Radiation Shielding material Any material provides some shielding Iron, concrete, lead, and soil. Shielding…

Download Radiation Shielding material Any material provides some shielding Iron, concrete, lead, and soil. Shielding…

Post on 10-Jul-2018

212 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • Radiation Shielding

    Agen-689Advances in Food Engineering

  • Factors that affect radiation dose

    Regulations and procedures have been developed and implemented to limit radiation dose by regulating the use, storage, transport, and disposal of radioactive material by controlling time,distance and shieldingTime

    The short the time spent near the source, the smaller the dose

    DistanceThe greater the distance the smaller the dose

    ShieldingUse of materials to absorb the radiation dose

  • Shielding materialAny material provides some shieldingIron, concrete, lead, and soil. Shielding ability of a material is determined by the thickness of the material required to absorb half of the radiation This thickness of the material is called the half-thickness Radiation that has passed through one half-thickness will be reduced by half again if it passes through another half-thickness (HT)The HT depends on the characteristics of the material and type and radiation energy

  • Types of radiation and shielding

    particlescan be stopped, or shielded, by a sheet of paper or the outer layer of skin.

    particles can pass through an inch of water or human flesh. can be effectively shielded with a sheet of Al 1/25 of an inch thick.

    rays can pass through the human body like x - rays.dense materials such as concrete and Pb can provide shielding

  • Gamma-ray shieldingTransmission of photons thru matter under conditions of good geometrySince -rays exhibit a log relation between thickness and intensity, only partial reduction of the radiation can be obtained

    Narrow beam

    R

    d

    R>>d

    xoeIxI

    =)(x

  • Gamma-ray shielding

    Narrow beam

    R

    d

    R>>d

    xoeII

    =x

    The particle flux for this situation is:

    The intensity from a point source radiation can be decreased by increasing the distance r from the source or the x of the absorberAn absorber with higher can reduce the thickness needed

    xernA

    = 24

  • Broad beamThe measured intensity is greater than that of the good geometryScattered photons will also be detectedSo, including a constant B:

    B = the building factors (B>1)Tables give values of B for different materials

    xdetector

    o& &

    hvo = &&

    x

    xoeBII

    =

  • Relaxation lengthThe thickness of a shield for which the photon intensity in a narrow beam is reduced to 1/e of its original valueOne relaxation length = 1/, the mean free pathDependence of B in tables on shield thickness is expressed by variation with number of relaxation lengths, x

  • Building factor for concreteCan be obtained from tables as the average of values for Al and Fe:

    [ ]FeAlconcrete BBB += 21

  • Example#1

    Calculate the thickness of a lead shield needed to reduce the exposure rate 1 m from a 10-Ci point source of K-42 to 2.5 mR/h.

  • Answer

    K-42

    Ca-42

    3.52 MeV82%

    2.00 MeV18%

    1.52 MeV

    With no shielding, the exposure rate at r=1 m is:

    An initial estimate of the shielding required is based on narrow-beam geometry. The number of relaxation lengths x is:

    hRCEX /37.1)52.118.0(105.05.0 ===&

    31.6

    1082.11370

    5.2 3

    =

    ==

    x

    e x

  • Answer, cont.The energy of the photons emitted by K-42 is 1.52 MeVFrom Table 15.1 (point source), for photons of this energy in lead and the thickness of 6.31 RLs, B = 3To keep the required reduction (1.82x10-3) the same when the buildup factor is used, the number of RLs in the exponential must be increasedThe number y of added RLs that compensate a B=3 is:

    10.13ln31

    ==

    =

    y

    e y

  • Answer, cont.Added to the initial value, the estimated shield thickness becomes:

    Inspection of Table 15.1 shows that B has increased to 3.5Thus a better guess is y= ln3.5 = 1.25, with an estimated sheild thickness of 6.31 + 1.25 = 7.56 RLsIt remains to verify a final solution numerically by try and error

    RLs41.710.131.6 =+

  • Answer, contFor x = 7.56, a 2-D linear interpolation in Table 15.1:

    The reduction factor with buildup included is:

    Which is the same value given before.

    4.483.662.0

    4.313.533.351.52

    3.743.021.0

    107.567

    356.7 1084.153.3 == eBe x

  • Answer, contThe mass attenuation coeff. is 0.051 cm2/g (Fig 8.8)With = 11.4 g/cm3 for lead, = 0.581 1/cmThe required thickness of lead shielding is:

    A shield of this thickness can be interposed anywhere between the source and the point of exposureUsually, shielding is placed close to a source to realize the greatest solid-angle protection

    cmx 13581.06.76.7

    ===

  • Photon with different energies

    Up to now, we have discussed monoenergetic photonsWhen photons of different energies are present, separate calculations at each energy are usually neededSince the attenuation coefficient and buildup factors are different

  • ExampleA 144-Ci point of Na-24 is to be stored at the bottom of a pool. The radionuclide emits 2 photons per disintegration with energies 2.75 MeV and 1.37 MeV in decaying by - emission to stable Mg-24.

