research article coupled hydromechanical model of two-phase...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 589452 8 pageshttpdxdoiorg1011552013589452

Research ArticleCoupled Hydromechanical Model of Two-Phase Fluid Flow inDeformable Porous Media

You-Seong Kim and Jaehong Kim

Department of Civil Engineering Chonbuk National University Jeonju 561-756 Republic of Korea

Correspondence should be addressed to Jaehong Kim woghdjfkgmailcom

Received 3 April 2013 Revised 13 August 2013 Accepted 15 August 2013

Academic Editor Piermarco Cannarsa

Copyright copy 2013 Y-S Kim and J KimThis is an open access article distributed under the Creative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A model of solid-water-air coupling in triphasic mixtures is compared with solid-water coupling in biphasic mixtures with anapplication to partially saturated porous media Based on thermodynamics the mathematical framework governing the behaviorof a partially saturated soil is derived using balance equations and the numerical implementation and drainage tests of a soilcolumn are carried out to validate the obtained formulationsThe role of the air phase in the hydro-mechanical behavior of triphasicmixtures can be analyzed from the interactions among multiple phases for the constitutive behavior of a solid skeleton and thetriphasicmixturemodel can be applied in geotechnical engineering problems such as CO

2sequestration and air storage in aquifers

1 Introduction

For the features of the hydromechanical behavior of multi-phase porous media regarding solid-water-gas interactionsthe soil model has two kinds of constituent volume fractionsin geotechnical engineering problems One type is whenthe air phase is considered active in the voids of a three-phase soil as a gas phase and the other is to assume thegas phase as a vacuum in a soil mixture In soil mechanicsldquosaturatedrdquo means the mixture of soil particles and waterbut other fields such as earth science and multi-phasemedia define the mixture concept differently as shown inFigure 1

Depending on the role of the air phase some soil hasoccluded air bubbles in triphasic mixture and another onehas continuous air bubbles in biphasic mixtureThe air phaseplays a role in the air-water interface of partially saturatedsoil or contractile skin of hydromechanical behavior ofdeformable soil as a four-phase system [1] When performingstress analysis on an element Fredlund and Rahardjo [1]assumed that partially saturated soil can be visualized asa mixture with soil particles and a contractile skin thatapproaches equilibrium under applied stress gradients andair and water phases that flow under applied stress gradientsThe contractile skin acts like a thin rubbermembrane pullingthe particles together when the pore water pressure gradually

becomes negative during shrinkage-type experiments involv-ing the drying of a small soil specimen as it is exposed toatmosphere

Recently in order to observe coupled solid-water-airphenomenon in detail many researchers have formulatedconservation equations for a three-phase system consisting ofa solid and two immiscible fluids liquid and gas [2ndash7] Theseresearchers also derived expressions for the effective stresstensor in multi-phase porous media exhibiting two porosityscales micro- and macroporosity during the course ofloading Figure 2 illustrates the concept of partially saturatedporous media with double porosity The three-phase mixtureis composed of porous media consisting of solid and fluidsin the micropores and consisting of only fluid constituentsliquid and gas in the macropores as shown in Figure 2

The developments presented in the literature regardingmixtures of solid liquid and gas have been proposed as aconstitutive framework of a coupled model to simulate waterand air flow in deformable soil Finite element analysis ofpartially saturated soil is generally treated as a biphasic (ietwo field phases) mixture state in geotechnical engineering

2 Balance Equations andConstitutive Equations

Partially saturated soil which has governing equations basedon mixture theory consists of solid (119904) water (119908) and air (119886)

2 Mathematical Problems in Engineering

Unsaturated

GasVoids

(empty space)

Water

Solidgrains

Solidgrains

Water

UnsaturatedSaturated Unsaturated

Soil mechanicsMultiphase media

Figure 1 Schematic description of three-phase mixtures [14]

Equations regarding the volume and mass of a mixture aredefined as the mathematical relations [8 9] The volume of amixture is V = V

119904+ V119908+ V119886 and the corresponding total mass

is119898 = 119898119904+ 119898119908+ 119898119886 Similarly for the 120572 phase119898

120572= 120588120572119877V120572

(nearly homogeneous) where 120572 = 119904 119908 119886 and 120588120572119877 is the truemass density of the 120572 phaseThe volume fraction occupied bythe 120572 phase is given by 119899120572 = V

120572V and thus for water air and

solid

119899119904+ 119899 = 119899

119904+ 119899119908+ 119899119886= 1 (1)

where the porosity 119899 = (V119908+ V119886)V = 119899119908 + 119899119886 If a material

is homogeneous 119899120572 = V120572V whereas if it is heterogeneous

the volume fraction at a material point 119899120572 = dV120572dV for a

differential volume of the mixture The partial mass densityof the 120572 phase is given by 120588120572 = 119899120572120588120572119877 and thus

120588119904+ 120588119908+ 120588119886= 120588 (2)

where 120588 = 119898V is the total mass density of the mixtureAs a general notation phase designations in the superscriptform (eg 120588120572) pertain to average or partial quantities andthose in the superscript form with 119877 (eg 120588120572119877) to intrinsicor real quantities Based on the current configuration of themixture (for small strains theoretically not different fromthe reference or current configurations) the mass balanceequations describe the motions of the water and air phasesrelative to the motion of the solid phase In addition mixturetheory assumes that the three phases are smeared together at aspatial position in the current configuration thus limiting thetheory to a continuum representation of a partially saturatedsoil it is large enough length scale that the soil behaves as acontinuum

