ML050030061

Psat 3056CT.QA.04, Calculation Component Effect of Indian Point 3 Engineered Safety Feature Leakage Iodine Release with Boundary Layer, Revision 1

ML050030061
Person / Time
Site:	Indian Point
Issue date:	12/22/2004
From:	Entergy Nuclear Operations
To:	Office of Nuclear Reactor Regulation
References
FOIA/PA-2006-0179, NL-04-162 PSAT 3056CT.QA.04
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Text

ATTACHMENT 4 TO NL-04-162 POLESTAR NON-PROPRIETARY CALCULATION (42 PAGES)

ENTERGY NUCLEAR OPERATIONS, INC INDIAN POINT NUCLEAR GENERATING UNIT 3 DOCKET 50-286

PSAT 3056CT.QA.04 Non-Proprietary Page I of II Rev.0 N 2 3 4 CALCULATION TITLE PAGE CALCULATION NUMBER: PSAT 3056CT.QA.04 CALCULATION TITLE: Calculation of Indian Point 3 Engineered Safety Feature Component Leakage Iodine Release with Boundary Layer Effect ORIGINATOR CHECKER IND REVIEWER Print/Sign/Date Print/Sign/Date Print/Sign/Date REVISION: 0 Jun Li Rudolph Sher James Metcalf 1

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4 REASON FOR REVISION: Nonconformance Rpt 0 - Initial Issue N/A 1 - Compliance with I OCFR2.2790, Jan, 2003. Proprietary N/A designation removed from all pages except first page and those pages actually containing proprietary information.

Proprietary information on each proprietary page designated in ((bracketslJ On Page 2 of Attachment 2, errors have been found in the row of Temperature (K) from the 2nd value to the 7 th. They should be 370.9, the same as the first value in that row, instead of 397.0. They are corrected. Since the calculation used only the first value in that row, the errors and the correction have no impact at all on the results in the report.

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PSAT 3056CT.QA.04 Page 2 of 11 Rev.0 N 2 3 4 Table of Contents Section Pge Purpose 2 Methodology 2 Assumptions 3 References 6 Calculation 7 Results 10 Conclusions 11 : XL Spreadsheet Calculation of Pool Evaporation Rates (2 pages) : XL Spreadsheet Calculation of Iodine Concentration in Bulk Gas Phase with BL Effect, Hallway Outside (4 pages)

Purpose In accordance with the requirements of Section 4.2, Reference [1], the purpose of this calculation is to determine the Indian Point 3 (IP3) engineered safety feature (ESF) component leakage iodine release to the environment with and without consideration of the effect of the gas boundary layer that exists at the interface between the ESF liquid leakage and the bulk gas in the room into which the ESF leakage occurs (i.e., liquid-gas interface).

Methodology The approach is to apply the Reference [2] Standard Review Plan (SRP) guidance that if the calculated flash fraction is less than 10% or if the water is less than 212 F, then an amount of iodine smaller than 10% of the iodine in the leakage may be used if justified by the actual sump pH history and ventilation rates. In addition, the effect of the gas boundary layer existing at the liquid-gas interface on the bulk gas iodine concentration has been included.

The steps in the calculation are as follows:

Evaluate the elemental iodine concentration in the ESF liquid. This is a function of the core inventory of iodine, the iodine release from the core during the DBA LOCA (this iodine is assumed to go into solution in the RCS and containment sump liquid which is in turn recirculated through certain primary auxiliary building (PAB) rooms as ESF liquid),

the total ESF liquid mass, and the ESF liquid pH.

PSAT 3056CT.QA.04 Page 3 of 11 Rev.0 E1 2 3 4

• Assume that ESF liquid leaks onto the floors of certain PAB rooms during the recirculation and cumulates to form a pool. [I 11

• Evaluate the gaseous iodine concentration at the pool surface. [I 11

• Evaluate the resistance to the mass transport of iodine across the boundary layer, which is referred to as decontamination factor (DF) in this report. if 11

• Using the volumetric flow of gas from the ESF gas space to the environment and gaseous iodine concentration in the ESF gas space (with or without consideration of boundary layer DF), calculate iodine release to the environment.

The ESF liquid is circulated outside containment beginning at 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> after the initiating event per Reference [3], items 4.2-4.4. Per Assumption 4, it is assumed that leakage can occur in the RHR cells, the Sump Tank Room, the Hallway to the Sump Room, the Hallway Outside (which includes the Sump Room), the Piping Penetration Area, and the High Head Pump Room.

Assumptions Assumption 1: Flashing need not be considered.

Justification: The peak post-recirculation ESF liquid temperature is about 208 F at 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> (Reference [3]). This is less than 212 F (saturation temperature at atmospheric pressure). Thus, flashing wvill not occur.

Assumption 2: Iodine partitioning to the gas phase is assumed to take place instantaneously under equilibrium conditions at the ESF leakage pool surface. The partition coefficient depends only on the temperature of the liquid.

