Text

Structural Safety Evaluation of Cracked Fort St Vrain Core Support Block During Loss of Forced Circulation W/Firewater Cooldown
	ML20003B036
Person / Time
Site:	Fort Saint Vrain
Issue date:	12/31/1980
From:	James Kim, Vollman R; GENERAL ATOMICS (FORMERLY GA TECHNOLOGIES, INC./GENER
To:	;
Shared Package
ML20003B033	List: ML20003B032; ML20003B036;
References
	GA-C16190, NUDOCS 8102100250
	Download: ML20003B036 (70)
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STRUCTURAL SAFETY EVALUATION OF A s

CRACKED FORT ST. VRAIN CORE SUPPORT BLOCK DURING h

LOFC WITH FIREWATER COOLDOWN i

w by JOE KD1 and RUSSELL E. VOLDIAN i

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n ir t

Prepared under i

Purchase Order N-2938 f

For Public Service Company of Colorado L

l, hr i

P00R ORGNAL k

E GENERAL ATOMIC PROJECT 2970 I

DECEMBER 1980 p

l GENERAL ATOMIC COMPANY i10210.0 ge

2 GA-C16190 s

STRUCTURAL SAFETY EVALUATION OF A CRACKED FORT ST. VRAIN CORE SUPPORT BLOCK DURING LOFC WITH FIREWATER C00LDOWN AUTHORS j,

WM~

Joe Kim and Russell E. Vollman

[

REVIEWER

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Fu H. Ho i

APPROVALS

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H. Jones

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S O Manager, Core Design W. Simon

, g.g _se Manager, Core Engineering gg lh O hD %

/ 36[8' G. Bramblett/J. Lopez 5

Project Manager, FSV Project U DECEMBER 1980 Me P00R BRIGINAL

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SLTIARY The purpose of this study is to determine the safety consequences of a cracked core support block on the subsequent cooldown of the FSV-HTCR during a hypothetical LOFC with firewater cooldown after a 90 minute delay.

A very conservative cracking pattern is derived f rom thermal stress fields predicted by LASL personnel (Ref.1) f rom ORNL ORECA calculations of core thermal performance during the hypothetical event.

The results show that the cracked core support block remains in place with minimal disarray of the fuel columns.

It is concluded that the cracked core support block does not interfere with safe cooldown of the reactor core.

O h

111 I-

CONTENTS

SUMMARY

iii 1.

INTRODUCTION 1-1.

2.

GEOMETRIC CONFIGURATION.

2-1 2.1.

Core Support Structure 2-1 2.1.1.

Core Support Posts 2-1 2.1.2.

Core Support Block.

2-3 3.

CORE BARRP JD GAP ANALYSIE 3-1 3.1.

Normt Operation Gap 3-1 3.2.

Core Barrel Temperatt.re During Loss of Forced Circulation Event 3-1 3.3 Gap Due to Core Barrel Thermal Expansion 3-1 4.

CORE SUPPORT BLOCK CRACKING ANALYSIS-4-1 s

4.1.

Assumption of Analysis 1 4.2.

Cracking Pattern 4-2 5.

ANALYSIS OF CRACKED ELOCK - WITH KEYS.

'5-1 5.1.

Force Equilibrium

.5-1 5.2.

Stress Results 5 6.

ANALYSIS OF CRACKED BLOCK - WITHOUT' KEYS 1-6.1.

Force Equilibrium. 1-6.2.

Stress Results 6-1 6.3 Jammed Core' Support Block.

6-1:

i 7.

SUMMARY

OF RESULTS 7......................

i 8.

CONCLUSION 8-1 i

9.

REFERENCES 9-1 r

10.

ACKNOWLEDGMENT 10-1 APPENDIX A: WEIGHT CALCULATION-A-li l

I t

APPENDIX B: KINEMATICS OF CSB.

-B-1:

APPENDIX C:

BEARING SURFACE AREA C-11 i

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I APPENDIX D:

STATIC EQUILIBRIUM OF CENTER PIECE OF CSB.

D-1 APPENDIX E: ANALYSIS OF CRACKED BLOCK - WITH KEYS.

E-1 i

APPENDIX F:

ANALYSIS OF CRACKED BLOCK - WIIHOUT KJY1 F-1.

l APPENDIX G: CORE BARREL THERMAL EXPANSION.

C-1 1

FIGURES 1.

Core support arrangement 2 >

2.

Core barrel temperatures and maximum PCRV side wall thermal i

barrier surface temperatures during accident 3-3 l

3.

Core support floor configuration 5 4.

LOFC gaps before cracking.

3-6 I

. 4-3 5.

Cracking pattern of core support block 6.

Cracking pattern of core support block

. 4-4 7.

Cracking pattern of core support block.

4............

.[

8.

Cracking pattern of core support block 4-7' s

4-8 9.

Cracked (ore support block 10.

Core support block before crack.

4-9 i

11.

Loads en cracked block before key sheared o#f.

5-2 l

12.

Loads on cracked block with friction-forces before key-sheared off - case 1 3 13.

Loads on cracked block with friction forces before' key-l sheared off - case '

5-4 4

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14.

Loads on block after key sheared off 6-2 15.

Jammed core support block.

6-3 1

16.

Cracked and tilted 'alock 6-6 A-1 17.

Dislocation of center piece of CSB.

18.

A part of center piece of CSB (upside down).

........... D-1 19.

Force equilibrium without friction' forces.

E-1 20.

Force ~ equilibrium with friction forces - case 1..

- E-5

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21.

Force equilibrium with friction forces - case:2.

E-6 I

22.

Plane. view of_ core' support block

.F-4 23.

Profile view of Fig. 22a-a F-5 24.

A center piece of cracked CSB.

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Summary of core barrel and gap analysis 3-2 2.

Stress results.

6-4 3.

Summary of static equilibrium analysis - with keys E-5 4.

Summary of static equilibrium analysis with friction forces -

with keys, case 1 E-7 5.

Summary of static equilibrium analysis with friction forces -

with keys, case 2 E-9 6.

Swnmary of static equilibrium analysis - without keys F-4 l

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1.

INTRODUCTION During operation of the Fort St. Vrain High Temperature Gas-Cooled Reactor (HIGR) a Loss of Forced Circulation (LOFC) event could hypothet-ically occur. The purpose of this study is to determine the safety conse-quence of an LOFC with subsequent cooldown on one Pelten wheel-driven circulator powered by the firewater system aided by booster pumps on a PGX graphite core support (CS) block found to' experience rather'high thermal stresses during the' cooldown transient.

The overall graphite and coolant temperature history was predicted by (

\\'

Dr..Syd Ball using the ORECA code at ORNL.

4 The results of the above thermal analysis were used as thermal bound-ary conditions for a thermal stress analysis. of fche affected CS block by Dr, Tom Butler of Los Alamos Scientific Laboratory - (LASL). - The results of this analysis indicated high thermal stresses in the keyways of the

~

CS blocks.

