PG
gz
F (P)
vp
G

FICKV_PV
vp
F (P)
vp
vp

D_FV_T
F T
vp (T)
T


D_FV_PG
gz
F
vp (P)
gz
P
gz

FICKA_T
F T (T)
AD

FICKA_S
F S (S)
AD

FICKA_PA
AD
F (P)
AD
AD

FICKA_PL
lq
F (P)
AD
lq

D_FA_T
F T
vp (T)
T


LAMB_T
T
(T)
T


T
T
()

DLAMB_T
T

T




T

(

LAMB_PHI
)




T
()

DLAMBPHI





LAMB_S
T
(S)
S




T
(S)

DLAMBS
S
S




LAMB_CT
T

CT

Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 46/52

Note:

For modelings utilizing it thermal, and for the calculation of the specific heat
homogenized, the relation is used: C 0 = (1 -)
S
p
C + C M

+ 1 - S C + C.
S
lq lq
lq
(lq) (p
p
vp
vp
have
have)
In this formula, one confuses with his initial value 0
whose value is read under the key word
S
S
RHO of the key word factor ELAS.

Appendix 3 Dérivées of the pressures according to the deformations
generalized

One details here the calculation of derived from pressure according to the generalized deformations. It is pointed out that
dpvp
dpw
dT
the equation [éq 3.2.6.3] is
=
+ L
with
m
m
L = H - h. En outre


T
vp
W
vp
W
R
RT
dp = dp - dp =
p dT +
dp and dp = dp - dp. By combining these equations one obtains:
AD
lq
W
have
have
K
K
have
gz
vp
H
H

RT

dT
RT

dp
L
p
1 dp
dp
vp
- W = (-
W +
AD)
+
-
gz +






C
K

T
K
H
vp

H






vp RT

vp
p
RT

dp
1
LR
dT
1 dp
dp
W
- = -
+ AD
+
-
gz +








C

K
K
T
K
W
H

H

H


One can thus write the derivative partial of water and the vapor according to the generalized deformations:



-
vp
LR
+ p
RT
AD
- 1
p
K
T
p
K
p
W
H
W
H

1
=
;
=
;
W =

T



vp RT
p
RT
p
RT
gz
vp

- 1
- 1
C
vp
- 1
K
K
K
W
H
W
H
W
H
RT - 1
p

- L + p
1
p

K
p

vp
(
1
W
AD)
vp
vp
H
=
. ;
=
;
=

T

RT

T
p

RT

p

RT

W
gz
W
C
W
-
-
-
K

K

K

H
vp
H
vp
H
vp

The relations dp
= dp - dp and dp = dp - dp - dp makes it possible to derive all the pressures,
have
gz
vp
AD
gz
C
W
since one will have
p

p

p

p

p

p

have
vp
have
vp
have
vp
= -
;
= 1
;
= -

T

T

p

p

p

p

gz
gz
C
C
and

p

p

p

p

p

p

AD
W
AD
W
AD
W
= -
;
= 1
;
= - 1

T

T

p

p

p

p

gz
gz
C
C
Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 47/52

Appendix 4 Dérivées seconds of air and steam pressures
dissolved according to the generalized deformations

One calculates here the derivative partial of the second command of the steam pressure necessary to the section [§5.5.2].
One will note thereafter:

RT

RT

W
A1 =
-
and 2
With = W
- 1
K






K
H
vp
vp
H

1
p
vp 1
1
p
W
A3 =
- -
-
3

W
p
T

T
K
T

vp
W
1
p
vp 1
1
p
W
A4 = -
+ +
-
3

W
p
T

T
K
T

vp
W

Derived seconds from the steam pressure:

p
A2 1
p
1
p
vp



=

W -
vp
2

p p
A1
K
p
p
p
C gz
W.C.
vp C
p
A2 1
p
1
p
vp



=

W -
vp
2

p
p
A1
K
p
p
p
gz gz
W gz
vp gz
pvp
R
1 RT

R


=
-
- 1
- W 4
With
2




T p
K
1
With
1
With

gz
H
K H

K H
vp


p

1 1
p
vp

1
p
W
vp

= -

-
W
2

p
p
A1 p
p
K
p
C C
vp
vp C
W.C.
p

1 1
p
vp

1
p
W
vp


= -

-
W
2

p
p
A1 p
p
K
p
gz C
vp
vp gz
W gz
pvp
1 R


= -

- W 4
With
2

T p
1
With

C
K H
vp


p

1 1
p


1
p
1
p
vp




W
W
W




= -
A1 1
1
(+ L
) + (p
L)

2
AD -

W
W
-
vp
p
T
T A1
p
K

K
p
p
p
C







C
W

vp
W.C.
vp



C
p

1 1

p


1
p
1
p
vp



W
W
W




= -
With
1 1
1
(+ L
) + (p
L)

