Radiation Safety and Shielding
Radiation Safety and Shielding NE 404
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Date Created: 10/15/15
RADM HON SAFETY and SHIELDIIVG NE 404504 Fall 2008 g Lecture 9 Gerdd Wit5 CHP Objectives I Perform gamma exposure rate calculations without shielding for various photon sources and geometries I Perform gamma exposure rate calculations with shielding for various photon sources geometries and shield materials General Equations EXPOSURE XE klmEp F E E k 6606 E5 for Rh if zpE is used k 1835 E8 for R if uence ltIgtE is used E phoinn energy in MeV p2nEp Fquot mass energy absorption coef cient in air in cm2 zpE is the phoinn ux density for energy E at the point of interest in cm392s1 E DE I MMEP 15015 k 577 E7 for Gyh or 577 E5 for radh ifltpE is used k 1602 E10 for Gy if uence ltIgtE is used p2nEp mass energy absorpu39on coef cient for energy E in the medium of interest in cm2g DOSEEQUIVALENT 15 DEQ DE far 01 uniis are Sv or rem and DE must be dose in tissue General Shielding Equation R R0 B expux Where R is response B is the buildup factor for R x is shield slant path eg exposure with buildup XE k PmE P air E ME 3 eXPUX Dose Limits for Shielding Variable dose limils for shielding based on I PL Type offaclllty n I Practical factors wno is ekpoeed 7 public or radiation Workers Regulatory based limils include I 10 CFR 20 based val S lt 2 mremh a for lhe general publlc soning post ALAM clean Lp e elow Radiauon Area lt 100 mremh to be below High Radiauon Area 49 CFR Transportation Radioactive package tirnits Surface ofpackage lt0 5 mremn lt 50 mremn lt 200 mremn lt 1000 mremn 1 meter from package lt 1 mremn lt 10 mremn 2 meters from vehde lt 10 mremn Vehlde cab lt 2 mrem 10 CFR 50 Appendix 1 nuclear power plant lirnits 10 CFR 61 Waste disposal 10 CFR 71 Transportation especlally for Type B packages A E Facility Radiation Hotection Rogram may contain additional limils Estimation of Gamma Exposure POINT SOURCE X 6C En d2 WHERE X m C Activity in g E Energy in m n Fraction of Occurrence d Distance from the source in E 6 Unit conversion factor Estimation of Gamma Exposure Example Consider the exposure at 6 feet from aradioactive source Cs137 at 143 mCi Assume the halflife is 30 years the gamma energy is 0662 MeV and n is 85 BCEn x 0392 X 60143 Ci0662 MeV085 6 Feet2 x 92 00134 Rhr or 134 mRhr Gamma Constant POINT SOURCE X 9 D2 F Gamma Constant in RcmZIhrmCi 2 5EMeV Rm2hrCi 05E SVmZIhrBq 12x10397E X Exposure Rate Q Activity D Distance Gamma Constant Example Consider the exposure at 6 feet from aradioactive source Cs137 at 143 mCi Assume the halflife is 30 years the gamma energy is 0662 MeV 323 Rcm2hrmCi and n is 85 39 Q x r 2 143 mCi R m2 m X 323 0 hr CI 6 mew48 C 012 X 00138 Rh 138 mRh Note When distance is in feet you must convert to cm GammaRay FqumDose Rate Conversion Fuclurs I nlynmniul Coef cients in Analytic Form ABxCx1hd crgy in McV and x In E mum huge xnd39l39mhcyUHNLRSIC D In E rem h cm2 s E Photun En 4M 39 Spw m 9AMquot k FEquot P 9M I I lmmn Energy 0 J A n C F 5 1 1 00 m 003 zu 77 7 H454 1 0113 m 0 5 1 bquot i 5739 I7 10954 114397 t7r r 03 In 5 u 07100 7 umsooa 5010 150 i 12 79 28309 0 H1373 Mil Pvababm k 394 m A r 0 Gamma DOSE Convevsian Factor Alla Ungev and Tmbey A n ma 0142 m39k W W 1503 Hwy aleM4117 IaM 954 ML rapl 39gt 1EDA sz m m lt remh per MeVsq cm sp Svh per emh per a 39o m 1506 Q remh per photonsqvcm s 1E07 D 0 Energy MeV Gamma Duse Cunversiun Curves POINT KERNEL Point Source and PointKernel Method For a point source 0 S47n392 where r is distance where 0 is the uncollided ux density ie o Uul L eu cros sectional shielding considered Other Geometries A collection ofpoint sources can be used to make up aline disk re gu ar area sphere cylinder rectangular volume etc ISOTROPIC POINT SOURCE Point isotropic source Probably the most