In order to solve for three unknowns (u 119901119908 119901119886) using

three equations given in (3)ndash(5) based on the formulationof Coussy [8] Borja [10] and de Boer [9] we can write thebalance of linear momentum for a triphasic mixture and the

Unsaturated

Water

Solid



UnsaturatedUnsaturated

Gas

Water

Gas

Solid

Voids(empty space) Water

Gas

Voids(empty space)

Porous media with double porosity

Solid and micropores

DescriptionMacropores

na

nw

ns

=na1

nw1

ns

+

na2

nw2

Figure 2 Schematic description of a mixture with double porosity[7]

balance of the mass of the solid water phase and air phase as[11]

div 120590 + 120588g = 0 (3)

119899

120597119878

120597119904

(119886minus 119908) + 119878 div v = minus div v119908 (4)

[minus119899

120597119878

120597119904

+

119899 (1 minus 119878)

119870119886

] 119886+ 119899

120597119878

120597119904

119908+ (1 minus 119878) div v

= minus

1

120588119886119877

div (120588119886119877v119886) (5)

where the total stress is written in terms of the effective stress1205901015840 (positive in tension) as 120590 = 1205901015840 minus 119901

1198861 + 119878(119901

119886minus 119901119908)1 [10 12]

and the degree of saturation 119878 is defined in the classical formof [13]

119878 = 119878119903+ 119878119890(1 minus 119878

119903)

119878119890= [1 + (120572119904)

119899]minus119898

(6)

where 119878119890is the effective degree of saturation 119878

119903is the

residual degree of saturation 119901119908and 119901

119886are the pore water

pressure and the pore air pressure (positive in compression)respectively and 120572 119899 and 119898 are the soil water characteristiccurve parameters (SWCC) g is the gravity accelerationvector v119908 = 119899119908(v

119908minusv119904) is the Darcy seepage velocity of water

where v119908is the true velocity of water and v

119904is the velocity of

the solid skeleton The constitutive equations are the linearisotropic elasticity for the effective stress and the generalizedDarcyrsquos law for the Darcy seepage velocity of water writtenrespectively as

1205901015840= c119890 120598

v120572 = 119896120572(119878 119899) [minusnabla119901

120572+ 120588120572119877g]

(7)

where 120572 is two constituents (water and air) the fourth-orderelastic modulus tensor is c119890 = 1205821 otimes 1 + 2120583I the Lameparameters are 120582 and 120583 120598 = symnablau is the symmetricsmall strain tensor and 119896

119908and 119896

119886are the partially saturated

hydraulic conductivity function written as

Mathematical Problems in Engineering 3

(1) water flow in a partially saturated porousmedium [13]

v119908 = 119896119908(119899 119878) [minus

120597119901119908

120597x+ 120588119908119877g ]

119896119908(119899 119878) =

120600 (119899)

120578119908

119896119903119908(119878)

119896119903119908(119878) = radic119878(1 minus (1 minus 119878

1119898)119898)

2

(8)

(2) air flow in a porous medium [8]

v119886 = 119896119886(119899 119878) [minus

120597119901119886

120597x+ 120588119886119877g]

119896119886(119899 119878) =

120600 (119899)

120578119886

119896119903119886(119878)

119896119903119886(119878) = radic1 minus 119878(1 minus 119878

1119898)

2119898

120588119886119877= exp [

119901119886

119870119886

] 120575 (119899) =

1198993

1 minus 1198992

(9)

where the material property 120600 is the intrinsic permeability ofthe soil skeleton and the function of the porosity 119899 119896

119903120572is the

relative permeabilities related to respectively water and airand 120578120572is the dynamic water and air viscosity The density of

air 120588119886119877 is approximately 12 kgm3 at sea level and at 20∘C andthe air bulk modulus 119870

119886is 105 Pa at a constant temperature

The relative permeabilities 119896119903119908

and 119896119903119886as a function of 119878 are

given in Figure 3It can be shown that120600(119899) = 1198972120575(119899) where 1198972 is a parameter

of dimension area (m2) and 120575(119899) by the Kozeny-Carmanrelation (pore space formed by regular packing of spheres)[8] for representing the porosity dependence of hydraulicconductivity The porosity 119899 is a function of the volumetricstrain of the solid skeleton

3 Weak Form and Coupled FiniteElement Formulation

It is assumed that the whole domain of the body 119861 is partiallysaturated Applying the method of weighted residuals [11 15]the coupled weak form for a triphasic mixture is written as

int

119861

[nablaw (1205901015840 minus 1198781199011199081 minus (1 minus 119878) 119901

1198861)] dV

= int

119861

120588w sdot g dV + intΓ119905

w sdot t d119886

int

119861119908

120578119899

120597119878

120597119904

(119886minus 119908) dV + int

119861119908

120578119878 div vdV minus int119861119908

nabla120578 sdot v119908dV

= int

Γ119908

119904

120578119878119908d119904

minus int

119861119886

120593119899

120597119878

120597119904

119886dV + int

119861119886

120593

119899 (1 minus 119878)

119870119886

119886dV + int

119861119886

120593 119899

120597119878

120597119904

119908dV

+int

119861119886

120593 (1minus119878)div v dV+int119861119886

120593

119870119886

nabla119901119886sdot v119886dV minus int

119861119886


= int

Γ119886

119904

120593119878119886d119904

(10)

where w is the weighting function for the displacementu t is the applied traction 120578 and 120593 are the weightingfunction for the pore water pressure (119901

