geologic repository of high-level nuclear waste · 2016-07-04 · geologic repository of high-level...

Geologic repository of

high-level nuclear waste

-Constitutive modeling and boundary

value problems-

Ho CHO, Nagoya Institute of Technology

Taiwan-Japan joint Workshop, May 22-23, 2016, Taiwan

Background

Disposal of high-level radioactive waste (HLRW) becomes one of

the largest environmental issues in 21st century

Nuclear waste contains radioactive substances that are dangerous to human being

Radioactivity usually

decays with long time

During the decaying, it can

generate huge amount of heat

How to deal with HLRW, safely, certainly and

permanently without care of human being ?

Moreover

Background

Technological barrier &engineered

barrier

Natural barrier

Geologic repository

Geologic repository of HLRW is

usually executed at the place below

300m in stable geologic environment.

One of the most important factors in

geological disposal of HLRW, is to understand

the thermo-hydraulic-mechanical-air (THMA)

behavior of natural barrier or host rock

Multiple barrier

THMA coupling process

mechanical

heat

hydraulic

conv

ectio

n

visco

sity

pore pressure

porosity

diffusiontherm

al stress

I. Thermal elastoplastic model

for saturated/unsaturated

geomaterials

State parameters of unsaturated geomaterials

1. Net stress

Basic Barcelona model: BBM 2. Suction

3. Void ratio

State variables:

1. Skeleton stress

2. Degree of saturation

3. Void ratio

Smooth transition between saturated and unsaturated state

Easy to incorporate with other physical states, such as density, structure, and anisotropy

Simple framework with a small number of parameters with definite physical meaning

Compared with BBM

" t

ij ij ijU

(1 )r w r aU S u S u

" ( )t n

ij ij a ij r a w ij ij r iju S u u S s

n t

ij ij a iju a ws u u

U : mean pore pressure

s : suction

Sr : degree of saturation

: net stress

: skeleton stress

Skeleton stress tensor here after:

n

ij"

ij

"

ij ij

0.0

0.2

0.4

0.6

0.8

1.0

0 20 40 60 80 100

1

2

3

4

5

6

7

c = S r

Co

effi

cien

t c

Degree of saturation Sr [%]

1: compacted boulder clay (Bishop et al, 1960)

2: compacted sale (Bishop et al, 1960)

3: breahead silt (Bishop & Donald, 1961)

4: silt (Jennings & Burland, 1962)

5: silty clay (Jennings & Burland, 1962)

6: sterrebeek silt (Zerhouni, 1991)

7: white clay (Zerhouni, 1991)

⇒Bishop’s effective stress（1959）：

" = - ua+c (ua - uw) = net+c s

Sr=c

,

Triaxial compression device for unsaturated/saturated

sample with double-cell type volume meter

(Axial translation method)

① Sample

② Pedestal

③ Load cap

④ Membrane

⑤ Inner cell

⑥ Load cell

⑦ Pore water pressure

meter

⑧ Axial displacement

transducer

⑨ Dual-tube bullet

⑩ Pressure difference

meter for volumetric

stain

⑪ Axial actuator (Up)

⑫ Axial actuator (Down)

⑬ Air supplying tank

⑭ Reference water level

for inner cell

⑮ Air pressure meter

0.40

0.60

0.80

1.0

1.2

1.4

1.6

1 10 102 103 104

Initial Sr=15.7%Initial Sr=21.3%Initial Sr=25.9%Saturated

Void

rat

io e

Mean skelton stress p (kPa)

0.40

0.60

0.80

1.0

1.2

1.4

1.6

1 10 102 103 104

Initial Sr=15.7%Initial Sr=21.3%Initial Sr=25.9%saturated

Voi

d ra

tio e

Mean skelton stress p (kPa)

(a) s=73.5 kPa (b) s=294 kPa

Influence of degree of saturation on e-ln p relation

(Results of oedometer test for unsaturated soils

extracted from the work by Honda, 2000)

N.C.L.

C.S.L.

Mean skeleton stress p

Void

rat

io

e

N

G

pr

eN1e

N.C.L.SC.S.L.S

p0

e0

pN　 1 pN1e

rs

r

ln p

( )rSG

( )rS

e-lnp relation considering moving up of

N.C.L. and C.S.L. due to instauration

( ) ln ( 0)r

r

p qe N S

p p

( ) ln ( )r

r

p qe S M

p pG

( , ) lnr

r

pe S

pc

2 2

2

( ) ( )( ) ln ln

ln 2

r rr

r

N S S M pe N S

M p

G

2 2

2

0 0 0 0

( ) ( ) 1 1 1ln ln 0

(1 ) ln 2 1 1

ps er rv

p p p p

N S Sp Mf

p C e M e C e C C

r rG

0 ( ) ( )=( )ln2p

v r r

M

fd N S S

p

G

2 2

2

0 0 0

1 1 1ln ln 0

1 1

p

vs e

p p p

p Mf

p M e C e C C

r r

Associated flow rule:

Hardening parameter

p

ij

ij

fd

0

( ) ,1

ee sd a b

e p

r rr r r

s rd QdSr r

s r

r r

N NQ

S S

Evolution of density:

Evolution of saturation :

( )p

ij ijkl ijkl ij ijkl

kl

fd E E d AE

/p

ijkl ijqr mnkl

mn qr

f fE E E D

σ

0

1 1

1p

r

QA dS

C e D

p

p

mnkl

mn kl

h f fD E

C

p

mm mm

fh

r

On condition that elastic strain caused bychange of temperature obeys Hooke’s theory,equivalent stress is then defined as;

Thermal effect described with the concept of equivalent stress （Zhang & Zhang, 2009）

Change of temperature may also cause elastic and plastic strain

0mp

0mp

mp

mp

T0T

0mp

0mp

mp

mp

Temperature change

（pm0=const.）

Stress change

（Ｔ=const.）equivalent

stress

Concept of equivalent stress is

used to consider the influence of

temperature and is expressed as,pm0: real stress

K: bulk modulus

s: thermal expansion coefficient

T: current temperature

T0: reference temperature (15C)

Only one parameter, thermal

expansion coefficient , T , is

added to the constitutive model!