popular source geometry involved in many calculations is the point isotropic source While no real source is a true oint man sources are sufficiently small in dimensions that they can be treated mathematically as point sources In pr ic i th distance from source to ose point exceeds about three times the maximum source 9 mma radiation that emits S gamma rays per second and that is situated at a distance 139 cm from the dose point Further we shall assume a shield of thickness T cm through which the gamma radiation passes before reaching the dose point see sketch below Point isotropic source shielding configuration T o 0 Source Dose point Shield Point Source Unshielded dose rate The unshielded dose rate at the dose point is given by l kSE t1 p 4727392 where E is the photon energy MeV penp is the mass energy absorption coefficient for the material at the dose point cm2 9quot values also available at NIS an k is a collective constant to convert e rgy fluence rate to dose rate it the dose rate is in grayhour k will have a value 0 576 X10 Point Source Shielded primary photon dose rate The primary photon dose rate is attenuated exponentially and the dose rate from primary photons taking account of the shield is given by HE A PM p D T a where ill is the linear attenuation coef Cient for the photons in the shield material This expression does not account for the buildup of secondary radiation and wil g erally underestimate the true dose rate especially for thick shields and when the dose point is close to the shield surface Point Source Shielded dose rate accounting for buildup The added effect of the buildup is taken into account by incorporating a point isotropic source close buildup factor B into equation 3 ME Be 3 4mg 4 I m p The magnitude of the buildup factor depends on the photon energy the shield material and thickness the source and s ield geometry and the distance from the shield surface to the dose point in most cases dose buildup factors for point isotropic sources have been determined under the assumption that both the source and the dose point reside within an infinite volume of the shield material As a consequence shielded doses evaluated using such buildup factors tend to be conservative for most practical situations in which the dose point Is outside the shield and not subject to backseatterlng from shield material behind the dose point Buildup Factor Among the most popular is an expression referred to as Taylor s form of the buildup factor given by B Ale39m 17 Ale 7 5 where A on and 12 are constants for a given energy and shield material Tabulations of these parameters can be found in various engineering and shielding sources eg Shultis Taylor s form has the advantage that it has only exponential terms in uT and when it is used in an equation that expresses the shielded dose rate the form of the ultimate solution is fundamentally the same as the solution for the primaw photons alone except that it will have twice as many terms because of the two exponential terms in the buildup factor Point Source When the expression for E from equation 5 is inserted into equation 4 we obtain IrSELWAlequot 39I 1 v 4 e a 13 7 6 or k5 Hm Aminealm 1 41ei ra1j f D 7 4mquot Extended Source Geometries Extended geometries Once we have an expression for the point isotropic source we can write reasonable expressions that will apply to other nonpoint source geometries by recognizin so rc 39 form algebraic solution to shielding problems We can however write the differential equations that describe the dose rate from one generalized differential element in the source and then by numerical integration add up the contributions from all such elements to obtain final dose rates We will demonstrate this through a line source application Let us assume that we have a gammaemitting source distributed uniformly