ML19317H382

Analysis of Hypothetical Accidents,Technical Evaluation Rept,Submerged Demineralizer Sys, App 1

ML19317H382
Person / Time
Site:	Crane
Issue date:	05/23/1980
From:	Metropolitan Edison Co
To:
Shared Package
ML19317H381	List: ML19317H380 ML19317H382
References
NUDOCS 8006030327
Download: ML19317H382 (15)

Text

_. -

..(}.

~

b.

4.,.

l APPENDIX I r

i t-l AN LYSIS OF HYPOTHETICAL ACCIDENTS 4

1 i

I I

L i

i i.

i 1

t TECHNICAL EVALUATION REPORT SUBMERGEDDEMINERALIZERSiSTEM i

MAY 23, 1980 4

h i

h-t aeoso'se321 }p

r Inadvertent pumping of containment water into the fuel storage pool.

Assumptions:

1)

The effluent line from the final filter develops a leak which is not detected immediately.

Contaminated water is released into the pool at a rate of 30 gpm for a period of 15 minutes (450 gallons) at concentrations present in the containment sump.

2)

It is assumed that the toal activity of concern is made up of the cesium isotopes 134 and 137.

The reported concentration in the contaminated sump is 40 uCi/ml of Cs-134 and 176 pCi/ml of Cs-137.

3)

Assume the volume of water in the fuel storage pool is 233,000 gallons.

Calculate the expected exposure rate for an individual on the walkway at a point 6 feed above the surface of the water.

Concentration of isotopes in the fuel pool:

50 gal. = 1.928 x 10-3 Dilution Factor 23 Cs-134 40 pCi/ml x 1.928 x 10-3 = 0.077 pCi/ml Cs-137 176 pCi/ml x 1.928 x 10-3 = 0.34 pCi/ml The total activity in the 233,000 gallon fuel pool is:

3 5

Cs-134 0.077 pCi/ml x 3.79 x 10 ml/ gal x 2.33 x 10 gal =68 Ci 3

5 Cs-137 0.34 pCi/ml x 3.79 x 10.ml/ gal x 2.33 x 10 gal =300 Ci Approximation of the dose rate at a point "P" which is a distance "a" above the surface of an infinite slab of thick-ness "h" can be calculated from the following equation (Goldstein).

AI-l 4

.,?

2 (b ) - E2 (b )

Sv E

. 4p = 5 i

3 3

Where cp = flux at point P Sy = volume source strength u = macroscopic cross section of source s

b=by+uh 3

s by=pa a

y, = macroscopic cross section of air Evaluate pa and u assuming a 0.66 MeV photon for both isotopes 3

Mass attenuation coefficient 2

Water = 0.090 cm fg 2

Air

= 0.080 cm /g Density 3

Water = 1.0g/cm Air

= 1.293 x 10-3 g/cm3 2

3 p = (0.090 cm /g)(1.0 g/cm ) = 0.09 cm-I 3

= (0.080)(1.293 x 10-3) = 1.0 x 10-4

-1 p

cm a

l Calculate b and b3 (same for both isotopes) y by = pa, a = 6 feet = 133 cr a

~

-1)(183 cm) y =.(1.0 x 10-4 b

cm

-2 by = 1.83 x 10

~

3 b3 = 9 h + b, h = 38 feet = 1.15 x 10 3

y cm b3 = (0.09 cm-1)(1.16 x 103 cm) + 1.83 x 10-2 b3 = 104 AI-2 w.

Sv

• p " T2T(0.09 cm-1) b (1.83 x 10~ ) + E 2

2 (104)

From Goldstein's Fundlmentals of Shielding

.E for 1.83 x 10-2 i_ 0.913 2

2 for 104 is less than 3.83 x 10-6 E

Sv

&p = 0.18 (0.913) cm 4 = 5.07 Sv cm p

Calculate Sv, volume source strencth For Cs-137:

0 Sv = 0.34 pCi/mi = 1.25 4 10 cis/:

.s Prccons 0.652 MeV,85, E eff.= 0.56 MeV/ dis 4

3 Sv = (1.26 x 10 dis /cm.s)(0.55 Ft!/ dis) 3 3

Sv = 7.06 x 10 MeV/cm.s For Cs-134:

3 dis /c 3.s Sv ='0.077 pCi/ml = 2.85 x 10 Photons 0.570 MeV 23f 0.610 MeV 98f.