119908) and pore air

pressure (119901119886) respectively and 119878119908 is the positive inwardwater

seepage on the boundary Γ119904 Assuming amixed finite element

formulation as indicated by the quadrilateral elements in theexample mesh in Figure 4 the discretized displacement uℎ isinterpolated biquadratically and the porewater pressure (119901ℎ

119908)

and pore air pressure (119901ℎ119886) bilinearly [15]

Introducing the shape functions and expressing inmatrixform the coupled nonlinear finite element form is written as

A119899el119890=1(c119890)119879 sdot [k119890119889119889 sdot d119890 minus k119890119889120579 (120579119890 120577119890) sdot 120579119890

minus k119890119889120577 (120579119890 120577119890) sdot 120577119890

= f119890119889119891(d119890 120579119890 120577119890) + f119890119889

119905]

A119899el119890=1(120572119890)119879

sdot [k119890120579120577 (d119890 120579119890 120577119890) sdot 120577119890

minus k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120579119889 (120579119890 120577119890) sdot d119890

+ k119890120579120579 (d119890 120579119890 120577119890) sdot 120579119890

= f119890120579 (d119890 120579119890 120577119890) + f119890120579119904]

A119899el119890=1(120573119890)119879

sdot [k119890120577120579 (d119890 120579119890 120577119890) sdot 120577119890

+ k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120577119889 (120579119890 120577119890) sdot d119890

+ k119890120577120577 (d119890 120579119890 120577119890) sdot 120577119890

= f119890120577 (d119890 120579119890 120577119890) + f119890120577119904]

(11)

where A119899el119890=1

is the element assembly operator for element 119890over the number of elements 119899el c119890 is the element nodalweighting function values for 120596ℎ 120572119890 and 120573119890 are the elementnodal weighting function values for 120578ℎ and 120593ℎ respectively(both of which are arbitrary except where they are zero atthe boundaries with essential boundary conditions) d119890 is theelement nodal displacement vector and 120579119890 and 120577119890 are theelement nodal pore water and air pressure vector

After applying the boundary conditions and assemblingthe finite element equations the matrix form is obtained


0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01

Degree of saturation (S)

Relat

ive p

erm

eabi

lity

of w

ater

(krw

)

(a)

0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of ai

r (kra

)

(b)

Figure 3 Relative permeabilities 119896119903119908

(a) and 119896119903119886(b) plotted against 119878 when varying119898 from 01 to 1

1 m

1 m

20 elements

z

Figure 4 Finite element mesh (20 elements) for numerical simula-tion

as the monolithically coupled nonlinear first-order ordinarydifferential equation to solve forD as

C (D) sdot D + K (D) sdotD = F (D) (12)

where

D = [[[

d120579

120577

]

]

]

D = [[

d120579

120577

]

]

C (D) = [[

0 0 0K120579119889 (D) minus K120579120577 (D) K120579120577 (D)K120577119889 (D) K120579120577 (D) K120577120579 (D)

]

]

Table 1 Soil parameters for triphasic mixture implementation

Soil parameters ValueYoungrsquos modulus 119864 13MPaPoisson ratio ] 04Solid real density 120588sR 2000 kgm3

Bulk modulus of air 119870119886

01MPaWater real density 120588wR 1000 kgm3

Air real density 120588aR 12 kgm3

Initial porosity 119899 02975Intrinsic permeability 120600 45 times 10

minus13m2

Water viscosity 120578119908

10 times 10minus3 Pas

Air viscosity 120578119886


K (D) = [[[

K119889119889 minus K119889120579 (D) minus K119889120577 (D)0 K120579120579 (D) 00 0 K120577120577 (D)

]

]

]

F (D) = [[[

f119890119889119891(D) + f119890119889

119905

f119890120579 (D) + f119890120579119904

f119890120577 (D) + f119890120577119904

]

]

]

=[

[

[

F119889 (D)F120579 (D)F120577 (D)

]

]

]

(13)

where C is the combination of the damping matrix and thestiffnessmatrix of the dof vector time derivative andK is thestiffness matrix Then the location matrix (LM) can be usedto assemble the individual 26 times 26 and 26 times 3 contributionsto the global ldquodampingrdquo matrix C the stiffness matrix K andthe forcing vector F and the matrix form uses generalizedtrapezoidal integration to solve transient equations [15]

For consolidation analysis the generalized trapezoidalrule is used to integrate transient problems by FE coupledbalance of mass and linear momentum equations at time119905119899+1

and derived from difference formulas forD119899+1

andV119899+1


Gravity load

Flow

Sand column

Flow out

Impermeable

Flow in No flow

No flow

Gravity load

Flow out

02 m

05 m

Undrainedcondition

At time = 0minus

At time = 0minus

At time = 0+

At time = 0+

(minus)

1 m

Figure 5 Diagram of the numerical analysis and the experimental test

10

08

06

04

02

00

Dep

th (m

)

minus10 minus8 minus6 minus4 minus2 0Pore water pessure (kPa)

5 min (coupled code)20 min (coupled code)60 min (coupled code)120 min (coupled code)

5 min (Schrefler 2001)20 min (Schrefler 2001)60 min (Schrefler 2001)120 min (Schrefler 2001)