along the length of a line In reality many sources that have one straight line dimension much greater than any ther dimensions may be treated as a line source We shal ssume a source of length L with the dose point opposite the end 0 the line source and along a line perpendicular to the source The gammaray emission rate per unit length of source will be given by S which has units of gammas per cm per second The shield at unitorm thickness T will be between the source and the dose point as shown below Line Source Line source shielding configuration Dose Point Line source Shield Let us select a small differential length element dl of the line source located at a distance I from the lower end of the source at the point in the above diagram where the oblique line p from the dose point meets the source line The angle between R and p we shall call 6 From the geometry shown we can specify the followtng R s c 6 R tan 9 and by differentiation dl R secQB d6 2 radiation path length through shield T sec 9 and Sd gamma emission rate from differential source element dl Line Source The shielded differential dose rate at the dose point from primary photons emitted from the differential source element may then be written39 it39s NE W 5W IrSR sec2 6185 m D p p yum AnsE luau eeursecad p 7 397 7 i 8 4150 471R sec 5 471R To obtain the total primary photon dose rate from all differential source elements along the line source we have simply to integrate the above expression over the range of the variable 6 ie from zero to the angle whose tangent is LR H I m 4L A515 tau E 39 P when D 2 15 4er o r 9 Line Source The integral is of a form usually identified as the Sievert integral or the secant integral An exact solution is not available for this integral out it is easin solved using available computer software or a programmable calculator that has integration capability There are also tables available that yleld acceptable approximate solutions of the integral for given values of S and T One advanta e to erformin the above calculation for the dose rate at a point opposite the end of the line source is that if the dose point in another case is opposite some other part of the line source the result for such case can be readil obtained by adding together the dose rates from two line sources where the dose point is opposite the end of each line source Thus for a situation where the dose point was on a line perpendicular to the line source and the source length above the dose point was L and the source length below the dose point was L1 the primary photon dose rate would be kle quotW tanquot 4 7 P UTSCEi fume D 7 7 e d6 7 e 16 MR i 10 ln order to account for the added dose from buildup from a single line source again with the dose point opposite the end of the line we would insert Taylor s expression for B into equation 8 and then proceed as earlier to obtain kSIEIA we muquot 7 7 11 u 5 I 7 91 1 5 D74 R I AM1 1 546 J17A1e M ads 11 D 0 it is clear that the form of the solution with buildup is fundamentally the same as that for the primary photons except that there are two similar terms that arise from the two terms in the buildup factor and where the parameter p appeared in the equation for primary photons the parameters Harm and 1O2p appear in each of the respective terms in the solution when buildup is considered For the case when the dose point is opposite some other point on the line source we would obtain v H 45 75 mg l m R mu R 39 P 41ml HH 70410117966 D A 39 75 l A 39 W 47 l 1 1 l 12 The summation from i 1 to i 2 accounts for the two line segments L and L3 Again the solution is of the expected form now with four terms two for the dose rate contribution from each line segment lrr r1 Men f1 a1AL or f A3049 TmQ A uQ WW L e mms guild 54hr 3 122013 Evin Form pqf39 W7 Mf Are e mm 42 A Col204231 rm W 3 WW dame