SHIELDING CALCULATIONS-Mahender D.pptx

Shielding thickness Calculation- Working
out room layout and shielding calculations
for external beam, brachytherapy and
simulator (Physical/CT) installations
MAHENDER DAMERA MSc(Physics),Dip.R.P
Medical Physicist cum RSO
AOI-Bommidala Cancer center Guntur

Reference books
• AERB Safety Code No.
AERB/RF-MED/SC-1 (Rev. 1)
Radiation Therapy Sources,
EquipmentAnd Installations
• NCRP Report No. 151 and IAEA
SRS No. 47

Establishing a Radiotherapy Facility
🠶To establish a Radiotherapy facility, the user institution must go through the
Regulatory requirements as mentioned in:
 Atomic Energy (Radiation Protection) Rules, 2004 &
 AERB Safety Codes (AERB/SC/MED-1 and 3).
🠶 No regulatory clearance is issued for establishing the radiotherapy facility by AERB,
unless the user complies with the regulatory requirements, specified in these
documents.
🠶 The first step is to submit the layout plan of the radiation installation and get it
approved from AERB from radiation safety standpoint.

Radiotherapy Installations
🠶 Siting of the Installation
 Any radiation therapy installation shall be located in the hospital at a place
where occupancy is the minimum
 Shall have the approval of the competent authority
 Closer to the related facilities:
 Simulator Room, Mould Room, Patient Waiting Area, Treatment Planning
System Room, Radiation Oncologist(s) Room, Medical Physicist (s) Room etc.
AERB Safety Code No.AERB/RF-MED/SC-1 (Rev. 1) Radiation Therapy Sources, EquipmentAnd Installations

🠶Doors and Passages
 The doors, passages and turnings of a radiation therapy installation shall
permit:
 Safe and easy transport of the equipment
 Source transfer flask
 Patients on wheel chairs &trolleys.

🠶 Openings and Discontinuities
 Service and dosimetry openings in the shielding shall be provided with:
 Shielded baffle or overlap of shielding
 Openings pose no significant radiation hazard under normal working
conditions

🠶 Safety Interlocks and Warning Lights
 Shall have appropriate door interlock and warning light
 At/or near the entrance to the treatment room
 To warn against inadvertent entry of persons during irradiation
 The color of the warning light shall be:
 Green when the beam is in the OFF position
 Red in the Transit position
 Yellow in the beam ON position

🠶 Radiation warning symbol

Radiotherapy Facilities -
Shielding Requirements

Basic Principles of Radiation Safety
🠶 Time
🠶 Distance
🠶 Shielding
 Not much control over time and distance for staff
 Therefore adequate shielding design is essential

Basic Principles of Radiation Safety
 Radiation exposure at 1 m
= {10 Ci x 1.3 Rh-1Ci-1m2 x 1000 mR/hr} / 1 m2 = 1300 mR/hr
 Radiation worker can occupy only 22 sec for 8 mR/day
 Need to maintain a distance of 36 m to achieve 1 mR/hr
 Hence a shielding thickness of 3 TVT + 0.38 HVT is required for 1 mR/hr
Co-60
Source
1 m
Point of Interest
1 mR/hr
8 mR/day
40 mR/wk
Radiation
Worker
10 Ci

Aim of Structural Design Planning
🠶 To provide the walls of sufficient thickness so that the radiation level outside
the room is well within the permissible limit
 To limit radiation exposure to members of the public and radiation
worker to an acceptable level

🠶 Controlled Areas
 Limited access area in which the occupational exposure of personnel to
radiation or radioactive material is under supervision of an individual in-charge
of radiation protection (e.g.. Control console, treatment rooms, etc.)
🠶 Uncontrolled Areas
 All other areas other than controlled areas in the hospital and the surrounding
environments (e.g. Office, examination room, rest room, etc.)
Basic concepts

Basics of structural design planning
Source details ?
Nature of radiation source
Energy of radiation source
Irradiation time
Distance?
Dimension of the room
Point of
Interest
Nature of occupancies
Who will occupy?
(Public or worker)
How long will occupy?
Permissible dose ?
Workload in cGy/wk at
1 m from the target

🠶 Shielding thickness of a wall depends on:
 Workload (W)
 Use factor (U)
 Occupancy factor (T)
 Permissible dose (P)
 Distance of the wall from the source of radiation (d)
Basis of structural design planning

🠶 R.F = WUT / Pd2
 R.F = reduction factor
 W = Workload of machine (cGy/Wk); U = Use factor; T = Occupancy factor
 P = Permissible limit; d = distance of the wall from the source
🠶 Barrier thickness, T = log10(RF) x TVT
🠶 Transmission factor is the reciprocal of required attenuation
 B = 1/R.F
 Barrier thickness, T = log10(RF) x TVT
Basics of structural design planning

Workload (W)
🠶 Time integral of the absorbed-dose rate determined at the depth of the
maximum absorbed dose, 1 m from the source
🠶 It is stated in terms of weekly dose (cGy/wk) delivered at 1m from the source
🠶 It should consider all uses, such as:
🠶 Use of unit during treatment
🠶 Use of unit for special techniques (IMRT, TBI ..)
🠶 Use of unit for QA purposes

Workload (W)
 W = 1.5 * 105 cGy/wk at 1 m (or) 1.5 * 108 mR/wk at 1 m for 60Co
 W = 1 * 105 cGy/wk at 1 m for accelerators up to 10 MV
 W = 5 * 104 cGy/wk at 1 m for high energy accelerators


