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%?45'3 4E1 D50.1= ( @a5 53$ I @a5A J3 4E1 31M4 =41/G 4E1 4'21= .34'0 4E1 $1:611 %C 0'>.'$ =54.654'%3

&6'416'5 561 C.0C'001$ '3 :0%?50 ;Q 2%$10= S54 &53'=416 2'$9E1':E4T 561 &%2/561$ L'4E 4E1 4'21= C%6 LE'&E 4E1 /61==.61 &6'416'5 %C #.CC( 53$ #.CCI 561 C.0C'001$ C%6 4E1 0%&50 ;Q@ 2%$10=A #5=1$ %3 4E1

%.4&%21 %C 4E1 &%2/56'=%3G 4E1 $1:611 %C 0'>.'$ =54.654'%3 &6'416'5 561 1D50.541$ 53$ 61C'31$A

B =&E1254'& =V14&E '00.=4654'3: 4E1 61054'%3 ?14L113 4E1 .=1$ 2%$10= '3 4E'= =4.$7 '= =E%L3 '3 \':.61 (A ;% 4E1 01C4 %C 4E1 D164'&50 E54&E1$ 0'31 L1 E5D1 2%$10= %3 5 :0%?50 =&501A J3 4E1%67G 4E1 2%=4 :131650 2%$10 '= 5 ;Q@92%$10 SLE'&E L1 E161 $% 3%4 E5D1 5&&1== 4%TG 53$ =%21LE54 01== :131650 561 4E1 5D5'05?01 /5'6= %C ;Q92%$10= L'4E '22%?'01 21&E53'&50 61/61=13454'%3= SJ3'4'50 %6 Q%2%:13'[1$T /61=1341$ '3 SUV1==%3 14 50A I<(<G &E5/416 -TA J3 4E1 C%00%L'3:G 4E1 /5'6= SIJG IQT 53$ S+JG +QT L'00 ?1

&%3='$161$A ;% 4E1 6':E4 %C 4E1 D164'&50 E54&E1$ 0'31 '3 \':.61 ( 4E1 0%&50 =&501 ;Q@92%$10= 561 '00.=46541$A B= '3$'&541$ '3 \':.61 (G 4E1 0%&50 2%$10= L'00 %307 ?1 61/61=13454'D1 C%6 5 $'=& %C ?.CC16 54

&53'=416 2'$9E1':E4G 'C 4E1 /61=&6'?1$ #H= 561 61/61=13454'D1 C%6 4E1 Z:0%?50 &%3$'4'%3=f 54 4E'=

/%='4'%3A

Figure 1. Illustration of the relations between models used in this study. Global models are visualized to the left. To the right, a schematic local model geometry, to represent an axisymmetric disc at canister mid-height, is visualized. To the right of the schematic local geometry the states considered are indicated.

;E1 =%0.4'%3= %C %31 /5'6 %C :0%?50 ;Q92%$10=G SIJG IQTG 561 .4'0'[1$ C%6 $1=':3'3: #H= .=1$ '3 4E1 0%&50 2%$10=A ;E1 :0%?50 ;Q92%$10= 561 $1=&6'?1$ '3 $145'0 '3 SUV1==%3 14 50A I<(<G &E5/416 -AI 53$

-A-TA J3 =E%64G 4E1=1 2%$10= 561 &%3='$161$ 5 /5'6 %C ?5=1 &5=1 2%$10= L'4E%.4 1M/0'&'4 C65&4.61

61/61=13454'%3 53$ 5 0%L16 &E%'&1 %C E7$65.0'& 6%&V &%3$.&4'D'47 SK j (<^9(I2W=TA ;E1 1M465&41$ #H= 54

&53'=416 2'$9E1':E4 561X S(T 65$'50 E154 C0.M 54 4E1 &53'=416 =.6C5&1G SIT 412/1654.61 54 4E1 E%01 L500G 53$ S-T 0'>.'$ /61==.61 54 4E1 E%01 L500G 500 4E611 561 =E%L3 '3 \':.61 IA

(9)

, B= &53 ?1 =113 '3 \':.61 IG 4E1 &E%=13 &%3$'4'%3= C%6 4E1 4L% 2%$10= 561 >.'41 50'V1A ;E.=G 4E1 21&E53'&50 61/61=13454'%3= %C 4E1 ?.CC16 =112 3%4 4% 5CC1&4 4E1 ;Q9/6%&1==1= %3 4E1 Z%.4='$1f %C 4E1

?.CC16 4% 5 :6154 $150A K.1 4% 4E1 ='2'056'4'1= ?14L113 4E1 2%$10=^ &%3$'4'%3=G %307 4E1 61=/%3=1= %C IQ 561 .=1$ LE13 $1=':3'3: #H=A J3 \':.61 IG =72?%0= '3$'&541 /%'34= ?14L113 LE'&E 0'3156 '3416/%054'%3 L5= 25$1 LE13 C%62'3: 4E1 #H=A

Figure 2. Responses for the global models and corresponding boundary conditions for local models: radial heat flux at the canister surface (top), temperature at the hole wall (mid), and liquid pressure at the hole wall (bottom).

