X- AND GAMMA RAY EXPOSURE CALCULATION OF SHIELDING

Transcription

1 MINT/1/1996,164 MY *»'' X- AND GAMMA RAY EXPOSURE CALCULATION OF SHIELDING T f - «MALAYSIAN INSTITUTE FOR NUCLEAR TECHNOLOGY RESEARCH 7 8HE 1 INSTITUT PENYEUDIKAN TEKNOLOG! NUKI.EAR MALAYSIA BANGI, KAJANG, MALAYSIA TEL : FAX : i.

2 BANGI, 430O0 KAJANG SELANGOR DARUL EHSAN SUBJECT DOCUMENT NO ISSUING DATE NO OF PAGES X- AND GAMMA-RAY EXPOSURE ROOM ES MINT:P&P/9003)/(2)/2.6 1 December Prepared By DR. ABD NASSIR IBRAHIM (HEAD, NONDESTRUCTIVE EVALUATION PROGRAM) (MINT)

3 ISSUE DATE: 1/9/96 PAGE: 1 1. Scope This document presents the calculation of shielding thickness for an exposure room to be used for X- and gamma radiography using the following radiation source and radiation machine : - Maximum of 100 Ci Iridium-192 Source for gamma radiography, and -7mA and 300kV directional x-ray machine This calculation is for the purpose of fulfilling the requirement of the Atomic Energy Licensing Board (AELB) as a precondition for the approval of the design and construction of an exposure room in IKATAN. 2. Relevant Documents 2.1 ICRP Publication.3, 15 and Code of Practice on Radiation Protection in Industrial Radiography, AELB, Location The exposure room is located at the PUSAT LATIHAN INSTITUT KEJURUTERAAN TEKNOLOGI TENAGA NASIONAL (IKATAN), Mukim Dengkil, Bangi Selangor. Detail location of the exposure room known as Lab. 5 is given in Appendix 1, 2 and 3 4. Shielding Calculation 4.1 For Gamma Source (Iridium -192,100 Ci source) This calculation is based on the following considerations: -The exposure room is located at the first floor -The area/room around the exposure room, including the room immediately above it, is considered as clean area,thus accessible to the member of public. -Concrete to be used shall have a minimum density of 2.35 g/cm-*

4 ISSUE DATE: 1/9/96 PAGE Primary Barrier a) Thickness of walls sorrounding the source -Maximum allowable transmission (B) is given by the following formula: PdT B= (ICRP Publ 15and21) (1) m/t -P (Maximum permissible exposure just outside the shielding) = 0.01 rem/week (for member of public) -d (distance in metres from source to position occupied) -the source will be about the centre, thus the nearest distance between the source to position occupied may be taken as 1 metres (This will allow the source to be positioned at any location as long as the distance to the shielding does not exceed 1 metre) -W (weekly workload in R/week at 1 meter) -For 100 Ci Ir-192, dose rate at 1 meter is 50 R/hr. -Maximum total working hour/week is 40 hr. -Thus W= 2000 R/week -U Use factor = 1 for all 4 walls and ceiling (ICRP Publ. 3)-These walls and ceiling will be routinely exposed to useful beam. -T (occupancy Factor) = 1 ( according to ICRP Publ 3) Under this circumstances: Transmission factor (B) = (0.01 X 1 X 1 )/ (2000 X 1 X 1) = 2X10-6

5 ISSUE DATE: 1/9/96 PAGE: 3 According to the transmission curve provided by ICRP 15 and 21, (See Fig. 1), thickness required to provide the protection is 80cm. Thus, the thickness of primary barrier (for all 4 walls) is 80 cm of concrete. b) Ceiling thickness For this calculation, equation 1 will be used with all parameters are similar except in this case it is assumed that the distance between the source to the ceiling at all time shall not less than 2 meters In this case: Transmission factor (B) = (0.01 X 2 X 2 )/ (2000 X 1 X 1) = 2 X 10-5 From Fig. 1, this transmission factor corresponds to the concrete ceiling thickness of 70.0 cm Thus, the required thickness of ceiling is 70.0 cm of concrete Secondary Barrier a. Scattered Radiation. This calculation is relevant only for door thickness. Maximum percentage of scattering is taken (as precaution). This occur when both the incident angle and angle of scattering are 45 (See Fig. 2, from ICRP 15 and 21). Energy from Iridium-192 is assumed to be equivalent to about 0.6 MV x-ray.

