Environmental Scatter: Problem 6 -...

Environmental ScatterWorkshop in Bologna July 14-16 2003

P6 1QUADOSQUADOS

Problem 6Environmental Scatter

Jean-Louis Chartier1 and Bernd R.L.Siebert2

Intercomparison on the Usage of Computational Codes in Radiation Dosimetry

Workshop in Bologna-Italy July 14 -16 2003

(2) Physikalisch-Technische Bundesanstalt, Braunschweig, Germany

(1) Institut de Radioprotection et de Sûreté Nucléaire, Fontenay aux Roses, France


P6 2QUADOSQUADOS

Table of ContentThe problemNomenclatureResults for task 1, 2, 3 and 4Physics of the problemParticipation and approachesDetailed discussion of computing

• directional fluence and• contributions from walls, floor and ceiling.

OralsSummary and Conclusion

If there is a questionor the needto discuss,

please interrupt,anytime!

Orals will be given by Tom McLean and Miloslav Králik


P6 3QUADOSQUADOS

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 1010.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8Direct

241AmBe(α,n)

252Cf(sf) 252Cf(sf), moderated (15 cm D2O) + 1 mm Cd

E ϕ E

/ ϕ

E / MeV

Environmental Scatter: Workplace Fields

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 1010.00

0.05

0.10

0.15

0.20Inscattered

241AmBe(α,n)

252Cf(sf) 252Cf(sf), moderated

(15 cm D2O) + 1 mm Cd

E ϕ E

/ ϕ

E / MeV

Inscattered: 241AmBe (α,n) 252Cf(sf) 252Cf(sf),moderated

Work place neutron fieldsfor the Calibrationof Personal Dosimeterscan be “mocked up” in the laboratory

Sources, direct: 241AmBe (α,n) 252Cf(sf) 252Cf(sf),moderated


P6 4QUADOSQUADOS

Environmental Scatter: Workplace Fields

Shadow ConesShadow Cones

“BUNKER” operated by (Photos by Urbach)

Monitor LB6411

For information on calibrations:[email protected]

Position controlling


P6 5QUADOSQUADOS

Problem 6: Environmental Scatter

Shadow coneShadow conewith two layerswith two layers

Position 1Position 1

Position 2Position 2

252Cf- isotropic252Cf- isotropicpoint source inpoint source incentrecentre

y

xx

A bare 252Cf neutron source is located at the centre of a concrete walled calibration room.

Task 1 :Compute thefluence as function of energyin positions 1 & 2

Task 2:Compute thefluence as function of energyand angle in positions 1 & 2

Task 3: Compute thecontributions from air , walls,floor and ceiling in positions1 & 2Task 4: Compute the influence of a change in the water content in concrete in positions 1 & 2

Task 1: Compute thefluence as function of energyin positions 1 & 2

Task 2: Compute thefluence as function of energyand angle in positions 1 & 2Task 3: Compute thecontributions from air , walls,floor and ceiling in positions1 & 2

Figure is not scaled

Task 4: Compute the influence of a change in the water content in concrete in positions 1 & 2


P6 6QUADOSQUADOS

Nomenclature (1/5)

ICRU Report 60: Fundamental Quantities and Units for Ionizing Radiation BIPM : Le Système international d´unités (7e édition 1998)

The particle number, N, is the number of particles that are emitted, transferred, or received. The unit is 1.Note: The distribution, NE , of the particle number with respect to energy is given by NE =dN/dE

The fluence , Φ, is the quotient of dN by da , where dN is the number of particles incident on a sphere of cross sectional area da, thus The SI-unit is m-2,but cm-2 is ok, too. a

Nd

d=ΦNote: The distribution, ΦE , of the fluence with respect to energy is given by ΦE = dΦ/dE, the unit is m -2 eV -1.

The flux , N, is the quotient of dN by dt, where dN is the increment of the particle number in the time interval dt, thus N = dN/dt ; unit is s-1.

.


P6 7QUADOSQUADOS

ϑϑ

t=h/|cos(ϑ)|t=h/|cos(ϑ)|

hh

Surface sSurface ss=2πr2s=2πr2

2r2r

track lengthtrack length

ϕϕs=πr2

Nomenclature (2/5) Relation between thenumber of particles, N, that cross a surface and the fluence in a cell

Practical definition of fluence in a cell for use with transportcalculations: The fluence, Φ , in a cell is the quotient of thesum of track lengths ti and the volume, Vcell, of that cell, thus

∑=

=N

iitV 1cell

1Φ

Since Vcell= h.s

∑=

=N

i icoss 1

11ϑ

Φ

cos|

|

hti

i ϑ= ( ) [ ]1

1+

=

− ∈= ∑ jji

n

iij s,|

sϑϑϑϑΦ cocoscos wherecos|1

1


P6 8QUADOSQUADOS

( )[ ] [ ] ( )|cocos|21 co,coscos 1 cosE 1 1 cos ++ϑ ϑ+ϑϑϑ∈ϑ=ϑ ∑ jj

n

ijjij ss

ng

j

Nomenclature (3/5)