 




 days   1
.  . 
day Week PDD (d 10 cm)
 Patient 
. 
 Dose (cGy)
W  
 patient
 
 week 
W   cGy  at1m distance

Workload (W)
🠶Mechalakos et al., 2004 study
 450 Gy week–1 for a 6 MV single-energy
 400 Gy week–1 for 6 and 15 MV dual energy machine
 30 patients per day (high energy photons)
 Primary barrier thickness depends on energy
 Proportion of high-energy x-ray usage on a dual-energy machine ranges from
40 to 70 %

Workload calculation
🠶 Special procedures: TBI
 WTBI = DTBI x d2
(TBI)
 Average one TBI patient per week
 12 Gy at a location 4 m from the x-ray target,
 TBI irradiation increases W at 1 m by 192 Gy week–1.
 i.e. 12 Gy x 42 (4 m treatment distance to isocenter 1 m)

Workload calculation
🠶 Special procedures: IMRT / SRS
 Small field sizes of the large number of beamlets
 Total monitor units (MU) are much higher
 Does not significantly increase the workload for the primary barrier, or for the
patient and wall-scattered radiation components of the secondary barrier.
o Treatment doses for IMRT and conventional RT are similar
 However, the contribution to the leakage-radiation workload is significantly
higher

Use factor (u)
🠶 Fraction of time the primary beam is directed at a particular wall
🠶 For accelerators and Co-60 units
 U = ¼ for primary walls (for each rotational directions)
 U = 1 for gantry pointing down (If basement)
 U = 1 for secondary walls (Scatter and leakage radiation)

Occupancy factor (T)
🠶 Average fraction of time an individual is present beyond the barrier while
the beam is ON.
🠶 Fraction of working hours in the week that the individual would occupy
the area
 T =1 for full occupancy
 T = ¼ for partial occupancy
 T = 1/16 for occasional occupancy

Permissible dose limits
🠶 Radiation worker
 20 mSv/yr
 2000 mR/ 50 wk = 40 mR/wk or 0.4 mSv/wk
🠶 General public
 1 mSv/yr
 2 mR/Wk or 0.02 mSv/Wk

Distance
🠶 Distance (d) of the point to be shielded
 For primary radiation
 Source to the point of calculation
 For leakage and scatter radiation
 Isocenter to the point of calculation

Protective Barrier
🠶 Primary barrier
 It is a wall, ceiling or floor that will intercept radiation emitted directly from the
source (primary or useful radiation)
 In other word, it is an attenuation structure that faces the direct radiation
beam.
🠶 Secondary barrier
 It is a wall, ceiling floor or other structure that will intercept the secondary
radiation.
 Secondary radiation consist of leakage radiation from the source housing
and scattered radiation from the patient

Secondary Radiation Sources
🠶 Leakage
 Depends on the design, typically limited to 0.1% of the primary beam
 Originates from the target, not necessarily through isocenter
🠶 Scatter
 Scattered radiation from the patient
 Scattered leakage radiation from the walls
 Difficult to calculate – use largest field size for measurements
 Lower the radiation energy, greater the scatter

Typical Assumptions
Barrier thickness often assume perpendicular incidence of the radiation
🠶 Attenuation of the primary beam by the patient is neglected (up to 30%)
🠶
🠶 Leakage radiation from the equipment is assumed to be at the maximum value
recommended by IEC.
🠶 The recommended occupancy factors for uncontrolled areas are conservatively
high
🠶 The “two-source rule” is applied (also called as “Add HVL Rule”)
 If the thickness required for two types of radiation are comparable, (not
more than 1 TVT) then add 1 HVT to the larger thickness
 If thicknesses differ by 1 TVT or more, the larger barrier thickness is used

Primary Barrier
🠶 Thickness of the primary barrier
 Tpri = log {WUT/Pd2} . TVTpri
 Where WUT/Pd2 is the reduction factor
🠶 Required barrier thickness
 T = TVL1 + (n-1) TVLe
 The first TVL1 and equilibrium TVLe of the
desired material are used to account for the spectral
changes in the radiation as it penetrates the barrier
NCRP Report No. 151

Primary Barrier
Primary barrier TVLs for ordinary concrete (2.35 g/cm3), steel (7.87 g/cm3) and lead (11.35 g/cm3)
NCRP Report No. 151

Primary Barrier width
🠶 1 foot (30 cm) margin on each side of the beam rotated 450
🠶 Barrier width required assuming 40 cm x 40 cm field size
 W = 0.42dN + 60 cm
Field size
Maximum field
dimension

🠶 Safety margin to account 200 scatter
NCRP Report No. 151

🠶 Field typically nor perfectly square (corners are clipped)
🠶 35 cm x 35 cm field size is used to account for clipped edges
 W = 0.352dN + 60 cm

Primary Barrier - precaution
🠶 Laser lights are used to align the patient in the primary beam
🠶May be recessed in the concrete.
🠶Recess thickness can be  1 HVL for high-energy radiation
🠶 Equivalent thickness of steel or lead plate may be provided
🠶Lasers generally require a mounting plate to allow for lateral adjustments to
position the laser unit
🠶Two functions can be combined in one plate

Secondary Barrier calculation
🠶 Secondary barriers need to be designed to adequately protect individuals beyond
the accelerator room from:
 Leakage radiation
 Scattered radiation from the patient
in the accelerator head or in scattering
 Scattered radiation from the walls
 Secondary radiations produced
throughout the room
 Including photo-neutrons and neutron capture gamma rays