In discussions of the results it could not be conclusively proven whether or not the possible cracks that could form would propagate leading to. collapse of.the CS block.

Thus, Lit'was decided to perform an

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analysis of the CS block to determine the consequences to the CS function of thermal stress cracking.

Furthermore, as a basis for conservatism, the cracks would be assumed to propagate to a very conservative pattern that could be inferred f rom the stress field' predicted by Dr.' T. Butler.

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GEOMETRIC CONFIGURATION 2.1.

CORE SUPPORT STRUCTURE i

The CS consists of the concrete CS floor, the graphite CS floor

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assembly, and the lateral restraint assembly.

The' graphite'CS floor is assembled from graphite blocks and posts to provide a stable horizontal

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surface upon which the core and reflector rest, as shown in Fig. 1. ~The modular design allows for independent thermal movements between the refuel-ing regions and the permanent side reflectors (PSR). LThree core support posts support each CS block-in a stool fashion. The post' ends'are spheri :

cal and roll on spherical cups to accommodate relative movement between the i

core and concrete CS floor. :In this'way stressesLdue'to differential move-ments are eliminated, is i

The core and CS are restrained laterally by-a steel core' barrel which.

j-is keyed radially to the PCRV. The CS floor and the top: of the PSR 'are l

keyed to the core ~ barrel.

In this way' radial and vertical-movements are.

permitted but not rotation about a vertical axis.

~

l The CS posts and CS blocks will be described in more' detail in the-following paragraphs.

. 2.1.1.

Core Support Posts-4 Primary functions of support' posts are as follows:

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1.

To support-the_CS blocks, reactor-core, top'and bottom'reflec-I.

i tors,J PSR's, Land top plenum. orifice elements.

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D 2.

To.provideJa: core outletzcoo' ant mixing ' plenum lbefore distribut-l ing:the gas to the heat exchangers.-

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Core support arrangement s

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e The posts have hemispherical ends. This hemispherical shape allows s

j the posts to rock in the CS blocks and lower bearing seats without bending and sliding.

In this way, differential movements between the blocks and floor (Ref. 1) can be accommodated without interference.

2.1.2.

Core Support Block The primary function of the core support block is to support top and bottom reflectors, reactor-core, and top-plenum-orifice-elements.

In addi-tion, its function is to mix the core-outlet-coolant gasses exiting from a refueling region in the core-support-block and measure the refueling region coolant mixed mean outlet temperature.

Each CS block is radially keyed to the surrounding six adjacent blocks.

The peripheral CS blocks are keyed to permanent-side reflector I

support blocks which are in turn keyed to the steel core barrel. As the core barrel radially expands 'due to thermal expansion, the permanent side reflector support blocks are pulled out with the core barrel'. The CS blocks keyed to the PSR support blocks are also pulled out in the radial direction to take up the core barrel radial displacement. These movements result in an increase in the nominal gaps between CS blocks during 100%

power normal operation. This subject will be discussed more in Section l

3 on gap analysis.

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CORE BARREL AND GAP ANALYSIS The total gap between adjacent CS blocks is the sum of the normal-operation gap and an additional gap due to thermal expansion of the core barrel during the LOFC.

The parameters used for gap analysis and results are summarized in Table 1.

3.1.

NORMAL OPERATION GAP The CS blocks are installed with nominal gaps between them of 0.25 in.

As the core and core barrel heat up to 100% power steady state operation these gaps are estimated to grow to 0.399 in. (Refs. 2 and 3).

~

3.2.

CORE BARREL TEMPERATURE DURING LOSS OF FORCED CIRCULATION EVENT Figure 2, taken from the FSV FSAR (Ref. 4), illustrates the core bar--

rel temperature near core support keys and core midplane.

It also shows that the core barrel temperature near the support keys reaches ll50*F (621*C) 10 hours1.157407e-4 days 0.00278 hours 1.653439e-5 weeks 3.805e-6 months after LOFC occurred.

The core-barrel-temperature reaches a maximum (=2300*F or 1260*C) near core midplane about 150 hours0.00174 days 0.0417 hours 2.480159e-4 weeks 5.7075e-5 months after the start of the LOFC accident.

To be conservative in the gap prediction, the core barrel thermal expansion is calculated using the maximum temperature during LOFC event.

This surely yields an upper bound value for the gap estimate.

3.3.

GAP DUE TO CORE BARREL THERMAL EXPANSION The radial core barrel thermal expansion is 2.09 in. based on the 2300*F core barrel temperature.

This expansion is divided into four gaps 3-1

s TABLE 1 SLM!ARY OF CORE BARREL AND GAP ANALYSIS Thermal expansion coefficient at 2500*F (ASTM 387 Grade B as 8 x 10~ /*F 1.0 Cr-0.5 Mo alloy and carbon steel)

Temperature of core barrel at normal operation 750*F Maximum temperature of core barrel at LOFC/FWCD 2300*F Diameter of core barrel (Ref. 3) 28.4 ft Radial thermal expansion of core barrel as2.1 in.

Maximum gap between CS blocks at normal operation

==0.40 in.

Maximum gap between CS blocks used for this analysis

= 1.0 in.

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9 2400 MAXIMUM CORE BARREL N

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Cure barrel temperatures and maximum PCRV side wal1 tiiermal Sarrier surface temperatures during accident (Ref. 5)

b from the center of core to the PSR (see Fig. 3) resulting in a 1.0 in, gap between blocks.

Figure 4 shows the result of gap analysis for the normal operation and LOFC.

Further detailed calculation is presented in Appendix G.

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CORE SUPPORT BLOCK CRACKING ANALYSIS 3

4.1.

ASSUMPTIONS OF ANALYSIS The following assumptions were made in this analysis:

i 1.

Cracks start at the corners of the keyways and propagate to coolant channals. This assumption is based on the results of the 4

thermal stress analysis performed by LASL personnel which pre-dicted maximum tensile stresses of 1600 psi in the corner of the top of the keyway.

The maximum principal stress drops rapidly away from the corner to about 300 psi then-rises again towards the a

edge of the coolant hole to about 600 psi.

s 2.

The web between coolant holes cracks along the minimum ligament thickness.

In order for-the block to break into more than'one piece the webs between coolant holes must crack.

It is reasonable to assume that cracking will occur at a location along the plane of minimum web thickness.

3.

Both frictionless and' frictional contact surfaces are analyzed separately.

Contact surfaces without friction allow surfaces to slide relative to one another'so that pieces of the CS block will-

~

achieve their maxiwum movements before they jam between adjacent CS blocks.

Contact surfaces with friction forces will yield the.

proper' equilibrium position. The problem'is then bounded by-these-Jnalyses.

4.

Gaps between; blocks are based'on 2300'F core barrelEtemperature.

This~ allows maximum size gaps'to form between blocks-so that the CS block disarray is a maximum resultingEin greatest' fuel column 1

4-1

=

s covements.

The assumption is that if the region support holds e

together for this upper bound gap size then it will also hold together for lower core barrel temperatures.

5.