    How deep must be the water if the exposure rate at a point 6 m directly above the source is not to exceed 20 mR/h?What is the exposure rate at the surface of water right above the source?

  • Solution

    lengths relaxation 62.5550020:mR/h 20 to thisreduce to

    50536

    7521445050m 6 da of MeV2.75 from rate exposure

    MeV)1.37(for 061.0

    ;2.75MeV)(for 043.0

    1

    2

    12

    11

    1 ==

    ===

    ==

    =

    xe

    R/h..**.dCE.X

    cm

    cm

    x

    &

  • Sol, cont.

    RLsofnumber larger evena need weso6 Bshows Table 7, for RL

    41.779.162.5

    79.1)6ln(is buildup

    ofamount for this compensate that RLof # The RL)5.62 thickness thisof shieldwater

    afor photons MeV2.75for 6 (B 15.1 tableFrom

    1

    1

    >>=+=

    ==

    =

    xsoy

  • Sol, cont.

    824 Table, From9.10)0430.0/061.0(70.7

    :coef. att. of ratio by thelarger is RLsin thickness thephotons, MeV1.37for

    /5.18550044.7Xphotons 2.75for 447

    gives Table in ioninterpolat theso ,70.7 try slet' so added, be toneeded still

    photonsenergy -lower thefrom rate exposure The

    2

    2

    70.72.75

    1

    1

    .Bx

    hmRe.B

    x

    ===

    ==

    ==

    &

  • Sol, cont.

    hmR

    hmRXXX

    hmReX

    hR..X

    .

    .

    /22080)19.8(600/1:is surface at the level exposure the

    cm 1807.70/0.043: thenis water of depth needed the

    mR/h) (20 figure design the toclose/8.1925.15.18

    :is rate exposure totalthe/25.127408.24

    :24.8)(B shield with

    /74.26

    37114450:is shieldingwithout

    photons for these m 6at rate exposure The

    2

    37.175.2

    9.10371

    2

    2371

    =

    =

    =+=+=

    ==

    =

    =

    =

    &&&

    &

    &

  • Shielding in X-Ray installations

    SecondaryProtective barrier Leakage

    Radiation

    LeakageRadiation

    ScatteredRadiation

    PrimaryProtective barrier

    Usefulbeam

    subjectX-raytube

    Primary protective barrierLead-lined wallDirection of the beamReduces exposure rate

    Other locations exposed to photons

    Leakage radiation from X-ray housingScattered photons from exposed objects in primary beamFrom walls, ceilings, etc

    Secondary protective barriers needed to reduce exposure rates outside the X-ray area

  • Shielding in X-Ray installations

    SecondaryProtective barrier Leakage

    Radiation

    LeakageRadiation

    ScatteredRadiation

    PrimaryProtective barrier

    Usefulbeam

    subjectX-raytube

    Structural shielding designed to limit average dose equivalent to individuals outside and X-ray room

    to 1 mSv/wk in controlled areasTo 0.1 mSv/wk in uncontrolled areas

    Dose equivalent the product of absorbed dose Dand a dimensionless quality factor Q (fnc of LET) the unit is the siervet (Sv)

  • Primary Protective BarrierAttenuation of primary X-ray beams thru different thickness of various materials have been measuredThe primary beam intensity transmitted thru a shield depends strongly on the peak operating voltage but very little on the filtration of the beamThe total exposure per mA min is independent of the tube operating current itselfSo X-ray attenuation data for a given shielding material can be presented as a family of curves at different kVp valuesMeasurements are referred to a distance of 1 m from the target of the tube with different thicknesses of shield interposed

  • Primary Protective Barrier

    10-5

    10-4

    10-3

    10-2

    10-1

    100

    101

    K(R

    /mA

    min

    ) at 1

    m

    0 1 2 3 4 5 6 7Sheilding Thickness (mm)

    #kVp

    Attenuation curves for different peak voltages (kVp) are plotted with the ordinate as K (the exposure in R/mA min) and the abscissa gives the shield thicknessSee Fig.15.3 for 2 mm of lead, the exposure 1m from the target of an X-ray machine operating at 150 kVp is 0.001 R/mA minIf the machine operates with beam current of 200 mA for 90 s, I.e., 200*1.5= 300 mA min, so the exposure will be 300*0.001 = 0.3 R behind the 2 mm lead shield

  • Primary Protective BarrierThe value of K in a specific application will depend on several other circumstances:

    The max permissible exposure rate, P (=0.1 R/wk or 0.01 R/wk)The workload, W (weekly amount of use of machine in mA min/wk)The use factor U (fracti

Recommended

View more >