Neutron Door Designg

Robert J Barish, Ph.D.

Alternate form vendor information (Siemens)

Mazed Vault

d1

S0

d

0

Sd2 S1

Starting point for door calcs. (easy way)g p ( y y)Varian 10 MV 0.00004 Sv/Gy (at 1 m)

Example: 500 Gy x 0.00004 Sv/Gy = 20 mSv

(50 000 cGy x 0 00004 cSv/cGy = 2 cSv = 20 mSv)(50,000 cGy x 0.00004 cSv/cGy = 2 cSv = 20 mSv)

But consider 50% IMRT with “IMRT factor” = 4But consider 50% IMRT, with IMRT factor 4

250 Gy x 0.00004 Sv/Gy = 10 mSvy y

250 Gy x 0.00004 Sv/Gy x 4 = 40 mSv

Total: 50 mSv (at 1 m)

Kersey MethodVarian 0.00004 Sv/Gy (at 1 m)

A “typical” maze length is about 6 m. Its height is 3 m and width is 2 mA 6 2 SArea = 6 m2 = S1

A typical gap between the maze and the far wall is 2.5 m and room height is also 3 m Area = 7 5 m2 = Sheight is also 3 m . Area = 7.5 m2 = S0

Kersey’s formula gives the neutron dose at the door as the neutron dose y gat the isocenter x (1/d1)2 x (S0/ S1) x 10-(d2/5)

Using the previous value of 50 mSv and aUsing the previous value of 50 mSv and a “typical” d1 value of 5.5 m, obtain the f ll i (1/5 5 )2 (7 5 2 / 6 2 ) 10 (6/5)following: (1/5.5 m)2 x (7.5 m2 / 6 m2 ) 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 151Φ βQ 5 4 βQ 1 3 QΦ = βQn + 5.4 βQn + 1.3 Qn

4πd2 2πSr 2πSrr r

Β = transmission through headgQn = neutrons per Gy of x-raysS = total surface area of roomSr = total surface area of room2π accounts for scattered and thermal neutrons entering maze

Neutron fluence in NCRP 151, Table B.9

ΦA = βQn + 5.4 βQn + 1.3 QnΦA βQn 5.4 βQn 1.3 Qn4πd2 2πSr 2πSr

Β =1 for Pb; d=5.5 m; Sr = 250 m2

10Qn = 6 x 1010 neutrons/GyResult:Result:ΦA = 4.14 x 108 neutrons at maze

Modified Kersey Method10Varian: neutron fluence (10 MV) = 6 x 1010/Gy

Φ = 4.14 x 108 neutrons at mazeHn,d = 2.4 x 10-15 x 4.14 x 108 x (S0/S1)1/2 x [1.64 x 10-(d2/1.9) + 10-(d2/TVD)]

where TVD = 2.06 (S1)1/2where TVD 2.06 (S1)

Equivalent dose = H x WL

Using the same values of d S and W as previous we obtain a value ofUsing 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

And with W = 250 Gy + 1000 Gy H d = 92 µSvAnd with WL 250 Gyconv + 1000 GyIMRT , Hn,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 Use modified forx 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 p

the entrance, 50 µSv each1 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.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 doorNeed 2.54 cm BPE for mazed door

Once the neutrons are dealt with, there is the photon dose from three sources to be handled:sources to be handled:(1) Direct leakage(2) Capture gamma radiation(2) Capture gamma radiation(3) Room and patient scatter

Capture-gamma radiation (thermal n’s)

Boron: 420 keV Polyethylene: 2 2 MeV Boron: 420 keV Polyethylene: 2.2 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 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)g ( , p )

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.

F hi h t t For high-energy maze rooms neutron capture gammas are the dominant component and patient

ll h t tt b i d wall photon scatter can be ignored.

