ML20235N369

Thin Wall Pipe Integrity Assessment for Calvert Cliffs Unit 1 - Fracture Evaluation

ML20235N369
Person / Time
Site:	Calvert Cliffs
Issue date:	03/31/1987
From:	Nair P SOUTHWEST RESEARCH INSTITUTE
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ML20235N256	List: ML20235N253 ML20235N316 ML20235N369 ML20235N433
References
NUDOCS 8710060495
Download: ML20235N369 (14)

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s SOUTHWEST RESEARCH INSTITUTE Thin Wall Pipe Integrity Assessment for Calvert Cliffa Unit 1 - A Fracture Evaluation b7 P. K. Nair, Ph.D., P.E.

H. G. Pennick t 4.

i SwRI Project 17-4772-861 i

March, 1987 M

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2 SOUTHWEST RESE ARCH INSTITUTE 4

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Calvert Cliffs Unit 1 - A Fracture Evaluation i

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6 TABLE OF CONTENTS Page Objective 1 Approach 1 Given 1 Assumptions 1 Analysis 3 Calculate Longitudinal Stress 3 Critical Crack Size Calculation Based on Linear 3 Elastic Fracture Methods (ASME Section XI Appendix A)

Estimation of Required Pipe Wall Thickness for 4 Limit Load Failure Calculations 4 Estimate Crack Growth Over 2 Years and 30 Years 6 Summary & Conclusions 10 References 11 i

Southwest Research Institute l'

Thin Wall Pioe Integrity Assessment l

i Objective: To determine the long term integrity of the reduced section l

thickness of the steam pipe based on fracture and/or rupture considerations.

Approach: The reduced thickness of the pipe is limited to a circumferential segment in the base material near the weld. The reduced thickness pipe is conservatively assumed, for fracture mechanics evaluation purposes, to have a surface connected indication with a depth .

of average pipe thickness minus the reduced pipe thickness. In the analysis linear elastic fracture principles consistent with ASME Section XI, Appendix A, and fully plastic solutions for cracked piping are used.

Given: 1) Maximum anticipated pipe internal pressure = 1693 psi (Hydro test),

2) Pipe outer radius, Ro = 17 in.

Pipe average thickness = 1.08 in.

Pipe mininum thickness = 0.85 in.

3) Pipe yield stress, a o

= 28,620 psi.

4) Table I. Expected Plant cycles over 40 years.

Assumptions: 1) Maximum pipe internal pressure is 1693 psi.

2) Lower bound on fracture toughness for piping material = 100 KsiY in. (Reference ASME Meeting

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. TABLE I. Expected Plant Cycles Over 40 Years

a. 500 Heat Up and Cool Down cycles.

b. 15,000 Power Changes 15-100% 0 5%/ min 532F - 572.5F 900psig - 850 psig

c. 2000 Power Changes 10-905/100-20%

Same temperature & pressure ranges

d. 1,000,000 cycles of +/- 6F

e. 400 trips from 100%

Same temperature & pressure ranges ,

f. 40 cycles of loss of turbine at 100%

Same temperature & pressure ranges

g. 40 cycles of loss of reactor coolant flow Same temperature and pressure ranges

.h. 5 cycles of loss of secondary pressure i

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3 minutes task group on pipe flaw evaluation, October 27, 1986)

3) Crack growth is suberitical crackgrowth.

4) Initial crack depth (at) = 0.23

5) Maximum crack depth after 30 years = 0.30 in.

6) Table II plant cycles over next 2 years.

7) Assume all stress is membrane.

ANALYSIS:

Calculate Longitudinal Stress PR I (1693) (16.15) o g= = -- = 16083.5 psi = 16.10 Ksi 2t 2 (0.85)

This is the stress used in the_ pipe evaluation.

Critical Crack Size Calculation Based on Linear Elastic Fracture Methods (ASME Section XI Apoendix A) 2

~2 Q K IC 0.95 100 a = = = 4.29 in, w

Mo m -

n 1.65 (16.10) -

where Q = 0.95,' flaw shape factor. .

l M3 = 1.65, Magnification factor for membrane stress.

o= 16.10 Ksi, maximum longitudinal stress.

The average wall thickness is given as 1.08 in. which is less than L_____________ j

-i 4

the calculated critical crack-size of 4.29 in. This implies that the thin-walled pipe will leak before it can break. The through thickness

. leak can develop only when the thin pipe section reaches limit load conditions.

Estimation of Recuired Pipe Wall Thickness for Limit Load Failure. An estimation of the pipe wall thickness required to support the anticipated maximum axial load (Pg ) due to the calculated 16.10 kai longitudinal stress is given below. The calculations are based on setting the limit load (P o) equal to the anticipated maximum axial load (Pg ) and then solving that quadratic equation for the required section" thickness (C).

Calculations 2 2 PR g P

t

= n (R g -R1) 2t 1

l w (28.18) (1693) (16.15) 2(0.85)

= 1,423,873.6 lbs where i P = 1693 psi, internal pressure.

l R t = 16.15 in., pipe inner radius at the reduced section.

l.

t = 0.85 in., pipe thickness at reduced section.

1 .

~ 1.

'_. .j 5

( .,

then~,

l:

, , (g2 -R)2 EPRI REPORT NP-1931 <1

'O 0 P=-[2. ..

0 3

.and let

, Pg=Po

/

2 P

L= ao w (Ro2 - (Ro - c)2}

r 2

2

= a o w' [ R o

-R o2 + 2Ro C - C2 )

0 2 2 .,

=

ao w (2RoC - C ) -

P L[ = 2R C -l;.2 ,

2agu .)