Gaps between core support blocks are uniformly distributed. This is a reasonable ass'amption since the blocks are designed to move along the key-keyway constraints between blocks.

The core barrel pulls radially outward uniformly.

There are gaps between keys and keyway of about 0.060 in, which will result in some non-uniformity of the gaps in question. However, the very large gaps resulting from the 2300*F core barrel should overshadow any tolerance buildup.

4.2.

CRACKING PATTERN The f racture pattern scenario is divided into four steps. The four steps are as follows:

1.

The crack initiates at one corner of the keyway due to the presence of thermal stress exceeding the ultimate tensile strength of PGX graphite as shown in Fig. 5.

The crack initiates here (points A in Fig. 5) because this is where the coldest adjacent block is located creating the highest thermal stress.

2.

The initiated crack is assumed to propagate to the coolant channel and vertically downward to the bottom of the core support block.

This. process is shown in Fig. 6.

Actually it is not clear that the crack will propagate since it is initiated in a rapidiv decaying tensile stress field.

3.

Carrying the scenario farther it is assumed that the other keyways in the same block also crack, even though they are at a lower stress value.

At this stage, all cracks have initiated from the key-way-corners and propagated to the coolant channels and to 4-2

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a-the bottom of the block (Fig. 7).

Thus, the center control-rod-column is supported solely by webs between coolant channels of the support block.

4.

The center control rod colu=n and other surrounding standard fuel columns impose bending and shear on the webs.

That resulting stress state is assumed to induce tensile stresses large enough to exceed the ultimate strength of PGX graphite. The webs crack, as shown in Fig. 8. resulting in the block breaking into four pieces. The completed cracking pattern is shown in Fig. 9.

Fig-ure 10 shows the block before cracking for comparison with the cracked block.

It is to be noted that the core outlet thermocouple assemblies are assumed structually non-significant duing this hypothetical accident.

In addition, the thermocouple graphite sleeve breaks in two as the center cf the block drops.

Upon completion of step 4, the control rod colu=n and the center piece of CS block move downward 2.63 in, pushing the post pieces 1 in, in the radially outward direction until flat face to face contact occurs with the adjacent blocks.

Slight rotation of the block pieces occur within the limits of the key-keyway dimensions ultimately resulting in static equilibrium of the assembly. The dislocation of the center piece is illustrated in Fig. 9.

Kine =atics of the cracked CS block is shown in deta.1 in Appendix B.

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I 5.

ANALW IS OF CRACKED BLOCK - WITH KEYS f

At the end of the cracking scenario outlined in the' previous section

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the core support block is left in four pieces as shcun in Fig. 9.

As the l

pieces rotate within the confine of the key-keyway system the keys are loaded by a force couple at the top and bottom of the key. Once the key jams in the keyway no further relative rotation can occur.. The static

]

analysis that follows presents calculated forces on the block and resultant stresses.

i 5.1.

FORCE EQUILIBRIUM 4

The loads on the CS block pieces come' f rom the weight of the fuel' l

element, flow cr.atrol valve, and the weight of the block pieces themselves.

^

Each outer fuel column weighs 2550 lb which the center control rod column weighs 2215 lb.

Figure 11 shows all static equilibrium forces without friction forces activg on the cracked core support 'alock.

The detailed steps.to obtain-this result are given -in Appendix E.

In addition friction forces are-l considered for two special cases-in order to compare the'results with-j

- friction and those without friction. The cases-are as follows.

A ii 1.

Complete'. flat, facr.-to-face' contact between ^ cracked.;and adjacent CS, blocks.

.ip;are,r12 +shows the results of case 1.

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2.

Contact between the cracked CS block and an adjacent CS' block

. occurs'at the top. of-' the blocks'.., The results of - this case are -

)

shown in Fig.~ 13.

s i

e

5-1

.u..

m.-

. m.

m 1.-

m.m m m.

m m

e m________,__.mk__-

1 Af?iEd I

j,,. g Lod 2,550 (ib) 2227(lb) 129 C F60 2.FFottb)

APP ted y

l L0d

/ R g(!b)

NE#dIO" y ?,

/

Cs= s54 PS' T5 343 P58

~

}

l.04 Icth)

Reac{gon (6 si)

P p

1,04s(gb)

Readlon

,/L 1515ab)

Reaction sm)

?00R ORIGINAL 127(fSi).

fos {-

Reada'on Fig 11.

Loads on cracked block before key sheared off-friction forces are neglected 5-2

4 Aff ied l

Agylted bad Leo.4 2 b500b) 2227(ib) 12-) P$s)

man, Arelied r

-Q,a/,,

me-ua N

%s9?si

,,,ao9 T = B'0 fu

, s-sn UL) b lh fr.rle a.ib) c d

l

\\

84'$<l9ges,,,

\\

64 Psi) g s.g p

b }

k Ob)

Reactr g ds o

/

2321 aby Rdron wwee as')

P00R ORIGINAL dSb R e. A d o n Fig. 12.

Loads on cracked block with friction forces before key sheared off - case 1 5-3

O MelA Arrhed wl 1.od gggo(Ib) 22270b)

I23 IP50 f

" e\\t &

Lt:.Ottb)

Q Arpired g,,d:on

,3,23 @s) s

)3 9

3

$44(!b) w')

(6C Cod" l

l f fetc}tos p&oll0 d~ :*. : '. l b )

Radion ao ae)

P00R ORIGINAL vymn on foSI-Ef.A

'0 "

Fig. 13.

Loads on cracked block with friction forces before key sheared off - case 2 i

3-4 l

e s

5.2.

STRESS RESULTS Of significance to the accident scenario is the integrity of the CS post under each outer piece of the block and the integrity of the key which is holding the block in rotational equilibrium.

It is also of interest to determine if the bearing stress between the center piece and each of the outer pieces at the cracked web surface is of any consequence.

The maximum stress anywhere in the block is found to be the tensile stress associated with shear force on the interblock keys.

This maximum principal tensile stress is estimated to be 343 psi which is 30% of the minimum tensile strength of PGX.

It can be concluded from this analysis that the keys will not fail and the equilibrium position of the block pieces will be established by the keys jamming in their respective keyway. The amount of block rotation is less than 0.2* which is imperceptible with respect to fuel column tilting.

s i

l l

l f

l l

1 I

l i

5-5 i

e 4

6.

ANALYSIS OF CRACKED BLOCK - WITHOUT KEYS An even more conservative step in the scenario is to assume that the cracked block keys shear of f and do not provide rotational restraint to the outer pieces of the CS block even though the stress analysis showed the keys do not fail. Once the keys are sheared off, equilibrium can only be main-tained by the pieces of the CS block rotating until they jam between the adjacent blocks as shown in Figs. 14 and 15.

Only the case without friction-force is considered herein. This scenario could potentially result in a more serious disarray of the fuel columns.

6.1.

FORCE EQUILIBRIUM Results of forces on the CS block with the key sheared off are shown s

for the cracked block in Fig. 14.

See Appendix F for the detailed calculations.

6.2.

