Radiation shielding basic topic covered.

1. GOVERNMENT ENGG. COLLEGE, DAHOD
HEAT TRANSFER
RADIATION SHIELDING
BASIC

2. Contents :
 Introduction
 Shielding materials
 Gamma ray protection
 Shielding in X-ray installation
 Primary protective barrier
 Secondary protective barrier

3. Introduction :
• What is Radiation Shielding ?
• Radiation Shielding is define as protection against harmful radiation by
absorbing radiation using radiation absorbing materials.
• It is also known as Radiation protection. Almost any materials acts as a
radiation shield against x-rays or gamma rays.
• Most of ironizing radiations are harmful for human body and tissues, which
must be controlled or shielded.
• For example mobile phones uses wireless communication which uses Radio
frequency and micro wave bandwidth. This radiation was absorbed by
mobile phone users and converted into heat.

4. Shielding material :
• Any material provides some shielding „
• Iron, 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

5. Types of Radiation :
1) Alpha particles :
• Alpha particles are generally helium-4 nucleus. Can be stopped and shielded
by sheet of paper and outer layer of skin.
2) Beta particles :
• Beta particles are high speed & high energized electrons. Can pass through
inch of water and human flash. Effectively shielded with a sheet of 1/25 inch
of thick aluminium.
3) Gamma particles :
• Gamma particles
are high energized photons
Decayed from atom
nucleus. Dense material such as
Concrete and pb provide shield.

6. Gamma-ray shielding
• Transmission of photons
thru matter under
conditions of ‘good’
geometry
• Since -rays exhibit a log
relation between thickness
and intensity, only partial
reduction of the radiation can
be obtained
Narrow beam
d
R
R>>d
I(x)  I ex
o
x

7. Narrow beam
d
R
R>>d
oI  I ex
x
 The particle flux for this
situation is:
 
nA
ex
4r2
 The intensity from a point
source radiation can be
decreased by increasing the
distance r from the source or
the x of the absorber
 An absorber with higher  can
reduce the thickness needed

8. Shielding in X-Ray installations
Secondary
Protective
barrier Leakage
Radiation
Leakage
Radiation
Scattered
Radiation
Primary
Protective
barrier
Useful
beam
subject
X-ray
tube
 Primary protective barrier
 Lead-lined wall
 Direction of the beam
 Reduces exposure rate
 Other locations exposed to
photons
 Leakage radiation from
X-ray housing
 Scattered photons from
exposed objects in
primary beam
 From walls, ceilings, etc
 Secondary protective
barriers needed to reduce
exposure rates outside the
X-ray area

9.  Structural shielding designed to limit average dose equivalent to
individuals outside and X-ray room
 to 1 mSv/wk in controlled areas
 To 0.1 mSv/wk in uncontrolled areas
 Dose equivalent – the product of absorbed dose D and a
dimensionless quality factor Q (fnc of LET) – the unit is the siervet
(Sv)

10. Primary protective Barriers :
 Attenuation of primary X-ray beams thru different thickness
of various materials have been measured
 The primary beam intensity transmitted thru a shield
depends strongly on the peak operating voltage but very
little on the filtration of the beam
 The total exposure per mA min is independent of the
tube operating current itself
 So X-ray attenuation data for a given shielding material can be
presented as a family of curves at different kVp values
 Measurements are referred to a distance of 1 m from the
target of the tube with different thicknesses of shield
interposed

12.  The value of K can be computed as:
 With P [R/wk], d [m], W [ mA min/wk], so K
[ R /mA min] at 1 atm
WUT
Pd2
K 

13. Secondary Protective Barrier
 Designed to protect areas not in the line of the useful
beam from leakage and scattered beam
 Shielding requirements are computed separated for
leakage and scattered radiations
 The final barrier thickness is the summ of each one
 Assume that leakage and scattered radiations are
isotropic (so U = 1)

14. Contents

15. Thank you