_ j 2

C - 2R C -

PLV3 =0 1

l

. 2a w .,

J where i R

e R1+a a = Ro -R1 - c, crack depth, in.

R c

= R o -R1+R1-c R

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o

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c = section' thickness required to support longitudinal stress of 16,083.5 psi.

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.q- ,A s, = ..

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,[

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g

.(,

4

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' (- .'p p ,

3 fe

' R- _ thend,o)3ingthequadraticequation y,

c. ,

\

,g s

. .cr ' Pt 1.723,'-

) ,V -(-2Rg ) * (-2Rn )2 - H 1)l( 2a w I s' \ O /

c=-

o-2 (1) l

'1.10 Pt-l.' 34 1156.0 - ~

lt -

.s o 34 .t 33.19

=

!r. '

U 2 2

?

i's = 0.405 Jn.

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I' ,

l Hence, a pipe wall thickness of 0.405 in or less is required to support L

/

the assumed' axial icad./ Tbis' approach is simi[ar to I'43-3640 section XI

./

(

f winter addencum ,1985.-

0. g . 1983 This addendum )' uplies primarily to s .

t.

/ austenitic piping. , Currently, the Section 3I Working Group is i

. j- evaluating carbon steel pipes,  ;

t Estimate Crack Growth Over 2' Years ar.d 30 Years a .

1 See Table II for cycle esAlmates and' stress intensities.

3 3 SETION XI APPENDIX A  !

= 3.795 x 10~'0 AK .726 dN f (water environment)

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aa a 3.795,x 'O-10 aK ,726 AN 3 '

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.' o TABLE II., Assumed Plant Cycles Next Two. Years and' Thirty Years Number Number of. 'of Stress Cycles Cycles Change

'2 years 30 years (ksi) Transient Cycle 50 750 16.10 Heat Up.cooldown

. K.

750- 11250 0.81 Power Changes 15% - 100n 100 1500' O.81 Power changes 10-90%/100-20 3.

+/- 6 'F 50,000 750000 ' O.81

1. 20.0 300 0.81 Trips from 100% ,,

i.

2.0 30' O.81 Loss of Turbine 0 100%

2 ~. 0 30 0.81 Loss of Reactor' Coolant Flow 0.25 3.75 3.22 Loss of Secondary Pressure i

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d p

l f

_ _ _ _ _ _ _ _ _ _ _ _ . . . _ _ l

~_ -----.-=-_------_--- _ - _ _

s.

8 aK = ae~gM - hu a/Q, stress intensity factor change

. Mg = 1.65,-magnification factor for membrane stresses

.c Q = 0.95,- flaw shape factor See Table III, Listing of Algorithm Used.

.For both the two year case and the thirty year case, the crack was grown t-in two transient load blocks. For the two year case, there were 50 f'

total heat - up and cooldown cycles - applied in the first transient load'

+

block. Then 50,874 cycles of load changes were applied in the second I

transient load block. The total crackgrowth over both load blocks was 0.00237 inches. Hence, an initial crack with a depth of 0.23 in. could grow to a ' depth of 0.232 in, due to the application of the two transient load blocks. In the thirty year crackgrowth estimates, there were 750 total heat up and cooldown cycles applied' in the first transient load block. Then 763,114 cycles of load changes were applied in the second transient load block. In this case, the total crackgrowth over both load blocks was 0.04029-in.

Hence, these results imply that crackgrowth is not significant during the thirty year time period when the initial crack depth is 3.23

/ .in. Therefore, failure during this time period must be through net 1

d section yielding or pipe collapse.

er 3

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TABLE III. Crackgrowth Algorithm 10 CLS 20 PRINT "*+w*.++* F R0GRA M RUNN 1 N G + ++ + * * * * " '

30 LLIST 40 MM= .65 l 30 D5c= 16.10 60 AI=.23 70 DS= DSS 00 A=AI 30 Qu. /$

100 DDA=0 -

110 FOR I = 1 TO 750 120 DA= (3. 795E-10) 4 (DS 'MM+50R ( A*3.1415;?/ 01 ) ^ 3. 726 130 A=A + DA

, 140 CDA=DDA

• CA 150 NEXT '

160 LFR:NT "r= " A," DDA= ";0DA 170 05=.05+ DSS i 160 FOR I = 1 TO 7e 114!

190 DA= ( T. 79 $E-1 O > + ( 05 eMM. SOR ( A +3.1 a 1 = ? 'O) ) '3. 726 2v0 A=A + DA 210 DDA=CA + DDA 220 NEXT 250 LFPINT "A a ":A." 'gDA= ' DDA l

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10 Sur: mary &

Conclusions:

A detailed long term structural integrity

,I assessment was performed for the thin section of the pipe. It has been f determined that a pipe thickness of 0.405 inch or smaller will be required before the pipe can leak under any anticipated accident overload condition. The future inservice degradation of thickness,

! based on cyclic fatigue crack growth calculations, demonstrate no i

significant crack growths within a two year inspection schedule. End of t

! life (of plant - 40 years) degradation rates are also small, l

The rupture load is approximately two times greater than the maximum i,

anticipated load used in the pipe limit load calculations.

3 <..

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11

. REFERENCES

1) ASME Boiler and.Presure Vessel Code,Section XI, Appendix A, 1983

2) V. Kumar, M. D. German, C. F. Shih,."An' Engineering Approach for Elastic-Plastic Fracture Analysis", EPRI Report NP-1931, July 1981.

3) ASME Boiler and Pressure Vessel code,Section XI, IWB-3640, 1983 Winter Addendum 1985.

4) ASME Meeting Minutes Task Group on Pipe Flaw Evaluation, l

October 27, 1986,. San Diego, CA.

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