STRESS RESULTS Stress results are identical with the results given in Table 2 except that the key stresses are not relevent to this scenario.

The 2000-lb loads on opposite corners will cause some local crushing but do not result in any i

other significant stresses.

l 6.3.

JA. ED COP" SUPPORT BLOCK T

l In this hypothetical scenario, the CS block piece rotates about the post due to the weight of the fuel columns until it contacts the side of an adjacent CS block near the top.

Then it slides down until the opposite bottom corner touches the other adjacent CS block.

These movements cause one side of the block to drop 1.4 in. as shown in Fig. 15.

The angle of 5.* t e i rg rp

'M r*

6-1

^WA 2c,..R I

g Aprked Lead Lug R'*dia n 2,550(lb) s */,/

2,227d b) l Aylted l.0ad 2.5Coub)

O

<O i1

'8

(?I

\\

f y

..I

.I h,

-(

l l

s\\

\\

l i

~

h 1048 Cib) u\\

05P5i) p\\.-

s

)

. y, /

N 1,041 cab) i i

EeacEto n

,N'

.,D

5L

b)

I betten G413(ib)

' ( 2 2 7,p5:) 0OSt' Readion Fig. 14.

Loads on block af ter key sheared of f 6-

e 0

/

U l

0

/

~

/ / //

/

a

~

t

.s G M 2.

O g

T

~w, __$.

e.

o k

E

@ @q) @

~

v c

s m

A g

V)

,3 t

(

x o

d h

N O

r a

o av vwo f

A C Q @v fC o

-cO&

\\'\\ \\

\\

\\ \\ \\ \\ \\'

co s

\\

62 N

N o

s I

P90R ORIGINAL

~

-ie

- ~

ww w

p-eve r--

e e-r i --

-m-9 e

j l

TABLE 2 STRESS RESULTS (psi)

Minimum ultimate tensile strength of PGX graphite (Ref. 6) 1160 Minimum ultimate compressive strength of PGX graphite (Ref. 6) 4460 Minimum ultinate compressive strength of ATJ graphite (Ref. 6) 7204 Results from frictionless surface assumption Bearing stress on the slanted surface of web 129 Bearing stress on a flat surface 6

Maximum shear stress on a key 343 Maximum bending stress on a key 154 s

Compressive stress in a post (ATJ graphite) 227

~

Results from the consideration of friction forces on cracked CS block Bearing stress on a flat surface 4.4 Maximum shear stress on a key 310 liaximum bending stress on a key 139 Compressive stress.in a pest (ATJ graphite) 234 h h h's 13,;?lh$ h, 6-4

'liu

>b

O s

rotation of the block is 2.7*.

Detailed calculation of these movements are contained in Appendix F.4.

It should be noted that the fuel column which overhangs the adjacent core support block on the drapped side is lifted and tilted radially inward.

The fuel columns in this mode of tilt lean against the central fuel colu=n of the region without " jawing" of the horizontal joints between blocks.

The dowels pull free from the CS block.

At this point in the scenario the peripheral fuel columns are leaning against the central column much like a stack of rifles. The angles of tilt are less than 0.2'.

Figure 16 is a scale drawing of the tilted CS block pieces and the bottom block in each fuel column. The central section has dropped down about 2.6 in, along with the central column. The disarray is difficult to see in this drawing because the movements are very small com-pared to the dimensions of the blocks and the height of the fuel columns.

The coolant flow through the fuel columns, whether it be upward or downverd, is not inhibited by the cracked and tilted block.

The large coolant holes in the bottom reflector blocks intersect the triangular shaped space in the center of the cracked block. This triangular-shaped block in turn intersects the large coolant channels in the CS block which communicate to the lower plenum.

Inus, Uie flow path is changed, but, unimpaired.

i 5 0 $;f 0 {]

?I ?

3 5

I e

6-3

no ig er a

n i

n mu loc leu f

hcae

/

J /

fo k

/

~

ll co

/

l b

%/-

mo t

t o

o b

d

/' N

'~

n

/

a t

kc N\\/

o

.N l

/L s

b W

tr

~

s op p

u

\\. N er g

f o

C d

e t

l i

t dna de kcar C

$8. gs.

a-6 1

8 i

l

t e

L i

s 1

7.

SUMMARY

OF RESULTS i

The results of this analysis can be summarized as follows.

1.

The extent of cracking assumed is extremely conservative, since the stresses calculated by LASL resulting from thermal gradients l

and are highly peaked in the corners of the keyways.

Furthermore, r

cracks initiated in these corners are not expected to propagate through the block from top to bottom.

2.

Caps between blocks are assumed large, ince core barrel expan-sion determines these gaps.

Tb* core barrel' temperature was i

assumed to be 2300*F instead of about 1150*F to be conservative.

3.

Core support block key stresses are low. Thus,.the region dis-array resulting from the " sheared key scenario" is considered to be an extreme..

4 4.

Core support block pieces remain jammed between adjacent core-support blocks. This precludes region collapse keeping signifi-l cant amounts of debris from falling into the lower plenum.

5.

Disarray of fuel blocks is minimal and' occurs.enly at the core support floor / bottom reflector interface. Thus,Jcoolant flow

~

through the fuel blocks'is' unimpaired.

1 6.

The core support block continues to function in the cracked condition.

)

e -

7-l'

. -. - -. - - - -. - - - - - - ~ -. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - --- --

^^- - ^ ^ ' ^^^^

-^

~ ~ ' ' ^ ^ ~ ~ ^ ~ ~ ~

~

8.

CONCLUSION From the results of this analysis it can be concluded that cracking of a core support block in the conservative manner assumed does not interf ere with safe cooldown of the reactor core.

8-1

f 1

a 1

i 4

a 9.

REFERENCES i

1.

" Fort St. Vrain Nuclear Generating Station Final Safety Analysis Report " Vol. I, 3.3,2.2.

I 2.

Bradley, E., " Interim Report en Gap Comparison Study - Core 4

Fluctuation Investigation " General Atomic Document No. 904658.

3.

" Reactor Internals Assembly - FSV." General Atomic Drawing No.

90-SK-5523.

4.

"FSV Nuclear Generating Station Final Safety Analysis Report," Vol.

IV Fig. D.1-23.

5.

"FSV Nuclear Generating Station Final Safety Analysis Report " Vol.

I, Table 3.1-1.

6.

Salavaccioglu, R., " Generic - Graphite Design Material Properties,"

4 General Atomic Document No. 904434, pp.'73 and 119.

l

}

Y e

G.

-Tse+9-e,%-

Ag-c

++,- -

w-pgg",-

w+e, 9

ry.,-

9-e

_-$~.'

we w+

-s p

--y'-

g -*f w ier - ~

,49 f

y

-t g

-v4

---c-'

..e y

e-d te d-g s-tyW=y

9

o 4

i 10.