@%$10 H8;kN?G $1=&6'?1$ '3 &E5/416 *A) %C SUV1==%3 14 50A I<(<TG L5= .=1$ 5= 4E1 =4564'3: /%'34 C%6 4E1 0%&50 2%$10=A J3 4E1=1G E%L1D16G 4E1 /56521416= %C 4E1 21&E53'&50 25416'50 2%$10 %C 4E1 /10014 =0%4 25416'50 E5$ 4% ?1 504161$G 5= &%2/561$ 4% H8;kN?G 4% /6%$.&1 5 61/61=13454'D1 /%6%='47 /6%C'01 54 C.00 L5416 =54.654'%3G 6101D534 C%6 =0%L L5416 =54.654'%3 /6%&1==1=A ;% %?45'3 1=4'2541= %C 61/61=13454'D1 5D165:1$ /%6%='4'1= C%6 4E1 ?0%&V 53$ /10014 =0%4G 4E1 :65/E '3 \':.61 *9() '3 SUV1==%3 14 50A I<(<TG 61/1541$ ?10%L '3 \':.61 -G L5= .=1$A ;E1 :65/E /61=1341$ '3 \':.61 - L5= /6%$.&1$ .='3: 4E1 535074'&50 2%$10 $1=&6'?1$ '3 &E5/416 *A- %C SUV1==%3 14 50A I<(<TA ;E1 '3/.4 4% 4E1 535074'&50 2%$10 L5= &E%=13 5=c ?0%&V D%'$ 654'% e^bj <A,I9<A,-G /61==.61 654'% j <A+9<ARG 53$ 53 5==.2/4'%3 %C 5 Z/5650010 L144'3: /6%&1==fG LE'&E &53 ?1 &%3='$161$ 61/61=134'3: 5 =0%L L144'3: /6%&1==A ;E1 %?45'31$

%.4/.4G 5 653:1 '3 /10014 =0%4 D%'$ 654'%G ?1&521 e^pj <A,*9<A,+A ;L% 0%&50 6150'[54'%3=G 5= $1C'31$

5&&%6$'3: 4% ;5?01 IG L161 $1D10%/1$ '3 2%$10= $13%41$ ;Pak31Lk(I 53$ ;Pak31Lk()A

0 20 40 60 80 100

0 20 40 60 80 100 120 140 160 180 200

Radial heat flux (W/m2)

Time (yr)

2I q_T(r = r_heater) 2H q_T(r = r_heater) BC: q_T(r = r_heater)

0 10 20 30 40 50 60 70

0 20 40 60 80 100 120 140 160 180 200

Temperature (C)

Time (yr)

2I T(r = r_hole) 2H_T(r = r_hole) BC: T(r = r_hole)

-50 -40 -30 -20 -10 0 10

0 20 40 60 80 100 120 140 160 180 200

Liquid pressure (MPa)

Time (yr)

2I p_l(r = r_hole) 2H p_l(r = r_hole) BC: p_l(r = r_hole)

(10)

+

#1C%61 5$$61=='3: 4E1 &6'416'5 4653=054'%3 C6%2 /61==.61 4% $1:611 %C =54.654'%3G L1 45V1 4E1

%//%64.3'47 4% =4.$7 4E1 21&E53'&50 /6%&1== '3 4E1 2%$10= '3 %6$16 4% D'=.50'[1 4E1 E%2%:13'[54'%3 /6%&1== 53$ 50=% LE161 53$ LE13 #.CC( 53$ #.CCI 561 C.0C'001$ '3 4E1 2%$10=A J4 &%.0$ 50=% ?1

2134'%31$ 4E54 5 21=E $1/13$13&7 =4.$7 L5= /16C%621$ C%6 2%$10 ;Pak31Lk(I ?7 .='3: 4E611 4'21=

4E1 3.2?16 %C 1012134= 65$'5007 '3 4E1 /10014 =0%4 53$ ?0%&V D%0.21= '3 5 31L 2%$10

S;Pak31Lk(IkC'31TA _% =':3'C'&534 21=E $1/13$13&7 &%.0$ ?1 =113 5= =E%L3 '3 B//13$'M B A

Figure 3. Compilation of solutions using the analytical model described in chapter 5.4 of ( kesson et al. 2010). The black thick arrows indicate the pellet slot void ratios, e

^p

= 0.75 and e

^p

= 0.78, obtained from applying the input (e

^b

= 0.72, = 0.9, parallel wetting) and (e

^b

= 0.73, = 0.8, parallel wetting), respectively. This graph is a part of Figure 5-14 in ( kesson et al. 2010).

C:! =')#% <'/&% .&56'*5&5

;E1 :65/E 54 4E1 4%/ '3 \':.61 ) =E%L= D%'$ 654'% /6%C'01= %?45'31$ 5C416 &%2/0141$ ='2.054'%3=

4%:14E16 L'4E 4E1 '3'4'50 D%'$ 654'% /6%C'01A B0=%G 4L% 566%L= '3$'&541 4E1 &E53:1 '3 /%='4'%3 53$ D%'$

654'% C%6 4L% /%'34= '3 4E1 ;Pak31Lk(I 2%$10A H%2/56'3: L'4E 4E1 '3'4'50 =4541G 4E1 ?.CC16 E5=

1D%0D1$ '34% 5 &%3='$165?07 2%61 E%2%:13'[1$ =4541A ;E1 /10014 =0%4 25416'50 E5= ?113 &%2/61==1$ ?7 4E1 Z2%61 =46%3:07f =L100'3: ?0%&V 25416'50A ;E1 01C4 566%LG =E%L'3: '3'4'50 53$ C'350 =4541 %C 5 /%'34 '3 4E1 ?0%&V &0%=1 4% 4E1 '3'4'5007 %/13 '3316 =0%4G '3$'&541= 4E1 =L100'3: &E565&416 '3 4E'= /564 %C 4E1

