A finite volume scheme for the transport of radionucleides in porous media

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A finite volume scheme for the transport of radionucleides in porous media

Eric Chénier, Robert Eymard, Xavier Nicolas

To cite this version:

Eric Chénier, Robert Eymard, Xavier Nicolas. A finite volume scheme for the transport of ra- dionucleides in porous media. Computational Geosciences, Springer Verlag, 2004, 8 (2), pp.163-172.

�10.1023/B:COMG.0000035077.63408.71�. �hal-00695261�

E. Chenier, R.Eymard and X.Niolas

UniversitedeMarne-la-Vallee

Abstrat. Thispaperpresentstheuseofanitevolumeshemeforthesimulation

of theCOUPLEX1Testase.We rstshowthat theresultsof thesimulation an

bemainlypreditedthankstoananalysisofthedata.Wethengivetheformulation

of anitevolumesheme, yieldingaurateand stable resultsfor alow omputa-

tionalost.Wenallypresentsomeofthenumerialresults,omparingtheexpliit

MUSCL shemeandtheimpliit shemewithvariableupwindingaordingtothe

loaldiusion.

Keywords:COUPLEX1,nitevolumesheme,variablePeletnumber

AMS Subjet Classiations::35K65,35K55

1. Introdution

TheCOUPLEX-1Testase(ANDRA,2001)isabenhmarkofnumer-

ialtehniquesdesignedforthesimulationofthetransportofontami-

nantsbythewaterowingthroughaporousmedium.Sinethegoalis

more to ompare numerial methods thanto improve the engineering

study, the geologial onguration is simplied in a four-layer ross-

setion. Nevertheless, within the framework of an engineering study,

the numerial results are neessarily ompared to simple alulations

without a omputer,even under oarseapproximations. Thus we rst

presentinSetion 2an approximateanalysiswhihgivessome india-

tionsonthequalitativeresultswhihanbeaprioriexpetedfromthe

dataof theproblem. Weseondlygive,inSetion3,a shortdisussion

on the appropriate numerial shemes. We then present in Setion 4

some of thenumerial resultswhih have beenobtained using, on the

one hand, an expliit MUSCL sheme for the transport part of the

problem,ontheotherhand,animpliitshemewithvariableupwinding

aording to theloaldiusion.

Laboratoired'EtudedesTransfertsd'EnergieetdeMatiere,UMLV

(http://www-letem.univ-mlv.fr) 5, boulevard Desartes - Champs-sur-Marne -

77454 Marne-la-ValleeCedex2,FRANCE

2. Approximate analytial study

We rst begin witha global studyof the problem,mainly usinghand

alulations. Suha proedureis ommonlyfollowedinindustrialon-

texts. The data are realled, for the sake of ompleteness, in the ap-

pendixof thispaper.

2.1. Charateristis of the veloityfield

Sine the geologial desription of the domain redues to two highly

permeablethiklayers(limestoneanddogger)separatedbytwoweakly

permeablethinlayers(marlandlay),(seegure1)thehydraulihead

eld an be approximately evaluated usingthevolumiowonserva-

tion ina 1Dmediumforeah ofthepermeablelayers:

thisyieldsaonstantheadgradientinthedoggerlayer,leadingto

a linearlydereasingheadfrom thebottom right vertialboundaryto

the bottom left vertialboundary,

thisalsoyieldsthefollowingequationforthehydrauliheadinthe

limestonelayer(sinethelowerboundaryofthislayeristilted,thearea

of the vertial setions is given by a linear funtion of the horizontal

positionx)

300

300 245

25000 x

H

x

=onstant ; (1)

withx=0attheleftboundaryandx=25000mattherightbound-

ary. Therefore a logarithmi head prole is available in the limestone

layer, dereasingfromthe right to theleft.

Figure1. Aprioriontourlevelsofthepiezometrihead

It yields the following values for the hydrauli head in the dogger

layer:

H(x)=286+

289 286

25000

x; (2)

also available at the bottom horizontal boundary of the lay layer,

and

H(x)=200+

110

ln(245=300) ln

1

300 245

30025000 x

: (3)

in the limestone layer, also available at thetop tilted boundaryof

the laylayer.

Therefore, using equations (2) and (3), the piezometri head H is

expeted to beapproximatelyequal to H

b l

=288:2 m at thebottom

boundary of the lay layer and H

t l

= 278:9 m at the top boundary

of the lay layer at the level of the left side of the repository (x =

18440m).Inthesameway,thepiezometriheadapproximatelyequals

H

b r

= 288:6 m at the bottom boundary and H

t r

= 294:0 m at

the topboundaryof the lay layer at the levelof the right sideof the

repository(x=21680 m).

