rpdir-l12 shielding web

IAEAInternational Atomic Energy Agency

RADIATION PROTECTION INDIAGNOSTIC AND

INTERVENTIONAL RADIOLOGY

L12: Shielding and X Ray room design

IAEA Training Material on Radiation Protection in Diagnostic and Interventional Radiology

IAEA 12: Shielding and X Ray room design 2

Introduction

• Subject matter: the theory of shielding design and some related construction aspects.

• The method used for shielding design and the basic shielding calculation procedure


Topics

Equipment design and acceptable safety standards

Use of dose constraints in X Ray room design

Barriers and protective devices


Overview

• To become familiar with the safety requirements for the design of X Ray systems and auxiliary equipment, shielding of facilities and relevant international safety standards e.g. IEC.


Part 12: Shielding and X Ray room design

Topic 1: Equipment design and acceptable safety standards



Purpose of Shielding

• To protect:• the X Ray department staff

• the patients (when not being examined)

• visitors and the public

• persons working adjacent to or near the X Ray facility


Radiation Shielding - Design Concepts

• Data required include consideration of:• Type of X Ray equipment• Usage (workload)• Positioning• Whether multiple tubes/receptors are being

used• Primary beam access (vs. scatter only)• Operator location• Surrounding areas


Shielding Design (I)

Equipment

• What equipment is to be used?• General radiography

• Fluoroscopy (with or without radiography)

• Dental (oral or OPG)

• Mammography

• CT


Shielding Design (II)

The type of equipment is very important for the following reasons:• where the X Ray beam will be directed

• the number and type of procedures performed

• the location of the radiographer (operator)

• the energy (kVp) of the X Rays


Shielding Design (III)

Usage• Different X Ray equipment have very

different usage.• For example, a dental unit uses low mAs

and low (~70) kVp, and takes relatively few X Rays each week

• A CT scanner uses high (~130) kVp, high mAs, and takes very many scans each week.


Shielding Design (IV)

• The total mAs used each week is an indication of the total X Ray dose administered

• The kVp used is also related to dose, but also indicates the penetrating ability of the X Rays

• High kVp and mAs means that more shielding is required.


Shielding Design (V)

Positioning

• The location and orientation of the X Ray unit is very important:• distances are measured from the equipment

(inverse square law will affect dose)

• the directions the direct (primary) X Ray beam will be used depend on the position and orientation


Radiation Shielding - Typical Room Layout

A to G are pointsused to calculate shielding


Shielding Design (VI)

Number of X Ray tubes

• Some X Ray equipment may be fitted with more than one tube

• Sometimes two tubes may be used simultaneously, and in different directions

• This naturally complicates shielding calculation


Shielding Design (VII)

Surrounding areas• The X Ray room must not be designed

without knowing the location and use of all rooms which adjoin the X Ray room

• Obviously a toilet will need less shielding than an office

• First, obtain a plan of the X Ray room and surroundings (including level above and below)


Radiation Shielding - Design Detail

Must consider:• appropriate calculation points, covering all

critical locations• design parameters such as workload,

occupancy, use factor, leakage, target dose (see later)

• these must be either assumed or taken from actual data

• use a reasonable worst case more than typical case, since undershielding is worse than overshielding



Topic 2: Use of dose constraints in

X Ray room design



Radiation Shielding Parameters (I)

P - design dose per week

• usually based on 0.3 mSv per year


Radiation Shielding Parameters (II)

• Film storage areas (darkrooms) need special consideration

• Long periods of exposure will affect film, but much shorter periods (i.e. lower doses) will fog film in cassettes

• A simple rule is to allow 0.1 mGy for the period the film is in storage - if this is 1 month, the design dose is 0.025 mGy/week


Radiation Shielding Parameters (III)

• Remember we must shield against three sources of radiation

• In decreasing importance, these are:• primary radiation (the X Ray beam)

• scattered radiation (from the patient)

• leakage radiation (from the X Ray tube)


U - use factor

• fraction of time the primary beam is in a particular direction i.e.: the chosen calculation point

• must allow for realistic use

• for all points, sum may exceed 1

Radiation Shielding Parameters (IV)


Radiation Shielding Parameters (V)

• For some X Ray equipment, the X Ray beam is always stopped by the image receptor, thus the use factor is 0 in other directions

• e.g.: CT, fluoroscopy, mammography

• This reduces shielding requirements


Radiation Shielding Parameters (VI)

• For radiography, there will be certain directions where the X Ray beam will be pointed:• towards the floor• across the patient, usually only in one direction• toward the chest Bucky stand

• The type of tube suspension will be important, e.g.: ceiling mounted, floor mounted, C-arm etc.