?.CC16A ;E1 6':E4 566%LG ?10%3:'3: 4% 4E1 /10014 =0%4G '3$'&541= 4E1 &%2/61=='%3 45V'3: /05&1 '3 4E'= /564

%C 4E1 ?.CC16A

;E1 0%L16 /564 %C \':.61 ) '= 5 ?0%L9./ %C 4E1 C'350 =4541 '3 4162= %C D%'$ 654'%A #1='$1 4E1 3.216'&50 2%$10= 61=.04=G :'D13 5= /6%C'01= S=72?%0=T 53$ D%0.21 5D165:1= S?05&V 0'31=TG

<A, <A,* <A+ <A+* <AR

<A,

<A,*

<A+

<A+*

<AR

<AR*

(

#0%&V D%'$ 654'%

a10014 =0%4 D%'$ 654'%

= 0.8 = 0.9 = 1.0

= 1.0

= 0.9

= 0.8

= 0.7

= 0.7 Serial wetting (w_ref^block= 0.17, w_ref^pellets= 0.64)

Parallel wetting (wrefpellets= wrefblock) A

C B

D

Assumption of constant volume

CRT

(11)

R VA( ) ≡ 1

( ) ,

%D16 4E1 ?0%&V 53$ /10014 =0%4G 50=% 4E1 535074'&50 =%0.4'%3=G %?45'31$ C6%2 .='3: 4E1 '3C%6254'%3 '3

\':.61 -G 561 :'D13 S:617 4E'&V 0'31=TA ;E1 /10014 =0%4 61/61=13454'%3= 5= $1=&6'?1$ '3 ;5?01 I L161

&E%=13 5= 4% :1316541 D%0.21 5D165:1= '3 &0%=1 5:6112134 L'4E 4E1 535074'&50 =%0.4'%3=G 53$ 5= &53 ?1

=113 4E'= '= 50=% 4E1 &5=1A

F31 2%61 &E1&V %C 4E1 5:6112134 ?14L113 4E1 535074'&50 53$ 3.216'&50 =%0.4'%3= &53 ?1 /16C%621$ ?7

&%2/56'3: 4E1 /61==.61 654'% j ppWpbA J3 4E1 535074'&50 =%0.4'%3 4E1 /61==.61 654'%G :'D13 ?7 4E1 654'%

%C D%0.21 5D165:1= %C /61==.61 %D16 4E1 /10014 =0%4 53$ ?0%&VG L5= &E%=13 5= jl<A+G <ARmA \%6 ?%4E 0%&50 2%$10= j<A+* LE'&E &015607 5:611= L100 L'4E 4E1 '3/.4 4% 4E1 535074'&50 2%$10A

(12)

(<

Figure 4. Void ratio profiles obtained for the two local models at full simulation time. The initial profile (hatched line) and the two solutions (symbols) are shown at top where the arrows indicate initial and final states for two different nodes belonging to TEP_new_12. The bottom graph shows a close up at the final state for both models (symbols) with the analytical solution (grey thick line) also indicated as well as the volume averages (thin lines) over block and pellet slot for the model solutions.

;% D'=.50'[1 4E1 &0%=.61 %C 4E1 %/13 :5/ &0%=1 4% 4E1 &53'=416 53$ 4E1 E%2%:13'[54'%3 /6%&1== %C 4E1

?.CC16G \':.61 * =E%L= 4E1 /54E %C 25416'50 /564'&01= S:617 0'31=TG '3'4'5007 /%='4'%31$ L'4E 1>.50

$'=453&1=G C%6 ?%4E 0%&50 2%$10=A ;E1 '3416C5&1= S'3316 12/47 :5/G ?0%&VT 53$ S?0%&VG /10014 =0%4T 561 50=% '3$'&541$ S=%0'$ ?05&V 0'31=TA B3 '=%0'31 &%3='=4'3: %C S4'21G /%='4'%3T9/5'6=G C%6 LE'&E 4E1

$1C%6254'%3 '= <ARR %C 4E1 C'350 $1C%6254'%3 '= 50=% :'D13 '3 \':.61 * SE54&E1$ 0'31 L'4E =72?%0=TA

!%0.4'%3 $545 561 :'D13 '3 ;5?01 -A

eE13 =4.$7'3: 4E1 /564'&01 /54E=G '4 &53 ?1 =113 4E54 4E1 $1C%6254'%3 %C 4E1 ?0%&V 25416'50 &53 ?1 4E%.:E4 %C 5= ?10%3:'3: 4% 4L% 2%$1=X S(T =L100'3: '34% 4E1 /10014 =0%4 53$ SIT '3316 =0%4 &0%=.61A J3'4'5007G 53 '3=4534 65/'$ &%2/61=='%3 %C 4E1 /10014 =0%4 &53 ?1 =113 LE13 4E1 %.416 /564 %C 4E1 ?0%&V