Sine the vertial dimension of the domain is muh smaller than

thehorizontalone,thevertialowwillpredominate inthelaylayer,

and theDaryveloityan therefore beomputedusingthedierene

of pressure between the top boundary and the bottom boundary on

a vertial line. Thus, the hange of sign of this dierene along the

repository onrms that its loation has been hosen suh that the

veloityeldvanishes at thelevelof therepository.

Taking this result into aount and sine the thikness of the lay

layervariesbetween135:6mand142:7mat theleveloftherepository,

the maximum upward veloity at the left side of the repository is

about u

max l

= 3:1510

6 278:9 288:2

135:6

' 2:210 7

m=year and

the maximum downward veloity at its right side is about u

max r

=

3:1510

6 294:0 288:6

142:7

' 1:210 7

m=year.

The x-oordinate at whih the vertial pressure gradient in the

lay is vanishing is then given by x = 18440 +(21680 18440)

2:2

2:2+1:2

'20500m.Theaurayofthisvalueshouldbedisussed,sine

asmallunertaintyonthepiezometriheadleadstoamore important

unertaintyon thisposition.

2.2. Charateristi times for theiodine transport

From the previous values of piezometri heads (H

b l

and H

t l ) and

veloities (u

max l and u

max r

), one an guess dierent harateristi

time values:

let d

t

= 85:6 m be the shorter distane between the repository

and the top boundary of the lay layer; the onvetive time of iodine

to thisboundaryist

onv t i

=

!

i R

i dt

ju

max l j

,whihisabout 10

3

185:6

2:210 7

=

3:910 5

years;

let d

b

= 44 m be the distane between the repository and the

bottom boundary of the lay layer; the onvetive time of iodine to

this boundary is t

onv b i

=

!

i R

i d

b

jumax rj

, whih is about 10

3

144

1:210 7

=

3:610 5

years;

thediusiontimeinthelaylayeralulated fromtherepository

to the top boundary is about t

diff t i

=

!

i R

i d

2

t

D

i

= 10

3

185:6 2

9:510 7

=

7:710 6

years;

thediusiontimeinthelaylayeralulated fromtherepository

to thebottom boundaryis about t

diff b i

=

!

i R

i d

2

b

D

i

= 10

3

144 2

9:510 7

=

2:010 6

years;

the onvetive time in the dogger layer from the repository to

the leftbottom boundaryofthe omputationaldomain(distaned

l

=

18440 m) is about t

onv dog i

=

!

i R

i d

l

K

dog rH

=

0:1118440

25:2(288:2 286)=18440

=

6:110 5

years;

the onvetive time inthe limestone layerfrom the repository to

the left middleboundaryof the omputationaldomain (distaned

l

=

18440 m) is about t

onv l im i

=

!iRid

l

K

lim rH

=

0:1118440

6:3(278:9 200)=18440

=

6:810 4

years.

This rst estimationof theharateristi times of theiodine trans-

portinthelaylayershowsthatt

diff t i

>t

diff b i

>t

onv t i

>

t

onv b i

. Therefore, in this layer, the diusive transport is slower

than the onvetive transport and iodine reahes the bottom bound-

ary of the lay layer before thetop boundary. Furthermore, sine the

harateristi times in the lay layer and the onvetive times in the

limestoneanddoggerlayersaremuhsmallerthanthehalflifeofiodine

(1:5710 7

years),nearlythefullamountoftheiodinewhihwillreah

thelimestoneanddoggerlayerswillpropagateuntiltheleftboundaryof

theomputationaldomain, beforethedeayoftheiodineradioativity

begins.