Radiation Shielding Parameters (VII)

T - Occupancy

• T = fraction of time a particular place is occupied by staff, patients or public

• Has to be conservative

• Ranges from 1 for all work areas to 1/20 for toilets and 1/40 for unattended car parks


Occupancy (NCRP 147)

Area• Administrative or clerical

offices; laboratories, pharmacies and other work areas fully occupied by an individual; receptionist areas, attended waiting rooms, children indoor play areas, adjacent X ray rooms, film reading areas, nurse stations, X ray control rooms

• Room used for patient examinations and treatments

• Corridors, patients rooms, employee lounges, staff rest rooms

Occupancy factor T1

1/2

1/5


Occupancy (NCRP 147)

Area• Corridor doors • Public toilets, unattended

vending areas, storage rooms, outdoor areas with seating, unattended waiting rooms, patient holding areas

• Outdoor areas with only transient pedestrian or vehicular traffic, unattended parking lots, vehicular drop off areas (unattended), stairways, unattended elevators

Occupancy factor T1/81/20

1/40


Radiation Shielding Parameters (VIII)

W - Workload

• A measure of the radiation output in one week

• Measured in mA-minutes

• Varies greatly with assumed maximum kVp of X Ray unit

• Usually a gross overestimation

• Actual dose/mAs can be estimated


Workload (I)

• For example: a general radiography room

• The kVp used will be in the range 60-120 kVp

• The exposure for each film will be between 5 mAs and 100 mAs

• There may be 50 patients per day, and the room may be used 7 days a week

• Each patient may have between 1 and 5 films

SO HOW DO WE ESTIMATE W ?


Workload (II)

• Assume an average of 50 mAs per film, 3 films per patient

• Thus W = 50 mAs x 3 films x 50 patients x 7 days

= 52,500 mAs per week = 875 mA-min per week• We could also assume that all this work

is performed at 100 kVp


Examples of Workloads in Current Use (NCRP 147)

Weekly Workload (W) mA-min at:

100 kVp 125 kVp 150 kVp

General Radiography 1,000 400 200

Fluoroscopy (including spot films) 750 300 150

Chiropractic 1,200 500 250

Mammography 700 at 30 kVp (1,500 for breastscreening)

Dental6 at 70 kVp (conventional intra-oral

films)

More realistic values include CT: see ref. Simpkin (1997)


Workload - CT

• CT workloads are best calculated from local knowledge

• Remember that new spiral CT units, or multi-slice CT, could have higher workloads

• A typical CT workload is about 28,000 mA-min per week


Tube Leakage

• All X Ray tubes have some radiation leakage - there is only 2-3 mm lead in the housing

• Leakage is limited in most countries to 1 mGy.hr-1 @ 1 meter, so this can be used as the actual leakage value for shielding calculations

• Leakage also depends on the maximum rated tube current, which is about 3-5 mA @ 150 kVp for most radiographic X Ray tubes


Radiation Shielding Parameters


Room Shielding - Multiple X Ray Tubes

• Some rooms will be fitted with more than one X Ray tube (maybe a ceiling-mounted tube, and a floor-mounted tube)

• Shielding calculations MUST consider the TOTAL radiation dose from the two tubes



Topic 3: Barriers and protective devices



Shielding - Construction I

Materials available:• lead (sheet, composite, vinyl)

• brick

• gypsum or baryte plasterboard

• concrete block

• lead glass/acrylic


Shielding - Construction Problems

Some problems with shielding materials:

• Brick walls - mortar joints

• Use of lead sheets nailed to timber frame

• Lead inadequately bonded to backing

• Joins between sheets with no overlap

• Use of hollow core brick or block

• Use of plate glass where lead glass specified


Problems in shielding - Brick Walls & Mortar Joints

• Bricks should be solid and not hollow

• Bricks have very variable X Ray attenuation

• Mortar is less attenuating than brick

• Mortar is often not applied across the full thickness of the brick


Problems in shielding - Lead inadequately bonded to backing

• Lead must be fully glued (bonded) to a backing such as wood or wallboard

• If the lead is not properly bonded, it will possibly peel off after a few years