0.796 MeV 99?:

1.370 MeV

3. 4 ',

1.040 MeV

1. 0 '.

1.170 MeV

1. 9 '.

E eff = 1.60 MeV/ dis 3-dis /cm s)(1.60 Me'./ dis) 3 Sv = (2.85 x 10 3

3 Sv = 4.56 x 10 MeV/cm - s Total Sv-3 3

3 Sv = (7'06 x 10

- 4.'55 x 10 ) MeV/cm. s 3

Sv = 1.16 x 10 MeV/cm.s AI-3

4

' 4.=--(5.07 cm)(1.1'6 x 10 MeV/ce.3 s) p 4

2 4 = 5.88 x 10 'MeV/cm -s-P

- Calcu%ce the exposure rate given the flux-at point P.

The mass-absorption coefficient for air for 600 KaV photons is 0.C3 2

cm /g (Morgan and Turner, p.108).

Exposure Rate (ER) at point P.

4 2

o ER.= (5.SS x 10 MeV/cm. W C. 3 c /c)(1.6 x 10 ~c er s/"ed S7.7 er;s/g R

-5

-ER = 3.22 x 10 R/s ER =-116 mR/h NOTE:

This exposure rate - GR). is lower than the value of 430 mr/hr reported in the TER.

The lower value results from a realistic approximation of activi:y levels, decayed to October.1, 1980, the anticipated date for initiation of SDS operation.

s J

N l

AI-A -

f r

y e-n r

-,q

=y-+

r*--*

Off-Site Direct Dose from Inadvertent Pumping of Containment Water into the Fuel Storage Pool.

Assumptions:

1) The isotopes of concern are Cs-134 and 137.

2) The concentrations of Cs-134 is 0.077 pCi/mi and Cs-137 is 0.34 Ci/r.l.

3) The distance to the closest off-site point is approximately 200 me ers.

4) The fuel pool is a point source for exposure estimates at a distance of 200 meters.

5) There 'is no source self-absorption.

6)

The fuel pool wall is approxima al;.1.5 meters thick of reir.f:rced concrete.

7) The walls of the fuel handling building are approximately 1.5 r.e ers thick of reinforced concrete.

Calculate the exposure 200 meters from the source, assuming no shieldir.;.

The gamma exposure factors for the cesium isotopes are (Radiological Fealth Handbock, p. 131):

Cs-134 0.87.R/hr Ci at 1 meter Cs-137 0.33 R/hr Ci at 1 meter The exposure at 200 meters from both isotopes is:

(0.87 R/hr Ci x 68 Ci) + (0.33 R/hr. Ci x 300 Cil = 4 x l'~

Jnr (200)9-This exposure will be greatly reduced by the two separate 1.5 meter tnick shields of reinforced concrete. Tne standard gamma absorption equaticn for wide-beam radiation will be utilized to estimate this reducticn (.a:icic;ical Health Handbook, p. 30).

AI-5

I = Blo -UX e

Where lo = original radiation exposure rate I = attenuated radiation exposure rate p = linear absorption coefficient X = absorber thickness B = " buildup" factor 2

p/p = 0.03 cm /g (Morgan & Turner, p. 108) 3 o = 2.3 g/cm x = 3.0 m.

Io = 4'x 10-3 R/hr 2

3 Find p: p/p- = 0.03 cm /g, p = 2.3 g/cm 2

3 p =-0.03 cm /g x 2.3 g/cm u = 0.07 cm-1 The dose build-up factor (B) is found on page 145 of the Radiological Health Handbook.

The number of relaxation lengths is:

= 0.07 cm-1 x 300 cm pX UX = 20.7 The shielding properties of. concrete are similar to those of aluminum.

From the tables, a 0.5 MeV photon'and 20_ relaxation lengths will yield a dose build-up factor of 141. This calculation will assume a build-up factor of 150.