( )jjj ϑ−ϑ⋅π⋅= + coscos2∆ 1Ω

j

J

jj

ΩΦΦ Ω ∆1

⋅= ∑=

( ) [ ]11

+− ∈= ∑ jji

n

iij s,|

sϑϑϑϑΦ cocoscos wherecos|1

Presentation per sr

jjjΩΦΦΩ

-1∆⋅=

]E[cos

j

jj s

nϑ

Φ⋅

≈ ΦTally 2 ≈ ΦTally 4 or 5

if fluence is isotropic then

∑=

⋅=J

jjn

s 1

2Φ

< cos ϑ >=0.5

[ ] jcos,coscos nnjj

:=+∈ 1ϑϑϑ

nj is given in MCNP by tally F1

if fluence is isotropic in that bin

E is expectation value

=


P6 9QUADOSQUADOS

Nomenclature (4/5)

Generally for MCNP and MCNPX:• a fluence estimator (F4 or F5) should be used for tasks 1,3 and 4• and a particle estimator (F1) [or a fluence estimator (F2)] in conjunction with cos ϑ sorting for task 2• all results should be normalised to one source neutron.

Mean track length for parallel incidence on sphere is 4/3 r ⇒ Φ = 1/ π r 2 but every track is counted twice (entering and leaving the sphere).

Most participants used MCNP or MCNPX, one used TRIPOLI

Some comments on MCNP (1/2)

F4 and F5 compute track length sum divided by the specified volume.

F1 (with surface): normalise by dividing by area and mean cosϑ

F1 (with sphere): normalise by dividing by area = 2 π r 2 with F2 only by 2

WHY ?


P6 10QUADOSQUADOS

Nomenclature (5/5)

MCNP-convention: upper bin energy, group fluence !Several particpants were not aware of this.Group fluence to lethargy representation via ln(Ei / Ei-1).

Most participants used MCNP or MCNPX, one used TRIPOLI

Cumulative distribution

Flagging allows to partially reconstruct the geometrical history of a neutron

Multiple scattering contributions can be separated, one should alwaysat least view the uncollided fraction for any tally.

Some comments on MCNP (2/2)

( ) ( ) EEEE

E

J

jjJ d

01∫∑ =≈=

=

ΦΦϕΦ( ) ( )

jj

jjj EE

EE

′+′

′+′′ −

−=

1

1 ΦΦϕ

can be used to re-bin the results.


P6 11QUADOSQUADOS

Results for Task 1 (1/7) Task 1:Compute the fluence as functionof energy in position 1

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

EΦE(

E ) /

cm2

ZZCCGGBBDDFFHHIIJJL without thermalL without thermalMM

with equal bin size

G

D


P6 12QUADOSQUADOS


10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

EΦE(

E) /

cm2

ZZCCGGBBDDFFHHIIJJL without thermalL without thermalMM

with equal bin size=original

D

J


P6 13QUADOSQUADOS


10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

EΦE(

E ) /

cm2

ZZE * 4,2E * 4,2KKN * 4N * 4A * 4 πA * 4 π

with equal bin size


P6 14QUADOSQUADOS


10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

10-6

2.10-6

3.10-6

4.10-6

5.10-6

Φ (E

) / c

m2

ZZCCGGBBDDEthermal: 1 binEthermal: 1 binFFHthermal: 1 binHthermal: 1 binIIJJ

KKLwithout thermalLwithout thermalMMNfrom tally F2Nfrom tally F2AA

with equal bin size

( ) ( ) EEEE

E

J

jjJ d

01∫∑ =≈=

=

ΦΦϕΦ


P6 15QUADOSQUADOS

Task 1:Compute the fluence as functionof energy in position 1

Results for Task 1 (5/7)

per bintotal

E n 0 .4 1 4 1 1 0 1 0 0 1 1 5e V k e V k e V M e V M e V M e V

A .1 2 .0 9 .0 9 .0 8 .0 7 .0 9B 1 .0 0 1 .0 1 1 .0 0 1 .0 1 1 .0 2 1 .0 3C .9 7 1 .0 0 .9 8 1 .0 0 1 .0 0 1 .0 1D .9 9 .9 5 1 .0 1 .8 7 .9 5 .9 5E .7 7 .3 7 .3 0 .2 2 .1 9 .2 4F 1 .0 0 1 .0 1 .9 8 .9 9 1 .0 1 1 .0 0G 1 .0 6 1 .0 2 .9 8 1 .0 0 1 .0 0 1 .0 0H 1 .0 0 1 .0 0 .9 8 1 .0 0 1 .0 0 1 .0 0