1 1( ) ln[( OCR) / ]e N Np pr

New void ratio difference,

considering the influence

of temperature, is given.

0 0

0

( ) ln[( OCR) / ]

( ) ,1

e N

e

N

e s

p

d a be

p

r

rr r

rr

p

Associate flow rule:

( ) ( ); ( )

( ) ( );

s rrr r r r rs r

r r

s rs r r r s r

r r

s r

N NN S N S S N N S

S S

N NN S N Q S S Q

S S

d QdS

r

r

p

ij

ij

fd

r

rS s

rSand

2 2

2

0 0 0

1 1 1ln ln 0

1 1

psv

p p p

ep Mf

p M e C e C C

r

r

Elastoplastic model for saturated/unsaturated geomaterials

(i) Evolution of is dependent on：

Saturation , Overconsolidation and

temperature

(ii) Evolution of rs is dependent on:

Saturation

Overconsolidation termSaturation term

: residual saturation 100% saturation

Compared with Cam-clay model for saturated soils, only saturation is added

Equivalent stress

er

Mean skeleton stress p

Dd

evia

tory

str

ess

q

p0 pN　 1 pN1e p

(p* , q*)

(p , q)

Normal yielding surface

Subloading yielding surface

Extension of subloading concept (Hashiguchi and Ueno,

1977) to unsaturated soil in skeleton stress space

(1) Initial drying curve110 0 12

( ) tan (( 1) / )dc sc ss s r

r r r rS S S S e e

(2) Main drying curve

1112( ) tan (( 1) / )dc sc ss s r

r r r rS S S S e e

2212( ) tan (( 1) / )wc sc ss s r

r r r rS S S S e e

Skeleton curve

Scanning curve

1 1 1

0 1s s sk k k

1 1 3

1(1 )s

s s

rk k c

r

2

1

/ 0

/ 0

dsr

ds

Parameters：

Sd: drying AEV; Sw: wetting AEV

Srr: residual Sr; Sr

s; saturated Sr

c1 c2 c3: scanning parameters

ks0 gradient at r =0

rdS ds - 1sk

MCC

Degree of saturation Sr

Suc

tion

s

Srr Sr

s Sr

s 0=1

Drying A.E.V.

Primary drying curve

Secondary drying curve

Wetting curve

W.E.V.

Main wetting curve

Initial drying curve

Main drying curve

Current 1

Moisture characteristic curve (MCC)

Tangential stiffness of suction-

saturation relation

(3) Main wetting curve

Parameters involved in constitutive model

Compression index 0.050 Swelling index 0.010 Critical state parameter Μ 1.0 Void ratio N (p’=98 kPa on N.C.L.) 1.14 Poisson’s ratio ν 0.30

Parameter of overconsolidation a 5.00 Parameter of suction b 5.50 Parameter of overconsolidation 1.0 Void ratio Nr (p’=98 kPa on N.C.L.S.) 1.28

Parameters involved in MCC

Saturated degrees of saturation s

rS 0.82

Residual degrees of saturation r

rS 0.64

Parameter corresponding to drying AEV (kPa) Sd 550

Parameter corresponding to wetting AEV (kPa) Sw 320

Initial stiffness of scanning curve (kPa)e

spk 200000

Parameter of shape function c1 0.008



0

200

400

600

800

1000

1200

60 65 70 75 80 85 90 95 100

Secondary drying curve

Wetting curve

Primary drying curve

Scanning curve

Suction (kPa)

Degree of saturation (%)

Starting point (0.715,450)

Simulated moisture characteristics curve

of unsaturated fictional silt

Test and simulated moisture characteristics

curve of unsaturated rockfill (test data from the

work by Kohgo et al, 2007)

Verification of proposed model

Verification of the model by drained and exhausted triaxial

compression tests for a rockfill with submergence process (test

data from the work by Kohgo et al, 2007)


Verification of the model by drained and exhausted

triaxial compression tests for a rockfill with submergence

process (test data from the work by Kohgo et al, 2007)



1. Consolidation tests

2. Triaxial compression tests

under different temperature .(Test data from Uchaipichat and Khalili , 2009)

Material parameters of silt

Saturated degrees of saturation 1.00

Residual degrees of saturation 0.30

Parameter corresponding to drying AEV(kPa) Sd

18.0

Parameter corresponding to wetting AEV(kPa) Sw

5.0

Initial stiffness of scanning curve (kPa) 90.0




s

rSr

rS

e

spk

Parameters of silt in MCC

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1 10 100 1000

Drying curve

Wetting curve

Suction s (kPa)

Deg

ree

of

satu

rati

on

Sr

Line: simulation

Dot: experiment

MCC

Compression index 0.09

Swelling index 0.006

Critical state parameter Μ 2.45

Void ratio N (p’=98 kPa on N.C.L.) 0.638

Poisson’s ratio ν 0.30

Parameter of overconsolidation a 60.00

Parameter of suction b 0.00

Parameter of overconsolidation 2.0

Void ratio Nr (p’=98 kPa on N.C.L.S.) 0.665

0.48

0.50

0.52

0.54

0.56

0.58

0.60

10 100 1000

T=25o

T=40o

T=60o

Vo

id r

atio

e

Mean effective stress '' (kPa)

Line: simulation

Dot: experiment

0.48

0.50

0.52

0.54

0.56

0.58

0.60

10 100 1000

T=25o

T=40o

T=60o

Void

rat

io e


Line: simulation

Dot: experiment

0.48

0.50

0.52

0.54

0.56

0.58

0.60

10 100 1000

T=25o

T=40o

T=60o

Vo

id r

atio

e


Line: simulation

Dot: experiment

0.50

0.52

0.54

0.56

0.58

10 100 1000

T=25o

T=40o

T=60o

Vo

id r

atio

e


Line: simulation

Dot: experiment

s=0 kPa s=10 kPa

s=50 kPa s=100 kPa

Consolidation tests under different temperature

Verification of proposed model (Test data from Uchaipichat and Khalili , 2009)