I = Blo ~DX e

I = (150)(4 x 10-3 R/hr) e-0.07 x 300 I = 4.5 x 10-10 R/hr The direct dose to an individual at the site boundary from the inadvertent pumpingoi0 containment water into the fuel storage pool will be approximately 4.5 x 10 R/hr.

4 AI-6

,4 airknrn 'nff site Relea'ses from inadvertent Pumoino of Contain-ant Mater into the fuel Storace Pool Assumptions:

1) Any activity spilled into the pool will be evenly distributed irrediately.

2) Tha " Pool Clean-up Leakage Containment Ion-Exchanger System" will remove activity from the fuel pool.

This system will treat the pool water at the rate of 100 GFM. This flow rate results in a turnover rate of once per 39 hour4.513889e-4 days <br />0.0108 hours <br />6.448413e-5 weeks <br />1.48395e-5 months <br />. period.

3) The inventory of isotopes available to become airborne will not decrease for one week. At the-end of the one week period, seven purification half lives have occurred and no activity remains to become airborne.

5 4)

The volume of the fuel pool is 2.33 x 10 gallons.

This volume turns cver to the s'urface as tne rate of tha "Ico-Exchanger System" (100 GPil is available for suspension).

5) The entrainment factor for activity in water to be suspended in air is 10-6,

6) The air frca the fuel pool building will be filtered by 2 banks of HEPA filters resulting in a DF of 10*.

Calculate the amount of activity for one isotope that will be released in air during the one week pericd required to clean the pool water.

State it as a fraction of activity released during normal SDS operation.

The activity in the fuel pool for Cs-137 from this incident has been calculated to be 0.34 uti/r.l.

Calculate the activity of Cs-137 released off-site in one week.

100 cal / min. x 3785 ml/ cal x 0.3 eCi/ml x 10-6 (entr. facti = 1.3 x 10-3 102 (DF)

1. 3 x 10-5 Ci/ min. x 60 min./hr. x 16S hr./wk. = 13.0 uCi/wk.

It was calculated that 347 uti of Cs-137 will be released frca the normal operation of the SDS.

This incident could add 13.0 2Ci X 100 = 3.75;

. v. ]

.C i The amount of potential airborne activity released off-site from this incident is 3.755 of the amount projected from th= normal operation of the SDS for a year.

The fraction 3.75:1 will hold for the other isotopes as well.

AI-7

Pipe Rupture on Filter Inlet Line (above water level)

Assumptions:

1) Contaminated water sprays into the air from around the 1 cad shielding. _ Approximately 675 gallons _ of water is released into the pool and 75 gallons spreads onto a surface area of 200 ft.2 2)

It is assumed that the total activity of concern is made up of the cesium isotopes 134 and 137. The reported concentrations in the contaminated effluent is 40 pCi/ml of_Cs-134 and 176 pCi/ml of Cs-137.

Calculate the exposure rate on the walkway six feet above the pool and the exposure rate three feet above the center of the cortaminated area.

Calculation of the exposure rate on the walkway can be done by ratioing the volume of contaminated water released into the pool in the previous accident to.the amount of water released in this accident.

Application of this ratio to the exposure rate cciculated in the previous accident will yield the expected exposure rate from this accident.

Ratio = 675 callons = 1.50 450 gallons Exposure Rate = (116 mR/h)(1.50)

Exposure Rate = 174 mR/hr Calculation of the exposure rate three feet above the centar of the i

.contam nated area can best be approximated by assuming the area is a circle with an area of 200 ft.

Area = 200 ft.2 = 1.86 x 105 2

cm Radius = 243.3 cm Approximation of the exposure rate at a point "P" which is a distance "a" above the surface of a circular disk source of radiation can be calculated from the following equation (Goldstein).

