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CERF simulation Mitsu 14th Feb 2013. Simulation components Production Transportation Detector response <- already done.

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Presentation on theme: "CERF simulation Mitsu 14th Feb 2013. Simulation components Production Transportation Detector response <- already done."— Presentation transcript:

1 CERF simulation Mitsu 14th Feb 2013

2 Simulation components Production Transportation Detector response <- already done

3 Transportation

4 Geometry Concrete (2.4 m x 4.8 m x 0.8 m) Cu target Mixed beam 1.08 m ECC (10 x 12.5 x 8.6 cm 3 ) on the roof  = 2.30 g/cm 3 2.30 x 80 cm = 184 g/cm 2 60 cm 10 cm (= 12~13 km altitude in atmosphere) 2 x 10 5 neutrons (= 10 PICs)

5 Neutron incident angle Z direction in ECC From underflow of momentum components

6 Neutron momentum / energy MomentumEnergy 400 MeV/c 65 MeV/c 200 MeV/c

7 Cross section in 16 O Oxygen is main component in concrete (50%) and air (21%). Data from ENDF/B VII (US) Order of ten MeV/c Neutron wavelength ~ O nucleus size (4 fm) 65 MeV/c 200 MeV/c 400 MeV/c

8 Comparison with another experiment ● : Measured data from Nakao et al., NIMB.266(2008)93 Histogram : GEANT4 MC

9 Another simulator Nakao et al., NIMB.266(2008)93 100 MeV

10 Machine power Neutrons per p.o.t : 2.1 x 10 3 / 2 x 10 5 = 1.1 x 10 -2 Recoil protons per neutron : 1055 / 1 x 10 6 = 1.1 x 10 -3 (including reconstruction efficiency. Please see Tomoko’s slide.) Rate of recoil protons : 1.1 x 10 -5 / p.o.t Required time in my machine : 2.2 s / p.o.t -10 2 recoil protons : 9.0 x 10 6 p.o.t (230 days) -10 3 : 9.0 x 10 7 p.o.t (6 years) -10 4 : 9.0 x 10 8 p.o.t (60 years)

11 Summary I’m performing the simulation of neutron transportation in concrete. Neutron spectrum is determined by cross section of Oxygen. CERF can be used to estimate the radiation exposure in aircraft. MC spectrum and measured data by another experiment are good agreement except higher energy region (E k > 100 MeV).

12 Neutron cross section in 16 O JENDL 4.0 (JAEA)

13 Rate estimation (very rough) Living probability : e -t/ = e -80 cm/28.32 cm = 0.059 - nuclear collision length : 28.32 cm (concrete) - concrete thickness : 80 cm Number of incident neutrons : 3.8 x 10 -2 / pot - fraction of solid angle (10 cm x 10 cm) : 1.5 x 10 -4 - Multiplicity from MC : 250 n / p.o.t Neutrons per event : 2.3 x 10 -3 n / p.o.t This estimation take no thought of the effects of large angle scattering : - sky shine - ground shine …


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