• Not all glues are suitable for lead (oxidization of the lead surface)


Problems in shielding - Joins between sheets with no overlap

• There must be 10 - 15 mm overlap between adjoining sheets of lead

• Without an overlap, there may be relatively large gaps for the radiation to pass through

• Corners are a particular problem


Problems in shielding - Use of plate glass

• Plate glass (without lead of specified quantity as used in windows, but thicker) is not approved as a shielding material

• The radiation attenuation of plate glass is variable and not predictable

• Lead glass or lead Perspex must be used for windows


Radiation Shielding - Construction II

• Continuity and integrity of shielding very important

• Problem areas:• joins

• penetrations in walls and floor

• window frames

• doors and frames


Penetrations

• “Penetrations” means any hole cut into the lead for cables, electrical connectors, pipes etc.

• Unless the penetration is small (~2-3 mm), there must be additional lead over the hole, usually on the other side of the wall

• Nails and screws used to fix bonded lead sheet to a wall do not require covering


Window frames

• The lead sheet fixed to a wall must overlap any lead glass window fitted

• It is common to find a gap of up to 5 cm, which is unacceptable


Shielding of Doors and Frames


Shielding - Verification I

• Verification should be mandatory

• Two choices - visual or measurement

• Visual check must be performed before shielding covered - the actual lead thickness can be measured easily

• Radiation measurement necessary for window and door frames etc.

• Measurement for walls very slow


Shielding Testing


Records

• It is very important to keep records of shielding calculations, as well as details of inspections and corrective action taken to fix faults in the shielding

• In 5 years time, it might not be possible to find anyone who remembers what was done!


Summary

• The design of shielding for an X Ray room is a relatively complex task, but can be simplified by the use of some standard assumptions

• Record keeping is essential to ensure traceability and constant improvement of shielding according to both practice and equipment modification


Practical Questions for shielding calculationsRadiography

• Find the necessary amount of lead to protect personnel sitting in an adjacent office, for a dose constraint of 0.30mSv per year [1], taking into account the following assumptions and data. You must calculate and consider all 3 types of radiation (Primary, Scattered and Leakage).

IAEA

ASSUMPTIONS DATA KNOWN

Leakage dose is 1mGy per hour at a distance of 1.0 meter. (Worst Case Scenario)

Workload of 100 patients per day

3 films per patient

20mAs per film

7 days per week

Scatter fraction (Sf) from patient is 0.0025 at 125kVp and 135 degrees, (Worst Case Scenario)

Unit maximum current 2.0mA

Average dose per unit workload of 4.72mGy [2,3] per week at 1 meter distance.

Field size of 1000cm². Relate this to the standard 400cm²

Focus to skin distance 80cm. (Needed to calculate

the scattered dose) Use factor U=0.25.

Occupancy factor T=1.0.

Critical distance d=2.5 m


Solution (a) 1/8

Step1. Calculate the Workload (W)

in mAmin/week

W = (mAs/film) x (patients/day) x (films/patient) x (days/week)

= 20 x 100 x 3 x 7

= 42,000 mAs/week

= 700mAmin/week


Solution (a) 2/8

Step 2. Calculate the primary dose per week at 1 meter

P1 = (mAmin/week) x (Average dose per unit workload in

mGy/week)

= 700 x 4.72

= 3,304mGy/week


Solution (a) 3/8

Step 3. Calculate the Primary dose per week at the critical distance.

P = (P1 x U x T)/d²

= (3,304 x 0.25 x 1)/2.5²

= 132.16mGy/week


Solution (a) 4/8

Step 4. Calculation of scattered dose per week at the critical distance

S = (P1 x T x Sf x 1000)/(400 x d x 0.8²)

= (3304 x 1 x0.0025 x 1000)/(400 x 2.52 x 0.8²)

= 5.2mGy/week


Solution (a) 5/8

Step 5. Calculation of the leakage radiation per week at the critical distance

a) Tube on time = (W)/(Tube current)

for leak. calc. = 700/(2 x 60)

= 5.83 hours

b) Leakage dose (L) = (Time x U x T)/d²

= (5.83 x 1 x 0.25)/2.5²

= 0.233mGy/week


Solution (a) 6/8

Step 6. Add the three sources of radiation together

Total dose = P+S+L

= (132.16 + 5.2 + 0.233)

= 137.59mGy/week


Solution (a) 7/8

Step 7. Calculate the required attenuation

If the required attenuation is 0.006mGy/week (0.3mSv/year¹), then the required attenuation () would be:

= 0.006/137.59

= 0.000044


Solution (a) 8/8 Shielding Calculation From graph below a lead of thickness approximately 2.6mm is necessary.