Justification: The assumption of equilibrium conditions is made to simplify the calculation, and in addition is conservative since the gaseous iodine concentration is actually lower before reaching equilibrium condition.

Assumption 3: The liquid is assumed to always be wvell-mixed and homogeneous.

PSAT 3056CT.QA.04 Page4of II Rev.0 J2 3 4 Justification: This assumption is conservative, since it over-estimates aqueous phase iodine concentration near the pool surface and therefore over-estimates the release to the gas phase.

Assumption 4: The continuous ESF leakage (i.e., 4 gph) will be assumed to be distributed in all rooms in which ESF components are located.

Justification: Distributing the ESF leakage involves a greater gas phase volume and thus effectively increases the partitioning Of i2 to the gas phase, since per Assumption 2, the gaseous iodine concentration at the pool surface is in equilibrium with the aqueous phase iodine, and so the larger the gas volume, the greater the gaseous '2 content that is necessary to maintain the equilibrium gaseous concentration. In addition to resulting in a larger gaseousi2 content, distributing the ESF leakage will increase the volumetric exchange rate to the environment. The combination of these effects will therefore lead to a conservative iodine release to the environment.

Assumption 5: The volumetric exchange with the environment will be assumed to be the sum of the exchange rates of all ESF rooms into which leakage can occur with the rooms being woell-mixed.

Justification: As mentioned above it is conservative to assume that the flow from all ESF rooms into thhich leakage can occur is available since maximizing the exchange with the environment will maximize the iodine release. It is also conservative to assume well-mixed rooms since time for mixing would slow the exchange with the environment relative to the well-mixed case.

Assumption 6: Per Assumption 8 belowv, the ESF liquid leakage is expected to pooi on the floor of the area into which the leakage occurs. However, for purposes of quantifying the effect of the gas-liquid boundary layer, the ESF leakage is assumed to be in the form of a liquid film on the floor, walls, and structural surfaces of the ESF area into which the leakage occurs.

Justification: In considering the effect of the boundary layer, it is conservative to maximize the area of the liquid surface which will in turn increase the mass of iodine transported across the boundary layer. Thus, while it is expected that the liquid will be pooled on the floor, this assumption provides significant conservatism.

Assumption 7: Transient effects will be neglected in calculating the iodine concentration in the bulk gas region.

Justification: In the boundary layer calculation, both the steady state and the transient equation for bulk gas region iodine concentration are solved and it is shown that equilibrium is reached quickly compared to the time intervals used to characterize temperature.

PSAT 3056CT.QA.04 Page 5 of II Rev.0 2 3 4 Assumption 8: The ESF leakage is assumed to pool on floor surfaces. 1I 11

PSAT 3056CT.QA.04 Page 6 of 11 Rev.0 El 2 3 4 it 11 References Reference 1: PSAT 3056CT.QA.02, "Implementing Procedure for Design Control for Calculation of Indian Point 3 Engineered Safety Feature Component Leakage Iodine Release," Rev. 0.

PSAT 3056CT.QA.04 Page 7of II Rev.0 N 2 3 4 Reference 2: NUREG-0800, NRC Standard Review Plan, Section 15.6.5, Appendix B, "Radiological Consequences of a Design Basis Loss of Coolant Accident:

Leakage from Engineered Safety Features Components Outside Containment".

Reference 3: PSAT 3056CT.QA.03,"Plant-Specific Design Input for Calculation of Indian Point 3 ESF Component Leakage Iodine Release", Rev. 0.

Reference 4: II 1]

Reference 5: D. Powers and S. Burson, "A Simplified Model of Aerosol Removal by Containment Sprays," NUREG/CR 5966, June, 1993.

Reference 6: L. Soffer et al., "Accident Source Terms for Light-Water Nuclear Power Plants," NUREG-1465, February, 1995.

Reference 7: R. Sher and J. Jokiniemi, "NAUAHYGROS 1.0: A Code for Calculating the Behavior of Aerosols in Nuclear Plant Containments Following a Severe Accident," EPRI Report TR-1 02775, July, 1993.

Reference 8: lo 1I Reference 9: I[ 1I Reference 10: "Calculation of Indian Point 2 Engineered Safety Feature Component Leakage Iodine Release with Boundary Layer Effect", PSAT 126.02CT.QA.05, Rev.1, Polestar Applied Technology, Inc.

Calculation The basic calculational approach is described in detail in Reference [10], and only requires minor numerical changes to account for differences in some of the input data for IP3, e.g. core inventory of iodine, assumed pH, peak average pool temperature, and geometry and flow data. [1 As previously noted, external recirculation begins at 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> after the initiating event. There are 7 rooms into which ESF leakage can occur and through which there is bulk gas flow. The geometry and flow characteristics of these rooms, taken from Reference [3], are shown in Table

1. They are the volume (V), the volumetric flow rate to environment (V), the total surface area (A, taken to be the sum of all wall surfaces, except for the ends since the ends are open, plus additional area equal to twice the floor area to conservatively account for equipment surfaces),

the length of the room (L), the cross-section area of the room (AQ), and flow velocity (U, derived from V divided by A,). The volumetric exchange rate with the environment, V / V, is (4000+4000+200+12000+7500+11700+4000) = 33400 cfm or 2.004E6 cfh divided by the total volume of 1.02857E5 ft3 = 19.48 vol/hr.