I 1 .0 0 1 .0 0 .9 9 1 .0 0 1 .0 0 1 .0 0J 1 .0 5 1 .0 6 1 .0 5 1 .0 6 1 .0 6 1 .0 6K .0 0 .8 2 .7 8 .7 6 .7 6 1 .4 8L .0 0 1 .0 1 .9 9 1 .0 0 1 .0 1 1 .0 0

M 1 .0 1 1 .0 0 .9 8 1 .0 0 1 .0 0 1 .0 0N .8 1 .3 9 .3 2 .2 3 .2 0 .2 5Z 1 8 3 2 6 3 0 .3 2 2 6 .1 2 9 9 .7 6 9 4 .6 3 4 9 .0Z 1 8 3 2 2 4 6 3 2 6 8 9 2 9 8 8 3 6 8 3 4 0 3 2 Z .109


P6 16QUADOSQUADOS

E n 0 .4 1 4 1 1 0 1 0 0 1 1 5e V ke V ke V M e V M e V M e V

B 1 .0 0 1 .0 0 1 .0 1 1 .0 0 1 .0 0 1 .0 1C .9 7 .9 9 .9 9 .9 9 .9 9 1 .0 0D .9 8 .9 4 .9 6 .8 6 .8 8 .9 9E .7 5 .3 4 .2 8 .2 1 .1 6 .2 0F .9 9 1 .0 1 1 .0 0 1 .0 0 1 .0 0 1 .0 1G 1 .0 6 1 .0 2 1 .0 0 1 .0 0 1 .0 0 1 .0 0H 1 .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0I 1 .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0J 1 .0 5 1 .0 6 1 .0 6 1 .0 6 1 .0 6 1 .0 5K .0 0 .8 2 .7 9 .7 7 .7 7 1 .5 0L .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0 1 .0 0

M .9 9 1 .0 2 1 .0 6 1 .0 2 1 .0 0 1 .0 1N .7 9 .3 6 .3 0 .2 2 .1 7 .2 1Z 1 8 1 8 6 1 9 .3 2 1 6 .4 2 8 7 .3 6 4 1 .2 3 0 6 .0Z 1 8 1 8 2 4 3 7 2 6 5 3 2 9 4 1 3 5 8 2 3 8 8 8


per bintotalZ .109


P6 17QUADOSQUADOS

Results for Task 1 (7/7) Task 1:Compute the fluence as functionof energy in position 1&2

1E-7 1E-5 1E-3 0.1 101E-13

1E-11

1E-9

1E-7

1E-5

1E-3

0.1

10

Flue

nce

(par

ticle

s cm

-2M

eV-1)

Neutron energy (MeV)

Det. 1 Det. 2 Det. 1 w/ cone Det. 2 w/ cone

Participant I Thedifferencebetween

thevariousspectra

is used tomock up

work placespectra

(environmental)[see task 4, F25]


P6 18QUADOSQUADOS

Results for Task 2 (1/3) Task 2:Compute the fluence as functionof energy and angle in position 1

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

EΦE(

E) /

(cm

2 sr)

(Z / 170 cm)Tally F5(Z / 170 cm)Tally F5cosϑ ε [-1,-0.866]cosϑ ε [-1,-0.866]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.5,0]cosϑ ε [-0.5,0]cosϑ ε [ 0, 0.5]cosϑ ε [ 0, 0.5]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.866, 0]cosϑ ε [ 0.866, 0](Z / 170)Tally F1(Z / 170)Tally F1

Total Fluence and Total Fluence and Fluence in Angular Bins as ComputedFluence in Angular Bins as Computed

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-8

4.10-8

6.10-8

8.10-8

EΦE(

E) /

(cm

2 sr)


Total Fluence and Total Fluence and Fluence in Angular Bins per SteradianFluence in Angular Bins per Steradian

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-8

4.10-8

6.10-8

8.10-8

E ΦE(

E) /

(cm

2 sr)

(I / 170 cm)Tally F5(I / 170 cm)Tally F5cosϑ ε [-1,-0.866]cosϑ ε [-1,-0.866]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.5,0]cosϑ ε [-0.5,0]cosϑ ε [ 0, 0.5]cosϑ ε [ 0, 0.5]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.866, 0]cosϑ ε [ 0.866, 0](I / 170)Tally F1(I / 170)Tally F1


10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / M eV

0

2 .10-8

4 .10-8

6 .10-8

8 .10-8

EΦE(

E ) /

(cm

2 sr)

(Z / 170 cm )T ally F 5(Z / 170 cm )T ally F 5cosϑ ε [-1 ,-0 .866]cosϑ ε [-1 ,-0 .866]cosϑ ε [-0 .866,-0 .5]cosϑ ε [-0 .866,-0 .5]cosϑ ε [-0 .5 ,0]cosϑ ε [-0 .5 ,0]cosϑ ε [ 0 , 0 .5]cosϑ ε [ 0 , 0 .5]cosϑ ε [ 0 .5 , 0 .866]cosϑ ε [ 0 .5 , 0 .866]cosϑ ε [ 0 .866, 0]cosϑ ε [ 0 .866, 0](Z / 170)T ally F1(Z / 170)T ally F1