🠶 The leakage radiation and scattered
radiation are of different energies
 Secondary-barrier thickness are
computed separately
 Compared to arrive at the final
thickness.
Layout showing distances associated with patient
scattered (dsca, dsec) and leakage radiation (dL)
NCRP Report No. 151

🠶 Reduction factor for radiation scattered by the patient
(RFps) is given by:
🠶 dsca = distance in meters from the x-ray target to the
patient or scattering surface
🠶 dsec = distance in meters from the scattering object to
the point protected
🠶  = scatter fraction (Table B4, Appendix B, NCRP 151)
🠶 F = field area at mid-depth of the patient at 1 m (cm2) NCRP Report No. 151
400
P d2 2
sca sec


d  

.

 
   F
RFps   W T

🠶 The value 400 assumes the scatter fractions are normalized to those measured
for 20 cm x 20 cm field size.
🠶 The use factor for patient-scattered radiation is 1
🠶 The scattered-radiation energy is significantly degraded (beyond 20 degree) from
that of the primary beam
🠶 TVL values in concrete and lead, for radiations scattered from the patient at
different scattering angles and beam energies are given in Tables B.5a and B.5b
(NCRP Report No. 151).

🠶 Reduction factor for leakage radiation (RFL)
alone is given by:
🠶 The factor 10-3 arises from the assumption
that leakage radiation from the accelerator
head is 0.1 % of the useful beam.
🠶The use factor is one
🠶 Distance d is measured from the isocenter
2
P dL



 
10-3
W T 
RFL  

🠶 After secondary-barrier transmission factor is determined for both leakage and
scattered radiation,
 Required shielding thickness for each contribution can be determined by the
use of tenth-value layers
🠶 If the two thickness is about the same for each secondary component, 1 HVL is
added to the larger of the two barrier thicknesses.
or more, the larger barrier thickness is
🠶 If the two thicknesses differ by a TVL
used.
 This is often referred to as the “two-source rule”.
🠶 In high energy accelerators, a secondary barrier adequately designed for leakage
radiation component is highly adequate for the scattered radiation.

Doors and Mazes
🠶 The maze design is treated under two separate headings:
 Low-energy accelerators (=10 MV)
 High-energy accelerators (>10 MV)
 Due to major differences in the
secondary radiation types

Reduces the need for a heavy shielding door
🠶 Reduces the radiation dose near the entrance
🠶
🠶 Route for ventilation ducts and electrical conduits without compromising the
shielding
◾Maze should be as long and width as small cross-section as possible
◾The minimum width may be determined by
 The dimensions of the treatment unit
 Access for a hospital bed
Doors and Mazes

Doors and Mazes
🠶 The radiation reaching the maze door is due to scattering of photons from the room
surfaces and patient as well as direct penetration of head leakage radiation through
the inner maze wall.
🠶 These components are given as follows:
 HS = dose equivalent per week due to scatter of primary beam from the room
surfaces
HLS
 = dose equivalent per week due to head leakage photons scattered by the
room surfaces
 HLT = dose equivalent per week due to leakage radiation transmitted through the
inner maze wall
 Hps = dose equivalent per week due to primary beam scattered from the patient

 Total dose equivalent (HG) at maze door
Doors and Mazes
 The use factors for major beam directions
(0, 90, 180 and 270 degrees) are taken as
one-quarter
 The total dose equivalent (HTot) at the
maze door from photon leakage radiation
and scattered radiation is not simply 4HG,
(use factor for 4 beam directions – 0, 90,
270 and 180) but is estimated as 2.64 HG
HTot = 2.64 HG
Section 2.4, NCRP Report No 151

Doors and Mazes
🠶 For high energy accelerators
 The estimate of dose equivalent from photons scattered through the maze can
be obtained using above formulas.
 However the average energy of neutron capture gamma rays from concrete is 3.6
MeV.
 A maze and door that provide sufficient shielding for the neutron capture gamma
rays will be adequate for the scattered photons.

Doors and Mazes
🠶 For high energy accelerators
 If maze from A to B is >2.5 m, the photon field is
dominated by neutron capture gamma rays
 Scattered photon component can be ignored.
 The door shielding in high-energy rooms is
usually dominated by the neutron capture
gamma ray and photo-neutron requirements.

Door shielding
🠶 The scattered and leakage dose equivalents are generally relatively low compared
with the other components
 Photo-neutrons and neutron capture gamma rays
🠶 Becomes energetically possible at higher energies.
🠶 Accelerator rooms with maze lengths of 8 m or greater
 Requires 0.6 to 1.2 cm of lead and 2 to 4 cm of BPE (borated polyethylene) for
door shielding

Door shielding
🠶 The suggested arrangement of lead and borated polyethylene (BPE) is:
 lead, BPE, lead.
 Lead on the source side of the BPE is to reduce the energy of the neutrons by non-
elastic scattering
 Makes the BPE more effective in neutron shielding
 Lead on the outside of the BPE will serve to attenuate the neutron capture gamma
rays from the BPE
🠶 Often, outside lead will not be necessary when the maze is long enough to attenuate the
neutrons sufficiently before they encounter the door

Special Considerations

Special considerations
🠶Skyshine radiation
 Some radiation facilities are designed with
little shielding in the ceiling above the
accelerator
 A problem may then arise as a result of
the radiation scattered by the atmosphere
to points at ground level outside the
treatment room.
 Stray radiation of this type is referred to
as Skyshine.

