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IAEAInternational Atomic Energy Agency
RADIATION PROTECTION INDIAGNOSTIC AND
INTERVENTIONAL RADIOLOGY
Part 12.1 : Shielding and X-ray room design
Practical exercise
IAEA Training Material on Radiation Protection in Diagnostic and Interventional Radiology
IAEA 12.1 : Shielding and X-ray room design 2
Overview / Objectives
• Subject matter : design and shielding calculation of a diagnostic radiology department
• Step by step procedure to be followed
• Interpretation of results
IAEAInternational Atomic Energy Agency
Part 12.1 : Shielding and X-ray room design
Design and shielding calculation of a diagnostic radiology department
Practical exercise
IAEA Training Material on Radiation Protection in Diagnostic and Interventional Radiology
IAEA 12.1 : Shielding and X-ray room design 4
Radiation Shielding - Calculation
• Based on NCRP 147
• Assumptions used are very pessimistic, so overshielding is the result
• Various computer programs are available, giving shielding in thickness of various materials
IAEA 12.1 : Shielding and X-ray room design 5
Shielding Calculation - Principle
• We need, at each calculation point, the dose per week per mA-min, modified for U and T, and corrected for distance
• The required attenuation is simply the ratio of the design dose to the actual dose
• Tables or calculations can be used to estimate the shielding required
IAEA 12.1 : Shielding and X-ray room design 6
Shielding Calculation - Detail
Dose per week - primary
• Data being used for NCRP 147 suggests that for :• 100 kVp, dose/unit workload = 4.72 mGy/mA-
min @ 1 meter
• 125 kVp, dose/unit workload = 7.17 mGy/mA-min @ 1 meter
IAEA 12.1 : Shielding and X-ray room design 7
Shielding Calculation - Detail
• Thus if the workload were 500 mA-min/week @ 100 kVp, the primary dose would be :
500 x 4.72 mGy/week @ 1 meter = 2360 mGy/ week
IAEA 12.1 : Shielding and X-ray room design 8
Sample Shielding Calculation
• Using a typical x-ray room, we will calculate the total dose per week at one point
Office
2.5 m
Calculation Point
IAEA 12.1 : Shielding and X-ray room design 9
Shielding Calculation - Primary
If U = 0.25, and T = 1 (an office) and the distance from the x-ray tube is 2.5 m, then the actual primary dose per week is :
(2360 x 0.25 x 1)/2.52 = 94.4 mGy/week
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Shielding Calculation - Scatter
• Scatter can be assumed to be a certain fraction of the primary dose at the patient
• We can use the primary dose from the previous calculation, but must modify it to the shorter distance from the tube to the patient (FSD, usually about 80 cm)
• The “scatter fraction” depends on scattering angle and kVp, but is a maximum of about 0.0025 (125 kVp @ 135 degrees)
IAEA 12.1 : Shielding and X-ray room design 11
Shielding Calculation - Scatter
• Scatter also depends on the field size is simply related to a “standard” field size of 400 cm2 - we will use 1000 cm2 for our field
• Thus the worst case scatter dose (modified only for distance and T) is :
(2360 x 1 x 0.0025 x 1000)
-------------------------------- = 3.7 mGy
(400 x 2.52 x 0.82)
IAEA 12.1 : Shielding and X-ray room design 12
Shielding Calculation - Leakage
• Leakage can be assumed to be at the maximum allowable (1 mGy.hr-1 @ 1 meter)
• We need to know how many hours per week the tube is used
• This can be taken from the workload W, and the maximum continuous tube current
• Leakage is also modified for T and distance
IAEA 12.1 : Shielding and X-ray room design 13
Shielding Calculation - Leakage
• For example: if W = 300 mA-min per week and the maximum continuous current is 2 mA, the “tube on” time for leakage calculation
= 300/(2 x 60) hours
= 2.5 hours
• Thus the leakage = 2.5 x 1 x 0.25 / 2.52 mGy
= 0.10 mGy
IAEA 12.1 : Shielding and X-ray room design 14
Shielding Calculation - Total Dose
• Therefore the total dose at our calculation point:
= (94.4 + 3.7 + 0.1) = 99.2 mGy / week
• If the design dose = 0.01 mGy / week
then the required attenuation
= 0.01/99.2
= 0.0001
IAEA 12.1 : Shielding and X-ray room design 15
Shielding Calculation - Lead Required
• From tables or graphs of lead shielding, we can find that the necessary amount of lead is 2.5 mm
• There are tables or calculation formula for lead, concrete and steel at least
• The process must now be repeated for every other calculation point and barrier
IAEA 12.1 : Shielding and X-ray room design 16
Shielding Calculation
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
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Radiation Shielding Parameters
IAEA 12.1 : Shielding and X-ray room design 18
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 all tubes
IAEA 12.1 : Shielding and X-ray room design 19
CT room design
• General criteria:• Large room with enough space for:
• CT scanner • Auxiliary devices (contrast media injector, emergency bed and
equipment, disposable material containers, etc) • 2 dressing-rooms
• Other spaces required: • Console room with large window large enough to see the patient
all the time• Patient preparation room • Patient waiting area• Report room (with secondary imaging workstation) • Film printer or laser film printer area
IAEA 12.1 : Shielding and X-ray room design 20
Room shielding
• Workload
• Protective barriers
• Protective clothing
2.5 Gy/1000 mAs-scan
Typical scatter dose distribution around a CT scanner
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• Workload (W): The weekly workload is usually expressed in milliampere minutes.• The workload for a CT is usually very high
• Example:
6 working day/week, 40 patients/day, 40 slices/patient,
200 mAs/slice, 120 kV
• Primary beam is fully intercepted by the detector assembly. Barriers are interested only by scattered radiation
mAmin/week32000W 60200.40.40.6
Protective barriers
IAEA 12.1 : Shielding and X-ray room design 22
Scattered radiation
Typical maximum scatter radiation around a CT : Sct= 2.5 Gy/mAmin-Scan @ 1 meter and 120 kV.
This quantity may be adopted for the calculation of protective barriers
The thickness S is otained from the attenuation curve for the appropriate attenuation material assuming scattered photons with the same penetrating capability of those of useful beam
Example: 120 kV; P = 0.04 mSv/week,
dsec= 3 m, W= 32000 mAmin/week, T= 1
Requires 1.2 mm of lead or 130 mm of concrete
TWS)(dP
uX ct
2secK
0.0045(1)(0.0025)(32000)(3.0)0.04
uX
2
K
Secondary barrierdsec
Computation of secondary protective barriers
IAEA 12.1 : Shielding and X-ray room design 23
Where to Get More Information
• National Council on Radiation Protection and Measurements “Structural Shielding Design for Medical X Rays Imaging Facilities” 2004 (NCRP 147)