2.3. Charateristi times for theplutonium transport

The harateristi timevaluesforplutoniumarequitedierent:

theonvetive timeof plutoniumto thetopboundaryofthelay

layerisaboutt

onv t p

=

!pRpdt

u

max l

= 0:210

5

85:6

2:210 7

=7:810 12

years;

the onvetive time of plutonium to the bottom boundary of

the lay layer is about t

onv b p

=

!pRpd

b

umax

r

= 0:210

5

44

1:210 7

= 7:3

10 12

years;

the diusiontime of plutonium in the lay layer alulated from

the repository to the top boundary is about t

diff t p

=

!pRpd 2

t

D

p

=

0:210 5

85:6 2

4:410 4

=3:310 11

years;

the diusiontime of plutonium in the lay layer alulated from

therepositorytothebottomboundaryisaboutt

diff b p

=

!pRpd 2

b

Dp

=

0:210 5

44 2

4:410 4

=8:810 10

years;

Therefore, the onvetive times, diusive times and half life time

(3:7610 5

years)ofplutoniumareverydierent.Themainmehanism

isthereforethedeayofplutoniumbeforeitisonvetedordiused.No

signiativeamountofradioativeplutoniumanreahtheboundaries

of thelaylayer.

2.4. Conlusion of this first survey

Wehavebeenabletogiveoarsepreditionsoftheamountofradionu-

leideswhih reahestheleft boundaryofthedomain, usingonlyvery

simple alulations. This is mainly due to the fat that the data are

here very simple, ompared to realisti ones. However, from ourpoint

ofview,suhaproedure mustalwaystakeplae beforeanynumerial

study,beause it gives the physialkey points whih help to validate

the numerial results.

3. Numerial shemes

Thefollowinganalysisoftherequirednumerialshemesanbemade.

The problem is a onvetion-diusion problem with a heteroge-

neousanisotropidiusionmatrix(duetotheexpressionofthedisper-

sionmatrix).ThereforesomeadvantagesanbedrawnfromaP1-nite

element formulation (lineareldson triangles),viewed as a nitevol-

ume methodonthedualmesh(givenbyorthogonalbissetors)forthe

onvetionterms(Eymard, Gallouet, Herbin, 2000).

Thehighratiobetweenthehorizontalandthevertialdimensions

impliesto usemeshes designedforthispurpose.

Thedispersionterms areinompetitionwiththenumerialdiu-

sionprovidedbytherstorderupstreamweightednitevolumesheme

fortheonvetionterm(a entered shemeannotbeusedeverywhere

beause of the onstrast between the dierent rok properties within

thedomain).Thelowvertialsizeofthedispersionmatriximpliesthat

thisterman be auratelyhandledinthe vertialdiretion,butthat

it isneessarily partly inreasedin thehorizontal diretion.

We have therefore used the followingshemes. A P1-nite element

shemeis usedto solvetheequation

divrH =0; (4)

in whih we denote by isotropi heterogeneous value of the per-

meability.We writethisshemeasfollows

X

L2N

K T

KL (H

L H

K

)=0; (5)

where K is a vertex of the mesh, and N

K

isthe set of the verties

of all triangleshavingK asavertex (see gure2,left).

K

N = { } _K

K

Figure2. Triangularmesh:neighboursofavertex(left),dualmesh(right).

In equation (5), we denote byT

KL

the termof the rigiditymatrix,

given by

T

KL

= Z

r'

K

(x)(x)r'

L

(x)dx; (6)

denoting by '

K

the P1 basis funtion, whih is linear in eah tri-

angle, ontinuous, whose value is 1 at the vertex K and 0 at all the

other verties. Asweremarked above,ru(andtherefore ofD(ru)) is

onstant ineah triangle. Note that theP1 niteelement sheme an

also be seen as a nite volume sheme on the dual mesh: see gure 2

(right), whih shows theVorono meshrelated to the triangular mesh

of gure 2(left).

We an nowwrite, inthesame way,a nitevolumesheme forthe

onentrations, setting the volumiow between two ontrol volumes

around vertiesK and Lby

Q

K ;L

=T

KL (H

K H

L

) (7)

and setting initialvaluesforonentrations

(0)

K

=0: (8)

The sheme writes,at agiven timestep n,

!

K m

K h

( (n+1)

K

(n)

K

)=t+ (n+1)

K i

+

X

L2N

K

"

(m)

K ;L Q

+

K ;L

(m)

L;K Q

+

L;K

D

KL (

(m)

L

(m)

K )

#

=f (n)

K

(9)

where the diusive oeÆient D

KL

is given using the dispersion

matrix by

D

KL

= Z

r'

K

(x)D(rH)(x)r'

L

(x)dx: (10)

In(9),m

K

istheareaoftheontrolvolumeK aroundthevertexK

(itisinfattakenas1=3oftheareasofalltheneighbouringtriangles),

!