🠶 Side-scattered photon radiation
 If radiation facilities designed with
thin roof barriers
 Radiation may be scattered
laterally from the thin roof barrier
to the adjacent structure

🠶Groundshine radiation
 May be a problem when a high-energy
treatment room is constructed of thin
laminated barriers
 To rectify this problem a lead slab (dashed
lines) can be added to the floor of the
occupied space or the lead and polyethylene
wall can be extended into the floor.
 Barriers composed of thin concrete-lead
laminates can also exhibit this problem.

🠶Activation
 High-energy accelerators (>10 MV), expose both patients and personnel to
radionuclides created by neutron and gamma-ray activation of materials within
the treatment room.
 Rawlinson et al. (2002) indicates that at 18 MV the principal radionuclides
produced result from (n,γ) reactions.
 The sources for these radionuclides are:
 Aluminum (28Al) in the couch frame,
 Antimony (122Sb) in the lead shielding in the gantry head and
 Multiple sources throughout the room (56Mn and 24Na)

🠶Activation
 Rawlinson et al. (2002) make the following specific recommendations in order to
ensure compliance with international effective dose limits:
 IMRT treatments should be delivered at low x-ray energies
 Manufacturers should design accelerators to minimize neutron production
 Avoid the use of aluminum and other materials which have a high neutron-
capture cross section
 Irradiations involving high energy x-rays, (QA measurements) should be
scheduled at the end of the day in order to allow for overnight decay of the
longer lived radionuclides.

🠶 Ozone production
 Electron beams are much more efficient producers of ozone than photon beams
 It is the electron interaction with the oxygen molecule that produces ozone
 McGinley (2002) concluded that for normal clinical use of electron beams, a room
ventilation rate of about three room changes per hour is more than adequate for
health protection.
 Air changes per hour, is a measure of the air volume added to or removed from a
space divided by the volume of the space

🠶Tomotherapy
 Robinson et al. (2000) conclusions:
 Primary shielding barriers for tomotherapy may be up to 10 times narrower
than that required by a conventional field size, while the primary barrier
requires one added TVL over its central portion.
 Secondary shielding barrier should be at least 2 TVLs thicker than the
conventional treatments as a result of the modulation factors as well as indexing
needed to treat the whole length of the tumor volume

Shielding Materials

Shielding materials
In case of space constraint, higher density shielding material can be used
🠶 Commonly used shielding material is Concrete (least expensive)
🠶
🠶 For therapy installation,
 Compton interaction dominates; Depends on the density of the material
Shielding material Density (g/cc)
Concrete 2.35
Heavy concrete > 2.35
Lead (Pb) 11.35
Steel 7.9
Earth 1.6

Shielding materials
🠶 Concrete (Ordinary)
 Most commonly used material
 Good shielding material for photons and neutrons
 Good structural strength
🠶 Heavy Concrete
 Concrete with density > 2.35 g/cc can be considered as heavy concrete
 Drawbacks: cost and handling difficulties

Shielding materials
🠶 Lead
 Excellent shielding material for photons
 Drawbacks:
 Lack of structural integrity as it is malleable. So needs to be sandwiched
between concrete or steel
 High neutron production cross section for high energy photons
 Low absorption cross section for neutrons
 It is toxic
o These restrict the use of lead as a shielding material for accelerator
operated above 10 MV

Shielding materials
🠶 Steel
Relatively expensive than concrete
Not toxic like lead
Its shielding value is between lead and concrete
🠶 Good structural material
🠶
🠶
🠶
🠶 It is nearly transparent to neutrons but reduce neutron energy

Shielding materials
🠶 Polyethylene and Paraffin
🠶 Both materials are used for neutron shielding
🠶 Polyethylene is the best neutron shielding material
 Reasonably expensive
 Available in pure as well as added with varying % of boron
🠶 Paraffin has the same percentage of hydrogen (14.3%) as polyethylene and is
less expensive
 It has lower density and is flammable
 It is avoided in any permanent barriers

Shielding materials
as shielding material for partially or entirely
🠶 Earth
 Earth is also commonly used
underground
 Earth is not a well defined material and its density can vary
 It is sufficient to consider the density as 1.5 g/cc
🠶 Wood
 It can be used for temporary shielding for neutron (4 cm of wood ~ 1 cm of
polyethylene

Shielding materials
NCRP Report No. 151

Shielding materials
🠶 Neutron shielding
 For therapy installation operating above 10 MV, shielding against neutrons must
both the
be considered because of production of Photoneutrons (, n) in
accelerator head and the room shielding.
 Photoneutrons are produced when the primary photons have energies above the
neutron binding energy of roughly 8 MeV for most nuclides.
 Photoneutron yields from most of the accelerator do not become significant until
the incident energy exceeds 10 MeV.
 Concrete contains a relatively high hydrogen content and is therefore efficient at
shielding against fast neutrons.

