ML19241B705
| ML19241B705 | |
| Person / Time | |
|---|---|
| Site: | Crane  |
| Issue date: | 04/25/1979 |
| From: | Butler W Office of Nuclear Reactor Regulation |
| To: | Tedesco R Office of Nuclear Reactor Regulation |
| Shared Package | |
| ML19241B703 | List: |
| References | |
| NUDOCS 7907230174 | |
| Download: ML19241B705 (39) | |
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UNITED STATES NUCLEAR REGULATORY CCMf.usslJN
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.MSHINGTON, D. C. 20555
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",7 APR 2 51979 MEMORANDUM FOR:
R. L. Tedesco, As?istant Director for Reactor Safety, DSS FROM:
W. R. Butler, Chief, Containment Systems Branch, DSS
SUBJECT:
THREE MILE ISLAND, UNIT 2:
SELECTED CONTAINMENT RELATED ISSUESANALYSIS The purpose of this memorancun is to document the results of analyses and evaluations performed by the Containment Systams Branch for several containment-related issues in support of the incident response team.
As evidenced by the following discussior, the primary concern with restect to the con +.ainment centered around:
1) the potential sources of and rates of hydrogen and oxygen production within the reactor vessel and containment building; 2) the potential for and effects of explosion or detonation of hydrogen within the reactor vessel and cea*2inment building; and 3) methods for collection and recomoinat containment pressure when considering the depressurization effects of containment spray cooling, 1.
Reactor Vessel Non-Condensible Bubble On Friday, March 30, 1979, we were asked to perform a best estimate type analysis to determine the possible, contents of a reported non-condensible bubole of about 1000 fta in the reactor vessel upper head and the possible sources of the non-condensibies.
were also asked to estimate the projected growth rate of the bubble We due to formation of hyorogen and oxygen by radiolysis of core water As a basis for the calculation, we performed an analysis with the COGAP concuter code (see SRP 6.2.5) with the following assumptions:
al 1% failed fuel with 10% of the solids and 50% of the halogens from the failed fuel going into the coolant; b ) G-values of 0.44 molecules /100 EV in the core and 0.3 molecules /100 EV in the sump solutions; anc c) all radiolysis products thus generated were contained in tne reactor vessel.
Centact:
W. Milsteac, CSB Ext. 27711 b
voore30/75-
R. L. Tedesco APR 2 ~ :07c Using the results of this COGAP run, the reported bubble size, and the reactor coolant pressure and temperature, we estimated the contents of the bubble on March 29, March 30 and March 31, 1979.
Based on the bubble size measurements taken on Maren 31, 1979 (765 ft3 91024 psia and 280 F), we prepared the at* ached figure (Figure 1) showing mole fraction of Og, steam, and, by subtraction, H2 in the bubble, and the projected size of the bubble as a fur-tion of time. The e lculations perfomed are provided in Appendix "A".
During the course of our evaluation of the bubble, we contacted GE personnel at KAPL regarding the potential for radiclysis in the reactor vessel.
KAPL performed analyses of the radiolysis potential in the containment using DNR codes and concluded tnat in spite of the low water temcerature the net radiolysis should be zerc.
Their analysis indicated that the potent.al for H /02 recembination in the 2
reactor coolant water under the influence or tne gama ray field was sufficiently high to recombine essentially all the Q formed by radiolysis of coolant water, a long as the water in the core is not boiling.
KAPL also indicated that their analysis would indicate a high fraction of H2 insolution (up to 2000 SCC /kg) and cautioned that any depressuri-
- ation of the RCS should be very slow because of the potential for large H2 releases from the solution which might result in growth of the bubble and the Dotential for boiling in the core which wouId increase the rate of radiolysis in the core region and result in subsequent additions of 02 to the bubble.
Resul s of both the KAPL analysis of the bubble and our estimates of t
bubble content indicate that the bubble was predcmimantly hydrocen, the major fraction of which must have been the result of a circoniub/ water reaction at high temperature in the core.
