x- and gamma ray exposure calculation of shielding · this calculation is for the purpose of...

MINT/1/1996,164

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X- AND GAMMA RAY EXPOSURECALCULATION OF SHIELDING

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M A L A Y S I A NI N S T I T U T EFOR NUCLEARTECHNOLOGYR E S E A R C H

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INSTITUT PENYEUDIKAN TEKNOLOG! NUKI.EAR MALAYSIABANGI, 43000 KAJANG, MALAYSIATEL : 6 - 03 - 825 0510 FAX : 6 - 03 - 825 8262 \

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MALAYSIAN INSTITUTE FOR NUCLEAR TECHNOLOGY RESEARCHBANGI, 430O0 KAJANG

SELANGOR DARUL EHSAN

SUBJECT

DOCUMENT NO

ISSUING DATE

NO OF PAGES

X- AND GAMMA-RAY EXPOSURE ROOMCALCULATION OF SHIELDING THICKNESSES

MINT:P&P/9003)/(2)/2.6

1 December 1995

12

Prepared By

DR. ABD NASSIR IBRAHIM(HEAD, NONDESTRUCTIVE EVALUATION PROGRAM)

(MINT)

MALAYSIAN INSTITUTE FOR NUCLEARTECHNOLOGY RESEARCH

X- AND GAMMA RAY EXPOSURE ROOMCALCULATION OF SHIELDING THICKNESS

DOCUMENT NO:MINT:P&P/(003)(2)/2.6ISSUE DATE: 1/9/96PAGE: 1

1. Scope

This document presents the calculation of shielding thickness for an exposure room to be usedfor 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 EnergyLicensing Board (AELB) as a precondition for the approval of the design and construction ofan exposure room in IKATAN.

2. Relevant Documents

2.1 ICRP Publication.3, 15 and 21.2.2 Code of Practice on Radiation Protection in Industrial Radiography, AELB, 1995

3. Location

The exposure room is located at the PUSAT LATIHAN INSTITUT KEJURUTERAANTEKNOLOGI 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, isconsidered as clean area,thus accessible to the member of public.-Concrete to be used shall have a minimum density of 2.35 g/cm-*



DOCUMENT NO:MINT:P&P/(003)(2)/2.6ISSUE DATE: 1/9/96PAGE 2

4.1.1 Primary Barrier

a) Thickness of walls sorrounding the source

-Maximum allowable transmission (B) is given by the following formula:

PdTB= (ICRP Publ 1 5 a n d 2 1 ) (1)

m/T

-P (Maximum permissible exposure just outside the shielding) = 0.01 rem/week (formember 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 toposition occupied may be taken as 1 metres (This will allow the source to bepositioned at any location as long as the distance to the shielding does not exceed1 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 andceiling 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




According to the transmission curve provided by ICRP 15 and 21, (See Fig. 1), thicknessrequired to provide the protection is 80cm.

Thus, the thickness of primary barrier (for all 4 walls) is 80 cm ofconcrete.

b) Ceiling thickness

For this calculation, equation 1 will be used with all parameters are similar except in this caseit is assumed that the distance between the source to the ceiling at all time shall not less than 2meters In this case:

Transmission factor (B) = (0.01 X 2 X 2 )/ (2000 X 1 X1) = 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.

4.1.2 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 incidentangle 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.

MALAYSIAN INSTITUTE FOR NUCLEAR

TECHNOLOGY RESEARCH

X- AND GAMMA RAY EXPOSURE ROOM

CALCULATION OF SHIELDING THICKNESS

DOCUMENT NO:MINT:P&P/(003)(2)/2.6

ISSUE DATE: 1/12/95

PAGE: 4

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I transmission of gamma rays from various radionuclideslensity of 2.35 g/cc



DOCUMENT NOMINT:P&P/(003)(2)/2.6ISSUE DATE: 1/12/95PAGE: 5

0.24

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potential, MV

Fig. 2 Variation with potential of the absorbed dose rate measured in air due to x-rayscattered at 90o from various materials The beam is obliquely incident on the thickscatterer. Percent scatterer is related to primary beam measurements in free air at thepoint of incidence. (ICRP Publ. 15 and 21)




