12. Shielding and X-ray facility design

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Transcript 12. Shielding and X-ray facility design

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 and 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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12.1 : Shielding and X-ray room design
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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, Structural Shielding
Design for Medical X-Ray Imaging Facilities.
• Assumptions used are conservative, so
over-shielding is typical
• Computer software is available, giving
shielding in thickness of various materials
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12.1 : Shielding and X-ray room design
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Radiation Shielding– Calculation
• NCRP 147
• Provides methods to calculate the thickness of
shielding needed to decrease the kerma in a
shielded area to P/T
• P = the weekly permitted kerma in the occupied area
• T = the occupancy factor, the average fraction of time
that the maximally exposed individual is present while
the x-ray beam is on.
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Radiation Shielding– Calculation
• NCRP 147 provides
• K1, the average kerma expected for a patient
procedure
• 1 m from the x-ray tube in the primary beam (Table
4.5), and,
• 1 m from the patient from secondary (scatter and
leakage) radiation (Table 4.7)
• Then, the unshielded weekly kerma, K0 at
distance d for N patient procedures per week is
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K1 N
K0  2
d
12.1 : Shielding and X-ray room design
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Radiation Shielding– Calculation
• Then the acceptable thickness, x, of a
shielding barrier will be that which provides
transmission, B, not in excess of
P / T (P / T ) d 2
B( x) 

K0
K1 N

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
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Radiation Shielding– Cardiac Cath
• Consider a wall in a cardiac cath lab, with
d=4 m, P = 0.02 mGy wk-1, T=1, 90° scatter,
N=25 patients wk-1
• From NCRP 147, Table 4.7, for a cardiac
cath lab: K1 = 2.7 mGy patient-1
• Total unshielded weekly kerma is then
2.7 mGy pat 1  25 pat wk 1
1
K0 

4
.
22
mGy
wk
( 4m) 2
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Radiation Shielding– Cardiac Cath
• Required transmission is
P / T 0.02 mGy wk 1
B

 0.0047
1
K0
4.22 mGy wk
• Look at transmission curve for secondary
radiation from Cardiac Cath Lab (NCRP 147,
Fig. C.2) Requires 1.2 mm Pb.
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Shielding Calculation– Cardiac Cath
B=0.0047
x =1.2 mm Pb
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Radiation Shielding– R-F Room
• For general radiographic or fluoroscopic rooms,
the x-ray tube(s) may generate primary beams
directed at a number of barriers, as well as scatter
and leakage radiations from these beams
• NCRP 147 § 4.5 specifies thicknesses required for
primary and secondary barriers in these rooms as
a function of NT/(Pd2)
• Note that NCRP 147 § 4.5 accounts for all primary
and secondary radiation sources in the room.
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Radiation Shielding– R-F Room
• Consider the floor in a general
radiographic room.
• Let N = 125 patients wk-1, d =
3.8 m, P = 0.02 mGy wk-1,
T = 1.
• Then NT/Pd2 = 432 mGy-1 m-2 ,
which, from Fig. 4.6a, requires
110 mm of concrete.
• This ignores attenuation in the
image receptor and its
supports.
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12.1 : Shielding and X-ray room design
3.8 m
Fully occupied
uncontrolled
area
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Shielding Calculation– R-F Room
NCRP 147
Fig 4.6a
NT/(Pd2) = 432
mGy-1 m-2,
requiring a floor
thickness of 110
mm standard
density concrete
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Radiation Shielding– CT Scanner
• Computed tomography (CT) scanners will
generate scatter and leakage radiation to the
environs of the room.
• For every 10 mm of x-ray beam width, the
intensity of this secondary radiation at 1 m is
a fraction, , of the peripheral CTDI100.
• Head scans:  = 910-5
• Body scans:  = 310-4
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Radiation Shielding– CT Scanner
• Since the product of the CTDI used for each
patient and the thickness of the patient imaged is
the Dose Length Product, DLP, the unshielded
kerma at 1 m from each patient is:
K
1
CT head
(mGy)   head  DLPhead
1
K CT
body ( mGy)  1.2   body  DLPbody
• The DLP values can be read off of the scanner, or
from European Commission Guidelines:
• DLP = 1,200 mGy cm for heads
• DLP = 550 mGy cm for bodies
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Radiation Shielding– CT Scanner
• Consider a barrier in a CT scanner room,
with P = 0.02 mGy wk-1, T = 1, d=3 m, 200
patients wk-1 (125 bodies + 75 heads), with
40% of patients having scans both pre- and
post-contrast medium injection
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Radiation Shielding– CT Scanner
• The unshielded kerma per head patient at 1 m is:
K
1
CT head
5
1
 1.4  (9 10 cm )  1200 mGy cm  0.15 mGy
• The unshielded kerma per body patient at 1 m is:
1
4
1
K CT

1
.
4

1
.
2

(
3

10
cm
)  550 mGy cm  0.28 mGy
body
• So total unshielded weekly kerma at 1 m is
mGy
K 10  75 head pat  0.15 head
pat
mGy
 125 body pat  0.28 body
pat
 46.3 mGy
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Radiation Shielding– CT Scanner
• The unshielded weekly kerma at 3 m is
46.3 mGy
K0 
 5.14 mGy
2
(3 m)
• The transmission required in this barrier is
0.02 mGy
B
 3.9  10 3
5.14 mGy
• which, from NCRP 147 Figs. A2 and A3, at 140
kVp, is achieved by
• 1.52 mm Pb, or,
• 150 mm standard density concrete
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Shielding Calculation– CT Scanner
1E+0
8
6
Transmission of CT Scanner
Secondary Radiation Through Pb
4
2
140 kVp
1E-1
8
6
Fitting parameters to Equation B.2
Transmission
4
kVp (mm-1(mm-1
120
2.246
5.73
0.547
140
2.009
3.99
0.342
120 kVp
2
1E-2
8
6
4
2
1E-3
8
6
4
2
1E-4
0
0.5
1
1.5
2
2.5
3
Lead Thickness (mm)
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Shielding Calculation– CT Scanner
1E+0
Transmission of CT Scanner Secondary
Radiation Through Concrete
8
6
4
2
1E-1
Fitting parameters to Equation B.2
8
6
kVp (mm-1(mm-1
120 0.0383 0.0142 0.658
140 0.0336 0.0122 0.519
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Transmission
2
1E-2
8
6
4
2
140 kVp
1E-3
8
6
4
2
1E-4
120 kVp
8
6
4
2
1E-5
0
50
100
150
200
250
300
Concrete Thickness (mm)
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Radiation Shielding– CT Scanner
• Note that, should the ceiling require added
shielding, wall shielding should be extended
up to the ceiling in order to cover the gap
above the normal shielding height.
Additional Pb required on ceiling
ADD Pb to wall
above 2.1 m
CT Scanner
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Where to Get More Information
• National Council on Radiation Protection
and Measurements, Report 147, Structural
Shielding Design for Medical X-Ray Imaging
Facilities, NCRP, Bethesda, MD. 2004
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