0.50 0.70 0.90 1.10 1.30 1.50 1.70 1.90

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

Void ratio [-]

Radial position [m]

e(TEP_new_12)

e(TEP_new_14)

0.70 0.72 0.74 0.76 0.78 0.80 0.82

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

Void ratio [-]

Radial position [m]

e(TEP_new_12) aver(e(TEP_new_12)) e(TEP_new_14) aver(e(TEP_new_14))

(13)

((

45V1= ./ L5416 53$ 1M/53$=A eE13 L5416 615&E1= C.64E16 '3G 4E1 %/13 '3316 =0%4 ?1:'3= 4% &0%=1 53$

C'35007 4E1 ?0%&V E5= 1M/53$1$ 500 4E1 L57 4% 4E1 &53'=416A ;E1 4L% Z$1C%6254'%3 2%$1=f 0'=41$ 5?%D1 5&4 '3 $'CC16134 $'61&4'%3=G 4E161?7 4E1 D'='?01 V'3V %3 4E1 /564'&01 /54E= 54 4E1 4'21 LE13 4E1 '3316 =0%4

&0%=1=A

B= C%6 4E1 &E565&416 %C 4E1 C'350 =45:1 %C $1C%6254'%3 '4 &53 ?1 =%21LE54 .3$16=4%%$ ?7 =4.$7'3: 4E1 5//15653&1 %C 4E1 '=%0'31A ;E1 2%=4 %?D'%.= C154.61 '= 4E54 4E1 25416'50 &0%=1 4% 4E1 &53'=416 '= 4E1 C'6=4 4% &15=1 $1C%62'3: L'4E 4E1 1M46121 54 4E1 '3316 =0%4 &0%='3: 5C416 (( 9 () 76A B4 65$'' ` <ANI* 2 4E161 '=G L'4E '3&615='3: 65$'.=G 5 :65$.50 E504 %C 4E1 $1C%6254'%3 C6%2 (( 9() 76 4% 5?%.4 (*) 76G LE'&E 50=%

'= 4E1 25M'2.2 D50.1 C%6 4E1 134'61 ?.CC16A \%6 65$'' b <ANI* 2 4E1 E504 %C 4E1 $1C%6254'%3 '= 45V'3:

/05&1 Z2%61 ='2.04531%.=fG '3 5 4'21 653:1 5?%.4 (I< n ()* 76A

;Pak31Lk(I ;Pak31Lk()

Figure 5. Local models deformation evolution (solid lines) and isoline consisting of (time, position)-pairs for which the deformation is 0.99 of the final is indicated (hatched),i.e. tS|u

r

|T≈

0.99 |u

r

St=200T|.

;% D'=.50'[1 LE13 53$ LE161 #.CC( 53$ #.CCI 561 C.0C'001$ '3 4E1 2%$10=G \':.61 N =E%L= '=%0'31=

L'4E 1012134= S4'21G '3'4'50 65$'50 /%='4'%3T C%6 LE'&E p ≈ ( @a5 S=%0'$T 53$ p ≈ I @a5 SE54&E1$TA

!%0.4'%3 $545 61:56$'3: 4E'= 53507='= 561 50=% :'D13 '3 ;5?01 -A

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 50 100 150 200

Radial position (m)

Time (yr)

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 50 100 150 200

Radial position (m)

Time (yr)

(14)

(I

;E1 2%=4 &%3=/'&.%.= C154.61 %C ?%4E '=%0'31= '= 4E1 $'=4'3&4 $'CC1613&1 C%6 ?0%&V 53$ /10014 =0%4A ;E1 /10014 =0%4 25416'50 615&E 4E1 /61==.61 01D10= =':3'C'&53407 05416 5= &%2/561$ 4% 31':E?%6'3: /%'34= '3 4E1 ?0%&V 25416'50A JC 4E'= '= '3 5&&%6$53&1 L'4E 6150'47 %6 Y.=4 5 253'C1=454'%3 %C 4E1 5$%/41$ 25416'50 61/61=13454'%3 '=G E%L1D16G 3%4 V3%L3A B3%4E16 %?D'%.= :131650 C154.61 '= 4E1 1D%0.4'%3 %C 5445'3'3:

4E1 /61==.61 01D10=G C'6=4 %&&.66'3: '3 4E1 %.416 /564 %C 4E1 ?0%&V 25416'50 53$ 4E13 /6%:61=='3: '3L56$=A B= &53 ?1 =113G 4E161 '= =%21 $'D16:13&1 C6%2 4E'= 4613$ C%6 p ≈ ( @a5 '3 4E1 '3316 /564 %C 4E1 ?0%&V 25416'50 &0%=1 4% 4E1 '3'4'5007 %/13 '3316 =0%4A

