ML20081J200
| ML20081J200 | |
| Person / Time | |
|---|---|
| Site: | Limerick  |
| Issue date: | 11/04/1983 |
| From: | Kemper J PECO ENERGY CO., (FORMERLY PHILADELPHIA ELECTRIC |
| To: | Schwencer A Office of Nuclear Reactor Regulation |
| References | |
| NUDOCS 8311080348 | |
| Download: ML20081J200 (5) | |
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PHILADELPHIA ELECTRIC COMPANY 23O1 M ARKET STREET P.O. BOX 8699 1881 -1981 NOV 041983 PHILADELPHIA. PA.19101 (215)841 4502 VICE PRESIDENT ENG4NE E MtNG AND WE.SE ARCH Mr. A. Schwencer, Chief Docket Nos. 50-352 Licensig Branch No. 2 50-353 Division of Licensing U. S. Nuclear Regulatory Camission Washington, D.C. 20555
Subject:
Limerick Generating Station, Units 1 and 2 Feedwater Contaiment Isolation Check Valves
References:
(1) Letter fran J. S. Kenper to A. Schwencer dated August 29, 1983.
(2) Telecon between Renee Li (NFC), Boyd Harper and Tan Hutson (Bechtel) and 'Rxn Shannon and Jerry Phillabaum (PECO) on Septenber 19, 1983.
File:
GOVT l-1 (NPC)
Dear Mr. Schwencer:
The reference (2) telephone call requested additional information on the Evaluation of Feedwater Containment Isolation Check Valves for a Hypothetical Pipe Rupture Condition for Limerick Generating Station which was transmitted by reference (1).
In response to reference (2) the enclosure provides clarification and additional information on the integrity of feedwater contaiment isolation check valves following a hypothetical pipe rupture to close Limerick SER Open Issue 4.
Sincerely, OK4 JLP/gra/S-1 Copy to: See Attached Service List 8311080348 831104 PDR ADOCK 05000352 i,
E PDR I
cc: Judge Lawrence Brenner (w/ enclosure)
Judge Peter A. Morris (w/ enclosure)
Judge Richard F. Cole (w/ enclosure)
Troy B. Conner, Jr., Esq.
(w/ enclosure)
Ann P. Hodgdon, Esq.
(w/ enclosure)
Mr. Frank R. Romano (w/ enclosure) l Mr. Robert L. Anthony (w/ enclosure)
Mr. Marvin 1. Lewis (w/ enclosure)
Judith A. Dorsey, Esq.
(w/ enclosure)
Charles W. Elliott, Esq.
(w/ enclosure)
Jacqueline I. Ruttenberg (w/ enclosure)
Zori G. Ferkin, Esq.
(w/ enclosure)
Mr. Thomas Gerusky (w/ enclosure)
Director, Pennsylvania Emergency Management Agency (w/ enclosure)
Mr. Steven P. Hershey (w/ enclosure)
Angus Love, Esq.
(w/ enclosure)
Mr. Joseph H. White, III (w/ enclosure)
David Wersan, Esq.
(w/ enclosure)
Robert J. Sugarman, Esq.
(w/ enclosure)
Martha W. Bush, Esq.
(w/ enclosure)
Spence W. Perry, Esq.
(w/ enclosure)
Jay M. Gutierrez, Esq.
(w/ enclosure)
Atomic Safety and Licensing Appeal Board (w/ enclosure)
Atomic Safety and Licensing Board Panel (w/ enclosure)
Docket and Service Section (w/ enclosure)
3 1.
Was the pipe rupture location the worst case?
The pipe rupture location was assumed to be located upstream of the check valve.
However, to be conservative with respect to allowing maximum flow, and hence maximum valve disk closing velocity, friction and inertial pressure losses between the check valve and pipe rupture locations were assumed to be zero. Thus, the worst location was chosen.
2.
What is basis for assuming that the seat is stressed to 50% of yield at 2132 psi?
At 2132 psi the calculated stress for the seat is 18 kai.
For the SA216 material used for the major portions of the seat and valve body, the minimum yield stress is 36 kai.
Therefore, the valve seat is designed for 50% of yield at 2132 psi.