Liakopoulos data

(a)

10

08

06

04

02

00

Dep

th (m

)

0 2 4 6 8 10Matric suction (kPa)



(b)

Figure 6 Comparison of the numerical results and experimental data with pore water pressure and matric suction

where velocity V119899+1

is D(119905119899+1) and 120572 is the time integration

parameter

C (D119899+1) sdot V119899+1+ K (D

119899+1) sdotD119899+1

= F119899+1(D119899+1)

D119899+1= D119899+ Δ119905V

119899+119886

V119899+119886= (1 minus 120572)V

119899+ 120572V119899+1

(14)

The form of (14) allows us to consider a semi-implicitintegration scheme of a linear form which is written as

C (D119899) sdot V119899+1+ K (D

119899) sdotD119899+1= F (D

119899) (15)

4 Numerical Results for Two-Fluid Flow in aDeformable Porous Medium

As previously mentioned since there is no exact solution forthe problem of water and air flow in deformable partially


10

08

06

04

02

00

Dep

th (m

)

minus7 minus6 minus5 minus4 minus3 minus2 minus1 0 1Pore air pressure (kPa)


5 min (Gawin 1997)30 min (Gawin 1997)60 min (Gawin 1997)120 min (Gawin 1997)

(a)

10

08

06

04

02

00

Dep

th (m

)




(b)

Figure 7 Comparison of numerical results of air pressure 119901119886

10

08

06

04

02

00

Dep

th (m

)

minus6 minus5 minus4 minus3 minus2 minus1 0 1Pore air pressure (kPa)


2 min (sand)5 min (sand)8 min (sand)10 min (sand)

(a)

10

08

06

04

02

00

Dep

th (m

)



5 min (weathered soil)20 min (weathered soil)60 min (weathered soil)

(b)

Figure 8 Comparison between the coupled code and the experimental air pressure data

saturated soils numerical modeling of the experimentalresults of the drainage of a sand column is performed Forvalidation and application of a couplingmodel of solid waterand air in partially saturated soils based on thewater drainageexperiment of a sand column described by Liakopoulos [16]the numerical solutions given by Schrefler and Scotta [17] andGawin et al [18] are compared to various results obtainedfrom the coupled model The mesh of this example whichis composed of a column of 20 nine-node isoparametric

Lagrangian elements of equal size was employed for allnumerical simulations Numerical integration was semi-implicit and the triphasic model associated with linearelasticity used mesh of 2D plain strain of nine integrationgauss points

Figure 5 shows the initial and boundary conditions (left)for numerical analysis and the procedure of the experimentaltest (right) A soil column test with a column 05m in heightis carried out to investigate the approximate value of pore


Schrefler and Scotta (2001)Gawin et al (1997)Coupled code

00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)

Figure 9 Displacement at the top surface of a drainage test in atriphasic mixture

air pressure for the validation of numerical results eventhough the simulation used a soil column 1m in height forthe numerical analysis

The physical experiment consisted of a soil column1m in high and a constant flow through the soil columncorresponding to a water pressure gradient initially equal tozero initially At the starting time steps the water inflow iscut at the top of the soil column and the water is flowedout at the bottom Air pressure is equal to atmosphericpressure at both the top and bottom of the column with zerovertical load at the top and no deformation at the bottomand on the lateral walls of the column The gravity-governedchanges in the constituent volume fractions only depend onthe soil and water parameters In the numerical test withthe same properties and boundary conditions implementedby Schrefler and Scotta [17] the coupled model also usesthe relationship of Brooks and Corey [19] for the relativepermeability of gas pressure and the experimental functionof Schrefler and Scotta [17] for the hydraulic properties of thesoil as shown in (16) The material properties used for thenumerical test are summarized in Table 1

One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121

119878 = 1 minus 19722 times 10minus11sdot 11990424279

119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)

where 120578120572is the dynamic viscosity and 119896

119903120572is the relative

permeability of the 120572 phase which depends on the relativesaturation 119878

120572through the experimental relationship 119896

119903120572=

119896119903120572(119878120572) 120600 is the intrinsic permeability and the respective

degrees of saturation 119878119908and 119878

119886sum to one 119878

119908+ 119878119886=

1 Even if the data of the mechanical behavior and theparameters of the Del Monte sand used by Liakopoulos[16] were missing and unpublished his solutions have beenobtained numerically by trial and error techniquesThus 120582 is01 and the residual saturation 119878

119903is 006689 for sand [17 18]

The triphasicmixture analysis of Schrefler and Scotta [17]which was results of the numerical solutions based on theLiakopoulos [16] are compared to those from coupled codeas shown in Figures 6ndash9 As no measurement of pore airpressure was made by Liakopoulos [16] the numerical resultsare plotted and also compared to other results [17 18] inFigure 7The evolution of air pressure is more sensitive to theanalysis method than that of water pressure The comparisonof pore water pressure in Figure 6(a) is similar to that ofSchrefler and Scotta [17] but the results (Figure 6(b)) of thecoupled code showed suction increases slower than thosefound by Schrefler and Scotta [17] since the air pressureresponse from the methods applied is sensitive as shown inthe suction evolution in Figure 7

Comparing with the two previous numerical results theair pressure profiles from the coupled model fit closer to thatof Gawin et al [18] than that of Schrefler and Scotta [17]These differences are produced by choosing different sets ofgoverning equations and numerical algorithms In particularthe averaged density of the mixture 120588 = (1 minus 119899)120588119904119877 + 119899119878120588119908119877 +(1 minus 119899)(1 minus 119878)120588