4 = Sa E1 (va) - E1(u R2+a2l 2

Where & = flux at point P Sa=areasource-strength (Ci/cN) u = linear attenuation coefficient (cm_1) a = distance from source (cm)

R = Radius of source 4

AI-8

For this calculation R.=.243.3 cm

a = 91.4 cm p = 1.05 x'10~4

-1 cm, for 660 kev photon in air Calculate Sa for both isotopes Fo'r Cs-137:

.Sa = (75 gals)(3785 ml/ gal)(176 uCi/ml) 1.86 x.100 cm4-6 2

Sa = 268.5 uCi/cm = 9.9 3 x 10 dis /cm, 3 E eff = 0.56 MeV/ dis-0 2

Sa = (9.93 x 10 dis /cm s)(0.56 MeV/ dis) 6 2

Sa = 5.55 x 10 MeV/cm. s For Cs-134:

Sa = (76 gals)(3785 ml/ gal)(40 uCi/ml) 0 2

1.86 x 10 cm Sa = 60.9 pCi/cm2 = 22.59 x 105 2

dis /cm s

E eff = 1.6 MeV/ dis 5

2 Sa = (22.59 x 10 dis /cm,. s)(1.6 MeV/ dis) 2 Sa ='3.6'x 106 MeV/cm, 3 Total Sa 6 + 3.6 x 10 ) MeV/cm s

6 2

Sa = (5.5E x ?9 6

2 Sa = 9.15 x 10 MeV/cm. s 9

f

_d

-AI-9

e-f.

Calculate E I I) l E (pa) = E -(1.05 x 10-4.x 91.4) = E -(9.6 x 10-3) 1 1

i E (yk + a ) = E1 (1.05 x 10~4/(243.3)2 + (91.4)2) = El (2.73 x 10-2) 2 l

From Goldstein's Fundamentals of Shieldina E f r 9.6 x 10-3 is 4.04 l

E for 2.73 x 10-2 is 3.08 y

Calculate the' flux t = SJ (4.04 - 3.03) t = (24g)[g,73 x 10 gg77cg, 3 )

- 5 2

6 2

4 = 4.38 x 10 MeV/cm,3 Calculate the exposure rate given the flux at point P.

The mass absorption coefficient for air for 600 kev photons ?s 2

0.03 cm /g (Morgan and Turner, p. 103).

Exposure Rate (ER) at point P.

6 2

2 ER =-(4.38 x 10 MeV/cm.s)(0.03 cm /c)(1.6 x 10-6 ercs /MeV) 87.7 ergs /g R ER =.24 x 10'4 R/s

~

ER = 8.64 R/h AI-10

Off-Site Direct Dose from Pipe Ruoture on Filter Inlet Line (above water level)

Assumptions:

1) - The isotopes of concern.are Cs-134 and Cs-137.

2) The distance to the closest off-site point is approximately 200' meters.

3) The fuel pool'is a point source for exposure estimates at a distance of 200 meters.

4)' There is no source self-absorption.

5).The fuel pool wall is approximately 1.5 meters thick of reinforced. concrete.

6) The walls'of the fuel handling ouilding are approxicately 1.5 meters thick of reinforced concrete.

7) 675 gallons is. spilled into the pool with concentrations for Cs-134 of 40 pCi/ml and for Cs-137 of 176 pCi/ml.

8) The 75 gallons sp#11ed out of the fuel pool will be cleaned up immediately.

Calculate the exoosure 200 meters from the source.

This calculation will'be the same as for the inadvertent pumping of containment water into the pool except for the quantities of water.

The ratio of 0.uantities of water spilled into the pool can be multiplyed by the' off-site exposure rate calculated for the previous accident to yield the exposure rate from this incident.

675 cal.

= 1.5 450 gal.

-10 Dose from previous incident = 4.5 x 10 R/hr.

(1.5)(4.5 x 10-10 R/hr) = 6.75 x 10 R/hr

-10 The direct dose to an. individual at the site boundary from the pipe rupture on the filter inlet line will be approximately 6.75 x 10-10 R/hr.

4:

AI-11

43 Airborne Off Site Releases from the Pice Ruoture on Filter Inlet Line

. (above water level!

. Assumptions:

'1) Any: activity spilled into the ; ci will be evenly distributed immediately.

2). The " Pool' Clean-up Leakage Ccr.tainment Ion-Exchanger Systen" will remove activity from the fuel : col.