Over the period between March 31 and April 2, i979, the bubble in the core was removed either through repeated pressurication and venting of the reactor coolant system through the pressurizer or the nydrogen in the bubble went into solution in the RCS water.
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Radiclysis Calculations We performed several calculations using the COGTP computer code, varying input values of sump G-value and core release fractions for the halogens and solid fission products, to predict the rate af hydrogen generation from sump radiolysis and accumulation within che containment. The results of our calculations were compared to p.'edictions by RSR based on activity measurements from a H O sample 2
taken on March 30, 1979. We found best agreement with the RSR calculations by assuming a G (H,,) value of 0.5, and a core release of 7". of the halog nr
..id 2.5% of th.- solid fission products.
The results of all our CO2AP analyses and the RSR analifsis indicated that the rate of H2 evolution due to radiolysis in t' e containment sump is small (on the order of 600 to 1000 SCF per aay) in comparison to the capacity of the hydrogen recombiner (3200 SCF/ day). We have also conch,ed that sump radiolysis was a minor contributor to the hydrogen inventory in the containment building.
3.
Centainment Hydrogen ~.1alysis We performed a number of calculations to try to esti:r. ate the total hydrogen production in order to obtain an estimate of the extant of metal-watcr reaction in the reactor.
The potential sources of hydrogen investigated were core and sump radiolysis, metal-water reaction in the core, corrosion of metals in the containment and decomposition of organic coatings dut to radiation.
As previously shown in our COGAP analyses, core and sump radiolysis is a minor contributor. We have estimated from 0.0 lb, moles of H9 (cue to reccabination effects? up to a maximum of about 27.0 15-moles of H2 from radiolysis on the fourth day following the accicent (Marcn 31,1979). We have estimated that cn March 31, 1979, the 765 ft3 bubble contained a minimum of 89.4 lb-moles of H2 (76.2 lb-moles of wnich resulted from metal-water reaction) to a maximum of 95.5 lb-moles of hydrogen (all of wnich resulted from metal-water reaction). We have not estimated any hydrogen generation due to corrosion of metals in the containment due to the low containment temperature and the absence of any significant amount of containment spray. We have also discounted decomposition of organic coatings as being a potential ;ource of hydrogen because of the low containment tem:erature and the absence of any sig.iificant amounts of chemical additives in the ':ontainment.
Tile only data point which p-- Q n
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R. L. Tedesco we could find indicates that radiation exposures iin the containment dome were at least three orders of magnitude less than the threshold for significant damage for epoxy paints (109 rad.).
Even if it were assumed that all painted surfaces in the containment (150,000 ft2 estimate based on the Davis-Besse SAR), did decompose, and that all the gasses released were hydrogen (-20 SCC /gm) that would only yield about 5 lb-moles of hydrogen.
At about 10 hours1.157407e-4 days <br />0.00278 hours <br />1.653439e-5 weeks <br />3.805e-6 months <br /> after the accident, a rapid containment pressure spike occurred. The containment pressure went fremi approximately 2 psig to approximately 28 psig and back to 2 psig in about.10 minutes.
All a' ailaole evidence indicates that this spike was caused by the bu 14'1g of hydrogen in the containment building.
'.We have estimated the amount of hydrogen consumed in this event by t.wo different methods. In one instance we calculated the amount of hydrogen neces-sary to reduce the 02 concentration in the contair ment to the 18.9%
recorded on March 31, 1979.
In the other metnod, we calculated the amount of energy recuired to raise the containment pressure frcm 2 psig to 28 psig. Both methods indicate that aoout. 226 lb-moles of hycrogen was burned.
The measured hydrogen concentration on March 31,1979, was 1.7% which indicates the presence of about 94 lb-moles of H 7 in the containment.