119886119877 and the bulk modulus of the solid grains(106MPa) and the water (2 times 103MPa) used by Schrefler andScotta [17] are different from those used by Gawin et al [18]and the coupled code Gawin et al [18] and the coupled codeboth derived the mass balance equation assuming the bulkmoduli (119870

119904and 119870

119908) are infinite due to the large values The

averaged density of the mixture is 120588 = (1 minus 119899)120588119904119877 + 119899119878120588119908119877 +119899(1 minus 119878)120588

119886119877For sand and weathered soil types experimental tests

are performed to investigate the pore air pressure at 5 cm10 cm and 15 cm place from top surface of soil sampleFigure 8 shows that experimental results are similar to thoseof the coupled code although the air pore pressures are justmeasured by three sensors at the top portion of the soilcolumn

As shown in Figure 9 the vertical displacements at thetop surface of the soil sample show little difference with timebut the final vertical displacements coincide with those ofSchrefler and Scotta [17] and Gawin et al [18] under identicalinitial conditions Because the coupled code has the hydraulicconductivity 119896

119908 which is the function of porosity the vertical

displacement of the coupled code deforms little faster thanthose of other numerical solutions at early time step

5 Conclusions

We have implemented a numerical integration algorithm(semi-implicit solution) for solid-water-air coupling finite


element formulation using balance equations Based onLiakopoulos [16]rsquos experimental results Gawin et al [18] andSchrefler and Scotta [17] presented numerical simulations forthe behavior and the diffusion of air pressure in a drainage testof a soil column In this study the developed coupled finiteelement model for a deformable partially saturated soil basedon linear isotropic elasticity describes the poromechanicalbehavior of a soil column by linking solid displacementpore water pressure and air pressure simultaneously Theresults of the coupled model approach the simulation ofdrainage test because it uses partially saturated permeabilitywhich is the function of porosity The numerical resultsof the coupled model are more similar to those of Gawinet al [18] rather than to those of Schrefler and Scotta[17] regarding the diffusion and dissipation of air pressurematric suction and vertical displacementThe coupledmodelwas validated through comparisons with the literature andthrough laboratory tests of the drainage of a soil columnand the results of two fluids flow obtained by semi-implicitlinear solution also demonstrate the stability of the solutionby comparing nonlinear models of Gawin et al [18] andSchrefler and Scotta [17]

Acknowledgments

This work was supported by the Energy Efficiency ampResources of the Korea Institute of Energy TechnologyEvaluation and Planning (KETEP) Grant funded by theKorea Government Ministry of Knowledge Economy (no20122020200010) and by research funds of ChonbukNationalUniversity in 2013

References

[1] D G Fredlund andH Rahardjo Soil Mechanics for UnsaturatedSoils John Wiley amp Sons New York NY USA 1993

[2] B A Schrefler and Z X Zhan ldquoA fully coupled model for waterflow and airflow in deformable porous mediardquoWater ResourcesResearch vol 29 no 1 pp 155ndash167 1993

[3] N Khalili and S Valliappan ldquoUnified theory of flow anddeformation in double porous mediardquo European Journal ofMechanics A vol 15 no 2 pp 321ndash336 1996

[4] N Khalili R Witt L Laloui L Vulliet and A Koliji ldquoEffectivestress in double porous media with two immiscible fluidsrdquoGeophysical Research Letters vol 32 no 15 Article ID L153092005

[5] W G Gray and B A Schrefler ldquoThermodynamic approach toeffective stress in partially saturated porous mediardquo EuropeanJournal of Mechanics A vol 20 no 4 pp 521ndash538 2001

[6] WG Gray and B A Schrefler ldquoAnalysis of the solid phase stresstensor in multiphase porous mediardquo International Journal forNumerical and Analytical Methods in Geomechanics vol 31 no4 pp 541ndash581 2007

[7] R I Borja and A Koliji ldquoOn the effective stress in unsaturatedporous continua with double porosityrdquo Journal of the Mechanicsand Physics of Solids vol 57 no 8 pp 1182ndash1193 2009

[8] O Coussy Poromechanics pp 45ndash51 157ndash168 John Wiley ampSons New York NY USA 2004

[9] R de Boer Trends in Continuum Mechanics of Porous MediaTheory and Applications of Transport in Porous Media SpringerNew York NY USA 2005

[10] R I Borja ldquoCam-Clay plasticity part V a mathematicalframework for three-phase deformation and strain localizationanalyses of partially saturated porous mediardquo Computer Meth-ods in Applied Mechanics and Engineering vol 193 pp 5301ndash5338 2004

[11] J Kim Plasticity modeling and coupled finite element analysis fopartially-saturated soils [PhD thesis] University of Colorado atBoulder Boulder Colo USA 2010

[12] R W Lewis and B A Schrefler The Finite Element Method inthe Deformation and Consolidation of Porous Media pp 6ndash20chapter 2 John Wiley amp Sons New York NY USA 1987

[13] M T van Genuchten ldquoClosed-form equation for predicting thehydraulic conductivity of unsaturated soilsrdquo Soil Science Societyof America Journal vol 44 no 5 pp 35ndash53 1980