Radiotherapy Facilities –
Shielding thickness calculation

TYPICAL LAYOUTOF TELECOBALTINSTALLATION

PRIMARY BARRIER
🠶 W = 1.5 x 105 cGy/wk
= 1.5 x 106 mGy/wk
🠶 U = ¼; T = 1;
🠶P = 0.02 mSv/wk
🠶 d = (2.5 + 1.3 + 0.8)m = 4.6 m
🠶 1 TVT of concrete = 20.3 cm
RF =WUT/Pd2; T=log(RF) x TVT

PRIMARY BARRIER
🠶 RF = (1.5 x 106 x ¼ x 1)/ (0.02 x 4.62)
= 8.86105 x 105
🠶 T = log(RF) x TVT
= log(8.86105 x 105) x 20.3 cm
= 5.947 x 20.3
= 121 cm = 130 cm
RF =WUT/Pd2; T=log(RF) x TVT

PRIMARY BARRIERWIDTH
🠶 Max field size: 35 x 35 cm2 at 80 cm
🠶 @ d = 2.5 + 0.8 = 3.3 m
= 35 x (3.3/0.8) = 144.4 cm
 Safety margin of 60 cm
= 144 + 60 = 204 cm
 200 cm = 100 cm + 100 cm

SECONDARY BARRIER1
Only 900 scattering &
🠶 No primary beam
🠶
🠶 ON position leakage
 0.1% of the useful beam
🠶 900 scattering insignificant,
hence only ON position leakage
is considered
RF =0WUT/Pd2

SECONDARY BARRIER1 🠶 RF = {10-3} x{(1.5 x 106 x 1 x 1)/
(0.02 x (3.75)2)}
= 5333.33
= log(5333.33) x 20.3 cm
=3.73 x 20.3 = 75.6 cm
= 75 cm
RF =0WUT/Pd2

SECONDARY BARRIER2 🠶 RF = {10-3} x{(1.5 x 106 x 1 x 1)/ (0.4
x (3.5)2)}
= 306.12
= log(306.12) x 20.3 cm
=2.485 x 20.3 = 50.46 cm
= 50 cm
RF =0WUT/Pd2

SECONDARY BARRIERCORNER 🠶 ON position leakage {0= 10-3 }
 (0.1% of the useful beam)
🠶450 scattering
 {45= 3.6 x 10-3; Us = ¼}
🠶 d= ((d1)2 + (d2)2)
 d1= 3 + 0.75 = 3.75 m
 d2= 2.5 + 1.3 = 3.8 m
🠶 d= ((3.75)2 + (3.8)2) = 5.34 m
RF =(Su +0u) W
T/Pd2

SECONDARY BARRIERCORNER
🠶 RF = {(3.6 x10-3 x ¼) + (10-3 x 1)} x (1.5 x 106 x
1)/ (0.02 x (5.34)2)
= 4997.2
= log(4997.2) x 20.3 cm = 75 cm
🠶 Oblique correction
= 75 x Cos450 = 55 cm
RF =(Su +0u) WT/Pd2

MAZE WALL
🠶 d= 0.8 + 2.5 + 0.85 + 1.8 + 0.7 = 6.65 m
🠶 RF = (1.5 x 106 x ¼ x 1)/ 0.02 x (6.65)2
= 4.24 x 105
= log(4.24 x 105) x 20.3 cm
=5.63 x 20.3 = 114 cm
RF =WUT/Pd2

MAZE WALL
🠶 Out of 114 cm, maze thickness = 85 cm
 ~ 4 TVT thickness
🠶 Primary barrier width same as that of the
other primary barrier
 i.e. 200 cm (100 cm + 100 cm)
🠶 Toreduce scatter contribution at maze an
additional margin of 20 cm is added
 i.e. 220 cm (100 cm + 120 cm)
RF =WUT/Pd2

MAZE WALL CORNER
🠶 ON position leakage {0= 10-3 }
 (0.1% of the useful beam)
🠶300 scattering
 {30= 6 x 10-3; Us = ¼}
🠶 d= ((d1)2 + (d2)2)
 d1= 2.5 + 0.85 + 1.8 + 0.7 = 5.85 m
 d2= 3 + 0.75 = 3.75 m
🠶 d= ((5.85)2 + (3.75)2) = 6.95 m

🠶 RF = {6 x10-3 x ¼ + 10-3 x 1} x
{(1.5 x 106 x 1)/ (0.02 x (6.95)2)}
= 3881.7
= log(3881.7) x 20.3 cm = 72.8 cm
🠶 Oblique correction
= 72.8 x Cos300 = 63 cm = 70 cm
MAZE WALL CORNER

🠶 Radiation reaching maze door is due to:
 Scattering of photons from room surfaces
 Scattering from Patient
 Direct penetration of head-leakage
radiation through inner maze wall
 {Thickness at other corner = 70 cm
 maze wall thickness at corner = 50 cm
 = 70 – 50 cm = 20 cm  25 cm}
MAZE WALL CORNER

CEILING
🠶Ceiling (Primary barrier)
🠶 Ceiling primary barrier width
🠶 Ceiling tapering (scattering + ON
position leakage
(Similar to primary barrier calculations)

TYPICAL LAYOUTOF 6MV LINEARACCELERATOR

PRIMARY BARRIER
= 1 x 106 mGy/wk
🠶 U = ¼; T = 1; P = 0.02 mSv/wk
🠶 d = (3.0 + 1.9 + 1.0)m = 5.9 m
🠶 TVT of concrete:
 TVL1= 37 cm; TVLe= 33 cm
RF =WUT/Pd2;
T=log(RF) x TVT
🠶 W = 1 x 105 cGy/wk