T otal F luence and T otal F luence and F luence in A ngu lar B ins per S terad ianF luence in A ngu lar B ins per S terad ian

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / M eV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

EΦE(

E) /

(cm

2 sr)


Total Fluence and Total Fluence and Fluence in Angular Bins as ComputedFluence in Angular Bins as Computed

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-8

4.10-8

6.10-8

8.10-8

E ΦE(

E) /

(cm

2 sr)

( G/ 170 cm)Tally F5( G/ 170 cm)Tally F5cosϑ ε [-1,-0.866]cosϑ ε [-1,-0.866]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.5,0]cosϑ ε [-0.5,0]cosϑ ε [ 0, 0.5]cosϑ ε [ 0, 0.5]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.866, 0]cosϑ ε [ 0.866, 0](G / 170)Tally F1(G / 170)Tally F1



P6 19QUADOSQUADOS

Task 2:Compute the fluence as functionof energy and angle in positions 1 & 2

Angle Participantsdegree B C G I L Z

180-150 .93 .94 .26 .94 .89 .95150-120 1.04 1.05 .39 1.05 1.05 1.05120- 90 .89 .91 1.71 .91 .88 .89 90- 60 1.07 1.07 1.85 1.06 1.08 1.08 60- 30 .99 1.00 .38 1.00 1.03 1.00 30- 0 1.10 1.02 .27 1.03 1.05 1.03sum/F4 1.00 1.00 1.07 1.00 1.00 1.00


Angle Participantsdegree B C G I L Z

180-150 .78 .77 .21 .78 .61 .78150-120 .90 .90 .35 .93 .86 .91120- 90 1.12 1.11 1.73 1.05 1.09 1.12 90- 60 1.02 1.01 1.72 .99 .98 1.03 60- 30 1.08 1.09 .41 1.10 1.15 1.08 30- 0 1.06 1.05 .28 1.06 1.09 1.05sum/F4 1.02 1.01 1.03 1.00 1.00 1.02

Position 1 (170 cm)

Position 2 (300 cm)

B,C,Z: F1 (surface)

I,L : F1 (volume)

G : F2 (volume)

?


P6 20QUADOSQUADOS

Angular dependence of neutron fluence at position 1

1,0E-12

1,0E-11

1,0E-10

1,0E-09

1,0E-08

1,0E-07

1,0E-06

1,0E-05

1,0E-10 1,0E-09 1,0E-08 1,0E-07 1,0E-06 1,0E-05 1,0E-04 1,0E-03 1,0E-02 1,0E-01 1,0E+00 1,0E+01 1,0E+02

Neutron Energy (MeV)

E p

hi(E

) per

sou

rce

neut

ro

150-180 degrees120-150 degrees 90 -120 degrees 60- 90 degrees 30- 60 degrees 0- 30 degrees

Note : Data not corrected for double counting I.e. divide by 2 to get absolute f luence in each cosine bin

Taken from participant G

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

10-9

10-8

10-7

10-6

EΦE(

E) /

(cm

2 sr)

cosϑ ε [-1,-0.866]cosϑ ε [-1,-0.866]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.5,0]cosϑ ε [-0.5,0]cosϑ ε [ 0, 0.5]cosϑ ε [ 0, 0.5]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.866, 0]cosϑ ε [ 0.866, 0]

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

10-9

10-8

10-7

10-6

E ΦE(

E ) /

(cm

2 sr)

cosϑ ε [-1,-0.866]cosϑ ε [-1,-0.866]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.866,-0.5]cosϑ ε [-0.5,0]cosϑ ε [-0.5,0]cosϑ ε [ 0, 0.5]cosϑ ε [ 0, 0.5]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.5, 0.866]cosϑ ε [ 0.866, 0]cosϑ ε [ 0.866, 0]

Results for Task 2 (3/3) Task 2:Compute the fluence as functionof energy and angle in position 1

Angular dependence of neutron fluence at position 1

1,0E-12

1,0E-11

1,0E-10

1,0E-09

1,0E-08

1,0E-07

1,0E-06

1,0E-05

1,0E-10 1,0E-09 1,0E-08 1,0E-07 1,0E-06 1,0E-05 1,0E-04 1,0E-03 1,0E-02 1,0E-01 1,0E+00 1,0E+01 1,0E+02

Neutron Energy (MeV)

E p

hi(E

) per

sou

rce

neut

ro

150-180 degrees120-150 degrees 90 -120 degrees 60- 90 degrees 30- 60 degrees 0- 30 degrees

Note : Data not corrected for double counting I.e. divide by 2 to get absolute f luence in each cosine bin

G ?