0.0

1.0

2.0

3.0

4.00 5 10 15 20 25

Axial strain (%)

Volu

metr

ic s

train

(%

)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Calculated result

0.00

50.0

100

150

200

250

0 5 10 15 20 25

Dev

iato

r st

ress

(k

Pa)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Experimental result

(a)

(b)

0.00

50.0

100

150

200

250

0 5 10 15 20 25

Dev

iato

r st

ress

(k

Pa)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Calculated result

0.0

1.0

2.0

3.0

4.00 5 10 15 20 25

Vo

lum

etr

ic s

train

(%

)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Experimental result

Conventional triaxial compression tests under different constant suction

and temperature conditions (initial mean net stress= 50 kPa)

(a) experimental results (Uchaipichat and Khalili, 2009)

(b) calculated results

s=50kPa, T=25o

s=50kPa, T=40o

s=50kPa, T=60o

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

Conventional triaxial compression tests under different constant suction

and temperature conditions (initial mean net stress= 100 kPa)

0.00

50.0

100

150

200

250

300

350

0 5 10 15 20 25

Axial strain (%)

Dev

iato

r st

ress

(k

Pa)

0.0

2.0

4.0

6.0

8.00 5 10 15 20 25

Volu

metr

ic s

train

(%

)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Experimental result

Calculated result

(a)

(b)

0

50

100

150

200

250

300

350

0 5 10 15 20 25

Dev

iato

r st

ress

(k

Pa)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Calculated result

0.0

2.0

4.0

6.0

8.00 5 10 15 20 25

Vo

lum

etr

ic s

train

(%

)

Axial strain (%)

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

Experimental result

(a) experimental results (Uchaipichat and Khalili, 2009)

(b) calculated results

s=100kPa, T=25o

s=100kPa, T=40o

s=100kPa, T=60o

s=300kPa, T=25o

s=300kPa, T=40o

s=300kPa, T=60o

-2

0

2

4

6

8

10

12 -6

-5

-4

-3

-2

-1

0

10 1 2 3 4 5

dev

iato

r st

ress

a-

r (

MP

a) volu

metric strain

v (%)

axial strain a (%)

○ 20℃△ 40℃□ 60℃× 80℃

-2

0

2

4

6

8

10

12 -6

-5

-4

-3

-2

-1

0

10 1 2 3 4 5

dev

iato

r st

ress

a-

r (

MP

a) volu

metric strain

v (%)

axial strain a (%)

○ 20℃△ 40℃□ 60℃× 80℃

-2

0

2

4

6

8

10

12 -6

-5

-4

-3

-2

-1

0

10 1 2 3 4 5

dev

iato

r st

ress

a-

r (

MP

a) volu

metric strain

v (%)

axial strain a (%)

○ 20℃△ 40℃□ 60℃× 80℃

3=0.49MPa 3=0.98MPa 3=1.47MPa

Stress-strain-dilatancy relation

Triaxial compression test results

Soft sedimentary rock (saturated state)

(Test data from Zhang, Nishimura and Kageyama, 2013 & 2014)

soft sedimentary rock (saturated state, 3=0.49MPa)

-2

0

2

4

6

8

10

12 -6

-5

-4

-3

-2

-1

0

1

0 1 2 3 4 5

dev

iato

r st

ress

a-

r (M

Pa) v

olu

metric strain

v (%)

axial strain a (%)

○ 20℃△ 40℃□ 60℃× 80℃

-2

0

2

4

6

8

10

12 -6

-5

-4

-3

-2

-1

0

10 1 2 3 4 5

dev

iato

r st

ress

a-

r (

MP

a) volu

metric strain

v (%)

axial strain a (%)

○ 20℃△ 40℃□ 60℃× 80℃

Tests Simulation


(Test data from Zhang, Nishimura and Kageyama, 2013 & 2014)

There is a basic and intrinsic relationship among the state

variables, void ratio e, degree of saturation Sr and gravimetric

water content w ,

r

s

S ew

G

s r rG dw edS S de

Which means that even if dw=0 (constant water content), if

void ratio e changed, degree of saturation will also change!

A simple finite deformation scheme for MCC

Simulated behavior of soil at the process of drying-isotropic loading to full

saturation-suction reduction

0

50

100

150

200

250

300

350

1 10 100 1000 10000

Suct

ion (

kP

a)

Mean net stress (kPa)

A

B C

D

(a)

Dry

ing

Net stress loading

Wet

tin

g0.2

0.4

0.6

0.8

1.0

1.2

1 10 100 1000 10000


Void

rat

io

A

B

C. D

(b)

NCL---saturated soil

Dry

ing

Net stress loading

Saturation point

0.2

0.4

0.6

0.8

1.0

1 10 100

Suction (kPa)

Vo

id r

atio

(c)A

B

CD

Drying

Net

str

ess

load

ing

Wetting

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100

Suction (kPa)

Deg

ree

of

satu

rati

on A

D

B

C

(e)

Drying

Net

str

ess

load

ing

Wetting

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100 1000 10000

Vo

id r

atio

Mean skeleton stress (kPa)

(f)A

B

C.D

Saturation point

Drying

Net stress loading


0.2

0.4

0.6

0.8

1.01 10 100 1000 10000

Deg

ree

of s

atur

atio

n


A

B

C. D

(g)

Dry

ing

Net stress loading

Saturation point

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100 1000 10000

Deg

ree

of

satu

rati

on


A

B

C

D

(d)

Dry

ing

Net stress loading

Saturation point

0.00

0.01

0.02

0.03

0.04

0.05

0.06

1 10 100 1000 10000

void

rat

io o

f po

ssib

le c

olla

pse


(h)

Simulated behavior of soil at the process of drying-isotropic

loading-wetting collapse

0.4

0.6

0.8

1.0

1.2

1 10 100 1000


Void

rat

io

(a)

