ML19312C734
| ML19312C734 | |
| Person / Time | |
|---|---|
| Site: | Oconee  |
| Issue date: | 07/24/1969 |
| From: | Long C US ATOMIC ENERGY COMMISSION (AEC) |
| To: | Deyoung R US ATOMIC ENERGY COMMISSION (AEC) |
| References | |
| NUDOCS 7912190943 | |
| Download: ML19312C734 (4) | |
Text
_
x JUL 2 41969 Richard C. DeYoung, Deputy Assistant Director for Reactor Technology Division of Reactor Licensing EXCURSION ENERGY CALCULATION FOR THE OCONEE NUCLEAR UNIT (DOCKET NOS. 50-269, 50-270, AND 50-287)
The rod-ejection analysis in the Oconee FSAR (paragraph 14.2.2.2, pp 14-26 through 14-35) concludes that 1.52% 4.k/k is required to release 3.37 x 108 cal energy to the coolant. B&W used a point-kinetics model, which appears to be nonconservative with respect to WIGL2 for rod worths above 0.6% ak. Please ce==ent on the B&W procedures for calculating rod-ejection consequences.
A damage calculation is provided in the FSAR for the reactor coolant system, paragraph 14.2.2.2.4, pages 14-33 through 14-35. The calculation purports to show margin between the rod-ejection accident and the thresh-old of failure for the reactor coolant systen. The calculation uses a forimala that correlates a damaging value of explosive versus dimensions and properties of the pressure vessel. The correlation is a form of equation 9.1 (p 81) of :ICLTR63-140 (Reference 12 of the FSAR, Chapter 14).
I The FSAR assumed an ultimate strain 6% of 26%, and half of that value, or 13%, was used in their calculations. They also used 3 = 55,000, C = 80,000g and vessel temperature = 6000 F.
Their conclusion was that 1410 lb TNT would strain the mid-meridian ring up to 50% 6w,i.e.
13 strain.
I have the following comments:
l 1.
A simplified calculation (by RP) for excursion energy is provided l
in Enclosure A, which RT may wish to comument on.
2.
According to NOLTR63-140, p 83, real vessels can be expected to withstand safe strains only up to
(: d3, instead of 6d2. This would reduce the damaging-charge value.
3.
Also according to NOLTR63-140 the properties of weldments must be considered, and the properties 6y
, C.
(u,must be evaluated for each vessel coeponent (base, HAZ, weld) in light of temperature and radiation effects.
-l OFFfCE >
SURNAME >
OATE>
Form ABC.318 (Rev.9-63) u.s.sovenmupT mmme wrics :ip2 442s 7 912190 7% d I
g 2 4 1963 Richard C. DeYoung 2
Your assistance is requested in answering the following questions:
1.
What is a conservative correlation for excursion energy versus step insertion for this accident?
2.
For the B&W design, what are the proper values of og
, Q, and (w, considering temperature and radiation effects?
3.
Do you conclude that the applicant's procedure for evaluating a damage threshold has technical merit? Please note that the applicant used an equivalent vessel thickness of 10.78 in.,
although the vessel is only 8-7/16 in, thick, plus 1/8 in. clad.
4 What quantity of TNT do you conclude represents a conservative value for evaluation of the rod-ejection accident?
It is possible that the assistance of Jim Proctor, our consultant from NOL, may be required to complete your review. If so, please keep :ne informed.
/5/
\\
Charles G. Long, Chief Raaetor Project Branch No. 3 Division of Reactor Licensing As stated above Dis tribution:
L. Forse A. Schwencer C. G. Long D. F. Ross (2)
RPB-3 Reading RPB-3/DRL RPB-3/DRL omcr,
M in (j :',.t DFRoss:pt4 CGLong SURNAME >
.7/.3/69 I/j,[ 7//d/69 o4Tr >
~-~.-
h
---,e r
ENCLOSURE A ESTIMATE OF. ENERGY PRODUCTION FOLLOWING A 1.5%a K STEP INSERTION FROM EOT CRITICAL FOR A B&W REACTOR 1.
System Properties a.
Mass of UO = 94,100 kg = 207,000 lb.
2 bjo~
b.
C of UO = 0.08 2
(
see
= 17.4 c.
MC
=
y d.
., = 7.1 x 10-
= 1 $ (p 14-31 of FSAR)
Ir
)=2.48x10 sec (p 14-31 of FSAR)
-5 e.
-3 f.
_A =
/
' 3.49 x 10
=
-3 7.1 x 10
- 1
- 1 g.
K = -1.17 x 10~ A k/k-F = -
D/ F g
7.1 x 10
-3 K = -1.65 x 10 D/ F D
(D in dollars) 2.
Step Approximation E = energy generated (See Technology of Nuclear f
Reactor Safety, pp 417 et seq.)
~
E
=
f y
where C/, = initial reactivity injection in dollar units of super-prorpt criticality
2 CYo = D/st.
b = b'/j K
D b' = energy feedback function = M C P
In our case b' = 1*65 x 10-'
-5 9.5 x 10
$/Mw-sec
=
17.4 D = 1 $ (approxinately)
Cio= 1/3.49 x 10-
= 286 b = b'/_3
=
2.72 x 10
=
_3 3.49 x 10 2cto ~ (2)(286)
= 21,000 :tw-see E
=
g b
-2 2.72 x 10 If die nuclear energy is only 1/5 as "ef fective" as TNT, a equivalent energy is approximately 4000 Mw-sec, which is in turn equivalent to about -
2000 lb TNT (B&U got 1400 lb TNT).
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