PRIMARY BARRIER
🠶 RF = (1.0 x 106 x ¼ x 1)/ (0.02 x 5.92)
= 359092.2
n = log(359092.2) = 5.555
T = log(RF) x Tpri
Tpri = TVL1 + (n-1) TVLe
T = 37 + (5.555 – 1) x 33 = 1.87 m
 190 cm

🠶 Clipped field size:
 35 x 35 cm2 at 100 cm
 W = 0.352dN + 60 cm
🠶 d = 3 + 1 + 0.75 = 4.75 m
= (0.35 2 x 4.75) +0.6 = 295.7 cm
 300 cm
= (150 + 150 cm)
AERB Calculations: d =4.75; FS =40 x 40 cm2; Col angle 450 not considered
(conservative approach) =(40 x 4.75) +0.6 =250  260 cm =130 cm +130 cm

SECONDARY BARRIER1
🠶 RF =0WUT/Pd2
= (10-3 x 1 x 106 x 1 x 1) / (0.02 x (4.65)2
t = log(2311) x 34 cm
= 3.36 x 34 = 114.4 cm
= 115 cm
{NCRP: t = 37 + (3.36-1) x 33 = 115.01}
RF =0WUT/Pd2

SECONDARY BARRIER2
🠶 450 scattering
 {45= 1.39 x 10-3; Us = ¼}
🠶 d= ((d1)2 + (d2)2)
 d1 = 3 + 1.9 = 4.9 m
 d2 = 3.5 + 0.8 = 4.3 m
 d = ((4.9)2 + (4.3)2) = 6.51 m
RF =(Su +0u) W
T/Pd2

SECONDARY BARRIER2
🠶 RF = ((1.39 x 10-3 x ¼ + 1 x 10-3) x 106)/
(0.02 x (6.51)2)
= 1589.2
🠶t = log(1589.2) x 34
= 3.201 x 34 cm
= 108.8
🠶 Oblique thickness = 108.8 x Cos 450
= 76.96  80 cm
RF =(Su +0u) W
T/Pd2

TYPICAL LAYOUTOF 15MV LINEARACCELERATOR

PRIMARY BARRIER
= 5 x 105 mGy/wk
🠶 U = ¼; T = 1; P = 0.02 mSv/wk
🠶 d = (3.0 + 2.2 + 1.0)m = 6.2 m
🠶 TVT of concrete:
 TVL1= 44 cm; TVLe= 41 cm
RF =WUT/Pd2;
T=log(RF) x TVT
🠶 W = 5 x 104 cGy/wk

PRIMARY BARRIER
🠶 RF = (5.0 x 105 x ¼ x 1)/ (0.02 x 6.22)
= 162591.05
T = log(RF) x Tpri
Tpri = TVL1 + (n-1) TVLe
n = log(162591.05) = 5.211
T = 44 + (5.211 – 1) x 41 = 2.16 m
 220 cm

🠶 Clipped field size:
 35 x 35 cm2 at 100 cm
 W = 0.352dN + 60 cm
🠶 d = 3 + 1 + 0.8 = 4.8 m
= (0.35 2 x 4.8) +0.6 = 297.5 cm
 300 cm
= (150 + 150 cm)
AERB Calculations: d =4.8; FS =40 x 40 cm2; Col angle 450 not considered
(conservative approach) =(40 x 4.8) +0.6 =252  250 cm =125 cm +125 cm

SECONDARY BARRIER1
🠶 RF =0WUT/Pd2
= (10-3 x 5 x 105 x 1 x 1) / (0.02 x (4.9)2
t = log(1041.233) x 44 cm
= 3.017 x 44 = 132 cm
 140 cm
{NCRP: t = 44 + (3.017-1) x 41 = 115.01} RF =0WUT/Pd2

SECONDARY BARRIER2
🠶 450 scattering
 {45= 8.64 x 10-4; Us = ¼}
🠶 d= ((d1)2 + (d2)2)
 d1 = 3 + 2.2 = 5.2 m
 d2 = 3.5 + 0.85 = 4.35 m
 d = ((5.2)2 + (4.35)2) = 6.78 m
RF =(Su +0u) W
T/Pd2

SECONDARY BARRIER2
🠶 RF = ((8.64 x 10-4 x ¼ + 1 x 10-3) x 5x105)/
(0.02 x (6.78)2)
= 662.30
🠶 t = log(662.30) x 41
= 2.821 x 41 cm
= 115.6
🠶 Oblique thickness = 115.6 x Cos 450
= 81.78  85 cm
RF =(Su +0u) W
T/Pd2

TYPICAL LAYOUTOF HDR192IrBRACHYTHERAPYINASTALLATION

Brachytherapy Facilities
🠶 HDR brachytherapy treatment rooms are designed with similar constraints as are
linac and teletherapy rooms
🠶 All walls are considered as primary barriers
 Source can be positioned anywhere in the room
 Radiation is emitted isotropically
🠶 The attenuation in the patient is not considered in primary barrier transmission
calculations.
🠶 The workload specification is given in terms of air kerma in air per week

🠶 Typical workload for HDR 192Ir facility is determined using the following data:
 Maximum source activity: 370 GBq (10 Ci)
 Maximum number of patients treated: 5 per day
 Number of working days (or treatment days) per week: 5 days/week
 Maximum treatment time: 10 min (for 10 Ci) per patient
 Air kerma rate constant for 192Ir: 0.111 Gy.m2. h-1. MBq-1
 Workload, W = RAKR * A * t * n