J4 '= 5:5'3 =461==1$ 4E54 4E1 ?1E5D'%6 $1=&6'?1$ 5?%D1 &%3&163= 2%$10=G 53$ 4E54 '4 '= .3V3%L3 4% LE54

$1:611 4E'= 5:611= L'4E LE54 %&&.6 '3 6150'47A

Figure 6. Local model isolines consisting of (t,r

0

)-pairs for which p ≈ 1 MPa (solid) and p ≈ 2 MPa (hatched)A

\'35007G '3 %6$16 4% $1D10%/ 31L #.CC( 53$ #.CCI &6'416'5 1M/61==1$ '3 4162= %C $1:611 %C 0'>.'$

=54.654'%3G 4E1 61054'%3 %C 4E'= D56'5?01 4% /61==.61 E5= 4% ?1 1=45?0'=E1$A ;% C5&'0'4541 4E'=G 5 /6%&1$.61

='2'056 4% LE54 L5= $%31 LE13 /6%$.&'3: 4E1 :65/E= '3 \':.61 N '= .4'0'[1$A ;E'= '= ?5=1$ %3

D'=.50'[54'%3 %C LE13 53$ LE161 #.CC( 53$ #.CCI 561 C.0C'001$ '3 4E1 2%$10=G 5:5'3 .='3: '=%0'31= C%6 LE'&E p ≈ ( @a5 53$ p ≈ I @a5G ?.4 4E'= 4'21 4E1 '=%0'31= &%3='=4 %C S$1:611 %C 0'>.'$ =54.654'%3G '3'4'50 65$'50 /%='4'%3T9/5'6=G =11 \':.61 ,A J3 4E1 :65/E= 4E1 '3'4'50 =4541 %C 4E1 ?0%&V 53$ /10014 =0%4 25416'50 '= 50=% '3$'&541$ C%6 61C1613&1A !%0.4'%3 $545 61:56$'3: 4E'= 53507='= 561 :'D13 '3 ;5?01 -A B:5'3G 5= 50=% L5= 4E1 &5=1 LE13 =4.$7'3: 4E1 :65/E= '3 \':.61 NG 4E1 ?0%&V 53$ /10014 =0%4 25416'50=

=E%L $'=4'3&4 $'CC1613&1=A J3 \':.61 , 4E1 /10014 =0%4 25416'50 5445'3= 4E1 &6'416'5 54 5 &%3='$165?01 0%L16

$1:611 %C =54.654'%3 5= &%2/561$ 4% 4E1 ?0%&V 25416'50 53$ 4E1 '=%0'31= 50=% =/53 %D16 5 056:1 653:1 LE'&E '3$'&541 4E54 Sl'= 3%4 53 '$150 '3$'&54%6 C%6 /61==.61 '3 4E1=1 =7=412=A ;E1 0%L16 /61==.61 '=%0'31

%C 4E1 ?0%&V 25416'50 E5= 5 /1&.0'56 C154.61 '3 4E1 '3316 /564A ;E1 0%L16 /61==.61 01D10 S( @a5T '=

5&4.5007 %?45'31$ C%6 Sl1>.50 4% 4E1 '3'4'50 D50.1 Sl<A @%=4 0'V107G 4E'= &%21= C6%2 4E1 &E565&416 %C 4E1 E%2%:13'[54'%3 /6%&1== LE161 4E1 $67 53$ =4'CC '3316 /564 %C 4E1 ?0%&V '= &%2/61==1$ ?7 =L100'3: %C

%.416 /564= 53$ 4E161C%61 4E1 /61==.61 '3&615=1= $1=/'41 3% 0%&50 L5416 ./45V1 '3 4E1 '3316 /564A ;E'=

50=% '3$'&541= 4E54 Sl'= 3%4 53 '$150 '3$'&54%6 C%6 /61==.61 '3 4E1=1 =7=412=A

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 5 10 15 20 25 30

Radial position (m)

Time (yr)

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 5 10 15 20 25 30

Radial position (m)

Time (yr)

(15)

(-

Figure 7. Local model isolines consisting of (S

l

,r

0

)-pairs for which p ≈ 1 MPa (solid black) and p ≈ 2 MPa (hatched) and initial state (grey).

"#$%& 2: 3'%0+1'* /#+# ('. +9& %')#% <'/&%5:

H6'416'%3 @%$10X ;Pak31Lk(I @%$10X ;Pak31Lk()

#0%&V a10014 =0%4 #0%&V a10014 =0%4

purp≈ <ARR qpurStjI<<Tp t j l()G (*)m 76 t j l(I)G ()<m 76 t j l((G (*)m 76 t j ()< 76 p ≈ ( @a5