A

B

C

D

Dry

ing

Net stress loading


Wet

tin

g

0.4

0.6

0.8

1.0

1.2

1 10 100 1000


Void

rat

io

(b)

A

B

C

D

Drying

Net stress loading

Wet

tin

g

0.4

0.6

0.8

1.0

1 10 100

Suction (kPa)

Vo

id r

atio

(c)A

B

CD

Drying

Net

str

ess

load

ing

Wetting collapse

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100 1000

Deg

ree

of

satu

rati

on


(d)

A

B

C

D

Dry

ing

Net stress loading

Wet

tin

g

0.2

0.4

0.6

0.8

1.0

1.2

1 10 100

Deg

ree

of

satu

rati

on

Suction (kPa)

(e)

A

B

C

D

Drying

Lo

adin

g

Wetting

Wetting

0.2

0.4

0.6

0.8

1.01 10 100 1000

Deg

ree

of

satu

rati

on


(f)

A

B

C

D

Net stress loading

Wet

tin

g

II. Field equations for THMA

analysis

1. Stress2. Saturation3. Water head4. Air pressure5. Temperature

( )0

s ss

i

i

u

t x

rr

( )0

w ww

i

i

u

t x

rr

( )0

a aa

i

i

u

t x

rr

Conservative equations for all phases

0ij

i

j

bx

r

1. Equilibrium equation

2. Continuum equations (solid + liquid)

2 10

s wwwii r

w w

i i r

p Skp

n x x K S

3. Continuum equations (solid + air)

2 10

1

s aaaii r

a a

i i r

p Skp

n x x K S

Boundary condition

1. Stress2. Displacement3. Water head4. Air pressure5. Temperature

Discretiza

tion

in sp

ace

an

d tim

e d

om

ain

4. Energy conservation equation

2

( ) (1 )( )w w a a

r i r i t

i i i i

T T T Tc nS c v n S c v k Q

t x x x xr r r

Field

equ

atio

ns

Soil-water-

air coupling

Neglecting energy from plastic deformation

Static loading

Initial condition

Discretization of governing equations

Equilibrium equation

| | | | | | |

w w a a w w a a

N t t Sat dE t t Sat dE t t t t T N t t Sat dE t Sat dE tK u I p I p F F T I p I p

| |

1

| |[ ]

mT w a w

r t v sr dE sr dE t t i idEN t t t t t t

i

w a

sr dE sr dE t WT N t tt

S K u A F p F p p

A F p F p K T

| |

1

| |

(1 )

[ ]

mT w a a

r t v sr dE t t sr dE i idEN t t t t t t

i

w a

sr dE t sr dE AT N t tt

S K u F p B F p p

F p B F p K T

Continuum equations（solid + liquid）

Displacement, temperature：4 node

Pressures of Water and air: center of gravity

2D isoparametric element

Continuum equations（solid + air）

t t t

N t N tC T K T f

Discretization in space

(Galerkin)

N t t N t N t N t t N tT T t T t T T

1t t t t

N t t N t N tC t K T F K T t T

Tt

V

C c N N dVr

( )

(1 S )( )

T

Tt w

t r

i i iV V

Ta a

r i

iV

N N NK k dV nS c N dV

x x x

Nn c v N dV

x

r

r

w

iv

T Tt

V S

f N EdV q N dS

Discretization in time

（Newmark ）

En

ergy

con

serva

tion

equ

atio

n


Overall stiffness matrix

| |

1

|

|

1

||

0

(1 ) [ ]r

w w aN t tSat Sat

mT w w w w

r v sr sr dE t t i idE t t

iaT wdEt t mv sr sr

a

i idE t t

i

w w

t t ST N t t at d

uK H H

S K A F F h h

pS K F B F

p

F H hF T

| |

| |

|

|

( )

( )

w a a

E t Sat dE t

w w a

sr dE sr dE WT N tt t

w w a

sr dE t sr dE t

t

AT N t t

H p

A K T

K

F h F p

F h F TB p

| | |1t t t t

N t t N t N tC t K T f K T tT

Displacement, water head, air pressure and

temperature as unknown variables


III. Slope failure in unsaturated

Sirasu ground

THMA coupling analysis in FE-FD

scheme (Temperature=Const.)

Triaxial compression test of silty clay under undrained

and unvented condition (Test data from Oka et al., 2011)

MCC

One element mesh and boundary conditions

30= 450kPa

pa0= 250kPa

S0 =10kPa, 30kPa

50kPa, 100kPa

0= 0.5%/min

Temp. = 20oC

Parameters of MCC of silty clay

Material parameters of silty clay

0

20

40

60

80

100

70 75 80 85 90 95 100

Suct

ion (

kP

a)


.

Comparison between test and calculation

Test

Calcu

latio

n

Pore water pressureAir pressureStress-strain relation

Test data from Oka et al., 2011

Model test (Kitamura et al., 2007)

180cm180cm

90cm

Test box

Rain falling device

90cm

Water tank

Water injection (Width:

60cm, Depth: 180cm)

Flow meter

10cm

Detail of rain fall tank

Sirasu

Drainage

Water h

ead

Water tank

Test conditions

Physical properties of Sirasu

Position of Water injectionCase 1

Case 2

Case 3

Drained

condition is the

same as test

FEM mesh and loading condition

Location of tensiometers and

water injection pattern

Drainage conditions

No.1

No.2

No.3

No.4

No.5

No.6

No.7

No.8

No.9

No.10

No.11

No.12

No.13

No.14

No.15

Range of water injection (Case 3)

Range of water injection (Case 1)

Ran

ge

of

wat

er

inje

ctio

n (

Ca

se 2

)

