NCRP #147 Examples - Radiation Shielding for Medical Instalations

Example 5.1: Cath Lab Wall• N=25 patients/wk• d=4 m• Uncontrolled, fully-occupiedarea– P = .02 mGy/wk, T = 1• Let’s use Model 2 with theCardiac Angio workloaddistributionConservatively assume30.5 cm diameterimage intensifier, 180°– K 1 sec = Kerma per patient at 1 m scatter, all primary= 38 3.8 mGy/patient tfrom Table 4.7 47radiation stopped byimage receptor3

Example 5.1: Cath Lab Wall• Total weekly unshielded kerma from thissecondary radiation is then• The required barrier transmission is• On Figure C.2. look up required lead thicknessthat gives this transmission4

Figure C.2Example 5.1Cath Lab WallB=0.0034x=1.3 mm Pb5

Example 5.1: XRAYBARR Output6

Example 5.1: XRAYBARR OutputSimply illustrates self-consistency,since the data in Table 4.7 are basedon XRAYBARR results7

Example 5.2: Dedicated Chest Unit• Need to shield an uncontrolled,fully-occupied area: P = .02mGy/wk, T = 1, in the primarybeam (U=1)• N=300 patients/wk• d=3 m• Let’s use Model 2 with the ChestRoom workload distribution– K 1 pri = Kerma per patient at 1 m =1.2 mGy/patient from Table 4.58

Example 5.2: Chest Unit itP Primary Barrier• Total weekly unshielded kerma from thisprimary radiation is then• The required barrier transmission is• Look up required lead thickness that gives thistransmission on Figure B.2.9

Fig B.2 Example 5.2Chest UnitPrimary WallB=0.0005Total primarybarrier thicknessx=2.2 22mm Pb10

Example 5.2: Chest Unit Primary Barrier• Consider the shielding afforded in the chestunit’s image receptor and its structure– x pre = 0.85 mm lead from Table 4.6• Therefore, the thickness of the lead in thewallneedstobe22mm2.2 – 085mm=140.85 1.4mm lead– Note: this x pre value is a single conservativevalue from the list in Dixon & Simpkin (1998)and not exact for the Chest Room workload11

Example 5.2: XRAYBARR Output“More correct”12

Example 5.2: Dedicated Chest Unit• Now shield the secondary wall(U=0), P/T=0.02 mGy/wk• N=300 patients/wk• d=2.1 m• Let’s use Model 2 with theChest Room workloaddistribution– K 1 sec = Secondary kerma perpatient at 1 m = 2.7 ×10–3 fromTable 4.7xAssume F=1,535cm 2 (at 1.83 mSID),90° scatter13

Example 5.2: Chest Unit itS Secondary Barrier• Total weekly unshielded kerma from thissecondary radiation is then• The required barrier transmission is• Look up required lead thickness that gives thistransmission on Figure C.2.14

Figure C.2B=0.11Example 5.2 ChestRoom Sec Wallx=0.42 mm Pb15

Example 5.3 The Radiographic Room16

Example 5.3 The Radiographic Room• Let N = 125 patients/week• Assume the workload distribution and usefactors follow that from the RadiographicRoom from the AAPM-TG9 survey(Simpkin, 1996)17

Example 5.3.1.1 Floor Under Table• Initially, only consider the primaryradiation from the overhead x-ray tubeposition• Assume the tube usage follows the RadRoom/floor other barriers workloaddistribution• The survey found U = 0.89, but let U =1 to be conservative• From Table 4.5, kerma aper patient t= 5.2mGy patient –1 at 1 m for thisdistributionx18

Example 5.3.1.1 Floor Under Table• Then unshielded kerma is• To shield to P/T = 0.0202 mGy/wk, needtransmission• Fig B.3 shows the primary transmissionthrough concrete, requiring 107 mmconcretex19