K

gathers the eets of the eetive porosity and of theretardation

fator, f (n)

K

denotes the soure term, and in (10), D(rH)(x) denotes

the dispersionmatrix in thedomain asa funtion of theapproximate

gradientofhydraulihead.Anexpliiteshemeisobtainedwithm=n

(the timestep t isthen boundedto respetaCFL value:thisbound

is approximately 77 years for the mesh whih is used below) and an

impliite one is given by m = n+1 (in this ase, the time step t is

ajusted along the simulation to yield maximum variations of onen-

tration equal to 10 4

). We now have to dene the way of omputing

the valuesof theinterfae onentration (m)

K ;L

used in(9).

In the expliite ase we use a MUSCL sheme. We rst ompute

an approximate value forr (n)

inall theontrol volumes around the

verties. We then limit this gradient in order to obtain that all the

values

(m)

K ;L

= (n)

K +

1

2

~

r (n)

K

~

KL (11)

bebetween (n)

K and

(n)

L

,foranypairofvertiesofthesametriangle

(in (11), we denote by

~

r (n)

K

thelimited gradient).

Intheimpliitease,thefollowingupwindingshemean beused:

(m)

K ;L

= (n+1)

K

: (12)

Sheme(12)willbereferred,inthefollowing,asthe\upwinding"im-

pliitsheme.TakingintoaountsomedisussionsduringtheCouplex

Workshop,we introduedthesheme

(m)

K ;L

=(1

K ;L )

(n+1)

K

+

K ;L

(n+1)

L

; (13)

forall pairsK ;L ofneighbouringverties.In (13),

K ;L

is given by

Q +

K ;L

K ;L

=min

D

KL

; 1

2 Q

+

K ;L

(14)

Sheme (13)-(14) thus only adds the minimum needed numerial

diusionforthestabilityofthesheme(whihthussatisesaondition

on the loal Pelet number). Note that (13)-(14) an even add some

diusion ifD

KL

<0 (whih an ourwith a non diagonal dispersion

matrix),yieldingastrongrespetoftheloalmaximumpriniple.This

sheme will be referred, in the following, as the \entered" impliit

sheme.

Using a diret Gauss band solver for all linear systems, we have

obtained the following run times on a PC omputer (lok frequeny

500 MHz):

usingtheimpliitsheme,therun-timeshavebeenofabout1hour

and 10 minutes for23906verties,

a 5 hours omputing time has been neessary for the MUSCL

expliitsheme on thesame grid.

4. Numerial results

Wepresent hereafterthemainresultsonerningthepiezometrihead

eld andtheIodinetransport(inagreement withsetion2,thePluto-

niumtransportisnotnumeriallysigniant).

4.1. Numerial results: thepiezometri head field

Figure 3 shows the grid used, and ontour levels for the piezometri

head (a value equal to 180 m is given at the upper left orner, then

theontourlevels 196,212,228,244,260,from 276to 292 withstep1,

308, 324 m are shown from the left to theright, and a value equal to

340 m is givenat the upperright orner).

The obtained values of H are very losed to that whih have been

given in Setion 2 in the limestone and dogger layers (the dierene

with equations(2)and (3)being lowerthat 1:5m).

4.2. Numerial results for Iodine transport

There is no ontour levels at time = 200 years. The ontour levels

=10 12

;10 10

;10 8

;10 6

;10 4

forthe times 10110, 50110, 10 6

,10 7

yearswithinthethreeshemes(expliite,upwindingimpliite,entered

impliite) are given in Figures 4 to 7 (dereasing from therepository

to theboundaryofthedomain).The expliitMUSCLsheme and the

Figure3. Contourlevelsofthepiezometrihead.

entered impliitsheme seemto be less diusivethan theupwinding

impliitsheme.

Figure 4. Contour levels of iodine onentration at time 10110 years (expliit

MUSCL(left),upwindingimpliit(middle)andenteredimpliit(right))

Figure 5. Contour levels of iodine onentration at time 50110 years (expliit

MUSCL(left),upwindingimpliit(middle)andenteredimpliit(right))

In Figure 8, we have shown separately the four umulative Iodine

amounts obtained by integration of the uxes with respet to time

(fromthelaylayerto thelimestoneanddoggerlayers,arossthetop

Figure6. Contourlevelsofiodineonentrationattime10 6

years(expliitMUSCL

(left),upwindingimpliit(middle)andenteredimpliit(right))

Figure7. Contourlevelsofiodineonentrationattime10 7

years(expliitMUSCL

(left),impliitupwindingupwindingimpliit(middle)andenteredimpliit(right))

and bottom left boundaries), in order to hek the qualitative results

based on theharateristi times.The obtainedresultsthusappearto

be infullagreement withthe rstsurveyofthe problem.