ACKN0'4LEDGMENT The authors deeply wish to thank Dr. F. Ho for his independent review.

l d

4 e

10-1

s APPENDIX A WEIGHT CALCULATION VER. ICAL SURFAG T

SLANTED SURFAd

$[

u

'~

g z.63 -

a o

o w

o o

m

\\.QD-v\\ r C

t

,~

as.

fast @tet t

V

\\

/

,/.}

N6 e

j.

V A'

P00RORGH1 a.

l t

Fig. 17.

Dislocation of center piece of CSB l

A-1 I

wr w-r

A.l.

NET VOLUSE OF CENTER PIECE e

Approximate volume is only obtained in the following way:

Net volume of center piece = Volume *(opQLMN)

-3 x [ Volume *(ABJ'K'A'B')]

-3 x [ Volume of coolant channel]

-3 x [ Volume *(LJSOGR)]

Volume *(opQLMN) = 1 (35.84" x 35.84 sin 60) x 8" a.

7

= 4449.66 = 4450 in.3 1

b.

Velume *(ABJ'K'A'B') = 7 (18.05" x 3.03") x 8" 2 265 in.3 c.

Volume of Coolant Channel = 7 x (4.037" x 3.775) x 8"

= 373 in.3

. L.

b b = 3.775" a

Arci = rab

.. - y-

% g

}.-.-

a = 3.775/cos 20.75*

d.

Volume *(LJSOCR) = 1- (5.062 x 5.062 sin 60) x 8"

= 88.76 in.3

.. Net volume of center piece

= a - 3b - 3c - 3d

= 2378.05 in.3

' ' " t.c f y

See Fig. 18.

A-2

a s

A.2.

WEIGHT OF CENTER PIECE Weight of center piece is obtained as follows:

We = 5 x Vc where S = weight density of PGX graphite.

= 0.064 lb/in.3, Vc = volume of center piece, We = weight of center piece.

We = 0.064 lb/in.3 x 2378.05 in.3 We = 152 lbf A.3.

WEIGHT OF POST PIECE o

Weight of each post piece:

Weight of CSB - Weight of Center Piece 3

= (1723* - 152)/3 = 523.6 lbf

0btained from the following way.

(21.56" x cos 30*) x 21.56" x 3 x 23" 2

-(n 7.55 /4) x 6 x 8"/cos 20.45*

-(

212/4) x 15" - seats etc.

t

=220.00 in.3

=12.73 ft3 12.73 x 110 lb/ft3 = 1400 lb i

i a.

Core weight including CSB = 19.238 lb (Ref. 8) b.

Weight of core excluding CSB =

Wt of 6 fuel columns + Wt of control rod column (6 x 2550 lb)

(2515 lb)

(Ref. 9)

= 17,515 lb

. a-b = 19,238 - 17,515 = 1723 lb e

e A-3

s APPENDIX B KINEMATICS OF CSB Center Piece t

4 5 H--

\\

?

I l

--./

/

Post Piece h

/

I l

/ l I

[

/

i 7

'A 9

I I

l e

o s

I w

where 6 = radial displacement by P-Piece,

= 1.0 in. (calculated in Appendix G.2),

L = vertical displacement by center piece, L = 6/ tan 20.75*,

= 1 in./ tan 20.75* = 2.63 in.

P00R ORIGINAL e

l B-1

4 APPENDIX C BEARING SURFACE AREA Each bearing surface between center and post piece is approximately.

shown below.

5"

- i D2 28*

'\\

8.$'

N s

xx\\

y.,r N'

From Fig. 16 I

JL s

1-A $-2*-4 4

s 1

isearing Surface Area = 7 (4 + 2) 5.75 = 17.25

= 17.25 in.2 P00R ORMIL 9

C O

L APPENDIX D STATIC EQUILIBRIUM OF CENTER PIECE OF CSB in

'Wfg s

t E I

j;t

(

{

NL/

8 l

8,,

t

~ 4 W

j l

l JO t

3 y

I l

j g

- % Rq-i-

\\

/3 g.-

25,yf NP Fig. 18.

A part of center piece of CSB (upside down)

Assumption:

Magnitudes of reaction forces acting normal to slanted surfaces are identical.

D.l.

FORCE ON SLANTED SURFACE W = Weight of control element column and center piece Weight of control element column = 2,215 lbf POORORIGINAl.

-1

A Weight of center piece = 152 lbf W = 2,215 + 152 = 2,367 lbf.

3 = Reaction force resultants acting on slanted surfaces.

1 (i = 1 to 3) so IE ! " l3 l " 5 l = R i

3 3

c

=R cos 69.25*

R

=

c R

=(

W)/cos 69.25*

= 2227 lbf D.2.

BEARING STRESS ON SLANTED SURFACE Assuming uniformly distributed Bearing stress = 2227 lbf/17.25 in.

= 129.1 psi Maximum bear stress may be higher by a factor of 2 or 3, due to non-uniform distribution, say 350 psi maximum.

k 9

tuitiiU i:ua t em e

o 4

APPENDIX E ANALYSIS OF CRACKED BLOCK - WITH KEYS E.1.

FORCE EQUILIBRILH

'dno s

v - {,..

.H

- y, s.

/

ps

.a

'E M w.

~

+

x N

1 NY v

r 9.7E' c'

23" k

E.

N

/,

.F^~ W:

- \\6

~

as

~4 P00R ORGINAL l

Fig. 19.

Force equilibrium without friction forces E-1

\\

W Nomenclature s

- a.

Forces W

= Weight of post piece

~cg W,W2 = Weight of standard fuel column y

Reaction forces acting on top,and bottom key R and R

=

3 4

respectively R and'R = Reaction forces acting on a flat face of. post piece 2

R = Reaction force by post

~s

~R

= Reaction force normal to.the slanted. surface.

~c b.

Distance Vector 4-Y

= Distance vector-from the origin.0,to_where W is

~cg cg_

acting Y

and Y

= Distance vectors from the origin 0 to.where.Wy.and gy W2 W,,- are acting Y3 "" 14 = Distance vectors from the origin _0 to where R3

"d' R4_are acting l

Y and Y = Distance vectors from the origin'0 to where R and p

2 y

R are acting 2

l i-

.Y

= Distance vector from-the origin 0 to where-R is

(.

~s-s L

acting-l Y = Distance vector = from the' origin. O to' where R is

~C C

acting

\\$

b W ' = 523.6K.

+

f

~cg; O *

<O-E-2 T

m

..i m

.m.

t

-o:

s.

M

=Y xW 3

-o

~cg.

~cg

= -2104.87i - 5644.413 (2) Y

= 16.21i - 6.193 W1 i.

W = 2550K

~1 M=y xW = -15784.5i - 41335.53

~o

~W1

~1 I

(3) 'y

= 1.531 - 8.043 1

~W2 A

W,

= 2550K 4

t M = Y,, x W

-o

~W~

~2 n

n

= -20502I - 3901.5J (4) Y = 13.283 + 1.125J + 3.2K

^

~3 s

n R = -R I

~3 3

n n-

-3.2 (R JJ + 1.125_(R )K' M

=y xR

=

~o

~3

~3 3

3 i

n

.. n.