PSAT 3056CT.QA.04 Page 8 of II Rev.0 N 2 3 4

]1 Table 1. Geometry & flow characteristics of rooms into which ESF leakage can occur V V-dot3 V-dot/\ A L A, U Room (M3 ) (m /s) (hr') (M2 ) (m) (M2 ) (mis)

RHRRoom31 75 1.89 90.7 118 5.79 13 0.146 RHR Room 32 113 1.89 90.7 155 5.79 19.5 0.097 Sump Tank Room 55.6 0.094 6.1 77.7 3.35 16.6 0.0057 Hallway to Sump Room 34 5.66 599 85.5 6.1 5.57 1.016 Hallway Outside (inc. Sump Room) 617 3 . 5 4a 20.7 619 13 .7 b 45 0.079 Penetration Area 1389 1.89 4.9 858 13 .0 b 107 0.018 High Head Pump Room 629 5.52 31.6 689 17.9 35 0.157 a The supplied flow rate includes the flow in the RHR cells, which has been subtracted to obtain V-dot.

b Irregular shape; values chosen for length and width are arbitrary and preserve the floor area.

Table 2. 1[

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PSAT 3056CT.QA.04 Page 9 of 11 Rev.0[323 4 11

)) The details of the iodine release calculation are presented in Appendix B and the results are given in Table 3, which is the same as the table in Appendix B.

Table 3. Fraction of incoming iodine released to the environment during external recirculation beginning at 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br />.

Time period pH = 7.9 pH = 7.61 pH = 7.27 0 to 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> N/A N/A N/A 6.5 to 8 hours9.259259e-5 days <br />0.00222 hours <br />1.322751e-5 weeks <br />3.044e-6 months <br /> 0.0268 0.0881 0.2652 8 to 24 hours2.777778e-4 days <br />0.00667 hours <br />3.968254e-5 weeks <br />9.132e-6 months <br /> 0.0278 0.0921 0.2787 1 day to 4 days 0.0274 0.0907 0.2746 4 to 30 days 0.0265 0.0881 0.2697 The boundary layer DFs are calculated in Attachment 2, following the method used in ref.

[10]. The results are shown in Table 4 below and the minimum DF is 10.

Table 4. The boundary layer DF for all PAB rooms in question V V-dot V-dot/V A L A, U DF Room (M3 ) (M3/s) (hr') (m2 ) (m) (mi) (m/s)

RHRRoom31 75 1.89 90.7 118 5.79 13 0.146 1[

RHR Room 32 113 1.89 90.7 155 5.79 19.5 0.097 Sump Tank-Room 55.6 0.094 6.1 77.7 3.35 16.6 0.0057 Hallway to Sump Room 34 5.66 599 85.5 6.1 5.57 1.016 Hallway Outside 617 3 .5 4 a 20.7 619 13 . 7 b 45 0.079 Penetration Area 1389 1.89 4.9 858 13 . 0 b 107 0.018 High Head Pump Room 629 5.52 31.6 689 17.9 35 0.157

PSAT 3056CT.QA.04 Page lOof 11 Rev.0 N 2 3 4 Results In Table 5 below, the fractional releases with boundary layer effects are shown. For pH =

7.27 and 7.61, a DF of IO was applied to the releases shown in Table 3 above. For pH = 7.9, the result for DF = 1 is shown. The results are in units of fraction of the incoming iodine in the ESF leakage into the ESF component room(s) (i.e., fraction of the total dissolved iodine in the 4 gph leakage water). In the first 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> (23400 seconds) the release is zero since recirculation has not yet started. The fractional releases depend strongly on the pH, and for the same DF, differ by an order of magnitude between pH = 7.9 and pH = 7.27. [1 HI For pH = 7.27, the average release rate (with DF = 10) is about 2.7% of the total incoming rate of iodine in the leakage water. For pH = 7.61, the average release rate (with DF = 10) is about 0.9%, while for pH = 7.9, the limiting value with DF = 1 is about 2.7%.

It is noted that the results reported here are quite conservative with respect to several effects in addition to the cooling due to heat transfer, as has been discussed previously: II 11 Table 5. Fraction of incoming iodine released to the environment during external recirculation beginning at 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> with boundary layer effects.