P6 21QUADOSQUADOS

Results for Task 3 (1/3)Task 3: Compute the contributions fromair , walls, floor and ceiling in pos. 1

B F I M Zair 13.78 13.40 14.42 1.59 13.65near sensor 25.58 25.80 25.83 9.36 25.83opp. ssensor 6.63 6.57 6.49 3.68 6.48side 1 12.20 12.22 14.32 10.53 12.25side 2 12.16 12.18 14.32 10.53 12.23floor 14.27 14.29 12.31 6.44 14.34ceiling 14.25 14.33 12.32 6.45 14.32

Normalised to total fluence for each author

Most results that usedflagging agree very well.Results for position 2 are of the same quality, in general.

See pages 38 & 39 andAppendix

Side 1 and 2 andfloor and ceiling

are by virtue of thesymmetry of the problemexpected to be identical


P6 22QUADOSQUADOS


10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E ) /

(cm

2 sr)

Z airZ airZ wall near sensorZ wall near sensorZ wall opposite sensorZ wall opposite sensorZ side wallsZ side wallsZ floor and ceilingZ floor and ceiling


P6 23QUADOSQUADOS

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B wall near sensorB wall near sensorF wall near sensorF wall near sensorI wall near sensorI wall near sensorM wall near sensorM wall near sensorZ wall near sensorZ wall near sensor

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E ) /

(cm

2 sr)

B airB airF airF airI airI airM airM airZ airZ air

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B wall opposite sensorB wall opposite sensorF wall opposite sensorF wall opposite sensorI wall opposite sensorI wall opposite sensorM wall opposite sensorM wall opposite sensorZ wall opposite sensorZ wall opposite sensor

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B side wallsB side wallsF side wallsF side wallsI side wallsI side wallsM side wallsM side wallsZ side wallsZ side walls

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B floor & ceilingB floor & ceilingF floor & ceilingF floor & ceilingI floor & ceilingI floor & ceilingM floor & ceilingM floor & ceilingZ floor & ceilingZ floor & ceiling


10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B airB airF airF airI airI airM airM airZ airZ air

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / M eV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B w all opposite sensorB w all opposite sensorF w all opposite sensorF w all opposite sensorI w all opposite sensorI w all opposite sensorM w all opposite sensorM w all opposite sensorZ w all opposite sensorZ w all opposite sensor

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B side wallsB side wallsF side wallsF side wallsI side wallsI side wallsM side wallsM side wallsZ side wallsZ side walls

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV

0

3.10-7

EΦE(

E) /

(cm

2 sr)

B floor & ceilingB floor & ceilingF floor & ceilingF floor & ceilingI floor & ceilingI floor & ceilingM floor & ceilingM floor & ceilingZ floor & ceilingZ floor & ceiling

10-9 510-8 510-7 510-6 510-5 510-4 510-3 510-2 510-1 5100 5101

E / MeV 0

3.10-7

EΦE(

E) /

(cm

2 sr)

B wall near sensorB wall near sensorF wall near sensorF wall near sensorI wall near sensorI wall near sensorM wall near sensorM wall near sensorZ wall near sensorZ wall near sensor


P6 24QUADOSQUADOS

Task 4: Compute the influence ofa change in the water content inconcrete in positions 1 & 2


Normalised to normal water content for each authorH2O B C F G I L Z -1% 103.90 - - 103.78 - 107.66 104.06 +1% 96.41 96.78 96.46 96.74 97.04 92.54 96.47 +2% - 93.99 93.77 94.02 94.29 88.16 93.80 +3% - 91.46 91.62 91.66 - - 91.13

H2O B C F G I L Z -1% 104.27 - - 103.91 - 110.11 104.11 +1% 96.83 96.64 97.12 96.72 96.81 85.81 96.68 +2% - 93.50 93.95 94.76 94.42 80.77 93.52 +3% - 90.91 91.27 91.26 - - 90.93

Normalised to normal water content for each author


P6 25QUADOSQUADOS

Z .106

Normalised to Z within each energy bin-1 0.414 eV 1 keV 10 keV 100 keV 1 MeV 16 MeV B 1.00 1.00 1.01 1.01 1.02 1.03 G 1.06 1.02 .99 .99 1.00 1.00 L .00 1.00 1.00 1.00 1.01 1.01 Z 1.82 0.68 0.24 0.33 0.75 0.37+1 B 1 . 00 1 . 01 1 . 03 1 . 01 1 . 02 1 . 03 C . 97 1 . 00 1 . 02 1 . 01 1 . 01 1 . 01 F 1 . 00 1 . 00 1 . 01 . 99 1 . 01 1 . 00 G 1 . 06 1 . 02 1 . 02 1 . 00 1 . 00 1 . 00 I 1 . 00 1 . 01 1 . 02 1 . 01 1 . 01 1 . 01 L . 00 . 99 1 . 01 . 99 1 . 00 1 . 00 Z 1 . 83 0 . 59 0 . 20 0 . 28 0 . 65 0 . 33