6 ISSUE DATE: 1/12/95 PAGE: 4 1. i i i i i i i i i i i i i i i i i i - z - IO II transmission > o IO- 5 V = - I0" L_ i i i i i i i i concrete, cm Fig. 1: Broad-bearrI transmission of gamma rays from various radionuclides through concrete, clensity of 2.35 g/cc

7 DOCUMENT NO ISSUE DATE: 1/12/95 PAGE: potential, MV Fig. 2 Variation with potential of the absorbed dose rate measured in air due to x-ray scattered at 90o from various materials The beam is obliquely incident on the thick scatterer. Percent scatterer is related to primary beam measurements in free air at the point of incidence. (ICRP Publ. 15 and 21)

8 ISSUE DATE: 1/9/96 PAGE: 6 According to ICRP 15 and 21 Maximum Allowable Transmission (B s ) for scattered radiation is given by the following formula: 00 Pd 2 P and T are equal to those in equation (1), i.e. 1. W may be taken as 2000 R/week (considering source to scatterer distance of 1 meter), S (% of absorbed dose rate scattered at 1 meter taken from Fig. 2) =0 085% d (the distance between the wall (opposite to the door-known as scatterer) to the door itself is taken as 3 metres Thus, 100x0.01x3x3 8 1 = = x1x0.085 By referring to Fig.3 (as given by ICRP Publ. 15 and 21), the minimum thickness of lead door required for the transmission B of = 20 mm. b) Leakage Radiation Leakage radiation is not relevant while gamma projector is in use. While not in use the maximum allowable leakage radiation for the projector (class P), in accordance to the Code of Practice on RadiationProtection in Industrial Radiography, Part 11, Article is 2 mrem/hr. This value is too small and in this case the radiation will be taken care by the primary and secondary barrier. 4.2 For X-ray Machine This calculation is based on the precondition that the x-ray machine is of directional type and it shall always be used with its primary beam directed toward the floor.

9 ISSUE DATE: 1/12/95 PAGE 7 1 Mill K i i i i , MM' I I 1 1 M transmission,o- IO- 2 irr* M V > s w "Au To IQ-5 v X : io i i i i > () ' lead, cm " > ; Fig: 3 Broad beam transmission of gamma rays from various radionuclides through lead, density 11.35g/cc

10 ISSUE DATE: 1/9/96 PAGE: Primary Barrier Since the useful beam will be directed to the floor and the room is located at the ground floor, then the primary barrier need not to be caclulated ( Floor is the only primary barrier) Secondary Barrier a. Scattered Radiation. This calculation may be divided into two sections; a.i. Thickness of walls sorrounding the machine: According to ICRP 15 and 21 Maximum Allowable Transmission (B s ) for scattered radiation is given by the following formula: B 100 Pd: WTS P and T are equal to those in equation (1), i.e. 1. W may be taken as ma min/week (machine capacity is 7mA, total working minute in a week is considered as 2400 mins.) S is taken as a maximum % incident absorbed dose rate scattered to 1 meter per 100 cm2 irradiated area. In this case the worst case is taken where maximum scattering occur at angles between 160 to 180.According to Fig. 4, this value is 0.073% (backscattered). d (the distance between the floor( known as scatterer) to the wall is taken as 1 metres. Thus, 100x0.01x1x1 Bs = = 8.2x OOxlx.O73

11 ISSUE DATE: 1/12/95 PAGE 9 1 ' 1 ' i i i I i i i i i i i irradiated area scattering s angle cm c t beam > '. m per sttered to A /K)0to300 kv 0.06 : ; absorbed dose c % incide : /, / / /. 6MV Yi i i 90 i i i i i i i i i K scattering angle, degrees Fig 4. Scattering patterns of diverging x-ray and gamma ray beams normally incident on a concrete shield. Percent scatter is related to primary beam measurements in free air at the point of incidence (ICRP Publ. 15 and 21)