3 5 7 Log10(time (years))

0 1 2 3

Cumulative Iodine amount

source term clay−>limestone clay−>dogger limestone−>out dogger−>out

Figure8. CumulativeIodineamount

5. Conluding remarks

On the COUPLEX-1 Test ase, a rst hand omputation survey of

theproblemangivetheessentialresults:nearly alltheiodinereahes

the left boundary, and nearly no amount of plutonium an reah the

boundaries of the lay layer. This shows that if the main point of

engineering studies is the unertainties evaluations, simple analytial

modelsmustalso beusedfor theevaluationof themasstransfers.

A more aurate is nevertheless usefull.For thispurpose, thenite

volume sheme that we have used here appear to be eÆient, stable

andheap.However,COUPLEX-1problemhasbeenarefullydesigned

in order that the mathematial aspets were well posed. This an be

dierent withmore realistidata.

Referenes

ANDRA:http://www.andra.fr/ouplex/.

Eymard, R., Gallouet, T., Herbin, R.: `Finite Volume Methods', Handbook of

NumerialAnalysis,P.G.CiarletandJ.L.Lions.eds.,VIIpp723{1020,2000.

Appendix: Data of the Couplex1 Test as

C OUPLEX 1 Test Case Nuclear Waste Disposal Far Field Simulation

January 25, 2001

Abstract

This first C OUPLEX test case is to compute a simplified Far Field model used in nu- clear waste management simulation. From the mathematical point of view the problem is of convection diffusion type but the parameters are highly varying from one layer to another.

Another particularity is the very concentrated nature of the source, both in space and in time.

1 Introduction

The repository lies at a depth of 450m (meters) inside a clay layer which has above it a layer of limestone and a layer of marl and below it is a layer of dogger limestone. Water flows slowly (creeping flow) through these porous media and convects the radioactive materials once the con- tainers leak; there is also a dilution effect which in mathematical terms is similar to diffusion.

The problem has two main difficulties:

1. The radioactive elements leak from the containers, into the clay, over a period that is small compared with the millions of years over which convection and diffusion are active.

2. The convection and diffusion constants are very different from one layer to another; for instance, in the clay layer there is almost no convection while, in the other layers, diffusion and convection are both important.

2 The Geometry

In this first test case, the computation is restricted to a 2D section of the disposal site. Thus, the computational domain is in a rectangle

O=(0;25000)(0;695)

in meters. The layers of dogger, clay, limestone, and marl are located as follows (with the origin taken at the bottom left corner of the rectangle):

dogger

0<z<200

1

clay lies between the horizontal line

z=200

and the line from

(0;295)

to

(25000;350)

limestone lies between the line from

(0;295)

to

(25000;350)

and the horizontal line

z=

595

marl

595<z<695

.

The repository, denoted by

R

, is modeled by a uniform rectangular source in the clay layer:

R=f(x;z)2(18440;21680)(244;250)g

The geometry is summarized on figure 1 below. For this domain the computation should be carried for

t2(0;T)

with

T

=107

years.

000000000000000000 000000000000000000 000000000000000000 000000000000000000 111111111111111111 111111111111111111 111111111111111111 111111111111111111

000000000000000000 000000000000000000 000000000000000000 000000000000000000 000000000000000000 000000000000000000 000000000000000000 000000000000000000

111111111111111111 111111111111111111 111111111111111111 111111111111111111 111111111111111111 111111111111111111 111111111111111111 111111111111111111

000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000

111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111

000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111 111111111111111111111111111

x 295 m

595 m

350 m z

695 m

200 m

0 25000 m

repository

Dogger Clay Marl

Limestone

Figure 1: Geometry of computational domain

3 The Flow

It is assumed that all rock layers are saturated with water and that boundary loads are stationary so that the flow is independent of time. Darcy’s law gives the velocity

u

in terms of the hydro- dynamic load

H=P=g+z

:

u=KrH

(1)

2

where the permeability tensor

K

, assumed constant in each layer is given in Table 1,

P

is the pressure and

g

is Newton’s constant. Conservation of mass (

r(u)=0

, with the density

assumed constant) implies that

r(KrH)=0

in

O

(2)

Marl Limestone Clay Dogger

K

(m/year) 3.1536e-5 6.3072 3.1536e-6 25.2288 Table 1: Permeability tensor in the four rock layers On the boundary, conditions are:

H=289

on

f25000g(0;200);

H=310

on

f25000g(350;595);

H=180+160x=25000

on

(0;25000)f695g;

H=200

on

f0g(295;595);

H=286

on

f0g(0;200);

H

n

=0

elsewhere.