Results for Task 4 (2/4)D

ecrease with increasing energy


Z 0.414 eV 1 keV 10 keV 100 keV 1 MeV 16 MeV -1% 1.80 .67 .24 .32 .70 .32 +1% 1.82 .62 .22 .29 .64 .31 +2% 1.83 .58 .20 .27 .60 .29 +3% 1.81 .52 .18 .23 .53 .27


P6 26QUADOSQUADOS


10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

E ΦE(

E ) /

cm2

-1%-min-1%-min-1%-max-1%-max 0%-min 0%-min 0%-max 0%-max+1%-min+1%-min+1%-max+1%-max

Influence of H20 -content in concreteInfluence of H20 -content in concrete(Sensor position 1 in problem P6)(Sensor position 1 in problem P6)

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

E / MeV

0

2.10-7

4.10-7

6.10-7

8.10-7

10-6

E ΦE(

E) /

cm2

-1%-min-1%-min-1%-max-1%-max 0%-min 0%-min 0%-max 0%-max+1%-min+1%-min+1%-max+1%-max

Influence of H20 -content in concreteInfluence of H20 -content in concrete

(Sensor position 1 in problem P6)(Sensor position 1 in problem P6)

( ) ?∆

∆∆ OH2

=

⋅=

Φ

αΦ

uc

( ) ( )( )( ) ?∆

∆∆ OH2

=

⋅=

EucEE

E

E

Φ

αΦ



P6 27QUADOSQUADOS


0 2 4 6 8 10 12 14 16 18 20 222.4

2.6

2.8

3.0

3.2

3.4

3.6

3.8

4.0

4.2

4.4

4.6

Flue

nce

x10-6

(par

ticle

s cm

-2)

Moisture (%)

Det.1 Det.2

Participant I Almost all14 participants

providedstochastic

uncertainties.

There waslittle to nonediscussionon other

sources ofuncertainties.



P6 28QUADOSQUADOS

Generals• ‘Realistic’ neutron spectra• Concept in radiation protection dosimetry• Replicate workplace neutron fields• Enable calibration of area and personal dosimeters in

conditions close to those of their use• Calibration implies :

– Use of reference radiation fields– Knowledge of energy and angle distributions of

fluence—> operational quantities dose equivalent : H*(d) ; Hp(d,a)


P6 29QUADOSQUADOS

Problem 6 : Environmental scatter

• 14 participants + author• Codes :MCNP versions 4B2, 4C, 4C2, 5, X-2.4k,

X-2.5B and Tripoli 4.3• Task 1 : 14 solutions• Task 2 : 8 solutions ( remark 1)• Task 3 : 8 solutions ( remark 2)• Task 4 : 6 solutions


P6 30QUADOSQUADOS

Non-exhaustive list of computational techniquesinvolved in problem 6 (1)

• Task1• Biased isotropic source : SI,SP,SB• Cylindrical shape : EXT,SI,SP• Spectral distribution : ERG,SI,SP• Thermal treatment S(a,ß) in CH2 (shadow cone)

• Thermal treatment S(a,ß) in concrete (~H in H2O)

• Tally F2 (positions 1 & 2) ; disc perp./Ox ; r variable• Tally F4 (positions 1 & 2) ; sphere ; r variable• Tally F5 (positions 1 & 2) ; point detector ; Ro variable• DXTRAN spheres (RIi , ROi) at positions 1 and 2


P6 31QUADOSQUADOS

Non-exhaustive list of computational techniquesinvolved in problem 6 (2)

• Task 2• Tally F1 + Cn cosine card (6 angular bins) reference direction for the cosine binning = normal to the scoring disc surface• Tally F1 + FTn ‘FRV’ (tally special treatment) reference direction for the cosine binning = vector FRV (V1,V2,V3) for a scoring spherical

surface


P6 32QUADOSQUADOS

Non-exhaustive list of computational techniquesinvolved in problem P6 (3)

• Task 3• Cell-flagging : F4 + CFn ‘cell numbers’• Cell-flagging : F5 + FTn ‘ICD’ + FUn ‘cell numbers’

• Task 4• Several ‘task 1 runs’ with different concrete materials• PERTurbation cards with different concrete materials

(only 1 run)


P6 33QUADOSQUADOS

Task 2 : Directional distribution of fluence (1)

• Tally F1 + cosine binning —> Neutron current in thedifferent angular bins crossing disc or sphere surface

• Determination of the neutron fluence in the differentangular bins :- current / (disc surface • mean cosine in angular bin)- current / (sphere surface • 0.5)

