1. IAEA International Atomic Energy Agency RADIATION PROTECTION IN DIAGNOSTIC 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

2. IAEA 12.1 : Shielding and X-ray room design2 Overview / Objectives Subject matter : design and shielding calculation of a diagnostic radiology department Step by step procedure to be followed Interpretation of results

3. IAEA International 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

4. IAEA 12.1 : Shielding and X-ray room design4 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

5. IAEA 12.1 : Shielding and X-ray room design5 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

6. IAEA 12.1 : Shielding and X-ray room design6 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

7. IAEA 12.1 : Shielding and X-ray room design7 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

8. IAEA 12.1 : Shielding and X-ray room design8 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

9. IAEA 12.1 : Shielding and X-ray room design9 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)/2.5 2 = 94.4 mGy/week

10. IAEA 12.1 : Shielding and X-ray room design10 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)

11. IAEA 12.1 : Shielding and X-ray room design11 Shielding Calculation - Scatter Scatter also depends on the field size is simply related to a “standard” field size of 400 cm 2 - we will use 1000 cm 2 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 2.5 2 x 0.8 2 )

12. IAEA 12.1 : Shielding and X-ray room design12 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

13. IAEA 12.1 : Shielding and X-ray room design13 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 / 2.5 2 mGy = 0.10 mGy

14. IAEA 12.1 : Shielding and X-ray room design14 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

15. IAEA 12.1 : Shielding and X-ray room design15 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

16. IAEA 12.1 : Shielding and X-ray room design16 Shielding Calculation 1 2 3 4 5 6 7 8 mm 10 5 10 4 10 3 10 2 10 Lead Required Reduction factor 5075 kV100150 200 kV 250 300 kV

17. IAEA 12.1 : Shielding and X-ray room design17 Radiation Shielding Parameters

18. IAEA 12.1 : Shielding and X-ray room design18 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

19. IAEA 12.1 : Shielding and X-ray room design19 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

20. IAEA 12.1 : Shielding and X-ray room design20 Room shielding Workload Protective barriers Protective clothing 2.5  Gy/1000 mAs-scan Typical scatter dose distribution around a CT scanner

21. IAEA 12.1 : Shielding and X-ray room design21 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 60 200.40..6  Protective barriers

22. IAEA 12.1 : Shielding and X-ray room design22 Scattered radiation Typical maximum scatter radiation around a CT : S ct = 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, d sec = 3 m, W= 32000 mAmin/week, T= 1 Requires 1.2 mm of lead or 130 mm of concrete TWS )(dP uX ct 2 sec K  Secondary barrier d sec Computation of secondary protective barriers

23. IAEA 12.1 : Shielding and X-ray room design23 Where to Get More Information National Council on Radiation Protection and Measurements “Structural Shielding Design for Medical X Rays Imaging Facilities” 2004 (NCRP 147)