4 The Radioactive Elements

We are considering two species of particular interest, Iodine 129 and Plutonium 242. Both escape from the repository cave into the water and their concentrations

Ci;i=1;2

is given by two independent convection-diffusion equations:

Ri!(

Ci

t

+iCi) r(DirCi)+urCi=fi

in

O(0;T)i=1;2:

(3) where

Ri

is the latency Retardation factor, with value 1 for

¹²⁹

I,

10

5

for

²⁴²

Pu in the clay and 1 elsewhere for both Iodine and Plutonium;

the effective porosity

!

, is equal to 0.001 for

¹²⁹

I, 0.2 for

²⁴²

Pu in the clay layer and 0.1 elsewhere for both;

i=log2=Ti

with

Ti

being the half life time of the element :

^1:57107

for

¹²⁹

I,

^3:76105

for

²⁴²

Pu (in years);

The effective diffusion/dispersion tensors

Di

for any species

i=1;2

depend on the Darcy velocity as follows:

Di=deiI+juj[liE(u)+ti(IE(u))℄

3

with

Ekj(u)= ukuj

juj2 :

and with the coefficients, assumed constant in each layer, given in Table 2 below.

129

I

²⁴²

Pu

de1

(m

²

/year)

L

(m)

T

(m)

de2

(m

²

/year)

L

(m)

T

(m)

Dogger 5.0e-4 50 1 5.0e-4 50 1

Clay 9.48e-7 0 0 4.42e-4 0 0

Limestone 5.0e-4 50 1 5.0e-4 50 1

Marl 5.0e-4 0 0 5.0e-4 0 0

Table 2: Diffusion/dispersion coefficients for the radioactive elements in the 4 layers In this test case, the values of the source terms

fi;(i=1;2)

in the repository

R

are given in tabulated form in separately provided data files. The source terms are assumed to be spatially uniformly spread out in all the repository

R

. It is assumed that there is no source outside the repository (

fi;(i=1;2)

in

On

R

). The dependence in time is shown on figure 2, for illustrative purposes. The structure of the data file is described in appendix A.

0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.0016

1000 10000 100000

’iode.txt’

0 1e-05 2e-05 3e-05 4e-05 5e-05 6e-05 7e-05 8e-05 9e-05 0.0001

1000 10000 100000 1e+06

’plutonium.txt’

Figure 2: Release of Iodine and Plutonium as a function of time

4.1 Initial and Boundary Conditions

We call time zero the time when the containers begin to leak and the radioactive elements to spread, hence the initial values of the concentration

Ci

are zero at time zero.

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Boundary conditions for the transport of any nuclide

i=1;2

are

Ci

n

=0

on

f0g(295;595)

Ci

n

=0

on

f0g(0;200)

DirCinCiun=0

on

(0;25000)f0g Ci=0

elsewhere on the boundary.

where

n

is the outward normal to the vertical line

f0gf0;695g

5 Output requirements

The following output quantities are expected from the simulations(both tables and graphical representations):

Contour levels of

Ci

at times 200, 10110, 50110,

10 6

,

10

7

years (the following level values should be used:

10

12;1010;108;106;104

);

Pressure field (10 values uniformly distributed between

180

and

340

;

Darcy velocity field, along the 3 vertical lines given by

x=50

,

x=12500

,

x=20000

, using 100 points along each line;

Places where the Darcy velocity is zero;

Cumulative total flux through the top and the bottom clay layer boundaries, as a function of time;

Cumulative total fluxes through the left boundaries of the dogger and limestone layers;

The discretization grid of the domains and the time stepping used in the simulations should also be given.

A Descritpion of the data file

The file source.datcontains data needed to compute the source term

fi

in eq. (3). These data come from a Near Field computation. The file has 212 lines, and each line contains three numbers

t

p

;

~

f p

1;

~

f p

2;p=1;:::212

, where

t

p

is the time, and the source term

fi(t p

)

is related to

~

f p

i

by:

^fi(tp)=f^~ p

i=S

, where

S

is the surface of the repository.

The times

t

p

are in years, and the numbers

f^~ p

i

are in units of mole

=year

.

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