40 cm

25

cm

60 cm

45o

80

cm

140cm

10

cm

10

cm

10

cm

10

cm

10

cm

20cm 20cm 20cm

Case1 Case2 Case3

Top drained drained drained

Bottom undrained undrained undrained

Back side undrained undrained undrained

Slope face drained drained drained

Case1 Case2 Case3

Density of soil

particle (g/cm3)2.45 2.40 2.45

Water content

in nature (%) 25.6 23.3 23.1

Void ratio 1.57 1.47 1.57

Total density of

soil (g/cm3)1.20 1.20 1.17

Material parameters

Element behavior of Sirasu

10-7

10-6

10-5

10-4

10-3

10-7

10-6

10-5

10-4

10-3

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Water permeability

Air permeability

Wat

er p

erm

eabi

lity

(m/s

) Air perm

eability (m/s)


Relationship between permeability of

water & air with saturation

Gravitation stress field+5kPa

-10

-5

0

5

10 -2

-1

0

1

20 2 4 6 8 10

q (suction=8kPa)

q (suction=0kPa)

v (suction=8kPa)

v (suction=0kPa)

1

3 (

kP

a)

v (%

)

a (%)

3=10kPa

Calculation conditions










Case 1 Case 2 Case 3

Saturated degrees of saturation 0.87 0.89 0.95

Residual degrees of saturation 0.20 0.25 0.20

Parameter corresponding to drying AEV(kPa) Sd

12.0 15.0 12.0

Parameter corresponding to wetting AEV(kPa) Sw

0.07 0.10 0.17

Initial stiffness of scanning curve (kPa) 90.0 20.0 50.0

Parameter of shape function c1 0.30 0.30 0.33



s

rSr

rS

e

spk

Parameters of MCC

Soil water characteristic curve (MCC)


0

5

10

15

20

25

30

20 30 40 50 60 70 80 90 100

ExperimentSimulation


Su

ctio

n (

kP

a)

0

5

10

15

20

25

30

20 30 40 50 60 70 80 90 100



Suct

ion (

kP

a)

0

5

10

15

20

25

30

20 30 40 50 60 70 80 90 100


Su

ctio

n (

kP

a)


Initial condition

Suction: 8kPa

Saturation: 38%

Case 1 （底面注水）

calculated experimental

No.6

No.7

No.8

No.6

No.7

No.8

No.10No.9

No.10

No.9

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)


No.11

No.12

No.13

No.11

No.12

No.13

No.15No.14

No.15

No.14

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)


No.1

No.2

No.3

No.1

No.2

No.3

No.4

No.5No.4

No.5

No.1

No.2

No.3

No.4

No.5

No.6

No.7

No.8

No.9

No.10

No.11

No.12

No.13

No.14

No.15

45o

Comparison of EPWP (Case 1)

Position where

water is injected

Calculation is on the whole coincident

with test. The failure time is also the same

in calculation and test. In slope surface,

however, the lose of suction in calculation

is much slower than that at test

Case 2 （背面注水）


No.6

No.7

No.8

No.6

No.7

No.8

No.10No.9

No.10

No.9

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)


No.11

No.12

No.13

No.11

No.12

No.13

No.15No.14

No.15

No.14

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)


No.1

No.2

No.3

No.1

No.2

No.3

No.4

No.5No.4

No.5

-12

-10

-8

-6

-4

-2

0

0 50 100 150 200 250 300

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)

No.1

No.2

No.3

No.4

No.5

No.6

No.7

No.8

No.9

No.10

No.11

No.12

No.13

No.14

No.15

45o


Position where

water is injected

Calculation is on the whole coincident

with test. In slope surface, however, the

lose of suction in calculation is much

slower than that at test

Case 3 （上面注水）


No.11

No.12

No.13

No.11

No.12

No.13

No.15No.14

No.15

No.14


No.6

No.7

No.8

No.6

No.7

No.8

No.10No.9

No.10

No.9


No.1

No.2

No.3

No.1

No.2

No.3

No.4

No.5No.4

No.5

-10

-8

-6

-4

-2

0

0 20 40 60 80 100 120Neg

ativ

e P

WP

(kP

a)

Elapsed Time (min)

-10

-8

-6

-4

-2

0

0 20 40 60 80 100 120

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)

-10

-8

-6

-4

-2

0

0 20 40 60 80 100 120

Elapsed Time (min)

Neg

ativ

e P

WP

(kP

a)

No.1

No.2

No.3

No.4

No.5

No.6

No.7

No.8

No.9

No.10

No.11

No.12

No.13

No.14

No.15

45o


Case 3 上面降雨Position where

water is injected

Calculation is on the whole coincident with test. The

failure time is also the same in calculation and test.

Change of degree of saturation

Case 1 Case 2

T=100 min

T=200 min

T=300 min

T=100 min

T=200 min

T=300 min

Case 3

T=13.3min

T=53.3min

T=93.3min

T=26.6min

T=66.6min

T=106.6min

T=40.0min

T=80.0min

T=120.0min

Change of degree of saturation

T=300 min T=300 min T=120 min

T=300 min T=300 min T=120 min

22 pI ( : 2nd invariant of deviatoric plastic strain)

The calculations are totally the same as the tests, that is, Case 1 and

Case 3 are much easier to fail than Case 2

Shear strain2

pI

Distribution of deviatoric plastic strain

Distribution of displacement vector


IV. THMA coupling analysis

of heating test for saturated

geomaterials (pa=constant, Sr=1.0)

Layout of test site

Stage Start:

day

End:

day

Duration

: day

Activity

1 0 90 90 Heating

(325W/heater)

2 91 339 248 Heating

(975W/heater)

3 340 518 180 Cooling

Multi stages of the heating tests

Heating test for saturated rock (Gens et al.,2007)

Heater (1m)

Separation:0.8m

3D FEM mesh

Initial condition:

Water pressure: 1 MPa

Mean stress: 5 MPa

Temperature: 15oC

Boundary conditions:

Up, back and right planes:

undrained and heat insulated

Other planes:

Drained and fixed temperature (15oC)

3D THMA analysis conditions

x

zy

5

10

15

20

-1 -0.5 0 0.5 1 1.5 2

Model

Model

Experiment

Experiment

σ1-σ3(MPa)

ε3 (%) ε

1(%)

10-13

10-12

10-11

10-10

10-9

10-8

10-7

10-6

0 20 40 60 80 100 120

Per

mea

bil

ity

(m

/s)

Temperature (o

C)

Element simulation of triaxial

compression test (38 MPa)

Relation between permeability

and temperature

Thermal expansion coefficient αt(K-1) 8.0e-06

Thermal expansion coefficient αw(K-1) 2.1e-04

Thermal conductivity kt (kJ m-1 K-1Min-1) 0.18

Specific heat C (kJ Mg-1 K-1) 840.