t j l(AIG RAIm 76 Slj l<A+*G <A+Rm iBSSlTj <A++

t j () 76

Slj l<A-)G <A))m iBSSlT j <A-R

t j l(AI G *ANm 76 Slj l<A+*G <A+Rm iBSSlT j <A++

t j (( 76

Slj l<A--G <A)Im iBSSlT j <A-+

p ≈ I @a5

t j lIARG I<m 76 Slj l<AR<G <ARIm iBSSlT j <AR(

t j I* 76

Slj l<A-RG <A*)m iBSSlT j <A),

t j lIA-G (,m 76 Slj l<A+RG <ARIm iBSSlT j <AR(

t j I- 76 Slj l<A-RG <A*)m iBSSlT j <A)*

C:, I&?&%'6<&*+ '( *&J ).1+&.1#

\'6=4G Y.=4 4% 612'3$ 4E1 615$16G 4E1 %D16500 %?Y1&4'D1 '= 4% $1C'31 '3416D50= '3 $1:611 %C =54.654'%3

&%./01$ 4% &%3$'4'%3=W1D134= 4E54 /6%2%41 0%3: 4162 =5C147A ;E1 =5C147 C.3&4'%3= $1C'31$ ?7 !"# 257

?1 610541$ 4% &6'416'5 1M/61==1$ '3 4162= %C /61==.61A ;E1 45=V E161 4E13 '= 4% 4653=0541 4E1 /61==.61

&6'416'5 4% &6'416'5 :'D13 '3 4162= %C $1:611 %C =54.654'%3A

e'4E 4E1 C%6216 $'=&.=='%3 610541$ 4% 4E1 61=.04= '3 \':.61 , '3 2'3$G &015607G 4E161 1M'=4 /%4134'50

%?Y1&4'%3= C%6 .='3: Sl5= 5 :131650 '3$'&54%6 %C /61==.61A B %3194%9%31 &%661=/%3$13&1 &1645'307 3%4 1M'=4=A J3 %6$16 4% 615&E 4E1 %?Y1&4'D1 =14 %.4 C%6G 5 &%3&1/4 %C '3416D50= '3 $1:611 %C =54.654'%3

&%661=/%3$'3: 4% 5 &1645'3 25:3'4.$1 '3 /61==.61 L'00 ?1 .=1$A

!'3&1 4E1 :0%?50 53$ 0%&50 2%$10= $% 3%4 &%661=/%3$ 1M5&407 '3 4162= %C '3'4'50 $1:611 %C =54.654'%3G Sl<G 61054'D1 $1:611 %C =54.654'%3G

≡ −

1 − ,

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Radial position (m)

Degree of water saturation (-)

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Radial position (m)

Degree of water saturation (-)

(16)

()

L'00 ?1 .=1$ LE13 C%62.054'3: 4E1 &6'416'5A J3 %6$16 4% %?45'3 &6'416'5 2535:15?01 LE13 1D50.54'3: 4E1 :0%?50 2%$10=G 4E17 L'00 ?1 C%62.0541$ '3 4162= %C D%0.21 5D165:1= %C 61054'D1 $1:611 %C =54.654'%3G VA G %D16 4E1 ?0%&V 25416'50 53$ /10014 =0%4 25416'50A K.6'3: 4E1 '3D1=4':54'%3 '4 L5= C%.3$

&%3D13'134 4% L%6V '3 4162= %C VA( )G :'D13 '3 ;5?01 -G 53$ '3=415$ %C VA .='3: 4E1 61054'%3G

VA = 1

1 − VA( ) − .

;E1 &E%=13 &6'416'5 C%62.0541$ '3 4162= %C 5 653:1 '3 VA $1C'31$ ?7 2'3'2.2 53$ 25M'2.2 D50.1=A ;E1=1 D50.1= 561 %?45'31$ C6%2 1D50.54'3: 4E1 L61=/%3=1 %C 4E1 0%&50 2%$10=A B ='3:01

&6'416'%3 '= $1C'31$ ?7G

criterion ≡ {minimum,maximum}

minimum ≡ VA − max − VA

maximum ≡ max

eE13 &50&.054'3: 4E1 &6'416'5 ?5=1$ %3 4E1 1M/61=='%3= 5?%D1G 4E1 D50.1= :'D13 '3 ;5?01 ) 561 %?45'31$A

"#$%& 8: M1.5+ ?&.51'* '( *&J N0((! #*/ N0((, ).1+&.1# &K6.&55&/ 1* +&.<5 '( .&%#+1?& /&7.&& '( %1O01/

5#+0.#+1'* ('. +9& $%')E #*/ 6&%%&+ 5%'+ <#+&.1#%5:

H6'416'%3 1M/61==1$ '3 p H6'416'%3 1M/61==1$ '3 iBSSlrelT

#0%&V a10014 =0%4

#.CC( ≈ ( @a5 {0.145, 0.278} l0.165, 0.285}

#.CCI ≈ I @a5 l0.302, 0.487} l0.225, 0.414}

\':.61 + =E%L= 4E1 &6'416'5 :65/E'&5007G '3$'&541$ ?7 4E1 :617 =%0'$ 53$ :617 E54&E1$ 0'31=G C%6 #.CC(

S=%0'$ :617T 53$ #.CCI SE54&E1$ :617TG 61=/1&4'D107A ;E1 &6'416'5 561 =E%L3 4%:14E16 L'4E 4E1 =521 '3C%6254'%3 5= \':.61 ,G ?.4 3%L 1M/61==1$ '3 4162= %C A B= 4E1 &6'416'5 561 $1=':31$G 4E17 1M&0.$1 4E1 Z45'0f %C 0%L D50.1= %C =54.654'%3 '3 4E1 ?0%&V 25416'50 &0%=1 4% 4E1 &53'=416A ;E1 &6'416'5 561 &%3='$161$ %3 4E1 &%3=16D54'D1 ='$1 '3 4E1 =13=1 4E54 4E17 2%=4 0'V107 %D161=4'2541 4E1 4'21 .34'0

#.CC( 53$ #.CCI 561 C.0C'001$A

(17)

(*

Figure 8. Local model isolines consisting of (S

lrel

,r

0

)-pairs for which p ≈ 1 MPa (solid black) and p ≈ 2 MPa (hatched black). Also, the adopted minimum and maximum limits are

indicated in solid and hatched grey lines, for Buff1 and Buff2, respectively.

