Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
1/8

Organization (S): EDF-R & D/AMA

Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
Document: V7.32.106

WTNP106 - Chauffage of a désaturé porous environment
with dissolved air

Summary:

One heats a porous environment of which the pores are filled with a mixture of water (liquid and vapor) and of air (dry and
dissolved in water). Initial saturation in liquid is 50%, the loading is a uniform heat flux
on the edges of the field. The modeling made by only one element corresponds to the modeling of one
homogeneous problem in space.

The reference solution is an approximate analytical solution.
Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
2/8

1
Problem of reference

1.1 Geometry

y
D
C
1
m
F
E
X
Z
h=
With
B
1m


Co-ordinates of the points (m):

To - 0,5 - 0,5
C
0,5 0,5
B
0,5 - 0,5
D - 0,5 0,5

Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
3/8

1.2
Properties of material

One gives here only the properties whose solution depends, knowing that the command file
contains other data of material (thermal conductivity, moduli of elasticity…) who finally
do not play any part in the solution of the dealt with problem.

Liquid water
Density (kg.m-3)
103
Heat with constant pressure (J.K-1)
4180
thermal dilation coefficient of the liquid (K-1) 0.
Dynamic viscosity of liquid water (Pa.s)
0.001
Permeability relating to water
Kr

W (S) = 1
Vapor
Specific heat (J.K-1)
1900
Initial enthalpy (latent heat of vaporization) 2,5E6.
J/kg
0,018
Mass molar (kg.mol-1)
Gas
Specific heat (J.K-1)
1900
Mass molar (kg.mol-1)
0,018
Permeability relating to gas
Kr

gz (S) = 1
Viscosity of the gas (kg.m-1.s-1)
1,8E-5
Dissolved air
Specific heat (J.K-1)
1900
Constant of Henry (Pa.m3.mol-1)
50000
Skeleton
Heat-storage capacity with constant constraint (J.K-1) 1050
Initial State
Porosity
0,3
Temperature (K)
300
Gas pressure (Pa)
1E5
Steam pressure (Pa)
3700
Initial saturation in liquid (Pa)
0,5
Constants
Constant of perfect gases
8,315
Coefficients
Homogenized density (kg.m-3)
2200
homogenized Isotherme of sorption
S (
=
- -
-
-

C
P)
12
0 5
.
10
(
0
0
C
P
vp
P
C
P)
With 0
vp
P = 3700
0
P

C = 0

1.3
Boundary conditions and loadings

On all the edges:
Heat flux
6
Q .n
ext.
= 10
Hydraulic flow no one

Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
4/8

2
Reference solution

2.1
Method of calculation

2.1.1 Calculation of the steam pressure starting from the temperature

We suppose the linear curve of saturation. It is thus written:

S = S0 + S
CP
éq
2.1.1-1
The equation [éq 2.2.3.3-2] of the reference document [R7.01.11] gives then:
m

= S P

W
W
C
m

= - 0 0
1 - S

0 - S 0
0
P

vp
(vp vp) (
)
vp
C
éq
2.1.1-2
m

= - 0
0
S

0 + S 0
0
P

AD
(AD AD)
AD
m

= - 0 0
1 - S

0 - S 0
0
P

have
(have ace) (
)
have
C

It is written that the total water mass and the total mass of air are preserved (because there is no flow
from gas water nor at the edge) and one obtains:

m
m

W +
vp = 0

(



éq
2.1.1-3
W -
vp) S
PC + (
0
vp -
vp) (1 - S
=
0)
0
m
m

AD +
have = 0

(





éq 2.1.1-4
AD -
have) S
PC + (
0
have -
have) (1 - S) + (
0
AD -
S
AD
=
0
) 0
0

[R7.01.11] [éq 4.1.4-1] gives in addition:

p
ol
ol
ol
ln
M
1
1
M
M
vp = vp (
-
)
0
0
(p p0
p
p0
p
p0
gz
-
) + vp
gz
0
(vp -) - vp
vp
0
+ (C -
c)+


p

RT
K
K
RT
vp
W
H
W
H
W
éq
2.1.1-5
ol
MR. R
T
M
vp
ln
0
(
ol
T
dT
p
p
H
H
vp
-
)

+ vp
gz
0

(mvp - MW)


K
T
R
2
0
T
W
H


T

Coupling of the equations [éq 2.1.1-3], [éq 2.1.1-4] and [éq 2.1.1-5], for which it is necessary to add the equation of
perfect gases for the vapor, the dry air and the dissolved air as well as the law of Henry are strongly a system
nonlinear that we will solve in small disturbances, which makes it possible to linearize it.
Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
5/8

All made calculations, one obtains:


ol
0
P
1 S M







vp (W -
vp
RT
0
vp)
(- 0)

S

+
- (W - 0 S P
P
S
0
vp)


W +


0


have (W - vp)


. 1
+






RT



K H



ol
0

RP
MR. P
- (

W -
0) S. have
vp
- vp vp
S
T
2
(1 - 0)

=

0

0


K H
RT


0
P





vp (
RT
ol
S
S
0
AD -
0
have) S)
- (0ad - 0as)

S
Pw + P



0
0
1
0
-

0
have (AD - have)
(
)