[14] L Laloui G Klubertanz and L Vulliet ldquoSolid-liquid-air cou-pling in multiphase porous mediardquo International Journal forNumerical and Analytical Methods in Geomechanics vol 27 no3 pp 183ndash206 2003

[15] T J Hughes The Finite Element Method pp 1ndash51 57ndash75Prentice-Hall Upper Saddle River NJ USA 1987

[16] A C Liakopoulos Transient flow through unsaturated porousmedia [PhD thesis] University of California Berkeley CalifUSA 1965

[17] B A Schrefler and R Scotta ldquoA fully coupled dynamic modelfor two-phase fluid flow in deformable porous mediardquo Com-puter Methods in Applied Mechanics and Engineering vol 190no 24-25 pp 3223ndash3246 2001

[18] D Gawin L Simoni and B A Schrefler ldquoNumerical modelfor hydro-mechanical behaviour in deformable porous mediaa benchmark problemrdquo in Proceedings of the 9th InternationalConference on Computer Methods and Advances in Geomechan-ics pp 1143ndash1148 Wuhan China November 1997

[19] R N Brooks and A T Corey ldquoProperties of porous mediaaffecting fluid flowrdquo Journal of Irrigation Draining Division vol92 pp 61ndash88 1966

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Unsaturated

GasVoids

(empty space)

Water

Solidgrains

Solidgrains

Water



Figure 1 Schematic description of three-phase mixtures [14]

Equations regarding the volume and mass of a mixture aredefined as the mathematical relations [8 9] The volume of amixture is V = V

119904+ V119908+ V119886 and the corresponding total mass

is119898 = 119898119904+ 119898119908+ 119898119886 Similarly for the 120572 phase119898

120572= 120588120572119877V120572

(nearly homogeneous) where 120572 = 119904 119908 119886 and 120588120572119877 is the truemass density of the 120572 phaseThe volume fraction occupied bythe 120572 phase is given by 119899120572 = V

120572V and thus for water air and

solid

119899119904+ 119899 = 119899

119904+ 119899119908+ 119899119886= 1 (1)

where the porosity 119899 = (V119908+ V119886)V = 119899119908 + 119899119886 If a material

is homogeneous 119899120572 = V120572V whereas if it is heterogeneous

the volume fraction at a material point 119899120572 = dV120572dV for a

differential volume of the mixture The partial mass densityof the 120572 phase is given by 120588120572 = 119899120572120588120572119877 and thus

120588119904+ 120588119908+ 120588119886= 120588 (2)

where 120588 = 119898V is the total mass density of the mixtureAs a general notation phase designations in the superscriptform (eg 120588120572) pertain to average or partial quantities andthose in the superscript form with 119877 (eg 120588120572119877) to intrinsicor real quantities Based on the current configuration of themixture (for small strains theoretically not different fromthe reference or current configurations) the mass balanceequations describe the motions of the water and air phasesrelative to the motion of the solid phase In addition mixturetheory assumes that the three phases are smeared together at aspatial position in the current configuration thus limiting thetheory to a continuum representation of a partially saturatedsoil it is large enough length scale that the soil behaves as acontinuum

In order to solve for three unknowns (u 119901119908 119901119886) using

three equations given in (3)ndash(5) based on the formulationof Coussy [8] Borja [10] and de Boer [9] we can write thebalance of linear momentum for a triphasic mixture and the

Unsaturated

Water

Solid



UnsaturatedUnsaturated

Gas

Water

Gas

Solid

Voids(empty space) Water

Gas

Voids(empty space)

Porous media with double porosity

Solid and micropores

DescriptionMacropores

na

nw

ns

=na1

nw1

ns

+

na2

nw2

Figure 2 Schematic description of a mixture with double porosity[7]

balance of the mass of the solid water phase and air phase as[11]

div 120590 + 120588g = 0 (3)

119899

120597119878

120597119904

(119886minus 119908) + 119878 div v = minus div v119908 (4)

[minus119899

120597119878

120597119904

+

119899 (1 minus 119878)

119870119886

] 119886+ 119899

120597119878

120597119904

119908+ (1 minus 119878) div v

= minus

1

120588119886119877

div (120588119886119877v119886) (5)

where the total stress is written in terms of the effective stress1205901015840 (positive in tension) as 120590 = 1205901015840 minus 119901

1198861 + 119878(119901

119886minus 119901119908)1 [10 12]

and the degree of saturation 119878 is defined in the classical formof [13]

119878 = 119878119903+ 119878119890(1 minus 119878

119903)

119878119890= [1 + (120572119904)

119899]minus119898

(6)

where 119878119890is the effective degree of saturation 119878

119903is the

residual degree of saturation 119901119908and 119901

119886are the pore water

pressure and the pore air pressure (positive in compression)respectively and 120572 119899 and 119898 are the soil water characteristiccurve parameters (SWCC) g is the gravity accelerationvector v119908 = 119899119908(v

119908minusv119904) is the Darcy seepage velocity of water

where v119908is the true velocity of water and v

119904is the velocity of

the solid skeleton The constitutive equations are the linearisotropic elasticity for the effective stress and the generalizedDarcyrsquos law for the Darcy seepage velocity of water writtenrespectively as

1205901015840= c119890 120598

v120572 = 119896120572(119878 119899) [minusnabla119901

120572+ 120588120572119877g]

(7)

where 120572 is two constituents (water and air) the fourth-orderelastic modulus tensor is c119890 = 1205821 otimes 1 + 2120583I the Lameparameters are 120582 and 120583 120598 = symnablau is the symmetricsmall strain tensor and 119896

