Lecture 15-1Shielding of Broad Beam, Point and Line Sources

Radiation ShieldingType of shields Biological shields Thermal shieldsApparatus shieldsBiological Shields Shielding for reducing the radiation exposure to persons in the vicinity of radiation sources is called biological shielding

Biological Shields Shielding for reducing the radiation exposure to persons in the vicinity of radiation sources is called biological shieldingApparatus Shields The shields used to protect the delicate apparatus (electronic components) against gamma rays and neutrons.

GAMMA RAY SHIELDING BROAD BEAM Let, o = flux (intensity of mono-directional ray [s/cm2-s] Eo = energy of gamma rays [MeV] = mass attenuation coefficient of air The exposure rate (without shield) at point P is given by

or

Monodirectional beam of rays incident on slab shield

Energy spectrum of incident ray beam

Thin shield (Uncollided flux)With the shield in place

whereo = flux without shield [s/cm2-s]u = attenuated un-scatted fluxa = thickness of the shield [cm] = total attenuation coefficient at the energy Eo [cm-1]The relation for u is valid for thin shields (i.e. when a 1/ or a 1 mfp)

Thick Shields (Buildup flux)Unscattered and Compton scattered rays both reach the detector. There is increase in flux

The increased flux is called buildup

Energy spectrum of ray emerging from shield

where= exposure rate in the absence of the shield = total attenuation coefficient at energy Eoa = shield thickness a = Number of mean free paths = exposure buildup factor for mono-direction beamSignificance of buildup factor At higher values of a, the magnitude of Bm increases many folds, therefore it can not be ignored in calculation of flux

The values of Bm are tabulated and given as a function of energy for several materials.

As

by same analogy

where

ProblemA monodirectional beam of 2 MeV rays of intensity 106 rays/ cm2-sec upon a shield 10 cm thick. Calculate at the rear side of the shield: (a) the uncollided flux; (b) the build up flux; (c) exposure rate [Example 10.1; Lamarsh]

SolutionEo = 2 MeV o = 106 rays/ cm2-seca = 10 cm

a)b)

c)

POINT SOURCETwo casesPoint source is surrounded by a shield i.e. no space between source and shieldPoint source followed by a shield and point of measurement is away from shieldCase 1Point source is surrounded by a shield of radius Rwhere

Uncollided flux

Collided flux

Case 2Point source is followed by a shield and point of measurement is away from the shield

The computed value of Bp are tabulated as a function of energy for different materials

Determination of Shield Thickness

Among the various activities, yields and energies; maximum source strength and maximum energy are chosenRadius is calculated for that set of energy and source strengths using the following relation

A graph is plotted RHS versus RFrom R corresponding to the RHS = 1, the value of R is estimated

ProblemAn isotropic point source emits 108 rays/s with an energy of 1 MeV. The source is to be shielded with a spherical iron shield. What must the shield radius be if the exposure rate at its surface is to be 1 mR/hr? [Example 10.2; Lamarsh]SolutionS = 108 rays/s Eo = 1 MeV= 1 mR/hr

ProblemA 3.7x104 MBq (1 Ci) source of 137Cs is to be stored in a spherical lead container when not in use. How thick must the lead be it the dose equivalent rate at a distance of 1 m from the source is not to exceed 25 Sv/h (2.5 mrem/h)? Assume the source to be sufficiently small to be considered as a point source. [Example 10.4; Cember 1997]SolutionS = 3.7x104 MBq, Eo = 0.662 MeV, x = 1 m, = 25 Sv/h

ProblemDesign a spherical lead storage container that will attenuate the exposure rate from 1 Ci of 24Na to 10 mR/h at a distance of 1m from the source [Example 10.5; Cember 1997]Solution

Lecture 15-2Shielding of Line source

LINE SOURCEAssuming thatL = length of the source R= radius of cylindrical shield around the source SL= rays emitted isotropically per unit length (s/cm-sec)Let the point of observation is at point PUncollided Flux

Line Sourceconti.Isotropic line source imbedded in a cylindrical source

Line Sourceconti.Changing integration variables from z to r = R sec , z = r tan , dz = R sec2 d

The integral can not be evaluated analytically, it can be expressed in terms of Sievert Integral function which is defined as follows:

Line Sourceconti.Graphs of F (,x) vs are available in literatureFor large values of and x, F(,x) can be computed from the formula

It should be noted that

The un-collided flux is therefore given as

The function f(,b)

Line Sourceconti.Buildup FluxUsing the Taylors form of buildup factor

Line Sourceconti.When observation point is beyond the surface of shield Let a = Shield thicknessx = Distance between source and point of observationUNCOLLIDED FLUX

BUILDUP FLUX

Line Sourceconti..ShieldaxPLineSourcea

Line Sourceconti.Shield Located at Bottom of Source Uncollided flux at Point P

Changing integration variable from z to r = R sec , rs = a sec, dz = - R cosec2 dIntroducing in the Eq. of u

Shield at end of line source

Line Sourceconti.Which yields the uncollided flux

Buildup Flux at point P

Limiting caseAs P P/, 1 and 2 0, then

Line Sourceconti.For small value of , u behaves as

Similarly

Line Sourceconti.Formulas derived are used to designCylindrical fuel casksFuel coffinsThe calculations done for line source are used for calculation of thickness of Cylindrical sidesEnd sides

Line Sourceconti.Problem: Cobalt source is frequently used to measure neutron flux in a nuclear reactor. Calculate exposure rate at a distance 1 foot from the mid point of a 60.96 cm section No. 22 gauge wire (diameter 0.0253 in) which has been exposed for 100 h to a neutron flux of 1014 n/cm2.s. How much would the exposure rate be reduced if a 5 cm thick sheet of lead is interposed between the cobalt wire and the point of observation. The wire may be assumed vertically suspended.Data:(1) Cross section for reaction 59Co(n,)60Co, th = 19 b(2) T1/2 of 60Co = 5.26 y(3) Pb for E 1.17 MeV = 0.72 cm-1(4) Pb for E 1.33 MeV = 0.65 cm-1