S. 1
+ Mr.
vp
+

+






0


K
K
RT



H
H


ol

RP
MR. P
- (0


AD -
0) S. have
have
- vp have
S
T
2
(1 - 0)

= 0


0


K H
RT


ol


1
M vp

P

vp -
+
Pw +

0
0

P

vp
RT
W

ol
ol
m
m


M vp P
M
H
H
have
R
vp
vp
W
T
0
0 (1 -
)
-


+
= 0


2
0

K T
R
W
H
T








éq 2.1.1-6

2.1.2 Calculation of the temperature

The equation [éq 3.2.4.3-1] of the reference document [R7.01.11] gives:
Q
m
= - T p

+ C0
3
T


éq
2.1.2-1
gz
gz
(since the other dilation coefficients are null).

The equation [éq 3.2.4.3-2] gives:
1
m
(Slq)
=
gz

éq
2.1.2-2
T
3
One thus obtains:
Q
= - (- S + + 0
1

éq
2.1.2-3
lq) (p
p
vp
have)
C T

In this problem, Q
is anything else only the heat brought per unit of volume.
By calling Vol the total volume of the part and Surf its side surface and T
the time of application
flows:

Surfing
Q
= T

Q
N
.
ext.
éq
2.1.2-4
Flight
Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
6/8

2.1.3 System to be solved


1
ol
S M
ol
0
0
M
(
RP
RT
W
- 0vp
) (0) vp
S+
- (W
- 0
vp
have
S
S p
S
S
0
vp)

- (1 - 0) 0
.
vp
- (W - 0
2
vp)
(W 0vp) 1.(-)



RT
0
RT
H
K
H
K





0
ol 0
RP
MR. P

0
RT
have
vp have

0
0
0
0
0
0
0
0
ol
S
1
0
- S

(AD
- have
)
S
- (AD
- have
)
S
- (AD
- have
)
S.
-
(1 - 0S) S
M
2
(AD have)
(0)

. 1 - + vp
K
0



.
+
0




H
RT

H
K
H
K
RT



1
ol
ol 0
ol m
vp
M
vp
M have
P
vp
M
vp
H - m
H


-
1
(-)
R +
W
0


0
0
0
0
2
P
RT
K T
R
0
vp
W
W H

T


0

- (
1 - lq
S)
0
C
- (
1 - lq
S)



vp
P
0




W
P
0

×
= Surfing

T T
Q.
ext.
N

Flight
P

have
0

éq
2.1.2-5

S

0
S
0
T
0
p


vp
0
vp
H
0
vp (calculated)
lq
5,00E-01 - 1,00E-12 3,00E+02 3,70E+03 2,50E+06 2,67E-02 1,00E+03








0
R
0

S (calculated)
S
C
p
C
C
C
lq L
p
vp
0
(calculated)
2,20E+03 3,00E-01 2,93E+03 1,05E+03 4,18E+03 1,90E+03 2,78E+06







Q
N
.
ext.

T

Surfing
Flight



1,00E+06
10
400
1,00E+04


The following results are obtained:

After resolution of this system, one obtains:

Pvp
4
.
29




Pw -

99500

=

T 144
.
0

P
7
.
45

have


What gives in term of result Aster (increment):

PRE1 PRE2
DT PVP
(V3)
9.95E4 7.5E1 1.44E-1 2.94E1

2.2 Uncertainties

Uncertainties are rather large because the analytical solution is an approximate solution of
fact of the linearization of the equations.
Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
7/8

3 Modeling
With

3.1
Characteristics of modeling A

Modeling in plane deformations. A Q8 element.
Discretization in time: only one step of time: 10 S.

3.2 Functionalities
tested

Order Option


AFFE_MODELE
D_PLAN_THH 2D


DEFI_MATERIAU
THM_LIQU


THM_GAZ
THM_VAPE_GAZ
THM_AIR_DISS
THM_DIFFU
THM_INIT
ELAS
AFFE_CHAR_MECA FLUX_THM_REP
FLUN_HYDR1

STAT_NON_LINE COMP_INCR RELATION
KIT_THH


RELATION_KIT
ELAS

LIQU_AD_GAZ_VAPE
HYDR_UTIL

3.3 Values
tested

Node Urgent Field Component
(S)
Reference
Aster Difference
(%)
(analytical)
NO1
DEPL
TEMP 10 S 0.1440 0.1439
0.08%

NO1
DEPL
PRE1
10 S
9.95 104 9.95
104 0.02%

NO1
DEPL
PRE2
10 S
75
73.3
2.21%

NO1
VARI_ELNO_ELGA V3
10 S
29.4
29.5
0.2%

Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

Code_Aster ®
Version
7.2
Titrate:
WTNP106 - Chauffage of a porous environment désaturé with dissolved air
Date:
13/10/04
Author (S):
S. GRANET, C. CHAVANT Key
:
V7.32.106-A Page:
8/8

4
Summary of the results

Solution ASTER is in very good agreement with the analytical solution except for the gas pressure.
The weak differences are due to the linearization.
Handbook of Validation
V7.32 booklet: Thermo hydro-mechanical in porous environment unsaturated
HT-66/04/005/A

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