Time period pH = 7.9 pH = 7.61 pH = 7.27 DF=1 DF=10 DF=10 0 to 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> N/A N/A N/A 6.5 to 8 hours9.259259e-5 days <br />0.00222 hours <br />1.322751e-5 weeks <br />3.044e-6 months <br /> 0.0268 0.00881 0.02652 8 to 24 hours2.777778e-4 days <br />0.00667 hours <br />3.968254e-5 weeks <br />9.132e-6 months <br /> 0.0278 0.00921 0.02787 1 day to 4 days 0.0274 0.00907 0.02746 4 to 30 days 0.0265 0.00881 0.02697

PSAT 3056CT.QA.04 Page 11 of 11 Rev.0 [ 2 3 4 Conclusions The conclusions from this calculation are as follows:

• The iodine release from ESF leakage for IP3 is very sensitive to the assumed pH and the boundary layer DF. The fractional release for pH = 7.27 is approximately 2.7%. For pH =

7.61 it is about 0.9%, while forpH = 7.9 it is about 2.7% (DF = 1). I

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PSAT 3056CT.QA.04 Page 1 of 13 Revision 1 Appendix B. Calculation of 1P3 ESF Component Leakage Iodine Release Definition of variables lion concentration of Iodine ion in liquid pool 12L concentration of elemental iodine in liquid pool 12C concentration of elemental iodine in gas volume Nil total mole of Iodine in the system in question pH := 7.S pH valule kgmolc:= lOO1moic Kilogram mole definition mpl := Male mole/litter, the same as kgmolelm3 liter

,BpH) := I10 p~l rp. H+ concentration as a function of pH d := 6.051I lO rp constants used in iodine conversion equation ha:= L.47 10 9 mw 1 := 130 kg.kgmolc I Molecular weight of Iodine pS(T) := 83.12 kg Water density 3

0 .262 II+[ l-T (647.29-K)- ]023072] m Eq. Ia is rewritten as: 2 fd(pH) : = H 12L= lion .fi(pH)

PCXT):= 1 O(6.29-0149-T-K- I) Partition coefficient as a function of temperature (K).

Eq. 2 is rewritten as:

12G= I2L PC(T) = lion .f`(pH)-P0(T)

Other variables and given conditions are mH2 0 := 3.1-16 -lb total mass of water in containment sump (Item 3.1, Ref. [1])

Itot := 26.121.kg total core inventory of iodine f:=0.4 iodine release fraction from core to containment, Ref. [2]

PSAT 3056CT.QA.04 Page 2 of 13 Revision 1 co(T) ItOt.jT lion concentration in cont. sump co(396K) = 5.31718x 10 5mpl mxI.mH2 0 gph := gal hr l Volumetric flow rate unit - gallon per hour AV := 4*gph Leak rate of cont. sump water into the pool, Item 4.1, Ref. [1]

tL:= 23400sec Time to initiate the ESF leakage, Item 4.3, Ref. [1]

VL(t):= W.(t-tL) liquid pool volume due ESF leakage 0m 3 if t tL VG- 2913m3 Total gas volume after 6.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> (sum of volumes from Item 5.2 VG: 9m through 5.4, Ref. [1] )

A:= 19.4Shr Exchange rate between gas volume & environment > 6.5 hrs (ratio of the sum of flow rates from Item 5.2 through 5.4, Ref. [1]

to the sum of volumes VG)

Derivation of iodine release to environment 1]

PSAT 3056CT.QA.04 Page 3 of 13 Revision 1

[1 1]

PSAT 3056CT.QA.04 Page 4 of 13 Revision 1 11 From 23400 to 28800 sec

= 23400sec Tf:= 28800sec x:= O-mpl Frac:= RF(Ti,rfEX,T.pHVGX)

FracT = (2.88x 104 5.553035x 10j 7.9 315.2 0.026789) RFCH Frac 4

PSAT 3056CT.QA.04 Page 5 of 13 Revision 1 From 28800 to 30000 sec

-ri:= 28800sec rf:= 30000scc x:= Frac1 mp T:=315.7K Frac:= RF(rqjf, f,X,T,pH,VG,x) E := 0.00f FracT =(3x 10 5.552959x 105 7.9 315.7 0.027501) RFCH = Frac 4 From 30000 to 40000 sec

-y=30000sec tf:=40000scc x:= FracI'rp T:=315.9K Frac:= RF(TiifcX,TPH,VGx)

FracT =(4x 10 5.551489x 10 7.9 3]5.9 0.027683) RFCH = Frac4 From 40000 to 50000 sec

-ri:= 40000scc cf:= SOOOOsec x:= Frac ImP: T:=3161K Frac := RF(T ,Tf,, X,TpHVGX)

FracT = (x 10~ 5.550767x 10 7.9 316 0.02777Z) RFCH := Frac 4 From 50000 to 60000 sec r:= 50000 sec tf:= 60000 sec x:= FracI mp T:=316K Frac:= RF(rj,rf,c,XT,pH,VGx)

FracT = (6 x 10 5.550446x 10 7.9 316 0.027768) RFCH := Frac4 From 60000 to 70000 sec