🠶 Workload Calculation
 W = RAKR * A * t * n
 RAKR – Air kerma rate constant (Gy.h-1.MBq-1.m2)
 A – Activity of the source (MBq)
 t – treatment time per patient in hours
 n – number of patients per week (5 pat/day x 5 days/wk)
= 0.111 (Gy.h-1.MBq-1.m2) x 370 x 103 MBq x (10 min/(60 min/h)) x 25 pt/wk
= 0.17 Gy/wk at 1 m
W = 0.2 Gy/wk at 1m

🠶 Workload Calculation (old method)
 W = x * A * t * n
 x – Exposure rate constant (R.h-1.Ci-1.m2)
 A – Activity of the source (Ci)
 t – treatment time per patient in hours
 n – number of patients per week (5 pat/day x 5 days/wk)
= 0.48 (R.h-1.Ci-1.m2) x 10 Ci x (10 min/(60 min/h)) x 25 pt/wk
W = 20 R/wk at 1m

BARRIER1
🠶 W = 0.2 Gy/wk = 200 mGy/wk
🠶 T = 1; P = 0.02 mSv/wk; d = 2.45 m
🠶 RF = 200 / (0.02 x 2.452) = 1665.95
🠶 T = log (1665.95) x TVT = 3.2216 x 13.5 cm
= 43.5 cm  45 cm
RF =WT/Pd2; T=log(RF) x
TVT 1 TVT =13.5 cm

BARRIER2 - Maze 🠶 W = 0.2 Gy/wk = 200 mGy/wk
🠶 T = 1; P = 0.02 mSv/wk;
🠶 d = (2 + 0.3 + 1.5 + 0.4) m = 4.2 m
🠶 RF = 200 / (0.02 x 4.22) = 566
🠶 T = log (566) x TVT = 2.75 x 13.5 cm
= 37 cm  40 cm
🠶 Maze wall, thickness = 2 TVT  30 cm

BARRIER3 - Door 🠶 W = 0.2 Gy/wk = 200 mGy/wk
🠶 T = 1; P = 0.02 mSv/wk;
🠶 d1 = (2 + 0.4) m = 2.4 m
🠶 d2 = (2 + 0.3 + 1.5 + 0.4) m = 4.2 m
🠶 d = (2.42 + 4.22) = 4.84 m
🠶 RF = 200 / (0.02 x 4.842) = 426.8
🠶 T = log (426.8) x TVT = 2.63 x 13.5 cm = 35.5 cm
🠶 Oblique thickness = 35.5 cm x Cos 450 = 25.1 cm

Simulator
• Operates with
same geometry as
LINAC
• Radiation source is
diagnostic x-ray
tube
• Capable of
radiographic and
fluoroscopic
functions

Simulator
• Most exposures made in fluoroscopy
mode
• X-ray beam collimated, always incident
on image intensifier (II)
• Primary beam significantly attenuated
by patient and II

Shielding materials
• Lead (Pb) backed gypsum board (dry wall)
• Shielding provided to height of 7 feet
unless space above is occupied
• Viewing window with lead glass is used
at console area

Types of barriers
• Primary barriers
– Attenuate primary (direct)
beam
• Secondary barriers
– Leakage
– Patient scatter
– Wall scatter

Simulator room
isocenter
Lead glass window
Target rotational plane
Simulator control area
shielded
door
primary
secondary

Primary beam
• Barrier thickness depends on:
– Distance to POI from source (d)
– Target dose rate (P)
– Workload (W)
– Occupancy (T)
– Usage (U)
*Patient and table attenuation not taken into account

Basic situation
1 m
d
s
source isocenter

Primary barrier
• At isocenter max FS is 40 x 40 cm2
• Largest dimension is diagonal (56 cm)
• At barrier this will project to larger size
at barrier ~ 200 cm
at iso ~ 56 cm

Simulators: Primary beam
• Kux is transmission factor
• Expressed in (R/mA min) at 1m
• NCRP 49 (1976)
Kux =Pd2/WUT

Target dose rate P
Group
ICRP 60
Dose
limit
Exposur
e
rate
Exposu
re
rate
(mSv/y) (R/week) (R/y)
NEW 20 0.04 ~ 2
Public 1 0.002 ~ 0.1
*1 year has 50 weeks of 40 hrs/week or 2000 hr/year
** diagnostic X-ray installations are not licensed by CNSC but may fall under
provincial regulations

Workload
• W workload expressed in mA-min/wk:
• Radiography
– 50 patient/wk x 500 mAs/patient x 1 s/60 min = 400 mA
min/wk
• Fluoroscopy
– 50 patient/wk x 5 mA/patient x 1 min = 250 mA min/wk
W UT
Pd2
Kux =
W = 1000 mA-min/wk

Usage factor U
• U Accounts for beam orientation
• Isocentric units have same usage for
floors, ceiling, and walls.
• U = 0.25
Pd2
Kux =
W U
T

Occupancy factor T
1/16
1/4
Full
Offices, shops, labs, living area
Partial
Corridors, restrooms, parking
Occasional
Waiting room, stairway, janitor closet
1
Type of area
T
Pd2
Kux =
W U T

Simulators: Leakage
WT
• Assumption leakage is 0.1 R/hr at 1m
• Shielded to a factor of 600 per minute
600 I Pds
2
B =

Simulators: Leakage
WT
600 I Pds
2
B =
• B is the factor by which the intensity of
radiation (Po) must be reduced to achieve
the target dose rate P
Po
P
B =

