Neutron Door Design

Neutron Door Design
Robert J Barish, Ph.D.

Alternate form vendor information (Siemens)

Mazed Vault

d 1
S 0
d 2 S 1

Starting gp point for door calcs. (easy way)
Varian 10 MV 0.00004 Sv/Gy (at 1 m)
Example: 500 Gy x 0.00004 Sv/Gy = 20 mSv
(50,000000 cGy x 0.0000400004 cSv/cGy = 2 cSv = 20 mSv)
But consider 50% IMRT, with “IMRT factor” = 4
250 Gy x 0.00004 Sv/Gy = 10 mSv
250 Gy x 0.00004 Sv/Gy x 4 = 40 mSv
Total: 50 mSv (at 1 m)

Kersey Method
Varian 0.00004 Sv/Gy (at 1 m)
A “typical” maze length is about 6 m. Its height is 3 m and width is 2 m
A 6
2
S
Area = 6 m 2 = S 1
A typical gap between the maze and the far wall is 2.5 m and room
height is also 3 m . Area=75m 7.5 2 =S
0
Kersey’s formula gives the neutron dose at the door as the neutron dose
at the isocenter x (1/d 1 ) 2 x (S 0 / S 1 ) x 10 -(d 2/ 5)
Using the previous value of 50 mSv and a
“typical” d 1 value of 5.5 m, obtain the
following: (1/5.5 (1/55 m) 2 x (7.5 m 2 /6 m 2 ) x 10 -(6 / 5)
x 50 mSv = 0.13 mSv = 130 µSv at the door.

Starting point for door calcs. (hard way)
Neutron fluence in NCRP 151
Φ = βQn + 54βQ 5.4 βQn + 13Q 1.3 Qn
4πd 2 2πS r 2πS r
Β = transmission through head
Qn = neutrons per Gy of x-rays
S r = total surface area of room
2π accounts for scattered and
thermal neutrons entering maze

Neutron fluence in NCRP 151, Table B.9
Φ A = βQn + 5.4 βQn + 1.3 Qn
4πd 2 2πS r 2πS r
Β =1 for Pb; d=5.5 m; S r = 250 m 2
Q n = 6 x 10 10 neutrons/Gy
Result:
Φ A = 4.14 x 10 8 neutrons at maze

Modified Kersey Method
Varian: neutron fluence (10 MV) = 6 x 10 10 /Gy
Φ = 4.14 x 10 8 neutrons at maze
H n,d = 2.4 x 10 -15 x 4.14 x 10 8 x (S 0 /S 1 ) 1/2 x [1.64 x 10 -(d 2/ 1.9 )
+ 10 -(d 2/ TVD )
]
where TVD = 2.06 (S 1/2
1 )
Equivalent dose = H x W L
Using the same values of d, S, and W as previous, we obtain a value of
TVD = 5.05 m , H = 7.3 x 10 -8 Sv/Gy
AndwithW L = 250 Gy conv + 1000 Gy IMRT , H n,d = 92 µSv
Remember 0.004% for Varian production at 10 MV So one expects 0.4
x 330 µSv = 130 µSv using the Kersey Method
x 330 µSv = 130 µSv using the Kersey Method. Use modified for
only for non-standard rooms and save lots of work.

Typically shield to equal contribution
of neutron dose and photon dose at
the entrance, 50 µSv each
1 TVL = 4.5 cm borated poly for mazed rooms
1 TVL = 8.5 cm borated poly for direct-shielded rooms.
Material is available in sheets of 2.54 cm thickness in USA.
Using the values just obtained: 130 µSv
Need 2.54 cm BPE for mazed door
Once the neutrons are dealt with, there is the photon dose from three
sources to be handled:
(1) Direct leakage
(2) Capture gamma radiation
(3) Room and patient scatter

Capture-gamma radiation (thermal n’s)
Boron: 420 keV Polyethylene: 2.22 MeV
BPE: 478 keV
Typical intensity at the door for mazed rooms is 1/5 of the
neutron dose at t (Kersey, March 1980).
Note that he says mGy gammas = 1/5 mSv n’s
But Kersey references ICRP 26 (1977), Q for neutrons was
lower than at present: So there is an additional factor of 2 that
is sometimes ignored. (Also ICRP 21, 60 and 103 complicate)
mGy gammas really = 1/10 mSv n’s.
TVL= 6 mm Pb for BPE gammas.

Going back to our example for the mazed room, we
had 130 µSv at the door.
0.1 x 130 µSv = 13 µGy from capture gammas.
For high-energy h maze rooms neutron capture
gammas are the dominant component and patient
wall photon scatter can be ignored.
NCRP 151 uses a formula for the capture gamma
dose rate: h φ = K φ A 10 -(d 2 /TVD)
where K= 6.9 x 10 -16 Sv m 2 per unit neutron
fluence at point A, φ A was defined earlier, d 2 is
maze length, and TVD = ~5.4 54 for x-rays 18-25 MV
and ~3.9 m for 15 MV.