.(5) 34 = 8.281 + 1.125J + 19.8K R =Rf i

~4 4

= 19.8_ (R )3 - 1.125 - (R )k 4,

4 (6) -Y

= 17.41I + 11.5K Not necessary at the 7

centroid.of the: face

^-

R = -R J

~1 1

.. (near top).

M

=y xR

,o

~1

~1 1(no friction force)-

= 11'.5l(R )3~- 17.42 (R )K:

y y

i J

i (7) Y = 4~13I +-11.5K.

~2 i.

R _ = -R J 2

2

x i.

b,.

i E-3 t

, ~,.. _. _. _. _

... ~ -

M

=y xR 2

-o

~2

= 11.5 (R )I - 4.14 (R )

2 2

/.

A (8) y = 2.6I - 4.5J + 19.75K

~s a

R = -R K

~s s

M

=Y xR

~o

~s

~s a

a

= 4.5 (R )I + 2.6 (R )J s

(9) y = 10.78I - y J + 3.llK

~C C

R = 2227 (Cos 20.75J + sin 20.75K)

~C M

=y xR

~O

~C

= (-789 y - 6,476.7)1 - 8,505.53I

^

c s

+ 22,450K i

l I

e G

E-4

I Tall!.E 3

SUMMARY

OF STATIC EQUILiliRIUM ANAI.YSIS - WITil KEYS DIstmice Vecto Force Vector Momer t. Vect or (in.-lbf)

'(in.)

(Ibf)

(Distance Vector x Force Vector) n n

n n

a a

n I

.1 K

1

.I K

I

.I K

.t 10.78

-4.02 11.5 0

0 523.6

-2,104.87

-5,644.41 0

2 16.21

' -6.19 0

0-0 2,550

-15,784.5

-41,335.5 0

' - 1.,7 5 4 '

3.

-8.04 0

0 0

2,550

-20,502

-3,901.5 0

,4 3 J,28 1.125 3.2

-R3 0

0 0

-3.2 (R )

1.125 (R )

13 3

3

"(ip8 5

1.125 19.8 R4 0

0 0

19.8 (R4)

-1.125 (R4)

{1732 6

0 11.5 0

-R 0

11.5 (Rg) 0

-17.42 (RI) g

~7 Tl4' O

11.5 0

-R2 0

11.5 (R )

0

-4.14 (R )

2 2

8

+ 2. 6

.-4.5 19.75 0

0

-Ra 4.5 (Rs) 2.6 (Rs)

O pm wl

' 107)4 h -Yc 3.11 0

2,082.55 789.01

' -789ye - 6,476.7

-8,505.53 22,450 9

( \\

A A

A

E Fx. = 0 EMox=0 R1 = R' = 1,041.27 lbf yc = 10.781 - 10.06.1 + 3 llK 2

-EFy=0 EMoy=0 R3 = R4 = 2,573.14 lbf EF

=0 EMoz =.0

,!!3 = -6412.6K (Ibf) g

s E.1.1.

Force Equilibrium Considering Friction Forces - Case 1 e

Q I Ni 2:

i

/

<' $h,,

9 s

s.

r >'. ['

g, M

,f T

+

o Ng Il

,,y N

b 1

s

$+a %

,e.s 23 Y

.e g

/

$2.

j

/

~

$a

,/

L %43I cf

% [9 7

c 5

O Coefficient of Friction = 0.5 f = 0.5R f = 0.5R y

4 4

2 f = 0.5R

'5 = 0.5R 2

3 f = 0.5R 3

y POORBRIGlWl.

Fig. 20.

Force equilibrium with friction forces - case 1 E-6

i TABLE 4 l

SUMMARY

OF STATIC EQUILlBRIUM ANALYSIS WITil FRICTION FORCES - WITil KEYS, CASE 1 l-I Distance Vector (In.)

Force Vector (lbf)

Moment Vector (in.-lbf) i n

n n

n l

I (x)

J (y)

K (z)

I (x)

.I (y)

K (z)

I (x)

J (y)

K (z)

I L

10.78

-4.02 11.5 0

0 523.6

-2,104.87-

-5,644.41 0

l 2

'16.21

-6.19 0

0 0

2,550

-15,784.3

-41,335.5 0

l' 3

1.54

-8.04 0

0 0

2,550

-20,502

-3,901.5 0

.4 13.28

~1.125 3.2

-R3 0

0 0

-3.2 R3 1.125 R3 5

8.28 1.125 19.8 R4 0

0 0

19.8 R4

-1.125 R4 6

17.42 0

11.5 0

-R1

- 0 11.5 R1 0

-17.42 R j

g l

7 7

- 4.14 0'

11.5 0

-R2 0

11.5 R2 0

-4.14 R2

]2.6

-4.5 19.75 0

0

-R 4.5 R 2.6 R 0

8 3

3 3

l 9

l1Q.',78

-y 3.11 0

2,082.5 789.01

-789 Yc - 6,476.7

-8,505.5 22,450 c

. n.e 10 r;177,41 0

11.5 0

0

-0.5 R1 0

8.71 R1 0

n in-11' q*fj.;l 3 0

11.5 0

0

-0.5 R2 0

2.06 R2 0

12 8,.28 1.125 19.8 0

0 0.5 R4 0.56 R4

-4.14 R4 0

g4 1

n l

13 (QE28 1.125 3.2 0-0

-0.5 R3

-0.56 R3 6.64 R3 0

14 L10;78

-yc 3.11

-0

-395 1.042

-1,042 ye +-1,228.45

-11,232.76

-4,258.1

> g E{x = 0 R3=R4 EMoz=0 R1=R2 = 844 Efy = 0

Ry+R2 = 1687.55 EMoy=0 R3=R4 = 23'21

[

EF 0

R

=,6611

=

g 3

s E.1.2.

Force Equilibrium Considering Friction Forces - Case 2 e

b l$

t e.

v - 1.

s

, r, 8

4 f5 f

_,. 5 gf

_Y I

a T

{[

8 N

3 i

x h

x&

w'a y

23 sn' i,

t Coefficient of Friction = 0.5 Z,

[$

= 0.5R f = 0.5R 1

4 2

= 0.5R f = 0.5R 2

3 5

f = 0.5R 3

1 P00R BRGM Fig 21.

Force equilibrium with friction forces - case 2 E-8

4 1

R 3 R 2

1

)

R R

z 5

2 0 8 3

5 5

(

2 5 4

2 1

2 4,

2, 1

nK 1

7 E

1 1

4 2 4 S

0 0 0 1

- 0 0 0 0 0 2

A

)

C f

b 1

l 2

S 5

)

4 3,

. 5 3

4

)

5 4

Y 4

E n

)

5 4 R t

2 R 3

3

. 3 8 2 K

i y

4 3 1

R

(

R R R R 5 3

il

(

4 3 0 4

(

0 2,

=

8 1

7 1

4 6,

9, 2

5, T

r nJ 1

7 0

6 6 2 4 1

I o

5 4 3 9 3 4

8 1

R R W

t 1

- 0 0 8 2 6 2 c

= -

e V

}

3 S

R R E

t C

n 7

e R

e y

O m

6 F

o 7

2 M

4, 4

0 0 N

0 O

)