Using a value of 1670 SCC /kg, the maximum theoretical amount of hycrogen which could have been in solution in the eactor coolant (P =
1024 psig, T = 280*F) on March 31, 1979 is aDout 29 lb-moles.
Using the aoove estimates of hydrogen we have estim1ated the amount of metal-water reaction which occurred. Our estimates are a maximum of 48.3% and a minimum of 40.6%.
4 Exolosien or Detonation of Buoble in Reactor Vesse'l On April 1,1979, we were requested by Vince Noona-1/ DOR to evaluate tne containment consequences due to the explosion or detonation of a 500 ft3 or 150 ft3 bubble in the reactor vessel. Using the limits of flamability and detonation determ{ W ty the DNR at its KAPL facil-ity (Figure 2) and the 0, H an'
' tam f raction i-1 the reactor 5
7 vessel bubble previousl;' ets6e / Figure 1), we found that the limit of flamacility k e an 0, acle fraction of 0.05 and the limit for deto is/ '
,Q
- Jhed 3t an 0 cole fraction 9
of 0.12.
Data from t
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- _ incica tec~that explosions p[l
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- 3 of hydrocen rich H /02 mixtures near the lower 02 2
flammability limi t would result in pressure increases of about 2 to 2.5 times the initial pressure while unstable detonation at the lower limit for detonation would result in a pressure rise in the order of 20 to 50 times the initial pressure. This information was confirmed in a telecon with Westinghouse Electric Corporation personnel, who had posed the same questions to Dr. Bernard Lewis
- of Combustion and Explosives Research, Inc., of Pittsburgh, Pa.
Based on the above information we concluded that for explosion of the bubble near the lower limit of fla:m1 ability we would not expect failure of the reactor vessel hence there would be no significant release to the containment. However, for unstable detonation of the bubble in the reactor vessel, failure of the vessel seemed certain.
We analyzed the response of the containment to instantaneous rupture of the reactor vessel by unstable detonation of tne bubble.
For the purpose of the analysis we assumed instantaneous flashing of the entire reactor coolant inventory at a pressure of T000 psia and a temperature df 300 F.
We also added the energy available by combus-tion of the remaining hydrogen from the bubble upon release to the containment. Our calculations are provided in Appendix "B".
Results of our calculations indicated a peak containment pressure of 28.2 psig for the unstable detonation of a 500 ft3 bubble and 23.6 psig for the unstable detonation of a 150 ftd bubble. These results were trans-mitted to Vince Noonan by a telecon on April 3,1979.
5.
Containment Flooding The postulate was made that, althcugh extremely uriiikely, a detonable mixture of non-condensible gasses was developing in the reactor vessel head. The consequences of a detonation in the reactor vessel
- " Combustion, Flames and Explosions of Gases" by Bernard Lewis and Guenther von Elbe - Academic Press Inc., New York, London,1961, A
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.; 0 79 can be mitigated substantially if the containment were flooded with water up to the inlet / outlet nozzles of the reactor vessel.
Information was, therefore, developed as to the feasibility and potential consequences of flooding the containment up to the inlet /
outlet nozzles of the reactor vessel.
a.
Flooding Procedure Water can be added first to the borated water storage tank and mixed with a boron solution to the required concentration:
i.e., about 2000 ppm. The borated water can then be pumped into the containment by use of the containment spray pumps.
The pumping rate will be about 3000 gallons / minute.
b.
Containment Volume The inlet / outlet noz:les of the reactor vessel are at elevation 315.5 feet.
The containment floor at the top of the sump is at elevation 282.5 feet. Therefore, the water height will have to be 33 feet.
The containment inner diameter is 130 feet.
Therefore, the total cgntainment volume up to the inlet / outlet nozzles equals:
TT/4 D-H =.785 (130)2 (33) = 4.4 x 100 ftd. 5: quip;aent inside containment is estimated to occupy about 12% of this volume.