Let’s see how this complex approach compares with the simplistic one of
assuming capture gamma dose is 1/10 of neutron dose. Let us use a 15
MV accelerator as an example. First, the complex method:
Φ A = βQn + 5.4 βQn + 1.3 Qn
4πd 2 2πS r 2πS r
Β =1 for Pb; d=5.5 m; S r = 250 m 2 Q n = 76 x 10 10 neutrons/Gy
Φ A = 5.25 x 10 9 neutrons at maze entrance
h φ = K φ A 10 -(d 2 /TVD)
where K= 6.9 x 10 -16 Sv m 2 per unit neutron fluence at point A, φ A was
defined earlier, d 2 is maze length, and TVD = ~5.4 for x-rays 18-25 MV and
~3.9 m for 15 MV. (I’ll use 6 m maze as earlier.)
h φ = K φ A 10 -(d 2 /TVD)
= (6.9 x 10 -16 )(5.25 x 10 9 )10 -(6 /3.9) = 1.05 x 10 -7 Sv/Gy
1.05 x 10 -7 x W L = 1.05 x 10 -7 x 1,250 Gy = 131 µSv = 131 µGy

Next, the simple method:
At 15 MV, neutron production = 0.07% 0 of primary
1,250 Gy x 0.0007 = 0.875 Sv = 875 mSv
Using Kersey method, 875/(5.5 m) 2 x 7.5/6 = 36 mSv at
maze entrance. 36 mSv x 10 -(6/5) = 227mSv(n’s) 2.27 atdoor
2.27 mSv x 0.1 = 0.227 mGy = 227 µGy
Compare this with the 131 µGy just calculated.
Also note that the TVD in the “complex” formula is actually
~ 39 3.9 m as opposed dt to “exactly” 39 3.9 m, and you can easily
see that the results are quite comparable. Any difference is
covered by the use of a TVL of 6.1 cm Pb for the photons.

In addition to the capture gamma
component, there will also be direct
leakage and room scatter to the door. For
high-energy rooms, the room and patient
scatter is low compared with the capture
gamma radiation i and may be ignored.
Typically, a maze wall is designed to allow
50 – 100 µGy to impact the door from
direct head leakage. Let’s assume a value
of 100 µGy for the purpose of calculating
the door.

So what would this door look like
2.27 mSv of neutrons at the door must be reduced to 50 µSv.
2,270 µSv/ 50 µSv = 45.4x reduction
Log 45.4 4 = 1.65, so use 1.65 TVL = 7.5 cm (3”) BPE
The photon component, 227 µGy from capture gammas plus
100 µGy head leakage, must also be reduced to 50 µGy.
327 µGy/50 µGy = 6.54 x reduction
Log 6.54 = 0.82, so use 0.82 TVL = 5 cm of Pb.
The door is usually laminated between two steel sheets, each 6
mm thick, so the Pb can be reduced by 6 mm total but round to
“standard” materials.
For balance, put the 2.5 cm (1”) on each side of the BPE.

Direct-Shielded Vault

d 1

Direct-Shielded Door d 2 = 0
Varian: neutron fluence (10 MV) = 6 x 10 10 /Gy
Φ A = βQn = 1.92 x 10 8 (d= 5 m to door)
4πd 2
H n,d = 2.4 x 10 -15 x 1.92 x 10 8 x 1 x [1.64 x 1 + 1]
Equivalent dose = H x W L
H = 1.22 x 10 -6
Sv/Gy
And with W L = 250 Gy conv + 1,000 Gy IMRT , H n,d = 1,520 µSv
Using Kersey: 50 mSv x (1/5 m) 2 = 2000µSv 2,000 Again, using Kersey saves lots of time. NCRP 151
says it is “conservatively safe” to take this approach.

Photons At Door
Leakage radiation is dominant component.
Capture gammas, room and patient scatter
are also to be considered.
1) Direct leakage
Simple inverse square to door position
2) Capture gammas
h φ = K φ 10 -(d 2 /TVD) A where d 2 = 0 (So, simply py K φ A )
3) Room and patient scatter
Since scattering angle is ~ 90°, energy is < 0.3 MeV, so it can be ignored.

Example calculation: d = 6 m W L = 1,250 Gy x 0.001
001
Direct Leakage
1,250 Gy x 0.001 x 1/6 2 = 3.47 cGy = 34,722 µGy
Capture gammas
K φ A = 6.9 x 10 -16 Sv m 2 x 1.92 x 10 8 = 1.32 x 10 -7
1.32 x 10 -7 x 1,250 Gy = 165 µGy
Total = 34,887 µGy

So what would this door look like
1,390 µSv of neutrons at the door (50 mSv x 1/6 2 ) must be
reduced to 50 µSv.
1,390 µSv/ 50 µSv = 27.8x reduction
Log 27.8 = 1.44, so use 1.44 TVL = 12.3 cm (5”) BPE
The photon component , 34,887 µGy, must also be reduced
to 50 µGy.
34,887 µGy/50 µGy = 697x reduction
Log 697 = 2.84, so use 2.84 TVL = 17.3 cm of Pb.
The door is usually laminated between two steel sheets,
each 6 mm thick, and the 12.3 cm of BPE = 0.32 TVL (20
mm Pb equiv), so the Pb can be reduced by 26 mm total.
For balance, put the 7.5 cm (3”) on each side of the BPE.

Sliding Door problem

To isocenter
Sliding Door problem