6 1

=

I x

T

(

7 5 1

y, y

C 8

3

)

o o nI

. 4 2 4 R s

c -

M, H I

R 4 8 0 R

R y

E E F

0 7

5 6

( ' 9 28 l

1 6

5

, 5 0 8

5 5

2, i

2 1

2 T

0 7

I

- 0 0 0 0 0 0 0 4

1 W

5 5

5 S

t 2

3 EI R R 4 R 7

LS

)

6 0 0 R

2 8

Y.

z 5 5 5

5 5

6, 4

!Al

)

(

3 5,

5, 5

3 9 0,

TA f

2 0 0 0 R 8 1

N b

nK 5 2 2 0 0 0 0 0

7 1

1 A

=

1 l

(

M 6,

l 5

4 2

l r

5 R R 6 I

o

)

1 2

R t

y 0 0 0 0 0 R R 0 0 0 0 0 2 5

=. +

l c

(

8 9 l

I.

e 3

3 1

3 V

nJ 0,

R R R l

l 2

U e

Q c

)

E r

x o

(

4 3

C F

0 0 0 R R 0 0 0 0 0 0 0 0 0 0 0 0 I

nI T

=

A T

x y g S

)

5 1

1 ffg z

5 8

8 7

1 1

E E E 2

2 F

n

(

0 I

1 0 0 9 3 0 0 0 0 9 3 9 3 3

(

nK 1

1 1

1 Y

R r

A o

5 5 5 5 M

t M

c

.)

2 9 4 2 2 2 2 0

0 y

1 1

1 5

U e

(

e c

. S V

4 6 8 1

1 0 0 0 0 1

1 4 y y nJ e

cn a

)

8 1

3 8 8 1

3 1

3 8 8 8 8 t

x 7

2 5 2 2 4 1

1 2 2 6 7

7 4

s

(

1 0 6 1

8 3 7 4 7 4 8 3 2 0 0 0

nI 1

1 1

1 2 3 4 5 6 7 8 9 0 1

2 3 4 1

1 1

1

,S lfl l

l lllI

!l[!l

(

l l

E.2.

STRESS RESULTS E.2.1.

Shear Stress On a Key To be ccaservative, shear force transmits through area (= 5" x 2.25")

X l b a

ll43.6 3Yr*

I..L E*

T I

o o

e hI I'

5, 5'

Y+.-

=2 n,

s J'.

$ jL M-2.25*4 s - 2.25'

\\\\

k 6

Maximum Shear Stress V(max)(3) = 2573 3

a)

T

=

max A

2 (5,, x 2.25,,) (2) = 343 (psi) f

)

2321 b)

T 310 (psi) with friction forces

=

max (5,, x.,

,5,,)

E.2.2.

Maximum Bending Stress On a Key (2.25) x 2.5 1143.6 x 2

154 psi c

=

max 3

(2") 5 e

S O

4 E-10.

e s

l43.6 *fh t

i c

Considering a stress concentration h~

/

factor = 2 c=2 o

= 308 psi (w/c fr M ion) max 2"

/

c=2*c

= 279 psi (w/ friction) max J

6 i

l I

E-ll I

i

r s

APPENDIX F ANALYSIS OF CRACKED BLOCK - WITHOUT KEYS F.1.

FORCE EQUILIBRIUM gJ^

3

\\Re

/x y

9/

4 ::

.1 i

a 3

e T,1

y

-l

-so, y

1 N..

R a

sf,- ~

e

~

if I

VS

~

. ~

P00R ORIGNAL F-1

e A

A A

(1) y

= 10.781 - 4.02J + ll.5K e

~cg W

= 523.6$

~cg M

=y x TJ

~o

~cg

~cg n

= -2104.871 - 5644.4J

^

(2) y

= 16.21I - 6.19J

~wl W

= 2550K

~1 M

=y xW

-o

~wl

~1

= -15784.51 - 41335.53 (3) y

= 1.531 - 8.043

~w2 n

~~, = 2550K W

^

A

-205021 - 3901.5J M = y, x W

=

~2 g

~o

~.

A A

A (4) y

= 25.64I - 7.17J + 1.4K

~H R

"~R H

H M

=y x

~o

~H

~

H(

=-R

+

(5) y

= -4.74I - 7.17J + 22.7K

~B B

B M

=y xR

~o B

~3

~

n n

3 (22.7J + 7.17K)

=R n

(6) y

= 17.42I + 11.5K

~1 R

= -R J

~1 1

R I': ',JU3 791*,5 y

/

.> 4;R 'fil'. 51 - 17.42K) 1 F-2

o a

a (7) y = 4.141 + ll.5K s

3 n

R = -R J 2

2

[Io " 22

2 A

A

=R2 (11.51 - 4.14K) n n

n (8) y = 2.61 - 4.5J + 19.75K

-s A

R = -R K

-s s

M

=y xR

-o

-s

~s n

n

= R, (4.5I + 2.6J)

(9) y = 10.78i - Y J + 3.llK

-C C

^

R = 2082.5J + 789K

~C M

=y xR

~O

~C

~C n

n

= (-789 y - 6476.7)I - 8505.5J A

+ 22450K F.2.

BEARING STRESS ON A FLAT FACE 1041 lb/190 in.

=

5.5 = 6 psi (average)

=

(without friction) o 844 lb/190 in.'

<+.4 psi G(with friction)

=

l e

i O

e

-F-3

>;L e

s 1

g R

2

)

Rg g

R r

R 2 0

o nK 7

4 4

5 t

1 7

4, 1

c 1

7 e

7 1

4 2

) V 0 0 0 7

- 0 2 f

b e l c

- r 1

5 o

4 5

5 nF 5

4 g i

4 3

1 R R s 5 f

f

( x n I.

4 3,

0 R 0 b b 6,

9, 4

7 5

l l

rr 6

1 f

oo 5 4 3 1

2 8

1 5 b S

t t 2 0 0 2 4 0 l

Y cc 0,

0, E

ee 6

K VV 7

1 2

2 T

t e 6

=

1 U

nc 7

l en 2

g 4,

0 i 6 i

ma 4,

R R T

ot 6

Ms 7

5

=

IW i

8 D

4 2 3

2 3

t g

3

(

4 8 0 R

R R e

R R R

+I 0

7, 5,

y S

1 5 5 9 5

I

, 5 0

. 8 S

2 1

1 4 7 Y.

- O 0 1

1 I

A c

N 0

A 6

6

=

M 3

0 U

3 0 0 O 0 0 0 R 9 z

E. I nK 2

5 5 8

=

di lBR 5

5, 5, 7

zD AB T

l.

2 2 F

r E

=

I I

o U

t

=.

y Q

c o

F.

e) 5 yM Vf F E C

b 2

E I

eI 8

=_

T c(

1 2

0,

=

A r

n I.