In the valve seat stress analysis the design condition of 50% of yield at 2132 psi was used in determining the spring stiffness for each of the seat elements. This would be for static conditions of the valve disk in the closed position and 2132 psi existing downstream of valve. During the valve disk closing transient, where the disk is initially in the open position, the seat stress is caused by the disk closing impact. The seat is not stressed prior to disk impact, so that the full stress or energy absorbing capacity of the seat is available at the time of disk impact.
3.
What is the basis for using ductility ratio of 30?
The ductility ratio is the ratio of ultimate to yield displacements (or strain with a common length). For simple structures this can be simply the ratio of ultimate to yield stress strain for a material.
For complex structures the failure ductility ratio is difficult to determine without detailed calculations or test data taken from structure failure.
Generally, experience in structures is that failure ductility ratios range from 100 to 200. A value of 30 was chosen to be conservative. For the analysis performed for the check valve seat, a value of 30 was adequate because the seat retained its function for this conservative value chosen.
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e liowe ve r, to be more specific about the failure ductility ratio, a more detailed analysis has been performed. Using Article 56, Case 18 f rom Roark*, which is closely representative of the valve seat, a circular ring attached fully at its outer periphery with a force exert ed uniformly at its inside edge, the deflection at yield has been calculated to be 6.9 10-3 in.
The deflection leading to ultimate strain assuming a linear stress-strain relationship from the yield stress to the point of ultimate stress has been calculated to be 2.593 in.
The ratio of these two deflections gives a ductility ratio of 376 at failure. The material is SA216 with a yield stress of 36.0 ksi and an ultimate stress of 70.0 ksi.
(Actual failure has usually somewhat greater strain et ultimate but credit has not been taken for this.) Therefore, the assumed ductility ratio of 30 to represent failure is conservative by over a factor of ten.
l 4.
What is the basis for using a plastic section modulus of 1.57 The cross section of the seat portion of the valve body most closely resembles that of a rectangle. The ratio of the plastic section modulus to the section modulus for a rectangular cross section is 1.5.
Therefore, a value of 1.5 was assumed to be valid and was used in the seat stress analysis.
(The plastic section modulus is discussed and derived in Roark, on page 123:
R. J. Roark, " Formulas for Stress and Strain," Fourth Edition, McGraw-Hill Book Co., 1965.)
5.
Address stresses in the check valve disk pin.
The stresses in the disk pin were not addressed in the response submitted to NRC earlier. The pin stresses are caused by centrifugal force exerted on the disk during rapid closure, which in turn are transmitted to the pin resulting in shear stresses at each end of the pin.
For a disk closing velocity of 100 rad /sec, the maximum centrifugal force, occurring at impact, is 102400 lb.
This results in a force of 51200 lb at each end of the pin. The resulting shear stress is calculated to be 19.7 ksi. The
- R. J. Roark, " Formulas for Stress and Strain," Fourth Edition, McGraw-lilli Book Co., 1965, 2
f.
pin material is reputed by the Vendor to be of a Brinell liardnesa of 217 to 235. Typical Type 410 stainless steel with a Brinell liardness of 225 is found to have a yield stress (0.2% offset, tensile) of 85 ksi; for normal operations 42.5 ksi can be used for yield in shear based on the maximum shear stress theory. Even assuming no heat treatment at all ( ASTil A276, Type 410, Condition A), the minimum yield strength of 40 kai has a corresponding 20 ksi in shear, which is greater than the conservative shear stress value of 19.7 ksi calculated above. Additionally, if strain rate were taken into account, an additional margin of 50% would be available to give a yield in shear of 30 ksi.* Therefore, the pin will not fail during the disk closing transient of 100 rad /sec.
6.
What is the stress on the hinge pin as the disc contacts the seat and causes deformation?
At disc impact and coincident deceleration of the disc, the shear stress on the hinge pin decreases as the centrifugal force decreases due to the decrease in velocity of the disc.
Thus, as stated in paragraph 5 above, the maximum shear stress on the hinge pin occurs at impact of the disc and seat.
t
- M. J. Manjoine, " Influence of Rate of Strain and Temperature on Yield Stress of Mild Steel," J. Appl. Mech., December 1964 3
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