C:2 P?#%0#+1'* #*/ .&(1*&<&*+ '( *&J ).1+&.1#

;E1 C'6=4 D16='%3= %C 4E1 31L &6'416'5G 1M/61==1$ '3 VA G 561 E161 4% ?1 1D50.541$A ;'21 '3416D50=

LE13 4E1 31L &6'416'5 561 615&E1$ '3 4E1 :0%?50 2%$10=G IJ 53$ IQG 561 &%2/561$ L'4E 4'21 '3416D50= LE13 4E1 &6'416'5 1M/61==1$ '3 p 561 615&E1$ '3 4E1 0%&50 2%$10=G ;Pak31Lk(I 53$

;Pak31Lk()A ;E1 53507='= '= /16C%621$ 54 &53'=416 2'$9E1':E4 %C 4E1 :0%?50 2%$10= '3 %6$16 4% %?45'3 5= &0156 &%2/56'=%3 4% 4E1 0%&50 2%$10= 5= /%=='?01A

J3 \':.61 R 4E1 :0%?50 2%$10 61=/%3=1 '3 4162= %C D%0.21 5D165:1= %C 61054'D1 $1:611 %C 0'>.'$

=54.654'%3 C%621$ %D16 4E1 ?0%&V 53$ /10014 =0%4 54 &53'=416 2'$9E1':E4G 5= L100 5= 4E1 C'6=4 D16='%3

&6'416'5 5= 0'=41$ '3 ;5?01 ) 561 =E%L3A B= C%6 4E1 /10014 =0%4 '4 L5= $'=&614'[1$ .='3: 4L% 1012134=

65$'5007G 4E.= 4E611 3%$1= LE161 4L% ?10%3: 4% '3416C5&1= 53$ 4E1 2'$ 3%$1 ?10%3: 4% 4E1 /10014 =0%4 25416'50 =%0107A ;E1 D%0.21 5D165:1 '= 4E161C%61 45V13 5= 4E1 61=.04 54 4E1 2'$93%$1A ;E1 4'21 '3416D50= %?45'31$ C6%2 5//07'3: 4E1 &6'416'5 '3 ;5?01 ) %3 4E1 1D%0.4'%3 &.6D1= '3 \':.61 R 561 0'=41$

'3 ;5?01 *A

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Radial position (m)

Relative degree of water saturation (-)

0.525 0.575 0.625 0.675 0.725 0.775 0.825 0.875

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Radial position (m)

Relative degree of water saturation (-)

(18)

(N

#0%&Va10014 =0%4

Figure 9. Evolution of volume averages of relative degree of liquid saturation, for global models 2I and 2H are shown together with the first version of S

l

criteria. The volume averages over the pellet slot and buffer are formed at canister mid-height.

"#$%& C: Q$+#1*&/ +1<&5 F1* 4&#.5H J9&* +9& #/'6+&/ 3%).1+&.1# #.& (0%(1%%&/ ('. 7%'$#% <'/&%5 ,- #*/ ,R:

@%$10X IJ @%$10X IQ

#0%&V a10014 =0%4 #0%&V a10014 =0%4

#.CC( t={1.8, 3.4} t={5.8, 16} t={5.5, 12} t={0.058, 0.22}

#.CCI t={3.7, 6.8} t={13, 23} t={13, 22} t={0.26, 2.3}

a0%44'3: 4E1 4'21 '3416D50= LE13 4E1 p 53$ Sl&6'416'5 561 615&E1$ '3 4E1 ?0%&V 53$ /10014 =0%4 25416'50=

C%6 4E1 0%&50 2%$10= S'3 :617T 53$ :0%?50 2%$10= S4E'3 ?05&V 0'31=TG :'D1= 4E1 $'5:652= =E%L3 '3

\':.61 (<A J3 5$$'4'%3G 4E1 4E'&V ?05&V 0'31= '3$'&541 &%2?'31$ &%3='$1654'%3= %C ?%4E :0%?50 2%$10=G 'A1A %D16500 25M'2.2 53$ 2'3'2.2 D50.1= C%6 ?%4E :0%?50 2%$10=A

0 0.2 0.4 0.6 0.8 1

0 20 40 60 80 100 120 140 160 180 200

Relative degree of liquid saturation [ ]

Time [yr]

2I 2H Buff1 Buff2

0 0.2 0.4 0.6 0.8 1

0 20 40 60 80 100 120 140 160 180 200

Relative degree of liquid saturation [ ]

Time [yr]

2I 2H Buff1 Buff2

(19)

(,

\%6 4E1 ?0%&V 25416'50G 4E1 :0%?50 2%$10= 4%:14E16 L'4E 4E1 C'6=4 D16='%3 &6'416'5 /6%$.&1= 4'21

'3416D50= C%6 #.CC( 53$ #.CCIG LE'&E 561 '3 :%%$ 5:6112134 L'4E 4E%=1 %C 4E1 0%&50 2%$10=A ;E161C%61G 3% 5$Y.=42134= 561 25$1 4% 4E1 C'6=4 D16='%3 &6'416'5 C%6 4E1 ?0%&V 25416'50A

\%6 4E1 /10014 =0%4 25416'50G 4E1 :0%?50 2%$10= 4%:14E16 L'4E 4E1 C'6=4 D16='%3 &6'416'5 $% 3%4 254&E 4E1 61=.04= %C 4E1 0%&50 2%$10= L100 C%6 1'4E16 #.CC( %6 #.CCIA \%6 #.CC( 4E1 E%2%:13'[1$ :0%?50 2%$10 SIQT =1D16107 .3$161=4'2541= 4E1 4'21 .34'0 #.CC( '= 615&E1$G LE1615= 4E1 4'21 '3416D50 %C 4E1 '3'4'50