This systen will treat the pool water.'at the rate of 100 GF:1 This flow rate results in a turnover rate of once per 39 hour4.513889e-4 days <br />0.0108 hours <br />6.448413e-5 weeks <br />1.48395e-5 months <br /> period.

3) The inventory of isotopes available to become airborne will not decrease for one week.

At tr.e end of the one week period, seven purification half lives have occurred and no activity remains to tectme airt:rne.

4) The volume of the fuei p:3 is 2.33 x 10 galiens. This voi;te turns ' ova-to the surface as the rate of the "Icn-Exchar.ger Sys:er" (100 GP;1 is ceailable for sus ensicn).

5) The entrainment factor for activity in water to be suspended in air is 10-o.

6) The air frc the fuel pool butiding will be filtered by 2 ban %s of HEFA filters resulting in a CF of 10*.

7) The air fica across the fusi pool is 5500 cfn.

Calculate the arount of activity fcr or.e i ctope that will be released in air during the one week perict required to clean the p;ol water.

5: ate it

.as a fraction of activity released curing ncreal 505 c;eration.

This calculation was r,erforr.ed for a spill of 450 gallons into the occi.

By ratioing this amount to the amount spilled in this incident (675 callons) and multiplying the ratio by the previous result will yield the percenta:e for this incident.

~

675 callons = 1.5 450 gallons 3.75C x 1.5 = 5.63%

The amount of potential airborne activity released off-site from this incident is 5.63, of the amount projected from the nornal operation of the 505 for a year. Tne fraction 5.53; will. hold for the other isotepes as well.

a I

AI-12

+

t i

9

c

. -2

-Inadvertent lifting of prefilter above the pool surface Assumptions:

1) A failure in the control syste.m of the overhead crane system results in the filter being removed unshielded from the pool.

2)' The 'prefilter is loaded with 1000 curies (18% Cs-134, 82% Cs-137).

3) ~ The activity in the prefilter is a point source.

4) The distance to the point of concern is 4.57 meters.

.5) There is no source self-absorpt' ion.

LCalculate the exposure rate from the prefilter at a distance of 4.57 meters.

The gamma exposure factors for the cesium isotopes are (Radiologiccl Heait Handbook, p. 131):

n Cs-134 0.87 R/hr. Ci at 1 meter Cs-137 0.33 R/hr. Ci at 1 meter The number of curies for each isotope are:

Cs-134, (1000 Ci)(0.18) = 180 Ci Cs-137, (1000 C1)(0.82) = 820 '.i The exposure at 4.57 meters from both isotopes is:

(0.87 R/hr Ci x 180 Ci) + (0.33 R/hr Ci x 820 Ci) = 20.5 R/hr 4.572 AI-13

j.[..[*

.0ff-site Direct Dose from Inadvertent Liftinc of Prefilter Above

- the Pool-Surface Assumptions:

'1) A fail'ure in the control system of the overhead crane syste.-

results in the filter being removed unshielded fro 1 the pool.

2) The prefilter _is-loaded with 1000 curies (185 Cs-137, 82t Cs-117;.

3) -The activity in the prefilter is a point source.

'4) The _ distance to the closest off-site point is approximately 20 meters.

5) The walls of the auxiliary building are approximately 1.5 eters thick 'of reinforced concrete.

- Calculate the exposure rate at 203 maters from the source, assumin; n: snielcing.

-(0.87'R/hr Ci x.180 Ci)'+ (0.33 R/hr. Ci x 820 Ci) = 1.07 x 10~9- :Jnr 2

200 Utilizing the standard gamma absor; tion equation for wide-beam radiati:r.

and _the constants from a previous calculation will yield the ex osure ra e shielded by the wall. (Exception, E=40 for pX of 10).

-DX-I = BI e

g I = (40)(1.07 x 10~2 R/hr)(e-0.07 x 10, I = 1.1E x 10-5 R/hr The-direct dose to an individual at the site boundary from the inadvertent lifting of a prefilter out of tne pool will be approximately 1.2 x 10-

c nr.

e"

'AI-14

.