.ri:=60000scc f := 70000 sec x:= F rac 1-mp T:=316.2K Frac := RF(Ti.fAcXTpHVG0 X)

FracT 5

=(7x 10 5.549839x IC 7.9 316.2 0.027959) RFCH = Frac 4 From 70000 to 80000 sec

-ri:=70000scc ¶f:= 80000sec x:= F racI mp. T:=316 K Frac := RF(rirfc. ,T,pH,VGx)

FracT =(8 x 10 5.549803x 107 7.9 316 0.02776) RFCH := Frac4 From 80000 to 86400 sec

PSAT 3056CT.QA.04 Page 6 of 13 Revision 1

= 80000 sec Tf:= 86400scc x:= Frac I*mp T := 316K Frac := RF(Ti,¶f,E,XT,pH,VG,x)

FracT = (8.64x 10 4 5.549809x 10 7.9 316 0.027761) RFCH := Frac4 From 86400 to 90000 sec Tj:= 86400sec Tf:= 9 00 00 sec x:= Fracl-mp T:= 315.9 K Frac := RF(;, Tf,4 .X, T,pHVG X)

FracT = (9x 104 5.549832x 10 7.9 315.9 0.027661) RFCH = Frac4 From 90000 to 100000 sec Tj:= 90000sec Tf:= lOOOOOseC x:= Frac ImP: T := 315.9 K Frac := RF(¶irfEX,T,pH,VGx)

FracT = (Ix 10 5.54993 6x10 7.9 315.9 0.027664) RFCH := Frac4 From 100000 to 172800 sec E := o.of Tj:= lOOOOOscc tf:= 172800sec x:= Frac1 .mP- T :=316.3K Frac := RF(TrIf,c,,T,pH,VG,x)

FracT =(1.728x 10 5.54S301x 10 5 7.9 316.3 0.028044) RFCH  := Frac 4 From 172800 to 200000 sec Tj:= 172800sec Tf:= 2 000 00 sec x:= Fracl-mp T:= 315.9K Frac:= RF(Ti,¶f,cX,TpH,VGx)

FracT = (2 x 10 5.548714x 106 7.9 315.9 0.027651) RFCH = Frac 4 From 200000 to 300000 sec 3 000 00 T; := 200000sec Tf:= scc x:= Frai c -mp. T:= 315.2K Frac:= RF(rj,TfC,).,TpH,VG,X)

FracT = (3x 10 5.551902x 10C 7.9 315.2 0.026998) RFCH = Frac4 From 300000 to 345600 sec T;:= 300000sec Tf:= 345600sec x:= Fra cI' mp. T := 315 K Frac := RF(Ti,rf,E,X,T,pH,VG,x)

FracT = (3.456x 10 5.552916x 10 7.9 315 0.026827) RFCH = Frac4

PSAT 3056CT.QA.04 Page 7 of 13 Revision 1 From 345600 to 400000 sec Ti:= 345600sec -rf:=400000sec x:= F rac1 -mp. T:=314.S}K Frac := RF(TjtfC,XT,pHVGX)

FracT=(4x 10 5.554165x 10C 7.9 314.8 0.026652) RFCH  := Frac4 144 From 400000 to 500000 sec T;:= 400000sec xf:= 500000scc x:= F rac1.mp. T:=314.7K Frac:= RF(i, TfEX,T,pHVG,x)

Frac T=(5x 10 5.555847x 10 7.9 314.7 0.026572) RFCH = Frac 4 From 500000 to 2590000 sec £:= 0D.'

T;i:= 500000see rf:= 2590000scc x:= Frac I-mp T := 314.5 K Frac:= RF(r,.rf,EXT,pH,VGx)

FracT = (2.59x I1O 5.562135x 107 7.9 314.5 0.026437) RFCH = Frac4 Assuming that pH is 7.61, the release fraction of iodine atoms are:

c:= 2 10 3  := 19.4Shr pH := 7.61 T:=315.2K From 23400 to 28800 sec s;i:= 23400sec 'r := 28800sec x:= 0OmpJ Frac:= RF(jrfCs,X,TpHVGx)

FracT =(2.88x 1 5.190837x 10 5 7.61 315.2 0.08806) RFCM := Frac4 From 28800 to 30000 sec

-;i:= 28800sec Tf:= 30000sec x:= Frac 1 Mmp T:=315.7K Frac := RF(Ti,Tf,c,X,TpHVGX)

FracT =(3x 10 5.190364x 10 5 7.61 315.7 0.091329) RFCM I= Frac4 From 30000 to 40000 sec E := 0.00' Ti:=30000scc tf:= 40000scc x:= Frac C.mp T:=315.9 K Frac := RF(j.tic, XT.pHVGX)

FracT =(4x 10 5.186616x 10 7.61 315.9 0.09187) RFCM := Frac4

PSAT 3056CT.QA.04 Page 8 of 13 Revision 1 From 40000 to 50000 sec Tj := 40000sec rf:= 50000sec x:= Frac .m p T :=316K Frac := RF(ri1 ,fcXTpHVGx)