Simulators: Leakage
WT
• I is the tube current (mA)
• ds is the distance from source to POI
600 I Pds
2
B =

TVL - Tenth Value Layer
1
n = log ( )
B
HVL - Half Value Layer
1 TVL = 3.32 HVL

TVL and HVL
1 1
=
=
2x 10
2x 10
x log 2 = log
10
x = 3.32

Simulators: Scatter
• Scattered x-rays have same
barrier penetration as primary
beam
• NCRP 49 (1976)
400 P D2 d2 F
aWT
K =

Simulators: Scatter
• D is the distance from the source to
scatterer
• d is the distance from scatterer to POI
• F is the field area on patient
• a is the scatter fraction
400 P D2 d2 F
aWT
K =

Lead Glass
• Leaded glass may be used for
patient observation window
thicknes
s
(mm)
Lead 1.9 (1/16”) 2.6 (3/32”) 3.1 (1/8”)
Glass 8 11 14
X-ray kVp 150 150 200
Cost/m2 145 190 220

Doors
• Doors contain the lead equivalent
thickness required for secondary barrier
shielding
• 1 - 2 mm Pb in wood
• Make sure door is not in primary beam

Simulator (125 kVp) room
isocenter
Lead glass window
Simulator control area
Target rotational plane
shielded door
primary
A
B
secondary
Determine wall thickness
(concrete and Pb) required for
POI A and B. What would be the
thickness of the lead glass
required for the console area?
d iso to POI is 4m
A. is an office T =1
B.is a waiting room T = 1/16
U = 0.25 for simulators
W = 1000 mA min/week

• Determine target P
– At A , office with NEW (+ALARA ?)
– 20 mSv/year (ICRP 60)
– Target dose rate is 20 mSv/yr = 2 R/yr = 0.04 R/wk

WUT
• P = 0.04 R/week
• d = 4m
• W = 1000 mA min/week
• U = 0.25, T = 1
Pd2
Kux = = 0.00256

1000 mA min/wk x 0.25 x 1
0.04 R/wk x 42
Kux
=
Kux = 0.00256 R/mA min at 1m
~ 12 cm concrete or 1 - 2 mm Pb

• Leakage barrier (at B )
– I = 5 mA, T = 1/16, ds = 4m
600 I Pds
2
B =
WT

• Leakage barrier (at B)
600 I Pds
2
B =
WT
1000 x 1/16
600 x (5) x 0.002 R/wk x 42
= 1.536
=
= 0.186 TVLs or 0.618 HVLs

– Concrete @ 125 kVp = 0.186 x 6.6 cm = 1.2 cm
– Lead @ 125 kVp = 0.186 x 0.93 = 1.7 mm
– Lead glass equivalent = 8 mm

• Scatter barrier (at B )
– Equivalent to about 1 cm concrete so use leakage
calculation
FaWT
– F = 202 cm2, T = 1/16, D = 1m, d = 4m
– a = 0.002
400 P D2 d2
K = = 0.256

CT simulator room
• Dedicated CT scanner
for radiotherapy
• Flat table, lasers, big bore
• X-ray tube operating at 125
kVp and 250 mAs
• Primary beam is
inherently shielded and U
= 1

CT simulator room
control area
waiting room

CT simulator
• Workload
– W = 50 pt/wk x 100 slices/pt = 5000
slices/wk
• Isodose plots are provided from the
manufacturer to estimate the dose rate
in different parts of the room

CT scanner dose
• The workload at any unprotected point in
the room:
D = W Do T
• Do is the isodose value, T is the
occupancy

CT simulator room
• The required transmission is:
TR = P / D = P / W Do T
and TR = Xs/Xo
Xs = Xo TR
Xs = Xo P / W Do T
• Xs is the shielded intensity
• Xo is conversion R per mA min at 1m

CT simulator room
0.95
0.90
0.86
0.73
0.50
150
125
100
70
50
• R per mA min at 1 m from the x-ray
target
kV Xo

CT simulator room
Calculate the barrier
thickness required at point X.
10 patients are scanned a
day, 100 slices each patient.
The area to be protected is a
public access area with
occupancy T=1.
The scanner operates at 125
kV and 200 mA for 1.5s per
slice.
X
a) Calculate the Workload
b) Calculate the P
c) Determine the thickness of lead
required

• Workload W
– 10 pt/day x 100 slices/pt x 5 day/wk = 5000
slices/wk
• Target dose rate P
– Public limit 1 mSv/year is 0.02 mSv/wk
• Dose from isodoses Do
– Do = 0.03
• R per mA min conversion Xo
– Xo = 0.95

Xs = Xo P / W Do T
= (0.95 x 0.02 mSv/wk)
5000 slices/wk x 0.02 x 10-3 mGy
Xs = 0.19

Xs = Xo P / W Do
T
Xs = Xo P / W Do T
= (0.95 x 0.02 mSv/wk)
5000 slices/wk x 0.03 x 10-3
mGy
=
cm

SUMMARY
🠶Protocols
 Atomic Energy (Radiation Protection) Rules, 2004
 AERB Safety Codes (AERB/SC/MED-1 and 3)
 AERB/RSD/RT/Plan-Technote-1
🠶 Parameters required for shielding calculation
 Workload (W); Use factor (U); Occupancy factor (T);
Permissible dose limits (P) and Distance of the wall from
the source of radiation (d)
🠶NCRP 151 &49

SHIELDING CALCULATIONS-Mahender D.pptx

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