Heat transfer coefficient of air boundary ac

((kJ m-2 K-1 Min-1)236.

Specific heat of water Cw (kJ Mg-1 K-1) 4184.

Physical parameters

Material parameters

Material parameters and physical parameters










Comparison between test and calculation (temperature)

Change of temperature at different positions

D: distant of measuring point

to heater

Z: z coordinate of measuring

point

y

xz

0

20

40

60

80

100

120

0 100 200 300 400 500

Model

Experiment

Tem

peratu

re (

o

C)

Time (days)

X

Y

D=0m

Z=8m

heate

r

0

10

20

30

40

50

60

0 100 200 300 400 500

Model

Experiment

Time (days) T

em

peratu

re (

o

C)

X

Y

heate

r

D=0.872m

Z=8.000m

0

10

20

30

40

50

60

0 100 200 300 400 500

Model

Experiment

Time (days)

Tem

peratu

re (

o

C)

X

Y

heate

rD=1.068m

Z=8.000m

0

10

20

30

40

50

60

0 100 200 300 400 500

Model

Experiment

Time (days)

Tem

per

atu

re (

o

C)

X

Y

heaterD=1.200m

Z=6.580m

0

10

20

30

40

50

60

0 100 200 300 400 500

Model

Experiment

Time (days) T

emp

era

ture

(o

C)

heater

X

Y

D=2.200m

Z=5.040m

ヒータ

ヒータ

ヒータ

ヒータ

ヒータ

-2

-1

0

1

2

3

4

0 100 200 300 400 500

Model

Experiment

Ex

cess

Po

re W

ate

r P

ress

ure

(M

Pa

)

Time (days)

D=0.4m

Z=8m

heater

X

Y

-2

-1

0

1

2

3

0 100 200 300 400 500

Model

Experiment

Ex

cess

Po

re W

ate

r P

ress

ure

(M

Pa

)Time (days)

heater

D=0.00m

Z=5.04m

X

Y

-2

-1

0

1

2

3

0 100 200 300 400 500

Model

Experiment

Ex

cess

Po

re W

ate

r P

ress

ure

(M

Pa

)Time (days)

-2

-1

0

1

2

3

0 100 200 300 400 500

Model

Experiment

Time (days)

Exce

ss P

ore

Wate

r P

ress

ure

(M

Pa)

D=0.83m

Z=6.89m

heater

X

Y

-2

-1

0

1

2

3

0 100 200 300 400 500

Model

Experiment

Exce

ss P

ore

Wate

r P

ress

ure

(M

Pa)

Time (days)

heaterD=0.21m

Z=7.225m

X

Y

heater

X

Y

D=0.36m

Z=8m

y

xz

ヒータ

ヒータ

ヒータ

ヒータ

ヒータ

Comparison between test and calculation (EPWP)


to heater


point

Change of EPWP at different positions

-0.04

-0.03

-0.02

-0.01

0

0.01

100 200 300 400 500

Model

Experiment

Time (days)S

train

(%

)

X

Y

D=1.8m

Z=8m

heater

-0.04

-0.03

-0.02

-0.01

0

0.01

100 200 300 400 500

Model

Experiment

Time (days)

Str

ain

(%

)

X

Y

D=1.15m

Z=8m

heater

-0.04

-0.03

-0.02

-0.01

0

0.01

100 200 300 400 500

Str

ain

(%

)Time (days)

X

Y

D=0.0m

Z=8m

heater

Model

Experiment

heater

-0.04

-0.03

-0.02

-0.01

0

0.01

100 200 300 400 500

Str

ain

(%

)

Time (days)X

Y

D=0.18m

Z=8m

Model

Experiment

-0.04

-0.03

-0.02

-0.01

0

0.01

100 200 300 400 500

Model

Experiment

Str

ain

(%

)

Time (days)

X

Y

D=2.14m

Z=8m

heater

y

xz

ヒータ

ヒータ

ヒータ

ヒータ

ヒータ

Comparison between test and calculation (Strain)


to heater


point

Change of Deviatory strain at different positions

0

20

40

60

80

100

120

0 1 2 3 4 5

At the end of 1st phase

At the end of 2nd

phase

initial temperature

Tem

pera

ture

(o

C)

Distance From Heater (m)

Distribution of temperature (only calculation)

Distribution of temperature

Temperature only changed

in limited area (D<5m)

Reason ： Small thermal

conductivity of rock

V. THMA coupling analysis of

heating test for saturated/

unsaturated geomaterial (pa=0, Sr≤1.0)

Heating tests (Munoz, 2006)

Layout of heating test

Boring hole

（f 0.3m; Depth: 7m）

Rock

Bentonite buffer

Heater

（f75mm）

Stage Start (day) End (day) Time (day) Action

1 0 982 982 hydration phase

2 982 1522 540 heating phase

3 1523 1800 278 Cooling phase

Testing stages

Analytical conditions

heater

bentonite

5m

2m

20 m40

m

FEM mesh and boundary condition

MCC

Initial conditions

Unsaturated bentonite： s=136MPa; Sr=70%

Saturated soft rock： water head=40m

102

103

104

105

106

40 50 60 70 80 90 100


Suct

ion (

MP

a)

Nature rock

Bentonite

Line: calculation

Dot: experiment

20m40m

Dが大きいほど飽和に至るのが早い

約400日に渡って孔内は全部飽和した

Result: 1st stage

50 days 150 days 250 days 600 days

Distribution of saturation

Change of saturation in bentonite

and neighboring rock

D：distance

from heater

heater

bentonite

70

80

90

100

0 200 400 600 800 1000

D=0.05m

D=0.10m

D=0.15m

Sat

ura

tion (

%)