Figure B.3Example 5.3.1.1Rad Room FloorUnder TableB=5.1×10 -4x =107 mmconcrete20

Example 5.3.1.1 Floor Under Table• The image receptor in the radiographic tableand its support provide attenuation,– x pre = 72 mm concrete (Table 4.6)• Then the required floor thickness = 107 –72 = 35 mm concrete.21

Example 5.3.1.1 Floor Under Table• Now lets try an NT/Pd 2 calculationfor the floor• This will include all the othersources of radiation i in the room(e.g. scatter from exposuresagainst the chest board, cross tablelateral exposures, etc.)x22

Figure 4.6bExample53115.3.1.1Rad Room FloorUnder Tablex =37 mmconcreteNT/Pd 2 = 372 mGy –1 m –223

Example 5.3.1.1 Floor Under TableNTPD2 output24

Example 5.3.1.1 Floor Under Table• So a primary-only calculation l with U=1requires 35 mm concrete, while an NT/Pd 2calculation l (which h includes all othersources) requires 37 mm concrete.• (As expected, the vast majority of exposurebelow the table is due to the primary overtablebeam, and shielding against just theprimary beam is sufficient.)25

Example 5.3.1.2 Floor Not Under Table• Now let’s look at areas beneathroom, away from table. This willreceive secondary radiation only.• Let d = 3 m (vertical distancefrom patient on table)• Consider secondary radiationfrom the Rad Room (all barriers)workload distribution• 90° scatter• From Table 4.7, secondary kermaper patient = 3.4 ×10 -2 mGypatient–1 at 1 m for thisdistributionx26

Example 5.3.1.2 Floor Not Under Table• Then unshielded kerma is• To shield to P/T = 002 0.02 mGy/wk, need transmissioni• Fig C.3 shows the secondary transmission throughconcrete, requiring 33 mm concrete27

Figure C.3 Example 5.3.1.2B=4.3×10 -2 Rad Room FloorNot Under Tablex =33 mmconcrete28

Example 5.3.5 Cross-Table Lateral Wall2 m29

Example535Cross-Table 5.3.5 Lateral Wall• This wall needs to shield against– Primary radiation from cross-table exposures (but withU only = 0.09) following the Rad Room/floor otherbarriers workload distribution. We wish to account forimage receptor attenuation.– Secondary radiation for exposures directed down attable (& chest exposures) following the RadRoom/floor other barriers workload distribution• We can only guess at a solution using Model 2(the kerma per patient scheme) since there is noway to combine the two shielded doses with onetransmission curve30

Example535Cross-Table 5.3.5 Lateral Wall• But Model 3 (NT/Pd 2 scheme) should fit thiswell, since the NT/Pd 2 curves account for allsources in the room. Using d = 2.8m fromcross-table primary x-ray tube location,1NT 125 patients wk− × 1==7972 − 12Pd 0.02 mGy wk × (2.8m)mGy• Accounting for attenuation in the grid andimage receptor, from Fig 4.5b, 0.83 mm leadis required in the wall.−1m231

Figure 4.5bExample 5.3.5Rad RoomCross-TableLateral Wallx =0.83 mmleadNT/Pd 2 = 797 mGy –1 m –232

Example535Cross-Table 5.3.5 Lateral WallSetup XRAYBARR with 2x-ray sources33

Example535Cross-Table 5.3.5 Lateral WallNTPD2 output34

XRAYBARR• Performs first-principle extension of NCRP-49• Microsoft application for Windows written inVisual Basic• Freely-distributed shareware atwww.geocities.com/djsimpkin• Comes as a ~3MB zip file. Save zip on localdrive in a new folder. Unzip constituent ~2dozen files to this new folder. Double-clickSETUP.EXE to install program onto yoursystem.35

NTPd2• Performs calculations and lookup followingthe NT/Pd 2 model.• Microsoft application for Windows writtenin Visual Basic• Freely-distributed shareware• Just 4 files (executable, 2 help files, & iconfile). Keep them in a separate folder and noinstallation needed.36