• Additional information in T. Mc Lean’s oral


P6 34QUADOSQUADOS

Task 2 : Directional distribution of fluence (2)Position 1

Angular distribution of neutron current at 170 cm

degrees0 20 40 60 80 100 120 140 160 180

Cur

rent

/ co

sine

bin

0

1e-4

2e-4

3e-4

4e-4

5e-4Angular distribution of

neutron fluence at 170 cm

degrees0 20 40 60 80 100 120 140 160 180

Flue

nce

/ cos

ine

bin

0,0

2,0e-7

4,0e-7

6,0e-7

8,0e-7

1,0e-6

1,2e-6

1,4e-6


P6 35QUADOSQUADOS

Task 2 : Directional distribution of fluence (3)Position 2


degrees0 20 40 60 80 100 120 140 160 180

Cur

rent

/ co

sine

bin

0

1e-4

2e-4

3e-4

4e-4

5e-4

Angular distribution of neutron fluence at 300 cm

degrees0 20 40 60 80 100 120 140 160 180

Flue

nce

/ cos

ine

bin

0,0

2,0e-7

4,0e-7

6,0e-7

8,0e-7

1,0e-6

1,2e-6

1,4e-6


P6 36QUADOSQUADOS

Task 2 : Directional distribution of fluence (4)Angular binning ?? = 15°


degrees0 20 40 60 80 100 120 140 160 180

Cur

rent

/ co

sine

bin

0,0

5,0e-5

1,0e-4

1,5e-4

2,0e-4

2,5e-4Angular distribution of

neutron fluence at 170 cm

degrees0 20 40 60 80 100 120 140 160 180

Flue

nce

/ cos

ine

bin

0

2e-7

4e-7

6e-7


P6 37QUADOSQUADOS

Task 2 : Directional distribution of fluence (5)Angular binning ?? = 15°


degrees0 20 40 60 80 100 120 140 160 180

Cur

rent

/ co

sine

bin

0,0

5,0e-5

1,0e-4

1,5e-4

2,0e-4

2,5e-4

Angular distribution of neutron fluence at 300 cm

degrees0 20 40 60 80 100 120 140 160 180

Flue

nce

/ cos

ine

bin

0

2e-7

4e-7

6e-7


P6 38QUADOSQUADOS

Task 3 : Walls- and air-scattering fluence contributions

• 2 approaches were used:

1 - Cell-flagging Additivity should be verified, i. e. : Wall- and air scatter contributions (task 3) should

add up to total fluence (task 1)

2 - ‘Voiding’ technique (M.Králik’s oral)


P6 39QUADOSQUADOS

Task 3 : Walls- and air- scattering fluence contributions (2)

Air- and walls-scattering contributions

airsensors w.

oppos. w.side w. 1

side w. 2 floorceiling

Flue

nce

cont

ribut

ion

by

0,0

2,0e-7

4,0e-7

6,0e-7

8,0e-7

1,0e-6

1,2e-6

1,4e-6

1,6e-6

1,8e-6

at 170 cm

at 300 cm


P6 40QUADOSQUADOS

Oral: Tom McLean

Oral: Miloslav Králik


P6 41QUADOSQUADOS

Summary and Conclusions

The realistic problem attracted 14 participants Task 1: 10 solutions were in good agreement Task 2: 8 solutions Task 3: 8 solutions Task 4: 6 solutions were in good agreement

A wide range of techniques was used Basic differences were seen in computing the

Directional distribution of fluence (Task 2) and Contributions from walls, floor and ceiling (Task 3)

The numerical results wereby in large satisfactory.Several author interpreted theirresults adequately.

The uncertainty associatedwith the results was not *addressed. *Apart from providing stochastic uncertainties.


P6 42QUADOSQUADOS

Appendices

Table of Content

Cell-flagging and ‘Voiding’ techniques (1-4)

The use of calibration fields (1-3)


P6 43QUADOSQUADOS

Cell-flagging and ‘Voiding’ techniques (1)

Geometry 1 Geometry 2

Geometry 3 Geometry 4wall 2 wall 1

Air


P6 44QUADOSQUADOS

Cell-flagging and ‘Voiding’ techniques (2)

Geometry 1 Geometry 2

2 1 2 1

Geometry 2 —> Geometry 1 by voiding wall 1

Geom. 1: CF (air) = 6.40 10-8 = Total

Geom. 2: CF (air) = 8.50 10-8

CF (wall1) = 2.90 10-7 VT (wall1) : 3.75 10-7- 6,40 10-8= 3.11 10-7

Total = 3.75 10-7 Difference CF --> VT : +7.2%


P6 45QUADOSQUADOS

Cell-flagging and ‘Voiding’ techniques (3)Geometry 1 Geometry 3

2 1

Geometry 3 —> Geometry 1 by voiding wall 2

Geom. 1: CF (air) = 6.40 10-8 = Total

Geom. 3: CF (air) = 7.38 10-8

CF (wall1) = 7.45 10-8 VT (wall1) : 1.48 10-7- 6,40 10-8= 8.40 10-8

Total = 1.48 10-7 Difference CF --> VT : +12.7%

2 1


P6 46QUADOSQUADOS

Cell-flagging- and ‘Voiding’ techniques (4)