12 ISSUE DATE: 1/9/96 PAGE: 10 Once again by referring to Fig. 1 (as given by ICRP Publ. 15 and 21), the minimum thickness of these walls for the transmission B of 8.2x10"^= 48.0cm This thickness is much smaller than the thickness calculated for the walls to attenuate the radiation due to Indium-192 source, i.e.80.0cm a.ii. Thickness of ceiling As in the previous case, according to ICRP 15 and 21 Maximum Allowable Transmission (B s ) for scattered radiation is given by the following formula: B = wrs P and T are equal to those in equation (1), i.e. 1. W once again may be taken as ma min/week (machine capacity is 7mA, total working minute in a week is considered as 2400 mins.) S is taken as a maximum % incident absorbed dose rate scattered to 1 meter per 100 cm2 irradiated area (Fig.3), i.e % d (the distance between the floor( known as scatterer) to the wall is taken as 3 metres (room height). Thus, 100x0.01x3x3 Bs= = 7.3x10~ x1x073 By referring to Fig. 1 (as given by ICRP Publ 15 and 21), the minimum thickness of the ceiling for the transmission B of 7.3x10" 3 = 35.0cm This thickness is much smaller than the thickness calculated for the ceiling to attenuate the radiation due to Iridium-192 source, i.e. 70.0cm

13 ISSUE DATE: 1/9/96 PAGE: 11 b. Leakage Radiation Barrier for leakage radiation is given in ICRP 15 and 21 as follows: WJ -jtf (3) where Njyj is the number of tenth value thickness, T, d and p are similar to those in equation 1 (d is taken as 1 meter), and WL is the weekly leakage exposure rate. ICRP Publication 15 and 21, edition 1976 gives the value of Leakage radiation for an x-ray machine operating between 200kV to 400kV = 1 R/hr Thus WL= 40 R/week. 4O.vl jvnt rit = Lov = 3.6 * 10 l.vlxo.01 TVT for concrete when using x-ray source of 300 kv is given as 3.0cm. Thus the thickness required due to leakage radiation is 3x3 6cm=10.8 cm. This thickness is much less than the thickness required to attenuate scattered radiation, which in turn is much less than the thickness required to attenuate the radiation due to Iridium-192 source. 5. Conclusion Based on the calculation presented in Section 4 of this document, it is concluded that: 5 1 The wall thickness of Makmal 5 shall be equal to or more than 80.0 cm of concrete 5.2 The thickness of the door for Makmal 5 shall be equal to or more than 20.0 mm of lead 5.3. The thickness of ceiling for Makmal 5 shall be equal to or more than 70.0 cm of concrete. 5.4 Plan view of the exposure room is presented in Fig The thickness of labyrinth structure may be taken arbitrarily. Such a structure is meant to avoid the primary beam from reaching the door.

14 ISSUE DATE: 1/9/96 PAGE: 12 Wall thickness = 80 cm, ceiling thickness = 70 cm and lead door thickness = 20 mm. Figure 5: Plan View of The Exposure Room

15 ISSUE DATE: 1/9/96 PAGE: Condition of Usage The following conditions are applied for the use of this exposure room: 6.1 The room may be used for both x- and gamma radiography purposes. 6.2 Under no circumstances that both x- and gamma radiography will be executed in this room simultaneously. 6.3 Spaces sorrounding this room is defined as clean area, thus accessible to member of public. 6.4 While in operation, the distance between the iridium source/x-ray focal point and the wall shall not be less than 1 meter. 6.5 While in operation, the distance between the iridium source/x-ray focal point and the ceiling shall not be less than 3 meters. 6.6 While in operation, the primary beam of the x-ray machine shall always be directed to the floor 6.7 Concrete used for the construction shall have a minimum density of 2.35g/cc. 6 8 Lead used for door construction shall have a minimum density of g/cc.

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