wanna 4 723 WP 797 r9e I4Lamp 1 z aeoei 39 Io73ltc9 39IWLMTwB m e Axe 4929 i m Fe Io391MT Wk Al F 6 H ii g i H o zsx 4 39 W907 H a 7 T Ham m1 Elwqm p712 El MT a Mum 159 T LA 5 E1x1 Tgran mqaog 77 L177 777 77quot 7 T i JLLAJELUIHQWP Edwmuweo m W H 7 mom we EiquuM bw Indml Now u 3y N4 51 M 5117M 3 Iv b JAT and bgrlnllozl J I 7109 whmlx quot 57 Ffen in F 9115 line JM 777 1 f i 7177 3 Wm M12 new 41 New 12 H N V N 7 Wait 7 r 7 mp WC V S 1 m AFelmljb1 F91 zx351 4441 r 7 W quotW N 7 AJF9J IoJbz F6zl K ba F5 Wato me 724 23 Fm 918057 m HuH 7 quotcumofkwtem quot 9509 5 Fa 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J W M4 0 w 1 M C mu 39l w W r w r m W1 7 I yum s l 4 AM TRUNCATED CONE wow mm msL mi i 1an L 412mm nzwmn mu mm ms um mu 5 39 1 A quot4 mun ngt r lllkbuuw thumnn mm M W V nl m lur 1 mum m mm H W mh phhhim ii imam dumemmfmnawa nnhlzld xhh ml h camem is h expan he and mmnid mm wa m i l mine he Iy mm mm ams lads a mummiaypmm yi hi my w mhhmmhmimg hm myh he RO R14 A where RP is che resporse at point P R01 is che response at point P wichom a shield A is the a mua on factor or tnnmlission factor and epeh oh che phowh meigyahg1e ofincidcnce shieldmaten39al and shield chickness For a monomergetic photonpoint source at distance i m s 413 x swan Where mm is che responsefunc onforphotons ofenergy E For a point some wichmulh39ple gamma photon eheigies Rm Armi Where A is activity and 1quot is che speci c gamma ray constant for a givemadiohuelide eypieanyihazmz Cih Secant Integral and E1 and E2 Functions w Whn L 4 quotAm 1A numencally Sxevert orSecanLImegal Feb ofe dx equot 5 Engmeenng Functions E and E2 Enb bquot1 bfquot dx x e X where b sthe number or mumple ofthe number 0mm n15 typlcallyl orZ 1211 eb bE1b L lmear mw m u 7 mm man m n mm r In E I30 nevus rim 3th 30K r mquot 2quot g E m i E 1 a 3 m6 1 s m 391 17 z u f 7 7 m we 3 n 5 ID 15 2a 3 Neanr rarpam mm 1 musmnen chin simn urmminlegml FBb fn arw x a 5 w 5 2 Vaiuesocmc Modi ed Sievenlnlegral 185 w r 7 H L mark meaning a n m m 0 in so m 70 m m m Mudmed Exponential Integral Fundlon mm on mm mm moooo 10mm mm mom Immo 17mm Locum 5quot Em 53m BM 52 Elm mm nmosu 099 7 099m 09m ass152 rmam 095555 nme39n om 1mm 175752 quot5 um noun am 59635 mums 029317 N 4199995 09949 097944 Anson 091m mam ansm mum nmn mm mm an my a 5 mm 10 mm a m 095965 0510 034295 mm 05727 7 m mm am my quotMN m 3 am 94059 mm a 75m nssasz mm mm mum m Mm M5quot 0 0mm nnml 091m um vas was I 70 mm o 32 min a 4271 mm an m 7a m 22 73 En 9996 096054 a sssxs 0 7mm mm was 039512 013277 w am 0 W a o omsa assmi mm a 59255 055330 n wan 077m mm mm D mquot mm D as amt mi 2 031m V usmi 2927 mm 030577 25341 um I 111937 n 5 Drum mum 1mm umsx um 091509 u my 051797 aim 0 99994 091015 071277 cums 1le 0215622 nwaw mm mms mm 038965 runs wuss a 2273 mm F6b BerquotD E6b where E6b is the modified Sievert integral Enb belb E b where Enb the modified engineering function Secat Integral and E1 and E2 Functions Forlarge values ofe near um and b Feb can be approximated by Feb z 0511b 2 e b 1 0625b For small values oreFeb can be approximatedby F0b 0 e quot Forlarge values orb Enal can be approximated by Eb e 1lrIn nbn3 where for E110 18 error is 011 and for E120 the error is 004 E1b W1th ecom1 1b 110 maybe approxxmavedthh an accuracy of 1 partpermllhon by E1b e39hb ac ba1 baz b cc bC1 bcz b Where an 02372905 Cu 24766331 31 45307924 01 86660126 32 5 1266902 02 6 1265272 E1b w1thb lt 1 maybe approxxmatedwlth an accuracy of1 partperml Ion by E1b In 1 lb 057721 566 201 1 bn H where the summanon 1s camed out form from 1 to w E1b w1thb lt 01may be approxxmahed w1th1n1 n by E1b ln 1b 057721566 E2b w1thb lt 0 1 may be approxxmahed w1thm1 quot1 by 1211 cb b In b 05772 Internet References httpim rsiccornlgovl nhv atrr m hfml nrl 39 39 39 Ammunith quot 39n 39 Qh cc rad indexhtm m m A hmnmm rhtml m m A m 1 1