Geometry 4 —> Geometry 3 by voiding wall 1Geom. 4: CF (air) = 1.01 10-7 Geom. 3: CF(air) = 7.38 10-8

CF (wall1) = 3.31 10-7 CF (wall2) = 7.45 10-8

CF (wall2) = 8.42 10-8 Total = 1.48 10-7

Total = 5.17 10-7 VT (wall1) = 5.17 10-7 – 1.48 10-7= 3.69 10-7

Difference CF VT : +11.5%

Geometry 4

2 1 2 1

Geometry 3


P6 47QUADOSQUADOS

The quantity to be determined is a specific dose equivalent (DE), H, which is computed as theenergy integral over the product of ΦE (E) and the fluence-to-DE conversion factors, hΦ (E).

Let the subscript i=cal indicate the calibration field and i=wp the work place field. Then thecalibration factor, Ncal, is implicitly defined by the requirement: Hcal =: Ncal

.Mcal, and if thecalibrated dosemeter is used in a work place field then one obtains for the difference between thetrue value of the DE in that field to the measured one,

∆=Hwp - Ncal.Mwp,

The last Eqaution lends itself to derive the the design aims of a moderator assembly. One seeclearly, that the choice of the calibration field would be arbitrary, if the energy dependence of hΦis matched by that of RΦ.

The use of calibration fields (1/3)

( ) ( ) )(-)()(-)(d= cal

cal,w,

max

min

ERNEhEEEE

EEpE ΦΦΦΦ∆ ⋅⋅⋅∫

∫ ⋅⋅max

min i,i )( )(d=

E

EE EEhEH ΦΦ ∫ ⋅⋅

max

min i,i )( )(d=

E

EE EEREM ΦΦ


P6 48QUADOSQUADOS


10-3 10-2 10-1 100 101 102 103 104 105 106 107 108

Energy (eV)

10-4

10-3

10-2

10-1

Flue

nce

per

uni

t lo

g en

ergy

(Le

thar

gy)

Bare 252Cf without shadow coneBare 252Cf without shadow coneBare 252Cf with shadow coneBare 252Cf with shadow coneMod. 252Cf without shadow coneMod. 252Cf without shadow coneMod. 252Cf with shadow coneMod. 252Cf with shadow cone

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

En / MeV

0.1

0.2

0.3

( ΦE

. E )

/ cm

-2

Typ A : <h*Φ(10)> = 12,2 pSv cm2Typ A : <h*Φ(10)> = 12,2 pSv cm2

Typ B : <h*Φ(10)> = 23,9 pSv cm2Typ B : <h*Φ(10)> = 23,9 pSv cm2

Typ C : <h*Φ(10)> = 122 pSv cm2Typ C : <h*Φ(10)> = 122 pSv cm2

Typ D : <h*Φ(10)> = 213 pSv cm2Typ D : <h*Φ(10)> = 213 pSv cm2

Typical neutron spectra in nuclear power plants

10-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 101

En / MeV

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

( ΦE

. E )

/ cm

-2

252Cf-mod+Schatten : <h*Φ(10)> = 36,0 Sv cm2252Cf-mod+Schatten : <h*Φ(10)> = 36,0 Sv cm2

252Cf- mod + Cd : <h*Φ(10)> = 93,3 Sv cm2252Cf- mod + Cd : <h*Φ(10)> = 93,3 Sv cm2

252Cf : <h*Φ(10)> = 343 Sv cm2252Cf : <h*Φ(10)> = 343 Sv cm2

212Am-Be : <h*Φ(10)> = 381 Sv cm2212Am-Be : <h*Φ(10)> = 381 Sv cm2

"Broad calibration fields"


P6 49QUADOSQUADOS


10-3 10-2 10-1 100 101 102 103 104 105 106 107 108

Energy (eV)

100

101

102

103

Res

pons

e pe

r un

it flu

ence

(R

el. u

nits

)

Base (9)Base (9)

Planar (9)Planar (9)

Intermediate (10)Intermediate (10)

Fast (10)Fast (10)

Combination (11)Combination (11)

10-3 10-2 10-1 100 101 102 103 104 105 106 107 108

Energy (eV)

10-3

10-2

10-1

100

101

Res

pons

e pe

r u

nit

fluen

ce (

Rel

. uni

ts)

Eberline_2Eberline_2MODEL 930_2MODEL 930_2LB 6411LB 6411Eberline_3Eberline_3Model 930_3Model 930_3

Personal dosemeters

Survey instruments

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