2
AD -

W
W
-
vp
p
T
T A1
p
K

K
p
p
p
gz





gz
W
vp
W gz
vp




gz
p







vp
1
p

p
RT

R

AD
W
W
1
=

- L
- 3
W
-

-
+ T (
W
-
With)
4 (p - L)
T

T

T. 1
W
W




2
With
T

K
T



T. 12
AD
W
WITH K

K



W

H
vp
H
vp

Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 48/52

Derived seconds from the pressure of dissolved air:

p
RT A2 1 p
1
p
AD
vp

=

-
W
2

p p
K
A1
p
p
K
p
C gz
H
vp C
W.C.
p
RT A2 1 p
1
p
AD
vp


=

-
W
2

p
p
K
A1
p
p
K
p
gz gz
H
vp gz
W gz
2
p

1
2
AD
R
W
W
R WITH
= -
+

+

2 (1
A3 T
. )
T

p

K A1


K
A1
gz
H
vp
vp
H

p
RT
1 1 p
1
p
AD
W
vp

=

-
W
2

p
p
K A1
p
p
K
p
C C
H
vp
vp C
W.C.
p
RT
1 1 p
1
p
AD
W
vp


=

-
W
2

p
p
K WITH
p
p
K
p
gz C
H
vp
vp gz
W gz
p

R
1
AD
W
=
+

2 (1
WITH T
.
3)
T

p

K A1
C
H
vp

p

1 LR
p






AD
W
vp
vp
1 p
RT
1
LR
p
AD
W
AD
=

-
-
-
. 3
With
p

T

1
With


K
p
p

T p

K
12
WITH K
vp
T
C
vp
H
vp
C
C
H
vp
H


p

1 LR
p






AD
W
vp
vp
1 p
RT
1
LR
p
AD
W
AD
=

-
-
-
. 3
With
p

T

1
With


K
p
p

T p

K
12
WITH K
vp
T
gz
vp
H
vp
gz
gz
H
vp
H


p
1
LR 1
1
1

1
1
AD
W
vp
p


vp
pad
pad
RT

W
LR
p
AD



=

-

+
-
-

. 3

2
2
vp -
With +
T T
A1


K
p

1
vp
H
vp
T
T
T
T T







K
With
H
vp
K
T
H

T
Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 49/52

Appendix 5 Equivalence with formulations ANDRA

In order to be able to fit in platform ALLIANCE, it is necessary to be coherent with the formulations
posed by the ANDRA in the document [bib11]. We propose here an equivalence between the notations which
would be dissimilar. These differences relate to only the writing of:

· The equation of energy
· The law of Henry
· Diffusion in the liquid
· Diffusion in gas

Notice concerning the enthali:

It is required to have coherence between the two models which the user of Aster re-enters:
m
=
lq
H
= 0 and hm
L,
0
vp
0
0

A5.1 Equation of energy

The table above points out the two formulations:

Notations Aster
Notations ANDRA
m

lq
H
W
=

S N
W
W
m

vp
H
v
=

1
v (
Sw) N
m

have
H
have
=

1
have (
Sw) N
M
=
lq
W fw
M
=
have
have fas
M
=
vp
v fv

By rewriting the equation of energy of Aster with these notations, one finds:

D

F
F
F

+
W
have
v
Div
Div
W
+ have
+
Q
dt

S N
1 S N
1 S N
W
(-)
v
W
(- W) +


()

- (T - 0) D
T
([1 - N) C
S
S]
dt

D
dp
dp
+
2 dT
v

3 K T
3
3
9
0
0
+
m
C
T

T
TK
lq
- (mlq + mgz)
gz -
0
0
=
dt
dt
dt
dt
(F F F
W W +
v v +
have
have) G +

The first line being that of the ANDRA and others being a priori negligible.
Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 50/52

A5.2 Loi of Henry

ol
P M
In the formulation of the ANDRA, the formulation of Henry is given by
has
have
have
=
with
L
W
H M W
concentration of air in water that have it can bring back to a density such as A
=. H
L
AD
express yourself out of Pa.

In formulation ASTER, one recalls that the law of Henry is expressed in the form:

p
ol
AD
C =
with ol
have
C =
. K is expressed in Pa.m3.mol-1.
AD
ol
M
AD
K
H
AD
H

There is thus equivalence:
M W
K = H

H
W

A5.3 Diffusion of the vapor in the air

In formulation ANDRA the steam flow in the air according to the concentration of
steam d' in the air or the relative humidity is noted:

E
F
= - D
.

_

Diff
v
v
G
N
with the concentration defined as the molar report/ratio in gas: E

=
.
G
ng

In Aster, this same flow is written
: F
with the coefficient of Fick vapor
_ = F C
Diff
v
vp
vp
D


vp
F =
vp
and D
C 1
(- C)
vp the coefficient of diffusion of Fick of the gas mixture. Cvp is defined like
vp
vp
p
the report/ratio of the pressures such as:
vp
C =
.
vp
pgz

The law of perfect gases makes it possible to write that
E
C = thus
E
= C and F
.
_ = D
. C
vp
G
G
vp
Diff
v
v
vp
Thus equivalence ASTER/ANDRA is written simply:

F = D.
vp
v

A5.4 Diffusion of the air dissolved in water

In formulation ANDRA the flow of air dissolved in water is expressed

has
F

_
=D
.
_

ds E has
has
L

with A
AD
=
.
L
ol
M AD
Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 51/52

In Aster, this same flow is written: F
with the coefficient of air-dissolved Fick
_
_ = F
C
ds v has
AD
AD
D


AD
F =
AD
and D
C 1
(- C)
AD the coefficient of diffusion of Fick of the liquid mixture. Cad is definite such
AD
AD
that:
has
C = W. Thus:
AD
L
F = D.
AD
has

Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
HT-66/05/002/A

Code_Aster ®
Version
7.4
Titrate:
Models of behavior THHM


Date:
01/09/05
Author (S):
C. CHAVANT, Key S. GRANET
:
R7.01.11-B Page
: 52/52

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Handbook of Référence
R7.01 booklet: Modelings for Génie Civil and the géomatériaux ones
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