Therefore, the net frae volume that needs to be flooded equals 5 or 3.9 x 100 ft3 x 7.48 gallons /ft =
(0.88)6(4.2 x 100) ft 3 x 10 gallons.
c.
Ficoding Time The minimum time required to pump 3 million gai7 ens into the containment using the containment spray system f s (3 x 100 gallons)/(3 x 104 gallons / min) = 1000 minutes.
The flooding time, therefore, is (1000 minutes)/(60 minutes / hour) = 16-2/3 hours.
d.
Potential Consecuences Flooding of the containment to elevation 315.5 feet with an electrically conducting solution can create sno-t circuits in all the ficoded electrical ecuicment.
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of the reactor coolant pump motors (EL-323), containment fan colers (EL-315), and electrical penetrations cannot be assumed.
Information on elevation of the various penetration and surveillance instruments was not available for review.
e.
Conclusion If flooding of the containment became a requirement, it could be accomplished in about 16-E/3 hours by use of the borated water storage tank and the containment spray system.
- However, there exists an associated high probaoility for disabling the reactor coolant pumps, the containment fan coclers, the electrical penetrations, and surveillance inst:rumentation by electrical short circuits.
6.
Containment External Pressure Considerations At the recuest of Matt Chiramal, from the NRC group resident at Three Mile Island, we determined the maximum design external pressure of the containment building (-2.5 psig) and perfor:ned a series of calculations to allow site personnel to control the containment vacuum to a level which would not result in a depr'essurization of the containment building to less than the desigri external pressure of the building in the event that the containment spray system was inadvertently turned on. The culmination of these efforts was a headquarters tecnnical advisory ( Appendix "C") which was sent to the site by the NRR Technical Support Team on April 10, 1979.
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W. R. Butler, Chief Containment System:s Branch Division of Systems Safety
Enclosures:
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s APPEtiDIX "A" r-511 1/3
e CASE 1 l=, Failed Fuel 5% Release G = 0.44 core G = 0.30 Sump Initial Time = 3/31/79 Volume of Eubble
= 880 ft2 @ P = 875 psig T = 275*F
. @ P = 1024 psia, T = 275 F 3
V = 880 (
- 4) - 765 ft This is 4 days after accident SCF H2 (RAD) = 5036 = 5036 (5.227 x 10-3) = 26.32!
= 13.16# =ol es
- moles 1053.23 - 1059.37 = 6.09 # moles Vol. 0 9 P = 1024 - T= 725 2
3 V = MRT = 6.09(22)48.3.735 = a6.9 ft F
if24(i44)
Vol H 9 P = 1024; T = 735 2
3 V = 13.16( 2)( 767)735 100.6 ft
=
1024(144) 511 176
Steam 9 T = 275"F = 735'R v = 9. 401 f't / ?
3 Total Vol = 765 ft
- H O = 765 ( 1
) = 81.37 # steam 2
9.401 81.37 = 4.52 # moles steam 18 Vol H O @ P = 1024 ;
T = 735 2
3 V = 4.52(18)(735)(85.6) 3a.7 ft
=
(1024(144)