NCRP 151 uses a formula for the capture gamma NCRP 151 uses a formula for the capture gamma dose rate: hφ = K φA 10-(d2 /TVD)

h K 6 9 10 16 S 2 it t where K= 6.9 x 10-16 Sv m2 per unit neutron fluence at point A, φA was defined earlier, d2 is

l th d TVD 5 4 f 18 25 MV maze length, and TVD = ~5.4 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 assuming capture gamma dose is 1/10 of neutron dose. Let us use a 15 MV accelerator as an example. First, the complex method:ΦA = βQn + 5.4 βQn + 1.3 Qn

4πd2 2πS 2πS4πd2 2πSr 2πSr

Β =1 for Pb; d=5.5 m; Sr = 250 m2 Qn = 76 x 1010 neutrons/GyΦ 5 25 109 t t tΦA = 5.25 x 109 neutrons at maze entrance

hφ = K φA 10-(d2 /TVD)

where K= 6.9 x 10-16 Sv m2 per unit neutron fluence at point A, φA was defined earlier, d2 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.)

hφ = K φA 10-(d2 /TVD) = (6.9 x 10-16)(5.25 x 109) 10-(6 /3.9) = 1.05 x 10-7 Sv/Gy

1.05 x 10-7 x WL = 1.05 x 10-7 x 1,250 Gy = 131 µSv = 131 µGy

Next, the simple method:

1 0 0 % fAt 15 MV, neutron production = 0.07% 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) = 2 27 mSv (n’s) at doormaze entrance. 36 mSv x 10 (6/5) = 2.27 mSv (n s) at door

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 3 9 d t “ tl ” 3 9 d il~ 3.9 m as opposed to “exactly” 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 directcomponent, there will also be direct leakage and room scatter to the door. For high energy rooms the room and patienthigh-energy rooms, the room and patient scatter is low compared with the capture

di i d b i dgamma radiation and may be ignored.

Typically a maze wall is designed to allowTypically, a maze wall is designed to allow 50 – 100 µGy to impact the door from di t h d l k L t’ ldirect 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. µ

2,270 µSv/ 50 µSv = 45.4x reduction

Log 45 4 = 1 65 so use 1 65 TVL = 7 5 cm (3”) BPELog 45.4 = 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 µGy100 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

d1

Direct-Shielded Door d2 = 010Varian: neutron fluence (10 MV) = 6 x 1010/Gy

ΦA = βQn = 1.92 x 108 (d= 5 m to door)4 d24πd2

Hn,d = 2.4 x 10-15 x 1.92 x 108 x 1 x [1.64 x 1 + 1]

Equivalent dose = H x WL

H = 1.22 x 10-6 Sv/Gy y

And with WL = 250 Gyconv + 1,000 GyIMRT , Hn,d = 1,520 µSv

Using Kersey: 50 mSv x (1/5 m)2 = 2 000 µSvUsing Kersey: 50 mSv x (1/5 m) = 2,000 µSv

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 consideredare also to be considered.

1) Direct leakage) gSimple inverse square to door position

2) Capture gammashφ = K φA 10-(d2 /TVD) where d2 = 0 (So, simply K φA)φ φA 2 ( , p y φA)

3) Room and patient scatterSince scattering angle is ~ 90°, energy is < 0.3 MeV, so it can be ignored.

Example calculation: d = 6 m W = 1 250 Gy x 0 001Example calculation: d = 6 m WL = 1,250 Gy x 0.001Direct Leakage

1,250 Gy x 0.001 x 1/62 = 3.47 cGy = 34,722 µGy

Capture gammas

K φA = 6.9 x 10-16 Sv m2 x 1.92 x 108 = 1.32 x 10-7

1 32 x 10-7 x 1 250 Gy = 165 µGy1.32 x 10 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 1/62) must be µ ( )reduced to 50 µSv.

1,390 µSv/ 50 µSv = 27.8x reductionµ µ

Log 27.8 = 1.44, so use 1.44 TVL = 12.3 cm (5”) BPE

The photon component 34 887 µGy must also be reducedThe photon component , 34,887 µGy, must also be reduced to 50 µGy.

34 887 µGy/50 µGy = 697x reduction34,887 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 totalmm 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 problemg p

T i tTo isocenter

Sliding Door problemg p