119908and 119896

119886are the partially saturated

hydraulic conductivity function written as



v119908 = 119896119908(119899 119878) [minus

120597119901119908

120597x+ 120588119908119877g ]

119896119908(119899 119878) =

120600 (119899)

120578119908

119896119903119908(119878)


1119898)119898)

2

(8)


v119886 = 119896119886(119899 119878) [minus

120597119901119886

120597x+ 120588119886119877g]

119896119886(119899 119878) =

120600 (119899)

120578119886

119896119903119886(119878)


1119898)

2119898

120588119886119877= exp [

119901119886

119870119886

] 120575 (119899) =

1198993

1 minus 1198992

(9)


119903120572is the










int

119861


1198861)] dV

= int

119861


w sdot t d119886

int

119861119908

120578119899

120597119878

120597119904

(119886minus 119908) dV + int

119861119908

120578119878 div vdV minus int119861119908


= int

Γ119908

119904

120578119878119908d119904

minus int

119861119886

120593119899

120597119878

120597119904

119886dV + int

119861119886

120593

119899 (1 minus 119878)

119870119886

119886dV + int

119861119886

120593 119899

120597119878

120597119904

119908dV

+int

119861119886

120593 (1minus119878)div v dV+int119861119886

120593

119870119886


119861119886


= int

Γ119886

119904

120593119878119886d119904

(10)






119908)




minus k119890119889120577 (120579119890 120577119890) sdot 120577119890

= f119890119889119891(d119890 120579119890 120577119890) + f119890119889

119905]

A119899el119890=1(120572119890)119879

sdot [k119890120579120577 (d119890 120579119890 120577119890) sdot 120577119890

minus k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120579119889 (120579119890 120577119890) sdot d119890

+ k119890120579120579 (d119890 120579119890 120577119890) sdot 120579119890

= f119890120579 (d119890 120579119890 120577119890) + f119890120579119904]

A119899el119890=1(120573119890)119879

sdot [k119890120577120579 (d119890 120579119890 120577119890) sdot 120577119890

+ k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120577119889 (120579119890 120577119890) sdot d119890

+ k119890120577120577 (d119890 120579119890 120577119890) sdot 120577119890

= f119890120577 (d119890 120579119890 120577119890) + f119890120577119904]

(11)

where A119899el119890=1




0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of w

ater

(krw

)

(a)

0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of ai

r (kra

)

(b)



1 m

1 m

20 elements

z




where

D = [[[

d120579

120577

]

]

]

D = [[

d120579

120577

]

]

C (D) = [[


]

]







minus13m2





K (D) = [[[


]

]

]

F (D) = [[[

f119890119889119891(D) + f119890119889

119905

f119890120579 (D) + f119890120579119904

f119890120577 (D) + f119890120577119904

]

]

]

=[

[

[

F119889 (D)F120579 (D)F120577 (D)

]

]

]

(13)




andV119899+1


Gravity load

Flow

Sand column

Flow out

Impermeable

Flow in No flow

No flow

Gravity load

Flow out

02 m

05 m

Undrainedcondition

At time = 0minus

At time = 0minus

At time = 0+

At time = 0+

(minus)

1 m


10

08

06

04

02

00

Dep

th (m

)




Liakopoulos data

(a)

10

08

06

04

02

00

Dep

th (m

)




(b)




parameter

C (D119899+1) sdot V119899+1+ K (D

119899+1) sdotD119899+1

= F119899+1(D119899+1)

D119899+1= D119899+ Δ119905V

119899+119886

V119899+119886= (1 minus 120572)V

119899+ 120572V119899+1

(14)


C (D119899) sdot V119899+1+ K (D

119899) sdotD119899+1= F (D

119899) (15)




10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)


10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)







00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











v119908 = 119896119908(119899 119878) [minus

120597119901119908

120597x+ 120588119908119877g ]

119896119908(119899 119878) =

120600 (119899)

120578119908

119896119903119908(119878)


1119898)119898)

2

(8)


v119886 = 119896119886(119899 119878) [minus

120597119901119886

120597x+ 120588119886119877g]

119896119886(119899 119878) =

120600 (119899)

120578119886

119896119903119886(119878)


1119898)

2119898

120588119886119877= exp [

119901119886

119870119886

] 120575 (119899) =

1198993

1 minus 1198992

(9)


119903120572is the










int

119861


1198861)] dV

= int

119861


w sdot t d119886

int

119861119908

120578119899

120597119878

120597119904

(119886minus 119908) dV + int

119861119908

120578119878 div vdV minus int119861119908


= int

Γ119908

119904

120578119878119908d119904

minus int

119861119886

120593119899

120597119878

120597119904

119886dV + int

119861119886

120593

119899 (1 minus 119878)

119870119886

119886dV + int

119861119886

120593 119899

120597119878

120597119904

119908dV

+int

119861119886

120593 (1minus119878)div v dV+int119861119886

120593

119870119886


119861119886


= int

Γ119886

119904

120593119878119886d119904

(10)






119908)




minus k119890119889120577 (120579119890 120577119890) sdot 120577119890

= f119890119889119891(d119890 120579119890 120577119890) + f119890119889

119905]

A119899el119890=1(120572119890)119879

sdot [k119890120579120577 (d119890 120579119890 120577119890) sdot 120577119890

minus k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120579119889 (120579119890 120577119890) sdot d119890