According to ICRP 15 and 21 Maximum Allowable Transmission (Bs) for scatteredradiation is given by the following formula:

\ 00 Pd2

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 istaken as 3 metres

Thus,

100x0.01x3x38 = = 0.0531 2000x1x0.085

By referring to Fig.3 (as given by ICRP Publ. 15 and 21), the minimum thickness of leaddoor required for the transmission B of 0. 053 = 20 mm.

b) Leakage Radiation

Leakage radiation is not relevant while gamma projector is in use. While not in use themaximum allowable leakage radiation for the projector (class P), in accordance to the Code ofPractice on RadiationProtection in Industrial Radiography, Part 11, Article 12.1.1 is 2mrem/hr. This value is too small and in this case the radiation will be taken care by theprimary 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 itshall always be used with its primary beam directed toward the floor.




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4.2.1 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).

4.2.2 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 (Bs) for scattered radiationis given by the following formula:

100 Pd:B — WTS

P and T are equal to those in equation (1), i.e. 1.W may be taken as 16800 mA min/week (machine capacity is 7mA, total working minute in aweek is considered as 2400 mins.)

S is taken as a maximum % incident absorbed dose rate scattered to 1 meter per 100 cm2irradiated area. In this case the worst case is taken where maximum scattering occur at anglesbetween 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

168OOxlx.O73Bs = = 8.2x10 4




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Once again by referring to Fig. 1 (as given by ICRP Publ. 15 and 21), the minimum thicknessof 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 theradiation 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 (Bs)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 16800 mA min/week (machine capacity is 7mA, total workingminute in a week is considered as 2400 mins.)

S is taken as a maximum % incident absorbed dose rate scattered to 1 meter per 100 cm2irradiated area (Fig.3), i.e. 0.073%

d (the distance between the floor( known as scatterer) to the wall is taken as 3 metres (roomheight).

Thus,100x0.01x3x3

Bs= = 7.3x10~3

16800x1x073

By referring to Fig. 1 (as given by ICRP Publ 15 and 21), the minimum thickness of theceiling for the transmission B of 7.3x10"3= 35.0cm

This thickness is much smaller than the thickness calculated for the ceiling to attenuate theradiation due to Iridium-192 source, i.e. 70.0cm




b. Leakage Radiation

Barrier for leakage radiation is given in ICRP 15 and 21 as follows:

WJ-jtf (3)

where N j y j is the number of tenth value thickness, T, d and p are similar to those inequation 1 (d is taken as 1 meter), and W L is the weekly leakage exposure rate. ICRPPublication 15 and 21, edition 1976 gives the value of Leakage radiation for an x-ray machineoperating between 200kV to 400kV = 1 R/hr Thus W L = 40 R/week.

4O.vljVriT = Lov = 3.6n T *10 l.vlxO.01

TVT for concrete when using x-ray source of 300 kV is given as 3.0cm. Thus the thicknessrequired due to leakage radiation is 3x3 6cm=10.8 cm.

This thickness is much less than the thickness required to attenuate scattered radiation, whichin turn is much less than the thickness required to attenuate the radiation due to Iridium-192source.

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 concrete5.2 The thickness of the door for Makmal 5 shall be equal to or more than 20.0 mm of lead5.3. The thickness of ceiling for Makmal 5 shall be equal to or more than 70.0 cm ofconcrete.5.4 Plan view of the exposure room is presented in Fig.55.5. The thickness of labyrinth structure may be taken arbitrarily. Such a structure is meant toavoid the primary beam from reaching the door.




Wall thickness = 80 cm, ceiling thickness = 70 cm and lead door thickness = 20 mm.

Figure 5: Plan View of The Exposure Room




6. 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 roomsimultaneously.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 wallshall not be less than 1 meter.6.5 While in operation, the distance between the iridium source/x-ray focal point and theceiling 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 thefloor6.7 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 1 1.35g/cc.

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