FracT =(5x 0 5.184997x 10 7.61 316 0.092117) RFCM := Frac4 From 50000 to 60000 sec 50000sec SQ:= rf:= 60000sec x:= Fracl-m p T := 316K Frac  := RF(,tf, Ef,%,T,pH,VG,X)

FracT = (6x IO~ 5.184319x 10 7.61 316 0.09208I) RFCM := Frac4 From 60000 to 70000 sec Tji:= 60000scc rf:= 70000sec x:= FrnDc1.mp T:= 316.2K Frac := RF(T, Tfc,A,T,pH,VG,X)

FracT = (7 x I0 5.183004x 107 7.61 316.2 0.092694) RFCM := Frac4 From 70000 to 80000 sec tr:= 70000scc tf:= 80000scc x:= Fr;DacI'mp. T := 316K Frac := RF(r, Tf,E,XT,pHVGx)

FracT = (8 x 10 5.lS2975x 10 7.61 316 0.092023) RFCM := Frac4 From 80000 to 86400 sec

-ri:= SOOOOsec rf:= 86400sec x:= Fraic1-mp' T := 316K Frac := RF(tijrf.EXTpHVGx)

FracT = (8.64x 104 5.1829S2x 10 7.61 316 0.092027) RFCM := Frac4 From 86400 to 90000 sec ri:= S6400scc rf:= 90000scc x:= Fraic 1.mp' T:= 315.9K Frac:= RFtrj, ,e X,T,pH, VG,x)

FracT = (9 x IO 5.183075x 10 7.61 315.9 0.091697) RFCM 8:= Frac4 From 90000 to 100000 sec

PSAT 3056CT.QA.04 Page 9 of 13 Revision 1 T:= 90000sCc rf:= lOOOOOSCC x:= Fr rac 1 mp' T :=315.9K Frac := RF(T, f, E,X,T,pH,VG,x)

FracT = (I x 10 5.183332x 10 7.61 315.9 0.091709) RFCM = Frac 4 From 100000 to 172800 sec E := O.(

0 T:=316.3K Tr:= 100000scc Tf:= 17280 sec x:=Fr;rac 1 mp' Frac:= RF(Ti.fs,XT,pHVGx)

FracT = (1.728x 105 5.179914x 105 7.61 316.3 0.0929'31) RFCM  := Frac4 From 172800 to 200000 sec Tj:= 172800scc Tf:= 2 00000sec x:=Fr.rac 1 .mP T:= 315.9K Frac:= RF(i,rfC,X,T,pHVG,x)

FracT=(2x 1O 5.180831x 10 7.61 315.9 0.091613) RFCM  := Frac4 From 200000 to 300000 sec Ti:= 200000seccf:= 300000sec x:= Frac rI mp. T:=315.2K Frac:= RF(¶irf,c,X,T,pHVGx)

FracT = (3 x 105 5.18788x 10 57.61 315.2 0.08954) RFCM 12= Frac4 From 300000 to 345600 sec Tj:= 300000sec Tf := 345600sec x:= Frac, I-mp. T :=315K Frac := RF(T, fIsX,T,pH,VGx)

FracT =(3.456x 10 5.190106x 10 7.61 315 0.089053) RFCM  := Frac 4 From 345600 to 400000 sec Tj:= 345600scc -rf:= 40 0 00 0 scc x:= Fr ac ImP. T:= 314.8K Frac := RF(rirfc,X,T,pH,VGx)

FracT = (4 x 10 5.192704x 10 7.61 314.8 0.08852) RFCM = Frac4 From 400000 to 500000 sec Ti:= 400000sec If:= SOOOOOsec x:= Fr ac1 .mp. T:=314.7K Frac := RqFr,¶f, CEX,T, pH, VG, x)

RFCM = Frac4 FracT=(5x 105 5.196211x 10 7.61 314.7 0.088313)

PSAT 3056CT.QA.04 Page 10 of 13 Revision 1 From 500000 to 2590000 sec C:= U

-ri:= 500000scc -r := 2590000scc x:= Frac 1-mp! T :=314.5K Frac:= RqTj, rf,E,)X,T,pH,VGx)

FracT = (2.59x 10 5.208562x 10o 7.61 314.5 0.088067) RFCM = Frac4 Assuming that pH is 7.27, the release fraction of iodine atoms are:

c:= 2 10 1:= 19.4Shr pH := 7.2 From 23400 to 28800 sec T; := 23400 sec Tf:= 28800scc x:= O-mp! T := 315.2K Frac := RF(Tj,Tf,s,X,TpHVG,x)

FracT= (2.88x 10~ 4.143422x 106 7.27 315.2 0.26516Z) RFCb = -Frac 4 From 28800 to 30000 sec T;:= 28SOOscc Tf:= 30000sec x:= Frac I -mp T:=315.7K Frac:= RF(Trf,X,lT,pH,VG,x)