Time (days)

90

90

100

0 200 400 600 800 1000

D=2.00m

D=0.65m

D=0.20m

Sat

ura

tio

n (

%)

Time (days)

0.0

20.0

40.0

60.0

80.0

100.0

120.0

600 800 1000 1200 1400 1600

D=0.05 mD=0.65 mD=5.00 m

Tem

per

atu

re (

o

C)

Time (days)

Line: modelling

Dot : experimental data

0

20

40

60

80

100

0 1 2 3 4 5

990 days

1030 days

1200 days

Tem

per

atu

re (

o

C)

Distance from heater (m)

Line: modelling

Dot: experimental data

D=0.65m

D=1.65m

-100

0.00

100

200

300

400

500

800 1000 1200 1400 1600 1800

Ex

cess

ive

po

re w

ater

pre

ssu

re (

kP

a)

Time (days)

Line: modelling

Dot : experimental data

D：Distance

from heater

Distribution of temperatureChange of temperature

Change EPWP

Quick change of temperature can be

clearly simulated

No prominent change of temperature at

the place 5m away from the heat source

Change of EPWP due to temperature can

be clearly simulated

Result: 2nd and 3rd stages

VI. THMA coupling analysis

for geologic repository of

HLRW(pa=0, Sr≤1.0)

520m

210m

300m

20

m

7.6

0 [MPa]

20m520m

300m

x y

z

7.6

x

y

2D meshInitial stress field

THMA coupling analysis for geologic repository of HLRW

FEM mesh

Thermal, boundary and initial conditions• Boundary condition：fixed in side and bottom surfaces

• Hydraulic condition：only drained at top surface

• Thermal condition：all the surfaces far from the

heating source are at constant temperature of 15ºC

• Material: heating source is elastic

• Initial saturation: soft rock is saturated but heating

source and bentonite are unsaturated

3D meshInitial stress field

0

[MPa]

Time (year)

Q(t

)/Q

0

Case1(10~310 years)

Case2(100~410 years)

Background→ Constitutive model→FEM→ Conclusions

32

0 1 1( ) / (1 )tt

Q t Q e e

The relation of decay rate and time

Heat released from nuclear waste is given as follow

proposed by Thunvik and Braester (1991) :

The initial heat Q0 is given 28.28 kJ/m3 /min.

Heat emission as input in FEM

0

5

10

15

20

25

30

0 50 100 150 200 250 300

CASE 1 (10-310 years)


Heat

Q (

kJ/m

3/m

in)

Time (year)

100%

FEM---Setting of heat source

Background→ Constitutive model→FEM→ Conclusions

FEM---Setting of heat source

The schematic diagram of

the field project

The heat emission as input in FEM

1. Volume of HLRW only occupies about 1.6%

of total tunnel volume, correspondingly the heat

emission as input in FEM is given 1.6%.

2. Heat emission decreases greatly between 10

years and 100 years from the decay rate-time

relation. Two cases are considered starting from

10 years and 100 years after emplacing into

repository respectively. The calculated times are

300 years in two cases.0

0.1

0.2

0.3

0.4

0.5

0 50 100 150 200 250 300


CASE 2 (100-400years)

Hea

t Q

(k

J/m

3/m

in)

Time (year)

1.6%

3.7m

4.5

m

30m

Waste repository

R=0.4m

7m

4.5

m

3m

MCC and initial condition

MCC（Munoz et al, 2006)

0.1

1

10

100

1000

40 50 60 70 80 90 100

RockBentonite

Sucti

on(M

Pa)


10-14

10-13

10-12

10-11

10-10

10-9

10-8

40 50 60 70 80 90 100

BentoniteRock

Perm

eab

ilit

y o

f w

ate

r(m

/s)

Degree of saturation(%)

Rock Bentonite

Saturation [%] 100 86

Suction [MPa] 0 62

Total head [m] 520 -6945

Initial condition

Material parameters for constitutive model

Bentonite Rock

Parameter of overconsolidation a 5.000 2.000

Parameter of suction b 0.5 0.5

Parameter of overconsolidation β 1.00 1.10

Void ratio (p’=98 kPa on N.C.L) N 1.040 0.450

Void ratio (p’=98 kPa on N.C.L.S) Nr 1.060 0.530

Thermal expansion coefficient (1/K) αT -1.0×10-6 -2.5×10-5

Thermal conductivity (kJ m−1 K−1 Min−1) KtS 0.060 0.200

Specific heat (kJ Mg-1 K-1) CS 723 840

Thermal expansion coefficient of water (1/K) αT 2.1×10-4 2.1×10-4

Bentonite Rock

Compression index λ 0.050 0.010

Swelling index κ 0.010 0.001

Void ratio (p’=98 kPa on N.C.L) e0 1.040 0.500

Poisson’s ratio ν 0.300 0.120

Saturated degrees of saturation 𝑆𝑟𝑠 1.00 1.00

Residual degrees of saturation 𝑆𝑟𝑟 0.40 0.40

Parameter corresponding to drying AEV (kPa) 𝑆𝑑 11000 21000

Parameter corresponding to wetting AEV (kPa) 𝑆𝑤 800 1000

Initial stiffness of scanning curve (kPa) 𝑘𝑠𝑝𝑒 25000 90000

Parameter of shape function c1 0.000001 0.00003



Parameters for

MCC

Temperature change 2 (3D)

Case1

300year15year

0 [℃] 100

Case1

300year15year

0

50

100

150

200

0 50 100 150 200 250 300

0m(ベントナイト)2m10m200m

tem

per

ature

(℃

)

time (year)

熱源からの水平距離Case1

0

50

100

150

200

0 50 100 150 200 250 300

0m2m10m200m

tem

per

ature

(℃

)

time (year)

.