Vol H f#UU Z"IH O react = 763-46.9 - 100.6 - 34.'7 2
2
'. Vol H from Zr/H O 582.8 ft
=
2 2
13.16 # moles H
= 100.5 ft3@T&P 2
1 mole = 7.644 ft 3 d moles H2 (:r/H 0) = 76.24
- moles 2
1 I
511 170
t = 4 days 3
Vol Bubble 765 ft
=
3 Bubble is 46.9 ft -02 3
34.7 ft _g 0 2
3 633.4 ft -H 2 6.09
- T.01. - 0 2 4.52
- nol. - H 0 2
89.4
= moi. - H 2 100.01 0
M 1. Fraction 6.10%
=
2 H O Mol. Fraction 4. 5 0 ".
=
2 H
Mol. Fraction = 89.4",
2 511 1/%
5 days Assuming P t
1024; T = 735
=
=
34.7 ft" Steam 4.52 # mole
=
=
3 H2 (Z#ld 0)
= 76.24 # mole
= 582.8 ft 2
RADIOLYSIS
( aff.51)
SCF H 5839.61 13.16 15.26 # mole
=
=
=
2 g
! moles 0 = 1060.01 - 1053.28 = 7.13 y moles 2
Subble 4.52 # mole HO 4.38 Voli
=
=
2 7.13 # mole 0 6.91 Vol%
=
2 91.5
- mole H,
88.71 Vol%
=
c 103.15 3
Bubble Vol 788.5 ft
- 54. 5 ft"-0,-
=
34.7 ft*-H G c
699.5 ft -H
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i
6 days t
=
3 Steam
= 4.52 # moles
= 34. 7 ft 3
H (Zr)
=76.24 # moles
=582.8 ft 2
RADIOLYSIS SCFH 6575.79 13.16(6575.79) 17.18 # moles
=
=
2 5036.13
- mole 0
= 1053.28 - 1061.37 5.09 ! moles
=
2 Mole Fraction = Vol".
Bubble = 4.52 # mole H O 4.25 Vol LH O
=
2 2
8.09 # mole 0 7.63 Vol %0
=
2 2
93.42 # mole H 88.1 Vol%-H 2
2 106-.03 Subble Vol
= 106.03 (7.644)
= 310. 5 ft 3
34.7 ft -H O 2
3
- 61. 8 f t -02 0
71'.1 ft -H g 511 iUU
s
= '/ days t
3 Steam = 4.52 # moles 34.7 ft
=
3 H (Zr) =76.24 # moles
= 582.8 ft 2
PADI0 LYSIS 13.16(7251.17) = 18.97 # mole SCF H
= 7261.17
=
2 5036.13 4 mole 0
- 1062.25 - 1053.28
= 8.98 = noie 2
Mole Fracticn Bubble 4.52
- Mole H 0 4.16 Vol% H 3
=
=
2 2
8 98
- Mole 0 8.25 'Jol *; 0
=
2 2
95.21 a Mole H 07'5
'J O I ~* H 2
2 108.71 Bubble Vol
= 103.71 (7.644) = 830.98 ft"
- 34. 7 f t "-H.,0 58.6 ft"-0.
2 727. 8 f t -i, o,I
,n1 Ji 10
t = 10 days 3
Steam
= 4.52 # Mole 34.7 ft
=
3 H (Zr)
-76.24
= 528.8 ft 2
RADIOLYSIS SCF H
= 90/9.83 13.16 9079.83 23.73 # Mole
=
2 5036.13 e Mole 0
= 1064.41 - 1053.28 11.33 # Mole
=
2 Mole Fracti.on Buoble 4.52 # Mole H O
- 3. 90 Vol %-H O
=
=
2 2
11.33 # Mole 0 9.78 V 15-0
=
2 2