FracT = (3 x I10 4.139628x 10 7.27 315.7 0.278069) RFCLI= Frac4 From 30000 to 40000 sec x := 0.00'

-ri:= 30000scc Tf:= 40000sec x:= Frac I mP' T:=315.9 K Frac := RF(tjrfs,X,T,pHpVGX)

FracT=(4x 10 4.12693x 10 7.27 315.9 0.278577) RFC2 := Frac4 From 40000 to 50000 sec T;:= 4 0000scc rf:= 50000 scc x:= Frac 1 .mpf T:=3161K Frac:= RF(i,Tf,EXT,pH,VG, x)

FracT = (5 x 10 4.12324Sx 10 7.27 316 0.278745) RFCL3 := Frac4 From 50000 to 60000 sec T;:= 50000 sec rf:= 60000sec x:= FracI.mp T:= 316 K Frac := RF(T, f, , XT,pH,V G ,x)

PSAT 3056CT.QA.04 Page 11 of 13 Revision 1 FracT =(6x 104 4.122018x 10 7.27 316 0.278457) RFCL = Frac4 From 60000 to 70000 sec T;:= 60000sec rf:= 70000scec x:= Frac ImP. T := 316.2K Frac:= RF(ri,TfE,X,TpH,VG,x)

FracT =(7x I10 4.11913x 10 7.27 316.2 0.280131) RFCL - Frac 5 4 From 70000 to 80000 sec Tr;:= 70000scc Tf := 80000sec x:= Frac1ImP T:= 316K Frac := RF(-rixfc,X,T,pH,VG,x)

FracT = (8 x 10 4.119493x 10 7.27 316 0.278025) RFCL - Frac 6 F 4 From 80000 to 86400 sec T;:= 80000sCc Tf:= 86400sec x:= FraiI lMP' T:=316-K Frac:= RF(jTf, , XTpHVG,x)

FracT = (8.64x 1O 4.119662x 10 7.27 316 0.278071) RFCL = Frac 4 From 86400 to 90000 sec T;i:= 86400sec ¶f:= 9 00 0 0 sec x:= FraicI-mp' T:=315.9K Frac:= RF(Trsf, ,X,T,pH, VG, x)

FracT = (9 x I0 4.120024x 10 7.27 315.9 0.277104) RFCG:= Frac4 From 90000 to 100000 sec Tji:= 90000sec Tf:= lOOOOOsec x := Frac .mf' T:=315.91K Frac:= RF(,riTf,E,XTpHVG 0 x)

FracT=(lx 10 4.120831x 10 5 7.27 315.9 0.277203) RFC, := Frac4 From 100000 to 172800 sec E := 0.0o Tj:= lOOOOOsec Tf:= 172800sec x:= FracI'mP- T:=316.3K Frac:= RF(¶jpEf,)X,TPH,VGX)

FracT =(1.728x 10 4.11355x 10 7.27 316.3 0.280518) RFCL  := Frac4 From 172800 to 200000 sec

PSAT 3056CT.QA.04 Page 12 of 13 Revision 1 rj:= 172800scc ¶f:= 2 0 0 000 sec x:=FracI-mp- T:=315.9K Frac := RF(Ti,Tf,C,X,TpHVGX)

FracT =(2x 10 4.116181x 10 7.27 315.9 0.276476) RFCL = Frac 4 From 200000 to 300000 sec

,r := 200000sec -r:= 300000scc x:= Frac 1 .mp T:=315.21K Frac := RF(Ti,Trf,X,T,pHVG,x)

FracT =(3x 10 4.132961x 10 7.27 315.2 0.271233) RFCL  := Frac4 From 300000 to 345600 sec

= 300000scc Tf

= 345600sec x:= Frac 1*mp T:=315K Frac := RF(Ti~rf,C.,ATPHVGX)

FracT = (3.456x 10 4.137765x 10 7.27 315 0.270537) RFCL  := Frac 4 From 345600 to 400000 sec

,;i:= 345600sec rf := 400000scc x := Frai Iclmp' T:= 314.8K Frac:= RF(rtrfc.X,T,pHVGx)

FracT = (4 x 10 4.143152x 10 7.27 314.8 0.269334) RFCL  := Frac4 From 400000 to 500000 sec Ti:= 400000sec rf:= 500000seC x := Frai c' mp T:=314.7K Frac := RF.ri,.rf,EXT,pH,VGX)

FracT =(5x 10 4.150024x 10 7.27 314.7 0.269196) RFCL  := Frac4 154 From 500000 to 2590000 sec E:= 0.5 T 500000scc O:= T := 2590000sec x := Frac *mp T:=314.5K Frac := RF(TrTfE.XTPHVGX)