0 O 0 O 0 R R 0 x

T o

S F ~

2 x o F M I E F

O

'4 g

nI 0 O 0 R R 0 0 0 0 Y

R 0 0 AM 5

1

=

M 5

4 5

[ g 7

5 7 1

U nK S

r 1

O 0 1

2 1

1 9 3 I E o

1 2

1 1

1 t

ce V) 2 9'

4 7 7,

0 0

1 1

1 5

en a I.

c cI 4 6 8 7 7 0 0 4 y

n(

a t

s 8

1 3 4 4 2 4

8 i

D 7 2 5 6

4 6

7 1

7 nI 0 6 1

5 4 7 4 2 0 1

1 2

1 1

l' 2 3 4 5 6 7 8 9

~

1

'.j- ;

l i j, k

o s

F.3.

COMPRESSIVE STRESS ON A POST

=6413/f(36) c g

(w/o friction)

= 227 psi *

=6611/f(36)

Ccomp (w/ friction)

= 234 psi

F.4.

JAM 3ED CORE SUPPORT BLOCK A

/@

\\\\',

,//

l \\ '-- - -.'f }

e c

6 l

I l

Fig. 22.

Plane view of core support block (Flat face to face contact at 1 )

P00R ORIGINAL l

l t 14 fp p ry

~; p an-*

aNiti]onal; bending stress is introduced.

With 1 in. offset, The load capacity is reduced by 37M to:63% of'its original vertical load capacity.

l~

i See GA-D14137 for some component test results.

l F-5

s As shown in Fig. 22, blocks A and B contact at flat face (1), but face (2) re=ains without contact.

Due to this clearance, the cracked CSB without key rotates until it fills the gap at top and bottom. The analysis given below is approximate.

8 Mh 5

r C$6 g

C.S.B c 6 C

"A W la-S POST.

Fig. 23.

Profile view of Fig. 22 a-a where 5 = 0.58 in. original gap AB = radius about post ( =32.58 in.)

CB = 37.65 in.

The following are rotating steps of CSB.

1.

CSB rotates about point A until point B touches CSB (1) shown in Fig. 23.

2.

CSB sl, ices,d, y ug41, point C contacts with CSB (2) shown in Fig'. 23..ilhid p cfnbereplacedbytranslatingCSBuntil

~

.i,........-

it. touches CSB (2), then rotating about a point A.

In this case, the rotation radius will be 37.65 in.

P00R ORIGINAL j

F-6

O Those steps are shown in schematics below.

t 1.

a F) 8

,Gs)' 8'T '

A g]p l f AB = AB' = 32.58 in.

D E ',

AE.= 23.08 in.

l AD'= 23.66 in.

-1 3 08 0 = cos 44.89*

=

32. 8

-1 ^_s.66

~

$ = cos 43.43'

=

32.58 0 - $ = 44.89 - 43.4' = 1.46*

6

=

0.58" p b

g 8, '

- BB' = AB

6o (Rad) = 0.83

~

e

\\

(0.83)2 - (0.57)2

AL

=

a 0.59 in.

l 2.

a 6'.'4 ALL i

i 0' /, 9 f.

4 l

f f

/

f, CB= CB' = 37.65 g

l C

r o CE = 30.39 CD = 30.97' dbr. u-l l

-1 30.39 0

36.18'

= -cos

=

1 37.65 I

~1 30.97 C = cos 37.65 = 34.66*.

1 0 - 4.= 1.52'

=

1 1

1

~ F-7 y-2 e-

.c.

--v e

r

,.-ea--

,v

,.g

=e

-e o--

r

' ' + + - * -

e"

-'1

-r---

s b

o

C.51" N

B'B~ = (B'C) x aa1 (Rad)

Ob

\\

1 g~

r a l in.

B" 1~ - 0.57' = 0.81 in.

iL

=

7 Conbining the results in 1 and 2, Total Vertical Displacement = AL + LL y

2

= 0.59 + 0.81 = 1.4 in.

Total Rotated Angle

= au + Lay

= 1.46' e 1.52* = 2.98*

=r 3 ".

9 l

9 F-8

s s

APPENDIX G CORE BARREL THERE\\L. EXPANSION G.l.

RADIAL EXPANSION The core barrel thermal expansion is calculated as follows:

AR = R x a x AT where AR = radial displacement of core. barrel, R = radius of core barrel, j

a = thermal expansion coefficient of core barrel, AT = core barrel differential temperature between normal operation i

and LOFC maximum temperature.

~

l Using the parameters given in' Table 1.-

~

~0 *F) (2300 - 750)*F I"'

/

AR =

ft (8 x 10 2

g j

(ft j

= 2.09 in.

~

G.2.

GAP CALCULATION OF CSB Assuming that the radial di placement is evenly distributed;into four:

. spaces, then_the gap between CS blocks due to thermal expansion of core barrel is the following:

g = AR/4

.'f

'= 2.09/4 a 0.52 in. = 0.6 in'.

,0 0-i VG l

a Total Gap = Normal operation gap + Gap due to core barrel thermal expansion.

Gap

= 0.399 in. + 0.6 in, a 1.0 in.

g 37)

L 4

i 9

4 b

G-2

a s

O t

o o

-c.

y

. c.?.

r y,...

,. +d

'.s Let Lv a

Je

.9k O

O

. pg.g'.

c..i.e f'..a c.u.w.,p.:< :. :,h sa tu.. c

{^f..? E

' W

&. e',

A' yj. U l

W,.'.%,. ;7.;..

1*-

-).;: .

N,.-.'

i 0. ~ 0

]

. :.77..y. c.#;.

pi.'se, O

u-r 2f

^

'l., n;*v

{

.?,l',

f ;:..f

$,u, c::,

t

e g

.a

.3 o,s

,s... _

r

' y. uv.,

..~e.-

. f>...f-.S. ; yu.s= -m 4( 4..gsg;.

u

4

t.r ;<L. q;r.r,,f.tg:

L -I.J;'.f, ~.,.',. it.

G. v,.n* u.c,.,.'.!.:*.*..

.s..

r-

%.; 7.<l.e. '

.. g.y ;

h.

e

..e.

y-.,r-i

,r...,,.. g...

.cv.

m.

./6

,.';s; 3.$.. -.

.TD;D.;< J.;pc r

?

%c_ - ;.

4ttQ:~ Af'svu v.;:

L;::s.:t.r,s,f.:

.'..c.-

p i

4.4 r.s.. @f.G.

..., :. c _

uC Ts *,913 s, I, W.'

.q'. "g.. u..

P e

J.

e,,,

.,, e.

s:.,;.

v=

Me..s'..,.,,.p.-

N a.y*f. <-

l 1

h,'

t p-

\\

l P00R ORIGINAL o

4 Fig. 24.

A center piece of cracked CSB G-3

I t

e I

m' JNERAL ATOMN GENERAL ATOMIC COMPANY P. O. BOX 81608 SAN DIEGO, CALIFORNIA 92138 o

4 a-

- ~

,.6

.

ML20003B036

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