Transcript 89
IAEA Training Material on Radiation Protection in Diagnostic and Interventional Radiology RADIATION PROTECTION IN DIAGNOSTIC AND INTERVENTIONAL RADIOLOGY Part 12.1 : Shielding and X-ray room design Practical exercise IAEA International Atomic Energy Agency Overview / Objectives • Subject matter : design and shielding calculation of a diagnostic radiology department • Step by step procedure to be followed • Interpretation of results IAEA 12.1 : Shielding and X-ray room design 2 IAEA Training Material on Radiation Protection in Diagnostic and Interventional Radiology Part 12.1 : Shielding and X-ray room design Design and shielding calculation of a diagnostic radiology department Practical exercise IAEA International Atomic Energy Agency 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 4 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 5 Shielding Calculation - Detail Dose per week - primary • Data being used for NCRP 147 suggests that for : • 100 kVp, dose/unit workload = 4.72 mGy/mAmin @ 1 meter • 125 kVp, dose/unit workload = 7.17 mGy/mAmin @ 1 meter IAEA 12.1 : Shielding and X-ray room design 6 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 7 Sample Shielding Calculation • Using a typical x-ray room, we will calculate the total dose per week at one point Office Calculation Point 2.5 m IAEA 12.1 : Shielding and X-ray room design 8 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 IAEA 12.1 : Shielding and X-ray room design 9 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 10 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 11 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 12 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 13 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 14 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 15 Shielding Calculation Reduction factor 105 50 75 kV 100 150 200 kV 250 104 300 kV 103 102 10 Lead Required 1 IAEA 2 3 4 5 6 7 12.1 : Shielding and X-ray room design 8 mm 16 Radiation Shielding Parameters IAEA 12.1 : Shielding and X-ray room design 17 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 18 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 19 Room shielding • Workload • Protective barriers • Protective clothing 2.5 Gy/1000 mAs-scan Typical scatter dose distribution around a CT scanner IAEA 12.1 : Shielding and X-ray room design 20 Protective barriers • 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 W= 6 . 40 . 40 . 200 60 = 32000 mAmin/week • Primary beam is fully intercepted by the detector assembly. Barriers are interested only by scattered radiation IAEA 12.1 : Shielding and X-ray room design 21 Computation of secondary protective barriers KuX = 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 dsec P (dsec )2 WSctT Secondary barrier Example: 120 kV; P = 0.04 mSv/week, dsec= 3 m, W= 32000 mAmin/week, T= 1 KuX = 0.04 (3.0) 2 (32000) (0.0025) (1) = 0.0045 Requires 1.2 mm of lead or 130 mm of concrete IAEA 12.1 : Shielding and X-ray room design 22 Where to Get More Information • National Council on Radiation Protection and Measurements “Structural Shielding Design for Medical X Rays Imaging Facilities” 2004 (NCRP 147) IAEA 12.1 : Shielding and X-ray room design 23