1 2 3 4 5 6 7 8 mm

105

104

103

102

10 Lead Required

Reduction factor

50 75 kV 100 150 200 kV 250

300 kV


Question for shielding calculationComputer Tomography

A CT scanner is placed in a room as in figure 1. The height of the ceiling is 4.0m. The walls are made of lightweight concrete (1840kg/m³), with a minimum thickness of 110mm. The scanner isocentre is located 0.9m above floor level. Isodose curves have been provided for a 120kVp, 250mAs, 10mm slice on a 320mm diameter PMMA body phantom and a 350mAs 10mm slice on a 160mm head phantom.


Figure 1

IAEA

ASSUMPTIONS DATA KNOWN

1 The scatter dose per mAs from a 10mm slice through the head is half that from a slice through the body.

140 body examinations/week.

Average body examination comprises of 24 slices of 10mm width with a table feed of 14mm.

2 100 head examinations/week.

Average head examination consists of 10 slices of 10 mm width and 5 slices of 5mm width.


Solution (b) 1/9

Step1. Calculate the total workload per week

a)Number of 250mAs body slices per week = 140 body examinations x 22 slices per examination= 3080 slices per weekb)Number of 350mAs, 10mm equivalent head slices

per week.= (10 slices x 100) + [(5 slices x 100) x (5/10)]= 1250The division 5/10 was needed to normalize the five slices of 5mm to the equivalent of 10mm.


Solution (b) 2/9

c)= Equivalent number of 250mAs, 10mm head examinations.

= (1250/2) x (350/250)

(See assumption 1)

= 875 slices per week for head examination

Hence:

The total workload per week

= 3080 + 875

= 3955 body slices of 250mAs and 10mm width.


Solution (b) 3/9

Step 2. Calculation of the transmission factor, regarding wall B

From figure 1, distance from the isocentre is 2.5m and the dose contour is 1.5Gy

Hence:The dose per week from 3955 slices is equal to= 3955 x 1.5Gy= 5933Gy

The area behind wall B is an office, where the occupancy is estimated to be 100%

The required transmission for that barrier, BB = 0.3/(5.933mGy x 1 x 52)B = 3.2x10-3


Solution (b) 4/9

Step 3. Calculations of coefficients , and , based on Archers et al (1997) formula, interpolated for 10kVp

With reference to table 4.6 of BIR (2000)¹, the following coefficients have been calculated for lead material by interpolation for 120kVp:

= 2.2878 = 9.3780 = 0.7664


Solution (b) 5/9

Using the following formula the thickness of the material required (lead), x, to provide the desired transmission can be calculated

)/(1

)/ln

1 1

B

x

5703.01

5991.316)/(1 B

0991.5)/(1


Solution (b) 6/9

x = 0.5703 x ln(316.5991/5.0991)

x = 2.35 mm of lead


Solution (b) 7/9

With reference to table 4.6 of BIR (2000)¹, the following coefficients have been calculated for concrete by interpolation for 120kVp:

= 0.0359

= 0.0696

= 0.7302


Solution (b) 8/9

Using the following formula the thickness of the material required (concrete), x, to provide the desired transmission can be calculated

)/(1

)/ln

1 1

B

x

3805191

4338.314)/(1 B

9338.2)/(1


Solution (b) 9/9

x = 38.0519 x ln(314.4338/2.9338)

x =177.8727mm of concrete


References:

1. Radiation Shielding for Diagnostic X-Rays (2000), Ed. D.G. Sutton and J.R. Williams, Pub. BIR.

2. IAEA Training Material, Diagnostic Radiology, L.12.1, slide 16

3. National Council on Radiation Protection and Measurements “Structural Shielding Design for Medical X Rays Imaging Facilities” 2004 (NCRP 147)

4. Diagnostic X-ray shielding design, B. R. Archer, AAPM Monograph The expanding role of medical physics in diagnostic radiology, 1997


Where to Get More Information (I)

• New concepts for Radiation Shielding of Medical Diagnostic X Ray Facilities,

D. J. Simpkin, AAPM Monograph The expanding role of medical physics in diagnostic radiology, 1997

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