RADIATION PROTECTION IN DIAGNOSTIC RADIOLOGY by zhangyun

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									 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




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   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



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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


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 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


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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




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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



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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)

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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)



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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


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 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


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 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
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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


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Shielding Calculation

 Reduction factor
        50       75 kV       100 150           200 kV
  105
                                                                  250
  104
                                                                       300 kV
  103

  102

  10                                                              Lead Required

             1     2     3    4      5           6           7             8 mm
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Radiation Shielding Parameters




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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



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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

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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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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=                        = 32000 mAmin/week
                  6 . 40 . 40 . 200
                         60

• Primary beam is fully intercepted by the detector assembly.
  Barriers are interested only by scattered radiation



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Computation of secondary protective barriers

                                                                                      P(dsec )2
 Scattered radiation
 Typical maximum scatter radiation around a CT :
                                                                         KuX =         WSct T
 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
                                                                        Secondary barrier
 assuming scattered photons with the same                       d sec
 penetrating capability of those of useful beam

 Example: 120 kV; P = 0.04 mSv/week,
 dsec= 3 m, W= 32000 mAmin/week, T= 1
                   0.04 (3.0) 2
     KuX =     (32000) (0.0025) (1)   = 0.0045

 Requires 1.2 mm of lead or 130 mm of concrete


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Where to Get More Information

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




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