99.97 # Mole H 86.31 Vol';-H
=
2 115.82
~
g 3
Subole Vol 115.82 (7.644) 885.3 ft
=
=
3 34.7 ft -He0 3
85.5 ft -02 764.2 ft -H 2 511 192
15 days t
=
Steam
= 4.52 # Mole 34.7 ft
=
3 H2 (Zr)
=76.24 # Mole
= 528.8 ft RADIOLYSIS SCF H 11739.67 13.16 11739.67 30.68 # Mole
=
=
2 5036.13
- Mole 0 1068.06 - 1053.28 I4.78 # Mole
=
=
2 Mole..ection Bubble a.52 # Mole H O 3.68 Volf.-H O
=
=
2 2
14.78 = Mole 0 11.71 Y l,-0
=
2 2
106.92 # Mole H 84.71 VoM-H
=
2 2
126.22 = Mole 126.22 (7.644)
Bubble Vol 96a.32 ft
=
=
3
- 34. 7 ft -H,30 3
113.0 ft -0 817. 3 f t ~-H, 511 183
.s
_s 20 days t =
3 4.52 # Mole 34.7 ft Steam
=
=
3 H (Zr)
.G.24 # Mole 528.8 ft
=
=
2 RADIOLYSIS SCF H,3
= 14065.29 13.16 (14065.29) 36.75 ! Mole
=
51036.13 0, # Mole = 1071.08 - 1053.28 17.8 # Mole
=
Mole Fracticn Subble 4.52 # Mole H O
- 3. 34 Vol %-H O
=
=
2 2
17.8 # Mole 0 13.15 Vol';-0
=
2 2
112.99 # Mole H 83.50 Vol %-H
=
2 2
3 135.31 (7.644)
Bubble Vol 1034.3 ft
=
=
34.7 ft -H O 2
3 136.1 f c -0 2 3
863.7 ft _g,,c d
s CONVERSION FACTOR
- Mole to% H In Containcent 2
1 # Mole H 7.644 ft 0 P = 1024
=
2 T=
735 2
7.644 (1024) h-1 # Mole H 7 = 90 F
=
15
? = 15.0 psi C n Conditions 3
1 d Mole M 390.5 ft
=
2 1.952 x 10 ' V
"" " H2 (ContJ/? Mole 1 i Mole 390.5
=
=
6 2x 10 1.952 x 10-2 vol.% H /# MUI*
=
2 511 185
N If Bubble Released to Containment of t = x t = 4 days H
1.75% H 89.4 l' Mole
=
=
2 2
t = 5 days H
91.5 # Mole 1.73% H
=
=
2 2
t = 6 days H
93.42 # Mole 1.S2% H
=
=
2 2
t = 7 days H
9.21 # Mole 1.25% H
=
=
2 2
t = 10 days H
99.97 = Mole
- 1. 95'.' H
=
=
2 2
t = 15 days H
106.97 # Mole 2.CS% H
=
=
2 2
t = 20 cays H
112.99 = Mole 2.215 H
=
=
2 2
511 186
s
~.
APPENDIX "B" 511 187
s m
Vince Noonan Analysis Given Steam @ 1000 psi & 300*F Assume Flashes to 55 psia 6 f3 Vol of Containment = 2 x 10 3
v @ 55 psia = 7787 ft /d T @
287'F
=
h
= 256.3 BTU /#
h
= 9196 BTU /4 h = 1175.9 BTU /#
f fq q
h3 P = 1000 psia & T = 300*F h = 272.3 BTU /9 272.3 =hf+Xhfq
= 256.3 + 1175.9 X X = 272. 3 - 256. 3 16.0
.0136
=
=
1175.9 1175.9 511 186
m Fraction of R.C.
Flashing to 55 psia 1.36.