=4541 2%$10 SIJT '3&0.$1= 4E1 0%&50 4'21 '3416D50A \%6 #.CCIG E%L1D16G 4E1 4'21 .34'0 C.0C'02134 '=

=':3'C'&53407 .3$161=4'2541$ ?7 IQ 53$ =%21LE54 .3$161=4'2541$ ?7 IJA B3 5$Y.=41$ &6'416'%3 '=

4E161C%61 61:56$1$ 31&1==567 C%6 4E1 /10014 =0%4 25416'50A

#0%&Va10014 =0%4

Figure 10. Time intervals when the first version criteria are reached for the local models (grey) and global models, 2I (thin black solid) and 2H (thin black hatched). The thick black solid lines indicate the time intervals resulting from using the combined result of both global models. Note that the criteria generating the above are expressed in pressure for the local (THM) models and in degree of liquid saturation for the global (TH) models.

;% :14 ?14416 &%661=/%3$13&1 ?14L113 4E1 :0%?50 53$ 0%&50 4'21 '3416D50= C%6 4E1 /10014 =0%4 25416'50 L'4E%.4 &%2/0'&54'3: 4E'3:= 4%% 2.&EG 4E1 25M'250 &6'416'%3 %C #.CCI '= &E%=13 1>.50 4% 4E54 %C 4E1

?0%&V 25416'50 SVA b 0.487T 53$ %307 4E1 '3'4'50 =4541 2%$10 SIJT '= &%3='$161$ '3 4E1 Z&%2?'31$f &%3='$1654'%3A

;E1 61=.04 C6%2 4E1=1 5$Y.=42134= &53 ?1 =113 LE13 =4.$7'3: \':.61 (( 53$ 4E1 C'350 D16='%3 %C 4E1

&6'416'5G .=1$ '3 4E1 1D50.54'%3G '= =E%L3 '3 ;5?01 NA

;E1 5$Y.=42134= $1=&6'?1$ 5?%D1 2':E4 =112 6%.:EG ?.4 '4 =E%.0$ ?1 61212?161$ 4E54 4E1

E%2%:13'[1$ :0%?50 2%$10 E5= 5 D167 %D16='2/0'C'1$ 61/61=13454'%3 %C 4E1 /10014 =0%4 25416'50 53$ ?7

$'=61:56$'3: 4E'= %31 257 =57 4E54 5 E':E16 $1:611 %C 5&&.65&7 E5= ?113 %?45'31$A B0=%G LE13 .='3:

4E1 5$Y.=41$ &6'416'5G 4E1 :0%?50 4'21 '3416D50= '3&0.$1 4E1 &%661=/%3$'3: 0%&50 4'21 '3416D50= 53$

%?45'3 0%L16 0'2'4= &0%=16 4% 4E1 0%&50 2%$10 '3416D50=A

0 5 10 15 20 25 30 35 40

Time [yr]

Local 2H 2I

Global(2I & 2H)

Buff1Buff2

0 5 10 15 20 25 30 35 40

Time [yr]

Local 2H 2I

Global(2I & 2H)

Buff1Buff2 endering: DokumentID 1415873, Version 1.0, Status Godkänt, Sekretessklass Öppen

(20)

(+

a10014 =0%4

Figure 11. Time intervals when the adjusted pellet slot criteria limits are reached for the local models (grey) and global models, 2I (thin black solid) and 2H (thin black hatched). The thick black solid lines indicate the time intervals resulting from using the adjusted pelllet slot criteria where only the time interval of 2I is considered. Note that the criteria generating the above are expressed in pressure for the local (THM) models and in degree of liquid

saturation for the global (TH) models.

"#$%& B: M1*#% ?&.51'* '( #/S05+&/ N0((! #*/ N0((, ).1+&.1# &K6.&55&/ 1* +&.<5 '( .&%#+1?& /&7.&& '( %1O01/

5#+0.#+1'* ('. +9& $%')E #*/ 6&%%&+ 5%'+ <#+&.1#%5:

H6'416'%3 1M/61==1$ '3 p H6'416'%3 1M/61==1$ '3 iBSSlrelT

#0%&V a10014 =0%4r

#.CC( ≈ ( @a5 {0.145, 0.278} l0.165, 0.285}

#.CCI ≈ I @a5 l0.302, 0.487} l0.225, 0.487}

r F307 4E1 61=.04= %C 4E1 '3'4'50 =4541 2%$10 561 4% ?1 &%3='$161$ C%6 4E1 /10014 =0%4 25416'50A

0 5 10 15 20 25 30 35 40

Time [yr]

Local 2H 2I

Global(2I & 2H & Adj.)

Buff1Buff2 endering: DokumentID 1415873, Version 1.0, Status Godkänt, Sekretessklass Öppen

-A(A B3507='= %C =54.654'%3 '3416D50= &%3&163'3:

!.//012134567 25416'50 '3 5$$'4'%3 4% !89!'41 2%$10'3: 61/%64 ;89(<9(( 5= 61>.1=41$ ?7 !!@