Case1 熱源からの水平距離

Saturated condition(without bentonite)

Unsaturated condition

(with bentonite)

Bentonite

Distance form heat source


2D analysis

-10

0

10

20

30

40

50

60

70

0 10 20 30 40 50


su

ctio

n(M

Pa)

time (year)


-4

-3

-2

-1

0

0 10 20 30 40 50

su

ctio

n(M

Pa)

time (year)

Change of saturation Change of suction

-10

0

10

20

30

40

50

60

70

0 50 100 150 200 250 300


su

ctio

n(M

Pa)

time (year)

.


80

85

90

95

100

0 50 100 150 200 250 300


deg

ree

of

satu

rati

on(%

)

time (year)

.

熱源からの水平距離

Case1

80

85

90

95

100

0 10 20 30 40 50


deg

ree

of

satu

rati

on(%

)

time (year)

Case1熱源からの水平距離

Changes of saturation and suction

2D analysis

Bentonite → absorption

Rock → change from drainage to absorption

• Initial suction is -3MPa (300m beneath water level)

• Bentonite is much less in 3D than in 2D →water supplying from rock to bentonite is much less in 3D

3D analysis

Bentonite → absorption

Rock → always saturated

Bentonite


Bentonite




Bentonite

Bentonite

3D analysis

Bentonite→absorption

Changes of volumetric and deviatory strains (2D)

Saturated condition (without bentonite)

Unsaturated condition (with bentonite)

-25

-20

-15

-10

-5

0

5

0 50 100 150 200 250 300

0m(ベントナイト)2m10m

volu

met

ric

stra

in

v (

%)

time (year)

. Case1熱源からの水平距離

-0.10

-0.05

0.00

0.05

0.10

0 50 100 150 200 250 300

0m2m10m

volu

met

ric

stra

in

v (

%)

time (year)


0

1

2

3

0 50 100 150 200 250 300

2m10m

time (year)


0

0.005

0.01

0.015

0.02

0 50 100 150 200 250 300

2m10m

time (year)


Volumetric strain Deviatory strain

Volumetric expansion of bentonite is far larger than that of rock

mainly coursed by huge decrease of suction

Volumetric expansion near heating source

Thermal effect is confirmed


Bentonite




Substantial shear strain happened in rock when bentonite is involved !

Comparison of temperature between Case 1 and 2 (2D)

0

50

100

150

200

250

300

0 50 100 150 200 250 300


tem

per

ature

(℃

)

time (year)

.


0

50

100

150

200

250

300

0 50 100 150 200 250 300

0m(ベントナイト)2m

10m

200m

tem

per

ature

(℃

)

time (year)

.


Case1

Case2

300year15year

300year15year

0 [ºC] 180

Distribution of temperatureChange of temperature

Heat emission input in FEM

0

0.1

0.2

0.3

0.4

0.5

0 50 100 150 200 250 300


CASE 2 (100-400years)

Hea

t Q

(k

J/m

3/m

in)

Time (year)

1.6%



Bentonite

Bentonite

Change of saturation (2D)

80

85

90

95

100

0 50 100 150 200 250 300


deg

ree

of

satu

rati

on(%

)

time (year)

.


Case1

Bentonite saturated 70years later

80

85

90

95

100

0 50 100 150 200 250 300


deg

ree

of

satu

rati

on(%

)

time (year)

.

Case2


Bentonite saturated 80years later

Reason why saturated quickly when heating is strong

-7000

-6000

-5000

-4000

-3000

-2000

-1000

0

1000

0 20 40 60 80 100

Case1Case2

wat

er h

ead (

m)

time (year)

.

Change of total water head

strong heating results in quick increase of excess pore water pressure

quick decrease of suction (s=ua-uw)

Saturated quickly

Comparison of saturation between Case 1 and 2 (2D)



Bentonite

Bentonite

-30

-25

-20

-15

-10

-5

0

5

0 50 100 150 200 250 300

Case1Case2

volu

met

ric

stra

in

v (

%)

time (year)

.

In Case 2 (relative weak heating source), expansive of bentonite is larger, why?

Change of volumetric

strain of bentonite (2D)

s rd QdS r

Trace in absorption process of bentonite e-lnp

Taking more time to reach saturated state →Sr developed slower→also developed slowersr

sr

lnpMean skelton stress p

Void

rati

o e

Δe

Due to slower develop of , extra Δe happened in Case 2

Comparison of volumetric strain between Case 1 and 2 (2D)

sr

-30

-25

-20

-15

-10

-5

0

5

0 50 100 150 200 250 300

Case1Case2

volu

met

ric

stra

in

v (

%)

time (year)

.

0

0.1

0.2

0.3

0.4

0.5

0.6

0 50 100 150 200 250 300

2m10m

time (year)


0

0.1

0.2

0.3

0.4

0.5

0.6

0 50 100 150 200 250 300

2m10m

time (year)


• Deviatory strain in surrounding rock due to huge swelling of bentonite

• In case 2, volumetric strain in bentonite is much larger→larger deviatory stain

Comparison of deviatory strain between Case 1 and 2 (2D)

Change of volumetric

strain of bentonite (2D)

Change of deviatory strain of rock (2D)

dev

iato

ry s

tra

in (

%)

Distance form heat source Distance form heat source

A thermo-elastoplastic model for unsaturated/saturatedgeomaterials has been introduced based on theconcept of equivalent stress and subloading surface,which can not only describe properly thethermodynamic behavior, but also overconsolidatedbehavior of soft sedimentary rocks.

Verification by laboratory tests and field observationproved the validity of the proposed numerical method

Field equations in THMA FE-FD analysis has beenintroduced to treat thermo-soil-water-air couplingproblem. As a application, a field problem of geologicalrepository of high-level nuclear waste is simulated inunsaturated/saturated condition.

In 2D simulation, the calculated results shows that thetemperature has a great influence on THMA couplingbehavior of natural soft rock.

CONCLUSIONS

Thank you for your attention!

geologic repository of high-level nuclear waste · 2016-07-04 · geologic repository of high-level...

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