=
R.C. fiass = 512,796 a 512,796 (.0125)
= 6974.02 Mass steam # 55 psi
=
Vol occupied 6974 (7.787) 54,305 ft
=
=
3 V ; 9 2000 = 0.1878 Vol. of pressurizer dome = 434 ft a
'2 V 1000 =.4456 ft*/#
a
'/a 5
7.787 400 (7.787) = Vol expanded
6990 ft
.d456
)l l
')
2nd Try Assume P = 20 osia f
272.3 = 196.16 + X (960.1)
X = 272.3 - 196.16 0,0793
=
960.1 Mass Steam = 512,796 (0.0793) = 40,667 #
v 9 20 psia = 20.089 3
Vol Steam = 816,957 ft 3
Vol press. Steam = 400 (20.089)
= 18,033 ft
.4456 3
Total Vol 835,000 f t T = 227 F
.'. Steam Press less than 20 psia E 1 "l l9O Ji I e
3rd Try Assume P = 10 osia X = 272. 3 - 161.17
=.1125 982.1 Mass Steam = (.1126)(512,796)
= 57,748) v3 38.42
=
10 Vol = 2,218,714 Too Much Pst > 10 psia 511 191
Try Ps 12 psia
=
X = 272.3 - 170.05 0.1047
=
976.6 v @l2 = 32.394 Mst = (.1047)( 512,796)
= 53,689.7 #
3 Vol
= 1,739,?25 ft st Vol from Press Vapor = 400
=.394 = 29,079 ft
.4456 3
Vol Total Steam = 1,763,304 ft Mass Steam = 54,587 d Stean T = 201 F c 1,1 192 s
Vol. Air = 2 x 106 (14.7
) 460 + 201 14.7 + 12 460 + 90 3
Vol Air = 1,323,350 f t 0
Total Vol 3 x 10 Try P = 18 psia X = 272.3 - 190.66 0.0847
=
963.7 Mst - 512,796 (.0847) = 4 3,441 v = 22.168 3
Vol St (H O) = 963,013 ft 2
Vol from Press = 200 (22.168)
= 19,900 f t'
.4456 Vol Total 982,912 511 193
- -s S
T 222*
=
Vol. Air = 2 x lt (14.7
) #87 (18.3 + 18)
Vol Air
= 1,114,862 ft 982,912 Vol Total = 2,097,774 ok Cont Press from Steam 18.3 psia
=
18.3 Mst = 1,114,E52 18 22.165
- 35. 3
-14.7 Mst = 50,291 a 21.5 psig Mai r = 2,000,00G.14d l14.7) 53.25 550
= 144,282 i i O ['.
\\\\
.s T = 222*F Mt(G)(T-222 ) = Ma (.171)(T-222) + Mst (.336)(T-222)
E=
[Ma (.171) + Mst (.336)](T-222) 222 +
E T
=
f Ma (.171 ) + fis t (. 336 )
3 For V Bubble 500 ft Mass 1000. (144)S00 123.5 #
H
=
=
2 767. 760 Heat of Ccmbustion 61,000 BTU /#
E = 7,534,000 STU Sli 195
3 V Bubble
= 150 ft MH = 123.5.150
= 37. #
2 500 E = 37 (61,000) = 2,250,000 BTU I final 3
3 500 ft Bubble l
150 ft Bubble T = 222 + 7,534,000 7
144,282 (.171)'50,242 (.335)
T
= 222 + 54 f
T = 222 + 131 F T
= 276 F 7
f T = 403 F f
l i
l I
I Sil 196 i
l i
3 Initial Mass of Bubble 500 ft
=
117* O2 - 14.78 (5G4) 7.67 7 :4c'es
=
964 H, -106.92 (500) = 55.46 # Moles 964 Stoicniometric Surn (cetonate) 7.67 (2) = 15.3a 4 Mole
=
Final H2 = 40.12 d Moles = 30.24 #
E = 50.24 (61,000 BTU /#)
= 4,894,640 BT
T,
= 222 + 4,844,600
= 222 + ll6'F 144,282 (.171 ) + 50.291 (.336)
= 339"F 41570
= 13.3 (*60 + 339) = 21.4 psi D g 460 + 222 51i i97
~
~
3 Initial Mass of Bubble 150 ft
=
0 1 1. 7 '.'
- Mole 0 to Burn = 40.12 (150) = 12.045 2
2 500
- H2 = 24.09 #
E = 24.09 (61,000) - 1.409,500 BTU T = 222 + 1,469,500 = 222 + 35 = 257*F 7
49570
?.