+ k119890120579120579 (d119890 120579119890 120577119890) sdot 120579119890

= f119890120579 (d119890 120579119890 120577119890) + f119890120579119904]

A119899el119890=1(120573119890)119879

sdot [k119890120577120579 (d119890 120579119890 120577119890) sdot 120577119890

+ k119890120579120577 (d119890 120579119890 120577119890) sdot 120579119890

+ k119890120577119889 (120579119890 120577119890) sdot d119890

+ k119890120577120577 (d119890 120579119890 120577119890) sdot 120577119890

= f119890120577 (d119890 120579119890 120577119890) + f119890120577119904]

(11)

where A119899el119890=1




0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of w

ater

(krw

)

(a)

0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of ai

r (kra

)

(b)



1 m

1 m

20 elements

z




where

D = [[[

d120579

120577

]

]

]

D = [[

d120579

120577

]

]

C (D) = [[


]

]







minus13m2





K (D) = [[[


]

]

]

F (D) = [[[

f119890119889119891(D) + f119890119889

119905

f119890120579 (D) + f119890120579119904

f119890120577 (D) + f119890120577119904

]

]

]

=[

[

[

F119889 (D)F120579 (D)F120577 (D)

]

]

]

(13)




andV119899+1


Gravity load

Flow

Sand column

Flow out

Impermeable

Flow in No flow

No flow

Gravity load

Flow out

02 m

05 m

Undrainedcondition

At time = 0minus

At time = 0minus

At time = 0+

At time = 0+

(minus)

1 m


10

08

06

04

02

00

Dep

th (m

)




Liakopoulos data

(a)

10

08

06

04

02

00

Dep

th (m

)




(b)




parameter

C (D119899+1) sdot V119899+1+ K (D

119899+1) sdotD119899+1

= F119899+1(D119899+1)

D119899+1= D119899+ Δ119905V

119899+119886

V119899+119886= (1 minus 120572)V

119899+ 120572V119899+1

(14)


C (D119899) sdot V119899+1+ K (D

119899) sdotD119899+1= F (D

119899) (15)




10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)


10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)







00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of w

ater

(krw

)

(a)

0

02

04

06

08

1

0 02 04 06 08 1

m = 1

m = 01


Relat

ive p

erm

eabi

lity

of ai

r (kra

)

(b)



1 m

1 m

20 elements

z




where

D = [[[

d120579

120577

]

]

]

D = [[

d120579

120577

]

]

C (D) = [[


]

]







minus13m2





K (D) = [[[


]

]

]

F (D) = [[[

f119890119889119891(D) + f119890119889

119905

f119890120579 (D) + f119890120579119904

f119890120577 (D) + f119890120577119904

]

]

]

=[

[

[

F119889 (D)F120579 (D)F120577 (D)

]

]

]

(13)




andV119899+1


Gravity load

Flow

Sand column

Flow out

Impermeable

Flow in No flow

No flow

Gravity load

Flow out

02 m

05 m

Undrainedcondition

At time = 0minus

At time = 0minus

At time = 0+

At time = 0+

(minus)

1 m


10

08

06

04

02

00

Dep

th (m

)




Liakopoulos data

(a)

10

08

06

04

02

00

Dep

th (m

)




(b)




parameter

C (D119899+1) sdot V119899+1+ K (D

119899+1) sdotD119899+1

= F119899+1(D119899+1)

D119899+1= D119899+ Δ119905V

119899+119886

V119899+119886= (1 minus 120572)V

119899+ 120572V119899+1

(14)


C (D119899) sdot V119899+1+ K (D

119899) sdotD119899+1= F (D

119899) (15)




10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)


10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)







00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Gravity load

Flow

Sand column

Flow out

Impermeable

Flow in No flow

No flow

Gravity load

Flow out

02 m

05 m

Undrainedcondition

At time = 0minus

At time = 0minus

At time = 0+

At time = 0+

(minus)

1 m


10

08

06

04

02

00

Dep

th (m

)




Liakopoulos data

(a)

10

08

06

04

02

00

Dep

th (m

)




(b)




parameter

C (D119899+1) sdot V119899+1+ K (D

119899+1) sdotD119899+1

= F119899+1(D119899+1)

D119899+1= D119899+ Δ119905V

119899+119886

V119899+119886= (1 minus 120572)V

119899+ 120572V119899+1

(14)


C (D119899) sdot V119899+1+ K (D

119899) sdotD119899+1= F (D

119899) (15)




10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)


10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)







00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)


10

08

06

04

02

00

Dep

th (m

)




(a)

10

08

06

04

02

00

Dep

th (m

)




(b)







00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











00

minus05

minus10

minus15

minus20

minus25

minus30

minus35

minus40

Disp

lace

men

t (m

m)

0 20 40 60 80 100 120Time (min)




One has

119896119903119886= (1 minus 119878

119890)2

(1 minus 119878(2+3120582)120582

119890)

119878119890=

119878 minus 119878119903

1 minus 119878119903

119896119903119908= 1 minus 2207(1 minus 119878

119890)10121


119896120572

120600

120578120572

119896119903120572

120572 = 119908 119886

(16)





119903120572=



119886sum to one 119878

119908+ 119878119886=


119903is 006689 for sand [17 18]




119904and 119870








5 Conclusions




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article coupled hydromechanical model of two-phase...

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