FracT = (2.59x 106 4.169013x 10o 7.27 314.5 0.269715) RFCL  := Frac4 Iodine release fractions during X/Q time Intervals:

i:= 2..9 ti-2 := i 1000( t0 := 2880( t7 := 8640(

i := 9 . 11 t;- I := i-1I000( tl0 := 172801 i:= 12.. 14 t; 1 := (i - Io) 10000(

PSAT 3056CT.QA.0OI Page 13 of 13 Revision 1 ti+2 = (i - 8)- I0000( t13 := 34560( t16 := 2.59 106 Avc(i 0 ,ij,RFC,t):= I dummy - 0 for j E io.. il dummy *- dummy + RFC.(tj -t dummy t; -tI From 6.5 to 8 hour:

RFCjH = 0.0268 RFCN4 = 0.0881 RFCL = 0.26516 0

From 8 to 24 hour:

Avc(1,7,RFCy,t) = 0.02778 Avc(1,7,RFQM,t) = 0.09213 Ave(1,7,RFCLt) = 0.27869 From 1 to 4 days:

Ave(8,13,RF%,t) = 0.02737 Ave(8, i3,RFCMft) = 0.09074 Avc(8,13,RFCLt) = 0.27458 From 4 to 30 days:

Avc(14,16,RFCHt) = 0.02645 Ave(14,16,RFCr1 ,t) = 0.08809 Ave( 14,16, RFCL, t) = 0.26968 Summary of the results for iodine release fractions Time Interval (sec) pH = 7.9 pH = 7.61 pH = 7.27 23400 to 28800 0.0268 0.0881 0.2652 28800 to 86400 0.0278 0.0921 0.2787 86400 to 345600 0.0274 0.0907 0.2746 345600 to 2590000 0.0265 0.0881 0.2697

References:

1. PSAT 3056CT.QA.03, "Plant Specific Design Input for Calculation of Indian Point 3 ESF Component Leakage Iodine Release" Rev. 0.

2. L. Soffer et al., "Accident Source Terms for Light-Water Nuclear Power Plants, "

NUREG-1465, February, 1995.

PSAT 3056CT.QA.04 Page 1 of 2 Rev. 1

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PSAT 3056CT.QA.04 Page2af2 Attachment I Rev. 1

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11

PSAT 3056CT.QA.04 POLESTAR PROPRIETARY ((Trade Secrets Brackets)) Page 1 of 4 Rev. 1 Attachment 2 XL Spreadsheet Calculation of Iodine Concentration In Bulk Gas with BL Effect, Hallway Outside It 11

PSAT 3056CT.QA.04 Page 2 of 4 Rev. 1 Attachment 2 XL Spreadsheet Calculation of Iodine Hallway Penetratfo Pump Pump Sump Tank Hallway to High Head Concentration In Bulk Gas wf BL Effects Outside n Area Room 31 Room 32 Room Sump Room Pump Room Exchange flow (dim) 7500 4000 4000 4000 200 12000 11700 Case-Specific Parameter Inputs Volume I Exchg rate betw. vol 1 and 2 (m31s) 0 0 0 0 0 0 0 Totalroomvolume,V(m3) 617.4 1389.0 75.1 113.0 55.6 34.0 628.8 Exchg rale with env. (volhu) 20.6 4.9 90.5 60.2 6.1 600.0 31.6 Vol. flowto env.,V-dol (m3/s) 3.5 1.9 1.9 1.9 0.1 5.7 5.5 Cross sectional area, Ac(m2) 45.0 107.2 13.0 19.5 16.6 5.6 35.1 Temperature, T (K) 370.9 370.9 370.9 370.9 370.9 370.9 370.9 Pressure (alm) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Length scale, L (m) 13.7 13.0 5.8 5.8 3.4 6.1 17.9 AreaA(m2) 618.7 857.6 118.3 155.3 77.7 85.5 688.6 fl

PSAT 3056CT.QA.04 Page 3 of 4 Rev. I XL Spreadsheet Calculation of Iodine Concentration In Bulk Gas with BL Effect

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PSAT 3056CT.QA.04 Page 4 of 4 Rev. 1 XL Spreadsheet Calculation of Iodine Concentration In Bulk Gas with BL Effect Calculation of elemental iodine concentration at liquid edae of gas bounday laver 1[

ATTACHMENT 5 TO NL-04-162 COMMITMENTS ENTERGY NUCLEAR OPERATIONS, INC INDIAN POINT NUCLEAR GENERATING UNIT 3 DOCKET 50-286

NL-04-1 62 Docket 50-286 Attachment 5 Page 1 of 1 COMMITMENTS No. Commitment Date 1 Relocate the FSB and Containment Purge Testing requirements to the April15, FSAR and indicate they must be maintained to comply with 10 CFR 50, 2005 Appendix I.

2 The Technical Analysis in Attachment 1 said uA penalty has been Post Test applied to the Containment Spray System (CSS) flow rate to provide margin for future results from CSS pump testing." The analysis will be amended and submitted to the NRC staff for review and approval if it is determined that the test results of the containment spray system flow rate did not provide adequate margin.