= 18.3 (257 + a60) 19.3 osi
=
ai r 222 + 460 Pst 9 257*F v
222 F = 22.163 sat 3 T = 257 Sup Steam P
19 est a
=
3 T = 339 Sup Steam P
21.5 osia
=
Steam Charts plate 48 page 13 511 19B
Efi nal 3
3 500 ft Bubble 150 ft Bubble 21.5 19.3 21.4 19.0 42.9 38.3
-14.7
-14.7 28.2 osig 23.6 osim Sli 191
AppEMDIX "C" 511 200
5.0 CONTAINMENT The containment internal pressure has been slightly lower than arttient pressure for most of the time since the accident. At the present time, the containment is at approximately a 0.9 psi reverse pressure differen-tial. Since the design pressure is 2.5 psi, the currernt pressure is not of immediate concern. Current operating procecures f radicate that the water flow to the fan coolers should be terminated if the reverse pres-sure differential reacnes 2.0 psi. This action would. effectively ter-minate further cooldown of the containment atmospnere -thereby terminating the trans ~
.t.
In any case, this would be a rather si ow transient allowing sufficient time for proper action.
We believte, however, a more severe transient should also be considered.
This trantsient is the inadvertent operation of the containment sprays.
Init.iation of the sprays would result in rapid ccoling of the containmemt atmosphere causing a corresponding rapid decrease in containment pressure.
The lagnitude of the pressure decrease will depend upon trte inlet spray water temperature (SWST water temperature).
To assure that -the contair. ment does not exceed the oesign reverse differential pressurre of 2.5 psi, the containmen' parameters should be maintained aDove eninimum values as shown in the enclosed figure. The figure indicates that for a given iala spr / water temperature, the containment tempera ture as well as
- ntainment pressure should be maintained aoove minimurm values.
The
- ressure coulc be controlled by the addition of a non-condensible gas 511 201
- such as nitrogen or dry air.
This procedure has apparently been followed previously to decrease the reverse differential pressure to below 1 psi.
Control of containment temperature could also be achieved by terminating the vater to the fan coolers. Since the fan operation would continue, proper mixing of the atmosphere would be maintained while eliminating the heat removal mechanism. Since the consequences of exceeding the reverse design pressure differential is unknown, we believe it prudent to maintain containment conditions as indicated maove to allow for inadvertent spray o perati on.
6.0 RADI0 LOGICAL CCNSIDERATIONS The potential radiological consequences of loss of let down flow use of the RHR system, and steam generator leakage have been identified for consideration in this section.
6.1 PURIFICATION DEMINERALIZER HEATUP/ DEGRADATION SuDstantial radioactivity may have built up on the PJrification demineral-izer such that if the flow is stopped, the bed will heat up due to decay neat. Rough calculations indicate that the relief valve wil lift anc discnarge small amounts of water and possible traces of steam to the Reactor Coolant Holdup Sleed Tanks (RCHST) if the sytem is isolated. As 5ii 202
- long as some flow is maintained, there should not be any steam.
If water and traces of steam are relieved to the RCHBT, the eftfsite consequences should be nil because these tanks vent to the waste gas vent heacer which can be placed at a negative pressure by venting back - o containment.
Procedures sould exist for venting the waste gas vent header back to containment should this bec0ce a problem, 51i
'203 o02
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The Honorable John G. Kemeny Chaiman President's Ccs: mission on the Accident at Three Nile Island 2100 H St., N.W.
Washington, D.C.
20037
Dear Mr. Chairman:
My letter of June 25, 1979, included a respc-to question IV of the questions raised at our meeting of April E6. Enclosure 2 of that response containo:! notes and memoranda free the Office of Nuclearr Reactor Regulation concerning the production of hydrogen and oxygen in the reactor co11 ant systec: after the accident. A m o randum dated April 25,,1979 that provides further infomation on that subject was overlooked in tine staff's vriginal file search to assemble enclosure 2.
The mcnorandum is, enclosed here and should be added to enclosure 2 of the response to quest:fon IV.
Sin:erely, Joseph M. Hendrie!
Chaiman En:losure:
As stated DISTRIBUTION:
SECY-ED3 Rdg. File H. Denton, NRR R. Ryan, SP
': T/F L. Gossick
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F.. Mattson
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